Fundamentals of Engineering Economic Analysis [2nd ed.] 1119503043, 9781119503040

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Fundamentals of Engineering Economic Analysis Second Edition JOHN A. WHITE University of Arkansas KELLIE S. GRASMAN Missouri University of Science and Technology KENNETH E. CASE Oklahoma State University KIM LaSCOLA NEEDY University of Arkansas DAVID B. PRATT Oklahoma State University

VICE PRESIDENT & DIRECTOR

Laurie Rosatone

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Don Fowley

INSTRUCTIONAL DESIGN LEAD Thomas Kulesa INSTRUCTIONAL DESIGNER

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Ashley Patterson

MARKETING MANAGER

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MEDIA SPECIALIST

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EDITORIAL ASSISTANT

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COVER DESIGN

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Cataloging-in-Publication Data is on file at the Library of Congress. ISBN: 9781119503040

Preface Introduction We are pleased to present Fundamentals of Engineering Economic Analysis 2nd edition, a fully up-to-date text to serve an undergraduate engineering economics course. Building upon the successful award-winning first edition, the new text continues to offer a streamlined delivery of engineering econ fundamentals. In its first edition, the text was carefully optimized to serve a 1- semester, 1–3 credit-hour course without sacrificing rigor or essential content. The core content and approach of Fundamentals of Engineering Economic Analysis are built on the strong foundation of Principles of Engineering Economic Analysis, now in its sixth edition, by John A. White, Kenneth E. Case, and David B. Pratt. As such, the content has been thoroughly and successfully class-tested, and reflects decades’ worth of accuracy checking.

New to This Edition In revising Fundamentals of Engineering Economic Analysis, we have updated the text to reflect recent changes to the treatment of corporate income taxes and depreciation as dictated by the American Tax Cuts & Jobs Act of 2017. Further, we have improved in-chapter pedagogical features, aligning all text content and resources to learning objectives and offering concept check questions in-line with reading. Most significantly, this edition offers a fully integrated approach to link the traditional “textbook” content to a world of digital teaching and learning resources through an interactive e-text option, a fully New WileyPLUS environment, and hundreds of new problems for practice and assessment.

Three Big Things: 1. Interactive E-text with Media-Rich, Integrated Learning 2. New WileyPLUS: Complete “Out of the Box” Online Course

3. Abundant Problems for Practice and Assessment

1. Interactive E-text with Media-Rich, Integrated Learning The visual Interactive Pedagogical Framework section that follows shows how digital resources connect with reading content to provide a superior student-centered learning environment. The interactive e-text delivers all traditional textbook content along with just-in-time multi-media resources to engage students. While the interactive e-text brings traditional print content to life with interactive media, the WileyPLUS online course takes that a step further.

2. New WileyPLUS: Complete “Out of the Box” Online Course WileyPLUS combines the complete, dynamic interactive e-text of Fundamentals of Engineering Economic Analysis 2nd edition with all of the teaching and learning resources you need to deliver a complete online, hybrid, or traditional 1–3 credit-hour course, in one easy-to-use system. Completely redesigned for the 2nd edition, the New WileyPLUS environment offers instructors the option to get started with a complete “out of the box” course that may be implemented as is, or customized to meet your specific needs. You can access pre-created, customizable assignments for each chapter of the text. Assignments are based on the authors’ proven approach to teaching this specific content. The approach has been refined over many semesters, and offers efficiency to instructors while ensuring research-backed learning opportunities for students. Instructors may integrate the full set of assignments to create a complete online course for their students, or select specific types of assignments to incorporate with traditional classroom and/or hybrid delivery. Pre-Created Assignments Each chapter includes pre-created assignments, aligned with specific learning objectives, in each of the following categories:

Reading Review Students are prompted to read specific sections of the digital text, and then answer multiple-choice questions related to the text concepts. Lessons Students are prompted to view mini-lecture videos that present fundamental topics and simple examples, and then answer multiple-choice questions related to the topics. Excel® Lessons Students are presented videos that demonstrate how to use specific Excel® financial functions and tools in the context of a general example. Then they are asked to solve a similar problem using the function they just learned about. Examples Students are presented videos that demonstrate the solution to a textbook Example, and then solve a similar problem. Practice Problems Students are assigned problems, and offered structured support as they practice the problem-solving processes demonstrated in the text. Support may include a video solution of the problem or a GO Tutorial that outlines the problem-solving steps, along with answers and solutions displayed after two attempts are made. Homework Students are presented problems with limited support to test their knowledge of the problem-solving processes. Problems are algorithmic so each student sees different values. Utilizing the pre-created assignments creates a structured learning path for students. With so many resources available to support learning, the options can sometimes be overwhelming! Assignments guide students to specific resources, to interact with the material in a natural progression.

Reading Reviews introduce the text and encourage mastery of terminology and concepts. Lessons present the topics in video form, and Excel® Lessons present basics of Excel® functions. Examples encourage students to view a problem-solving process and model it in a similar assigned problem. Practice encourages more problem-solving—the key to success in the course—and offers abundant student support. Then Homework requires students to apply the problem-solving skills to new scenarios. If used sequentially, the pre-created assignments guide students through concepts and problem-solving in a progressive fashion. Instructors may choose to use all of the pre-created assignments, or choose specific categories based on their course needs. Further, assignments may be readily modified to add or remove questions. For instance, if instructors choose to omit a section of a chapter, they can easily remove related questions from a pre-created assignment. Each question is aligned to a learning objective and section from the text, so it is easy to see and add or remove questions. Assignment policies were carefully chosen and vary by category, and may also be modified to suit instructor preferences. The structured learning path approach and associated categories have been refined and class tested over many semesters—by the author—and offer a proven method to help students master engineering economy.

3. Abundant Problems for Practice and Assessment The pre-created assignments noted above utilize course-tested questions, many of which are available for students to view in the enhanced e-text. While the students may be familiar with the wording, problems are configured to present algorithmic values—so individual students receive a unique calculation scenario. Problems in the text and WileyPLUS course are categorized as Student Facing or Instructor-Only. The student facing problems are included in the digital text, and many are also available to assign to students in WileyPLUS. Student facing problems are generally intended to support student self-study or low-stakes formative assessment, and many are also included in pre-created assignments.

Abundant—and varied—problems are available. Over 1,200 problems and questions are available to students in the e-text; 22% new or updated for this edition, including a new category dedicated to student self-study of concepts. WileyPLUS includes a set of nearly 1,500 problems and questions instructors can use in auto-graded assignments, with 650 new to this edition. Student facing problems are categorized as follows: Student Facing (Practice) FE = Multiple-choice Fundamentals of Engineering exam prep questions Presented in student e-text with show/hide answers to evennumbered questions for self-study All available to assign in WileyPLUS, with algorithmic values CC = Multiple-choice concept check questions Presented in student e-text in-line with reading content, with show/hide answers to all questions, for self-study All available to assign in WileyPLUS PR = Numerical entry, end-of-chapter problems requiring calculations Presented in student e-text with show/hide answer and Video Solution for select problems, as indicated by “Video Solution” icon Many problems available to assign in WileyPLUS, as indicated by “WP” icon Select problems include GO Tutorials in WileyPLUS to guide solution process, as indicated by “GO Tutorial” icon The instructor-only problems are ONLY available to instructors, and have not been shared in any student facing environment. These problems have intentionally been reserved to reduce the proliferation of answers/solutions on internet sources. Instructor-only problems are generally intended to be

utilized in summative assessment or instructor-controlled assignments. All instructor-only problems are available to assign in WileyPLUS. Instructor-only problems are categorized as follows: Instructor-Only (Assessment) RR = Multiple-choice reading review questions Aligned to text sections and learning objectives Some included in pre-created assignments AP = Numerical entry, assessment problems involving calculation Algorithmic, and aligned to text sections and learning objectives Similar to end-of-chapter problems (PR), but fresh to students QU = Multiple-choice quiz questions Algorithmic, and aligned to text sections and learning objectives Shorter calculation problems, useful for quizzing All problems are numbered by chapter and section association. Each section of the text corresponds to a specific learning objective (LO). General numbering follows this convention: Ch#.LO# - XX001, where XX represents the problem category as described above. For example, 02.03AP004 represents assessment problem #4 related to Chapter 2, Section/LO 3.

Instructor Support The following instructor support materials are available; In addition to the pre-created assignments and abundant problems for online assessment creation in WileyPLUS, the title offers a wealth of other resources to support the way instructors wish to teach. Video Examples, Video Lessons, Excel® Video Lessons, and Video Solutions—created by the authors and experts in the field—are available for students to reference for self-study in both WileyPLUS

and the e-text. Additionally, instructors may utilize the proven video content to provide instruction in their online, hybrid, and/or traditional courses. Lecture slides in PowerPoint for each chapter facilitate teaching and learning from the book. They can be used for lecture and presentation purposes or serve as templates for instructors developing their own slides. All content is organized by learning objective, making it simple to create custom presentations based on specific topical coverage. Clicker Questions in PowerPoint for each chapter provide questions aligned by learning objective for use with student response systems (i.e., “clickers”). They can be integrated with lecture slides to provide active learning opportunities in the classroom or used as pre- or postclass quizzes to gauge student preparedness and understanding. Clicker Questions are unique, so students will not find them in their text or WileyPLUS assignments. New case study assignments demonstrate the application of engineering econ principles in complex, applied scenarios. Case studies describe scenarios for student analysis. Specific assignment guidelines and grading rubrics are provided, along with a video describing how one instructor uses similar cases in a high-enrollment course. The Instructor’s Solutions Manual includes solutions in Excel® spreadsheet form to make it easy to see the solutions, select particular cells to see the formula used, or convert the entire worksheet to see the formulas used in all cells. The Instructor’s Solutions Manual is accessible only to instructors adopting this text and contains solutions to textbook-based questions only. Answers to Discussion Questions provide sample answers to all discussion questions associated with chapter-opening vignettes. Useful Excel® spreadsheet utilities that assist in performing specific engineering economy analyses.

Approach of the Text

Fundamentals of Engineering Economic Analysis 2nd edition provides a rich and interesting learning experience for students as well as a strong teaching tool for instructors. This exciting new approach was carefully crafted to support different learning and teaching styles. An engaging visual design reinforces the key pedagogical aids throughout the text, such as learning objectives and key terms. Visual learning is also supported with numerous and carefully crafted illustrations, such as cash flow diagrams, that reinforce student understanding of the underlying concepts. Each chapter begins with a real-world vignette to demonstrate why the concepts to be addressed are important to a practicing engineer. A detailed summary at the end of each chapter succinctly captures the most important ideas to reinforce the students’ learning. Within the topical coverage, Fundamentals of Engineering Economic Analysis proceeds logically from time value of money concepts, to methods of comparing economic alternatives, to additional complexities that pervade real-world engineering economic analysis, such as taxes, depreciation, and inflation. A seven-step systematic economic analysis technique (SEAT) is introduced in Chapter 1 and revisited in the following chapters. This approach serves to remind students of the central role of comparing economic alternatives within the larger task of a full-blown engineering economic analysis. It also emphasizes the logical and methodical nature of such analyses. The book takes a cash flow approach throughout. We want students to recognize that a correct engineering economic analysis depends on the correct amount and timing of the relevant cash flows.

Use of Traditional Solution Methods and Excel® Spreadsheets Another important facet of our approach concerns the balance of traditional solution methods and Microsoft® Excel®. Throughout, the text supports the use of the basic formulas and interest factors for single sums, uniform series, gradient series, and geometric series. These have formed the heart of time value of money evaluations by engineers over the past 60 years. Traditional solution methods are followed by discussion of Excel® tools

and formulas, where they exist, to provide the balanced coverage that today’s students need. Indeed, the complexity of problems students will face in the real world practically demands that they develop some fluency with Excel® formulas and tools. But we introduce them only after careful presentation of the underlying math and traditional techniques. Fundamentals of Engineering Economic Analysis 2nd edition has been developed to support the different teaching styles of instructors and learning styles of students, including a range of approaches to the use of Excel® spreadsheets. For example, instructors and students can, if they choose, rely on hand calculations and ignore the material on spreadsheets. With a very few exceptions, the worked Examples in the text focus first on complete hand solutions, followed by the Excel® approaches where relevant. The exceptions to this approach are a handful of Examples that are sufficiently complex to require use of spreadsheets, and those Examples have titles that clearly indicate that requirement. For those instructors who wish to emphasize Excel® solutions, nearly all Examples and problems show a spreadsheet calculation. Further, most Video Lessons and Video Examples demonstrate solutions using a screen share of Excel® in addition to traditional solution methods. New for the 2nd edition, Excel® Video Lessons are linked in the e-text where Excel® functions are first introduced and provide students a brief primer on how to use the function. A complete library of Excel® Video Lessons is available to WileyPLUS users. Depending on the needs of the instructor, students, and/or the specific topic, a mixture of approaches can also be used.

Use of Solved Examples Solved Examples in the book feature several components that reinforce our approach to teaching engineering economics: Many of the Examples are linked with respect to context. That is, the same scenario is revisited in multiple Examples to implement progressive concepts and techniques. See, for instance, Examples 4.2, 5.3, and 6.4. This approach also facilitates comparison of results under different situations, such as no tax versus tax and various loan plans.

Most Examples are formatted with specific steps to aid students in developing a methodical problem-solving approach. Examples are first solved by traditional formula or factor techniques, followed by Excel® approaches, where relevant. The Excel® versions of solutions are distinguished by blue/boldface type. A few complex Examples are addressed with Excel® only. Examples reflect a mix of business/industry and personal finance contexts, to optimally motivate students’ interest and attention.

Content Summary Principles and methodology for engineering economic analyses considering the time value of money. Time value of money including all commonly used cash flow profiles. Time value of money calculations involving discrete and continuous cash flows, as well as discrete and continuous compounding. Borrowing, lending, and investing with business and personal applications. Measures of economic worth including present worth, capitalized worth, future worth, annual worth, internal rate of return, external rate of return, and benefit-cost. Depreciation including MACRS and its foundation methods—double declining balance and straight line. After-tax economic analysis, with and without borrowing. After-tax economic analysis, with bonus depreciation. After-tax analysis based on the current income tax rates for corporations. After-tax economic analysis, with Section 170 expense deductions. Before- and after-tax replacement analysis and optimal replacement interval. Inflation, including before- and after-tax analyses.

Supplementary analyses, including breakeven, sensitivity analysis, and risk analysis, as well as use of @ RISK simulation software. Capital budgeting to make sound overall investment under monetary constraints. Costs, estimation, accounting, and economic value added concepts.

Acknowledgments We would like to thank the following instructors who reviewed the manuscript at various stages, or reviewed or contributed to the WileyPLUS course: Kamran Abedini, California State Polytechnic University—Pomona Virgil Adumitroaie, University of Southern California Baabak Ashuri, Georgia Institute of Technology Swaminathan Balachandran, University of Wisconsin—Platteville Geza Paul Bottlik, University of Southern California Stanley F. Bullington, Mississippi State University Richard Burke, SUNY Maritime Karen M. Bursic, University of Pittsburgh Mark Calabrese, University of Central Florida John R. Callister, Cornell University Viviana I. Cesani, University of Puerto Rico Xin Chen, Southern Illinois University Edwardsville Oswald Chong, University of Kansas Donald Coduto, California State Polytechnic University—Pomona Bob Loo Craig Downing, Rose-Human Instutite of Technology William Foley, Rennselaer Polytechnic University Charles R. Glagola, University of Florida Craig M. Harvey, Louisiana State University Edgar A. Hollingsworth, Pennsylvania College of Technology Sung-Hee Kim, Southern Polytechnic State University Krishna K. Krishnan, Wichita State University

Leonard R. Lamberson, Western Michigan University John Lee, University of Wisconsin—Madison Bob Loo, University of Southern California, Missouri University of Science & Technology Alberto Marquez, Lamar University Gary Maul, Ohio State University R. Eugene McGinnis, Christian Brothers University Letitia M. Pohl, University of Arkansas Mamunur Rashid, Georgia Southern University Sarah Root, Pennsylvania State University Matthew S. Sanders, Kettering University Alex Savachkin, University of South Florida Dana Sherman, University of Southern California Surendra Singh, University of Tulsa Alice E. Smith, Auburn University Aimee Ulstad, The Ohio State University John M. Usher, Mississippi State University Tao Yang, California State Polytechnic University—San Luis Obispo Mehmet Bayram Yildirim, Wichita State University Obtaining input, counsel, examples, and data has been made much easier through the exceptional cooperation of the following individuals and others in their organizations (listed alphabetically by company): Abbot Laboratories: Miles D. White, Chairman and Chief Executive Officer. Intel: Dr. Craig R. Barrett, Chairman of the Board (retired). J. B. Hunt Transport Services, Inc: David G. Mee, Executive Vice President and Chief Financial Officer.

Motorola Solutions: Dr. Thomas F. Davis, Corporate Vice President and Chief Economist (retired). Gino A. Bonanotte, Executive Vice President and Chief Financial Officer list him before Tom Davis Wal-Mart Stores: Michael T. Duke, President and Chief Executive Officer (retired).

Contents Cover Title Page Copyright Page Preface Interactive Pedagogical Framework Contents CHAPTER 1: An Overview of Engineering Economic Analysis FE-Like Problems and Problems FE-Like Problems Problems Summary and Study Guide Study Resources 1.1: Time Value of Money 1.2: Engineering Economy Principles 1.3: Economic Justification of Capital Investments Readings Learning Objectives Introduction 1.1 Time Value of Money 1.2 Engineering Economy Principles 1.3 Economic Justification of Capital Investments CHAPTER 2: Time Value of Money Calculations FE-Like Problems and Problems FE-Like Problems Problems Summary and Study Guide

Study Resources 2.1: Cash Flow Diagrams 2.2: Single Cash Flows 2.3: Multiple Cash Flows: Irregular Cash Flows 2.4: Multiple Cash Flows: Uniform Series of Cash Flows 2.5: Multiple Cash Flows: Gradient Series of Cash Flows 2.6: Multiple Cash Flows: Geometric Series of Cash Flows 2.7: Compounding Frequency Readings Learning Objectives Introduction 2.1 Cash Flow Diagrams 2.2 Single Cash Flows 2.3 Multiple Cash Flows: Irregular Cash Flows 2.4 Multiple Cash Flows: Uniform Series of Cash Flows 2.5 Multiple Cash Flows: Gradient Series of Cash Flows 2.6 Multiple Cash Flows: Geometric Series of Cash Flows 2.7 Compounding Frequency CHAPTER 3: Equivalence, Loans, and Bonds FE-Like Problems and Problems FE-Like Problems Problems Summary and Study Guide Study Resources 3.1: Equivalence 3.2: Interest Payment and Principal Payments 3.3: Bond Investments 3.4: Variable Interest Rates Readings

Learning Objectives Introduction 3.1 Equivalence 3.2 Interest Payment and Principal Payments 3.3 Bond Investments 3.4 Variable Interest Rates CHAPTER 4: Present Worth FE-Like Problems and Problems FE-Like Problems Problems Summary and Study Guide Study Resources 4.1: Comparing Alternatives 4.2: Present Worth Calculations 4.3: Benefit-Cost Analysis 4.4: Discounted Payback Period 4.5: Capitalized Worth Readings Learning Objectives Introduction 4.1 Comparing Alternatives 4.2 Present Worth Calculations 4.3 Benefit-Cost Analysis 4.4 Discounted Payback Period 4.5 Capitalized Worth CHAPTER 5: Annual Worth and Future Worth FE-Like Problems and Problems FE-Like Problems Problems

Summary and Study Guide Study Resources 5.1: Annual Worth 5.2: Future Worth Readings Learning Objectives Introduction 5.1 Annual Worth 5.2 Future Worth CHAPTER 6: Rate of Return FE-Like Problems and Problems FE-Like Problems Problems Summary and Study Guide Study Resources 6.1: Internal Rate of Return Calculations 6.2: External Rate of Return Calculations Readings Learning Objectives Introduction 6.1 Internal Rate of Return Calculations 6.2 External Rate of Return Calculations CHAPTER 7: Replacement Analysis FE-Like Problems and Problems FE-Like Problems Problems Summary and Study Guide Study Resources

7.1: Cash Flow and Opportunity Cost Approaches to Replacement Analysis 7.2: Optimum Replacement Interval Readings Learning Objectives Introduction 7.1 Cash Flow and Opportunity Cost Approaches to Replacement Analysis 7.2 Optimum Replacement Interval CHAPTER 8: Depreciation FE-Like Problems and Problems FE-Like Problems Problems Summary and Study Guide Study Resources 8.1: Depreciation in Economic Analysis 8.2: Straight-Line and Declining Balance Depreciation Methods 8.3: Modified Accelerated Cost Recovery System (MACRS) 8.4: MACRS with Bonus Depreciation Readings Learning Objectives Introduction 8.1 Depreciation in Economic Analysis 8.2 Straight‐Line and Declining Balance Depreciation Methods 8.3 Modified Accelerated Cost Recovery System (MACRS) 8.4 MACRS with Bonus Depreciation CHAPTER 9: Income Taxes FE-Like Problems and Problems

FE-Like Problems Problems Summary and Study Guide Study Resources 9.1: Corporate Income-Tax Rates 9.2: After-Tax Analysis Using Retained Earnings (No Borrowing) 9.3: After-Tax Analysis Using Borrowed Capital 9.4: Leasing Versus Purchasing Equipment 9.5: After-Tax Analysis with Bonus Depreciation 9.6: After-Tax Analysis with a Section 179 Expense Deduction Readings Learning Objectives Introduction 9.1 Corporate Income-Tax Rates 9.2 After-Tax Analysis Using Retained Earnings (No Borrowing) 9.3 After-Tax Analysis Using Borrowed Capital 9.4 Leasing Versus Purchasing Equipment 9.5 After-Tax Analysis with Bonus Depreciation 9.6 After-Tax Analysis with a Section 179 Expense Deduction CHAPTER 10: Inflation FE-Like Problems and Problems FE-Like Problems Problems Summary and Study Guide Study Resources 10.1: The Meaning and Measure of Inflation

10.2: Before-Tax Analysis 10.3: After-Tax Analysis 10.4: After-Tax Analysis with Borrowed Capital Readings Learning Objectives Introduction 10.1 The Meaning and Measure of Inflation 10.2 Before-Tax Analysis 10.3 After-Tax Analysis 10.4 After-Tax Analysis with Borrowed Capital CHAPTER 11: Break-Even, Sensitivity, and Risk Analysis FE-Like Problems and Problems FE-Like Problems Problems Summary and Study Guide Study Resources 11.1: Break-Even Analysis 11.2: Sensitivity Analysis 11.3: Risk Analysis Readings Learning Objectives Introduction 11.1 Break-Even Analysis 11.2 Sensitivity Analysis 11.3 Risk Analysis CHAPTER 12: Capital Budgeting FE-Like Problems and Problems FE-Like Problems Problems

Summary and Study Guide Study Resources 12.1: The Classical Capital Budgeting Problem 12.2: Capital Budgeting Problem with Indivisible Investments 12.3: Capital Budgeting Problem with Divisible Investments Readings Learning Objectives Introduction 12.1 The Classical Capital Budgeting Problem 12.2 Capital Budgeting Problem with Indivisible Investments 12.3 Capital Budgeting Problem with Divisible Investments CHAPTER 13: Obtaining and Estimating Cash Flows FE-Like Problems and Problems FE-Like Problems Problems Summary and Study Guide Study Resources 13.1: Cost Terminology 13.2: Cost Estimation 13.3: General Accounting Principles 13.4: Cost Accounting Principles 13.5: Economic Value Added Readings Learning Objectives Introduction 13.1 Cost Terminology 13.2 Cost Estimation 13.3 General Accounting Principles

13.4 Cost Accounting Principles 13.5 Economic Value Added Appendix A: Time Value of Money Factors Appendix B Index End User License Agreement

List of Tables CHAPTER 2 TABLE 2.1 Tabular Solution to Example 2.2 TABLE 2.2 Derivation of Equation 2.3 CHAPTER 3 TABLE 3.1 Interest and Equity Payments in a Deferred Payment Loan CHAPTER 4 TABLE 4.1 Net Cash Flows for Example 4.9 TABLE 4.2 Present Worth and Cumulative Present Worth of Example 4.9 Cash Flows TABLE 4.3 Data For Highway Construction Costs CHAPTER 6 TABLE 6.1 Cash Flow Profile TABLE 6.2 Data for Example 6.5 TABLE 6.3 ERR Solutions for Various MARR Values in Example 6.7 CHAPTER 8 TABLE 8.1 Depreciation Allowances and Book Values for the SMP Investment TABLE 8.2 MACRS‐GDS Property Classes

TABLE 8.3 MACRS‐GDS percentages for 3‐, 5‐, 7‐, and 10‐year property are 200% DBSLH and 15... TABLE 8.4 CHAPTER 9 TABLE 9.1 Corporate Income-Tax Rates for Tax Years January 1, 1993 to December 31, 2017 TABLE 9.2 Before-Tax and After-Tax Analysis of the SMP Investment with SLN Depreciations TABLE 9.3 After-Tax Analysis of the SMP Investment with MACRS Depreciation TABLE 9.4 After-Tax Analysis of a $500,000 Investment in a Consulting Study TABLE 9.5 After-Tax Comparison of Investment Alternatives with Different Property Classes TABLE 9.6 After-Tax Comparison of Manual Versus Robotic Palletizing TABLE 9.7 After-Tax Analysis of the SMP Investment with $300,000 of Borrowed Capital Repai... TABLE 9.8 After-Tax Analysis of the SMP Investment with $300,000 of Borrowed Capital Repai... TABLE 9.9 After-Tax Comparison of Purchasing Versus Leasing Lift Trucks TABLE 9.10 After-Tax Analysis of the SMP Machine Investment with Bonus Depreciation TABLE 9.11 Applying the Section 179 Expense Deduction TABLE 9.12 Applying the Section 179 Expense Deduction and 50% Bonus Depreciation CHAPTER 10 TABLE 10.1 CPI Data 1913–2018 TABLE 10.2 Cash Flows and Present Worth for Example 10.2

TABLE 10.3 After-Tax Analysis of the SMP Investment; $300,000 Borrowed and Repaid with Plan... TABLE 10.4 After-Tax Analysis of the SMP Investment; $300,000 Borrowed and Repaid with Plan... TABLE 10.5 After-Tax Analysis of the SMP Investment; $300,000 Borrowed and Repaid with Plan... TABLE 10.6 Comparison of PWAT for Four Payment Plans, with and Without Inflation CHAPTER 11 TABLE 11.1 Optimistic, Pessimistic, and Most Likely Estimates for Four Scream Machine Param... TABLE 11.2 Means, Variances, Standard Deviations, and Probability Distributions for Annual ... TABLE 11.3 Computing the Expected Value and Variance for the Present Worth of the SMP Inves... TABLE 11.4 Risk Analysis for the Selection of the Scream Machine Design TABLE 11.5 Probability Distributions for Example 11.8 CHAPTER 12 TABLE 12.1 Characteristics of Five Investment Opportunities CHAPTER 13 TABLE 13.1 Cost Data for Automatic Tank Battery Operations TABLE 13.2 Cost Data for Manual Tank Battery Operation TABLE 13.3 Relationship Between Marginal Cost and Total Cost TABLE 13.4 Relationship Between Marginal Cost and Average Cost TABLE 13.5 Example Balance Sheet TABLE 13.6 Example Income Statement—Manufacturing Format

TABLE 13.7 Example Income Statement—Retail Format TABLE 13.8 Carson’s Cutlery Company Comparative Balance Sheet TABLE 13.9 Carson’s Cutlery Company Comparative Income Statement TABLE 13.10 Data for Example 11

List of Illustrations CHAPTER 2 FIGURE 2.1 A Cash Flow Diagram (CFD) FIGURE 2.2 CFDs for Alternatives A and B FIGURE 2.3 CFD of the Time Relationship Between P and F FIGURE 2.4 The Excel® FV Financial Functions Used with a Single Sum of Money FIGURE 2.5 The Excel® Goal Seek Tools Used to Solve Example 2.4 FIGURE 2.6 The Excel® Solver Tool Used to Solve Example 2.4 FIGURE 2.7 The Excel® PV Financial Function Used with a Single Sum of Money FIGURE 2.8 CFD of Multiple Cash Flows FIGURE 2.9 The Excel® NPV and FV Financial Functions Used to Solve Example 2.6 FIGURE 2.10 CFD of the Relationship Between P and A in a Loan Transaction FIGURE 2.11 CFD for Example 2.10 FIGURE 2.12 CFD for the Deferred Payment Example FIGURE 2.13 CFD of the Relationship Between A and F in an Investment Scenario

FIGURE 2.14 CFD of a Combination of Uniform and Gradient Series in an Investment Scenario FIGURE 2.15 CFD for Example 2.15 FIGURE 2.16 CFD for an Increasing Geometric Series FIGURE 2.17 Excel® Solution to Example 2.16 FIGURE 2.18 Excel® Solution to Example 2.17 CHAPTER 3 FIGURE 3.1 The Decreasing Gradient CFD for Example 3.1 FIGURE 3.2 Equivalent Gradient and Uniform CFDs for Example 3.1 FIGURE 3.3 Equivalent Gradient CFD and Converted Uniform CFD for Example 3.1 FIGURE 3.4 CFDs for Example 3.2 FIGURE 3.5 Using the Excel® SOLVER Tool to Solve Example 3.2 FIGURE 3.6 CFDs for Example 3.3 FIGURE 3.7 Using the Excel® SOLVER Tool to Solve Example 3.3 FIGURE 3.8 Solution to Example 3.3 FIGURE 3.9 CFD for Variable Interest Rate Example 3.9 CHAPTER 4 FIGURE 4.1 CFD for Example 4.1 FIGURE 4.2 Spreadsheet of Cumulative Present Worth Over the Planning Horizon FIGURE 4.3 CFDs for Example 4.2 FIGURE 4.4 Plot of Present Worths for Example 4.2 FIGURE 4.5 CFDs for Example 4.3

FIGURE 4.6 Costs and Benefits for Land Reclamation Project, i = 7% FIGURE 4.7 Costs and Benefits for Traffic Light FIGURE 4.8 Selecting Highway Safety Improvements Using B/C and B − C FIGURE 4.9 Selecting Driveway Approach from 3 Alternatives with Unequal Lives FIGURE 4.10 Summary of Data for Three Routes of Example 4.8 FIGURE 4.11 Benefit-Cost Calculations for Example 4.8 FIGURE 4.12 Plot of Cumulative Present Worths for Example 4.10 FIGURE 4.13 Excel® SOLVER Set Up to Calculate the DPBP in Example 4.11 with Gradient Decreas... FIGURE 4.14 Excel® SOLVER Solutions for the DPBP in Example 4.11 with Geometric and Gradient... FIGURE 4.15 CFDs for the Alternatives in Example 4.12 FIGURE 4.16 Excel® SOLVER Solution for DPBPC in Example 4.12 FIGURE 4.17 Discounted Payback Period Analysis for Example 4.12 FIGURE 4.18 CFD for Example 4.13 CHAPTER 5 FIGURE 5.1 CFD for Example 5.1 FIGURE 5.2 Analyzing the Effect on AW of Changes in MARR for Example 5.2 FIGURE 5.3 CFDs for Example 5.3 FIGURE 5.4 Set Up to Use the Excel® SOLVER Tool to Determine the Gradient Step Needed to Ac...

FIGURE 5.5 Excel® SOLVER Solution to Example 5.5 FIGURE 5.6 Impact of Annual Returns on an Investment Portfolio CHAPTER 6 FIGURE 6.1 CFD for Example 6.1 FIGURE 6.2 Plot of Future Worth for Example 6.2 FIGURE 6.3 IRR for a Natural Gas Investment FIGURE 6.4 CFDs for Example 6.4 FIGURE 6.5 Plot of Future Worth of Incremental Investment CF(2-1) for Example 6.5 FIGURE 6.6 ERR Solution to Example 6.6 FIGURE 6.7 Excel® SOLVER Setup and Solution to Example 6.7 FIGURE 6.8 Data and ERR Analysis for Example 6.9 CHAPTER 7 FIGURE 7.1 EUAC Components Used to Determine the Optimum Replacement Interval CHAPTER 8 FIGURE 8.1 Depreciation and Book Value Using DDB Switching to SLN CHAPTER 9 FIGURE 9.1 Pictorial Representation of Taxable Income FIGURE 9.2 Highest U.S. Federal Income-Tax Rate for Corporations (1909 thru 2018) FIGURE 9.3 Pictorial Representation of After-Tax Cash Flow (ATCF) FIGURE 9.4 After-Tax Analysis of the SMP Investment with $300,000 of Borrowed Capital Repai...

FIGURE 9.5 After-Tax Analysis of SMP Investment with $300,000 of Borrowed Capital Repaid Us... FIGURE 9.6 After-Tax Analysis of $500,000 Investment with 100% Borrowed Capital Repaid Usin... CHAPTER 10 FIGURE 10.1 CPI Annual Percent Changes 1913–2018 FIGURE 10.2 After-Tax, Then-Current Dollar Analysis of the SMP Investment with 4% Inflation FIGURE 10.3 After-Tax, Constant Dollar Analysis of the SMP Investment with 4% Inflation FIGURE 10.4 After-Tax Analysis of the SMP Investment with $300,000 Borrowed, Repaid with Pla... CHAPTER 11 FIGURE 11.1 Break-Even Analysis for the Annual Sales of a New Product FIGURE 11.2 Using the Excel® SOLVER Tool to Determine the Break-Even Value for the Planning ... FIGURE 11.3 Sensitivity Analysis for Example 11.3 FIGURE 11.4 CFDs for Example 11.4 FIGURE 11.5 Sensitivity Analysis for Example 11.4 FIGURE 11.6 Considering 81 Possible Combinations of 3 Estimates and 4 Parameters FIGURE 11.7 Normal Distribution FIGURE 11.8 Sample Beta Distributions FIGURE 11.9 PW Histogram from 100,000 Simulation Trials in Example 11.8. (The Figure was Gen... FIGURE 11.10 IRR Histogram from 100,000 Simulation Trials in Example 11.8. (The Figure was Ge...

FIGURE 11.11 Setup for @RISK Monte Carlo Simulation of Example 11.9. (The Figure was Generate... FIGURE 11.12a PW Histogram from 100,000 Simulated Investments in Design A FIGURE 11.12b PW Histogram from 100,000 Simulated Investments in Design B FIGURE 11.12c PW Histogram from 100,000 Simulated Incremental Investments Between Designs B an... CHAPTER 12 FIGURE 12.1 SOLVER Set Up for Example 12.1 FIGURE 12.2 SOLVER Parameters for Example 12.1 FIGURE 12.3 SOLVER Solution to Example 12.1 FIGURE 12.4 SOLVER Solution for Example 12.2, with SOLVER Parameters Shown FIGURE 12.5 SOLVER Parameters and Solution for Example 12.3 CHAPTER 13 FIGURE 13.1 Manufacturing Cost Structure FIGURE 13.2 Total Annual Cost as a Function of Mileage FIGURE 13.3 Revenue and Cost as a Function of Production Volume FIGURE 13.4 Break‐Even Points (a) Single, (b) Multiple FIGURE 13.5 Total Revenue, Total Cost, and Total Profit as a Function of Tons FIGURE 13.6 Cost of Detail in Estimating FIGURE 13.7 Relationship Between Balance Sheets and Income Statements FIGURE 13.10 Annual Worth and Economic Value‐Added Calculations for the SMP Machine

CHAPTER 1 An Overview of Engineering Economic Analysis

Chapter 1 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced e-text and WileyPLUS

FE-Like Problems 01-FE001 The fact that one should not add or subtract money unless it occurs at the same point in time is an illustration of what concept? a. Time value of money b. Marginal return c. Economy of scale d. Pareto principle 01-FE002 If a set of investment alternatives contains all possible choices that can be made, then the set is said to be which of the following? a. Coherent b. Collectively exhaustive c. Independent d. Mutually exclusive Correct or Incorrect? Clear

  Check Answer

01-FE003

Risks and returns are generally __________ correlated.

a. inversely b. negatively c. not d. positively

01-FE004 The “discounting” in a discounted cash flow approach requires the use of which of the following? a. An interest rate b. The economic value added c. The gross margin d. The incremental cost Correct or Incorrect? Clear

  Check Answer

01-FE005 Answering “what if” questions with respect to an economic analysis is an example of which step in the systematic economic analysis technique? a. Identifying the investment alternatives b. Defining the planning horizon c. Comparing the alternatives d. Performing supplementary analysis 01-FE006 If a student’s time value of money rate is 30%, then the student would be indifferent between $100 today and how much in 1 year? a. $30 b. $100 c. $103 d. $130 Correct or Incorrect? Clear

  Check Answer

01-FE007 Which of the following best represents the relationship between the weighted average cost of capital (WACC) and the minimum attractive rate of return (MARR)?

a. WACC and MARR are unrelated b. WACC is a lower bound for MARR c. WACC is an upper bound for MARR d. MARR ≤ WACC

Problems Section 1.1 Time Value of Money LEARNING OBJECTIVE 1.1 Apply the four discounted cash flow (DCF) rules to simple time value of money (TVOM) situations. 01.01-PR001 Wylie has been offered the choice of receiving $5,000 today or an agreed-upon amount in 1 year. While negotiating the future amount, Wylie notes that he would be willing to take no less than $5,700 if he has to wait a year. What is his TVOM in percent? 01.01-PR002 RT is about to loan his granddaughter Cynthia $20,000 for 1 year. RT’s TVOM, based upon his current investment earnings, is 8%. Cynthia’s TVOM, based upon earnings on investments, is 12%. a. Should they be able to successfully negotiate the terms of this loan? b. If so, what range of paybacks would be mutually satisfactory? If not, how far off is each person from an agreement? 01.01-PR003 RT is about to loan his granddaughter Cynthia $20,000 for 1 year. RT’s TVOM, based upon his current investment earnings, is 12%, and he has no desire to loan money for a lower rate. Cynthia is currently earning 8% on her investments, but they are not easily available to her, and she is willing to pay up to $2,000 interest for the 1-year loan. a. Should they be able to successfully negotiate the terms of this loan? b. If so, what range of paybacks would be mutually satisfactory? If not, how many dollars off is each person from reaching an agreement?

01.01-PR004 If your TVOM is 15% and your friend’s is 20%, can the two of you work out mutually satisfactory terms for a 1-year, $3,000 loan? Assume the lender has the money available and neither of you wants to go outside your acceptable TVOM range. Be explicit about who is lending and what is the acceptable range of money paid back on the loan. Sections 1.2 and 1.3 Engineering Economy Principles and Economic Justification of Capital Investments LEARNING OBJECTIVE 1.2 Identify the 10 principles of engineering economic analysis that can be used by all engineers in analyzing the economic performance of the products, processes, and systems they design. LEARNING OBJECTIVE 1.3 Identify the 7 steps of the systematic economic analysis technique (SEAT) used to perform engineering economic analyses. 01.02/03-PR001 The following stages of a project are each contingent upon the preceding stage. If the preceding stage is not performed (accepted), then none of the subsequent stages may be performed. Stage

Investment

1

A small FCC-licensed commercial radio station

2 3

Cost in Today’s $

Revenues in Today’s $

$180,000

$270,000

New antenna and hard-line from transmitter

$30,000

$50,000

$40,000

$30,000

4

New amplifier to boost from 5,000 watts to 30,000 watts New lightning-protection equipment

$30,000

$60,000

5

New control console

$15,000

$5,000

a. Remembering that no stages can be skipped, which set of the five stages do you recommend be purchased?

b. Of the ten principles, which one(s) is(are) well illustrated by this problem? c. Of the systematic economic analysis technique’s 7 steps, which one(s) is(are) well illustrated by this problem? 01.02/03-PR002 Barbara and Fred have decided to put in an automatic sprinkler system at their cabin. They have requested bids, and the lowest price received is $5,500 from Water Systems Inc (WSI). They decide to do the job themselves and obtain a set of materials (plastic pipe, nozzles, fittings, and regulators) from an all-sales-are-final discount house for $1,100. They begin the installation and rent a trencher at $80 per day. Unfortunately, they quickly hit sandstone in many places of the yard and require a jackhammer and air compressor at another $80 per day. They keep all the rental equipment for 5 days. By this time, Fred has hurt his knee, and Barbara is sick of the project. They again contact WSI, who tells them that they can use only some of the materials, reducing the cost by $500, and only some of the trenching, reducing the cost by another $500, bringing the total to $4,500, finished and ready to go. a. How much have they already spent? b. How much will they have spent when the project is over if they accept the new offer from WSI? c. A different contractor, Sprinkler Systems (SS), who heard of their situation approaches Barbara and Fred and recommends a design for a sprinkler system that would require a different set of materials and a new routing of the trenches. They offer to (1) backfill all existing trenches, (2) cut new trenches with their rock-impervious Ditch Witch, and (3) install the system. Their charge is $6,000 for a finished-and-ready-to-go project, and they correctly note that this is less than the total that will have been spent if Barbara and Fred go with WSI. Should Barbara and Fred go with WSI or SS? Why? d. Of the ten principles, which one(s) is(are) well illustrated by this problem? e. Of the systematic economic analysis technique’s 7 steps, which one(s) is(are) well illustrated by this problem?

01.02/03-PR003 List some nonmonetary factors in the alternative decision process that you should be prepared to address when presenting a proposal to management. Let your mind run free and think this out on your own, rather than trying to find words that fit from the text. a. Come up with 10 or more items. b. Of the ten principles, which one(s) is(are) well illustrated by this problem? c. Of the systematic economic analysis technique’s 7 steps, which one(s) is(are) well illustrated by this problem? 01.02/03-PR004 Four proposals (A, B, C, and D) are available for investment. Proposals A and C cannot both be accepted; Proposal B is contingent upon the acceptance of either Proposal C or D; and Proposal A is contingent on D. a. List all possible combinations of proposals and clearly show which are feasible. b. Of the ten principles, which one(s) is(are) well illustrated by this problem? c. Of the systematic economic analysis technique’s 7 steps, which one(s) is(are) well illustrated by this problem? 01.02/03-PR005 Five proposals (V, W, X, Y, and Z) are available for investment. At least two and no more than four must be chosen. Proposals X and Y are mutually exclusive. Proposal Z is contingent on either Proposal X or Y being funded. Proposal V cannot be pursued if either W, X, Y, or any combination of the three is pursued. a. List all feasible mutually exclusive investment alternatives. b. Of the ten principles, which one(s) is(are) well illustrated by this problem? c. Of the systematic economic analysis technique’s 7 steps, which one(s) is(are) well illustrated by this problem? 01.02/03-PR006 Three proposals (P, Q, and R) are available for investment. Exactly one or two proposals must be chosen; Proposals P and Q are mutually

exclusive. Proposal R is contingent on Proposal P being funded. List all feasible mutually exclusive investment alternatives. 01.02/03-PR007 AutoFoundry has contacted Centrifugal Casting Company about the purchase of machines for the production of (1) Babbitt bearings, and (2) diesel cylinder liners (engine sleeves). The cost of the machines prohibits AutoFoundry from purchasing both, so they decide to base their selection on which machine will provide the greatest net income on a “today” basis (also known later as a present worth basis). The bearing machine has a life of 5 years, after which it is expected to be replaced. It has “today” costs (considering first cost and 5 years of operating cost, maintenance, etc.) of $460,000 and provides new revenues over the 5 years of $730,000 in today’s dollars. The cylinder liner machine has a life of 9 years, with “today” costs and new revenues over the 9-year life of $650,000 and $990,000, respectively. a. What would you recommend that AutoFoundry do? b. Of the ten principles, which one(s) is(are) well illustrated by this problem? c. Of the systematic economic analysis technique’s 7 steps, which one(s) is(are) well illustrated by this problem? 01.02/03-PR008 Suppose you have been out of school and gainfully employed for 5 years. You have three alternatives available for investment with your own money. Each has some element of risk, although some are safer than others. Following is a summary of the alternatives, the risks, and the returns: Chance Returned to Chance Returned to You of You if Success, of You if Failure, Alternative Invest, $ Success $ Failure $ A B

$100,000 $100,000

95% 60%

$110,789.47 $150,000.00

5% 40%

$95,000 $50,000

C

$100,000

20%

$510,000.01

80%

$10,000

a. Which would you select? b. Why would you make this selection?

c. Of the ten principles, which one(s) is(are) well illustrated by this problem? d. Of the systematic economic analysis technique’s 7 steps, which one(s) is(are) well illustrated by this problem? 01.02/03-PR009 A Payne County commissioner has $20,000 remaining in the budget to spend on one of three worthy projects. Each is a one-time investment, and there would be no follow-on investment, regardless of which project is chosen. Project A involves the placement of gravel on a rough and often muddy road leading to a public observatory, providing net benefits (consider this as net revenue-in-kind) of $8,000 per year for 4 years, at which time the road will again be in disrepair. Project B involves the building of a water-retention dam to hold water during big rains, thereby lessening damage due to flash flooding; the benefits are expected to be worth $6,000 for each of 6 years, after which silt will have made the pond ineffective. Project C is to provide water, sewer, and electrical hookups for recreational vehicles at the fairgrounds; net benefits of $4,000 per year would be realized for 10 years, after which the system would need to be replaced. No matter which alternative is selected, once its useful life is over, there will be no renewal. a. What planning horizon should be used in evaluating these three projects? b. Of the ten principles, which one(s) is(are) well illustrated by this problem? c. Of the systematic economic analysis technique’s 7 steps, which one(s) is(are) well illustrated by this problem? 01.02/03-PR010 Reconsider the county commissioner’s evaluation of three projects in Problem 01.02/03-PR009. Take the facts as given, except now suppose the commissioner can commit the county to renewing these investments, even if a different commissioner is elected. So, after 4 years in project A, the road would be renewed with gravel or perhaps even paved. After 6 years, the water-retention dam could be dredged and renewed, or a new dam could be built. After 10 years, the RV hookups could be modernized and replaced. a. What are the considerations in selecting the appropriate planning horizon in this case?

b. Of the ten principles, which one(s) is(are) well illustrated by this problem? c. Of the systematic economic analysis technique’s 7 steps, which one(s) is(are) well illustrated by this problem?

Chapter 1 Summary and Study Guide Summary 1.1: Time Value of Money

Learning Objective 1.1: Apply the four discounted cash flow (DCF) rules to simple time value of money (TVOM) situations. (Section 1.1) The four DCF rules state that 1. Money has a time value. 2. Quantities of money cannot be added or subtracted unless they occur at the same point in time. 3. To move money forward one time unit, multiply by 1 plus the discount or interest rate. 4. To move money backward one time unit, divide by 1 plus the discount or interest rate. 1.2: Engineering Economy Principles

Learning Objective 1.2: Identify the 10 principles of engineering economic analysis that can be used by all engineers in analyzing the economic performance of the products, processes, and systems they design. (Section 1.2) The 10 principles of engineering economic analysis can be summarized as follows:

1. Money has a time value. 2. Make investments that are economically justified. 3. Choose the mutually exclusive investment alternative that maximizes economic worth. 4. Two investment alternatives are equivalent if they have the same economic worth. 5. Marginal revenue must exceed marginal cost. 6. Continue to invest as long as each additional increment of investment yields a return that is greater than the investor’s TVOM. 7. Consider only differences in cash flows among investment alternatives. 8. Compare investment alternatives over a common period of time. 9. Risks and returns tend to be positively correlated. 10. Past costs are irrelevant in engineering economic analyses, unless they impact future costs. 1.3: Economic Justification of Capital Investments

Learning Objective 1.3: Identify the seven steps of the systematic economic analysis technique (SEAT) used to perform engineering economic analyses. (Section 1.3) The 7-step systematic economic analysis technique (SEAT) used to perform engineering economic analysis can be summarized as follows: 1. Identify the investment alternatives. 2. Define the planning horizon. 3. Specify the minimum attractive rate of return (MARR), also known as the discount rate. 4. Estimate the cash flows. 5. Compare the alternatives.

6. Perform supplementary analyses. 7. Select the preferred investment.

Important Terms and Concepts Time Value of Money (TVOM) The value of a given sum of money depends on both the amount of money and the point in time when the money is received or paid. Discounted Cash Flow (DCF) The movement of money forward or backward in time. Planning Horizon The period of time or width of the “window” over which the economic performance of each investment alternative will be viewed. Minimum Attractive Rate of Return (MARR) Also referred to as the hurdle rate or discount rate, the MARR is the minimum rate of return on an investment that a decision maker is willing to accept given the associated risk and the opportunity cost of other forgone investments. Weighted Average Cost of Capital (WACC) The WACC is a value calculated to establish the lower bound on the MARR and to take into account that most firms have multiple sources of capital.

Chapter 1 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 1.1: Time Value of Money LO 1.1: Apply the four discounted cash flow (DCF) rules to simple time value of money (TVOM) situations. Video Lesson: Time Value of Money Video Lesson Notes: Time Value of Money 1.2: Engineering Economy Principles LO 1.2: Identify the 10 principles of engineering economic analysis that can be used by all engineers in analyzing the economic performance of the products, processes, and systems they design. 1.3: Economic Justification of Capital Investments LO 1.3: Identify the seven steps of the systematic economic analysis technique (SEAT) used to perform engineering economic analyses. Video Lesson: Systematic Economic Analysis Technique (SEAT) Video Lesson Notes: Systematic Economic Analysis Technique (SEAT) These chapter-level resources will help you with your overall understanding of the content in this chapter. Appendix A: Time Value of Money Factors Appendix B: Engineering Economic Equations Flashcards: Chapter 01 Excel Utility: TVM Factors: Table Calculator

Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values Excel Utility: Monthly Payment Sensitivity Excel Utility: TVM Factors: Discrete Compounding Excel Utility: TVM Factors: Geometric Series Future Worth Excel Utility: TVM Factors: Geometric Series Present Worth

CHAPTER 1 An Overview of Engineering Economic Analysis LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 1.1 Apply the four discounted cash flow (DCF) rules to simple time value of money (TVOM) situations. (Section 1.1) 1.2 Identify the 10 principles of engineering economic analysis that can be used by all engineers in analyzing the economic performance of the products, processes, and systems they design. (Section 1.2) 1.3 Identify the seven steps of the systematic economic analysis technique (SEAT) used to perform engineering economic analyses. (Section 1.3)

Engineering Economics in Practice Walmart In fiscal year (FY) 2018, Walmart (WMT) employed more than 2.2 million people worldwide, including 1.5 million in the United States. Its revenue surpassed $500 billion for the first time and it returned $14.4 billion to shareholders. As of January 31, 2018, Walmart had a total of 11,700 stores serving nearly 270 million customers a week operating in 28 countries including Walmart U.S., Sam’s Club U.S., and Walmart International. Walmart eCommerce sales grew 44% to $11.5 billion. A major factor in Walmart’s success has been the efficiency of its supply chain. Walmart was among the first to adopt electronic data interchange (EDI), allowing it to pay vendors for products purchased by customers in its stores. The real-time management of money across its thousands of stores has contributed significantly to Walmart’s economic performance. Walmart is an international leader in logistics, and its distribution network includes approximately 150 distribution centers and 50 transportation offices. It has one of the largest distribution operations in the world, servicing stores, clubs, and direct delivery to customers. Walmart transportation has a fleet of 6,100 tractors, 61,000 trailers, and more than 7,800 drivers. It has 6 disaster distribution centers, strategically located across the United States and stocked to provide rapid response in the event of a natural disaster. In FY 2018 alone Walmart responded to the devastating damage caused by hurricanes impacting Puerto Rico, Texas, and Florida, and an earthquake in Mexico with relief supplies to assist communities in distress. Walmart was a pioneer in the use of radio frequency identification (RFID) to do real-time tracking of products throughout the supply chain. In recent years, Walmart has targeted sustainability and worked with its vendor community to dramatically reduce the amount of waste arriving at and leaving from its stores in the form of packing and packaging materials. Its engineers work with vendors’ engineers to significantly reduce energy consumption in distributing the products it sells and to increase energy efficiency in its stores. Walmart continues to be innovative via digitizing experiences to make it easier and faster for customers to shop and more efficient for associates to manage inventory and serve customers. Its eCommerce business continues to expand. In FY 2018, it nearly doubled the number of stores offering online grocery pickup to more than 1,100 locations in the United States with plans to add another 1,000 in FY 2019. Online grocery shopping will expand also in Canada, Mexico and China. In FY

2018, Walmart has shifted its spending from opening new stores to remodels, eCommerce, supply chain, and technology to position itself for the future. Robert Thompson’s Financial Planning Robert Thompson is a recent engineering graduate who desires to invest a portion of his annual income each year. He is unsure about the kind of investments he should make—stocks, bonds, mutual funds, U.S. Treasury notes, certificates of deposit, rental property, land, and so on. Also, he is undecided about how much of his annual income he should set aside for investment. Further, he does not know what annual return he should expect to earn on his investments. Finally, he wonders how long it will take for him to achieve his financial goals. Discussion Questions 1. What do the preceding two examples have in common? 2. What role does engineering economic analysis play in these scenarios? 3. Is an engineering economic analysis just as relevant for the complicated business transactions at Walmart as for Robert Thompson’s much simpler transactions? 4. Compare the length of the financial planning horizon that each investor (Walmart and Robert) is considering. Are they the same or different?

Introduction The subject matter of this text is variously referred to as economic analysis, engineering economy, economic justification, capital investment analysis, or economic decision analysis. Traditionally, the application of economic analysis techniques in the comparison of engineering design alternatives has been referred to as engineering economy. The emergence of a widespread interest in economic analysis in public-sector decision making, however, has brought about greater use of the more general term economic analysis. We define engineering economic analysis as using a combination of quantitative and qualitative techniques to analyze economic differences among engineering design alternatives in selecting the preferred design. In this text we use a cash flow approach to the subject. A cash flow occurs when money actually changes hands from one individual to another or from one organization to another. Thus, money received and money dispersed (spent or paid) constitutes a cash flow. It is always tempting to skip the first chapter of an engineering textbook. It would be a big mistake to do so here, however. You will build on the foundation concepts introduced in this chapter throughout the rest of your study of engineering economics.

1.1 Time Value of Money LEARNING OBJECTIVE Apply the four discounted cash flow (DCF) rules to simple time value of money (TVOM) situations. Video Lesson: Time Value of Money Time Value of Money (TVOM) The value of a given sum of money depends on both the amount of money and the point in time when the money is received or paid. A fundamental concept underlies much of the material covered in the text: Money has a time value. This means that the value of a given sum of money depends on both the amount of money and the point in time when the money is received or paid. Just as in construction the placement of forces along a beam matters in designing structures, so the placement in time of money received or paid matters when evaluating the economic worth of an investment.

EXAMPLE 1.1 The Time Value of Money Illustrated To illustrate the time value of money (TVOM) concept, suppose a wealthy individual approaches you and says, “Because of your outstanding ability to manage money, I am prepared to present you with a tax-free gift of $1,000. If you prefer, however, I will postpone the presentation for a year, at which time I will guarantee that you will receive a tax-free gift of $X.” (For purposes of this example, assume that the guarantee is risk-free.) In other words, you can choose to receive $1,000 today or receive $X 1 year from today. Which would you choose if X equals (1) $1,000, (2) $1,050, (3) $1,100, (4) $1,500, (5) $2,000, (6) $5,000, (7) $10,000, (8) $100,000? Solution In presenting this situation to students in our classes, none preferred to receive $1,000 a year from now instead of receiving $1,000 today. Also, none preferred to receive $1,050 a year from now instead of $1,000 today. Gradually, as the value of X increased, more and more students switched from preferring $1,000 today to preferring $X a year from now. Not surprisingly, every student preferred to receive $100,000 a year from now to receiving $1,000 today. Also, it was no surprise that all students preferred to receive $10,000 a year from now to receiving $1,000 today. The greatest debate and uncertainty among the students concerning which amount to choose occurred when X was between $1,100 and $2,000. For each student, some value (or range of values) of $X exists for which the student is indifferent—that is, has no preference—about receiving $1,000 today versus receiving $X a year from today. If a student is indifferent when X equals $1,200, then we would conclude that $1,200 received 1 year from now has a present value or present worth of $1,000 for that particular student in his or her current circumstances. We would conclude that this student’s TVOM is 20%. From students’ responses to the questions posed in the preceding example, patterns have been noted in the choices they make. Several indicate a very strong need for money now, not later. Their personal circumstances are such that they do not believe they can wait a year to receive the money, even if they will receive significantly more at that time. Others are skeptical regarding the guarantee of the money being available a year later—they resort to the “bird in the hand, versus many birds in the bush” philosophy. Such responses occur in industry as well. Corporate executives exhibit similar tendencies when faced with current versus deferred choices.

1.1.1 Earning Power Versus Inflation Occasionally, students claim that inflation will make the future amount of money worth less to them. While it is certainly true that inflation, which represents a decrease in the purchasing power of money, will diminish the present worth of a future sum of money, money has time value even in the absence of inflation. (Chapter 10 will examine the effects of inflation on engineering economic decisions.) Why do we claim that money has time value in the absence of inflation? Because money has “earning power.” If you own money and someone else temporarily needs it, you can loan it to them and charge them interest. The interest rate you charge should be based on your TVOM. After all, if you loan it to someone, then you can no longer invest it; hence, you forego the opportunity to earn a return on your money. The lost opportunity should factor into how much you charge someone for using your money. Because of this, the TVOM is sometimes referred to as the opportunity cost of money. Other terms used to express the TVOM are interest rate, discount rate, hurdle rate, minimum attractive rate of return, and cost of capital. These terms will be used somewhat interchangeably throughout the text. In the first few chapters, however, interest rate and discount rate are used most frequently.

1.1.2 Discounted Cash Flows Another term used in financial circles and throughout the text is discounted cash flows, often referred to as DCF. Originally, DCF referred to the process of using the TVOM or discount rate to convert all future cash flows to a present single sum equivalent.1 Today, DCF tends to refer to any movement of money backward or forward in time. Discounted Cash Flow (DCF) The movement of money forward or backward in time. The fact that money has a time value changes how mathematical operations involving money should be performed. Simply stated, because money has a time value, one should not add or subtract money unless it occurs at the same point in time. Having introduced the notion of new rules for dealing with money, we can now summarize the four DCF rules: 1. Money has a time value. 2. Quantities of money cannot be added or subtracted unless they occur at the same point in time. 3. To move money forward one time unit, multiply by 1 plus the discount or interest rate.

4. To move money backward one time unit, divide by 1 plus the discount or interest rate.

EXAMPLE 1.2 Applying the Four DCF Rules Recall the previous example, where the student’s TVOM was 20%. Suppose the student is guaranteed to receive $1,100 one year from today, and nothing thereafter, if $1,000 is invested today in a particular venture. What is the return on the student’s investment? Solution It would be a mistake for the student to subtract the $1,000 investment from the $1,100 return and conclude that the investment yielded a net positive return of $100. Why? Rule 1 establishes that money has a time value; for this student, it can be represented by a 20% annual rate. Rule 2 establishes that the $1,000 investment cannot be subtracted from the $1,100 return, because they occur at different points in time. So, what should the student do? Apply either Rule 3 or Rule 4. Using Rule 3, the student would conclude that the future value or future worth of the $1,000 investment, based on a 20% TVOM, equals $1,000(1.20) or $1,200 one year later. Because the $1,000 was an expenditure or investment, it is a negative cash flow, whereas the $1,100 return on the investment was a positive cash flow. Hence, the net future worth of the investment is −$1,200 + $1,100, or −$100. Because the future worth is negative, the investment would not be considered a good one by the student. Using Rule 4, the student would conclude that the present value or present worth of $1,100 a year from now equals $1,100/1.20, or $916.67. Therefore, the $1,000 investment yields a negative net present value of $83.33. Again, the student should conclude that the investment was not a good one. Note that different conclusions would have occurred if the student’s TVOM had been 8% instead of 20%. By the way, if you did not understand the mathematics used here, don’t worry. The mathematics of TVOM operations are covered in Chapters 2 and 3.

Concept Check 01.01-CC001 Positive and negative cash flows occurring at varying points in time can simply be summed to determine a total cash flow. True or False?

Concept Check 01.01-CC002 Money has time value: a. Because it has “earning power” b. Because of inflation c. Because it earns interest d. All of the above

Concept Check 01.01-CC003 To move money forward one time unit: a. Divide by 1 plus the discount or interest rate b. Divide by the discount or interest rate c. Multiply by 1 plus the discount or interest rate d. Multiply by the discount or interest rate

1.2 Engineering Economy Principles

LEARNING OBJECTIVE Identify the 10 principles of engineering economic analysis that can be used by all engineers in analyzing the economic performance of the products, processes, and systems they design. Throughout this text, basic principles are presented that all engineers can use in analyzing the economic performance of the products, processes, and systems they design. No matter how impressive or how sophisticated an engineering design might be, if it fails to “measure up” economically, it usually is doomed to failure. The following 10 principles of engineering economic analysis provide a foundation for this text: 1. Money has a time value. This principle underlies almost everything we cover in the text. Due to the TVOM, we prefer to receive a fixed sum of money sooner rather than later; likewise, we prefer to pay a fixed sum of money later, rather than sooner. (Notice how many of the following principles are corollaries of the TVOM principle.) 2. Make investments that are economically justified. The second principle is captured in a succinct statement attributed to Henry Ford in the early 1900s: “If you need a new machine and don’t buy it, you pay for it without ever getting it.” The key to his quote is the word need; need indicates justification. The need can manifest itself in terms of cost reductions that will occur if a new machine is purchased, or the need can reflect the added business that will result from adding new manufacturing capacity or capability. Hence, if savings or revenues that will easily offset the purchase price of a new machine are foregone, then the new machine’s price is paid by continuing to incur higher costs than will occur with the new machine or by passing up the profits that will result from increased capacity or added capability. 3. Choose the mutually exclusive investment alternative that maximizes economic worth. This is a corollary of the first and second principles. The third principle addresses the situation when multiple investment alternatives exist and only one can be chosen. We refer to such investments as mutually exclusive. Note that the third principle considers only monetary aspects of the alternatives. Nonmonetary considerations may cause an alternative to be chosen that does not maximize economic worth. (The third principle is incorporated in Chapters 4, 5, and 6.) 4. Two investment alternatives are equivalent if they have the same economic worth. The fourth principle is an extension of the third, which states that for wellbehaved cash flow profiles, the equivalence holds only for the TVOM that equates their economic worths. (Chapters 2 and 6 examine this principle more closely.)

5. Marginal revenue must exceed marginal cost. The fifth principle comes from a first course in economics. Based on this principle, one should not make an investment unless the added revenues are greater than the added costs. Based on the first principle, the TVOM must be used in comparing marginal revenues and marginal costs if they occur at different points in time. 6. Continue to invest as long as each additional increment of investment yields a return that is greater than the investor’s TVOM. The sixth principle, a corollary of the fifth principle, was verbalized in a statement made in 1924 by General Motors’ chief financial officer, Donald Brown: “The object of management is not necessarily the highest rate of return on capital, but … to assure profit with each increment of volume that will at least equal the economic cost of additional capital required.” The sixth principle can also be stated as follows: Use someone else’s money if you can earn more by investing it than you have to pay to obtain it. Of course, one must consider the risks involved in borrowing money in order to make good investments. (More will be devoted to this principle in Chapters 6, 9, 10, and 12.) 7. Consider only differences in cash flows among investment alternatives. In performing engineering economic analyses, decisions are made between alternatives; hence, costs and revenues common to all investment alternatives can be ignored in choosing the preferred investment. (The seventh principle, which is also a corollary of the fifth principle, is at the heart of Chapters 4 through 11.) 8. Compare investment alternatives over a common period of time. The eighth principle is often violated. When alternatives have useful lives that differ in duration, there is a temptation to compare the life cycle of one investment with the unequal life cycle of another. As you will learn, it is important to compare the alternatives over the same length of time. (We address this principle in this chapter, Chapters 4 through 6, and in Chapter 12.) 9. Risks and returns tend to be positively correlated. The higher the risks associated with an investment, the greater the anticipated returns must be to justify the investment. 10. Past costs are irrelevant in engineering economic analyses, unless they impact future costs. The tenth principle relates to past costs or investments made previously. Past costs that have no carryover effect in the future, also called sunk costs, must be ignored when performing an engineering economic analysis. (Applications of this principle will occur throughout the text.) This text emphasizes the use of these 10 principles in choosing the best investment to make and addresses specific kinds of engineering investments. What kind of investments do we consider? When an existing production machine or process must be replaced, many alternatives are usually available as replacement candidates; but which one is best? Alternative candidates for replacement also exist when faced with replacing

bridges, transformers, telecommunications base stations, computers, road surfaces, sewers, chemical mixers, furnaces, and so forth. Likewise, investments in existing equipment—such as overhauling the equipment or adding new features to extend its useful life or to add new production capability—are included in the text. In designing a new product, many design decisions involve choosing from among alternatives. Some examples of choices include whether to use standardized parts that can be purchased or specially designed parts that must be produced; whether to enclose the product in a molded plastic case or a formed metal case; whether to use standard, replaceable batteries or a specially designed rechargeable battery; and whether to perform the manufacturing and assembly in-house versus outsourcing the manufacturing and assembly. Decisions must also be made regarding materials to use in construction and repair activities, as well as transportation alternatives for moving people and materials. Discounted cash flow methods also play a critical role in decisions regarding mergers, acquisitions, and disposition of manufacturing plants. Regardless of which branch of engineering is involved, numerous choices occur in designing and improving products, processes, and systems. Finally, while this text emphasizes engineering applications, the material it presents can be of great personal value. The principles provided can be used to identify the best engineering design, product, process, or system, and to assist in personal investing. This is particularly true for the material presented in Chapters 2 and 3.

Concept Check 01.02-CC001 Investment alternatives should be compared over the same length of time. True or False?

Concept Check 01.02-CC002 When choosing among mutually exclusive investments, multiple options may be selected. True or False?

Concept Check 01.02-CC003 Past costs (or sunk costs) are irrelevant in engineering economic analysis, unless they impact future costs. True or False?

1.3 Economic Justification of Capital Investments LEARNING OBJECTIVE Identify the seven steps of the systematic economic analysis technique (SEAT) used to perform engineering economic analyses.

Video Lesson: Systematic Economic Analysis Technique (SEAT) In performing engineering economic analyses, it is helpful to follow a consistent methodology. The following 7-step systematic economic analysis technique (SEAT) is recommended: 1. Identify the investment alternatives. 2. Define the planning horizon. 3. Specify the discount rate. 4. Estimate the cash flows. 5. Compare the alternatives. 6. Perform supplementary analyses. 7. Select the preferred investment. This text focuses particularly on step 5—comparing alternative investments in order to identify the best one—although several of the others also will be mentioned. Although typically not every step of this method is explicitly performed in the examples and problems in this text, you will need to understand the important concepts behind the steps. With this end in mind, the following sections take a closer look at each step.

1.3.1 SEAT Step 1 of 7: Identify the Investment Alternatives Generally, the aim is to select the best investment from a feasible set of mutually exclusive and collectively exhaustive investment alternatives. “Mutually exclusive” as used here means “either/or but not both.” “Collectively exhaustive” means that no other investment alternatives are available—all possible investments are considered. The collectively exhaustive assumption is critical. Care must be taken when forming the alternatives to ensure that all available alternatives are being considered. A “do-nothing” alternative frequently is included in the set of feasible investment alternatives to be compared. Such an alternative is intended to represent “business as usual” or “maintaining the status quo.” Business conditions rarely stand still, however. Doing nothing does not mean that nothing will be done; rather, it could mean that management has opted to forego the opportunity to influence future events. The “do nothing” alternative often is used as a baseline against which other investment alternatives are compared. In this text, as a matter of convenience, zero incremental costs often are associated with doing nothing. In practice, when the “do nothing” alternative is feasible, extreme care must be taken not to underestimate the cost of doing nothing. For many firms, business as usual is the most expensive alternative; “standing pat” for too long can be a disastrous course, because while the firm is doing nothing, its competitors are generally doing something.

1.3.2 SEAT Step 2 of 7: Defining the Planning Horizon As noted in the eighth principle of engineering economic analysis, it is important to compare investment alternatives over a common period of time. In this text, this period of time is referred to as the planning horizon. For investments in, say, equipment to perform a required service, the period of time over which the service is required might be used as the planning horizon. Likewise, with one-shot investment alternatives, the period of time over which receipts continue to occur might define the planning horizon. Planning Horizon The period of time or width of the “window” over which the economic performance of each investment alternative will be viewed. In a sense, the planning horizon defines the width of the “window” through which the economic performance of each investment alternative will be viewed. Using too short a planning horizon could preclude an investment that would yield very sizeable returns in the long run. Conversely, too long a planning horizon can result in the firm going out of business before realizing the promised long-term benefits.

When the lives of investment alternatives differ, five general approaches are used to determine the planning horizon’s length: 1. Set the planning horizon equal to the shortest life among the alternatives. 2. Set the planning horizon equal to the longest life among the alternatives. 3. Set the planning horizon equal to the least common multiple of the lives of the various alternatives. 4. Use a standard length horizon equal to the period of time that best fits the organization’s need, such as 10 years. 5. Use an infinitely long planning horizon. Approach #3, the least common multiple of lives, is a popular choice. When it is used, each alternative’s cash flow profile is assumed to repeat in the future until all investment alternatives under consideration conclude at the same time. Such an assumption might be practical for some applications, but it can prove untenable for others. For example, assume Alternative A has a useful life of 4 years, Alternative B has a useful life of 5 years, and Alternative C has a useful life of 6 years. What is the least common multiple of 4, 5, and 6? The answer is 60. It does not seem realistic to assume that identical cash flow profiles will repeat over a period of 60 years. Using the shortest life approach (#1) would establish the planning horizon’s length for Alternatives A, B, and C as 4 years. With a 4-year planning horizon, one must estimate the value of the one remaining year of useful service for Alternative B. Similarly, one must estimate the value of the 2 years of useful service available with Alternative C. The longest life approach (#2) would establish 6 years as the planning horizon for Alternatives A, B, and C. In this case, decisions must be made about the 2-year gap for Alternative A and the 1-year gap for Alternative B. Will identical replacements be made? If so, what will be their monetary values at the end of the planning horizon? Next, we consider the impact of a standard planning horizon of, say, 10 years (approach #4). As before, it is assumed that identical replacements will occur. For Alternative A, 10 years represents two complete life cycles and one-half a life cycle. For the latter, a salvage value would have to be estimated. For Alternative B, 10 years represents two complete life cycles. For Alternative C, 10 years represents one full life cycle and one partial life cycle; thus, requiring another salvage value estimate. Finally, we consider the infinitely long planning horizon (#5). Obviously, for many investments, it makes no sense to assume an indefinite planning horizon. When the least common multiple of lives is quite long, however, an infinitely long planning horizon will yield a reasonably good approximation.

1.3.3 SEAT Step 3 of 7: Specifying the Minimum Attractive Rate of Return The chapters that follow will examine many investment opportunities, always using an interest rate to compound (move forward in time) or discount (move backward in time) cash flows. This interest rate is commonly referred to as the minimum attractive rate of return or MARR. The value used for the minimum attractive rate of return matters —a lot! Minimum Attractive Rate of Return (MARR) Also referred to as the hurdle rate or discount rate, the MARR is the minimum rate of return on an investment that a decision maker is willing to accept given the associated risk and the opportunity cost of other forgone investments. Because any investment will consume some portion of a firm’s scarce resources, it is important for the investment to earn more than it costs to obtain the investment capital. In addition, the MARR should reflect the opportunity cost associated with investing in the candidate alternative as opposed to investing in other available alternatives. In fact, it is assumed that investment capital not committed to the candidate alternative is earning a return at least equal to the MARR. Generally, a company has multiple sources of capital: loans, bonds, stocks, and so on. Each has a different cost associated with it. Because firms have multiple sources of capital, they typically calculate the weighted average cost of capital (WACC) and use it to establish a lower bound on the MARR. It is only a lower bound because certain unprofitable investments often are required. For example, investments made to ensure compliance with governmental regulations, to enhance employee morale, to protect lives, and to prevent injuries have positive returns, particularly if social costs and social returns are included. It is difficult to quantify such returns, however. Even though the returns are less than the WACC, these investments still will be made. For this reason, optional investments must earn more than enough to cover the WACC. Weighted Average Cost of Capital (WACC) The WACC is a value calculated to establish the lower bound on the MARR and to take into account that most firms have multiple sources of capital. Capital available to a corporation can be categorized as either debt capital or equity capital. Examples of debt capital are bonds, loans, mortgages, and accounts payable; examples of equity capital are preferred stock, common stock, and retained earnings. Typically, capital for a particular investment consists of a mixture of debt and equity capital. Although many variations are available, a widely accepted WACC formula is

WACC = (E/V ) ie + (D/V )id  (1 − itr)

(1.1)

where E

=

a firm’s total equity, expressed in dollars

D

=

a firm’s total debt and leases, expressed in dollars

V

=

E + D, a firm’s total invested capital

ie

=

cost of equity or expected rate of return on equity, expressed in percent

id

=

cost of debt or expected rate of return on borrowing, expressed in percent

itr

=

corporate tax rate

The costs associated with debt capital are deductible from taxable income; however, the costs associated with equity capital are not deductible. As a result, (1 − itr) is associated with debt capital but not equity capital.

1.3.4 SEAT Step 4 of 7: Estimate the Cash Flows Once the planning horizon is determined, cash flow estimates are needed for each investment alternative for each year of the planning horizon. The Association for the Advancement of Cost Engineering International (www.aacei.org) defines cost estimating as a predictive process used to quantify, cost, and price the resources required by the scope of an asset investment option, activity or project. As a predictive process, estimating must address risks and uncertainties. The outputs of estimating are used primarily as inputs for budgeting, cost or value analysis, decision making in business, asset and project planning, or for project cost and schedule control processes. Cost estimating is not an exact science. Rather, it is an approximation that involves the availability and relevancy of appropriate historical data, personal judgments based on the estimator’s experience, and the time frame available for completing the estimating activity.

1.3.5 SEAT Step 5 of 7: Comparing Alternatives After the investment alternatives are identified, the planning horizon is defined, the discount rate is specified, and the cash flows are estimated, it is time to evaluate the alternatives in terms of their economic performance. When doing the comparison, it is necessary to select a criterion to use. Many options exist. In fact, two already have been presented. In Example 1.2, the investment was evaluated on the basis of its present value or present worth, as well as its future value or future worth. Later chapters will consider present worth analysis in more detail, as well as benefit-cost, payback period,

discounted payback period, capitalized worth, future worth, annual worth, and rate of return analyses. Depending on the particular type of investment as well as the country in which the investment is made, depreciation, income taxes, replacement, and inflation may need to be considered in comparing the alternatives. These topics also are covered in more detail in later chapters.

1.3.6 SEAT Step 6 of 7: Performing Supplementary Analyses The sixth step in comparing investment alternatives is performing supplementary analyses. The intent of this step is to answer as many “what if ” questions as possible. Up to this point, it has been assumed that the cash flow estimates, the length of the planning horizon, and the TVOM used were error free. Obviously, that will not always be the case. Conditions change, errors are made, and risks and uncertainties exist. In this step, risk and uncertainty are explicitly considered.

1.3.7 SEAT Step 7 of 7: Select the Preferred Investment Selecting the preferred investment is the final step in a systematic engineering analysis. Because many factors must be considered in making the selection, the preferred investment may not be the one that performs best when considered using only the economic criteria. Typically, multiple criteria exist rather than a single criterion of maximizing, say, present worth. The presence of multiple criteria coupled with the risks and uncertainties associated with estimating future outcomes makes the selection process quite complicated. To make this process easier, the engineer is encouraged to address as many of management’s concerns as possible in comparing the investment alternatives. If management’s concerns are adequately addressed, the selection decision will agree with the engineer’s recommendations. The text concentrates on economic factors throughout. Keep in mind, however, that management’s ultimate choice may be based on a host of criteria rather than a single monetary criterion. Despite attempts to quantify all benefits in economic terms, some intangible factors or attributes probably will not be reduced to dollars. Consider, for example, such factors as improved safety, reduced cycle times, improved quality, increased flexibility, increased customer service, improved employee morale, being the first in the industry to use a particular technology, and increased market visibility. Clearly, some of these factors are more readily measured in economic terms than are others.

Concept Check 01.03-CC001 When identifying the investment alternatives: a. Identify mutually exclusive alternatives b. Identify collectively exhaustive alternatives c. Frequently include the “do nothing” alternative d. All of the above

Concept Check 01.03-CC002 The period of time, or width of the “window,” over which alternatives will be compared is also known as the: a. Planning horizon b. Shortest life c. Minimum Attractive Rate of Return d. Do-nothing alternative

Concept Check 01.03-CC003 When specifying the minimum attractive rate of return (MARR): a. Use the value 10%, as it is an easy multiple to work with b. Exercise caution in calculating it, as once it is defined, you can use it indefinitely for all future investments c. Ensure that it reflects the opportunity cost associated with investing in the candidate alternative as opposed to investing in other available alternatives d. Calculate the weighted average cost of capital (WACC) to establish the upper bound on the MARR

Note 1. Strictly speaking, compounded cash flow means moving money forward in time and discounted cash flow means moving money backward in time; that is, using the TVOM to calculate a future value. This text, however, follows the generally accepted practice of using the term discounted cash flow or DCF to refer to the movement of money forward or backward in time.

CHAPTER 2 Time Value of Money Calculations

Chapter 2 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced etext and WileyPLUS

FE-Like Problems 02-FE001 If you want to triple your money at an interest rate of 6% per year compounded annually, for how many years would you have to leave the money in the account? a. 12 years b. 19 years c. 32 years d. Cannot be determined without knowing the amount invested. 02-FE002 Let F be the accumulated sum, P the principal invested, i the annual compound interest rate, and n the number of years. Which of the following correctly relates these quantities? a. F = P (1 + in) b. F

= P (1 + i)

c. F

= P (1 + n)

d. F

= P (1 + ni)

n

i

n−1

Correct or Incorrect? Clear

  Check Answer

02-FE003 Consider the following cash flow diagram. What is the value of X if the present worth of the diagram is $400 and the interest rate is 15% compounded annually? a. $246

b. $165 c. $200 d. $146

02-FE004 The plan was to leave $5,000 on deposit in a savings account for 15 years at 6.5% interest compounded annually. It became necessary to withdraw $1,500 at the end of the 5th year. How much will be on deposit at the end of the 15-year period? a. $11,359 b. $9,359 c. $12,043 d. $10,043 Correct or Incorrect? Clear

  Check Answer

02-FE005 A deposit of $3,000 is made in a savings account that pays 7.5% interest compounded annually. How much money will be available to the depositor at the end of 16 years? a. $8,877 b. $10,258 c. $9,542 d. $943 02-FE006 If you invest $5,000 three years from now, how much will be in h 15 f if 10% d d ll ?

the account 15 years from now if i = 10% compounded annually? a. $8,053 b. $15,692 c. $20,886 d. $27,800 Correct or Incorrect? Clear

  Check Answer

02-FE007 The maintenance costs of a car increase by $200 each year. This cash flow pattern is best described by which of the following? a. Gradient series b. Geometric series c. Infinite series d. Uniform series 02-FE008 Your company seeks to take over Good Deal Company. Your company’s offer for Good Deal is $3,000,000 in cash upon signing the agreement followed by 10 annual payments of $300,000 starting one year later. The time value of money is 10%. What is the present worth of your company’s offer? a. $3,000,000 b. $2,281,830 c. $4,843,380 d. $5,281,830 Correct or Incorrect? Clear

  Check Answer

02-FE009 What is the effective annual interest rate if the nominal annual interest rate is 24% per year compounded monthly? a. 2.00%

b. 24.00% c. 26.82% d. 27.12% 02-FE010 A young engineer calculated that monthly payments of $A are required to pay off a $5,000 loan for n years at i% interest, compounded annually. If the engineer decides to borrow $10,000 instead with the same n and i%, her monthly payments will be $2A. a. True b. False c. Cannot be determined without knowing the value of n and i d. Cannot be determined without knowing the value of n or i Correct or Incorrect? Clear

  Check Answer

02-FE011 The president of a growing engineering firm wishes to give each of 20 employees a holiday bonus. How much needs to be deposited each month for a year at a 12% nominal rate, compounded monthly, so that each employee will receive a $2,500 bonus? a. $2,070 b. $3,840 c. $3,940 d. $4,170 02-FE012 Under what circumstances are the effective annual interest rate and the period interest rate equal? a. Never b. If the number of compounding periods per year is one c. If the number of compounding periods per year is infinite d. Always Correct or Incorrect?

Clear

  Check Answer

02-FE013 A child receives $100,000 as a gift which is deposited in a 6% bank account compounded semiannually. If $5,000 is withdrawn at the end of each half year, how long will the money last? a. 21.0 years b. 15.5 years c. 25.0 years d. 18.0 years

Problems Section 2.1 Cash Flow Diagrams LEARNING OBJECTIVE 2.1 Construct a cash flow diagram (CFD) depicting the cash inflows and outflows for an investment alternative. 02.01-PR001 You are offered $200 now plus $100 a year from now for your used computer. Because the sum of those two amounts is $300, the buyer suggests simply waiting and giving you $300 a year from now. You know and trust the buyer, and you typically earn 5.0% per year on your money. So, is the offer fair and equitable? 02.01-PR002 You are offered $500 now plus $500 one year from now. You can earn 6% per year on your money. a. It is suggested that a single fair amount be paid now. What do you consider fair? b. It is suggested that a single fair amount be paid one year from now. What do you consider fair? 02.01-PR003 What words comprise the abbreviation “DCF”? Tell/describe/define what it means in 10 words or less. 02.01-PR004 State the four DCF rules.

02.01-PR005 Pooi Phan needs $2,000 to pay off her bills. She borrows this amount from a bank with plans to pay it back over the next four years at $X per year. Draw a cash flow diagram from the bank’s perspective. 02.01-PR006 A laser cutting machine is purchased today for $23,000. There are no maintenance costs for the next two years. Maintenance at the end of year 3 is expected to be $2,000, with each subsequent year’s maintenance costs exceeding the previous year’s by $1,000. An increase in revenues of $14,000 per year is expected. The planning horizon is 6 years. Draw the cash flow diagram. 02.01-PR007 Rodeo Jeans are stonewashed under a contract with independent USA Denim Company. USA Denim purchased two semiautomatic machines that cost $19,000 each at (t = 0). Annual operating and maintenance costs are $15,000 per machine. Two years after purchasing the machines, USA Denim made them fully automatic at a cost of $12,000 per machine. In the fully automatic mode, the operating and maintenance costs are $6,000 the first year, increasing by $1,000 each year thereafter. The contract with Rodeo Company is for 8 years. Draw the cash flow diagram for all of USA Denim’s investment and other costs assuming the contract will not be extended beyond 8 years. 02.01-PR008 Video Solution You rent an apartment for $550 per month, payable at the beginning of the month. An initial deposit of $450 is required. Utilities are an additional $150 per month payable at the end of the month. The deposit is refundable at the time you move out, assuming a clean apartment in good condition. Draw a monthly cash flow diagram, assuming you keep the apartment for 12 full months.

02.01-PR009 Kaelyn borrows $30,000 from her grandfather today to cover her college expenses. She agrees to repay the loan, with the first payment due 5 years from today in the amount of $2,000. No payment is made at the end of year 6. Starting 7 years from today, a series of 5 annual end-of-year payments is made, with the first in the amount of $X. Each subsequent payment is $1,500 greater than the previous payment. Draw the cash flow diagram of this transaction from the grandfather’s perspective. 02.01-PR010 Video Solution David is borrowing $150,000 from Hartford Bank to open Road and Off-Road Bicycle Shop. David expects it to

take a few years before the shop earns a sizeable profit, so he has arranged for no payments on the loan until the end of the fourth year. The first and second payments are due 4 and 5 years, respectively, from today in the amounts of $20,000 each. Starting at the end of year 6, a series of 4 annual end-of-year payments will be made. The first of these is $X. Each subsequent payment is $8,000 greater than the previous payment. Draw the cash flow diagram from David’s perspective.

02.01-PR011 Draw a cash flow diagram depicting the net cash flows associated with the purchase, operation, and disposition of a synthetic rubber blending machine. The cash flow components are shown below. Your CFD should have only one arrow at any given time period, reflecting the net of that period’s cash flows. At t = 0 (now), purchase blender for $62,000. At t = 0, install at cost of $8,000. At t = 1, savings generated by blender is $10,000. At t = 1, maintenance costs of $800. At t = 2, savings generated by blender are $12,000. At t = 2, maintenance costs of $1,200. At t = 3, savings generated by blender is $18,000. At t = 3, maintenance costs of $1,600. At t = 4, 5, 6, 7, 8, 9, 10, savings generated are $24,000 and maintenance is $4,000. At t = 10, the blender is sold for $8,000. At t = 10, blender removal costs are $1,600. 02.01-PR012 Video Solution Today you borrow $10,000 to pay for your expected college costs over the next four years, including a master’s degree. Two years from now, you determine that you need an additional $4,000 so you borrow this additional amount. Starting four years from the original loan (two years from the second loan), you begin to repay your combined debt by

making annual payments of $2,880. You will make these payments for 10 years. Draw a cash flow diagram of this situation from your perspective.

Section 2.2 Single Cash Flows LEARNING OBJECTIVE 2.2 Perform time value of money calculations for single cash flows with annual compounding. Section 2.2.1 Single Cash Flows—Future Worth Calculations 02.02-PR001 Video Solution If you deposit $5,000 4 years from today, how much will you be able to withdraw 10 years from today if interest is 8.5% per year compounded annually?

02.02-PR002 Video Solution How much will a $25,000 investment today be worth in 10 years if it earns 7% annual compound interest?

02.02-PR003 Use the six approaches from Example 2.4 to determine to the nearest year how long it takes for an investment to double if the interest is compounded annually at the following rates. a. 5% b. 7% c. 10% d. 15% e. 20% 02.02-PR004 On August 1, 1958, first-class postage for a 1-ounce envelope was 4¢. On August 1, 2007, a first-class stamp for the same envelope cost 41¢. What annual compound increase in the cost of first-class postage was experienced during the 49-year period?

02.02-PR005 What is the smallest integer-valued annual compound interest that will result in an investment tripling in value in less than or equal to 10 years? 02.02-PR006 Video Solution You purchase a quarter section (160 acres) of land for $176,000 today and sell it in exactly 9 years for $525,000 at auction. At what annual compound rate did the value of your land grow?

02.02-PR007 Video Solution With interest at 9% compounded annually, what is the fewest number of years (integer-valued) required for money to double in magnitude?

02.02-PR008 At what interest rate will money: a. Double itself in 10 years? b. Triple itself in 10 years? c. Quadruple itself in 10 years? 02.02-PR009 How much money can be withdrawn at the end of the investment period if: a. $1,000 is invested at 8%/year compounded annually for 10 years? b. $5,000 is invested at 11%/year compounded annually for 4 years? c. $13,000 is invested at 9%/year compounded annually for 7 years? d. $25,000 is invested at 10%/year compounded annually for 3 years? 02.02-PR010 If you invest $1,500 today and withdraw $2,500 in 3 years, what interest rate was earned? 02.02-PR011 What will be the amount accumulated by each of the following present investments? a. $3,000 invested for 7 years at 14% compounded annually. b. $1,600 invested for 17 years at 12% compounded annually.

c. $20,000 invested for 38 years at 16% compounded annually. d. $3,500 invested for 71 years at 8% compounded annually. e. $5,000 invested for 34 years at 11.5% compounded annually. 02.02-PR012 How long, to the nearest year, does it take an investment at 6% compounded annually to (approximately): a. Double itself? b. Triple itself? c. Quadruple itself? 02.02-PR013

What rate of interest compounded annually is involved if:

a. An investment of $10,000 made now will, 10 years from now, result in a receipt of $23,674? b. An investment of $2,000 made 18 years ago has increased in value to $15,380? c. An investment of $2,500 made now will, 5 years from now, result in a receipt of $4,212? Section 2.2.2 Single Cash Flows—Present Worth Calculations 02.02-PR014 Develop a single spreadsheet that allows you to calculate F|P and P|F factors. For each cell where the calculation is performed, place above it a small “control panel” that allows you to enter whatever numbers you need to perform the calculation. For example, for the F|P factor, your control panel needs to include i and n. Test each of your factors against the tables in Appendix A to ensure they work right. Keep this handy. It will be useful to you. 02.02-PR015 Charlotte wishes to accumulate $100,000 in a savings account in 10 years. If she wishes to make a single deposit today and the bank pays 4% compounded annually on deposits of this size, how much should Charlotte deposit in the account? 02.02-PR016 How much money would have to be deposited today to accumulate:

a. $10,000 after 6 years if the investment earns 5%/year compounded annually? b. $6,500 after 4 years if the investment earns 8%/year compounded annually? c. $3,400 after 12 years if the investment earns 6%/year compounded annually? d. $13,500 after 5 years if the investment earns 10%/year compounded annually? 02.02-PR017 What deposit today is required for it to be worth $150,000 in 25 years if the deposit earns 5% annual compound interest? 02.02-PR018 Video Solution What present amount of money must be deposited at 11% interest compounded annually to grow to $15,000 in 9 years? Give your answer to the nearest penny. a. Use the tables provided in Appendix A. b. Use the formula, either in a spreadsheet or using a calculator. c. Use the PV function in Excel®.

02.02-PR019 For your 21st birthday, your grandfather offers you a gift of $1,000 today. However, you have the choice of waiting 3 years and receiving $1,500 or waiting 5 years and receiving $3,000. If your money grows at a rate of 8% compounded annually, which alternative should you choose? 02.02-PR020 What is the present value of the following future receipts? a. $19,000 5 years from now at 9% compounded annually b. $8,300 12 years from now at 15% compounded annually c. $6,200 53 years from now at 12% compounded annually d. $13,000 18 years from now at 19.2% compounded annually

e. $5,000 10 years from now at 8% compounded annually 02.02-PR021 Calculate using the interest formula the factor (P|F11.5%,37). Compare that to the result obtained using Excel®’s PV function. 02.02-PR022 Jason takes out a loan at 10% compounded annually for 7 years. At the end of this period, he pays off the loan at a value of $23,384.61. What amount did he borrow? 02.02-PR023 How much money today is equivalent to $10,000 in 12 years, with interest at 10% compounded annually? 02.02-PR024 You want to withdraw a single sum amount of $6,000 from an account at the end of 7 years. This withdrawal will zero out the account. What single sum of money deposited today is required if the account earns 12% per year compounded annually? a. Use the tables provided in Appendix A. b. Use the P|F formula directly in Excel® or your calculator. c. Use an appropriate Excel® function. 02.02-PR025 If a fund pays 12% compounded annually, what single deposit now will accumulate $12,000 at the end of the 10th year? If the fund pays 6% compounded annually, what single deposit now is required in order to accumulate $6,000 at the end of the tenth year? Section 2.3 Multiple Cash Flows: Irregular Series Cash Flows LEARNING OBJECTIVE 2.3 Perform time value of money calculations for irregular series of cash flows with annual compounding. 02.03-PR001 The cash flow profile for an investment is given below and the interest rate is 6.5% compounded annually.

End of Year Net Cash Flow End of Year Net Cash Flow 0 1

$0 − $500

4 5

− $300 $500

2

$200

6

− $200

3

$400

7

$100

a. Find the future worth of this cash flow series using the actual cash flows. b. Find the present worth of this series using the actual cash flows. c. Find the present worth using the future worth. 02.03-PR002 If you invest $2,000 today, withdraw $1,000 in 3 years, deposit $3,000 in 5 years, deposit $1,500 in 8 years, and withdraw the entire sum three years after the final deposit, how much will you withdraw? Interest is 7%. 02.03-PR003 Video Solution An investment has the following cash flow series where interest is 8%: End of Year Cash Flow End of Year Cash Flow 0 $300 5 $0 1

$300

6

$800

2 3

$600 − $500

7 8

$700 $600

4

− $300

a. Determine the present worth of the series. b. Determine the future worth of the series at the end of the 8th year. c. Find the worth of the series at the end of year 2.

02.03-PR004 Ben deposits $5,000 now into an account that earns 7.5% interest compounded annually. He then deposits $1,000 per year at the end of the 1st and 2nd years. How much will the account contain 10 years after the initial deposit? 02.03-PR005 Renaldo borrows $8,000 from his aunt today to help pay for college expenses. He agrees to repay the loan according to the following schedule, at a rate of 6%/year compounded annually. End of Year Cash Flow End of Year Cash Flow 0 1

$8,000 $0

5 6

− $X − ($X + 100)

2

$0

7

− ($X + 400)

3 4

$0 $0

8

− $2X

a. Draw the cash flow diagram from Renaldo’s perspective. b. Find the value of X such that the loan is fully repaid with the last payment. c. What is the dollar amount of each of the 4 payments. 02.03-PR006 Video Solution The cash flow profile for an investment is given below and the interest rate is 8% compounded annually. End of Year Net Cash Flow End of Year Net Cash Flow 0 1

$0 $500

4 5

$300 − $100

2

− $200

6

$200

3

$600

a. Find the present worth of this series using the actual cash flows. b. Find the future worth of this cash flow series using the actual cash flows.

c. Find the future worth using the present worth. d. Find the worth of the series at EOY 4 using the individual cash flows. e. Find the present worth using the worth at EOY 4.

02.03-PR007 The manager at a Sherwin-Williams store has decided to purchase a new $30,000 paint mixing machine with high-tech instrumentation for matching color and other components. The machine may be paid for in one of two ways: (1) pay the full price now, less a 3% discount, or (2) pay $5,000 now, $8,000 one year from now, and $6,000 at the end of each of the next 4 years. If interest is 12% compounded annually, determine which way is best for the manager to make the purchase. 02.03-PR008 Ken loans his grandson Rex $20,000 at 5.5% per year to help pay for executive chef schooling in Florida. Rex requires 3 years of schooling before beginning to earn a salary. He agrees to pay Ken back the loan following the schedule below: End of Year Cash Flow End of Year Cash Flow 0 1

$20,000 $0

5 6

− $2X − $3X

2

$0

7

− $4X

3 4

$0 − $X

8

− $5X

a. Draw the cash flow diagram from Ken’s perspective. b. Find the value of X such that the loan is fully repaid with the last payment. c. What is the dollar amount of each of the 5 payments. d. Quite by surprise, following successful on-time completion of all payments, Ken gives back to Rex all interest paid. For how much does Ken write the check?

02.03-PR009 Maria deposits $1,200, $500, and $2,000 at t = 1, 2, and 3, respectively. If the fund pays 8% compounded per period, what sum will be accumulated in the fund at (a) t = 3, and (b) t = 6? 02.03-PR010 Juan deposits $1,000 in a savings account that pays 8% compounded annually. Exactly 2 years later he deposits $3,000; 2 years later he deposits $4,000; and 4 years later he withdraws all of the interest earned to date and transfers it to a fund that pays 10% compounded annually. How much money will be in each fund 4 years after the transfer? Section 2.4 Multiple Cash Flows: Uniform Series Cash Flows LEARNING OBJECTIVE 2.4 Perform time value of money calculations for uniform series of cash flows with annual compounding. 02.04-PR001 A debt of $1,000 is incurred at t = 0. What is the amount of four equal payments at t = 1, 2, 3, and 4 that will repay the debt if money is worth 10% compounded per period? 02.04-PR002 Develop a single spreadsheet that allows you to calculate any of the P|A, A|P F|A, or A|F, factors. For each cell where the calculation is performed, place above it a small “control panel” that allows you to enter whatever numbers you need to perform the calculation. For example, for the F|A factor, your control panel needs to include i and n. Test each of your factors against the tables in Appendix A to ensure they work right. Keep this handy. It will be useful to you. 02.04-PR003 Jason has been making equal annual payments of $7,500 to repay a college loan. He wishes to pay off the loan immediately after having made an annual payment. He has eight payments remaining. With an annual compound interest rate of 6%, how much should Jason pay? 02.04-PR004 Each and every year $7,500 is invested at 4% annual compound interest. a. What is the value of the investment portfolio after 20 years? After 25 years? After 30 years? b. Repeat part (a) if the investment is at 5% annual compound interest.

c. Based upon your answers to (a) and (b), what conclusions can be drawn regarding the impact of the interest earned versus the duration of the investment? 02.04-PR005 Five deposits of $500 each are made a t = 1, 2, 3, 4, and 5 into a fund paying 6% compounded per period. How much will be accumulated in the fund at (a) t = 5, and (b) t = 10? 02.04-PR006 Using a 5% annual compound interest rate, what investment today is needed in order to withdraw $5,000 annually: a. For 10 years? b. For 10 years if the first withdrawal does not occur for 3 years? 02.04-PR007 If money is worth more than 0% to you, would you rather receive $10,000/year for 5 years or receive $5,000/year for 10 years? What is your preference if you must pay these amounts, rather than receive them? 02.04-PR008 Video Solution Determine the present worth of 5 equal annual deposits of $1,200 at the end of years 1 through 5, followed by 4 equal annual withdrawals of $700 at the end of years 4 through 7. Note that both years 4 and 5 will have a deposit and a withdrawal. Interest is 5%.

02.04-PR009 Juan borrows $25,000 at 7% compounded annually. If the loan is repaid in five equal annual payments, what will be the size of Juan’s payments if the first payment is made one year after borrowing the money? 02.04-PR010 What equal annual deposits must be made at t = 2, 3, 4, 5, and 6 in order to accumulate $25,000 at t = 8 if money is worth 10% compounded annually? 02.04-PR011 Eight equal deposits of $1,000 are made at the end of each year into a fund paying 8%. a. What is the present worth, 1 year before the first deposit? b. What is the future worth, immediately after the last deposit? c. What is the future worth, 3 years after the last deposit?

02.04-PR012 Adriana wishes to accumulate $2,000,000 in 35 years. If 35 end-of-year deposits are made into an account that pays interest at a rate of 7% compounded annually, what size deposit is required each year to meet Adriana’s stated objective? 02.04-PR013 If annual deposits of $1,000 are made into a fund paying 12% interest compounded annually, how much money will be in the fund immediately after the 5th deposit? 02.04-PR014 An amount equal to $50,000 is borrowed at 7% annual compound interest. a. What size equal annual payment is required if the first of 5 payments is made one year after receiving the $50,000? b. What size payment is required if the first payment is not made until 4 years after receipt? 02.04-PR015 You purchase a house for $250,000 directly from the buyer who owns the home outright. You pay a 20% down payment. You sign a first mortgage and the buyer agrees to finance the remaining $200,000 at 7% annual compound interest with annual end-of-year payments over 12 years. How much is a single yearly payment? 02.04-PR016 You deposit $1,000 in a fund at the end of each year for a 10year period. The fund pays 5% compounded annually. How much money is available to withdraw immediately after your last deposit? 02.04-PR017 You take out a loan to buy a new audio system. Your equal annual payments are 20% of the amount you borrowed. The interest rate on the loan is 7% compounded annually. a. Determine the number of years you will be required to make payments. (This number may be a nonwhole one such as 4.791, for example.) b. If you make the same payment for an integer number of years, rounding up from your answer in part a, what interest rate will you be paying? 02.04-PR018 You take out a loan to build a swimming pool in the back yard of your new home. Your equal annual payments are 1/6th of the amount

you borrowed. If it will take you 7 years to fully repay the loan, what is the interest rate on the loan? 02.04-PR019 What uniform series over the interval [11,20] will be equivalent to a uniform series of $10,000 cash flows over the interval [1,10] based on: a. A 6% interest rate? b. A 10% interest rate? 02.04-PR020 What uniform series over the interval [1,8] will be equivalent to a uniform series of $10,000 cash flows over the interval [3,10] based on: a. A 6% interest rate? b. A 10% interest rate? 02.04-PR021 Janie deposits $10,000 in the bank today. Starting 3 years from now, she makes equal withdrawals of $1,000 for 5 years and then withdraws the remaining amount 10 years from now. How much will she be able to withdraw 10 years from now, assuming the bank pays 6% compounded annually? 02.04-PR022 You plan to open a retirement account. Your employer will match 50% of your deposits up to a limit on the match of $2,500 per year. You believe the fund will earn 12% over the next 30 years, and you will make 30 deposits of $5,000, plus 50% employer matching, totaling $7,500 per year. a. How much money will be in the account immediately after the last deposit? b. How much total money will you put into the fund? c. How much total money will your employer put into the fund? d. How much will the total investment earnings be? e. If you want the account to last for 30 years (30 withdrawals), starting 1 year after the last deposit, what amount will you be able to withdraw each year? f. If you want the account to last forever, what amount will you be able to withdraw each year? 02.04-PR023 How much money can be withdrawn at the end of 15 years if:

a. $2,000 is deposited at the end of each year and earns 5%/year compounded annually? b. $3,000 is deposited at the end of each year for 10 years and no deposits are made thereafter, where the fund earns 8%? c. $2,000 is deposited at the end of years 1 through 5, $4,000 is deposited at the end of years 6 through 10, and $6,000 is deposited at the end of years 11 through 15, with all deposits earning 8%? 02.04-PR024 How much money can be withdrawn at the end of the investment period if: a. $4,000 is invested at the end of each of 3 years at 5%/year compounded annually, with the lump sum then shifted into an investment paying 8%/year for 5 additional years? b. $12,000 is invested at the end of each of 10 years at 10%/year compounded annually, with the lump sum then shifted into an investment paying 5%/year for 3 additional years? c. $18,000 is invested at the end of each of 5 years at 9%/year compounded annually, with the lump sum then shifted into an investment paying 7%/year for 8 additional years? 02.04-PR025 Video Solution You decide to open an IRS approved retirement account at your local brokerage firm. Your best estimate is that it will earn 9%. At the end of each year for the next 25 years, you will deposit $4,000 per year into the account (25 total deposits). Three years after the last deposit, you will begin making annual withdrawals. a. How much money is in the account one year before the first withdrawal? b. If you want to make 30 annual withdrawals, what amount will you be able to withdraw each year? c. If you want the account to last forever, what amount will you be able to withdraw each year?

02.04-PR026 In planning for your retirement, you have decided that you would like to be able to withdraw $60,000 per year for a 10 year period. The first withdrawal will occur 20 years from today. a. What amount must you invest today if your return is 10% per year? b. What amount must you invest today if your return is 15% per year? 02.04-PR027 Determine the equivalent annual cash flow of this series at 10% interest: End of Year Cash Flow End of Year Cash Flow 0 − $2,500 5 $0 1

$3,000

6

− $1,000

2 3

$4,500 $0

7 8

$7,000 $3,000

4

− $5,000

02.04-PR028 Fishing Designs has arranged to borrow $15,000 today at 12% interest. The loan is to be repaid with end-of-year payments of $3,000 at the end of years 1 through 4. At the end of year 5, the remainder will be paid. What is the year 5 payment? 02.04-PR029 You deposit $X in an account on your 25th, 30th, and 35th birthdays. The account pays 9%. You intend to withdraw your savings in 10 equal annual withdrawals on your 41st, 42nd, . . . , 50th birthdays, just depleting your account. Just after making the withdrawal on your 45th birthday, you have $32,801.60 left in the account. What is $X? 02.04-PR030 You have $20,000 that you put on deposit on your 30th birthday at 5% compounded annually. On your 40th birthday, the account begins earning 6%. Then, on your 50th birthday it begins earning 7%. You plan to withdraw equal annual amounts on each of your 61st, 62nd, . . . , 70th birthdays. a. How much will be your annual withdrawal? b. On your way to the bank on your 65th birthday, you decide to withdraw the entire amount remaining. How much do you withdraw?

02.04-PR031 Develop a mathematical relationship for finding the accumulated amount F at the end of n years that will result from a series of n beginning-ofyear payments each equal to B if these payments are placed in an account for which the interest rate is i%/year. a. Express the relationship between F and B in terms of the factors listed in the tables of Appendix A. b. Express the relationship between F and B in terms of i and n. c. Demonstrate that your answers to (a) and (b) are equivalent by calculating the value of F using B = $1,000, n = 5, and i = 10% for each approach. Section 2.5 Multiple Cash Flows: Gradient Series Cash Flows LEARNING OBJECTIVE 2.5 Perform time value of money calculations for gradient series of cash flows with annual compounding. 02.05-PR001 On Juan’s 26th birthday, he deposited $7,500 in a retirement account. Each year thereafter he deposited $1,000 more than the previous year. Using a gradient series factor, determine how much was in the account immediately after his 35th deposit if: a. The account earned annual compound interest of 5%. b. The account earned annual compound interest of 6%. 02.05-PR002 Develop a single spreadsheet that allows you to calculate the P|G, F|G, and A|G factors. For each cell where the calculation is performed, place above it a small “control panel” that allows you to enter whatever numbers you need to perform the calculation. For example, for the P|G factor, your control panel needs to include i and n. Test the P|G and A|G factors against the tables in Appendix A to ensure they work right. Keep this handy. It will be useful to you. 02.05-PR003 John borrows $10,000 at 18% compounded annually. He pays off the loan over a 5-year period with annual payments. Each successive payment is $700 greater than the previous payment. How much was the first payment? 02.05-PR004 Solve problem 02.05-PR003 for the case in which each successive payment is $700 less than the previous payment.

02.05-PR005 Video Solution A small company wishes to set up a fund that can be used for technology purchases over the next 6 years. Their forecast is for $12,000 to be needed at the end of year 1, decreasing by $2,000 each year thereafter. The fund earns 8% per year. How much money must be deposited to the fund at the end of year 0 to just deplete the fund after the last withdrawal?

02.05-PR006 Video Solution Deposits are made at the end of years 1 through 7 into an account paying 6% per year interest. The deposits start at $5,000 and increase by $1,000 each year. How much will be in the account immediately after the last deposit?

02.05-PR007 Video Solution You want to be able to withdraw from a savings account $800 at the end of year 1, $900 at the end of year 2, $1,000 at the end of year 3, and so on over a total of 5 years. How much money must be on deposit right now, at the end of year 0, to just deplete the account after the 5 withdrawals if interest is 5% compounded annually?

02.05-PR008 Consider the following cash flow profile: EOY Cash Flow EOY Cash Flow 0 1

− $75,000 $3,000

5 6

$15,000 $18,000

2

$6,000

7

$21,000

3 4

$9,000 $12,000

8

$24,000

Using a gradient series factor, determine the present worth equivalent for the cash flow series using an annual compound interest rate of: a. 6% b. 7%

02.05-PR009 A person you trust foresees the need for a loan and suggests that you loan them $2,000 at the end of year 1, $1,000 at the end of year 2, nothing in year 3, and then they will pay you $1,000 in year 4, $2,000 in year 5, and $3,000 in year 6. They note that you will pay out a total of $3,000 to them, and then they will pay back $6,000 to you, allowing you to “double your money.” If you are able to make 12% per year on your investments, determine the present worth of this series of cash flows. 02.05-PR010 A $90,000 investment is made. Over a 5-year period, a return of $30,000 occurs at the end of the first year. Each successive year yields a return that is $3,000 less than the previous year’s return. If money is worth 5%, use a gradient series factor to determine the equivalent present worth for the investment. 02.05-PR011 A series of 10 end-of-year deposits is made that begins with $7,000 at the end of year 1 and decreases at the rate of $300 per year with 10% interest. a. What amount could be withdrawn at t = 10 b. What uniform annual series of deposits (n = 10) would result in the same accumulated balance at the end of year 10. 02.05-PR012 Consider the following cash flow profile: EOY Cash Flow EOY Cash Flow 0

− $45,000

5

$8,000

1 2

$12,000 $11,000

6 7

$7,000 $6,000

3

$10,000

8

$5,000

4

$9,000

With a compounded annual interest rate of 6%, what single sum of money at the end of the sixth year will be equivalent to the cash flow series? 02.05-PR013 In Problem 02.05-PR012, what uniform annual series over [4,7] will be equivalent to the cash flow profile if money is worth 6% compounded annually?

02.05-PR014 In Problem 02.05-PR012, suppose the positive-valued cash flows are replaced by a positive gradient series. If the cash flow at end of year 8 is $10,000, what gradient step is required for the cash flow profiles to be equivalent? 02.05-PR015 flows.

Consider a loan of $10,000 and the following pattern of cash

End of Year Cash Flow End of Year Cash Flow 0

− $10,000

3

$5,000

1 2

$3,000 $4,000

4

$6,000

a. What is the interest rate that makes the present worth equal to $0.00? b. Using the interest rate determined in part (a), and leaving the −$10,000 at year 0 in place, determine the equal annual incomes that are equivalent to the gradient series in years 1, 2, 3, and 4? 02.05-PR016 Consider the following cash flow profile: EOY Cash Flow EOY Cash Flow 0 − $50,000 5 $9,000 1 2 3

$13,000 $12,000 $11,000

4

$10,000

6 7 8

$8,000 $7,000 $6,000

What is the present worth equivalent for the cash flow series with an interest rate of 12%? 02.05-PR017 In Problem 02.05-PR016, using an interest rate of 10%, what uniform series over the closed interval [1,8] is equivalent to the cash flow profile shown? 02.05-PR018 In Problem 02.05-PR016, using an interest rate of 8%, what single sum of money occurring at the end-of-year 8 is equivalent to the cash flow profile shown?

02.05-PR019 In Problem 02.05-PR016, with an interest rate of 6%, what increasing gradient series is equivalent to the cash flow profile shown if the gradient series sought has a value of X at EOY = 1 and a value of 8X at EOY = 8? 02.05-PR020 A series of 25 end-of-year deposits is made that begins with $1,000 at the end of year 1 and increases at the rate of $200 per year with a 12% interest rate compounded annually. a. What amount can be withdrawn at t = 25? b. What uniform annual series of deposits (n = 25) would result in the same accumulated balance at t = 25? 02.05-PR021 Piyush has recently inherited 20 million INR (Indian rupees) from his late Uncle Scrooge. To keep Piyush from spending his money immediately, Scrooge made arrangements for the inheritance to be deposited at the time of his death into an account paying 5%. Further arrangements instructed the bank to pay Piyush 2 million INR at the end of the 1st year, 2 million + X INR at the end of the 2nd year, 2 million + 2X INR at the end of the 3rd year, 2 million + 3X INR at the end of the 4th year, and so on for a period of 10 years, just depleting the fund after the 10th payment. a. What is the value of X? b. How much is in the fund immediately after the 5th withdrawal? 02.05-PR022 Miller Machining needs to purchase a piece of machinery to be able to compete on a new contract with a first-tier automotive supplier. The machinery will cost $140,000 and the owner arranges to borrow the entire amount at 8% interest. The initial payment 1 year after purchase is $11,000 with successive payments increasing each year by $X. The last payment is to be made 6 years after the purchase. a. By how much ($X) does the payment increase each year? b. What is the amount of the final payment? c. Suppose that, at the last minute, the company decides to purchase the same machinery at the same rate (8%), with payments decreasing by $7,500 each year. How much is the first payment?

02.05-PR023 Below is an equation to compute the present value of a cash flow series. Determine the cash flow profile that is implied by the equation. P = −7,000 + [1,850 + 200 (A|G8%, 6)] (P |A8%, 6) (P |F 8%, 4)

02.05-PR024 A cash flow profile starts with $2,000 and increases by $1,000 each year up to $21,000 at time 20. Then, it starts again with $21,000 at time 21 and decreases by $1,000 each year to $2,000 at year 40. You desire to convert it to an equivalent gradient series beginning at year 1 with $X and continuing through year 40 with $500 increases each year (ending at $X + 19,500 at time 40). Interest is 8% compounded annually. What is X? 02.05-PR025 Your friend claims that the following series of payments is absolutely worthless because they add up to $0: End of Year Cash Flow End of Year Cash Flow 0 $100 6 − $20 1 $80 7 − $40 2 3 4

$60 $40 $20

5

$0

8 9 10

− $60 − $80 − $100

The time value of money is 18%. Determine the present value of these cash flows. 02.05-PR026 Below is an equation to compute the present value of a cash flow series. Determine the cash flow profile that is implied by the equation. P = 800 + 950 (P |A i%, 4) − 450 (P |G i%, 4) − 600 (P |A i%, 3) (P |F  i%, 4)

02.05-PR027 Land is purchased for $75,000. It is agreed for the land to be paid for over a 5-year period with annual payments and using a 12% annual compound interest rate. Each payment is to be $3,000 more than the previous payment. Determine the size of the last payment. 02.05-PR028 Solve Problem 02.05-PR027 for the case in which each successive payment is to be $3,000 less than the previous payment.

02.05-PR029 What single sum of money at the end of the 3rd year is equivalent to a payment series of $10,000 the 1st year, $9,000 the 2nd year, . . . , down to $6,000 the 5th year? Assume that money has a time value of 10%/year compounded annually. 02.05-PR030 Develop a mathematical relationship for finding the accumulated amount F at the end of n years of a geometric series where the interest is i%. Put differently, you already have access to a (P|G i%,n) factor. Develop an (F|G i%,n) factor. a. Express the (F|G i%,n) factor in terms of the existing factors listed in the tables of Appendix A. b. Express the (F|G i%,n) factor in terms of i and n. c. Demonstrate that your answers to (a) and (b) are equivalent by calculating the value of F using a first payment of $0, increasing by $1,000 each year with n = 5 and i = 10% for each approach. 02.05-PR031 Kim deposits $1,000 in a savings account. Four years after the deposit, half the account balance is withdrawn. Then, $2,000 is deposited annually for an 8-year period, with the first deposit occurring 2 years after the withdrawal. The total balance is withdrawn 15 years after the initial deposit. If the account earned interest of 8% compounded annually over the 15-year period, how much was withdrawn at each withdrawal point? 02.05-PR032 An easy payment plan offered by a local electronics store for your new audio system calls for end-of-year payments of $2,000, $2,500, $3,000, and $3,500 at the ends of years 1 through 4 respectively. Your money is well invested and earns a consistent 10% per year. a. What is the present worth of these payments? b. If you prefer to make equal annual payments having the same present worth, how much would they be? Section 2.6 Multiple Cash Flows: Geometric Series Cash Flows LEARNING OBJECTIVE 2.6 Perform time value of money calculations for geometric series of cash flows with annual compounding.

02.06-PR001 Solve Problem 02.05-PR003 for the case in which each successive payment is to be 10% greater than the previous payment. 02.06-PR002 Solve problem 02.05-PR003 for the case in which each successive payment is to be 10% less than the previous payment. 02.06-PR003 Solve Problem 02.05-PR027 for the case in which each successive payment is to be 12% greater than the previous payment. 02.06-PR004 Solve problem 02.05-PR027 for the case in which each successive payment is to be 20% less than the previous payment. 02.06-PR005 Consider the following cash flow profile: EOY Cash Flow EOY Cash Flow 0 − $45,000 5 $8,000 1 2

$12,000 $11,000

6 7

$7,000 $6,000

3 4

$10,000 $9,000

8

$5,000

Suppose the positive-valued cash flows are now replaced by a geometric series. If the cash flow at end of year 1 is $10,000, what geometric rate is required for the cash flow profiles to be equivalent? Interest is at a compounded annual rate of 6%. 02.06-PR006 Suppose you make 30 annual investments in a fund that pays 5% compounded annually. If your first deposit is $7,500 and each successive deposit is 5% greater than the preceding deposit, how much will be in the fund immediately after the 30th deposit? 02.06-PR007 In Problem 02.06-PR006, how much will be in the fund immediately after the 30th deposit if the fund pays 6% compounded annually and each successive deposit is 6% greater than the preceding deposit? 02.06-PR008 Develop a single spreadsheet that allows you to calculate the (P|A1 i%,j%,n), (F|A1, i%,j%,n), and (A|A1 i%,j%,n) factors. Note that the (A|A1 i%,j%,n) factor is not in Appendix A. For each cell where the calculation is

performed, place above it a small “control panel” that allows you to enter whatever numbers you need to perform the calculation. For example, for the (P|A1 i%,j%,n) factor, your control panel needs to include i, j, and n. Test the first two factors using i = 7%, j = 5% and n = 10 against the tables in Appendix A to ensure they work right. Then, using A1 = $1,000, i = 7%, and j = 5%, calculate A using the third factor and check it against (P|A1 i%,j%,n)*(A|P i%,n). Keep this handy. It will be useful to you. 02.06-PR009 A famous high-volume calculus text generates royalties beginning with $60,000 in the first year and declining each year by 40% of the previous year due to used sales and competition. The author is on a 4-year cycle of revision. Determine the present worth of one complete cycle of royalties if the author’s time value of money is 7%. 02.06-PR010 An easy payment plan offered by a local electronics store for your new audio system calls for end-of-year payments of $2,000 at the end of year 1, increasing by 15% each year thereafter through year 4. Your money is well invested and earns a consistent 10% per year. a. What is the present worth of these payments? b. If you prefer to make equal annual payments having the same present worth, how much would they be? 02.06-PR011 Video Solution A $90,000 investment is made. Over a 5-year period, a return of $30,000 occurs at the end of the first year. Each successive year yields a return that is 10% less than the previous year’s return. If money is worth 5%, what is the equivalent present worth for the investment?

02.06-PR012 You are preparing the business plan for a new company. A net revenue analysis covering the first 6 years is required for obtaining financing. Net revenue in year 1 is expected to be $50,000 and increase by 15% each year, thereafter. If i = 12% and the net revenue is assumed to be an end-of-year cash flow, what is the present value of the cash flow series over the 6 years? 02.06-PR013 On your child’s first birthday, you open an account to fund her college education. You deposit $2,000 to open the account. Each year, on her birthday, you make another deposit, with each being 10% larger than the

previous deposit. The account pays interest at 5% per year compounded annually. How much money is in the account immediately after the deposit on her 18th birthday? 02.06-PR014 A cash flow series is increasing geometrically at the rate of 6% per year. The initial cash flow at t = 1 is $1,000. The increasing payments end at t = 20. The interest rate in effect is 15% compounded annually. Find the present amount at t = 0 that is equivalent to this cash flow series. 02.06-PR015 A small company wishes to set up a fund that can be used for technology purchases over the next 6 years. Their forecast is for $9,000 to be needed at the end of year 1, increasing by 5% each year thereafter. The fund earns 10% per year. How much money must be deposited to the fund at the end of year 0 to just deplete the fund after the last withdrawal? 02.06-PR016 A boat is purchased by financing $50,000. The loan is to be paid for over a 5 year period with annual payments based on a 15% compounding rate per year. Each successive payment is scheduled to be 10% greater than the previous one. a. Determine the size of the smallest payment. b. Determine the size of the largest payment. 02.06-PR017 You want to be able to withdraw $1,000 from a savings account at the end of year 1, with withdrawals increasing by 10% each year thereafter over a total of 5 years. How much money must be on deposit right now, at the end of year 0, to just deplete the account after the 5 withdrawals if interest is 5% compounded annually? 02.06-PR018 Deposits are made at the end of years 1 through 7 into an account paying 5% per year interest. The deposits start at $4,000 and increase by 15% each year. How much will be in the account immediately after the last deposit? 02.06-PR019 Susan gets a job upon completion of her MSME degree with a mechanisms design firm. Her starting salary is $70,000; each successive year she gets a 5% raise. Assuming she deposits 10% of her salary each year into a fund earning 8% interest, how much money will she have in 10 years for donation to her university?

02.06-PR020 In a new, highly automated factory, labor costs are expected to decrease at an annual rate of 5%; material costs will increase at an annual rate of 4%; overhead costs will increase at 8%. The labor, material, and overhead costs at the end of the first year are $2 million, $3 million, and $1.6 million, respectively. The time value of money rate is 11% and the time horizon is 7 years. a. Determine the dollar value for each cost category (labor, material, overhead) for each year and the total cost for each year. (Hint: Use a spreadsheet) b. Determine the present worth of each cost category and the total cost. c. Determine the annual worth over 7 years that is equivalent to the present worth of the total cost. 02.06-PR021 In a new, highly automated factory, labor costs are expected to decrease at an annual compound rate of 5%; material costs are expected to increase at an annual compound rate of 6%; and energy costs are expected to increase at an annual compound rate of 3%. The labor, material, and energy costs the first year are $3 million, $2 million, and $1,500,000. a. What will be the value of each cost during each of the first 5 years? b. Using an interest rate of 10% compounded annually, what uniform annual costs over a 5-year period would be equivalent to the cumulative labor, material, and energy costs? 02.06-PR022 Video Solution On Juan’s 26th birthday, he invested $7,500 in a retirement account. Each year thereafter he deposited 8% more than the previous deposit. The account paid annual compound interest of 5%. How much was in the account immediately after his 35th deposit?

02.06-PR023 In Problem 02.06-PR022, if Juan decided to wait 10 years before investing for retirement, how much would he have to invest on his 36th birthday to have the same account balance on his 60th birthday? 02.06-PR024 In Problem 02.06-PR022, what uniform annual investment is required to achieve the same account balance?

02.06-PR025 Carlson Photography receives royalties based on the use of their photographs with a major client. Every year, the client makes deposits in Carlson’s bank account that earns 6% compounded annually. The client increases the amount they deposit into the Carlson account by 4%. If the client initially gives Carlson $15,000, how much will the account have in 5 years? Section 2.7 Compounding Frequency LEARNING OBJECTIVE 2.7 Perform time value of money calculations for multiple compounding periods per year. Section 2.7.1 Period Interest Rate 02.07-PR001 How many monthly payments are required to repay a loan of $12,000 with an interest rate of ¾% per month and end-of-month payments of $400? 02.07-PR002 A total of $50,000 is borrowed and repaid with 60 monthly payments, with the first payment occurring one month after receipt of the $50,000. The stated interest rate is 6% compounded monthly. What monthly payment should be made? 02.07-PR003 A refrigerator sold for $500. The store financed the refrigerator by charging 0.5% monthly interest on the unpaid balance. If the refrigerator is paid for with 30 equal end-of-month payments: a. What will be the size of the monthly payments? b. If the first payment is not made until one year after the purchase, what will be the size of the monthly payments? 02.07-PR004 You decide to open a retirement account at your local bank that pays 8%/year/month (8% per year compounded monthly). For the next 20 years you will deposit $400 per month into the account, with all deposits and withdrawals occurring at the end of the month. On the day of the last deposit, you will retire. Your expenses during the first year of retirement will be covered by your company’s retirement plan. As such, your first withdrawal from your retirement account will occur on the day exactly 12 months after the last deposit.

a. What monthly withdrawal can you make if you want the account to last 15 years? b. What monthly withdrawal can you make if you want the account to last forever (with infinite withdrawals)? 02.07-PR005 Wei Min opens a retirement account that pays 8%/year/month. For the next 30 years he deposits $300 per month into it, with all deposits occurring at the end of the month. On the day of the last deposit, Wei Min retires. As a benefit to retirees, the bank increases the interest rate to 12%/year/quarter from that time on. His first withdrawal will occur exactly 2 years after his last deposit. He then plans to make equal quarterly withdrawals from the account. a. What is the balance of the account immediately after the last monthly deposit? b. What is the balance of the account one quarter before the first quarterly withdrawal? c. What quarterly amounts can be withdrawn to last for 15 years? 02.07-PR006 Video Solution Your boss, who never took an engineering economy course is buying a new house and needs your help in answering some questions. The loan amount will be in the “jumbo loan” category of $600,000 at (1) 7.0% per year compounded monthly over 30 years, or (2) 6.625% compounded monthly over 15 years. There are no loan initiation fees, points paid, or other charges. Prepayment, if desired, can be done without penalty. a. What is the monthly payment for plan (1)? b. What is the monthly payment for plan (2)? c. What is the effective annual interest rate for plan (1)? d. What is the effective annual interest rate for plan (2)?

e. What is the total interest paid over the life of loan (1)? f. What is the total interest paid over the life of loan (2)?

02.07-PR007 Yavuz wishes to make a single deposit P at time t = 0 into a fund paying 15% compounded quarterly such that $1,000 payments are received at t = 1, 2, 3, and 4 (periods are 3-month intervals), and a single payment of $7,500 is received at t = 12. What single deposit is required? 02.07-PR008 Video Solution You invest $10,000 in a fund that pays 7% per year for 5 years. How much is in the fund at the end of 5 years if (forgetting leap years and making “convenient” assumptions): a. Compounding is annual? b. Compounding is quarterly? c. Compounding is monthly? d. Compounding is daily?

02.07-PR009 Video Solution How much money must be deposited now in order to be able to withdraw $10,000 in 4 years if interest is 5% compounded quarterly?

02.07-PR010 If you deposit $4,000 into an account paying 6% per year compounded semiannually, how much will you have in the account after 10 years?

02.07-PR011 Lynn borrows $5,000 at 15% per year compounded monthly. She wishes to repay the loan with 12 end-of-month payments. She wishes to make her first payment 1 month after receiving the $5,000. She also wishes that, after the first payment, the size of each payment be 10% greater than the previous payment. What is the size of her 6th payment? 02.07-PR012 Solve problem 02.07-PR011 for the case in which the size of each payment is $60 greater than the previous payment. 02.07-PR013 Mary Lib purchases a house for $450,000. She makes a down payment of $40,000 at the time of purchase, and the balance is financed at 6.0% compounded monthly, with monthly payments made over a 10-year period. a. What is the size of the monthly payments? b. If the loan period had been 20 years, what would have been the size of the monthly payments? 02.07-PR014 What equal monthly investment is required over a period of 40 years to achieve a balance of $2,000,000 in an investment account that pays monthly interest of ¾%? Section 2.7.2 Effective Annual Interest Rate 02.07-PR015 Develop a single spreadsheet that allows you to (1) calculate the effective annual interest rate given the nominal annual interest rate and the number of compounding periods per year, and (2) calculate the nominal annual interest rate given the effective annual interest rate and the number of compounding periods per year. Use the “control panel” approach that allows you to enter whatever numbers you need to perform the calculation. Keep this handy. It will be useful to you. 02.07-PR016 What is the effective annual interest rate for 10% compounded (a) semi- annually, (b) every four months, (c) quarterly, (d) every other month, (e) monthly? 02.07-PR017 Video Solution What is the effective annual interest rate for 5% compounded (a) semi-annually, (b) every four months, (c) quarterly, (d) every other month, (e) monthly?

02.07-PR018 Video Solution You borrow $2,000 from Gougo’s, a well-known loan consolidation outfit. The loan is an “unbelievably low” 2.5% per month compounded monthly. You have 2 years to pay back the loan. a. What is the nominal interest rate? b. What is the effective interest rate? c. If you wait until the end of year 2 to pay it off in one lump sum, how much must you pay? Use the “period interest rate” approach. d. If you wait until the end of year 2 to pay it off in one lump sum, how much must you pay? Use the “effective interest rate” approach. e. Of your payment in parts (c) or (d), how much is interest? f. Suppose you make equal end-of-month payments. How much is the monthly amount?

02.07-PR019 How much money must be invested in an account that pays 6% per year interest to be worth $20,000 at the end of 8 years if (forgetting leap years and making “convenient” assumptions): a. Interest is compounded annually? b. Interest is compounded semi-annually? c. Interest is compounded quarterly? d. Interest is compounded monthly? e. Interest is compounded weekly? f. Interest is compounded daily? g. Interest is compounded hourly?

h. Interest is compounded minutely? i. Interest is compounded secondly? 02.07-PR020 You have your eyes on a new automobile costing $25,000. If you had the $25,000 and wrote a check for that amount, you could drive off in your new car. You don’t have it, and must finance $20,000 through the dealership at 15%/year/month over a 5-year period. The dealer then proceeds to add on a 1.25% loan initiation fee of $250. Also, they have a prepaid loan closeout fee of another $250. Then there is the paperwork filing and storage fee of another $100 and another prepaid loan maintenance fee of only $8/month or $480. At this point, they are speaking very fast and assure you that these little “required” amounts are routine and can be rolled into your loan. They figure your monthly payment for the $20,000 loan as A = $21,080 (A|P 1.25,60) = $501.49. a. What is the monthly rate of “interest” you are really paying for the $20,000 loan? b. What is the nominal annual “interest” rate you are really paying for the $20,000 loan? c. What is the effective annual “interest” rate you are really paying for the $20,000 loan? 02.07-PR021 Naihui and Haiyan deposit $250 into their joint account at the end of each month. They want to have a total of $12,000 in their account after 40 months: a. What monthly rate of interest must they earn? b. What nominal annual rate of interest must they earn? c. What effective annual rate of interest must they earn? 02.07-PR022 Barbara makes four consecutive annual deposits of $2,000 in a savings account that pays interest at a rate of 10% compounded semiannually. How much money will be in the account 2 years after the last deposit? 02.07-PR023 Video Solution You are down on your luck and need a loan, quick! You locate Mr. Loa N. Shark who advertises weekly loans for “an almost imperceptibly small rate” of only 3%, prepaid at the time of the loan.

You sign over your federal tax refund for $1,000 to Mr. Shark, with proof that it is correct and will be forthcoming from the IRS in one week. a. How much money does Mr. Shark hand you? b. How much weekly interest are you really paying? c. What is the nominal annual interest rate? d. What is the effective annual interest rate?

Section 2.7.3 Differing Frequencies of Compounding and Cash Flows 02.07-PR024 Daniel deposits $20,000 into an account earning interest at 6% per year compounded quarterly. He wishes to withdraw $400 at the end of each month. For how many months can he make these withdrawals? 02.07-PR025 Daniel deposits $20,000 into an account earning interest at 6% per year compounded monthly. He wishes to withdraw $1,200 at the end of each quarter. For how many quarters can he make these withdrawals? 02.07-PR026 Mario and Claudia deposit $100 into their joint account at the end of each month. If their account earns 7%/year/quarter (7% per year compounded quarterly), how long will it take them to have a total of $15,000 in their savings account? 02.07-PR027 Video Solution A total of $50,000 is borrowed and repaid with 60 monthly payments, with the first payment occurring one month after receipt of the $50,000. The stated interest rate is 6% compounded quarterly. What monthly payment is required?

Section 2.7.4 Continuous Compounding and Continuous Cash Flows 02.07-PR028 What amount must be placed on deposit today to equal $15,000 in 4 years at 15% per year compounded continuously? 02.07-PR029 Semiannual deposits, beginning with $500 and increasing by $100 with each subsequent deposit, are made into a fund paying a nominal 10% per year compounded continuously. a. What will the fund amount to after 7 years? b. What is the present worth equivalent of the total set of deposits? c. What is the equal semiannual equivalent amount of the deposits? 02.07-PR030 A continuous uniform series of deposits totaling $1,000 per year are made into a fund paying 10% compounded continuously. a. What will the fund amount to after 7 years? b. What is the present worth equivalent of the total set of deposits?

Chapter 2 Summary and Study Guide Summary 2.1: Cash Flow Diagrams

Learning Objective 2.1: Construct a cash flow diagram (CFD) depicting the cash inflows and outflows for an investment alternative. (Section 2.1) The CFD is a powerful visual tool used to depict an investment alternative. The graphic depicts the timing, magnitude, and direction of the cash flow. Cash flows can represent present values, future values, uniform series, gradient series, or geometric series. Cash flows are positive or negative depending upon the perspective (such as borrower versus lender) from which they are drawn.

FIGURE 2.19 An Example Cash Flow Diagram 2.2: Single Cash Flows

Learning Objective 2.2: Perform time value of money calculations for single cash flows with annual compounding. (Section 2.2) In engineering economic analysis, it often is useful to transform the present sum of a single cash flow into a future sum or vice versa. The equations and factor notation for doing so are summarized in the table that follows. Single Cash Flows Present Worth of a Future Payment Future Worth of a Present Payment

P = F(1 + i)−n

Factor Notation P = F(P|F i%,n)

Excel® Notation =PV(i%,n,,−F)

F = P(1 + i)n

F = P(F|P i%,n)

=FV(i%,n,,−P)

Formula

Cash Flow Diagram

Cash Flow Diagram

2.3: Multiple Cash Flows: Irregular Cash Flows

Learning Objective 2.3: Perform time value of money calculations for irregular series of cash flows with annual compounding. (Section 2.3) Series of cash flows may exhibit a pattern or be irregular. If no regular pattern is encountered, the series is considered irregular. It is useful—and common—to transform series of cash flows for engineering economic analysis into some single equivalent cash flow, such as a present or future worth value. The equations and factor notation for making such transformations are summarized for all patterns in Appendix B. The relationships for irregular series of cash flows are summarized in the table that follows. Series Cash Flows Irregular Series Cash Flows

Formula P

=

A1 (1 + i)

−1

+A3 (1 + i)

+ A2 (1 + i)

−3

+An−1 (1 + i) +An (1 + i)

Factor Notation n

−2

P = ∑ At (P |F  i%,t)

+ …

t=1

Excel® Notation =NPV(i%,A1,A2,A3, …,An)

−(n−1)

−n

Cash Flow Diagram

2.4: Multiple Cash Flows: Uniform Series of Cash Flows

Learning Objective 2.4: Perform time value of money calculations for uniform series of cash flows with annual compounding. (Section 2.4) Series of cash flows may exhibit a pattern or be irregular. Recognizing patterns when they occur can facilitate analysis of a particular investment alternative.

The uniform cash flow pattern involves sums of equal (or uniform) values for each period. An example is the uniform payments of a monthly car loan or home mortgage payment. It is useful—and common—to transform series of cash flows for engineering economic analysis into some single equivalent cash flow, such as a present or future worth value. The equations and factor notation for making such transformations are summarized for all patterns in Appendix B. The relationships for uniform series cash flows are summarized in the table that follows. Series Cash Flows Uniform Series Cash Flows Present Worth of Uniform Series Uniform Series from Present Value

Formula

(1 + i)

n

Factor Notation

− 1

P = A[ i(1 + i)

i(1 + i) A = P[ (1 + i)

(1 + i)

Future Worth of Uniform Series

F = A[

Uniform Series from Future Value

A = F [

n

n

n

Excel® Notation

P = A(P|A i%,n)

=PV(i%,n,−A)

A = P(A|P i%,n)

=PMT(i%,n,−P)

F = A(F|A i%,n)

=FV(i%,n,−A)

A = F(A|F i%,n)

=PMT(i%,n,,−F)

]

n

] − 1

− 1 ]

i

i (1 + i)

n

] − 1

Cash Flow Diagram

2.5: Multiple Cash Flows: Gradient Series of Cash Flows

Learning Objective 2.5: Perform time value of money calculations for gradient series of cash flows with annual compounding. (Section 2.5)

Series of cash flows may exhibit a pattern or be irregular. Recognizing patterns when they occur can facilitate analysis of a particular investment alternative. The gradient series cash flow pattern involves cash flows that increase (or decrease) by a constant value, G, in each period. For example, maintenance costs for a piece of equipment might increase by a constant value each year. It is useful—and common—to transform series of cash flows for engineering economic analysis into some single equivalent cash flow, such as a present or future worth value. The equations and factor notation for making such transformations are summarized for all patterns in Appendix B. The relationships for gradient series cash flows are summarized in the table that follows. Single Cash Flows Gradient Series Cash Flows Present Worth of a Gradient Series Uniform Series Equivalent of a Gradient Series Future Worth of a Gradient Series

Formula

Factor Notation

⎡ 1 − (1 + ni)(1 + i) P = G⎢ i

⎣

(1 + i)

n

i [(1 + i)

n

n

i

2

P = G(P|G i%,n)

=NPV(i%,0,G,2G,…, (n−1)G)

A = G(A|G i%,n)

=PMT(i%,n, −NPV(i%,0,G,2G,…, (n−1)G))

F = G(F|G i%,n)

=FV(i%,n,, −NPV(i%,0,G,2G,…, (n−1)G))

⎦

] − 1]

− (1 + ni)

F = G[

⎤ ⎥

2

− (1 + ni)

A = G[

(1 + i)

−n

Excel® Notation

]

Cash Flow Diagram

2.6: Multiple Cash Flows: Geometric Series of Cash Flows

Learning Objective 2.6: Perform time value of money calculations for geometric series of cash flows with annual compounding. (Section 2.6) Series of cash flows may exhibit a pattern or be irregular. Recognizing patterns when they occur can facilitate analysis of a particular investment alternative. The geometric series cash pattern involves cash flows that increase (or decrease) by a constant percentage, j, each period. For example, labor costs might increase by a constant percentage each year.

It is useful—and common—to transform series of cash flows for engineering economic analysis into some single equivalent cash flow, such as a present or future worth value. The equations and factor notation for making such transformations are summarized for all patterns in Appendix B. The relationships for geometric series cash flows are summarized in the table that follows. Single Cash Flows

Factor Notation

Formula

Excel® Notation

Geometric Series Cash Flows Present Worth of a Geometric Series

⎧ ⎪ ⎪ P = ⎨ ⎪ ⎩ ⎪

n

1 − (1  +  j) (1  +  i) A1 [

] 

i − j

nA1 /(1 + i)

Future Worth of a Geometric Series

⎧ ⎪ ⎪ A1 [ ⎪ ⎪ ⎪ ⎪

1 − (1  +  j) (1  +  i)

i(1  +  i) (1  +  i)

(1  +  i) ⎧ ⎪ A1 [ F = ⎨ ⎩ ⎪

−n

nA1 (1 + i)

n

n

(1 + i)

n

n

] 

i  ≠  j

− 1

n−1

)

i = j

−  1

−  (1  +  j)

i − j

n−1

i(1 + i) ] [

i − j

A = ⎨ ⎪ ⎪ ⎪ ⎪ nA1 ( ⎪ ⎩ ⎪

i  ≠  j

i = j

n

Uniform Series Equivalent of a Geometric Series

P= =NPV(i%,A, A1(P|A1 (1+j)A, i%,j%,n) (1+j)2A,…, (1+j)n−1A)

−n

n

] 

i  ≠  j

i = j

A= =PMT(i%,n, A1(A|A1 −NPV(i%,A, i%,j%,n) (1+j)A, (1+j)2A,…, (1+j)n−1A))

F= =FV(i%,n,, A1(F|A1 −NPV(i%,A, i%,j%,n) (1+j)A, (1+j)2A,…, (1+j)n−1A))

Cash Flow Diagram

2.7: Compounding Frequency

Learning Objective 2.7: Perform time value of money calculations for multiple compounding periods per year. Although many engineering economic analyses assume annual compounding, in personal financing, compounding typically occurs more frequently. This compounding can be significant and should not be

ignored for economic decision-making. When cash flow frequency or compounding frequency (or both) is not annual, one of two approaches must be employed: the period interest rate approach or the effective interest rate approach. a. The interest rate using the period interest rate approach will match the interest rate to the compounding frequency, for example, the interest rate per month or quarter (where the monthly or quarterly designation represents the compounding frequency). The equation for the period interest rate is Nominal annual interest rate Period interest rate = Number of interest periods per year

b. The effective annual interest rate represents the annual interest rate that is equivalent to the period interest rate, thus annualizing the period interest rate. The formula for the effective annual interest rate is: ieff = (1 + r/m)

m

− 1 =EFFECT (r,m)

(2.43)

When cash flow and compounding frequencies differ, the effective interest rate per cash-flow period i is obtained by the equation i = (1 + r/m)

m/k

− 1

(2.45)

where r = the nominal annual interest rate for money, m = the number of compounding periods in a year, and k = the number of cash flows in a year. c. With continuous compounding at nominal annual rate, r, the effective annual rate is ieff = e

r

− 1

(2.47)

¯ ¯ ¯ When A dollars flow uniformly and continuously during a year, the annual equivalent end-of-year cash flow, A, is

r ¯ ¯ ¯ A (e − 1)

(2.50)

A = r

Important Terms and Concepts Capital Recovery Factor The amount of annual savings or recovery funds required to justify a capital investment over some period of time, n. Cash Flow Diagram (CFD) A diagram depicting the magnitude and timing of cash flowing in and out of the investment alternative. Compounding The practice of calculating interest on both the original principal and any interest accumulated to date. Effective Annual Interest Rate The annual interest rate that is equivalent to the period interest rate. Future Value The value of money at some future point in time. Geometric Series A series characterized by cash flows that increase or decrease by a constant percentage (j%) each period.

Gradient Series A series characterized by cash flows that increase by a constant amount (G) each period. Interest Rate The rate of change in the accumulated value of money. Nominal Annual Interest Rate The annual interest rate without adjustments for compounding. Period Interest Rate The nominal annual interest rate divided by the number of interest periods per year. Present Value The value of money at a present point in time. Sinking Fund Factor The amount of savings to deposit in order to accumulate a desired future amount. Uniform Series A series characterized by cash flows of equal (or uniform) value for each period.

Chapter 2 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 2.1: Cash Flow Diagrams LO 2.1: Construct a cash flow diagram (CFD) depicting the cash inflows and outflows for an investment alternative. Video Lesson: Cash Flow Diagrams Video Lesson Notes: Cash Flow Diagrams Video Solution: 02.01-PR008 Video Solution: 02.01-PR010 Video Solution: 02.01-PR012 2.2: Single Cash Flows LO 2.2: Perform time value of money calculations for single cash flows with annual compounding. Video Lesson: Simple and Compound Interest Video Lesson Notes: Simple and Compound Interest Video Lesson: Transformations—Single Cash Flow Video Lesson Notes: Transformations—Single Cash Flow Excel Video Lesson: FV Financial Function Excel Video Lesson Spreadsheet: FV Financial Function Excel Video Lesson: PV Financial Function Excel Video Lesson Spreadsheet: PV Financial Function Excel Video Lesson: GOAL SEEK Excel Video Lesson Spreadsheet: GOAL SEEK

Excel Video Lesson: NPER Financial Function Excel Video Lesson Spreadsheet: NPER Financial Function Excel Video Lesson: SOLVER Excel Video Lesson Spreadsheet: SOLVER Video Solution: 02.02-PR001 Video Solution: 02.02-PR002 Video Solution: 02.02-PR006 Video Solution: 02.02-PR007 2.3: Multiple Cash Flows: Irregular Cash Flows LO 2.3: Perform time value of money calculations for irregular series of cash flows with annual compounding. Video Lesson: Transformations–Multiple Cash Flows Video Lesson Notes: Transformations–Multiple Cash Flows Excel Video Lesson: NPV Financial Function Excel Video Lesson Spreadsheet: NPV Financial Function Video Example 2.6: Computing the Present Worth of a Series of Cash Flows Video Example 2.7: Determining the Future Worth of a Series of Cash Flows Video Example 2.8: Using the Excel NPV Function Video Solution: 02.03-PR003 Video Solution: 02.03-PR006 2.4: Multiple Cash Flows: Uniform Series of Cash Flows LO 2.4: Perform time value of money calculations for uniform series of cash flows with annual compounding. Video Lesson: Transformations–Uniform Series

Video Lesson Notes: Transformations–Uniform Series Excel Video Lesson: PMT Financial Function Excel Video Lesson Spreadsheet: PMT Financial Function Video Example 2.9: Computing the Present Worth of a Uniform Series of Cash Flows Video Example 2.10: Computing the Present Worth of a Delayed Uniform Series of Cash Flows Video Example 2.11: What Size Uniform Withdrawals Can Occur? Video Example 2.12: Determining the Size of Delayed Uniform Cash Withdrawals Video Solution: 02.04-PR008 Video Solution: 02.04-PR025 2.5: Multiple Cash Flows: Gradient Series of Cash Flows LO 2.5: Perform time value of money calculations for gradient series of cash flows with annual compounding. Video Lesson: Transformations–Gradient and Geometric Series Video Lesson Notes: Transformations–Gradient and Geometric Series Video Example 2.15: Determining the Present Worth of a Gradient Series (Sitting on Top of a Uniform Series) Video Solution: 02.05-PR005 Video Solution: 02.05-PR006 Video Solution: 02.05-PR007 2.6: Multiple Cash Flows: Geometric Series of Cash Flows LO 2.6: Perform time value of money calculations for geometric series of cash flows with annual compounding. Video Example 2.17: Determining the Future Worth of a Geometric Series

Video Solution: 02.06-PR011 Video Solution: 02.06-PR022/023/024 2.7: Compounding Frequency LO 2.7: Perform time value of money calculations for multiple compounding periods per year. Video Lesson: Effective Annual Interest Rate Video Lesson Notes: Effective Annual Interest Rate Video Lesson: Effective Rate per Payment Period Video Lesson Notes: Effective Rate per Payment Period Excel Video Lesson: EFFECT Financial Function Excel Video Lesson Spreadsheet: EFFECT Financial Function Video Example 2.19: Determining Car Payments Video Example 2.22: When Cash Flow Frequency Does Not Match Compounding Frequency Video Solution: 02.07-PR006 Video Solution: 02.07-PR008 Video Solution: 02.07-PR009 Video Solution: 02.07-PR017 Video Solution: 02.07-PR018 Video Solution: 02.07-PR023 Video Solution: 02.07-PR027 These chapter-level resources will help you with your overall understanding of the content in this chapter. Appendix A: Time Value of Money Factors Appendix B: Engineering Economics Equations Flashcards: Chapter 02

Excel Utility: TVM Factors: Table Calculator Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values Excel Utility: Monthly Payment Sensitivity Excel Utility: TVM Factors: Discrete Compounding Excel Utility: TVM Factors: Geometric Series Future Worth Excel Utility: TVM Factors: Geometric Series Present Worth Excel Data Files: Chapter 02

CHAPTER 2 Time Value of Money Calculations LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 2.1 Construct a cash flow diagram (CFD) depicting the cash inflows and outflows for an investment alternative. (Section 2.1) 2.2 Perform time value of money calculations for single cash flows with annual compounding. (Section 2.2) 2.3 Perform time value of money calculations for irregular series of cash flows with annual compounding. (Section 2.3) 2.4 Perform time value of money calculations for uniform series of cash flows with annual compounding. (Section 2.4) 2.5 Perform time value of money calculations for gradient series of cash flows with annual compounding. (Section 2.5) 2.6 Perform time value of money calculations for geometric series of cash flows with annual compounding. (Section 2.6) 2.7 Perform time value of money calculations for multiple compounding periods per year (Section 2.7)

Engineering Economics in Practice Kellie Schneider Analyzes Her Net Worth Immediately after receiving her engineering degree, Kellie Schneider began employment with a multinational company. Her initial salary was $80,000 per year. Kellie decided she would invest 10% of her gross salary each month. Based on discussions with the companyʼs personnel and with a person in human resources, she believed her salary would increase annually at a rate ranging from 3% to 15%, depending on her performance. After analyzing various investment opportunities, she anticipated she would be able to earn between 5% and 10% annually on her investment portfolio of mutual funds, stocks, bonds, U.S. Treasury notes, and certificates of deposit. Finally, Kellie believed annual inflation would vary from 2% to 5% over her professional career. With this information in hand, she calculated the range of possible values for her net worth, first after 30 years of employment, and second after 40 years of employment. She was amazed and pleased at what she learned. Discussion Questions 1. What do you suppose Kellie is attempting to do with her investment portfolio by selecting multiple forms of investments? 2. Although retirement may seem like a long time away, why is it important to start saving early for your “golden years”? 3. If interest rates end up being lower than Kellie assumes, what must she do to build her goal “nest egg”? 4. What impacts will a change in your family status (such as getting married or having a child) have on your investment decisions?

Introduction

In the previous chapter, we identified 10 principles of engineering economic analysis. Principle #1 is the subject of this chapter: Money has a time value. As noted in Chapter 1, TVOM considerations apply when moving money forward or backward in time. Recall, we referred to the movement of money forward or backward in time as discounted cash flow or DCF. Also, recall the following four DCF rules: 1. Money has a time value. 2. Quantities of money cannot be added or subtracted unless they occur at the same points in time. 3. To move money forward one time unit, multiply by 1 plus the discount or interest rate. 4. To move money backward one time unit, divide by 1 plus the discount or interest rate. In this chapter, we present the mathematics and basic operations needed to perform engineering economic analyses incorporating the DCF rules. These are the same techniques that Kellie Schneider used to analyze her future net worth. The only thing we will not cover in this chapter is how to incorporate the effects of inflation in such an analysis; we save that for Chapter 10. You will also learn how to determine the present worth and the future worth for three particular types of cash flow series. The material in this chapter serves as a foundation for the remainder of the text. Hence, a solid understanding of the mathematics and concepts contained in this chapter is essential.

2.1 Cash Flow Diagrams LEARNING OBJECTIVE Construct a cash flow diagram (CFD) depicting the cash inflows and outflows for an investment alternative. Video Lesson: Cash Flow Diagrams It is helpful to use cash flow diagrams (CFDs) when analyzing cash flows that occur over several time periods. As shown in Figure 2.1, a CFD is constructed using a segmented horizontal line as a time scale, with vertical arrows indicating cash flows. An upward arrow indicates a cash inflow or positive-valued cash flow, and a downward arrow indicates a cash outflow, or negative-valued cash flow. The arrows are placed along the time scale to correspond with the timing of the cash flows. The lengths of the arrows can be used to suggest the magnitudes of the corresponding cash flows, but in most cases, little is gained by precise scaling of the arrows. Cash Flow Diagram (CFD) A diagram depicting the magnitude and timing of cash flowing in and out of the investment alternative.

FIGURE 2.1 A Cash Flow Diagram (CFD) The CFD in Figure 2.1 depicts an expenditure of $4,000, followed by the receipt of $5,000, followed by an expenditure of $3,000, followed by a second receipt of $5,000, followed by a final expenditure of $2,000, followed by a final receipt of $5,000. This CFD is drawn from the investorʼs perspective: Downward arrows denote expenditures, and upward arrows denote receipts.

EXAMPLE 2.1 A Simple Illustration of the Time Value of Money As an illustration of how TVOM can impact the preference between investment alternatives, consider investment Alternatives A and B, having the cash flow profiles depicted in Figure 2.2. The CFDs indicate that the positive cash flows for Alternative A are identical to those for Alternative B, except that the former occurs 1 year sooner; both alternatives require an investment of $6,000. If exactly one of the alternatives must be selected, then Alternative A would be preferred to Alternative B, based on the time value of money.

FIGURE 2.2 CFDs for Alternatives A and B When faced with cash flows of equal magnitude occurring at different points in time, a corollary to the four DCF rules in Chapter 1 is when receiving a given sum of money, we prefer to receive it sooner rather than later, and when paying a given sum of money, we prefer to pay it later rather than sooner. Because we prefer to receive the $3,000 sooner, Alternative A is preferred to Alternative B.

2.1.1 End-of-Period Cash Flows In this chapter, we emphasize end-of-period cash flows and end-of-period compounding. With a finite number of periods occurring in, say, a year, we refer to the cash flows as discrete cash flows and the compounding as discrete compounding. Frequently, as a mathematical convenience, in finance courses in business schools, continuous cash flows and continuous compounding are assumed to occur. With few exceptions, in this book, we develop discrete mathematical models of financial compounding. Compounding The practice of calculating interest on both the original principal and any interest accumulated to date. Depending on the financial institution involved in personal finance transactions, savings accounts might not pay interest on deposits made in “the middle of a compounding period.” Consequently, answers obtained using the methods we describe may not be exactly the same as those provided by financial institutions. Beginning-of-period cash flows can be handled easily by noting that the end of period t is the beginning of period t + 1. For example, one can think of a payment made at the beginning of, say, March as having been made at the end of February. In this and subsequent chapters, end-of-period cash flows are assumed unless otherwise noted.

2.1.2 Benefits of CFDs

Drawing CFDs for economic transactions is important for at least two reasons. First and foremost, CFDs are powerful communication tools. A CFD presents a clear, concise, and unambiguous description of the amount and timing of all cash flows associated with an economic analysis. Usually, a well-drawn CFD can be readily understood by all parties of an economic transaction regardless of whether they have had any formal training in economic analysis. A second important reason for drawing CFDs is that they frequently aid in the identification of significant cash flow patterns that might exist within an economic transaction. One such pattern, for example, is a uniform series where all of the cash flows are equally spaced and have the same magnitude over a certain period of years. Uniform Series A series characterized by cash flows of equal (or uniform) value for each period.

Concept Check 02.01-CC001 In a cash flow diagram, an upward arrow indicates a cash inflow or positive-valued cash flow. True or False?

Concept Check 02.01-CC002 When faced with cash flows of equal magnitude occurring at different points in time, we prefer to: a. Receive it later than sooner b. Receive it sooner than later c. Pay it sooner than later d. Simplify the cash flows by summing them to determine the net cash flow

2.2 Single Cash Flows LEARNING OBJECTIVE Perform time value of money calculations for single cash flows with annual compounding. Video Lesson: Transformations—Single Cash Flow Our analysis of the time value of money begins with the simplest scenario: a single cash flow.

2.2.1 Future Worth Calculations (F|P) In nearly all business and lending situations, compound interest is used. For that reason, interest should be charged (or earned) against both the principal and accumulated interest to date. Such a process is called compounding. It is at the heart of everything else we do in the text. When compound interest is used, the interest rate (i) is interpreted as the rate of change in the accumulated value of money, and In is the accumulated interest over n years, given by n

In = ∑ iFt−1 t=1

(2.1)

Interest Rate The rate of change in the accumulated value of money. where t increments the years from 1 to n, F0 = P where P is the current or present value of a single sum of money, Fn is the accumulated value of P over n years, and (2.2)

Fn = Fn−1 (1 + i)

Present Value The value of money at a present point in time.

EXAMPLE 2.2 Compounding Interest for 5 Years Suppose you loan $10,000 for 1 year to an individual who agrees to pay you interest at a compound rate of 10%/year. At the end of 1 year, the individual asks to extend the loan period an additional year. The borrower repeats the process several more times. Five years after loaning the person the $10,000, how much would the individual owe you? Video Lesson: Simple & Compound Interest Key Data Given Loan amount = $10,000; interest rate = 10%/year; length of loan = 5 years Find Value of the loan amount after 5 years Solution As shown in Table 2.1, the $10,000 owed, compounded over a 5-year period at 10% annual compound interest, totals $16,105.10. TABLE 2.1 Tabular Solution to Example 2.2 Unpaid Balance at the Beginning of Year the Year 1 $10,000.00 2 $11,000.00 3 $12,100.00 4 $13,310.00 5 $14,641.00

Annual Interest $1,000.00 $1,100.00 $1,210.00 $1,331.00 $1,464.10

Payment $0.00 $0.00 $0.00 $0.00 $16,105.10

Unpaid Balance at the End of the Year $11,000.00 $12,100.00 $13,310.00 $14,641.00 $0.00

Excel® Data File

Calculating F with Equations The previous example involved two cash flows: an amount borrowed and an amount repaid. We can generalize the loan example and develop an equation to determine the amount owed after n periods, based on a compound interest rate of i%/period, if P is borrowed. As shown in Table 2.2, the future amount, F, owed is related to P, i, and n as follows:

F = P (1 + i)

(2.3)

n

TABLE 2.2 Derivation of Equation 2.3 (A)

(B)

End of Period Amount Owed Interest for Next Period

(C) = (A) + (B) Amount Owed for Next Period*

0

P

Pi

P + Pi = P(1 + i)

1

P(1 + i)

P(1 + i)i

P(1 + i) + P(1 + i)i = P(1 + i)2

2

P(1 + i)2

P(1 + i)2 i

P(1 + i)2 + P(1 + i)2 i = P(1 + i)3

3

P(1 + i)3

P(1 + i)3 i

P(1 + i)3 + P(1 + i)3 i = P(1 + i)4

⋮

⋮

⋮

⋮

n−1

P(1 + i)n−1

P(1 + i)n−1 i

P(1 + i)n−1 + P(1 + i)n−1 i = P(1 + i)n

n

P(1 + i)n

*Notice, the value in column (C) for the end of period (n − 1) provides the value in column (A) for the end of period n.

where i is expressed as a decimal amount or as an equivalent percentage. As a convenience in computing values of F (the future worth) when given values of P (the present worth), the quantity (1 + i)n is tabulated in Appendix A for various values of i and n. The quantity (1 + i)n is referred to as the single sum, future worth factor. It is denoted (F|P i%, n) and reads “the F, given P factor at i% for n periods.” The above discussion is summarized as follows. Let

P

=

the equivalent value of an amount of money at a present point in time, or present worth.

F

=

the equivalent value of an amount of money at a future point in time, or future worth.

i

=

the interest rate per interest period.

n

=

the number of interest periods.

Thus, the mathematical relationship between the future and present worth is given by Equation 2.3; the mnemonic representation is given by1 F = P (F |P  i%,n)

(2.4)

A CFD depicting the relationship between F and P is given in Figure 2.3. Remember, F occurs n periods after P.

FIGURE 2.3 CFD of the Time Relationship Between P and F Calculating F with Excel®

In addition to solving numerically for the value of (1 + i)n and using tabulated values in Appendix A of the single sum, future worth factor, (F|P i%,n), an Excel® financial function2 can be used to solve for the future worth. Specifically, the FV, or future value, function can be used. The parameters for the FV function are, in order of appearance, interest rate (i), number of periods (n), equal-sized cash flow per period (A), present amount (P), and type, which denotes either end-of-period cash flows (0 or omitted) or beginning-of-period cash flows (1). Future Value The value of money at some future point in time. Note that this factor notation will be used throughout the text, and often serves as a “shorthand” method for expressing the formulation of a problem. In some cases a pre-calculated tabulated value of the factor may be found in Appendix A. However, not all factors for all rates are tabulated in the Appendix. In cases where the function notation is presented in a problem formulation, one may always choose to solve using an equivalent formula and/or Excel® function. Reference Appendix B to view the equivalent Formula and Excel® Notation for a given Factor Notation. To solve for F when given i, n, and P, the answer can be obtained by entering the following in any cell in an Excel® spreadsheet: =FV(i,n,,−P). Notice there are no spaces between the equal sign and the closing parenthesis; also, two commas are placed between n and P in the function, because no equal-sized cash flow per period applies, and type is not included, because end-of-period cash flows apply. Finally, notice that a negative value is entered for P, because the sign of the value obtained for F by using the FV function will be opposite the sign used for P. The reason for the sign change in Excel® is simple: The FV function was developed for a loan situation where $P are loaned (negative cash flow) in order to receive $F (positive cash flow) n periods in the future when the money is loaned at i% interest per period. Excel® Video Lesson: The FV Function Recall Example 2.2, in which $10,000 was loaned for 5 years at 10% annual compound interest. An Excel® solution to the example is shown in Figure 2.4. As indicated, the value of i can be entered as a decimal or as a percentage, and the value of P can be entered as either a positive or a negative amount, because the sign for the future value obtained will be opposite that used for P. Finally, notice that dollar signs and commas are not used in denoting the value of P.

FIGURE 2.4 The Excel® FV Financial Functions Used with a Single Sum of Money Excel® Data File

EXAMPLE 2.3 Repaying a 5-Year Loan with a Single Payment Dia St. John borrows $1,000 at 12% compounded annually. The loan is to be paid back after 5 years. How much should she repay? Key Data Given Loan amount = $1,000; interest rate = 12% compounded annually; length of loan = 5 years Find Value of the loan amount after 5 years Solution Using the compound interest tables in Appendix A for 12% and 5 periods, the value of the single sum, future worth factor (F|P 12%,5) is shown to be 1.76234. Thus, F

=

P (F |P  12%,5)

=

$1,000(1.76234)

=

$1,762.34

Similarly, using Equation 2.3, F

n

=

P (1 + i)

=

$1,000(1 + 0.12)

=

$1,762.34

5

Excel® Solution Using the Excel® FV worksheet function, F =FV(12%,5,,−1000) = $1,762.34 Doubling the Value of an Investment A familiar application of a future worth analysis for a single cash flow is to wonder how long it would take for an investment to double in value if invested at i% compounded per period. It turns out there are several ways of answering this question: 1. Obtain an approximation by using the Rule of 72. 2. Consult the tables in Appendix A for the stated interest rate and find the value of n that makes the (F|P i%,n) factor equal 2, then interpolate as necessary. 3. Solve mathematically for the value of n that makes (1 + i)n equal 2. 4. Use the Excel® NPER worksheet function. Excel® Video Lesson: NPER Financial Function 5. Use the Excel® GOAL SEEK tool. Excel® Video Lesson: GOAL SEEK Tool 6. Use the Excel® SOLVER tool.

Excel® Video Lesson: SOLVER Tool Example 2.4 explores each of these methods.

EXAMPLE 2.4 Doubling Your Money How long does it take for an investment to double in value if it earns (a) 2%, (b) 4%, or (c) 12% annual compound interest? Solution Rule of 72 Approximation The first approach is to apply a rule of thumb called the Rule of 72. Specifically, the quotient of 72 and the interest rate provides a reasonably good approximation of the number of interest periods required to double the value of an investment: a. n ≈ 72/2 = 36 yrs b. n ≈ 72/4 = 18 yrs c. n ≈ 72/12 = 6 yrs TVOM Factor Table Solution The second approach is to consult Appendix A and, for each value of i, determine the value of n for which (F|P i%,n) = 2. a. For i = 2%, n is between 30 and 36; interpolating gives n ≈ 30 + 6(2.0000 − 1.81136)/(2.03989 − 1.81136) = 34.953 yrs. b. n ≈ 17 + (2.0000 − 1.94790)/(2.02582 − 1.94790) = 17.669 yrs; c. n ≈ 6 + (2.0000 − 1.97382)/(2.21068 − 1.97382) = 6.111 yrs. Mathematical Solution With the third approach, solving mathematically for n such that (1 + i)n = 2 gives n = log 2/log(1 +i). Therefore, the correct values of n (to 3 decimal places) are a. n = log 2/log 1.02 = 35.003 yrs; b. n = log 2/log 1.04 = 17.673 yrs; and c. n = log 2/log 1.12 = 6.116 yrs. Excel® Solutions The parameters of the Excel® NPER worksheet function are, in order of placement, interest rate, equal-sized cash flow per period, present amount, future amount, and type. As before, type refers to end-of-period (0 or omitted) versus beginning-of-period (1) cash flows. Letting F equal 2 and P equal −1, the NPER function yields identical results to those obtained mathematically: a. n =NPER(2%,,−1,2) = 35.003 yrs b. n =NPER(4%,,−1,2) = 17.673 yrs and c. n =NPER(12%,,−1,2) = 6.116 yrs To use the Excel® GOAL SEEK tool requires a spreadsheet, as shown in Figure 2.5. Letting x denote the row number in the spreadsheet, the parameters for GOAL SEEK are the following: Set cell: Cx To value: 2

By changing cell: Bx

FIGURE 2.5 The Excel® Goal Seek Tools Used to Solve Example 2.4 Excel® Data File Any number is entered in cell Bx, and the future value of $1, invested at interest rate Ax for Bx number of years, is calculated using the Excel® FV worksheet function. Then, the Excel® GOAL SEEK tool is used to determine the value of Bx that makes Cx = 2. The results obtained by GOAL SEEK are shown in Figure 2.5. Namely, a. n = 34.999 yrs b. n = 17.672 yrs c. n = 6.116 yrs The same spreadsheet used with GOAL SEEK can be used with the Excel® SOLVER tool, as shown in Figure 2.6. The SOLVER parameters are the following: Set Target Cell: Cx Equal To: ○ Max ○ Min ● Value of: 2 By Changing Cells: Bx

FIGURE 2.6 The Excel® Solver Tool Used to Solve Example 2.4 Excel® Data File As with GOAL SEEK, any number is entered in cell Bx, and the future value of $1, invested at interest rate Ax for Bx number of years, is calculated using the Excel® FV worksheet function. Then, the Excel® SOLVER tool is used to determine the value of Bx that makes Cx = 2, where x denotes the row number in the spreadsheet. The results obtained by SOLVER are shown in Figure 2.6. Namely,

a. n = 35.003 yrs b. n = 17.673 yrs c. n = 6.116 yrs Exploring the Solution The answers obtained using GOAL SEEK and SOLVER differ from each other and those obtained using the Excel® NPER worksheet function, especially if the calculation is carried out to 8 or 10 decimal places. GOAL SEEK and SOLVER use a search procedure that can end prematurely (i.e., before obtaining an exact solution). Of the six approaches to solving this example, only the mathematical one using logarithms, and the Excel® NPER function yielded exact solutions.

2.2.2 Present Worth Calculations (P|F) Because we can determine values of F when given values of P, i, and n, it is a simple matter to determine the values of P when given values of F, i, and n. In particular, from Equation 2.3, F = P (1 + i)

n

(2.5)

dividing both sides of Equation 2.5 by (1 + i)n, we have the relation, P = F (1 + i)

−n

(2.6)

or P = F (P |F  i%,n)

(2.7)

where (1 + i)–n and (P|F i%,n) are referred to as the single sum, present worth factor. In addition to solving numerically for the value of (1 + i)–n and using tabulated values of the single sum, present worth factor, (P|F i%,n), provided in Appendix A, an Excel® financial function can be used to solve for the present worth—specifically, the PV, or present value function. The parameters of the PV function, in order, are interest rate (i), number of periods (n), equal-sized cash flow per period (A), future amount (F), and type, which denotes either end-of-period cash flows (0 or omitted) or beginning-of-period cash flows (1). Calculating P with Excel® To solve for P when given i, n, and F, the following can be entered in any cell in an Excel® spreadsheet: =PV(i,n,, −F). Notice, as with the FV function, there are no spaces between the equal sign and the closing parenthesis; also, as before, because no equal-sized cash flow per period occurs, two commas are placed between n and F. Again, the sign of the value obtained for P when using the PV function will be opposite the sign of the value of F that is entered in the cell. Excel® Video Lesson: The PV Function

EXAMPLE 2.5 Saving Money To illustrate the computation of P given F, i, and n, suppose you wish to accumulate $10,000 in a savings account 4 years from now, and the account pays interest at a rate of 5% compounded annually. How much must be deposited today? Key Data Given F = $10,000; i = 5% compounded annually; n = 4 years Find P Solution Using the compound interest tables in Appendix A for 5% and 4 periods, the value of the single sum, present worth factor, (P|F 5%,4), is shown to be 0.82270. Thus, P

=

F (P |F  5%,4)

=

$10,000(0.82270)

=

$8,227.00

Similarly, using Equation 2.6, P

−n

=

F (1 + i)

=

$10,000(1 + 0.05)

=

$8,227.02

−4

Notice that the tabulated present worth factor is rounded to five decimal places, and produces an answer that is $0.02 less than the answer achieved by solving with equations. This is due to error introduced by rounding in the tables. Excel® Solution As shown in Figure 2.7, using the Excel® PV worksheet function, P =PV(5%,4,,−10000) = $8,227.02

FIGURE 2.7 The Excel® PV Financial Function Used with a Single Sum of Money Excel® Data File

Concept Check 02.02-CC001 With compound interest, we earn interest on the principal only. True or False?

Concept Check 02.02-CC002 The single sum, present worth factor: a. Can be depicted as (1 + i)−n b. Can be depicted as (P|F i%,n) c. Is represented as PV using the Excel® financial function with −1 inserted for the fv parameter d. All of the above

2.3 Multiple Cash Flows: Irregular Cash Flows LEARNING OBJECTIVE Perform time value of money calculations for irregular cash flows with annual compounding. Video Lesson: Transformations—Multiple Cash Flows So far we have considered present and future values of single cash flows. In this section we will extend our analysis to multiple cash flows. We begin with multiple cash flows that do not exhibit a pattern from one to another. We follow with cash flow series that form a pattern, allowing the use of shortcuts in determining the present worth and future worth. We will look at three such series as well: the uniform series, the gradient series, and the geometric series. Most engineering economic analyses involve more than a single return occurring after an investment is made. In such cases, the present worth equivalent of the future cash flows can be determined by adding the present worth of the individual cash flows. Similarly, the future worth of multiple cash flows can be determined by adding the future worth of the individual cash flows. Calculating P with Equations If we let At denote the magnitude of a cash flow (receipt or disbursement) at the end of time period t, then P = A1 (1 + i)

−1

+ A2 (1 + i)

−2

+ A3 (1 + i)

−3

+ ⋅ ⋅ ⋅ + An−1 (1 + i)

−(n−1)

+ An (1 + i)

−n

(2.8)

or, using summation notation, n

P = ∑ At (1 + i) t=1

or, equivalently,

−t

(2.9)

(2.10)

n

P = ∑ At (P |F  i%,t) t=1

EXAMPLE 2.6 Computing the Present Worth of a Series of Cash Flows Video Example Consider the series of cash flows depicted by the CFD given in Figure 2.8. Using an interest rate of 6% per interest period, what is the present worth equivalent of cash flows?

FIGURE 2.8 CFD of Multiple Cash Flows Key Data Given Cash flow series in Figure 2.8; i = 6% per interest period. Find P Solution The present worth equivalent is given by P

=

$300 (P |F  6%,1) − $300 (P |F  6%,3) + $200 (P |F  6%,4) + $400 (P |F  6%,6) + $200 (P |F  6%,8)

=

$300(1 + 0.06)

=

$300 (0.94340) − $300 (0.83962) + $200 (0.79209) + $400 (0.70496) + $200 (0.62741)

=

$597.02

−1

− $300(1 + 0.06)

−3

+ $200(1 + 0.06)

−4

+ $400(1 + 0.06)

−6

+ $200(1 + 0.06)

−8

Calculating F with Equations The future worth equivalent of a cash flow series is equal to the sum of the future worth equivalents for the individual cash flows. Thus, F = A1 (1 + i)

n−1

+ A2 (1 + i)

n−2

+ A3 (1 + i)

n−3

+ ⋅ ⋅ ⋅ + An−2 (1 + i)

2

+ An−1 (1 + i) + An

(2.11)

or, using summation notation, n

F = ∑ At  (1  +  i) 

n−t

(2.12)

t=1

or, equivalently, n

F = ∑ At (F |P  i%,n − t) t=1

(2.13)

Notice, in Equations 2.11 and 2.12, the exponent of the interest factor counts the number of periods between the cash flow and the time period where F is located. By convention, the future worth coincides in time with the nth or last cash flow; as such, it does not draw interest, as shown by An in Equation 2.11. Alternately, because we know the value of future worth is given by F = P (1 + i)

(2.14)

n

substituting Equation 2.9 into Equation 2.14 yields n

F = (1 + i)

n

∑ At (1 + i)

−t

t=1

Hence, n

F = ∑ At (F |P  i%,n − t) t=1

(2.15)

EXAMPLE 2.7 Determining the Future Worth of a Series of Cash Flows Video Example Given the series of cash flows in Figure 2.8 from Example 2.6, determine the future worth at the end of the eighth period using an interest rate of 6% per interest period. Also, determine the worth of the series at the end of the fifth period. Key Data Given Cash flow series in Figure 2.8; i = 6% per interest period. Find F, W5 Solution F

=

$300(F |P  6%,7) − $300(F |P  6%,5) + $200(F |P  6%,4) + $400(F |P  6%,2) + $200

=

$300(1 + 0.06)

=

$300 (1.50363) − $300 (1.33823) + $200 (1.26248) + $400 (1.12360) + $200

=

$951.56

7

− $300(1 + 0.06)

5

+ $200(1 + 0.06)

4

+ $400(1 + 0.06)

2

+ $200

Alternately, because we know the present worth is equal to $597.02, F

=

P (F |P  6%,8)

F

=

$597.02 (F |P  6%,8)

=

$597.02 (1.59385)

=

$597.02(1 + 0.06)

=

$951.56

8

Similar to finding the value of a series of cash flows at time 0 (a Present Worth) or finding the value of a series of cash flows at the end of the series (a Future Worth), we may also calculate the value of a series at any given point in time. Given the Future Worth is a value of the series at the end of the 8th period, we can “move” the F value back to end of year five to determine the worth of the series at the end of the 5th period. W5

=

F (P |F  6%,3)

=

$951.56 (P |F  6%,3)

=

$951.56(1.06)

=

$951.56 (0.83962)

=

$798.95

−3

Alternately, knowing the present worth is equal to $597.02 W5

Calculating P and F with Excel® NPV

=

$597.02 (F |P  6%,5)

=

$597.02 (1.33823)

=

$798.95

Obtaining the present worth and future worth equivalents of a cash flow series by summing the individual present worth and future worth, respectively, can be time consuming if many cash flows are included in the series. However, the Excel® NPV financial function is well suited for determining the present worth of a series of cash flows. This function computes the net present value or net present worth of a specified range of cash flows. Importantly, the value obtained occurs one time period prior to the first cash flow in the range given. Its parameters, in order, are interest rate (i), followed by the individual cash flows. Excel® Video Lesson: The NPV Function

EXAMPLE 2.8 Using the Excel® NPV Function Video Example Recall Example 2.6, depicted in Figure 2.8. The Excel® NPV worksheet function is well suited for such a problem. However, as shown in Figure 2.9, no blank cells can be included in the range of cash flows, and every time period must be accounted for. Finally, because the cash flow at time zero is, in fact, zero, there is no need to add the value of C3 to the NPV calculation. As before, the future worth can be obtained by using the NPV value. Notice, the value obtained for the present worth, $597.02, is within 2¢ of the value obtained in Example 2.6 using the interest tables in Appendix A, and the value obtained for the future worth, $951.56, is identical to the value obtained in Example 2.7 using the interest tables in Appendix A.

FIGURE 2.9 The Excel® NPV and FV Financial Functions Used to Solve Example 2.6 Excel® Data File

Concept Check 02.03-CC001 The present worth equivalent of an irregular series of cash flows may be calculated by: a. Summing the present worth of each individual cash flow in the series b. Summing each individual cash flow in the series c. Multiplying the number of cash flows by the number of years d. Converting each cash flow in the series to a future value

2.4 Multiple Cash Flows: Uniform Series of Cash Flows

LEARNING OBJECTIVE Perform time value of money calculations for a uniform series of cash flows with annual compounding. Video Lesson: Transformations—Uniform Series Calculating P Given A A uniform series of cash flows exists when all cash flows in a series are equally sized and spaced. In the case of a uniform series the present worth equivalent is given by (2.16)

n

P = ∑ A(1 + i)

−t

t=1

where A is the magnitude of an individual cash flow in the series. Letting X = (1 + i)−1 and bringing A outside the summation yields n

P

=

A∑ X

t

t=1 n

=

AX ∑ X

t−1

t=1

Letting h = t − 1 gives the geometric series (2.17)

n−1

P = AX ∑ X

h

h=0

Because the summation in Equation 2.17 represents the first n terms of a geometric series, the closed-form value for the summation is given by n−1

∑ X

h

h=0

1 − X =

(2.18)

n

1 − X

Replacing X with (1 + i)−1 yields the following relationship between P and A (1 + i)

n

(2.19)

− 1

P = A[ i(1 + i)

n

]

more commonly expressed as P = A (P |A i%,n)

where (P|A i%,n) is referred to as the uniform series, present worth factor and is tabulated in Appendix A for various values of i and n.

(2.20)

EXAMPLE 2.9 Computing the Present Worth of a Uniform Series of Cash Flows Video Example Troy Long wishes to deposit a single sum of money in a savings account so that five equal annual withdrawals of $2,000 can be made before depleting the fund. If the first withdrawal is to occur 1 year after the deposit and the fund pays interest at a rate of 5% compounded annually, how much should he deposit? Key Data Given A = $2,000; i = 5%; n = 5 Find P Solution Because of the relationship of P and A, as depicted in Figure 2.10, in which P occurs one period before the first A, we see that P

=

$2,000 (P |A 5%,5)

=

$2,000 (4.32948)

=

$8,658.96

FIGURE 2.10 CFD of the Relationship Between P and A in a Loan Transaction Similarly, we can solve using Equation 2.19 (1 + i)

n

(2.65)

− 1

P = A[ i(1 + i)

n

]

where A = $2,000, i = 0.05, and n = 5. Thus, if $8,658.96 is deposited in a fund paying 5% compounded annually, then five equal annual withdrawals of $2,000 can be made. After the fifth withdrawal, the fund will be depleted. Excel® Solution The Excel® PV function can be used to determine the present worth of a uniform series. Recall, when we used the PV function previously, we included two commas in consecutive order because equal-sized cash flows per period were not present for the application. Now, with a uniform series of cash flows being present, an entry is required for the third parameter (A). Therefore, when the present worth of a uniform series is desired, given i, n, and A, the following can be entered in any cell in an Excel® spreadsheet: =PV(i,n,−A). As before, a negative sign is used for A because the PV function reverses the sign of A when calculating the value of P. Using the Excel® PV worksheet function, P =PV(5%,5,−2000) = $8,658.95

EXAMPLE 2.10 Computing the Present Worth of a Delayed Uniform Series of Cash Flows Video Example In Example 2.9, suppose the first withdrawal does not occur until 3 years after the deposit. How much should be deposited? Key Data Given A = $2,000; i = 5%; n = 5 Find P at time 0 Solution As depicted in Figure 2.11, the value of P to be determined occurs at t = 0, whereas a straightforward application of the (P|A 5%, 5) factor will yield a single sum equivalent at t = 2, one period before the first A. Hence, to determine the present worth, the value obtained after using the (P|A 5%,5) factor must be moved backward in time 2 years. The latter operation is performed using the (P|F 5%,2) factor. Therefore, P

=

$2,000 (P |A 5%,5) (P |F  5%,2)

=

$2,000 (4.32948) (0.90703)

=

$7,853.94

FIGURE 2.11 CFD for Example 2.10 Excel® Solution Again, the Excel® PV function can be used to determine the present worth for this example. The example also provides an opportunity to show how an Excel® worksheet function can be embedded in another Excel® worksheet function. Specifically, for this example, the following can be entered in any cell in an Excel® spreadsheet: P =PV(5%,2,,−PV(5%,5,−2000)) = $7,853.93 Notice, two PV function calculations are performed; we will refer to them as the “outside” PV function and the “inside,” or embedded, PV function. By entering two commas in the outside PV function, the next entry is a single sum, future worth amount. (Notice, as before, the use of a negative sign.) Instead of entering a dollar amount for the single sum, future worth amount, we entered the inside PV function to calculate the present worth of the uniform series over time periods 3 through 7. Because the value obtained by the inside PV function occurred at the end of the second time period, a value of 2 was entered for n in the outside PV function. Calculating A Given P: The Capital Recovery Factor The reciprocal relationship between P and A can be expressed as

i(1 + i) A = P [ (1 + i)

n

(2.21)

n

] − 1

or as A = P (A|P  i%,n)

(2.22)

The expression (A|P i%,n) is called the capital recovery factor, because it provides the annual savings or recovery of funds (A) required to justify the capital investment (P). The (A|P i%,n) factor is used frequently in both personal financing and in comparing economic investment alternatives. Capital Recovery Factor The amount of annual savings or recovery funds required to justify a capital investment over some period of time, n.

EXAMPLE 2.11 What Size Uniform Withdrawals Can Occur? Video Example Suppose Rachel Townsley deposits $10,000 into an account that pays 8% interest compounded annually. If she withdraws 10 equal annual amounts from the account, with the first withdrawal occurring 1 year after the deposit, how much can she withdraw each year in order to deplete the fund with the last withdrawal? Key Data Given P = $10,000; i = 8% compounded annually; n = 10 Find A Solution Because we know that A and P are related by A = P (A|P  i%,n)

then A

=

$10,000(A|P  8%,10)

=

$10, 000(0.14903)

=

$1,490.30

Similarly, we can solve using Equation 2.21 i(1 + i) A = P [ (1 + i)

n

n

] − 1

where P = $10,000, i = 0.08, and n = 10. Excel® Solution The Excel® PMT worksheet function can be used to determine the uniform series equivalent to a single sum, present amount. (PMT is shorthand for “payment.”) This function computes the size of a payment when P is borrowed. The parameters of the PMT function, in order of occurrence, are interest rate, number of periods, present amount, future amount, and type. As with the PV and FV functions, type indicates end-of-period or beginning-of-period. To obtain the uniform series equivalent of a single sum, present amount, the latter two parameters are omitted. Excel® Video Lesson: The PMT Function Using the Excel® PMT worksheet function, A =PMT(8%,10,−10000) = $1,490.29

EXAMPLE 2.12 Determining the Size of Delayed Uniform Withdrawals Video Example Suppose, in Example 2.11, the first withdrawal is delayed for 2 years, as depicted in Figure 2.12. How much can be withdrawn each of the 10 years?

FIGURE 2.12 CFD for the Deferred Payment Example Key Data Given P = $10,000; i = 8% compounded annually; n = 10 Find Amount in the fund at t = 2(V2); then find A Solution The amount in the fund at t = 2 equals V2

=

$10, 000 (F |P  8%,2)

=

$10, 000 (1.16640)

=

$11, 664

Therefore, the size of the equal annual withdrawal will be A

=

$11,664 (A|P  8%,10)

=

$11,664 (0.14903)

=

$1,738.29

Excel® Solution The Excel® FV worksheet function can be used to determine the future worth of a uniform series. To solve the example problem, one Excel® worksheet function will be embedded in another. Specifically, the following can be entered in any cell in an Excel® spreadsheet: A =PMT(8%,10,−FV(8%,2,,−10000)) = $1,738.28 Notice, in this case, the FV function was embedded in the PMT function. The “inside” financial function calculated the future value of $10,000 moved forward 2 years at 8% annual compound interest. Instead of using the embedded approach, separate Excel® calculations could be performed, as shown below: V2 =FV(8%,2,,−10000) = $11,664.00 A =PMT(8%,10,−11664) = $1,738.28

Calculating F Given A The future worth of a uniform series is obtained by recalling that F = P (1 + i)

(2.23)

n

Substituting Equation 2.19 into Equation 2.23 for P and reducing yields (1 + i) F = A[

n

(2.24)

− 1 ]

i

or, equivalently, (2.25)

F = A(F |A i%,n)

where (F|A i%,n) is referred to as the uniform series, future worth factor.

EXAMPLE 2.13 Determining the Future Worth of a Uniform Series of Cash Flows If Luis Jimenez makes annual deposits of $1,000 into a savings account for 30 years, how much will be in the fund immediately after his last deposit if the fund pays 6% interest compounded annually? Key Data Given A = $1,000; i = 6% compounded annually; n = 30 Find F Solution F

=

$1, 000 (F |A 6%,30)

=

$1,000 (79.05819)

=

$79,058.19

Excel® Solution The Excel® FV worksheet function can be used to determine the future worth of a uniform series. Using the Excel® FV worksheet function, F =FV(6%,30,1000) = $79,058.19 Calculating A Given F: The Sinking Fund Factor The reciprocal relationship between A and F is easily obtained from Equation 2.24. Specifically, we find that A = F [

i (1 + i)

n

]

(2.26)

− 1

or, equivalently, A = F (A|F  i%,n)

(2.27)

The expression (A|F i%,n) is referred to as the sinking fund factor, because it is used to determine the size of a deposit to place (sink) into a fund in order to accumulate a desired future amount. As depicted in Figure 2.13, F occurs at the same time as the last A. Thus, the last A or deposit earns no interest.

Sinking Fund Factor The amount of savings to deposit in order to accumulate a desired future amount.

FIGURE 2.13 CFD of the Relationship Between A and F in an Investment Scenario

EXAMPLE 2.14 Who Wants to Be a Millionaire? Suppose Crystal Wilson wants to accumulate $1,000,000 by the time she retires in 40 years. If she earns 10% on her investments, how much must she invest each year in order to realize her goal? Key Data Given F = $1,000,000; i = 10%; n = 40 Find A Solution Applying Equation 2.27, A

=

$1,000,000(A|F , 10%,40)

=

$1,000,000(0.0022594)

=

$2,259.40/year

Excel® Solution Using the Excel® PMT worksheet function gives A =PMT(10%,40,,−1000000) = $2,259.41/year

Concept Check 02.04-CC001 When calculating the present worth of a uniform series using the relationship P = A(P|A, i%,n), the P value obtained occurs one time period prior to the first uniform cash flow. True or False?

2.5 Multiple Cash Flows: Gradient Series of Cash Flows LEARNING OBJECTIVE Perform time value of money calculations for a gradient series of cash flows with annual compounding.

Video Lesson: Transformations—Gradient and Geomtric Series A gradient series of cash flows occurs when the value of a given cash flow is greater than the value of the previous cash flow by a constant amount, G, the gradient step. Consider the series of cash flows depicted in Figure 2.14. The series can be represented by the sum of a uniform series and a gradient series. By convention, the gradient series is defined to have the first positive cash flow occur at the end of the second time period. The size of the cash flow in the gradient series occurring at the end of period t is given by At = (t − 1)G       t = 1, …, n

(2.28)

Gradient Series A series characterized by cash flows that increase by a constant amount (G) each period.

FIGURE 2.14 CFD of a Combination of Uniform and Gradient Series in an Investment Scenario As an illustration, if an individual receives an annual bonus and the size of the bonus increases by $100 each year, then the series is gradient. Also, operating and maintenance costs tend to increase over time because of inflation and a gradual deterioration of equipment; such costs often are approximated by a gradient series. The present worth equivalent of a gradient series is obtained by recalling

(2.29)

n

P = ∑ At (1 + i)

−t

t=1

Substituting Equation 2.28 into Equation 2.29 gives (2.30)

n

P = G ∑ (t − 1) (1 + i)

−t

t=1

which reduces to 1 − (1 + ni) (1 + i) P = G[ i

(2.31)

−n

]

2

or, equivalently, (2.32)

P = G (P |G i%,n)

where (P|G i%,n) is the gradient series, present worth factor and is tabulated in Appendix A. The uniform series equivalent to the gradient series, obtained by substituting Equation 2.31 for P in Equation 2.21, is A = G[

1 i

n −

(A|F  i%,n)]

(2.33)

i

or, equivalently, (2.34)

A = G (A|G i%,n)

where the factor (A|G i%,n) is referred to as the gradient-to-uniform series conversion factor and is tabulated in Appendix A. To obtain the future worth equivalent of a gradient series at time n, substitute Equation 2.31 for P in Equation 2.24 to obtain the (F|G i%,n) factor: (1 + i)

n

(2.35)

− (1 + ni)

F = G[ i

2

]

The (F|G i%,n) gradient series, future worth factor is not tabulated in Appendix A. Often a cash flow series is the sum or difference of a uniform series and a gradient series. To determine the present and future worth equivalents of such a composite, one can deal with each special type of series separately.

EXAMPLE 2.15 Determining the Present Worth of a Gradient Series (Sitting on Top of a Uniform Series) Video Example Maintenance costs for a particular production machine increase by $1,000/year over the 5-year life of the equipment. The initial maintenance cost is $3,000. Using an interest rate of 8% compounded annually, determine the present worth equivalent for the maintenance costs. Key Data For this problem, it is very helpful to sketch the cash flow diagram. As shown in Figure 2.15, the original sketch (composite series) can then be converted into two parts showing the uniform and gradient portions, respectively, of the series. Visualizing the flows in this way suggests an efficient approach to determining the solution.

FIGURE 2.15 CFD for Example 2.15 Given The maintenance costs—a composite series of cash flows as shown in Figure 2.15 Find P of the composite series Solution Note that the composite series may be converted to a uniform series plus a gradient series (see Figure 2.15). Converting the gradient series to an equivalent uniform series gives AG

=

$1,000 (A|G 8%,5)

=

$1,000 (1.84647)

=

$1,846.47

(Notice that n equals 5 even though only four positive cash flows are present in the gradient series.) Adding the “converted” uniform series to the “base” uniform series gives A = $1,846.47 + $3,000, or $4,846.47. Therefore, converting the uniform series to its present worth equivalent,

P

=

$4,846.47 (P |A 8%,5)

=

$4,846.47 (3.99271)

=

$19,350.55

To determine the future worth equivalent, we can use either the uniform series or present worth equivalent. Using the uniform series equivalent gives F

=

$4,846.47 (F |A 8%,5)

=

$4,846.47(5.86660)

=

$28,432.30

Similarly, using the present worth equivalent, the future worth equivalent is F

=

$19,350.55 (F |P  8%,5)

=

$19,350.55(1.46933)

=

$28,432.34

Excel® Solution Excel® does not have a special worksheet function for gradient series. However, the NPV worksheet function can be used. Specifically, the following entry can be made in any cell: =NPV(8%,3000,4000,5000,6000,7000) or =1000*NPV(8%,3,4,5,6,7). The result is a present worth of $19,350.56; once the present worth is known, the uniform series equivalent and future worth equivalent can be obtained using the PMT and FV worksheet functions: A =PMT(8%,5,−19350.56) = $4,846.47 F =FV(8%,5,,−19350.56) = $28,432.32

Concept Check 02.05-CC001 All of the following are true statements about a gradient series of cash flows except: a. The cash flows increase by the gradient step, G, every other period b. The series can be represented by the sum of a uniform series and a gradient series c. The gradient step, G, is a constant value d. The first gradient positive cash flow occurs at the end of the second time period

2.6 Multiple Cash Flows: Geometric Series of Cash Flows LEARNING OBJECTIVE Perform time value of money calculations for a geometric series of cash flows with annual compounding. A geometric series of cash flows, as depicted in Figure 2.16, occurs when the size of a cash flow increases (decreases) by a fixed percent from one time period to the next. If j denotes the percent change in a cash flowʼs size from one period to the next, the size of the tth cash flow can be given by At = At−1 (1 + j)         t = 2,…, n

or, more conveniently, At = A1 (1 + j)

t−1

        t = 1,…, n

(2.36)

Geometric Series A series characterized by cash flows that increase or decrease by a constant percentage (j%) each period.

FIGURE 2.16 CFD for an Increasing Geometric Series The geometric series is used to represent the growth (positive j) or decay (negative j) of costs and revenues undergoing annual percentage changes. As an illustration, if labor costs increase by 10% per year, then the resulting series representation of labor costs will be geometric. Calculating P for a Geometric Series The present worth equivalent of the cash flow series is obtained by substituting Equation 2.36 into Equation 2.9 to obtain n

P = A1 (1 + j)

−1

t

∑ (1 + j) (1 + i)

−t

(2.37)

t=1

This can be written as: ⎧ ⎪ ⎪ P =⎨ ⎪ ⎩ ⎪

n

1 − (1 + j) (1 + i) A1 [

(2.38)

−n

i − j

nA1 /(1 + i)

]

i ≠ j

i = j

or P = A1 (P |A1 i%,j%,n)

(2.39)

where (P|A1 i%,j%,n) is the geometric series, present worth factor and is tabulated in Appendix A for various values of i, j, and n.

EXAMPLE 2.16 Determining the Present Worth of a Geometric Series A company is considering purchasing a new machine tool. In addition to the initial purchase and installation costs, management is concerned about the machineʼs maintenance costs, which are expected to be $1,000 at the end of the first year of the machineʼs life and increase 8%/year thereafter. The machine toolʼs expected life is 15 years. Company management would like to know the present worth equivalent for expected costs. If the firmʼs time value of money is 10%/year compounded annually, what is the present worth equivalent? Key Data Given Machine maintenance costs – a geometric series with A1 = $1,000, j = 8%, i = 10%, n = 15 Find P – the present worth equivalent of all maintenance costs Solution P

=

$1,000 (P |A1  10%,8%,15)

=

$1,000 (12.03040)

=

$12,030.40

Excel® Solution Using the NPV worksheet function, the present worth for the increasing geometric series of maintenance costs is $12,030.40, as shown in Figure 2.17. Alternately, the present worth can be obtained by entering the following in any cell:

FIGURE 2.17 Excel® Solution to Example 2.16 Excel® Data File =NPV(10%,1000,1000*1.08,1000*1.08^2,1000*1.08^3,1000*1.08^4, 1000*1.08^5,1000*1.08^6,1000*1.08^7,1000*1.08^8, 1000*1.08^9,1000*1.08^10,1000*1.08^11, 1000*1.08^12, 1000*1.08^13,1000*1.08^14)

or

=1000*NPV(10%,1,1.08,1.08^2,1.08^3,1.08^4,1.08^5,1.08^6,1.08^7, 1.08^8,1.08^9,1.08^10,1.08^11,1.08^12,1.08^13,1.08^14)

The resulting present worth, $12,030.40, is identical to that obtained using the spreadsheet approach in Figure 2.17. (When more than 10 values are included in the range of cash flows, the spreadsheet approach is much preferred to entering the individual cash flow values in the NPV worksheet function.) Calculating A for a Geometric Series A uniform series equivalent to the geometric series is obtained by substituting Equation 2.38 for P in Equation 2.21 to obtain n

⎧ ⎪ ⎪ A1 [ ⎪ ⎪ ⎪ ⎪

1 − (1 + j) (1 + i)

−n

][

i − j

A = ⎨

i(1 + i) (1 + i)

n

(2.40)

n

]

i ≠ j

− 1

n−1

i(1 + i) ⎪ ⎪ ⎪ ⎪ nA1 [ ] ⎪ n ⎩ ⎪ (1 + i) − 1

i = j

or (2.41)

A = A1 (A|A1  i%,j%,n)

where (A|A1 i%,j%,n) is the geometric-to-uniform series conversion factor. This factor is not tabulated in Appendix A. Calculating F for a Geometric Series The future worth equivalent of the geometric series is obtained by substituting Equation 2.40 for A in Equation 2.24 and reducing to obtain ⎧ ⎪ A1 [

(1 + i)

F =⎨ ⎪ ⎩

nA1 (1 + i)

n

− (1 + j)

(2.42)

n

]

i ≠ j

i − j n−1

i = j

or F = A1 (F |A1  i%,j%,n)

where (F|A1 i%,j%,n) is the geometric series, future worth factor and is tabulated in Appendix A.

(2.43)

EXAMPLE 2.17 Determining the Future Worth of a Geometric Series Video Example Mattie Bookhout receives an annual bonus and deposits it in a savings account that pays 8% compounded annually. The size of her bonus increases by 10% each year; her initial deposit is $500 at the end of the first year. Determine how much will be in the fund immediately after her tenth deposit. Key Data Given A1 = $500, i = 8%, j = 10%, and n = 10 Find F Solution The value of F is given by F

=

$500 (F |A1 8%,10%,10)

=

$500 (21.74087)

=

$10,870.44

Excel® Solution Excel® does not have a worksheet function for geometric series. However, as with gradient series, the NPV worksheet function can be used to determine the present worth, uniform series, and future worth equivalents to a geometric series. For the example, the future worth for the geometric series is $10,870.44, as shown in Figure 2.18.

FIGURE 2.18 Excel® Solution to Example 2.17 Excel® Data File

Concept Check 02.06-CC001 Rather than changing by a constant amount each period as with the gradient series, the geometric series changes by a constant percentage each period. True or False?

2.7 Compounding Frequency

LEARNING OBJECTIVE Perform time value of money calculations for multiple compounding periods per year. Thus far, when referring to an interest rate, we stated it was x% compounded annually or x% annual compound interest. While it is true that practically all engineering economic analyses incorporate annual compounding, in personal financing, compounding typically occurs more frequently than once a year. For example, credit cards charge interest of, say, 1½% on the unpaid balance of the account each month. As a result, if you owe $1,000 and do not pay it by the monthly payment deadline, your balance owed increases to $1,015. If you do not make any payments, the interest owed the next month will be 1½% of $1,015. Hence, monthly compounding is at work. (In addition, with many credit cards, penalties are added for lack of payment and they, too, draw interest.) In the case of 1½% per month, an alternate way of expressing the interest rate is 1½% per month multiplied by 12 months in a year or 18% per annum compounded monthly, or 18% per year per month. When expressed in this form, 18% is known as the nominal annual interest rate. We designate the nominal rate3 as r. Nominal Annual Interest Rate The annual interest rate without adjustments for compounding. In the examples presented thus far, cash flows occurred on an annual basis and money was compounded annually. When cash flow frequency or compounding frequency (or both) is not annual, one of two approaches must be employed: the period interest rate approach or the effective annual interest rate approach. Period Interest Rate The nominal annual interest rate divided by the number of interest periods per year.

2.7.1 Period Interest Rate Approach To utilize the period interest rate approach, we must define a new term—the period interest rate: Nominal annual interest rate Period interest rate = Number of interest periods per year

When the interest period and the compounding period are the same (monthly), the factors in Appendix A can be applied directly. Note, however, the number of interest periods (n) must be adjusted to match the new frequency, as in Examples 2.18 and 2.19.

EXAMPLE 2.18 Determining Future Worth with Multiple Compounding Periods per Year Two thousand dollars is invested in an account. What is the account balance after 3 years when the interest rate is: a. 12% per year compounded monthly? b. 12% per year compounded semiannually? c. 12% per year compounded quarterly? Key Data Given P = $2,000; nominal interest rate = 12%; duration of investment = 3 years Find For each scenario: period interest rate; number of interest periods; F Solution a. Nominal annual interest rate = 12% /year Number of interest periods/year = 12 months/year 12%/year Period interest rate =

= 1%/month 12 months/year

Number of interest periods = 3 years (12 months/year) = 36 months F

=

$2,000 (F |P  1%,36)

=

$2,000 (1.43077)

=

$2,861.54

b. Nominal annual interest rate = 12% /year Number of interest periods/year = 2 semiannual periods/year 12%/year Period interest rate =

= 6%/semiannual period 2 semiannual periods/year

Number of interest periods = 3 years (2 semiannual periods/year) = 6 semiannual periods F

=

$2,000 (F |P  6%,6)

=

$2,000 (1.41852)

=

$2,837.04

c. Nominal annual interest rate = 12% /year Number of interest periods/year = 4 quarters/year 12%/year Period interest rate =

= 3%/quarter 4 quarters/year

Number of interest periods = 3 years (4 quarters/year) = 12 quarters

F

=

$2,000 (F |P  3%,12)

=

$2,000 (1.42576)

=

$2,851.54

EXAMPLE 2.19 Determining Car Payments Video Example Rebecca Carlson purchases a car for $25,000 and finances her purchase by borrowing the money at 8% per year compounded monthly; she pays off the loan with equal monthly payments for 5 years. What will be the size of her monthly loan payment? Key Data Given P = $25,000; nominal annual interest rate = 8% /year; duration of loan = 5 years; number of interest periods/year = 12 months/year Find Period interest rate; number of interest periods; A Solution 8%/year Period interest rate =

= 0.66667%/month 12 months/year

Number of interest periods = 5 years(12 months/year) = 60 months 0.0066667(1.0066667) A

=

$25,000 (A|P  0.66667%, 60) = $25,000  [ (1.0066667)

=

60

60

] − 1

$506.91/month

Excel® Solution Using the Excel® PMT worksheet function, A =PMT(0.08/12,60,−25000) = $506.91

2.7.2 Effective Annual Interest Rate The second approach to solving problems when compounding is not annual is the effective interest rate approach. The effective annual interest rate is the annual interest rate that is equivalent to the period interest rate as previously calculated. Video Lesson: Effective Annual Interest Rate Effective Annual Interest Rate The annual interest rate that is equivalent to the period interest rate. For example, if the interest rate is 12% per year compounded quarterly, then the nominal annual interest rate is 12%, and there are four interest periods per year. Thus, the period interest rate is 3% per quarter. Hence, $1 invested for 1 year at 3% per quarter has a future worth of F = $1 (F |P  3%,4) = $1 (1.12551) = $1.12551

To obtain the same value in 1 year requires an annual compound interest rate of 12.551%. This value is called the effective annual interest rate and is given by (1.03)4 − 1 = 0.12551, or 12.551%. Excel® Video Lesson: The EFFECT Function The Excel® EFFECT worksheet function can be used to determine the effective annual interest rate. This function has the following parameters: r (nominal rate) and m (number of compounding periods in a year). The general equation for the effective annual interest rate, ieff, is ieff = (1 + r/m)

m

− 1 =EFFECT(r,m)

where r is the nominal annual interest rate and m is the number of interest periods per year.

(2.44)

EXAMPLE 2.20 Calculating the Effective Annual Interest Rate Calculate the effective annual interest rate for each of the following cases: (a) 12% per year compounded quarterly; (b) 12% per year compounded monthly; (c) and 12% per year compounded every minute. Solution a. Twelve percent per year compounded quarterly: r = 12%; m = 4 From Equation 2.44, ieff

=

(1 + 0.12/4) 4

4

− 1

=

(1.03)

− 1

=

0.12551 = 12.551%

Using the Excel® EFFECT worksheet function gives the same result: ieff =EFFECT (12%,4) = 12.551%

b. Twelve percent per year compounded monthly: r = 12%; m = 12 From Equation 2.44, ieff

=

(1 + 0.12/12) 12

12

− 1

=

(1.01)

− 1

=

0.12683 = 12.683%

Using the Excel® EFFECT worksheet function gives the same result: ieff =EFFECT (12%,12) = 12.683%

c. Twelve percent per year compounded every minute: r = 12%; m = 525, 600 From Equation 2.44, to eight decimal places, ieff

525,600

=

(1 + 0.12/525, 600)

=

0.12749684 = 12.749684%

− 1

Using the Excel® EFFECT worksheet function gives ieff =EFFECT (12%,525600) = 12.749684%

EXAMPLE 2.21 Making Monthly House Payments Greg Brown borrowed $100,000 to purchase a house. He agreed to repay the loan with equal monthly payments over a 30-year period at a nominal annual interest rate of 6% compounded monthly. The closing cost on the loan is $2,000. a. What is Gregʼs monthly payment on the loan? b. What is Gregʼs monthly payment if he chooses to finance the closing cost along with the loan? c. What is the effective annual interest rate on the loan if closing costs are included? Key Data Given Loan amount = $100,000; closing cost = $2,000; nominal annual r = 6% (compounded monthly); n = 360 Find A, ieff Solution a. In order to calculate Gregʼs monthly payment we first need to determine the period interest rate: 6%/year period interest rate =

= 0.5%/month 12 months/year

Gregʼs monthly payment on the 30-year loan is calculated as follows: A

=

$100,000(A|P  0.5%,360)

=

$100,000(0.0059955)

=

$599.55

or, using the Excel® PMT worksheet function, A =PMT (0.06/12,360,−100000) = $599.55

b. If Greg chooses to finance the closing costs also, then his monthly payment will be A

=

$102,000 (A|P  0.5%,360)

=

$102,000 (0.0059955)

=

$611.54

or A =PMT (0.06/12,360,−102000) = $611.54

c. If Greg made 360 payments of $611.54/month for a loan of $102,000, then the effective annual interest rate on the loan including closing costs would be ieff =EFFECT (12*RATE (360,611.54,−102000),12) = 6.1678%

2.7.3 When Compounding and Cash Flow Frequencies Differ

In the previous example, the frequency of compounding coincided with the frequency of cash flows—for example, monthly compounding and monthly cash flows. What if they are not the same? Video Lesson: Effective Rate per Payment Period The approach used in this book assumes that money deposited during a compounding period earns interest regardless of when it is deposited. This approach is consistent with other DCF assumptions made throughout the text. Let r denote the nominal annual interest rate for money and m denote the number of compounding periods in a year; let k denote the number of cash flows in a year, and let i denote the effective interest rate per cash flow period. The value of i is obtained as follows: i = (1 + r/m)

m/k

(2.45)

− 1

Equation 2.45 results from setting the effective annual interest rate for the stated compounding frequency of money equal to that for the cash flow frequency: (1 + i)

k

− 1 = (1 + r/m)

m

(2.46)

− 1

and solving for i.

EXAMPLE 2.22 When Cash Flow Frequency Does Not Match Compounding Frequency Video Example What size monthly payments should occur when $10,000 is borrowed at 8% per year compounded quarterly and the loan is repaid with 36 equal monthly payments? Key Data Given P = $10,000; r = 8% (compounded quarterly); n = 36 months Find A Solution From Equation 2.45, r = 0.08, k = 12, and m = 4. Therefore, i

4/12

=

(1 + 0.08/4)

=

0.006623 or 0.6623%/month

− 1

Knowing the effective monthly interest rate, the monthly payment can be determined: A

=

$10,000 (A|P  0.6623%,36)

=

$10,000[(0.006623) (1.006623)

=

$313.12

36

/(1.006623)

36

− 1]

Using the Excel® PMT worksheet function, A =PMT (1.02ˆ(1/3)−1,36,−10000) = $313.12.

2.7.4 Continuous Compounding and Continuous Cash Flows

Thus far, we have assumed compounding and cash flows occur a discrete number of times during a year. The previous section dealt with the situation in which compounding frequency differs from cash flow frequency. Example 2.22 included compounding occurring four times in a year and equal-sized cash flows occurring 12 equally-spaced times in a year. Now, we extend our consideration of compounding and cash flow frequencies to include an infinite number of compounding periods and an infinite number of cash flows occurring in a year. Of course, this is an approximation, but for many situations it is a reasonable approximation, especially with business transactions occurring around the clock and around the world. Recall Equation 2.45, used to calculate the effective annual interest rate, based on a nominal annual rate of r and m compounding periods in a year. We now take the limit as m approaches infinity to obtain ieff =

lim (1 + r/m)

m

− 1 = e

r

(2.47)

− 1

m→∞

Therefore, for Example 2.20, the effective annual interest rate for 12% per year compounded continuously is e0.12 −1, or 12.749685%. From Equation 2.47, we can calculate values of P, F, and A by replacing i with er − 1 in the P|F, F|P, P|A, A|P, F|A, A|F, P|G, A|G, F|G, P|A1, A|A1, and F|A1 formulas provided in the chapter. (For the P|A1, A|A1, and F|A1 formulas, the geometric rate, j, is replaced with ec − 1, where c is the nominal compound rate of increase in the size of the annual cash flow.) Similarly, when using Excel® to perform calculations with continuous compounding, replace the rate parameter with EXP(r)−1 in PV, FV, PMT, NPV, and other financial functions. Now, suppose cash flows occur so frequently they can be approximated as occurring continuously and uniformly −

throughout, say, a year. Let A denote the magnitude of the flow during a year, k be the number of cash flows occurring during a year, and r be the nominal annual interest rate. From Equation 2.19, the present worth of the flow during a single year is equal to

P =

¯ ¯ ¯ A

(1 + (r/k))

k

− 1

[

k

(r/k) (1 + (r/k))

k

]

which reduces to

P =

¯ ¯ ¯ A

[

k

1

1

−

r

r(1 + (r/k))

k

]

Taking the limit of P as k approaches infinity gives ¯ ¯ ¯ lim P = A

e

r

(2.48)

− 1

re

k→∞

r

Letting n = 1 and i = er −1 in Equation 2.19, we obtain e P = A (e

r

r

(2.49)

− 1

− 1) e

r

For Equations 2.48 and 2.49 to yield the same result, (2.50)

r ¯ ¯ ¯ A (e − 1)

A = r −

Therefore, we can calculate values of P and F by replacing i with er − 1 and replacing A with A(e − 1)/r in Equations 2.19 and 2.24, respectively. Similarly, when using Excel® to perform calculations with continuous cash r

−

flows we replace the pmt parameter with A*(EXP(r)−1)/r in PV, FV, NPV, and other financial worksheet functions.

EXAMPLE 2.23 Performing DCF Calculations with Continuous Compounding and Continuous Cash Flows An electric utility plant operates around the clock every day of the year. It uses fuel at a continuous rate of $50 million per year. (a) Based on an interest rate of 8% compounded continuously, calculate the following values for a 5-year period: (i) the uniform annual equivalent of the fuel cost; (ii) the present worth of the fuel consumed over a 5-year period; and (iii) the future worth equivalent of the fuel cost. (b) For parts (ii) and (iii), what are the values based on $50 million end-of-year cash flow and 8% compounded continuously? (c) For parts (ii) and (iii), what are the values based on $50 million end-of-year cash flow and 8% compounded annually? a. Given

¯ ¯ ¯ A = $50 million

; r = 8% per year compounded continuously; n = 5 years

Find A, P, F i. To calculate the uniform annual equivalent of the fuel cost, we apply Equation 2.50: A A

= $50,000,000 (e

0.08

−1)/0.08 = $52,054,417 per year

=50000000*(EXP(0.08)−1)/0.08 = $52,054,417

ii. To calculate the present worth equivalent of the fuel cost, we apply Equation 2.19 and replace i with e0.08 −1 = 8.32871% and replace A with $52,054,417: P P

5

5

= $52,054,417 [(1.0832871) −1 ]/[ 0.0832871(1.0832871) ] = $206,049,971 =PV(EXP(0.08)−1,5,−50000000*EXP(0.08)−1)/0.08) = $206,049,971

iii. To calculate the future worth equivalent of the fuel cost, we apply Equation 2.24 and replace i with e0.08 −1 = 8.32871% and replace A with $52,054,417: F F

5

= $52,054,417 [(1.0832871) −1]/0.0832871 = $307,390,436 =FV(EXP(0.08)−1,5,−50000000*(EXP(0.08)−1)/0.08) = $307,390,436

b. Given A = $50 million; r = 8% per year compounded continuously; n = 5 years Find P, F ii. For part (b), end-of-year cash flows occur, with continuous compounding. To calculate the present worth equivalent of the fuel cost, we apply Equation 2.19 and replace i with e0.08 −1 = 8.32871%: P P

5

5

= $50, 000, 000[(1.0832871) −1]/[0. 0832871(1.0832871) ] = $197, 917, 854 =PV(EXP(0.08)−1,5,−50000000) = $197,917,854

iii. For part (b), end-of-year cash flows occur, with continuous compounding. To calculate the future worth equivalent of the fuel cost, we apply Equation 2.24 and replace i with e0.08 −1 = 8.32871%: F F

5

= $50, 000, 000[(1.0832871) −1]/0.0832871 = $295, 258, 743 =FV(EXP(0. 08)−1,5,−50000000) = $295, 258,743

c. Given A = $50 million; i = 8% compounded annually; n = 5 years Find P, F ii. For part (c), end-of-year cash flows occur, with annual compounding. To calculate the present worth equivalent of the fuel cost, we apply Equation 2.19: P P

5

5

= $50,000,000[(1.08) −1]/[0.08(1.08) ] = $199,635,502 =PV(0.08,5,−50000000) = $199,635,502

iii. For part (c), end-of-year cash flows occur, with annual compounding. To calculate the future worth equivalent of the fuel cost, we apply Equation 2.24: 5

F

= $50,000,000[(1.08) −1]/0.08 = $293,330,048

F

=FV(0.08,5,−50000000) = $293,330,048

Concept Check 02.07-CC001 As a borrower, would one prefer more or fewer compounding periods? a. More b. Fewer

Concept Check 02.07-CC002 When cash flow frequency or compounding frequency (or both) is not annual: a. We are unable to compare the alternatives b. We must annualize all cash flows and compounding frequencies c. We can use the period interest rate, which divides the nominal annual interest rate by the number of interest periods per year d. We must annualize all cash flows and use the nominal annual interest rate

Concept Check 02.07-CC003 Assuming that interest is paid at 10% compounded quarterly, the effective annual interest rate will be: a. Higher than 10% b. Lower than 10%

Notes 1. A comma may be placed between the factor identifier and the interest rate. Thus, (F|P, i%,n) and (F|P i%,n) are equivalent representations of (1 + i)n. 2. Microsoftʼs Excel® software is used throughout the text. Where a spreadsheet is used to generate a solution, Excel® financial, functions are shown in teal boldface type in the text. 3. In the text, we will consider nominal interest rates that are annual, and compounding periods that are either annual or more frequent than annual.

CHAPTER 3 Equivalence, Loans, and Bonds

Chapter 3 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced e-text and WileyPLUS

FE-Like Problems 03-FE001 What single sum of money at t = 4 is equivalent to receiving $5,000 at t = 1, $6,000 at t = 2, $7,000 at t = 3, and $8,000 at t = 4 if money is compounded at a rate of 8% per time period? a. $28,857 b. $26,000 c. $30,892 d. $33,363 03-FE002 If the interest rate is 10% per year, what series of equal annual payments is equivalent to the following series of decreasing payments: $5,000, $4,000, $3,000, $2,000, $1,000? a. $3,000 b. $3,000 (1 + 0.1) =$ 3,300 c. [$5,000(F |P 10%,4) +$ 4,000(F |P 10%,3) +$ 3,000(F | P 10%,2) + $2,000(F |P 10%,1) + 1,000 ]/ 5

d. $5,000

− $1,000(A|G 10%,5)

Correct or Incorrect? Clear

  Check Answer

03-FE003 You borrow $5,000 at 10% per year and will pay off the loan in 3 equal annual payments starting 1 year after the loan is made. The end-of-year payments are $2,010.57. Which of the following is true for your payment at the end of year 2? a. Interest is $500.00 and principal is $1,510.57 b. Interest is $450.00 and principal is $1,560.57 c. Interest is $348.94 and principal is $1,661.63 d. Interest is $182.78 and principal is $1,827.79 03-FE004 You borrow $10,000 at 15% per year and will pay off the loan in three equal annual payments with the first occurring at the end of the fourth year after the loan is made. The three equal annual payments will be $6,661.08. Which of the following is true for your first payment at the end of year 4? a. Interest = $6,661.08; principal = $0.00 b. Interest = $2,281.31; principal = $4,379.77 c. Interest = $1,500.00; principal = $5,161.08 d. Interest = $0.00; principal = $6,661.08 Correct or Incorrect?

Clear

  Check Answer

03-FE005 You purchase a $10,000 bond with a bond rate of 6% per year payable semiannually for 2 years. You pay $9,600 for the bond. Which statement is correct? a. Semiannual cash flows will be −$9,600, $300, $300, $300, and $9,900, and the bond will earn more than 10% b. Semiannual cash flows will be −$9,600, $300, $300, $300, and $9,900, and the bond will earn less than 10% c. Semiannual cash flows will be −$9,600, $300, $300, $300, and $10,300, and the bond will earn more than 10% d. Semiannual cash flows will be −$9,600, $300, $300, $300, and $10,300, and the bond will earn less than 10% 03-FE006

Consider a cash flow and interest profile as shown:

Year 0 Year 1 Year 2 Year 3 Cash Flow at End of Year −$1,000 $3,000 $2,000 $1,000 Interest Rate During Year

NA

6%

8%

10%

The worth at the end of Year 3 of these cash flows is: a. $5,000.00 b. $5,504.72 c. $5,994.56 d. $5,440.00 Correct or Incorrect? Clear

  Check Answer

03-FE007 A $200,000 bond having a bond rate of 8% payable annually is purchased for $190,500 and kept for 6 years, at which time it is sold. How much should it sell for in order to yield a 7% effective annual return on the investment? a. $168,000 b. $171,000 c. $174,000 d. $177,000 03-FE008 A house is to be purchased for $180,000 with a 10% down payment, thereby financing $162,000 with a home loan and mortgage. There are no “points” or other closing charges associated with the loan. A conventional 30-year loan is used at 7.5%, resulting in monthly payments of $1,132.73. The interest portion of the first monthly payment will be what? a. $1,012.50 b. $682.73 c. $120.23 d. The answer cannot be determined without more information.

Correct or Incorrect? Clear

  Check Answer

03-FE009 $150,000 is deposited in a fund that pays 5% annual compound interest for 2 years, 3% annual compound interest for 2 years, and 4% annual compound interest for 2 years. If uniform annual withdrawals occur over the 6-year period, what will be the magnitude of the annual withdrawals? a. $27,689.63 b. $28,614.29 c. $28,804.50 d. $29,552.62

Problems Section 3.1 Equivalence LEARNING OBJECTIVE 3.1 Compare the equivalence of two or more cash flow profiles. 03.01-PR001 If $7,000 is borrowed and repaid with four quarterly payments of $600 during the first year and four quarterly payments of $1,500 during the second year after receiving the $7,000 loan, what is the effective annual interest rate for the loan? 03.01-PR002 Quarterly deposits of $1,000 are made at t = 1, 2, 3, 4, 5, 6, and 7. Then, withdrawals of size A are made at t = 12, 13, 14, and 15. If the fund pays interest at a quarterly compounding rate of 4%, what value of A will deplete the fund with the fourth withdrawal? 03.01-PR003 Video Solution What uniform annual series of cash flows over a 12-year period is equivalent to an investment of $5,000 at t = 0, followed by receipts of $600 per year for 11 years and a final receipt of $1,600 at t = 12 if the investor’s time value of money is 6% per year?

03.01-PR004 Video Solution An automobile is “priced” at $7,000. A buyer may purchase the car for $6,500 now or, alternatively, the buyer can make a down payment of $1,000 now and pay the remaining $6,000 in 8 equal quarterly payments (over 2 years) at 8% compounded quarterly. a. If the buyer’s TVOM is 10% per year compounded quarterly, would the buyer prefer to pay the $6,500 outright or make the down payment and the quarterly payments? b. What is the effective annual interest rate at which these two payment options are equivalent?

03.01-PR005 Video Solution Consider the following two cash flow series of payments: Series A is a geometric series increasing at a rate of 8% per year. The initial cash payment at the end of year 1 is $1,000. The payments occur annually for 5 years. Series B is a uniform series with payments of value X occurring annually at the end of years 1 through 5. You must make the payments in either Series A or Series B.

a. Determine the value of X for which these two series are equivalent if your TVOM is i = 6.5%. b. Given the value of X from part a, if your TVOM is 8%, would you be indifferent between these two series of payments? If not, which do you prefer? c. Given the value of X from part a, if your TVOM is 5%, would you be indifferent between these two series of payments? If not, which do you prefer?

03.01-PR006 Kinnunen Company wishes to give its customers three options on payments for office equipment when the initial purchase price is over a certain amount. For example, the following three payment plans are options on a typical purchase, and Kinnunen wants to be sure they are equivalent at its TVOM of 14%. Determine the values of Q and R. End of Year Option 1 Option 2 Option 3 0 $0 $0 $0 1 1,800 Q R 2 1,800 2Q (1.1)R 3

1,800

3Q

(1.1)2R

4

1,800

4Q

(1.1)3R

5

1,800

5Q

(1.1)4R

03.01-PR007

Consider the following three cash flow series:

End of Year Cash Flow Series A Cash Flow Series B Cash Flow Series C 0 −$1,000 −$2,500 Y 1 X 3,000 Y 2 1.5X 2,500 Y 3 2.0X 2,000 2Y 4 2.5X 1,500 2Y 5 3.0X 1,000 2Y Determine the values of X and Y so that all three cash flows are equivalent at an interest rate of 15% per year compounded yearly. 03.01-PR008 Consider the following three cash flow series: End of Year Cash Flow Series A Cash Flow Series B Cash Flow Series C 0 $3.0X $1,000 2Y 1 2.5X 1,500 2Y 2 2.0X 2,000 2Y 3 4 5

1.5X 1.0X −1,000

2,500 3,000 −2,500

Y Y Y

Determine the values of X and Y so that you are indifferent among all three cash flows if your TVOM is 11% per year compounded yearly. 03.01-PR009 Zetterberg Builders is given two options for making payments on a brush hog. Find the value of X such that it will be indifferent between the two cash flow profiles if its TVOM is 12% per year compounded yearly. End of Year Series 1 Series 2 0 $150 $0 1 $200 $0 2 3 4 5

$250 $300 $0 $0

$35X $25X $15X $5X

Section 3.2 Interest Payments and Principal Payments LEARNING OBJECTIVE 3.2 Analyze immediate payment and deferred payment loans, including payment amount, remaining balance, and interest and principal per payment. 03.02-PR001 Video Solution In order to buy a car, you borrow $25,000 from a friend at 12%/year compounded monthly for 4 years. You plan to repay the loan with 48 equal monthly payments. a. How much are the monthly payments? b. How much interest is in the 23rd payment? c. What is the remaining balance after the 37th payment? d. Three and one-half years after borrowing the money, you decide to pay off the loan. You have not yet made the payment due at that time. What is the payoff amount for the loan?

03.02-PR002 Video Solution CTL (Concrete Testing Lab) borrowed $80,000 for new equipment at 8% per year, compounded quarterly. It is to be paid back over 3 years in equal quarterly payments. For each part below, use both the interest tables and the Excel® financial functions. Compare answers between the two. a. How much interest is in the 6th payment? b. How much principal is in the 6th payment? c. What principal is owed immediately following the 6th payment?

03.02-PR003 Med Diagnostics, Inc. borrowed $200,000 from a lender for a new blood analyzer module to improve the accuracy and consistency of its tests. The rate was 6.0%, 2% above the prime rate. The loan was to be paid back in equal monthly amounts over 7 years. a. How much is the monthly payment? b. Two years (24 months) of payments have been made. What is the principal remaining after 2 years? c. The prime rate has now risen, and the bank can make loans to other customers, if it has the available capital, for 8.5% payable in equal monthly payments. The bank contacts Med Diagnostics and offers to let it pay off the loan immediately for the amount owed at the end of year 2, less $X. What is the range of $X that will be beneficial to both the bank and Med Diagnostics? 03.02-PR004 Aerotron Electronics has just bought a used delivery truck for $15,000. The small business paid $1,000 down and financed the rest, with the agreement to pay nothing for the entire first year and then to pay $536.83 at the end of each month over years 2, 3, and 4. a. What nominal interest rate is Aerotron paying on the loan? b. What effective interest rate is it paying? c. How much of the 14th month’s payment is interest? How much is principal? d. How much of the 18th month’s payment is interest? How much is principal? e. How much of the 22nd month’s payment is interest? How much is principal? 03.02-PR005 BioElectroMechanical Systems (BEMS) is a startup company with high potential and little available cash. It obtains $500,000 for necessary technology from a venture capitalist, who charges them 24% compounded monthly. The agreement calls for no payment until the end of the first month of the 4th year, with equal monthly payments thereafter for 3 complete years (36 payments). a. How much are the monthly payments? b. What is the total interest paid to the lender? c. What is the total principal paid to the lender? d. If BEMS is doing incredibly well and would like to pay off the debt immediately after making the 24th payment in month 60, how much must it pay? 03.02-PR006 Video Solution $100,000 is borrowed at 6% compound annual interest, with the loan to be repaid with 10 equal annual payments. a. If the first payment is made 1 year after receiving the $100,000, how much of the third payment will be an interest payment? b. If the first payment is made 4 years after receiving the $100,000, how much of the first payment will be an interest payment? c. If the first payment is made 4 years after receiving the $100,000, how much of the last payment will be an interest payment?

03.02-PR007 $25,000 is borrowed at an annual compound rate of 8%. The loan is repaid with 5 annual payments, each of which is $500 greater than the previous payment. How much of the 2nd payment will be a principal payment? 03.02-PR008 $25,000 is borrowed at an annual compound rate of 8%. The loan is repaid with 5 annual payments, each of which is 10% greater than the previous payment. How much of the 2nd payment will be a principal payment? Section 3.3 Bond Investments LEARNING OBJECTIVE 3.3 Analyze investments in bonds and determine the purchase price, selling price, and return on such investments. 03.03-PR001 Five 15-year bonds each having a face value of $1,000 and a coupon rate of 6% per 6 months payable semiannually were purchased for $7,000 8 years ago, and the 16th coupon payment was just made. What can they be sold for now to a buyer if that buyer’s desired return is 4% per 6 months? 03.03-PR002 You have just purchased 10 municipal bonds, each with a $1,000 par value, for $9,500. You purchased them immediately after the previous owner received semiannual coupon payments. The bond rate is 6.6% per year payable semiannually. You plan to hold the bonds for 5 years, selling them immediately after you receive the coupon payment. If your desired nominal yield is 12% per year compounded semiannually, what will be your minimum selling price for the bonds? 03.03-PR003 You wish to purchase a $1,000 bond from a friend who needs the money. There are 7 years remaining until the bond matures, and interest payments are quarterly. You decide to offer $750.08 for the bond because you want to earn exactly 16% per year compounded quarterly on the investment. What is the annual bond rate of interest? 03.03-PR004 Video Solution Leann just sold ten $1,000 par value bonds for $9,800. The bond coupon rate was 6% per year payable quarterly. Leann owned the bonds for 3 years. The first coupon payment she received was 3 months after she bought the bonds. She sold the bonds immediately after receiving her 12th coupon payment. Leann’s yield on the bond was 12% per year compounded quarterly. Determine how much Leann paid when she purchased the bonds.

03.03-PR005 Eight bonds were purchased for $8,628.16. They were kept for 5 years and coupon payments were received at the end of each of the 5 years. Immediately following receipt of the 5th coupon payment, the owner sells each bond for $62.50 more than its par value. The bond coupon rate is 8%, and the owner’s money yields a 10% annual return. a. Draw a clear, completely labeled cash flow diagram of the entire bond transaction using dollar amounts where they are known and using $X to represent the face value of the bond. b. Determine the face value of each bond. 03.03-PR006 Video Solution Ten bonds are purchased for $9,855.57. They are kept for 5 years and coupon payments are received at the end of each of the 5 years. Immediately following the owner’s receipt of the 5th coupon payment, the owner sells each bond for $50 less than its par value. The bond coupon rate is 8%, and the owner’s money yields a 10% annual return. a. Draw a clear, completely labeled cash flow diagram of the entire bond transaction using dollar amounts where they are known and using $X to represent the face value of the bond.

b. Determine the face value of each bond.

03.03-PR007 Video Solution Twenty $1,000 municipal bonds are offered for sale at $18,000. The bond coupon rate is 6% per year payable semiannually. The bonds will mature and be redeemed at face value 5 years from now. If you purchase the bonds, the first premiums will be received 6 months from today. You have decided that you will invest $18,000 in the bonds if your effective annual yield is at least 8.16%. Will you buy these bonds? Why or why not?

03.03-PR008 Shannon purchases a bond for $952.00. The bond matures in 3 years, and Shannon will redeem it at its face value of $1,000. Coupon payments are paid annually. If Shannon will earn a yield of 12%/year compounded yearly, what is the bond coupon rate? 03.03-PR009 One hundred $1,000 bonds having bond coupon rates of 8% per year payable annually are purchased for $95,250 and kept for 6 years, at which time they are sold. Determine the selling price that will yield a 7% effective annual return on the investment. 03.03-PR010 One hundred $1,000 bonds having bond rates of 8% per year payable annually are available for purchase. If you purchase them and keep them until they mature in 4 years, what is the maximum amount you should pay for the bonds if you wish to earn no less than a 7% effective annual return on your investment? 03.03-PR011 One hundred $1,000 bonds having a bond rate of 8% per year payable quarterly are purchased for $97,500, kept for 4 years, and sold for $95,000. Determine the effective annual return on the bond investment. 03.03-PR012 Five $1,000 bonds having a bond rate of 8% per year payable quarterly are purchased for $4,940 and kept for 6 years, at which time they are sold. Determine the selling price that yields a 6% effective annual return on the investment. 03.03-PR013 Twenty-five $1,000 bonds having a bond rate of 8% per year payable quarterly are purchased for $22,250 and kept for 5 years. Assume that the bonds are sold immediately after receiving the coupon payments at the end of the 5th year. What must they be sold for in order to earn a 6% effective annual return on the investment? Section 3.4 Variable Interest Rates LEARNING OBJECTIVE 3.4 Calculate the worth of a cash flow profile with variable interest rates. 03.04-PR001 True or False: A single investment of $10,000 is to be made. Two investment alternatives exist. Alternative A pays interest rates of 1%, 2%, 3%, 4%, 5%, 6%, 7%, 8%, 9%, and 10% each of the next 10 years. The interest rates paid with Alternative B are the reverse order of those available for Alternative A. To maximize the value of the investment portfolio at the end of the 10-year period, you should not choose investment B. 03.04-PR002 True or False: Two investments are available; they require the same equal annual investments over a 10-year period. Investment A will pay interest rates of 1%, 2%, 3%, 4%, 5%, 6%, 7%, 8%, 9%, and 10% each of the next 10 years; the annual interest rates paid with Investment B are the reverse of those paid by Investment A. To maximize the value of the investment portfolio at the end of the 10-year investment period, you should choose investment A.

03.04-PR003 Based on the interest rates and cash flows shown in the cash flow diagram, determine the value of $X.

03.04-PR004 Based on the interest rates and cash flows shown in the cash flow diagram, determine the value of $X.

03.04-PR005 Video Solution Based on the interest rates and cash flows shown in the cash flow diagram, determine the value of $X.

03.04-PR006 Video Solution Maria deposits $2,000 in a savings account that pays interest at an annual compound rate of 3%. Two years after the deposit, the interest rate increases to 4% compounded annually. A second deposit of $3,000 is made immediately after the interest rate changes to 4%. How much will be in the fund 7 years after the second deposit?

03.04-PR007 In Problem 03.04-PR006, how much will be in the fund 7 years after the second deposit if the rates of interest are switched (i.e., 4% for 2 years and 3% for 7 years)? 03.04-PR008 You deposit $10,000 in a fund that pays compound annual interest equal to 3%. One year later the interest rate changes to 4%, 2 years later the interest rate changes to 5%, and 3 years later the interest rate changes to 6%. How much will you have in the fund after 4 years if you withdrew $2,500 at the end of each of the previous 3 years? 03.04-PR009 Charlie has $10,000 to invest for a period of 5 years. The following three alternatives are available to Charlie: Account I pays 4% for the 1st year, 6% for year 2, 8% for year 3, 10% for year 4, and 12% for year 5, all with annual compounding. Account II pays 12% for the 1st year, 10% for year 2, 8% for year 3, 6% for year 4, and 4% for year 5, all with annual compounding. Account III pays interest at the rate of 7.96294% per year for all 5 years. Based on the available balance at the end of year 5, which alternative is Charlie’s best choice? 03.04-PR010 You have $2,000 that you want to invest at the beginning of each of 5 years. The following alternatives are available to you: An investment that pays 7% for year 1, 6% for year 2, 5% for year 3, 4% for year 4, and 3% for year 5. An account that pays 3% for year 1, 4% for year 2, 5% for year 3, 6% for year 4, and 7% for year 5. An account that pays 5% per year each year. On the basis of available balance at the end of the 5th year, which alternative is the best choice? 03.04-PR011 Jimmy deposits $4,000 now, $2,500 3 years from now, and $5,000 6 years from now. Interest is 5% for the first 3 years and 7% for the last 3 years. a. How much money will be in the fund at the end of 6 years? b. What is the present worth of the fund? c. What is the uniform series equivalent of the fund (uniform cash flow at end of years 1 through 6)? 03.04-PR012

Consider the following cash flows and interest rates:

End of Year Interest Rate During Period Cash Flow at End of Period 0 1

10%

$0 $2,000

2 3

8% 12%

−$3,000 $4,000

a. Determine the future worth of this series of cash flows. b. Determine the present worth of this series of cash flows. c. Determine a 3-year uniform annual series that is equivalent to the original series.

Chapter 3 Summary and Study Guide Summary 3.1: Equivalence

Learning Objective 3.1: Compare the equivalence of two or more cash flow profiles. (Section 3.1) If two or more cash flow profiles have equal values, they are said to be equivalent. Determining whether cash flow profiles are equivalent—or not—enables us to decide whether we should prefer one over another. A common problem in engineering economic analysis is to determine the value of a parameter, often the interest rate, which makes two or more cash flow profiles equivalent. 3.2: Interest Payment and Principal Payments

Learning Objective 3.2: Analyze immediate payment and deferred payment loans, including payment amount, remaining balance, and interest and principal per payment. (Section 3.2) A loan payment consists of two parts, the principal and the interest. Recall that the first thing paid in repaying a loan is the interest. The portion of the payment that exceeds the interest charge will go toward paying the loan principal. Immediate payment loans begin paying the loan off immediately after the first interest period. With deferred loans, the loan payment begins more than one interest period after receipt of the principal. Interest charges can be deducted from taxable income; we will revisit this topic in Chapter 9, “Income Taxes.” 3.3: Bond Investments

Learning Objective 3.3: Analyze investments in bonds and determine the purchase price, selling price, and return on such investments. (Section 3.3)

Bonds are important financial instruments for raising capital to finance projects in both the public and private sectors. Analyzing bond investments includes establishing the purchase price or selling price for a bond and determining the rate of return on a bond investment given the purchase and sales prices. The general expression relating the terms associated with a bond is P = V r(P |A i%,n) + F (P | F i%,n)

where P

=

the purchase price of a bond

F

=

the sales price (or redemption value) of a bond

V

=

the par or face value of a bond

r

=

the bond rate (coupon rate) per interest period

i

=

the yield rate (return on investment or rate of return) per interest period

n

=

the number of interest (coupon) payments received by the bond holder

A

=

V r = the interest or coupon payment received per interest period

3.4: Variable Interest Rates

Learning Objective 3.4: Calculate the worth of a cash flow profile with variable interest rates. (Section 3.4) A simplifying assumption made for an engineering economic analysis often holds interest rates fixed over time. This is not necessarily a valid assumption, however, especially in cases where the time period is long. When interest rates vary over time, one must adjust the analysis to consider the varying interest rates.

Important Terms and Concepts Equivalence The state of being equal in value, such as with two cash flow profiles. Principal Payment The principal on a loan refers to the amount borrowed. Thus, the principal payment (or equity payment) is the portion of the loan payment that reduces the unpaid balance.

Immediate Payment Loans When the loan payment begins one interest period after receipt of the principal. Deferred Payment Loans When the loan payment begins more than one interest period after receipt of the principal. Bond A long-term note issued by a borrower to a lender, typically for the purpose of financing a large project. Face or Par Value The stated value (or face value) on an individual bond. Redeem To pay a bond holder a value as specified by the terms and conditions of the bond. A unit issuing a bond is obligated to redeem it. Maturity A specified period of time at which a bond reaches its par value. Bond Rate The amount that an issuing unit is obligated to pay on the par value of the bond during the interim between the date of issuance and the date of redemption.

Chapter 3 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 3.1: Equivalence LO 3.1: Compare the equivalence of two or more cash flow profiles. Video Lesson: Equivalence Video Lesson Notes: Equivalence Excel Video Lesson: PV Financial Function Excel Video Lesson Spreadsheet: PV Financial Function Excel Video Lesson: SOLVER Tool Excel Video Lesson Spreadsheet: SOLVER Tool Excel Video Lesson: GOAL SEEK Tool Excel Video Lesson Spreadsheet: GOAL SEEK Tool Excel Video Lesson: NPV Financial Function Excel Video Lesson Spreadsheet: NPV Financial Function Video Example 3.1: A Uniform Series Equivalency of a Decreasing Gradient Series Video Solution: 03.01-PR003 Video Solution: 03.01-PR004 Video Solution: 03.01-PR005 3.2: Interest Payment and Principal Payments LO 3.2: Analyze immediate payment and deferred payment loans, including payment amount, remaining balance, and interest and principal per payment.

Video Lesson: Loans—Interest and Principal Video Lesson Notes: Loans—Interest and Principal Excel Video Lesson: IPMT Financial Function Excel Video Lesson Spreadsheet: IPMT Financial Function Excel Video Lesson: PPMT Financial Function Excel Video Lesson Spreadsheet: PPMT Financial Function Excel Video Lesson: PMT Financial Function Excel Video Lesson Spreadsheet: PMT Financial Function Video Example 3.4: Purchasing a Car Video Solution: 03.02-PR001 Video Solution: 03.02-PR002 Video Solution: 03.02-PR006 3.3: Bond Investments LO 3.3: Analyze investments in bonds and determine the purchase price, selling price, and return on such investments. Video Lesson: Bonds Video Lesson Notes: Bonds Video Example 3.6: Determining the Selling Price for a Bond Excel Video Lesson: FV Financial Function Excel Video Lesson Spreadsheet: FV Financial Function Excel Video Lesson: RATE Financial Function Excel Video Lesson Spreadsheet: RATE Financial Function Excel Video Lesson: EFFECT Financial Function Excel Video Lesson Spreadsheet: EFFECT Financial Function Video Solution: 03.03-PR004 Video Solution: 03.03-PR006

Video Solution: 03.03-PR007 3.4: Variable Interest Rates LO 3.4: Calculate the worth of a cash flow profile with variable interest rates. Video Lesson: Variable Interest Rates Video Lesson Notes: Variable Interest Rates Video Example 3.9: A Cash Flow Series with Variable Interest Rates Excel Video Lesson: FVSCHEDULE Function Excel Video Lesson Spreadsheet: FVSCHEDULE Function Video Solution: 03.04-PR005 Video Solution: 03.04-PR006 These chapter-level resources will help you with your overall understanding of the content in this chapter. Appendix A: Time Value of Money Factors Appendix B: Engineering Economic Equations Flashcards: Chapter 03 Excel Utility: TVM Factors: Table Calculator Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values Excel Utility: Monthly Payment Sensitivity Excel Utility: TVM Factors: Discrete Compounding Excel Utility: TVM Factors: Geometric Series Future Worth Excel Utility: TVM Factors: Geometric Series Present Worth Excel Data Files: Chapter 03

CHAPTER 3 Equivalence, Loans, and Bonds LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 3.1 Compare the equivalence of two or more cash flow profiles. (Section 3.1) 3.2 Analyze immediate payment and deferred payment loans, including payment amount, remaining balance, and interest and principal per payment. (Section 3.2) 3.3 Analyze investments in bonds and determine the purchase price, selling price, and return on such investments. (Section 3.3) 3.4 Calculate the worth of a cash flow profile with variable interest rates. (Section 3.4)

Engineering Economics in Practice Samuel Washington Takes a Loan Fifteen years after graduating in electrical engineering and accepting employment with Texas Instruments, Samuel Washington decides to establish a consulting business. Although he has invested wisely for the past 15 years, the value of his investments is only $325,000. After developing a business plan, he realizes he will need $250,000 on hand initially, plus $150,000 each successive year, to cover the expenses of an office and an assistant. He is unsure about how much of his own money he should use and how much to borrow. In talking to the loan officer of a local bank, he learns that the bank will charge him annual compound interest of 6% for a 5-year loan period or 5.5% for a 10-year loan period. Over the past 10 years, Samuel earned an average of 5.25% annually on his investments; he believes he will continue to earn at least that amount on his investment portfolio. If he borrows money, he can repay the loan in several ways: pay accumulated interest monthly, plus pay the principal at the end of the loan period; make equal monthly payments; make monthly payments that increase like a gradient series; make monthly payments that increase like a geometric series; or make a lump sum payment at the end of the loan period. Because this is a business investment, any interest paid can be deducted from his taxable income. Discussion Questions 1. Discuss the quantitative (economic) tradeoffs that Samuel should consider when he decides how much money to use from his personal savings versus borrowing money from the bank. 2. What qualitative (noneconomic) factors should Samuel consider when he decides how much money to use from his personal savings versus borrowing money from the bank? 3. What types of assumptions is Samuel making when he determines his loan needs? What happens if these assumptions do not hold? 4. How might Samuel set out to secure funding for this proposed business venture?

Introduction This chapter builds on the foundation established in Chapter 2. With one exception, in this chapter you will learn how to analyze the financial needs Samuel Washington faces. How to incorporate income tax effects in his analysis is reserved for Chapter 9. In addition to learning how to perform the analysis Samuel performed, you will learn how to analyze various bond transactions, and how to determine the amount of each loan payment that is an

interest versus a principal payment. Throughout the chapter, you will see how to perform the analyses using tabulated values of compound interest factors and using Excel®’s financial functions.

3.1 Equivalence LEARNING OBJECTIVE Compare the equivalence of two or more cash flow profiles. Video Lesson: Equivalence In engineering economic analyses, equivalence means “the state of being equal in value.” The concept is primarily applied in the comparison of two or more cash flow profiles. A commonly used approach to determine equivalence is to compare the present worth of the cash flow profiles. If they are equal, then the cash flow profiles are equivalent. Equivalence problems, however, generally involve a parameter whose value is to be determined in order for the two cash flow profiles to be equivalent. Typically, the parameter is the interest rate or a cash flow. For example, are the following two cash flow profiles equivalent at 15% compounded annually? Equivalence The state of being equal in value, such as with two cash flow profiles. Cash Flow Profile 1: Receive $1,322.50 two years from today Cash Flow Profile 2: Receive $1,000 today. Computing the present worths for the two cash flow profiles yields PW(1)

= $1,322.50(P |F 15%,2) = $1,322.50(1.15)

−2

= $1,322.50(0.75614) = $1,000 =PV(15%,2,,−1322.5) = $1,000 PW(2)

= 1,000

Excel® Video Lesson: PV Financial Function Because the present worths are equal, the two cash flow profiles are equivalent. Notice, we did not obtain the value of the (P|F 15%,2) factor from Appendix A. Instead, we calculated the value of the factor to five decimal places in order to demonstrate that equivalency existed. (That is why the Excel® results were identical to the calculated results.) Based on the foundation established in Chapter 2, it should be obvious that the worths of the two cash flow profiles will be the same at any particular point in time, e.g., t = 2 and t = 6.

EXAMPLE 3.1 A Uniform Series Equivalency of a Decreasing Gradient Series Video Example Using an 8% discount rate, what uniform series over five periods, [1, 5], is equivalent to the cash flow profile given in Figure 3.1?

FIGURE 3.1 The Decreasing Gradient CFD for Example 3.1 Solution The cash flow profile in Figure 3.1 consists of the difference in a uniform series of $500 and a gradient series, with G = $100. A uniform series equivalent of the cash flow profile can be obtained over the interval [2, 6] as follows: A[2,6]  =

$500 − $100(A|G 8%,5)

=

$500 − $100(1.84647)

=

$315.35

Figure 3.2 shows the CFD for the uniform series that is equivalent to the original decreasing gradient series.

FIGURE 3.2 Equivalent Gradient and Uniform CFDs for Example 3.1 But we still need to convert the uniform series from interval [2, 6] to [1, 5]. To do so, the entire series must be shifted backward in time one time unit. From DCF Rule #4, we move money backward in time one time unit by dividing by 1 plus the interest rate. Hence, as shown in Figure 3.3, the equivalent worth over the interval [1, 5] is A[1,

5]

 =

$315.35(P |F 8%,1)

=

$315.35(0.92593)

=

$291.99

FIGURE 3.3 Equivalent Gradient CFD and Converted Uniform CFD for Example 3.1 Consequently, a uniform series of $291.99 over the interval [1, 5] is equivalent to the cash flow profile given in Figure 3.1. (If you have doubts concerning the equivalence, compare their present worths using an 8% interest rate.)

EXAMPLE 3.2 Determining an Equivalent Gradient Step Consider Figure 3.4. Determine the value of X that makes the two cash flow profiles equivalent using a TVOM of 15%.

FIGURE 3.4 CFDs for Example 3.2 Solution Equating the future worths of the two cash flow profiles at t = 4 gives $200(F |A 15%,4) + $100(F |A 15%,3) + $100 = [$200 + X(A|G 15%,4)](F |A 15%,4)

Eliminating the common term of $200(F|A 15%,4) yields $100(3.47250) + $100 = X(1.32626)(4.99338)

Solving for X gives a value of $67.53. Excel® Solutions This example offers an opportunity to use either Excel®’s SOLVER tool or Excel®’s GOAL SEEK tool. To use SOLVER, the data are entered in a spreadsheet as shown in Figure 3.5. The present worth of the cash flows to the left of the equality in Figure 3.5 is set equal to the present worth of the cash flows to the right of the equality. Excel® Video Lesson: SOLVER Tool Excel® Video Lesson: GOAL SEEK Tool

FIGURE 3.5 Using the Excel® SOLVER Tool to Solve Example 3.2 Excel® Data File Notice, the unknown is given by cell E10. The target cell, E9, is to be made equal to $826.71, which is the present worth of the equality’s left-hand side. The cash flows on the equality’s right-hand side appropriately include the value of cell E10. When SOLVER is applied, the value in cell E10 is changed to $67.53 with the result that the present worth in cell B9 equals the present worth in cell E9.

EXAMPLE 3.3 Determining an Equivalent Interest Rate For what interest (discount) rate are the two cash flow profiles shown in Figure 3.6 equivalent?

FIGURE 3.6 CFDs for Example 3.3 Solution Converting each cash flow profile to a uniform series over the interval [1, 5] gives −$4,000(A|P i%,5) + $1,500 = −$7,000(A|P i%,5) + $1,500 + $500(A|G i%,5)

or $3,000(A|P i%,5) = $500(A|G i%,5)

which reduces to (A|G i%,5) = 6(A|P i%,5)

Searching through the interest tables at n = 5, the value of the (A|G i%,5) factor is six times the value of the (A|P i%,5) factor for an interest rate between 12% and 15%. Specifically, with a 12% interest rate, (A|G 12%,5) − 6(A|P 12%,5) = 1.77459 − 6 (0.27741) = 0.11013

and, using a 15% interest rate, (A |G 15%,5) − 6(A| P 15%,5) = 1.72281 − 6(0.29832) = −0.06711

Interpolating for i gives i = 0.12 + (0.15 − 0.12)(0.11013)/(0.11013 + 0.06711)

or i = 0.13864

Therefore, using a discount rate of approximately 13.864% will establish an equivalence relationship between the cash flow profiles given in Figure 3.6.

Excel® Solutions Excel® can be used to determine the equivalency. As shown in Figure 3.7, we set the present worth of the equality’s left-hand side equal to its right-hand side. Excel®’s NPV function is used to compute the present worth of each cash flow stream. In cell E11 is entered the difference in the present worth of the left-hand side (B10) and the present worth of the right-hand side (E10). Each present worth is calculated using a value for the interest rate given in cell E12. Excel® Video Lesson: NPV Financial Function

FIGURE 3.7 Using the Excel® SOLVER Tool to Solve Example 3.3 Excel® Data File The Excel® SOLVER tool is used to set the target cell to E11. (Because we want the two present worths to be equal, E11 must equal zero.) To make E11 equal 0, cell E12’s value, the interest rate, must be changed. We initialize the underlying search performed by SOLVER by setting E12 equal to 10%. After applying SOLVER, as shown in Figure 3.8, a value of 13.8677% is obtained for the interest rate, which is quite close to the value obtained by interpolating values obtained from tables in Appendix A. (The Excel® GOAL SEEK tool also can be used to solve the example.)

FIGURE 3.8 Solution to Example 3.3

Excel® Data File

Concept Check 03.01-CC001 An investor considering two vastly different cash flows for alternate investments would be indifferent as to which alternative was selected when: a. The present worth of each investment was identical b. The investments were deemed equivalent from an economic analysis perspective c. Choice a, but not b d. Both a and b

3.2 Interest Payments and Principal Payments LEARNING OBJECTIVE Analyze immediate payment and deferred payment loans, including payment amount, remaining balance, and interest and principal per payment. Video Lesson: Loans—Interest and Principal In Chapter 9, we examine the after-tax effects of borrowing money. For businesses, interest charges can be deducted from taxable income. Similarly, for primary residences, interest payments are tax deductible. For these reasons, it is useful to know how to compute the amount of a loan payment that is interest, with the balance of the payment being a principal payment. Principal Payment The principal on a loan refers to the amount borrowed. Thus, the principal payment (or equity payment) is the portion of the loan payment that reduces the unpaid balance. What do we mean by a principal payment? The principal on a loan is the amount borrowed. The unpaid balance on a loan is the sum of the unpaid interest and the unpaid principal. As loan payments are made, over time the unpaid balance on a loan declines until the final payment reduces the unpaid balance to zero. In determining the amount of a loan payment that is interest, it is helpful to remember the following: The first thing paid in repaying a loan is interest. If the accumulated unpaid interest on a loan exceeds the magnitude of the loan payment, then all of the payment is an interest payment. Loan payments do not include a principal payment if the unpaid interest equals or exceeds the magnitude of the loan payment. Consider a loan in the amount of $10,000 at an interest rate of 1%/month. If the first loan payment is made one month after receiving the $10,000, then the amount of unpaid interest at the time of the first payment is $10,000(0.01) or $100. Hence, the amount of the first payment that will be a principal payment is equal to the magnitude of the loan payment less the $100 interest payment. (Principal payments are often referred to as equity payments. We use the terms interchangeably.) Suppose you borrow $P and repay the loan with n payments at times t through t + n − 1, where t ≥ 1. The payments to be made are denoted by At, At+1, …, At+n−1. Immediately after making the kth payment, you decide to pay off the loan. The amount you repay is given by the present worth of the n − k payments remaining to be paid. Likewise, if you decide to pay off the loan immediately after making the k + 1st payment, the amount owed is equal to the present worth of the n − k − 1 payments

remaining to be paid. The difference in the two present worth values is the amount of principal in the k + 1st payment. The segregation of a loan payment into interest and principal components can be performed for any set of loan payments. However, because the most commonly used payment schedule is one in which the set of payments is a uniform series, we will develop results for such a payment schedule. Two cases are considered: immediate payment loans and deferred payment loans.

3.2.1 Immediate Payment Loans Consider a loan in which the first payment is made one interest period after receipt of the loan principal. If $P is borrowed at t = 0 with a period interest rate of i%/period for n periods and is repaid with n equal end-of-period payments (starting at t = 1), then the magnitude of each payment, $A, is given by $A = $P(A|P i%,n). Such a loan is called an immediate payment loan. From the preceding discussion, the amount of principal remaining to be repaid immediately after making a payment at time t, Ut, is given by Ut = A(P |A i%,n − t)

(3.1)

where A, n, and i are defined as noted above. Immediate Payment Loans When the loan payment begins one interest period after receipt of the principal. A related quantity is the payoff quantity. Payofft is the total amount required to pay off the loan at time t, including both the current payment and the unpaid balance. Payofft = A + Ut = A + A(P |A i%,n − t)

(3.2)

= A[1 + (P |A i%,n − t)]

(3.3)

The interest accrued during any payment period is given by the product of the unpaid balance at the beginning of the period and the period interest rate. Letting It denote the portion of payment t that goes to pay interest, we have It = iUt−1

(3.4)

Substituting the results of Equation 3.1 into Equation 3.4, the following relationship can be derived. It = A[1 − (P |F i%,n − t + 1)]

(3.5)

The portion of the payment that does not pay accrued interest goes to principal reduction. Letting Pt denote the portion of payment t that is a principal payment (goes to reduce principal), we have Pt = A − I t

(3.6)

Pt = A(P |F i%,n − t + 1)

(3.7)

or

Excel® has two financial functions that apply directly to this material: IPMT and PPMT. The IPMT worksheet function determines the amount of a periodic payment that is interest and has the following parameters: interest rate, period for which the value is sought, number of periodic payments made, present amount, future amount, and type. The PPMT worksheet function determines the amount of a periodic payment that reduces the unpaid principal on a loan; it has the same parameters as IPMT. For conventional loans, the future amount and type parameters are not needed. Excel® Video Lesson: IPMT Financial Functions

Excel® Video Lesson: PPMT Financial Function

EXAMPLE 3.4 Purchasing a Car Video Example Sara Beth wants to purchase a used car in excellent condition. She has decided on a car with low mileage that will cost $20,000. After considering several alternatives, she identified a local lending source that will charge her an interest rate of 6% per annum compounded monthly for a 48-month loan: (a) What will be the size of her monthly payments? (b) What will be the remaining balance on her loan immediately after making her 24th payment? (c) If she chooses to pay off the loan at the time of her 36th payment, how much must she pay? (d) What portion of her 12th payment is interest? (e) What portion of her 12th payment is an equity payment? Key Data Given P = $20,000; i = 6% compounded monthly; n = 48 Find A, U24, Payoff36, I12, P12 Solution a. A = P(A|P i%, n) Period interest rate = 6%/12 months = 0.5%/month A

= $20,000(A|P 0.5%,48) = $20,000(0.02349) = $469.80 =PMT(0.5%,48,−20000) = $469.70

b.

Ut U24

= A(P |A i%,n − t)

n = 48

t = 24

n − t = 48 − 24 = 24

= A(P |A 0.5%,24) = $469.80(22.56287) = $10,600.04 =PV(0.5%,24,−PMT(0.5%,48,−20000)) = $10,597.79

Excel® Video Lesson: PMT Financial Function c.

Payofft n Payoff36

= A + A(P |A i%,n − t) = 48

t = 36

n − t = 48 − 36 = 12

= A + A(P |A 0.5%,12) = $469.80 + $469.80(11.61893) = $5,928.37 =PV(0.5%,12,−PMT(0.5%,48,−20000))+PMT(0.5%,48,−20000) = $5,927.12

d.

It

= iUt−1

n

= 48

U11 I12

t = 12

t − 1 = 12 − 1 = 11

= A(P |A 0.5%,37) = $469.80(33.70250) = $15,833.44 = 0.005($15,833.44) = $79.17 =IPMT(0.5%,12,48,−20000) = $79.15

e.

Pt t P12

= A − It = 12 = A − I12 = $469.80 − $79.17 = $390.63 =PPMT(0.5%,12,48,−20000) = $390.55

3.2.2 Deferred Payment Loans Repayment of loans does not always begin one interest period after receipt of the principal. For any number of reasons, some individuals and organizations arrange to delay the beginning of payment for one or more interest periods. Such loans are called deferred payment loans and are the subject of this section. Deferred Payment Loans When the loan payment begins more than one interest period after receipt of the principal. Determining the amount of a deferred payment that is interest versus equity is not as straightforward as it is for immediate payment loans. However, a tabular approach can be used in this situation, as illustrated in Example 3.5. (Importantly, until the accumulated interest resulting from deferring payments is paid, no principal payments occur.) To formalize the computation of interest payments and principal payments when deferred payments are made on a loan, we introduce the following notation. Let Princ

=

loan principal

IR

=

interest rate on the loan

UB

=

unpaid balance at the beginning of the interest period

UIB

=

unpaid accumulated interest immediately before making a payment

UIA

=

unpaid accumulated interest immediately after making a payment

AO

=

amount owed just before making a payment

Int

=

interest earned during the period = U B × I R

Ad

=

size of the deferred payment

IPmt

=

amount of the payment that is an interest payment

PPmt

=

amount of the payment that is a principal payment = Ad − IPmt

The amount of a payment that is interest is either the accumulated interest immediately before making a payment or the entire payment, whichever is the lesser, as expressed in Equation (3.8): I P mt = min(U I B; Ad )

(3.8)

The difference in the payment and the interest payment is the principal payment. Example 3.5 illustrates the use of this procedure with a deferred payment loan.

EXAMPLE 3.5 Interest and Equity in Deferred Payments The owner of a small business borrows $10,000 at 15% annual compound interest. Five equal annual payments will be made to repay the loan, but the first will not occur until 4 years after receipt of the principal amount. How much of each payment will be interest and principal? Key Data Given P = $10,000; IR = 15%; n = 5 Find Deferred payment amount (Ad); principal and interest amounts for each payment Solution The size of the new, deferred payments will be Ad

= $10,000(F |P

15%,3)(A|P

15%,5)

= $10,000(1.52088)(0.29832) = $4,537.09 =PMT(15%,5,−FV(15%,3,,−10000)) = $4,537.01

Table 3.1 summarizes the calculations to determine the amount of each payment that will be interest and equity. Because no payments are made until 4 years after receiving the $10,000, the amount owed immediately before making the first deferred payment equals $10,000(F|P 15%,4), or $17,490.10 (based on the compound interest values given in Appendix A) or $17,490.06 (using Excel®’s FV function). Note that the last column of Table 3.1, unpaid balance after payment (UBA), contains the value carried into the following payment period as unpaid balance before payment (UB). The first deduction from a payment is the interest on the unpaid balance. Because the amount owed ($17,490.06) includes $7,490.06 in accumulated interest, which is greater than a full payment ($4,537.01), all of the first payment is an interest payment. TABLE 3.1 Interest and Equity Payments in a Deferred Payment Loan Unpaid Balance Before Payment Year (UB) 1 $10,000.00

Interest Unpaid Interest During Year Before Payment Amount (Int) (UIB) Owed (AO) $1,500.00 $1,500.00 $11,500.00

Loan Payment (Ad) $0.00

2 3 4 5 6

$11,500.00 $13,225.00 $15,208.75 $12,953.06 $10,359.01

$1,725.00 $1,983.75 $2,281.31 $1,942.96 $1,553.85

$3,225.00 $5,208.75 $7,490.06 $4,896.01 $1,912.86

$13,225.00 $15,208.75 $17,490.06 $14,896.01 $11,912.86

7 8

$7,375.85 $3,945.22 UBt = UBAt−1

$1,106.38 $591.78 Intt = IRt × UBt

$1,106.38 $591.78 UIBt = Intt + UIAt−1

$8,482.23 $4,537.01 $4,537.01 $4,537.01 AOt = UBt + Adt Intt

Excel® Data File

$0.00 $0.00 $4,537.01 $4,537.01 $4,537.01

Interest Payment (IPmt) $0.00

P P

$0.00 $0.00 $4,537.01 $4,537.01 $1,912.86

$2,6

$1,106.38 $591.78 IPmt = min(UIBt; Adt)

$3,4 $3,9 PPm IPm

Likewise, the amount owed before making the second payment ($14,896.01) includes accumulated interest of $4,896.01, which is also greater than a full payment ($4,537.09). Therefore, all of the second payment is an interest payment. The amount owed at the time of the third payment ($11,912.86) includes accumulated unpaid interest totaling $1,912.86, which is less than a full payment ($4,537.01). Therefore, the $1,912.86 in accumulated interest is an interest payment; the $2,624.15 remaining in the $4,537.01 payment reduces principal. Just before making the fourth payment, $8,482.23 is owed, $1,106.38 of which is unpaid interest. Therefore, the $1,106.38 in interest earned during the year is deducted from the payment, and the $3,430.63 balance reduces principal. Finally, the amount owed just before the fifth and final payment is made totals $4,537.01, the size of a payment. The interest charged during the year equals $591.78. Therefore, the balance of $3,945.22 is a principal payment and reduces the unpaid balance to zero.

Concept Check 03.02-CC001 The first thing paid when repaying a loan is the principal. True or False?

Concept Check 03.02-CC002 A college loan that does not require the student to begin paying back until after graduation is an example of a deferred loan. True or False?

3.3 Bond Investments LEARNING OBJECTIVE Analyze investments in bonds and determine the purchase price, selling price, and return on such investments. Video Lesson: Bonds A bond is a long-term note issued by the borrower (normally a corporation or governmental agency) to the lender, typically for the purpose of financing a large project. This note specifies terms of repayment and other conditions. Bond A long-term note issued by a borrower to a lender, typically for the purpose of financing a large project. Bond investments are attractive vehicles for applying the DCF models developed in Chapter 2. Also, they prepare you for material to be covered in subsequent chapters. Individual bonds are normally issued in denominations such as $1,000. The stated value on the individual bond is termed the face or par value. The par value is to be repaid by the issuing organization at the end of a specified period of time, say 5, 10, 15, 20, or even 50 years. Thus, the issuing unit is obligated to redeem the bond at par value at maturity.

Face or Par Value The stated value (or face value) on an individual bond. Redeem To pay a bond holder a value as specified by the terms and conditions of the bond. A unit issuing a bond is obligated to redeem it. Maturity A specified period of time at which a bond reaches its par value. Furthermore, the issuing unit is obligated to pay a stipulated bond rate on the face value during the interim between date of issuance and date of redemption. This might be 10% per year payable quarterly, 8% per year payable semiannually, 9% per year payable annually, and so forth. For the purposes of the following examples, it is emphasized that the bond rate applies to the par value of the bond. Bond Rate The amount that an issuing unit is obligated to pay on the par value of the bond during the interim between the date of issuance and the date of redemption. We now employ the following notation: P

=

the purchase price of a bond

F

=

the sales price (or redemption value) of a bond

V

=

the par or face value of a bond

r

=

the bond rate (coupon rate) per interest period

i

=

the yield rate (return on investment or rate of return) per interest period

n

=

the number of interest (coupon) payments received by the bond holder

A

=

V r = the interest or coupon payment received per interest period

The general expression relating these terms is P = V r(P |A i%,n) + F (P | F i%,n)

Three types of bond problems are considered: 1. Given P, r, n, V, and a desired i, find the sales price F. 2. Given F, r, n, V, and a desired i, find the purchase price P. 3. Given P, F, r, n, and V, find the yield rate i that has been realized. Each of these cases is illustrated in the following examples.

(3.9)

EXAMPLE 3.6 Determining the Selling Price for a Bond Video Example On January 1, 2021, Austin plans to pay $1,050 for a $1,000, 12% semiannual bond. He will keep the bond for three years, receive six coupon payments, and then sell it. How much should he sell the bond for in order to receive a yield of 10% compounded semiannually? Key Data Given P = $1,050; V = $1,000; n = 6; r = 6%; i = 5% Find F Solution The present worth for the bond investment is given by P = V r(P |A 5%,6) + F (P | F 5%,6)

or $1,050 = ($1,000)(0.06)(5.07569) + F (0.74622)

Solving for F yields a value of $998.98. As long as the selling price at the end of three years is at least $998.98, the bond will yield a return of at least 10% compounded semiannually. Excel® Solutions The Excel® FV worksheet function is well suited for this example. Recall that the syntax for the FV worksheet function is FV(rate,nper,pmt,pv,type). Therefore, entering =FV(5%,6,60,−1050) in any cell in an Excel® spreadsheet will produce the following result: $998.99. (The 1¢ difference in results is due to round-off errors in the interest tables.) Excel® Video Lesson: FV Financial Function

EXAMPLE 3.7 Determining the Purchase Price for a Bond Emma plans to purchase a $1,000, 12% semiannual bond, hold it for three years, receive six coupon payments, and redeem it at par value. What is the maximum amount she should pay for the bond if she wants to earn at least 14% compounded semiannually on her investment? Key Data Given F = $1,000; V = $1,000; n = 6; r = 6%; i = 7% Find P Solution The present worth for the bond investment is given by P = V r(P |A 7%,6) + F (P |F

7%,6)

or P = ($1,000)(0.06)(4.76654) + $1,000(0.66634)

Solving for P yields a value of $952.33. As long as the purchase price is no greater than $952.33, Emma’s rate of return on her investment will be at least 14% compounded semiannually. Excel® Solutions The Excel® PV worksheet function matches perfectly with this bond calculation. Recall the PV syntax is PV(rate,nper,pmt,fv,type). Therefore, entering =PV(7%,6,−60,−1000) in any cell in an Excel® spreadsheet will give the answer sought: $952.33.

EXAMPLE 3.8 Determining the Rate of Return for a Bond Investment Charlotte purchased a $1,000, 12% quarterly bond for $1,020, kept it for three years, received 12 coupon payments, and sold it for $950. What was her quarterly yield on her bond investment? What was her effective annual rate of return? Key Data Given P = $1,020; V = $1,000; F = $950; r = 3% Find i, ieff Solution Setting the present worth of her investment equal to 0 gives −$1,020 + ($1,000)(0.03)(P |A i%,12) + $950(P |F i%,12) = $0

Solving by trial and error, we first let i = 2%/quarter (8% compounded quarterly): −$1,020 + $30(10.57534) + $950(0.78849) > $0 − $46.32570 > $0

Next, letting i = 2½%/quarter (10% compounded quarterly): −$1,020 + $30(10.25778) + $950(0.74356) < $0 − $5.88520 < $0

Linear interpolation gives 0.02 + 0.005($46.32570)/($46.32570 + 5.88520) = 0.02444

or 2.444%/quarter (9.776% compounded quarterly). The effective annual return is ieff = [(1 + 0.02444)

4

− 1]100% = 10.140%

Excel® Solutions The Excel® RATE and EFFECT worksheet functions are ideally suited for this example. Note, there will be some minor differences in values due to rounding. Entering the following in any cell of an Excel® spreadsheet will yield the quarterly rate: iqtr =RATE(12,30,−1020,950) = 2.442% Excel® Video Lesson: RATE Financial Function Embedding the RATE function in the EFFECT function allows us to calculate the effective annual return on the bond transaction: ieff =EFFECT(4*RATE(12,30,−1020,950),4) = 10.132% Excel® Video Lesson: EFFECT Financial Function

Concept Check 03.03-CC001 Bonds are short-term notes issued by the borrower to a lender usually to raise money for small projects. True or False?

3.4 Variable Interest Rates LEARNING OBJECTIVE Calculate the worth of a cash flow profile with variable interest rates. Video Lesson: Variable Interest Rates Thus far we have considered interest rates to be fixed over the duration of the financial transaction. Recent experience indicates that such a situation is not likely if the time period of interest extends over several years. Considering a single sum of money and discrete compounding, if it denotes the interest rate appropriate during time period t, the future worth equivalent for a single sum of money can be expressed as F = P (1 + i1 )(1 + i2 )…(1 + in−1 )(1 + in )

Extending the consideration of variable interest rates to cash flow series, the present worth of a series of cash flows can be represented as P = A1 (1 + i1 )

−1

+ A2 (1 + i1 )

−1

(1 + i2 )

−1

+ … + An (1 + i1 )

−1

(1 + i2 )

−1

…(1 + in )

−1

(3.14)

The future worth of a series of cash flows with variable interest rates can be given as F = An + An−1 (1 + in ) + An−2 (1 + in−1 )(1 + in ) + … + A1 (1 + i2 )(1 + i3 )…(1 + in−1 )(1 + in )

(3.15)

EXAMPLE 3.9 A Cash Flow Series with Variable Interest Rates Video Example Consider the CFD given in Figure 3.9 with the appropriate interest rates indicated. Determine the present worth, future worth, and uniform series equivalents for the cash flow series.

FIGURE 3.9 CFD for Variable Interest Rate Example 3.9 Solution Computing the present worth gives P  =

=

$200(P |F 10%,1) − $200(P | F +$300(P |F

8%,1)(P |F

+$200(P |F

12%,1)(P |F

$200(P |F

10%,1)(P |F

10%,1)(P |F 8%,1)(P |F

10%,1) − $200(P |F

+$200(P |F

12%,1)(P | F

10%,1)

10%,1) 8%,1)(P |F 10%,1)(P |F 10%,1)

10%,2) + $300(P |F

8%,2)(P |F

8%,1)(P |F 10%,2)

10%,2) = $372.63

The future worth is given by F  =

$200 + $300(F |P +$200(F |P

=

8%,1)(F |P

10%,1)(F |P

12%,1) − $200(F |P

8%,2)(F |P

8%,2)(F |P

12%,1)

12%,1)

$200 + $300(1.08000)(1.12000) − $200(1.16640)(1.12000) +$200(1.10000)(1.16640)(1.12000) = $589.01

The uniform series equivalent is obtained as follows: P  =

A(P |F

10%,1) + A(P |F

+A(P |F $372.63 =

8%,2)(P |F

10%,2) + A(P |F

10%,2) + A(P |F

8%,1)(P |F

12%,1)(P |F

10%,2)

8%,2)(P |F

10%,2)

A[(0.90909) + (0.82645) + (0.92593)(0.82645) + (0.85734)(0.82645) + (0.89286)(0.85734)(0.82645)] = 3.841958A

A =

$96.99

Thus, $96.99/time period for five time periods is equivalent to the original cash flow series. Excel® Solutions This example provides an opportunity to use an Excel® worksheet function not previously used in this text: FVSCHEDULE. Its syntax is FVSCHEDULE(principal, schedule), and it is applied to single cash flows. For the example,

F  =FVSCHEDULE(200,{0.1,0.08,0.08,0.12})−FVSCHEDULE(200,{0.08,0.08,0.12})        +FVSCHEDULE(300,{0.08,0.12})+200    = $589.01

Excel® Video Lesson: FVSCHEDULE Function

Concept Check 03.04-CC001 Interest rates are likely to vary in an economic analysis a. When the planning horizon expands over several years b. Depending on the riskiness of the investment c. During times of inflation or deflation d. All of the above

CHAPTER 4 Present Worth

Chapter 4 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced e-text and WileyPLUS

FE-Like Problems 04-FE001 When using present worth to evaluate the attractiveness of a single investment alternative, what value is the calculated PW compared to? a. 0.0 b. MARR c. 1.0 d. WACC 04-FE002 A natural gas well is projected to produce $200,000 in profit during its first year of operation, $190,000 the second year, $180,000 the third year, and so on, continuing this pattern. If the well is expected to produce for a total of 10 years, and the effective annual interest rate is 8%, which of the following most closely represents the present worth of the well? a. $1,770,000 b. $1,508,000 c. $1,253,000 d. $1,082,000 Correct or Incorrect? Clear

  Check Answer

04-FE003 The present worth of a multi-year investment with all positive cash flows (incomes) other than the initial investment is $10,000 at MARR = i%. If MARR changes to (i + 1)%, the present worth will be a. Less than $10,000

b. Equal to $10,000 c. Greater than $10,000 d. Cannot determine without the cash flow profile and a value for i 04-FE004 Consider the following cash flow diagram. Which of the expressions is not valid for the present worth?

a. P = 100(P|A 10%,6) + 50(P|G 10%,3) + 150(P|A 10%,3)(P|F 10%,3) b. P = 100(P|A 10%,3) + 50(P|G 10%,3) + 250(P|A 10%,3)(P|F 10%,3) c. P = 250(P|A 10%,6) − 50(P|G 10%,3) d. P = 100(P|A 10%,4) + 50(P|G 10%,4) + 250(P|A 10%,2)(P|F 10%,4) Correct or Incorrect? Clear

  Check Answer

04-FE005 Ivan, an industrial engineering student, is working on a homework problem for Engineering Economy. He needs to calculate the PW at 12% of a cash flow series with $1,000 at t = 3, $1,500 at t = 4, and $2,000 at t = 5. If Ivan uses the equation P = 1,000(P|A 12%,3) + 500(P|G 12%,3), where is the P now located? a. t = 4 b. t = 2 c. t = 1 d. t = 0 04-FE006 The owner of a cemetery plans to offer a perpetual care service for grave sites. The owner estimates that it will cost $120 per year to maintain

a grave site. If the interest rate is 8%, what one-time fee should the owner charge for the perpetual care service? a. $96 b. $120 c. $1,500 d. $12,000 Correct or Incorrect? Clear

  Check Answer

04-FE007 Consider a palletizer at a bottling plant that has a first cost of $150,000, operating and maintenance costs of $17,500 per year, and an estimated net salvage value of $25,000 at the end of 30 years. Assume an interest rate of 8%. What is the present equivalent cost of the investment if the planning horizon is 30 years? a. $335,000 b. $344,500 c. $360,000 d. $395,500 04-FE008 The heat loss through the windows of a home is estimated to cost the homeowner $412 per year in wasted energy. Thermal windows will reduce heat loss by 93% and can be installed for $1,232. The windows will have no salvage value at the end of their estimated life of 8 years. Determine the net present equivalent value of the windows if the interest rate is 10%. a. $412 b. $812 c. $1,044 d. $1,834 Correct or Incorrect? Clear

  Check Answer

04-FE009 An inline filter has an estimated life of 9 years. By adding a purifier to the filter, savings of $300 in annual operating costs can be obtained. Annual interest on capital is 8%. Compute the maximum expenditure justifiable for the purifier. a. $24 b. $33 c. $300 d. $1,875 04-FE010 The city council has approved the building of a new bridge over Running Water Creek. The bridge will cost $17,000 for initial construction and have an annual maintenance cost of $1,000. The council plans to withdraw money from the city’s Bridges and Highways account to open a special account to cover the initial construction and to fund a perpetuity to cover the maintenance costs forever. How much money must be withdrawn from the Bridges and Highways account if the city can expect to earn 5% on the special account? a. $1,000 b. $17,000 c. $18,000 d. $37,000 Correct or Incorrect? Clear

  Check Answer

04-FE011 Two projects, A and B, are analyzed using ranking present worth analysis with MARR at i%. It is found that PW(A) > PW(B). If MARR is changed to (i + 1)%, what will be the relationship between PW(A) and PW(B)? a. PW(A) > PW(B) b. PW(A) = PW(B)

c. PW(A) < PW(B) d. Cannot be determined without the cash flow profiles 04-FE012 A library shelving system has a first cost of $20,000 and a useful life of 10 years. The annual maintenance is expected to be $2,500. The annual benefits to the library staff are expected to be $9,000. If the effective annual interest rate is 10%, what is the benefit-cost ratio of the shelving system? a. 1.51 b. 2.24 c. 1.73 d. 1.56 Correct or Incorrect? Clear

  Check Answer

04-FE013 When using the benefit cost ratio (B/C) measure of worth, what benchmark is the calculated ratio compared to in determining if an individual investment is attractive? a. 0.0 b. MARR c. 1.0 d. IRR 04-FE014 When using a benefit-cost ratio (B/C) analysis to evaluate multiple alternatives, which of the following approaches is acceptable? a. Ranking approach only b. Incremental approach only c. Either incremental or ranking d. Neither incremental nor ranking Correct or Incorrect?

Clear

  Check Answer

04-FE015 Elm City is considering a replacement for its police radio. The benefits and costs of the replacement are shown below. What is the replacement’s benefit-cost ratio if the effective annual interest rate is 8%? Purchase Cost: $7,000 Annual Savings: $1,500 Life: 15 years a. 3.21 b. 1.83 c. 1.76 d. 1.34 04-FE016 When faced with alternatives having unequal lives, do the following when calculating the benefit-cost ratio. a. Use an Incremental B/C approach b. Compare all alternatives over a common planning horizon c. At the end of the planning horizon, provide estimates for residual benefits and salvage values for useful lives beyond the planning horizon d. All the above Correct or Incorrect? Clear

  Check Answer

04-FE017

What can be said about B/C and B − C?

a. When B/C > 1, B − C = 0 b. When B/C < 1, B − C = 0 c. B − C > 0 may be used to rank multiple alternatives d. B/C may be used to rank multiple alternatives 04 FE018

T o projects are being considered If B

$2 000 and C

04-FE018 Two projects are being considered. If B1 = $2,000 and C1 = $1,000 for B1/C1 = 2, and B2 = $10,000, what is the largest cost C2 that would cause projects 1 and 2 to be equally preferred? a. $9,500 b. $5,000 c. $9,000 d. $4,500 Correct or Incorrect? Clear

  Check Answer

Problems Section 4.1 Comparing Alternatives LEARNING OBJECTIVE 4.1 Describe the eight discounted cash flow methods used in comparing investment alternatives. 04.01-PR001 Match the measures of worth in the first column with an appropriate definition from the second column.

Measure of Worth

Definitions

(a) Annual worth

(1) Converts all cash flows to a single sum equivalent at t = (planning horizon) using i = MARR

(b) Discounted payback period (c) Capitalized worth (d) External rate of return (e) Future worth (f) Internal rate of return

(2) Converts all cash flows to a single sum equivalent at t = 0 using i = MARR

(g) Present worth

(7) Determines the PW when the planning horizon is infinitely long

(3) Converts all cash flows to an equivalent uniform series over the planning horizon (4) Determines an interest rate that yields a PW (or FW or AW) of 0 (5) Determines how long it takes for the cumulative present worth to be positive at i = MARR (6) Determines the interest rate that equates the future worth of invested capital to the future worth of recovered capital invested at i = MARR

04.01-PR002 GeoWorld Systems uses a subset of the following questions during the interview process for new engineers. For each of the following cases, determine if “the project” or “do nothing” is preferred. The value of MARR in each case is 14%. a. The present worth of the project is $1,367. b. The internal rate of return of the project is 12.9%. c. The annual worth of the project is −$632. d. The benefit-cost ratio of the project is 1.08. e. The future worth of the project is $3.75. f. The external rate of return of the project is 15.3%. g. The present worth of the project is −$47. h. The internal rate of return of the project is 14.7%.

i. The annual worth of the project is $6,775. j. The benefit-cost ratio of the project is 0.97. k. The future worth of the project is −$13,470. l. The external rate of return of the project is 3.7%. m. For the cases (if any) in which “do nothing” was preferred, what assumption is being made about the return generated by the “uninvested” funds? n. Is it possible that the values stated in (a) and (b) were correctly calculated on the same project? If not, why? o. Is it possible that the values stated in (c) and (d) were correctly calculated on the same project? If not, why? p. What do you know must be true about the present worth for the project in (j)? q. What do you know must be true about the internal rate of return for the project in (j)? 04.01-PR003 Match the measures of worth in the first column with an appropriate decision rule for preferring a project over “do nothing.” Measure of Worth

Decision Rule

(a) Annual worth (1) Measure of worth is greater than 0 (b) Benefit-cost ratio (2) Measure of worth is greater than 1 (c) External rate of return (3) Measure of worth is greater than MARR (d) Future worth (e) Internal rate of return (f) Present worth

LEARNING OBJECTIVE 4.2 Calculate a present worth (PW) converting all cash flows to a single sum equivalent at time zero for a given interest rate. Section 4.2.1 Present Worth of a Single Alternative

04.02-PR001 Video Solution DuraTech Manufacturing is evaluating a process improvement project. The estimated receipts and disbursements associated with the project are shown below. MARR is 6%/year. End of Year Receipts Disbursements 0 $ 0 $5,000 1 2 3

$ 0 $2,000 $4,000

$ 200 $ 300 $ 600

4 5

$3,000 $1,600

$1,000 $1,500

a. What is the present worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on present worth? c. Should DuraTech implement the proposed process improvement?

04.02-PR002 Bailey, Inc., is considering buying a new gang punch that would allow circuit boards to be produced more efficiently. The punch has a first cost of $100,000 and a useful life of 15 years. At the end of its useful life, the punch has no salvage value. Annual labor costs would increase $2,000 using the gang punch, but annual raw material costs would decrease $12,000. MARR is 5%/year. a. What is the present worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on present worth? c. Should Bailey buy the gang punch?

04.02-PR003 Repeat Problem 04.02-PR002 assuming that floor support and vibration dampening must be added for the gang punch. These one-time first costs are estimated to be $35,000. 04.02-PR004 Repeat Problem 04.02-PR002 assuming that the cost of capital has increased due to weak market conditions and MARR is now 6%/year. 04.02-PR005 Carlisle Company has been cited and must invest in equipment to reduce stack emissions or face EPA fines of $18,500 per year. An emission reduction filter will cost $75,000 and will have an expected life of 5 years. Carlisle’s MARR is 10%/year. a. What is the present worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on present worth? c. Is the filter economically justified? d. State at least one noneconomic factor that might influence this decision. 04.02-PR006 Repeat parts (a) through (c) of Problem 04.02-PR005 assuming that Carlisle’s board of directors has decided that because this action is based on a Federal government citation, no financial gain should be expected and the appropriate value of MARR is 0. 04.02-PR007 Fabco, Inc., is considering purchasing flow valves that will reduce annual operating costs by $10,000 per year for the next 12 years. Fabco’s MARR is 7%/year. Using a present worth approach, determine the maximum amount Fabco should be willing to pay for the valves. 04.02-PR008 Eddie’s Precision Machine Shop is insured for $700,000. The present yearly insurance premium is $1.00 per $100 of coverage. A sprinkler system with an estimated life of 20 years and no salvage value can be installed for $20,000. Annual maintenance costs for the sprinkler system are $400. If the sprinkler system is installed, the system must be included in the shop’s value for insurance purposes, but the insurance premium will reduce to $0.40 per $100 of coverage. Eddie uses a MARR of 15 %/year. a. What is the present worth of this investment?

b. What is the decision rule for judging the attractiveness of investments based on present worth? c. Is the sprinkler system economically justified? 04.02-PR009 Quilts R Us (QRU) is considering an investment in a new patterning attachment with the cash flow profile shown in the table below. QRU’s MARR is 13.5%/year. EOY Cash Flow EOY Cash Flow 0 −$1,400  8 $600 1 2

$0 $500

 9 10

$700 $800

3 4 5

$500 $500 $500

11 12 13

$900 −$1,000 −$2,000

6 7

$0 $500

14 15

−$3,000 $1,400

a. What is the present worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on present worth? c. Should QRU invest? 04.02-PR010 Galvanized Products is considering purchasing a new computer system for its enterprise data management system. The vendor has quoted a purchase price of $100,000. Galvanized Products is planning to borrow one-fourth of the purchase price from a bank at 15% compounded annually. The loan is to be repaid using equal annual payments over a 3-year period. The computer system is expected to last 5 years and has a salvage value of $5,000 at that time. Over the 5-year period, Galvanized Products expects to pay a technician $25,000 per year to maintain the system but will save $55,000 per year through increased efficiencies. Galvanized Products uses a MARR of 18%/year to evaluate investments. a. What is the present worth of this investment?

b. What is the decision rule for judging the attractiveness of investments based on present worth? c. Should the new computer system be purchased? 04.02-PR011 Jupiter is considering investing time and administrative expense on an effort that promises one large payoff in the future, followed by additional expenses over a 10-year horizon. The cash flow profile is shown in the table below. Jupiter’s MARR is 12%/year. EOY Cash Flow (K$) EOY Cash Flow (K$) 0

 −$2

 6

 $200

1

−$10

 7

 −$10

2 3

−$12 −$14

 8  9

 −$12  −$14

4

−$16

10

−$100

5

−$18

a. What is the present worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on present worth? c. Should Jupiter invest? 04.02-PR012 Imagineering, Inc., is considering an investment in CADCAM compatible design software with the cash flow profile shown in the table below. Imagineering’s MARR is 18%/year. EOY Cash Flow (M$) EOY Cash Flow (M$) 0 −$12 4 $5 1

 −$1

5

$5

2 3

  $5   $2

6 7

$2 $5

a. What is the present worth of this investment?

b. What is the decision rule for judging the attractiveness of investments based on present worth? c. Should Imagineering invest? 04.02-PR013 Home Innovation is evaluating a new product design. The estimated receipts and disbursements associated with the new product are shown below. MARR is 10%/year. End of Year Receipts Disbursements 0 1

  $0 $600

$1,000  $300

2

$600

 $300

3 4

$700 $700

 $300  $300

5

$700

 $300

a. What is the present worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on present worth? c. Should Home Innovation pursue this new product? 04.02-PR014 Video Solution Aerotron Electronics is considering purchasing a water filtration system to assist in circuit board manufacturing. The system costs $40,000. It has an expected life of 7 years at which time its salvage value will be $7,500. Operating and maintenance expenses are estimated to be $2,000 per year. If the filtration system is not purchased, Aerotron Electronics will have to pay Bay City $12,000 per year for water purification. If the system is purchased, no water purification from Bay City will be needed. Aerotron Electronics must borrow half of the purchase price, but they cannot start repaying the loan for 2 years. The bank has agreed to three equal annual payments, with the first payment due at the end of year 2. The loan interest rate is 8% compounded annually. Aerotron Electronics’ MARR is 10% compounded annually. a. What is the present worth of this investment?

b. What is the decision rule for judging the attractiveness of investments based on present worth? c. Should Aerotron Electronics buy the water filtration system?

04.02-PR015 Nancy’s Notions pays a delivery firm to distribute its products in the metro area. Delivery costs are $30,000 per year. Nancy can buy a used truck for $10,000 that will be adequate for the next 3 years. Operating and maintenance costs are estimated to be $25,000 per year. At the end of 3 years, the used truck will have an estimated salvage value of $3,000. Nancy’s MARR is 24%/year. a. What is the present worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on present worth? c. Should Nancy buy the truck? Section 4.2.2 Present Worth of Multiple Alternatives 04.02-PR016 Video Solution The engineering team at Manuel’s Manufacturing, Inc., is planning to purchase an enterprise resource planning (ERP) system. The software and installation from Vendor A costs $380,000 initially and is expected to increase revenue $125,000 per year every year. The software and installation from Vendor B costs $280,000 and is expected to increase revenue $95,000 per year. Manuel’s uses a 4-year planning horizon and a 10% per year MARR. a. What is the present worth of each investment? b. What is the decision rule for determining the preferred investment based on present worth ranking?

c. Which ERP system should Manuel purchase?

04.02-PR017 Parker County Community College (PCCC) is trying to determine whether to use no insulation or to use insulation that is either 1 inch thick or 2 inches thick on its steam pipes. The heat loss from the pipes without insulation is expected to cost $1.50 per year per foot of pipe. A 1-inch thick insulated covering will eliminate 89% of the loss and will cost $0.40 per foot. A 2-inch thick insulated covering will eliminate 92% of the loss and will cost $0.85 per foot. PCCC Physical Plant Services estimates that there are 250,000 feet of steam pipe on campus. The PCCC Accounting Office requires a 10%/year return to justify capital expenditures. The insulation has a life expectancy of 10 years. Determine which insulation (if any) should be purchased using a ranking present worth analysis. 04.02-PR018 Nadine Chelesvig has patented her invention. She is offering a potential manufacturer two contracts for the exclusive right to manufacture and market her product. Plan A calls for an immediate single lump sum payment to her of $30,000. Plan B calls for an annual payment of $1,000 plus a royalty of $0.50 per unit sold. The remaining life of the patent is 10 years. Nadine uses a MARR of 10%/year. a. What must be the uniform annual sales volume of the product for Nadine to be indifferent between the contracts, based on a present worth analysis? b. If the sales volume is below the volume determined in (a), which contract would the manufacturer prefer? 04.02-PR019 Quantum Logistics, Inc., a wholesale distributor, is considering the construction of a new warehouse to serve the southeastern geographic region near the Alabama–Georgia border. There are three cities being considered. After site visits and a budget analysis, the expected income and costs associated with locating in each of the cities have been determined. The life of the warehouse is expected to be 12 years and MARR is 15%/year.

City

Initial Cost Net Annual Income

Lagrange $1,260,000 Auburn $1,000,000

$480,000 $410,000

Anniston $1,620,000

$520,000

a. What is the present worth of each site? b. What is the decision rule for determining the preferred site based on present worth ranking? c. Which city should be recommended? 04.02-PR020 Tempura, Inc., is considering two projects. Project A requires an investment of $50,000. Estimated annual receipts for 20 years are $20,000; estimated annual costs are $12,500. An alternative project, B, requires an investment of $75,000, has annual receipts for 20 years of $28,000, and has annual costs of $18,000. Assume both projects have a zero salvage value and that MARR is 12%/year. a. What is the present worth of each project? b. Which project should be recommended? 04.02-PR021 DelRay Foods must purchase a new gumdrop machine. Two machines are available. Machine 7745 has a first cost of $10,000, an estimated life of 10 years, a salvage value of $1,000, and annual operating costs estimated at $0.01 per 1,000 gumdrops. Machine A37Y has a first cost of $8,000, a life of 10 years, and no salvage value. Its annual operating costs will be $300 regardless of the number of gumdrops produced. MARR is 6%/year, and 30 million gumdrops are produced each year. a. What is the present worth of each machine? b. What is the decision rule for determining the preferred machine based on present worth ranking? c. Which machine should be recommended? 04.02-PR022 Xanadu Mining is considering three mutually exclusive alternatives, as shown in the table below. MARR is 10%/year.

EOY A001 B002 C003 0 1

−$210 −$110 −$160 $80 $60 $80

2

$90

$60

$80

3

$100

$60

$80

4

$110

$70

$80

a. What is the present worth of each alternative? b. Which alternative should be recommended? 04.02-PR023 Two storage structures, given code names Y and Z, are being considered for a military base located in Sontaga. The military uses a 5%/year expected rate of return and a 24-year life for decisions of this type. The relevant characteristics for each structure are shown below. Structure Y Structure Z First Cost Estimated Life

$4,500 12 years

$10,000 24 years

Estimated Salvage Value

None

$1,800

Annual Maintenance Cost

$1,000

$720

a. What is the present worth of each machine? b. What is the decision rule for determining the preferred machine based on present worth ranking? c. Which structure should be recommended? 04.02-PR024 Several years ago, an individual won $27 million in the state lottery. To pay off the winner, the state planned to make an initial $1 million payment immediately followed by equal annual payments of $1.3 million at the end of each year for the next 20 years. Just before receiving any money, the person offered to sell the winning ticket back to the state for a one-time immediate payment of $14.4 million. If the state uses a 6%/year MARR, should it accept the offer? Use a present worth analysis.

04.02-PR025 Final Finishing is considering three mutually exclusive alternatives for a new polisher. Each alternative has an expected life of 10 years and no salvage value. Polisher 1 requires an initial investment of $20,000 and provides annual benefits of $4,465. Polisher 2 requires an initial investment of $10,000 and provides annual benefits of $1,770. Polisher 3 requires an initial investment of $15,000 and provides annual benefits of $3,580. MARR is 15%/year. a. What is the present worth of each polisher? b. Which polisher should be recommended? 04.02-PR026 Video Solution An environmental consultant is considering the installation of a water storage tank for a client. The tank is estimated to have an initial cost of $426,000, and annual maintenance costs are estimated to be $6,400 per year. As an alternative, a holding pond can be provided a short distance away at an initial cost of $180,000 for the pond plus $90,000 for pumps and piping. Annual operating and maintenance costs for the pumps and holding pond are estimated to be $17,000. The planning horizon is 20 years, and at that time, neither alternative has any salvage value. Determine the preferred alternative based on a present worth analysis with a MARR of 20%/year.

04.02-PR027 Dark Skies Observatory is considering several options to purchase a new deep-space telescope. Revenue would be generated from the telescope by selling “time and use” slots to various researchers around the world. Four possible telescopes have been identified in addition to the possibility of not buying a telescope if none are financially attractive. The table below details the characteristics of each telescope. A present worth ranking analysis is to be performed.

T1 Useful Life First Cost

T2

T3

T4

10 years 10 years 10 years 10 years $600,000 $800,000 $470,000 $540,000

Salvage Value $70,000 $130,000 $65,000 $200,000 Annual Revenue $400,000 $600,000 $260,000 $320,000 Annual Expenses $130,000 $270,000 $70,000 $120,000 a. Determine the preferred telescope if MARR is 25%/year. b. Determine the preferred telescope if MARR is 42%/year. 04.02-PR028 Value Lodges owns an economy motel chain and is considering building a new 200-unit motel. The cost to build the motel is estimated at $8,000,000; Value Lodges estimates furnishings for the motel will cost an additional $700,000 and will require replacement every 5 years. Annual operating and maintenance costs for the motel are estimated to be $800,000. The average rental rate for a unit is anticipated to be $40/day. Value Lodges expects the motel to have a life of 15 years and a salvage value of $900,000 at the end of 15 years. This estimated salvage value assumes that the furnishings are not new. Furnishings have no salvage value at the end of each 5-year replacement interval. Assuming average daily occupancy percentages of 50%, 60%, 70%, and 80% for years 1 through 4, respectively, and 90% for the fifth through fifteenth years, MARR of 12%/year, 365 operating days/year, and ignoring the cost of land, should the motel be built? Base your decision on a present worth analysis. 04.02-PR029 RealTurf is considering purchasing an automatic sprinkler system for its sod farm by borrowing the entire $30,000 purchase price. The loan would be repaid with four equal annual payments at an interest rate of 12%/year. It is anticipated that the sprinkler system would be used for 9 years and then sold for a salvage value of $2,000. Annual operating and maintenance expenses for the system over the 9-year life are estimated to be $9,000 per year. If the new system is purchased, cost savings of $15,000 per year will be realized over the present manual watering system. RealTurf uses a MARR of 15%/year for economic decision making. Based on a present worth analysis, is the purchase of the new sprinkler system economically attractive?

Section 4.2.3 Present Worth of One-Shot Investments 04.02-PR030 Consider the two one-shot investment alternatives shown in the table below. Neither alternative is expected to be available again in the future. MARR is 11%/year. Based on a present worth analysis, which alternative is preferred? EOY Alternative W Alternative X 0

−$100,000

−$150,000

1 2

$20,000 $20,000

$40,000 $45,000

3

$50,000

$50,000

4

$80,000

$55,000

5 6

$110,000

$60,000 $65,000

7

$70,000

04.02-PR031 Video Solution Two new opportunities are being considered for a venture capital firm. Both are one-time opportunities with no option for renewal. The firm uses a 12%/year expected rate of return for decisions of this type. The relevant characteristics for each option are shown below. Based on a present worth analysis, which option is preferred? Initial Investment Estimated Life

Option 1 Option 2 $100,000 $75,000 12 years

9 years

Expected Annual Return $16,500 $14,300

04.02-PR032 Technology Innovations is planning to purchase one of two chip insertion machines. Due to the pace of technological change in this area, it is realistic to assume that these are one-shot investments. The expected cash flows for each machine are shown below. MARR is 8%/year. Based on a present worth analysis, which machine is preferred?

E Series M Series Initial Investment Estimated Life

$40,000 $60,000 7 5

End of Life Salvage $10,000 Annual Income Annual Expense

$0

$19,400 $26,000 $10,000 $6,000

LEARNING OBJECTIVE 4.3 Perform an economic analysis of public investments utilizing the benefit-cost analysis for a given interest rate. Section 4.3.1 Benefit-Cost Calculations for a Single Alternative 04.03-PR001 Video Solution The Oklahoma City Zoo has proposed adding to their Web site a major segment providing a virtual tour of the grounds and animals, suitable for both routine enjoyment and educational purposes in classrooms. Survey data indicate that this will have either a neutral or positive effect upon actual zoo attendance. The Web site will be professionally done and have an initial cost of $325,000. Upkeep, refreshing the videos, and developing videos for scientific research and entertainment will cost another $80,000 per year. The zoo is expected to be in operation for an indefinite period; however, a study period of only 10 years for the Web site is to be assumed, with only a residual (salvage) value of $60,000 for the archival value being anticipated. Interest is 7%. An estimated 100,000 persons will visit the e-zoo in the first year, increasing by 30,000 each year, and they will receive, on the average, an additional $0.80 of benefit per visit when the new area is complete. On the basis of B/C analysis, should the Web site be supported for funding?

04.03-PR002 Ten cavemen with a remaining average life expectancy of 10 years use a path from their cave to a spring some distance away. The path is not easily traveled due to 100 large stones that could be removed. The annual benefit to each individual if the stones are removed is $6. Each stone can be removed at a cost of $1. The interest rate is 2%.

a. Compute the benefit-cost ratio for the individual who removes the 100 stones. b. Compute the benefit-cost ratio for the individual if the task is undertaken collectively, with each individual removing 10 stones. c. What maximum amount may be charged by a manager who organizes the group effort if the minimum acceptable benefit-cost ratio is 2? 04.03-PR003 The Logan Public Library in Iowa serves long-term residents, “bedroom community” residents who work in Omaha, and all of Harrison County. A renovation is planned, especially to include access to more electronic volumes, modernized computer facilities, and quicker check-in and checkout. In addition, two small meeting rooms with modernized e-access are needed. The cost of the renovation, including cabling, will be $33,000, the new equipment will cost $21,000, and e-volume access initiation will be $17,500. Maintenance is expected to run an additional $3,500 per year, plus $4,000 for renewed e-volume access. The interest rate is 8%, the planning horizon is 10 years, and the building renovation is expected to have a salvage value of 30%, with no salvage value for equipment or e-volume access. It is estimated that an additional 2,500 visits to the library will occur in the first year, increasing by 500 per year thereafter. It is estimated that the average benefits due to the new facilities, equipment, and access will be $2.00 per person per visit. a. Should the city government vote to approve the plans? Use PW and calculate B − C. b. What is the smallest benefit per person that will make this project desirable? 04.03-PR004 Lincoln Park Zoo in Chicago is considering a renovation that will improve some physical facilities at a cost of $1,800,000. Addition of new species will cost another $310,000. Additional maintenance, food, and animal care and replacement will cost $145,000 in the first year, increasing by 3% each year thereafter. The zoo has been in operation since 1868 and is expected to continue indefinitely; however, it is common to use a 20-year planning horizon on all new investments. Salvage value on facilities after 20 years will be 40% of initial cost. Interest is 7%. An estimated 1.5 million visits per year are made to the zoo, and the cost remains free year-round. How much

additional benefit per visit, on average, must the visitors perceive to justify the renovation? 04.03-PR005 The Boundary Waters Canoe Area Wilderness (BWCA) located in northeastern Minnesota, has 1 million acres of wilderness, 1,000 waterways, and 1,500 miles of canoe routes. While some youth, for example those in the Boy Scouts, are on high-adventure treks for 10 days at a time, it is common to have the rest of their family take advantage of the opportunity to also enjoy the BWCA area. A new area for this purpose is to be developed and will have an equivalent annual cost of $30,000, including initial cost (design, clearing, potable water, restrooms, showers, road, etc.), operating, upkeep, and security. Approximately 250 families will camp for 8 days each during the summer season. Another 2,200 persons will be admitted for single-day use of the facilities during the summer season. Although there are currently no fees charged, the average family camping in the area is willing to pay $12.00/night for the privilege, with some willing to pay more and some less. The average day user would be willing to pay $4.00/day, again some more and some less. a. What is the anticipated B/C ratio of this recreation area? b. What is the value of B − C? Section 4.3.2 Benefit-Cost Calculations for Multiple Alternatives 04.03-PR006 Recent development near Eugene, Oregon, has identified a need for improved access to Interstate 5 at one location. Civil engineers and public planners are considering three alternative access plans. Benefits are estimated for the public in general; disbenefits primarily affect some local proprietors who will see traffic pattern changes as undesirable. Costs are monetary for construction and upkeep, and savings are a reduction in cost of those operations today that will not be necessary in the future. All figures are relative to the present situation, retention of which is still an alternative, and are annualized over the 20-year planning horizon.

Alternative Benefits Disbenefits Costs Savings

A

B

C

$200,000 $300,000 $400,000 $37,000

$69,000 $102,000

$150,000 $234,000 $312,000 $15,000 $31,000 $42,000

a. What is the B/C ratio for each of these alternatives? b. Using incremental B/C ratio analysis, which alternative should be selected? c. Determine the value of B − C for each alternative. 04.03-PR007 A highway is to be built connecting Maud and Bowlegs. Route A follows the old road and costs $4 million initially and $210,000/year thereafter. A new route, B, will cost $6 million initially and $180,000/year thereafter. Route C is an enhanced version of Route B with wider lanes, shoulders, and so on. Route C will cost $9 million at first, plus $260,000 per year to maintain. Benefits to the users, considering time, operation, and safety, are $500,000 per year for A, $850,000 per year for B, and $1,000,000 per year for C. Using a 7% interest rate, a 15-year study period, and a salvage value of 50% of first cost, determine which road should be constructed. 04.03-PR008 The city of Columbus has identified three options for a public recreation area suitable for informal family activities and major organized events. As with most alternatives today, there are benefits, disbenefits, costs, and some savings. These have been estimated with the help of an external planning consultant and are identified in the table below. In each case, these are annualized over a 10-year planning horizon. Option 1 Option 2 Option 3 Benefits $400,000 $550,000 $575,000 Disbenefits $78,000 $125,000 $180,000 Costs Savings

$235,000 $390,000 $480,000 $25,000 $65,000 $90,000

a. Determine the B/C ratio for each project. Can you tell from these ratios which option should be selected? b. Determine which option should be selected using the incremental B/C ratio. c. Determine which option should be selected using B − C for each option. d. At what value of Option 2 costs are you indifferent between Option 1 and Option 2? 04.03-PR009 Video Solution An improvement to the roadway is desired from Philmont Scout Ranch to Springer in northeastern New Mexico. Alternative N (for north) costs $2,400,000 initially and $155,000/year thereafter. Route SA (for south, Alternative A) will cost $4,200,000 initially, and $88,000/year thereafter. Route SB is the same as SA with wider lanes and shoulders. It costs $5,200,000 initially with maintenance at $125,000/year. User costs considering time, operation, and safety are $625,000 for N, $410,000 for SA, and $310,000 for SB. The salvage values for N, SA, and SB after 20 years are 20% of initial cost, respectively. Using a MARR of 7% and a 20-year study period, which should be constructed? a. Use an incremental B/C analysis. b. Use a B − C analysis. c. Which route is preferred if 0% interest is used?

04.03-PR010 Video Solution A relocation of a short stretch of rural highway feeding into Route 390 northwest of Dallas is to be made to accommodate new growth. The existing road is now unsafe, and improving it is not an alternative. Alternate new route locations are designated as East and West. The initial investment by government highway agencies will be $3,500,000 for East and $5,000,000 for West. Annual highway maintenance costs will be $120,000 for East and $90,000 for the shorter location West. Relevant annual road user costs, considering vehicle operation, time en route,

fuel, safety, mileage, and so on, are estimated as $880,000 for East and only $660,000 for West. Assume a 20-year service life and i = 7%. a. Clearly identify the annual equivalent benefits and costs of route West over route East. b. Compute the appropriate B/C ratio(s) and decide whether East or West should be constructed.

04.03-PR011 Lynchburg has two old four-lane roads that intersect, and traffic is controlled by a standard green, yellow, red stoplight. From each of the four directions, a left turn is permitted from the inner lane; however, this impedes the flow of traffic while a car is waiting to safely turn left. The light operates on a 2-minute cycle with 60 seconds of green-yellow and 60 seconds of red for each direction. Approximately 10% of the 12,000 vehicles using the intersection each day are held up for an extra 2 full minutes and average three extra start-stop operations, solely due to the left-turn bottleneck. These delays are realized during 300 days per year. A start-stop costs 3 cents per vehicle, and the cost of the excess waiting is $18/hour for private traffic and $45/hour for commercial traffic. Approximately 3,000 of the vehicles are commercial, with the remainder being private. The potential benefit to the public is that the cost of extra waiting and start-stops can be reduced by 90% through a project to widen the intersection to include specific left-turn lanes and use of dedicated left-turn arrows. If the planning horizon is 10 years and the city uses a 7% interest rate, what is the most that can be invested in the project and maintain a B/C ratio of 1.0 or greater? There will be no additional maintenance cost. 04.03-PR012 During the past 5 years a local road has had 3 fatal crashes (0.6 per year), 10 injury crashes (2.0 per year) and 24 property damage crashes (4.8 per year). The city has identified two roadway projects to decrease these crashes and wants to determine the cost effectiveness of each using a B/C and a B − C approach based on present worth of benefits and costs. The overall crash cost to society (emergency response, medical cost, lost productivity, rehab, etc.) is $1.4 million for a fatal crash, $160,000 for an injury crash, and

$30,000 for a property damage crash. Accident costs are expected to increase each year by 5% of these current (“year 1”) costs. Option (1): Flatten the side slopes, reducing fatal, injury, and property damage crashes, each by 10%, at a first cost of $660,000. There will be no increase in annual maintenance cost, and the useful life will be 40 years. The increased benefits due to prevention of expensive accidents are calculated to be $6,520 per year. Option (2): A roadside barrier can be installed, reducing fatal crashes by 50%, injury crashes by 30%, and property damage crashes by 10% at a first cost of $275,000, with an added maintenance cost of $1,250 the first year, increasing by $37.50 each year, with a useful life of 20 years. The increased benefits due to prevention of increasingly expensive accidents are calculated to be $26,520 per year. The discount rate is i = 5%. Note that the useful lives are different. Option (1) is useful for 40 years and Option 2 is useful for 20 years. Due to rapid automotive technological improvements including autosensing, automatic control, driverless cars, and the uncertainty about increases in vehicle safety for the future, it is decided to use a planning horizon of 20 years. There will be residual benefits and costs at the end of the planning horizon. If Option 1 to flatten the side slopes is selected, there will be no costs during years 21–40; however, the benefits will continue over that time and should be credited to the PW of benefits. If the option to install the roadside barrier is selected, it will be removed at the end of 20 years at a cost of $35,000 and all benefits and maintenance costs will cease at that time. a. What is the B/C ratio for each alternative? b. Using an incremental B/C ratio, determine which alternative should be selected. Do consider the alternative to “Do Nothing.” c. Using B − C determine which alternative should be selected. 04.03-PR013 A family foundation donated $500 million for an extensive park to be placed on 100 acres, in an appreciative city. This park will be free for family members of all ages and backgrounds. While not a pure government project, it does involve city government resources for operations and upkeep, while the foundation’s enormous donation will fund design and construction of the park. Benefit-cost analysis is used to decide which of many proposed

themes to include in the park. These proposed themes include a skate park, bike park, swings, sports courts (basketball, volleyball, badminton, etc.), theme gardens, swing hill, a boathouse, water mountain, mist mountain, a lodge, adventure playground, reading tree, a slide vale, walking path, and much more. Several of the theme proposals have advanced sufficiently to merit a benefit-cost analysis using a consistent 10-year planning horizon and a discount rate of 6%. Following is a summary of present worth of net benefits and net costs for each proposed theme in thousands of dollars ($000’s omitted). Net benefits are realized by the public including enjoyment, exercise, safety, camaraderie, fun, and more. Disbenefits include time and travel to get to the park, the risk of scrapes, bumps, injuries, getting lost, and such. These benefits to be enjoyed, and disbenefits to be avoided, have influenced design features. Likewise, net costs to the foundation and the local government cover the entire 10 year planning horizon, including first cost, operations, planned and unanticipated maintenance, less any salvage or residual value at the end of 10 years. Theme

Net Public Net Fndn and Benefits ($000) Gov’t Costs ($000) B/C

1 Water, Restrooms, Snack Bar, First Aid; Required 2 Sports Court 3 Slide Vale

$20,000

$23,000

0.87

$65,000  $9,000

$22,000  $5,500

2.95 1.64

4 Skate Park (Basic and Advanced)

$42,000

$17,500

2.40

5 Bike Park (Basic and Advanced) $700 6 Swing Hill 7 Theme Gardens

$33,000

$21,000

1.57

 $9,000 $13,500

 $5,000 $11,000

1.80 1.23

The foundation and government planners would now like to answer the following questions. 1. Which of the themes offers the most positive impact? Use an Incremental B/C analysis.

2. At this point there is no cost constraint. They planners would also like to see a ranking of themes from highest to lowest impact. LEARNING OBJECTIVE 4.4 Calculate the discounted payback period (DPBP) for a given interest rate to determine how long it takes for the cumulative present worth (PW) to be positive. Section 4.4.1 Discounted Payback Period for a Single Alternative 04.04-PR001 Reconsider Problem 04.02-PR002 (repeated here). Bailey, Inc., is considering buying a new gang punch that would allow it to produce circuit boards more efficiently. The punch has a first cost of $100,000 and a useful life of 15 years. At the end of its useful life, the punch has no salvage value. Labor costs would increase $2,000 per year using the gang punch, but raw material costs would decrease $12,000 per year. MARR is 5%/year. a. What is the discounted payback period for this investment? b. If the maximum attractive DPBP is 3 years, what is the decision rule for judging the worth of this investment? c. Should Bailey buy the gang punch based on DPBP? 04.04-PR002 Video Solution Home Innovations is evaluating a new product design. The estimated receipts and disbursements associated with the new product are shown below. MARR is 10%/year. End of Year Receipts Disbursements 0   $0 $1,000 1 $600  $300 2 3 4

$600 $700 $700

 $300  $300  $300

5

$700

 $300

a. What is the discounted payback period for this investment?

b. If the maximum attractive DPBP is 3 years, what is the decision rule for judging the worth of this investment? c. Should Home Innovations buy the gang punch based on DPBP?

Section 4.4.2 Discounted Payback Period for Multiple Alternatives 04.04-PR003 Reconsider Problem 04.02-PR016 (repeated here). The engineering team at Manuel’s Manufacturing, Inc., is planning to purchase an enterprise resource planning (ERP) system. The software and installation from Vendor A costs $380,000 initially and is expected to increase revenue $125,000 per year every year. The software and installation from Vendor B costs $280,000 and is expected to increase revenue $95,000 per year. Manuel’s uses a 4-year planning horizon and a 10% per year MARR. a. What is the discounted payback period of each investment? b. Which ERP system should Manuel purchase if his decision rule is to select the system with the shortest DPBP? c. Does this decision agree or disagree with the results of the present worth analysis in Problem 04.02-PR018? 04.04-PR004 Video Solution Octavia Bakery is planning to purchase one of two ovens. The expected cash flows for each oven are shown below. MARR is 8%/year. Model 127B Model 334A Initial Investment Estimated Life End of Life Salvage

$50,000 10 $10,000

$80,000 5 $0

Annual Income Annual Expense

$19,400 $10,000

$26,000 $6,000

a. What is the discounted payback period for each investment? b. Which oven should Octavia Bakery purchase if they wish to minimize the DPBP?

04.04-PR005 Reconsider Problem 04.02-PR019 (repeated here). Quantum Logistics, Inc., a wholesale distributor, is considering the construction of a new warehouse to serve the southeastern geographic region near the Alabama–Georgia border. There are three cities being considered. After site visits and a budget analysis, the expected income and costs associated with locating in each of the cities have been determined. The life of the warehouse is expected to be 12 years, and MARR is 15%/year. City

Initial Cost Net Annual Income

Lagrange $1,260,000 Auburn $1,000,000 Anniston $1,620,000

$480,000 $410,000 $520,000

a. What is the discounted payback period for each location? b. Which city should Quantum Logistics select if they wish to minimize the DPBP? c. Is this recommendation consistent with a present worth analysis recommendation in Problem 04.02-PR019? Section 4.5 Capitalized Worth LEARNING OBJECTIVE 4.5 Calculate the capitalized worth (CW) of an investment for a given interest rate when the planning horizon is infinitely long. 04.05-PR001 A municipality is planning on constructing a water treatment plant at an initial cost of $10,000,000. Every 5 years, major repairs and cleanup are required at a cost of $2,000,000. Due to the necessity to remove

sludge and make minor repairs, annual costs of operating the treatment plant are estimated to be $700,000, $775,000, $850,000, $925,000, and $1,000,000 each year leading up to the 5-year major repair and cleanup. Based on a 4%/year TVOM, what is the capitalized cost for the water treatment plant? 04.05-PR002 Video Solution Two incinerators are being considered by a waste management company. Design A has an initial cost of $2,500,000, has annual operating and maintenance costs of $800,000, and requires overhauls every 5 years at a cost of $1,250,000. Design B is more sophisticated, including computer controls; it has an initial cost of $5,750,000, has annual operating and maintenance costs of $600,000, and requires overhauls every 10 years at a cost of $3,000,000. Using a 5%/year interest rate, determine the capitalized cost for each design and recommend which should be chosen.

04.05-PR003 Video Solution A flood control project at Pleasant Valley dam is projected to cost $2,000,000 today, have annual maintenance costs of $50,000, and have major inspection and upkeep after each 5-year interval costing $250,000. If the interest rate is 10%/year, determine the capitalized cost.

04.05-PR004 The gaming commission is introducing a new lottery game called Infinite Progresso. The winner of the Infinite Progresso jackpot will receive $1,000 at the end of January, $2,000 at the end of February, $3,000 at the end of March, and so on up to $12,000 at the end of December. At the beginning of the next year, the sequence repeats starting at $1,000 in January and ending at $12,000 in December. This annual sequence of payments repeats indefinitely. If the gaming commission expects to sell a minimum of 1 million tickets, what is the minimum price they can charge for the tickets to break even, assuming the commission earns 6%/year/month on its investments and there is exactly one winning ticket? 04.05-PR005 A generous benefactor donates $500,000 to a state university. The donation is to be used to fund student scholarships. Determine the dollar amount of scholarships that can be awarded each year under each of the

following conditions. The state university earns 4% per year on its investments. a. The donation is a quasi-endowment designed to last 20 years. b. The donation is a quasi-endowment designed to last 30 years. c. The donation is a quasi-endowment designed to last 50 years. d. The donation is an endowment designed to last forever. 04.05-PR006 Reconsider the data from Problem 04.05-PR005. Plot a graph of scholarship dollars versus number of years, where the number of years varies from 1 year to 100 years by 2-year increments. Dollars should be on the y-axis and years on the x-axis. 04.05-PR007 A prospective venture has the following cash flow profile over a 5-year horizon. What is the capitalized worth at 6%/year assuming the pattern repeats indefinitely? End of Year Receipts Disbursements 0 1 2

$100 $300 $500

$1,100 $50 $250

3 4 5

$400 $350 $250

$150 $100 $0

04.05-PR008 You decide to open an individual retirement account (IRA) at your local bank that pays 8%/year compounded annually. At the end of each of the next 40 years, you will deposit $4,000 into the account. Three years after your last deposit, you will begin making annual withdrawals. What annual amount will you be able to withdraw if you want the withdrawals to last a. 20 years? b. 30 years? c. Forever?

Chapter 4 Summary and Study Guide Summary 4.1: Comparing Alternatives

Learning Objective 4.1: Describe the eight discounted cash flow methods used in comparing investment alternatives. (Section 4.1) The eight DCF methods for comparing economic alternatives are: 1. Present worth (PW) 2. Benefit-cost ratio (B/C) 3. Discounted payback period (DPBP) 4. Capitalized worth (CW) 5. Annual worth (AW) 6. Future worth (FW) 7. Internal rate of return (IRR) 8. External rate of return (ERR) Present worth (PW), future worth (FW), annual worth (AW), and capitalized worth (CW) are all examples of ranking methods to determine economic worth. The alternative having the greatest worth value over the planning horizon is the most attractive. The discounted payback period (DPBP) method is another ranking method, but its goal is to identify the investment with the shortest discounted payback period. The incremental methods include the internal rate of return (IRR), external rate of return (ERR), and the benefit-cost ratio (B/C). The preferred alternative is that which satisfies Principle #6: Money should continue to be

invested as long as each additional increment of investment yields a return that is greater than the investor’s time value of money (TVOM). 4.2: Present Worth Calculations

Learning Objective 4.2: Calculate a present worth (PW) converting all cash flows to a single sum equivalent at time zero for a given interest rate. (Section 4.2) If the PW is positive, then the investment is recommended. When comparing multiple investments, the alternative with the highest PW is preferred. Mathematically, this is expressed as n

Maximize  PWj = ∑ Ajt (1 + MARR) ∀j

−t

(4.2)

t=0

4.3: Benefit-Cost Analysis

Learning Objective 4.3: Perform an economic analysis of public investments utilizing benefit-cost analysis for a given interest rate. (Section 4.3) Public investments such as cultural development, public protection, economic services, and protecting the environment usually involve large investments undertaken with tax dollars. In order to see if a public investment is attractive, we must ensure that the benefits for the public’s greater good exceed the costs of providing those benefits, or in other words the B/C ratio is greater than 1. Mathematically the B/C ratio is expressed as

n

∑ Bjt (1 + i)

−t

(4.3)

t=1

B/Cj (i) =

n

∑ Cjt (1 + i)

−t

t=1

4.4: Discounted Payback Period

Learning Objective 4.4: Calculate the discounted payback period (DPBP) for a given interest rate to determine how long it takes for the cumulative present worth (PW) to be positive. (Section 4.4) The DPBP shows the length of time for an investment to be fully recovered, including the time value of money (TVOM). When used as a stand-alone measure of economic worth, it is not always clear if the DPBP value obtained is acceptable or not. Therefore, DPBP should not be used to identify the investment that is to be made from a set of mutually exclusive investment alternatives. Instead, it is best used as a supplemental tool with other economic worth analyses. 4.5: Capitalized Worth

Learning Objective 4.5: Calculate the capitalized worth (CW) of an investment for a given interest rate when the planning horizon is infinitely long. (Section 4.5) Although an infinite series of cash flows would rarely be encountered in the real world, the CW method is a good approximation for very long-term investment projects such as bridges, tunnels, railways, dams, nuclear power plants and others. In these cases we assume that cash flows will extend indefinitely into the future. The alternative having the greatest CW is preferred. The equation for CW is

CW = A(P |A i%,∞) = A/i

(4.7)

Important Terms and Concepts Present Worth (PW) The value of all cash flows converted to a single sum equivalent at time zero using i = MARR. One-Shot Investment An investment that is available only once. Benefit-Cost Ratio (B/C) The ratio of the present worth of net public benefits to the present worth of net government costs using i = MARR. Benefits Minus Costs (B – C) The difference of the present worth of net public benefits and present worth of net government costs using i = MARR. Discounted Payback Period (DPBP) The length of time required for the cumulative present worth to become positive using i = MARR. Capitalized Worth (CW) The value of all cash flows converted to a single sum equivalent at time zero using i = MARR when the planning horizon is infinitely long.

Chapter 4 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 4.1: Comparing Alternatives LO 4.1: Describe the eight discounted cash flow methods used in comparing investment alternatives. Video Lesson: Project Analysis Methods Video Lesson Notes: Project Analysis Methods 4.2: Present Worth Calculations LO 4.2: Calculate a present worth (PW) converting all cash flows to a single sum equivalent at time zero for a given interest rate. Video Lesson: Present Worth Video Lesson Notes: Present Worth Excel Video Lesson: PV Financial Function Excel Video Lesson Spreadsheet: PV Financial Function Excel Video Lesson: NPV Financial Function Excel Video Lesson Spreadsheet: NPV Financial Function Video Example 4.2: Present Worth of Two Alternatives Video Example 4.3: Selecting from Among Multiple One-Shot Investments Video Solution: 04.02-PR001 Video Solution: 04.02-PR014 Video Solution: 04.02-PR016

Video Solution: 04.02-PR026 Video Solution: 04.02-PR031 4.3: Benefit-Cost Analysis LO 4.3: Perform an economic analysis of public investments utilizing the benefit-cost analysis for a given interest rate. Video Lesson: Benefit-Cost Analysis Video Lesson Notes: Benefit-Cost Analysis Video Example 4.4: Benefit-Cost Analysis for a Single Alternative Video Example 4.8: Benefit-Cost Analysis of Three Routes Video Solution: 04.03-PR001 Video Solution: 04.03-PR009 Video Solution: 04.03-PR010 4.4: Discounted Payback Period LO 4.4: Calculate the discounted payback period (DPBP) for a given interest rate to determine how long it takes for the cumulative present worth (PW) to be positive. Video Lesson: Payback Period and Discounted Payback Period Video Lesson Notes: Payback Period and Discounted Payback Period Excel Video Lesson: NPER Financial Function Excel Video Lesson Spreadsheet: NPER Financial Function Excel Video Lesson: GOAL SEEK Tool Excel Video Lesson Spreadsheet: GOAL SEEK Tool Excel Video Lesson: SOLVER Tool Excel Video Lesson Spreadsheet: SOLVER Tool

Video Example 4.9: Computing the DPBP of a Single Investment Opportunity Video Solution: 04.04-PR002 Video Solution: 04.04-PR004 4.5: Capitalized Worth LO 4.5: Calculate the capitalized worth (CW) of an investment for a given interest rate when the planning horizon is infinitely long. Video Lesson: Capitalized Worth Video Lesson Notes: Capitalized Worth Excel Video Lesson: PMT Financial Function Excel Video Lesson Spreadsheet: PMT Financial Function Video Example 4.15: Capitalized Cost for Water Delivery Video Solution: 04.05-PR002 Video Solution: 04.05-PR003 These chapter-level resources will help you with your overall understanding of the content in this chapter. Appendix A: Time Value of Money Factors Appendix B: Engineering Economic Equations Flashcards: Chapter 04 Excel Utility: TVM Factors: Table Calculator Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values Excel Utility: Monthly Payment Sensitivity

Excel Utility: TVM Factors: Discrete Compounding Excel Utility: TVM Factors: Geometric Series Future Worth Excel Utility: TVM Factors: Geometric Series Present Worth Excel Data Files: Chapter 04

CHAPTER 4 Present Worth LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 4.1 Describe the eight discounted cash flow methods used in comparing investment alternatives. (Section 4.1) 4.2 Calculate a present worth (PW) converting all cash flows to a single sum equivalent at time zero for a given interest rate. (Section 4.2) 4.3 Perform an economic analysis of public investments utilizing the benefit-cost analysis for a given interest rate. (Section 4.3) 4.4 Calculate the discounted payback period (DPBP) for a given interest rate to determine how long it takes for the cumulative present worth (PW) to be positive. (Section 4.4) 4.5 Calculate the capitalized worth (CW) of an investment for a given interest rate when the planning horizon is infinitely long. (Section 4.5)

Engineering Economics in Practice ConocoPhillips In 2017, ConocoPhillips was the world’s largest independent exploration and production company, based on proved reserves and production of liquids and natural gas. They explore for, produce, transport and market crude oil, bitumen, natural gas liquids and liquefied natural gas worldwide. Headquartered in Houston, Texas, ConocoPhillips operates in 17 countries. As of December 31, 2017, the company had 11,400 employees worldwide and assets of $73 billion. Its consolidated capital expenditures and investments in 2017 totaled $4.6 billion; total revenues were $32.6 billion. Its debt-to-capital ratio at the end of 2017 was 39%, compared to 44% at the end of 2016. ConocoPhillip’s operations are managed through six geographic regions including Alaska, Lower 48, Canada, Europe and North Africa, Asia Pacific and Middle East, and Other International. These market segments generally include a strong base of legacy production and inventory of low-cost supply investment opportunities. The company is also committed to conventional and unconventional exploration to support current and future growth opportunities. They are focused on increasing the pace of innovation through cutting-edge technology solutions. In April 2012, the company completed the separation of the downstream business into an independent, publicly traded company, Phillips 66. In its Form 10-K for 2017, the company provided the following comments regarding “standardized measure of discounted future net cash flows relating to proved oil and gas reserve quantities”: “In accordance with SEC and FASB requirements, amounts were computed using 12-month average prices (adjusted only for existing contractual terms) and end-of-year costs, appropriate statutory tax rates and a prescribed 10% discount factor. Twelve-month average prices are calculated as the unweighted arithmetic average of the first-day-of-the-month price for each month within the 12-month period prior to the end of the reporting period… For all years, continuation of year-end economic conditions was assumed. The calculations were based on estimates of proved reserves, which are revised over time as new data becomes available.” ConocoPhillips and numerous other organizations rely on present worth analysis in making investment decisions. To list the other firms that use present worth analysis would surely require listing every organization that uses discounted cash flow methods. Discussion Questions: 1. Do you think that small organizations rely on present worth analysis as well, or is this type of analysis important only for extremely large firms such as ConocoPhillips? 2. The statement about the calculations being based on estimates of proved reserves suggests to stakeholders that this is a somewhat volatile business. What measures can an engineering economist take to ensure that a rigorous analysis is performed to instill investor confidence? 3. Are you surprised to see the detailed explanation of how the numbers were calculated? 4. In what other parts of its business might Conoco-Phillips rely on present worth analysis?

Introduction Recall the fifth step in the 7-step SEAT process described in Chapter 1 for performing an engineering economic analysis: Compare the alternatives. Implicit in this step is a decision regarding the method(s) to be used in making the comparison. In this chapter, you will learn how results obtained from present worth analysis compare with other methods of measuring economic worth. Present worth analysis is the most popular DCF method of comparing investments. You will also learn about three methods that are special cases of the present worth method: benefit-cost ratio, discounted payback period, and capitalized worth.

Systematic Economic Analysis Technique 1. Identify the investment alternatives 2. Define the planning horizon 3. Specify the discount rate 4. Estimate the cash flows 5. Compare the alternatives 6. Perform supplementary analyses 7. Select the preferred investment

4.1 Comparing Alternatives LEARNING OBJECTIVE Describe the eight discounted cash flow methods used in comparing investment alternatives. Video Lesson: Project Analysis Methods This section summarizes the eight DCF methods that are covered in detail in Chapters 4–6. It also touches on several important considerations when comparing alternatives, regardless of method, including whether the analysis is before-tax or after-tax, whether the lives of the alternatives are equal or unequal, and the concept of the do-nothing alternative.

4.1.1 Methods of Comparing Economic Worth The eight DCF methods that can be used in comparing investment alternatives are present worth, benefit-cost ratio, discounted payback period, capitalized worth, annual worth, future worth, internal rate of return, and external rate of return. These methods may be described briefly as follows: 1. The present worth (PW) method converts all cash flows to a single sum equivalent at time zero using i = MARR. 2. The benefit-cost ratio (B/C) method determines the ratio of the present worth of benefits and savings to the present worth of investments and costs using i = MARR. A variant to this is to calculate benefits minus costs (B − C), the present worth of benefits and savings less investments and costs. Measures B/C and B − C surely differ and yet the analyses are equivalent and decisions are identical. 3. The discounted payback period (DPBP) method determines how long it takes for the cumulative present worth to be positive using i = MARR. 4. The capitalized worth (CW) method determines the present worth (using i = MARR) when the planning horizon is infinitely long. 5. The annual worth (AW) method converts all cash flows to an equivalent uniform annual series of cash flows over the planning horizon using i = MARR. 6. The future worth (FW) method converts all cash flows to a single sum equivalent at the end of the planning horizon using i = MARR. 7. The internal rate of return (IRR) method determines the interest rate that yields a future worth (or present worth or annual worth) of zero.

8. The external rate of return (ERR) method determines the interest rate that equates the future worth of the invested capital to the future worth of recovered capital (when the latter is computed using the MARR.)

4.1.2 Ranking and Incremental Methods of Economic Worth The eight DCF methods can be divided into two groups: ranking and incremental methods. Ranking Methods Present worth, future worth, annual worth, benefits minus costs (B − C) and capitalized worth are ranking methods; as such, the alternative having the greatest PW, FW, (B − C), AW, or CW over the planning horizon is the economic choice and would be recommended, absent nonmonetary criteria.1 DPBP also is a ranking method, but the goal is to identify the investment alternative with the shortest payback period. Incremental Methods Internal rate of return, external rate of return, and benefit-cost ratio B/C are incremental methods; as such, the preferred alternative is the one that satisfies Principle #6: Money should continue to be invested as long as each additional increment of investment yields a return that is greater than the investor’s TVOM. Incremental solutions that are equivalent to ranking solutions can be obtained using PW, FW, AW, (B − C) and CW. However, there seems to be little reason to employ the more cumbersome incremental solution, because it is much simpler to compute the value of the PW, FW, AW, (B − C) or CW and recommend the one having the greatest value over the planning horizon.

4.1.3 Equivalence of Methods All of these analysis methods are equivalent: PW, FW, AW, IRR, ERR, B/C and (B − C). That is, when applied correctly, all six of these methods will yield the same recommendation regarding investment alternatives. The CW and DPBP methods are not guaranteed to result in a recommendation identical to that obtained using any of the six equivalent economic worth methods. Examples in this and later chapters will illustrate why this is so.

4.1.4 Before-Tax Versus After-Tax Analysis In using a measure of economic worth to compare investment alternatives, you can employ either before-tax or after-tax cash flows. Be consistent, however! It is either/or but not both in the same analysis. If the comparison is based on before-tax cash flows, then use a before-tax MARR; likewise, if the comparison is based on after-tax cash flows, then use an after-tax MARR. Although we believe it is usually best to perform after-tax economic justifications, we will not cover tax issues until after discussing a variety of methods used to compare economic alternatives.2 Consequently, in this and the following two chapters, you may consider the analyses to be either before-tax (with before-tax cash flows and a before-tax MARR) or after-tax (with after-tax cash flows and an after-tax MARR).

4.1.5 Equal Versus Unequal Lives When comparing investment alternatives, they must be compared over a common time period, called the planning horizon. (Recall Principle #8: Compare investment alternatives over a common period of time.) If the duration of the planning horizon differs from the useful lives of the alternatives, then (at the end of the planning horizon) cash flow estimates must be provided for the terminal or salvage values for alternatives with lives greater than the planning horizon; for alternatives having useful lives less than the planning horizon, replacement decisions must be made and cash flow estimates must be provided for the replacements.

4.1.6 A Single Alternative “A single alternative” is an oxymoron. If there is no choice, then there is no alternative. When we consider “a single alternative,” however, we are considering doing something versus doing nothing. Assuming that the donothing alternative is feasible, you have two options when faced with the question “Should I invest in an opportunity or not?”: Invest, or don’t invest. In evaluating the “invest” option, we assume that the cash flow estimates reflect the differences in doing nothing versus doing something. On that basis,

when using PW, FW, AW, and (B − C) we choose to “do something” if the measure of economic worth has a value greater than zero;3 when using the B/C method, we choose to “do something” if the B/C ratio is greater than 1; and when using the IRR and ERR methods, we choose to “do something” if the measure of economic worth has a value greater than the MARR. For DPBP, the decision to invest or not depends on the prescribed acceptable value. Implicit in the criteria above is an opportunity cost assumption. Namely, we assume we are currently earning a return on our money equal to the MARR. Therefore, the decision to invest in a particular opportunity reduces to the following: Will we make more money by investing here or by leaving our money in an investment pool that earns a return equal to the MARR? If we will not make more money by investing in the opportunity in question, then we should leave our money in the investment pool. We also assume the investment in question has risks comparable to those of the investment pool.

Concept Check 04.01-CC001 Identify the one true statement. a. The B/C method determines the ratio of the present worth of benefits to the negative of the future worth of the investments b. The CW method determines the present worth using a finite planning horizon c. The IRR method determines the interest rate that yields a future worth of zero d. The ERR method determines the interest rate that yields a present worth of zero

Concept Check 04.01-CC002 Classify each method of comparing economic worth as a ranking method or as an incremental method. a. PW b. B/C c. DPBP d. CW e. AW f. FW g. IRR h. ERR

Concept Check 04.01-CC003 All eight of the analysis methods yield equivalent results. True or False?

Concept Check 04.01-CC004 It is preferred to perform after-tax economic justifications. True or False?

Concept Check 04.01-CC005 Identify the one true statement. a. Compare investment alternatives over a common planning horizon b. Investment alternatives with unequal planning horizons cannot be compared c. When the useful lives are not equal, use 10 years as the common time period d. When the useful lives are not equal, use infinity as the common time period

Concept Check 04.01-CC006 Doing nothing is a viable investment alternative. True or False?

4.2 Present Worth Calculations LEARNING OBJECTIVE Calculate a present worth (PW) converting all cash flows to a single sum equivalent at time zero for a given interest rate. Video Lesson: Present Worth In this section we focus on what is variously referred to as present worth, net present worth, present value, and net present value. Present worth analysis uses the MARR to express the economic worth of a set of cash flows, occurring over the planning horizon, as a single equivalent value at the current time, often referred to as “time now” or “time zero.” Present Worth (PW) The value of all cash flows converted to a single sum equivalent at time zero using i = MARR.

4.2.1 Present Worth of a Single Alternative When using the present worth method to evaluate whether an investment should be made, the decision depends on whether the present worth is positive. If so, then the investment is recommended. Recalling our work in Chapter 2, the present worth of an investment can be expressed mathematically as follows: n

PW(i%) = ∑ At (1 + i) i=0

Excel® Video Lesson: The PV Function

−t

(4.1)

EXAMPLE 4.1 Present Worth of a Single Investment Opportunity To automatically insert electronic components in printed circuit boards for a cell phone production line, a $500,000 surface mount placement (SMP) machine is being evaluated by a manufacturing engineer. Over the 10-year planning horizon, it is estimated that the SMP machine will produce annual after-tax cost savings of $92,500. The engineer estimates the machine will be worth $50,000 at the end of the 10-year period. Based on the firm’s 10% after-tax MARR, should the investment be made? Key Data Given The cash flows outlined in Figure 4.1; MARR = 10%; planning horizon = 10 years Find PW of the investment

FIGURE 4.1 CFD for Example 4.1 Solution From our work in Chapter 2, we compute the present worth for the investment: PW =

−$500,000 + $92,500(P |A

10%,10) + $50,000(P |F

=

−$500,000 + $92,500(6.14457) + $50,000(0.38554)

=

$87,649.73

10%,10)

or, using Excel®, Excel® Solution PW =PV(10%,10,−92500,−50000)−500000 = $87,649.62 Because PW > $0, the investment is recommended. Excel® Video Lesson: The PV Function Exploring the Solution With a present worth of $87,649.62, there can be little doubt that the investment is a good one for the company. However, it is useful to examine how the cumulative present worth behaves over the 10-year

period. Figure 4.2 shows that the cumulative present worth begins at −$500,000 with the purchase of the new machine and increases over the planning horizon to a final value of $87,649.62. Notice that the investment “loses money” (has a negative present worth) for the first 8 years. It is only in the 9th year that the cumulative present worth becomes positive. Therefore, if something unforeseen occurred, causing the company to abandon the investment before the 9th year, would the investment still be profitable? The answer to the question depends on the SMP machine’s salvage value at the time it is abandoned.

FIGURE 4.2 Spreadsheet of Cumulative Present Worth Over the Planning Horizon Excel® Data File We will examine cumulative present worth in more detail in Section 4.4, which discusses the discounted payback period.

4.2.2 Present Worth of Multiple Alternatives As a ranking method, PW is easily applied when choosing the preferred alternative from among several mutually exclusive alternatives: Choose the one with the greatest PW over the planning horizon. Mathematically, the objective is n

Maximize  PWj = ∑ Ajt (1 + MARR) ∀j

t=0

−t

(4.2)

EXAMPLE 4.2 Present Worth of Two Alternatives Video Example Entertainment Engineers, Inc., is an Ohio-based design engineering firm that designs rides for amusement and theme parks all over the world. Two alternative designs are under consideration for a new ride called the Scream Machine at a theme park located in Florida. The two candidate designs differ in complexity, cost, and predicted revenue. The first alternative design (A) will require an investment of $300,000 and is estimated to produce after-tax revenue of $55,000 annually over a 10-year planning horizon. The second alternative design (B) will require an investment of $450,000 and is expected to generate annual after-tax revenue of $80,000. A negligible salvage value is assumed for both designs. Theme park management could decide to “do nothing”; if so, the present worth of doing nothing will be zero. An after-tax MARR of 10% is used. Which alternative design, if either, should the theme park select? Key Data Given The cash flows outlined in Figure 4.3; MARR = 10%; planning horizon = 10 years Find PW of each alternative

FIGURE 4.3 CFDs for Example 4.2 Solution Letting A denote the alternative design for the $300,000 initial investment and B denote the other alternative, the present worth of each is as follows. Alternative A

PWA

= −$300,000 + $55,000(P |A 10%,10) = −$300,000 + $55,000(6.14457) = $37,951.35

>

$0.00 (therefore, A is better than doing nothing)

or, using Excel®, PWA

=PV (10%,10,−55000) −300000 = 37,951.19

>

$0.00 (therefore, A is better than doing nothing)

Alternative B PWB

= −$450,000 + $80,000(P |A 10%,10) = −$450,000 + $80,000(6.14457) = $41,565.60

>

$37,951.35 (therefore, B is better than A)

or, using Excel®, PWB

=PV (10%,10,−80000) −450000 = 41,565.37

>

$37,951.19 (therefore, B is better than A)

Based on the present worth analysis, the more expensive alternative (B) is recommended. It has the greater present worth. Exploring the Solution The theme park’s manager is interested in learning how the present worth for the two design alternatives is affected by the MARR value used. As shown in Figure 4.4, Alternative B has the greatest present worth for all values of MARR less than approximately 10.5%, but, thereafter, Alternative A has the greatest present worth. The manager also noticed that for MARR values greater than 13%, neither alternative has a positive present worth.

FIGURE 4.4 Plot of Present Worths for Example 4.2 Excel® Data File

4.2.3 Present Worth of One-Shot Investments Occasionally, investments are available only once. Such cases are called one-shot investments. When one-shot investments are being considered, the planning horizon is defined to be equal to the longest life among the investment alternatives. Then, the present worth is computed for each alternative. One-Shot Investment An investment that is available only once.

EXAMPLE 4.3 Selecting from Among Multiple One-Shot Investments Video Example Consider the two cash flow diagrams given in Figure 4.5. Both alternatives are one-shot investments. As such, we cannot predict what investment alternatives might be available in the future. However, the minimum attractive rate of return, 15%, reflects the opportunity to reinvest recovered capital. Which alternative is preferred?

FIGURE 4.5 CFDs for Example 4.3 Key Data Note that the one-shot investments in question have different durations. Therefore, we will use a 6-year planning horizon, with zero cash flows occurring in years 5 and 6 for Alternative 1. Given The cash flows outlined in Figure 4.5; MARR = 15%; planning horizon = 6 years Find PW of each alternative Solution Using present worth analysis, the following results are obtained: PW1 (15%)

= −$4,000 + $3,500(P |A 15%,4) + $1,000(P |F 15%,4) = −$4,000 + $3,500(2.85498) + $1,000(0.57175) = $6,564.18 =NPV(15%,3500,3500,3500,4500)−4000 = $6,564.18

PW2 (15%)

= −$5,000 + $1,000(P |A 15%,6) + $1,000(P |G 15%,6) = −$5,000 + $1,000(3.78448) + $1,000(7.93678) = $6,721.26 =NPV(15%,1000,2000,3000,4000,5000,6000)−5000 = $6,721.26

Excel® Video Lesson: The NPV Function Thus, we would recommend Alternative 2.

Concept Check 04.02-CC001 If the PW of an investment is negative, then we will recommend the investment. True or False?

Concept Check 04.02-CC002 When comparing multiple alternatives: a. Choose the alternative with the greatest PW over the planning horizon b. Choose the alternative with the greatest AW over the planning horizon c. Check our assumptions about the MARR, as this could impact our decision d. All of the above

Concept Check 04.02-CC003 In one-shot investments, define the planning horizon equal to the shortest life among the investment alternatives. True or False?

4.3 Benefit-Cost Analysis LEARNING OBJECTIVE Perform an economic analysis of public investments utilizing the benefit-cost analysis for a given interest rate. Video Lesson: Benefit-Cost Analysis Government units fund projects using money taken, usually in the form of taxes, from the public. They then provide goods or services to the public that would be infeasible for individuals to provide on their own. While they are not in business to make a profit, it is important that they make wise investment decisions. Projects should provide benefits for the public’s greater good that exceed the costs of providing those benefits. The most frequently used method in evaluating government (local, state, or federal) projects is benefit-cost analysis; however, engineers, particularly civil engineers, often also use the name “cost-effectiveness analysis.” Four classes cover the spectrum of projects that government pursues: cultural development, protection, economic services, and natural resources. Cultural development is enhanced through education, recreation, and historic and similar institutions or preservations. Protection is achieved through military services, police and fire protection, and the judicial system. Economic services include transportation, power generation, and housing loan programs. Natural resource projects entail wildland management, pollution control, and flood control.

Some projects belong in more than one area. For example, flood control is a form of safety for some, provides transportation and power generation for others, and relates to natural resource benefits. While benefit-cost analysis can be applied to most any economic evaluation, including use by engineers of any discipline, its use is predominant in evaluating government projects. For example, in the transportation sector, heavily populated with civil engineers, the AASHTO Highway Safety Manual (HSM) is the dominant directive. It requires proposed highway projects be analyzed with benefits measured in dollars saved through reduced fatalities and injuries, using federal estimates of the cost per fatality and per injury. These costs of fatalities and injuries can be enormous and are estimated annually by the National Highway Safety Administration. Some government projects are huge, having first costs of tens or hundreds of millions of dollars. They may have long lives, such as 50 years for a bridge or a dam. The concept of “multiple-use” is common, as in wildland management projects, where economic (timber), wildlife preservation (deer, squirrel), and recreation (camping, hiking) are each considered uses of importance. The benefits or enjoyment of some government projects are often completely out of proportion to the financial support available from specific individuals or groups. Also, multiple government agencies often have an interest in a project and will contribute to its planning, design, and implementation. For example, there is often support of state road projects involving federal, state, and municipal/county money and efforts. Finally, some public-sector projects are not easily evaluated due to difficulty in estimating benefits, disbenefits (negative benefits such as inconvenience, home displacement, reduction in ownership), costs, and negative costs (such as salvage values). Also, it may be many years before some benefits are realized.

4.3.1 Benefit-Cost Calculations for a Single Alternative While benefit-cost analysis using a benefit-to-cost ratio is preferred in many government project analyses, more traditional present worth methods can also be used, such as when benefits minus costs are used. Properly done, both should result in exactly the same decisions. The two most common forms of benefit-cost analysis are the benefit-cost ratio (B/C) or, equivalently, a measure of benefits minus costs (B − C). With either of these two forms, both the benefits (B) and costs (C) are typically expressed as present worth or annual worth monetary figures (e.g., $), using an appropriate discount rate for government programs or projects, and an appropriate planning horizon. Benefit-Cost Ratio (B/C) The ratio of the present worth of net public benefits to the present worth of net government costs using i = MARR. Benefits Minus Costs (B − C) The difference of the present worth of net public benefits and present worth of net government costs using i = MARR. To illustrate the mathematics of benefit-cost analysis, if Bjt

=

net public benefits (benefits minus disbenefits, expressed as money) associated with alternative j d

Cjt

=

net government costs (costs minus residuals such as salvage values) associated with alternative j d

i

=

appropriate interest or discount rate, and

n

=

length of the planning horizon,

then the B/C criterion may be expressed mathematically, using the present worth of all net benefits over the present worth of all net costs, as n

∑ Bjt (1 + i) B/Cj (i) =

−t

t=1 n

∑ Cjt (1 + i) t=1

−t

(4.3)

Although Equation 4.3 is a ratio of present worth of benefits to present worth of costs, it could just as well be a ratio of annual worth of net benefits to net costs. The net benefits minus net costs criterion (B − C) is expressed as n

(B − C) (i) = ∑ (Bjt − Cjt )(1 + i)

−t

(4.4)

j

t=0

which is like the present worth method described earlier in this chapter. It can be used directly to compare alternatives and will always be consistent with a proper B/C ratio analysis.

EXAMPLE 4.4 Benefit-Cost Analysis for a Single Alternative Video Example Consider a 10-year land reclamation project that will commit the government to a stream of expenditures appearing in the Cost column of Figure 4.6. Real benefits appear in the Benefit column. Discounting takes place at a rate of 7%. Is this project desirable?

FIGURE 4.6 Costs and Benefits for Land Reclamation Project, i = 7% Excel® Data File Solution The present value of benefits is $1,424,102, and the present value of costs is $1,063,987, so the net present value of benefits minus costs is B − C = $360,115. The project is desirable when considered alone. It is also common to calculate the ratio of benefits to costs, resulting in B/C = $1,424,102/$1,063,987 = 1.33. When the B/C ratio is greater than 1.00, the project is desirable when considered alone.

EXAMPLE 4.5 Benefit-Cost Analysis of a City Traffic Light A growing city of 60,000 has an intersection near the south end of the main business district. The north-south street has no traffic control at that intersection, and the east-west intersecting street has only stop signs. There have been a few accidents in recent years and it has been estimated that a traffic light can provide benefits of $55,000 per year in reduced property loss and disabling injuries, increasing by $2,500 per year. The disbenefits of the traffic light include cost of fuel and time delays, amounting to an estimated $24,000 per year, increasing by $2,000 per year. The installation cost of the traffic light is estimated to be $120,000 with maintenance and electricity running $6,000 in the first year and increasing by $300 per year over a time horizon of 10 years. It is anticipated that the intersection will need to be redesigned at that time and the traffic light removed and sold to a nearby small community for a net salvage value of $4,000. The discount rate is 7%. Is the traffic light justified? Key Data This is a single alternative and it is desired to know if it is cost effective using both a B/C and a B − C approach. The planning horizon is 10 years and the discount rate is 7%. Given The initial and increasing costs are shown, as well as the benefits and disbenefits over the planning horizon. Find The present worth of net benefits B and net costs C, and calculate B/C and B − C. Solution Note that B/C > 1 and B − C > $0. Therefore, the project is worthwhile. Figure 4.7 shows the calculations needed for the benefit-cost analysis. Because B/C > 1.0 and, equivalently, B − C > $0.00, the traffic light is justified.

FIGURE 4.7 Costs and Benefits for Traffic Light Excel® Data File Note that the government costs and salvage values as well as the public benefits and disbenefits in Figure 4.7 have been calculated separately to illustrate several different approaches to the analysis. Approach 1: Recommended Approach! In this book, it is recommended that the present worth of all public benefits and disbenefits be calculated as the present worth of net public benefits (PW benefits − PW disbenefits) in the numerator, and all government costs and salvage values be calculated as net government costs (PW costs − PW salvage value) in the denominator as follows:

B/C = (PW Benefits − PW Disbenefits)/(PW Costs − PW Salvage Value)

Note that a positive salvage value is treated as a negative cost. B/C = ($455,586 − $223,997)/($170,456 − $2,033) = 1.375

Because this is a number greater than 1.000, the project is worthwhile. An equivalent analysis is to calculate B − C. If B − C > $0, the project is worthwhile. B − C = $455,586 − $223,997 − $170,456 + $2,033 = $63,166

Approach 2 (Not recommended, but sometimes used in practice): In this approach, all of the “positives” (for example, public benefits and government salvage value) are put in the numerator and all of the “negatives” (government costs and public disbenefits) are put in the denominator as follows: B/C = ($455,586 + $2,033)/($170,456 + $223,997) = 1.160

Because this is a number greater than 1.000, even though different from the 1.375 above, it is a worthwhile project. An equivalent analysis is to calculate B − C. If B − C > $0, the project is worthwhile. B − C = $455,586 + $2,033 − $170,456 − $223,997 = $63,166

Note that this is simply reordering the present worth of benefits and costs, so naturally B − C will match the B − C calculated in Approach 1. Approach 3 (Not recommended, but sometimes used in practice): In this approach, the pure public benefits appear in the numerator and the government costs, salvage value, and public disbenefits are put in the denominator as follows: B/C = ($455,586)/($170,456 − $2,033 + $223,997) = 1.161

Because this is a number greater than 1.000, even though different from the 1.375 and the 1.160 above, it is a worthwhile project. Note that carrying the B/C ratio to three places after the decimal is highly unnecessary except to show that all three of the B/C ratios above are different. An equivalent analysis is to calculate B − C. If B − C > $0, the project is worthwhile. B − C = $455,586 − $170,456 + $2,033 − $223,997 = $63,166

Again, note that this is simply reordering the present worth of benefits and costs, so naturally B − C will match the B − C values calculated in Approaches 1 and 2. So, what is the point of these three approaches? It is simply to show that one should select and stay with a consistent method of calculating B/C. Especially if multiple alternatives are being considered for government investment, the use of different calculation approaches by different proposers can lead to inconsistent decisions. These three approaches also show that calculating B − C is always correct. Again, Approach 1 shows the recommended approach, keeping public benefits and disbenefits in the numerator and government costs and salvage values in the denominator. Recall that if B/C = 1.0, then the present worth of net benefits is equal to the present worth of net costs. Likewise, if B − C = $0.00, again the present worth of net benefits is equal to the present worth of net costs. To illustrate that each of the three solution methods illustrated above is equivalent, refer again to Example 4.5 and Figure 4.7. Although Approach 1 is recommended, consider increasing the initial cost of $120,000 by $63,166 in each of the three solutions. This will cause the present worth of costs to increase from $170,456 to $170,456 + $63,166 = $233,622. Now, recalculating B/C and B − C for each of the three solutions results in the following:

Solution 1 : B/C =

($455,586 − $223,997)/($233,622 − $2,033) = 1.000 and B − C = $455,586 −$223,997 − $233,622 + $2,033 = $0.00

Solution 2 : B/C =

($455,586 + $2,033)/($233,622 + $223,997) = 1.000 and B − C = $455,586 +$2,033 − $233,622 − $223,997 = $0.00

Solution 3 : B/C =

($455,586)/($233,622 − $2,033 + $223,997) = 1.000 and B − C = $455,586 −$233,622 + $2,033 − $223,997 = $0.00

Note that all three solutions have a B/C of 1.000 and a B − C of $0.00.

4.3.2 Benefit-Cost Calculations for Multiple Alternatives and Unequal Lives When two or more project alternatives are being compared using a B/C ratio, the analysis should be done using the Incremental B/C method. One common, but not necessary, approach is to first order all alternatives from the lowest to the highest present worth of net costs (government costs less salvage value, if any). The first in the list is Alternative 1 and the second is Alternative 2. “Do Nothing” should be considered Alternative 1 if that would be a viable decision; however, such reasons as safety, cost, or a requirement of law may preclude “Do Nothing” from consideration. Then, the present worth of the incremental benefits of Alternative 2 over Alternative 1, ΔB (i), are divided by the present worth of the incremental costs of Alternative 2 over Alternative 1, ΔC (i). That is, 2−1

2−1

(4.5)

n

∑ (B2t − B1t ) (1 + i) ΔB/C2−1 (i) =

ΔB2−1 (i) ΔC2−1 (i)

=

−t

t=1 n

∑ (C2t − C1t ) (1 + i)

−t

t=1

If the incremental benefits exceed the incremental costs, Alternative 2 is the “winner.” Otherwise, Alternative 1 is the “winner.” The “winner” then is compared to Alternative 3 and, again, a “winner” is selected. This continues throughout all available alternatives until a “grand winner” emerges. If instead of an Incremental B/C approach a B − C approach is used, the method is very simple; the present worth of net benefits minus the present worth of net costs is calculated using Equation 4.6 for each alternative and the alternative having the highest B − C value is selected. n

Δ(B − C)

2−1

(i) =

ΔB2−1 (i) − ΔC2−1 (i) = ∑ [B2t (i) − B1t (i)](1 + i)

−t

(4.6)

t=1 n

− ∑ [C2t (i) − C1t (i)](1 + i)

−t

≥ 0

t=1

The incremental basis of Equation 4.6 is not required when using the B − C criterion, if benefits and costs are known for each alternative. In this case, Equation 4.4 could be used directly for each alternative and the maximum value selected.

EXAMPLE 4.6 Selecting Highway Safety Improvements Using B/C and B − C A county within which a city of 1.5 million people reside is considering proposed highway safety improvement projects to initiate in the upcoming year and they also wish to determine a ranking of proposed projects. They have a good feel for initial costs and maintenance and have already consulted many published studies to determine estimates of the benefits of reducing minor, moderate, serious, severe, critical, and fatal accidents. Benefits and costs for each alternative are shown in Figure 4.8. Note the list of proposed projects is in increasing order of present worth of net government costs. a. Determine the overall best project based upon an Incremental B/C approach. b. Determine the overall best project based upon a B − C approach. c. Determine the ranking of these projects based upon a B − C approach.

FIGURE 4.8 Selecting Highway Safety Improvements Using B/C and B − C Excel® Data File Key Data Note that the planning horizon has not been declared, so we will assume that the horizon is the same for each project. Or, if they are different, we will assume the termination of each project has been properly adjusted according to Principle 8, again addressed following Figure 4.8.

Given The present worth of net benefits B and the present worth of net costs C have been calculated in the example. Find Beginning with the two projects with the lowest present worth of net cost C and the next lowest, calculate the incremental benefit-cost ratio, using the difference in net benefits divided by the difference in net costs. If the ratio is greater than one, the larger project number is the “winner.” The “winner” will now be compared in the same way to the project with the next higher C, pairwise, and either continue to be the “winner” or be replaced by a “new winner.” This pairwise incremental comparison will continue until there is a final overall winner. Note that it is simply “custom” to order the projects based on low to high present worth of net costs. They can be taken in any order. Solution Figure 4.8 shows that the overall winner is Project 5 – Chevrons. Another, and much easier, approach is to calculate B − C and order these values from high to low. Note that Project 5 is the “winner” with the decreasing order of preference being 5, 6, 3, 4, 2, 1. When faced with alternatives having unequal useful lives, the bottom line is the use of “Principal #8: Compare investment alternatives over a common period, the planning horizon.” For alternatives that have lives longer than the planning horizon, a “net residual benefit value” must be estimated to account for the benefits beyond the planning horizon. If there is any government salvage value or removal cost at the end of the planning horizon, it must also be calculated. For alternatives that have shorter lives than the planning horizon, replacement or renewal decisions must be made, and both public net benefits and government net costs must be documented throughout the remainder of the planning horizon.

EXAMPLE 4.7 Selecting Among Three Alternatives with Unequal Useful Lives A ranch has grown over the decades to become a major economic contributor to the county where it is located. The main approach from a rural road to the ranch’s wide driveway has been a source of trouble for several years. The approach was designed by the county long ago, and it is the responsibility of the county to maintain the entrance from the county road to the driveway. When the road is graded, the gullies along the road sides are deepened and cleaned so the rainwater will flow nicely alongside the road that runs downhill about a mile. Unfortunately, during heavy rains the water carries away any gravel buildup, creating a large and deep depression that is hard on regular motor vehicles as well as the large trucks that must deliver to, and pick up product from, the ranch. The disbenefit to the ranch, its employees, its suppliers, truckers, and customers costs about $18,000 per year. Three alternatives have been identified to lessen this inconvenience cost-effectively. The alternatives will be evaluated using the county’s discount rate of i = 5% and a planning horizon of 24 years with the assumption that any installation will be removed by the county at the end of the planning horizon. All three alternatives have different useful lives, and adjustments must be made to ensure each covers the full 24-year planning horizon. Alternative 1 is to excavate the gully at the entrance to the driveway and place a 3-foot-diameter, 45-footlong culvert pipe, covered with gravel. The first cost for the pipe, grading, placement, transportation, and supplies is $13,000. Yearly additional maintenance for the county will be $1,000. Benefits to the ranch will be $14,000 per year. The useful life will be 8 years with no salvage value, and there will be a $2,000 removal cost. Alternative 1 will be replaced at the end of years 8 and 16 to cover the 24-year planning horizon. Costs are expected to remain constant. Alternative 2 is to excavate the gully at the entrance to the driveway and install several joints of precast concrete box culverts in standard strengths and sizes. The first cost for this is $28,000 with only $400 additional yearly maintenance and a useful life of 16 years. Removal cost is $2,800. Benefits to the ranch will be $16,000 per year. For the remaining 8 years of the planning horizon, Alternative 1 will be used and its benefits and costs will apply. Alternative 3 is to divert the water flow upstream of the driveway, into the ranch property, to the first of a series of ponds with large holding capacity. This would be done underground with large corrugated nonmetallic piping. Benefits to the ranch will be $16,500. Mainly due to the deep excavation, distance, leveling, connecting, and installation care necessary, this would have a first cost of $60,000 and $0 additional maintenance per year and a useful life of 24 years with a $20,000 cost of removal. a. Determine the best alternative using the B − C approach. b. Determine the best alternative using the Incremental B/C approach. Key Data Note that these three alternatives have unequal lives of 8, 16, and 24 years. Because it makes sense to take a long-term view, a planning horizon of 24 years is selected, with Alternative 1 plus Alternative 2 making up the 24 years and Alternative 2 repeated three times to make up the 24 years. Alternative 3, the most permanent of the three, already meets the 24- year planning horizon. Given Net benefits by year have been estimated. Also, the initial cost, maintenance cost, and incremental increase in maintenance cost are shown. There may also be a cost of removal. The biggest challenge is to realize each of these benefits and costs at the right time. Find The best alternative using either B − C or the incremental benefit cost ratio in a pairwise manner. Solution As shown in Figure 9, Both approaches prefer Alternative 2. B − C is highest for Alternative 2. Also, ΔB/C prefers Alternative 2 and ΔB/C prefers Alternative 2. Both the B − C and incremental B/C approaches reach the same answer. 2−1

3−2

FIGURE 4.9 Selecting Driveway Approach from 3 Alternatives with Unequal Lives Excel® Data File

EXAMPLE 4.8 Benefit-Cost Analysis of Three Routes Video Example The state of Washington must decide among three highway alternatives to replace an old winding road. The length of the current route is 26 miles. Planners agree that the old road must be replaced or overhauled; they cannot keep it as it is. Alternative A is to overhaul and resurface the old road at a cost of $2 million/mile. Resurfacing will also cost $2 million/mile at the end of each 10-year period. Annual maintenance will cost $10,000/mile. Alternative B is to cut a new road following the terrain. It will be only 22 miles long. Its first cost will be $3 million/mile with resurfacing at 10-year intervals costing $2 million/mile with annual maintenance at $12,000/mile. Alternative C also involves a new highway to be built along a 20.5-mile straight line. Its first cost will be $4 million/mile with resurfacing at 10-year intervals costing $2 million/mile and with annual maintenance costing $20,000/mile. This increase over Routes A and B is due to additional roadside bank retention efforts. Traffic density along each of the three routes will fluctuate widely from day to day but will average 4,000 vehicles/day throughout the 365-day year. This volume is composed of 350 light trucks, 250 heavy trucks, and 80 motorcycles, and the remaining 3,320 are automobiles. The average operation cost for each of these vehicles is $0.70, $1.10, $0.30, and $0.60 per mile, respectively. There will be a time savings because of the different distances along each of the routes, as well as different speeds that each of the routes will sustain. Route A will allow heavy trucks to average 35 miles/hour, while the other vehicles can maintain 45 miles/hour. For each of Routes B and C, these numbers are 40 miles/hour for heavy trucks and 50 miles/hour for other vehicles. The cost of time for all commercial traffic is valued at $25/hour and for noncommercial vehicles, $10/hour. Twenty-five percent of the automobiles and all of the trucks are considered commercial. Finally, there is a significant safety factor to be included. An excessive number of accidents per year have occurred along the old winding road. Route A will reduce the number of vehicles involved in accidents to 105, and Routes B and C are expected to involve only 75 and 50 vehicles in accidents, respectively. The average cost per vehicle in an accident is estimated to be $18,000, considering actual property damages, lost time and wages, medical expenses, and other relevant costs. The Washington planners want to compare the three alternative routes using benefit-cost criteria, specifically the popular benefit-cost ratio. Solution The benefit-cost analysis includes the government first costs, resurfacing costs, and maintenance costs plus the public operational, time, and accident costs that follow. Annual public operational costs are: Route A: [(350)(0.70) + (250)(1.10) + (80)(0.30) + (3,320)(0.60)](26)(365) = $24,066,640/year Route B: [(350)(0.70) + (250)(1.10) + (80)(0.30) + (3320)(0.60)](22)(365) = $20,364,080/yr Route C: [(350)(0.70) + (250)(1.10) + (80)(0.30) + (3320)(0.60)](20.5)(365) = $18,975,620/yr

Annual public time costs:

Route A:

[(350)(1/45)(25) + (250)(1/35)(25) + (80)(1/45)(10) + (3,320)(0.25)(1/45)(25) +(3,320)(0.75)(1/45)(10)](26)(365) = $13,335,710/year

Route B:

[(350)(1/50)(25) + (250)(1/40)(25) + (80)(1/50)(10) + (3,320)(0.25)(1/50)(25) +(3,320)(0.75)(1/50)(10)](22)(365) = $10,119,808/year

Route C:

[(350)(1/50)(25) + (250)(1/40)(25) + (80)(1/50)(10) + (3,320)(0.25)(1/50)(25) +(3,320)(0.75)(1/50)(10)](20.5)(365) = $9,429,821/year

Annual accident costs per vehicle: Route A: (105)(18,000) = $1,890,000/year Route B: (75)(18,000) = $1,350,000/year Route C: (50)(18,000) = $900,000/year

All relevant government and public costs are summarized in Figure 4.10.

FIGURE 4.10 Summary of Data for Three Routes of Example 4.8 Excel® Data File Because planners have not yet defined the “benefits” per se, public benefits are taken as the incremental reduction in user costs between each pair of alternatives evaluated. Then, the incremental benefits are compared against the respective incremental costs needed to achieve the incremental benefits. The incremental benefits and costs for Route B as compared to Route A using a discount rate of 8% are given as follows:

ΔBB−A (8%)   =

Public costsA (8%) − Public costsB (8%) $39,292,350 − $31,833,888 = $7,458,462

ΔCB−A (8%) =

Government costsB (8%) − Government costsA (8%) $8,775,501 − $8,009,533 = $765,968

That is, for an incremental expenditure of $765,968/year, the government can provide added benefits of $7,458,462/year for the public. The appropriate benefit-cost ratio is then ΔBB−A (8%) ΔB/CB−A (8%) =

$7,458,462 =

ΔCB−A (8%)

= 9.74 $765,968

This clearly indicates that the additional funds for Route B are worthwhile, and Route B is desired over Route A. Using a similar analysis, the benefits, costs, and ΔB/C ratio may now be calculated to determine whether or not Route C is preferable to Route B: ΔBC−B (8%) =

Public costsB (8%) − Public costsC (8%) $31,833,888 − $29,305,441 = $2,528,447

ΔCC−B (8%) =

Government costsC (8%) − Government costsB (8%) $10,162,134 − $8,775,501 = $1,386,633

ΔBC−B (8%) ΔBC−B (8%) =

$2,528,447 =

ΔCB−A (8%)

= 1.82 $1,386,633

This benefit-cost ratio, being greater than 1.00, shows that the additional expenditure of $1,386,633/year to build and maintain Route C would provide commensurate benefits in public savings of $2,528,447/year. Of the three alternatives, Route C is preferred. Figure 4.11 illustrates the benefit-cost calculations using Excel®.

FIGURE 4.11 Benefit-Cost Calculations for Example 4.8 Excel® Data File

Concept Check 04.03-CC001 Identify the one true statement. a. Benefit-cost analysis is recommended for a formal economic analysis of private programs or projects b. A common benefit-cost analysis method is to compare the difference in benefits and costs, i.e., (B − C) c. The B/C ratio must use the present worth of all net benefits over the present worth of all net costs d. All of the above statements are true

Concept Check 04.03-CC002 When performing a benefit-cost analysis for multiple alternatives, calculate each B/C ratio, rank them, then pick the highest one. True or False?

4.4 Discounted Payback Period

LEARNING OBJECTIVE Calculate the discounted payback period (DPBP) for a given interest rate to determine how long it takes for the cumulative present worth (PW) to be positive. Video Lesson: Payback Period and Discounted Payback Period The time required for an investment to be fully recovered, including the time value of money, is the length of time required for the cumulative present worth to become positive using i = MARR; it is called the discounted payback period (DPBP) for an investment. Note that the DPBP may be a fractional or whole-year amount. Discounted Payback Period (DPBP) The length of time required for the cumulative present worth to become positive using i = MARR. In this section, we determine the DPBP for single and multiple investment alternatives.

4.4.1 Discounted Payback Period for a Single Alternative We begin with a relatively simple example of a DPBP analysis.

EXAMPLE 4.9 Computing the DPBP of a Single Investment Opportunity Video Example An initial $400,000 investment in new production equipment will yield annual positive cash flows of $150,000 in year 1. Annual cash flows will decrease by $25,000 each year thereafter. The new equipment has a useful life of 7 years. MARR is 10% per year. Determine the DPBP of this project. Key Data Given Net cash flows as shown in Table 4.1; MARR = 10%, planning horizon = 7 years Find DPBP

TABLE 4.1 Net Cash Flows for Example 4.9 EOY

Net Cash Flow

0 1

−$400,000 150,000

2 3

125,000 100,000

4

75,000

5 6

50,000 25,000

7

0

Solution To determine the DPBP for this investment, we must calculate the time required for the cumulative present worth to equal 0. As shown in Table 4.2, the cumulative PW reaches 0 and the investment is fully recovered between years 5 and 6. Assuming cash flows occur uniformly throughout the year, requires interpolation to calculate the DPBP of 5 + (2,927.02/14,111.85) = 5.21 years. If we assume an end-of-year cash flow convention, the DBPB is 6 years. TABLE 4.2 Present Worth and Cumulative Present Worth of Example 4.9 Cash Flows EOY Net CF (P/F 10%, n) PW of CF Cumulative PW of CF 0 1

−$400,000 150,000

1.00000 0.90909

−400,000 136,363.64

−400,000 −263,636.36

2

125,000

0.82645

103,305.79

−160,330.58

3 4

100,000 75,000

0.75131 0.68301

75,131.48 51,226.01

−85,199.10 −33,973.09

5 6

50,000 25,000

0.62092 0.56447

31,046.07 14,111.85

−2,927.02 11,184.83

7

0

0.51316

0

Excel® Data File To illustrate discounted payback period for a single alternative using Excel®, we revisit Example 4.1.

EXAMPLE 4.10 Using the Excel® NPER Tool to Determine DPBP of a Single Investment Opportunity In Example 4.1, suppose management asked the manufacturing engineer to determine how long it takes for the new SMP machine to recover fully its initial cost of $500,000, including the time value of money. Excel® Video Lesson: NPER Financial Function Key Data Given The cash flows shown in Figure 4.1; MARR = 10 %; planning horizon = 10 years Find DPBP using the Excel® NPER function Solution The Excel® NPER worksheet function can be used to determine how long it takes for the $500,000 investment to be recovered, based on an annual return of $92,500. What is not known, however, is what the SMP machine’s salvage value will be if its useful life is less than the 10-year planning horizon. To provide management with more information regarding the recovery period, the manufacturing engineer did the following analysis. First, the engineer used the Excel® NPER worksheet function, assuming a negligible salvage value: yrs to recover the investment =NPER(10%,92500,−500000) = 8.16 years

Next, the engineer estimated the salvage value if the machine was sold at any time before the end of the planning horizon. Not knowing exactly how the salvage value would decrease with use, the engineer made two approximations: The salvage value decreases as a geometric series, and the salvage value decreases as a gradient series. The engineer computed the geometric rate required to yield a $50,000 salvage value after 10 years and obtained a value of 20.6%; then, the engineer computed the gradient step required and obtained a value of $45,000. (As an exercise, you should verify the engineer’s calculations.) Using the salvage values shown in Figure 4.12, the engineer calculated the present worth, assuming the SMP machine was kept for 1, 2, …, 10 years.

FIGURE 4.12 Plot of Cumulative Present Worths for Example 4.10 Excel® Data File Based on the salvage value decreasing at a geometric rate of 20.6% each year, as shown in Figure 4.12, the investment in the new SMP machine is fully recovered during the seventh year. Based on the salvage value decreasing at a gradient rate of $45,000 per year, the investment is fully recovered during the third year. Management was much more comfortable making the $500,000 investment after learning that the present worth over the planning horizon was $87,649.62 and that, if things went badly, the investment would be fully recovered in no more than 7 years and, possibly, as quickly as 3 years, depending on the salvage value for the SMP machine. DPBP as a Supplemental Tool We do not recommend using DPBP to identify the investment that is to be made from among a set of mutually exclusive investment alternatives. Instead, we recommend it be used as a supplemental tool, just as it was used in the SMP example. If DPBP is used as a stand-alone measure of economic worth, it is difficult to understand how one would decide if the DPBP value obtained was acceptable or not. Such decisions would have to be arbitrary, because we know of no rational basis for saying, in the case of the previous example, that a DPBP value of 5 is acceptable but a value of 6 is not. DPBP and Excel® As illustrated in the previous example, if the salvage value for an investment of $P in an asset is negligible, regardless of how long the asset is used, then the Excel® NPER worksheet function can be used to determine the DPBP, when the annual returns are a uniform annual series of $A, DPBP =NPER(MARR,A,−P)

If, on the other hand, salvage value decreases with usage according to some mathematical relationship, then the Excel® SOLVER and GOAL SEEK tools can be used to determine the DPBP. In the previous example, salvage value was assumed to decline either geometrically or “gradiently.” The following example illustrates the calculations involved in determining the DPBP for the SMP investment. Excel® Video Lesson: GOAL SEEK Tool

EXAMPLE 4.11 Using the Excel® SOLVER Tool to Calculate the DPBP Use the Excel® SOLVER tool to determine the DPBP for the SMP investment. Excel® Video Lesson: SOLVER Tool Key Data Figure 4.13 shows a spreadsheet with the relevant input parameters.

FIGURE 4.13 Excel® SOLVER Set Up to Calculate the DPBP in Example 4.11 with Gradient Decrease in Salvage Value Excel® Data File Solution When salvage value decreases at an annual rate of 20.6%, the formula for salvage value after n years of use is $500,000(0.794)n. Likewise, when it decreases by $45,000 each year, the salvage value after n years is $500,000 − $45,000n. To determine the DPBP, the value of n that makes PW equal zero can be obtained using SOLVER, twice: once for the geometric decrease in salvage value and again for the gradient or linear decrease in salvage value. After entering a “trial value” in cell B2 for the DPBP when salvage value decreases as a geometric series, SOLVER is used to determine the value of B2 that makes present worth (cell B6) equal 0. As shown in Figure 4.14, SOLVER yields a value of 6.95 years when salvage value decreases geometrically over time.

FIGURE 4.14 Excel® SOLVER Solutions for the DPBP in Example 4.11 with Geometric and Gradient Decreases in Salvage Value Excel® Data File When salvage value decreases as a gradient series, a “trial value” for DPBP is entered in cell B8 and SOLVER is used to determine the value of cell B8 that makes present worth (cell B12) equal 0. As shown in Figure 4.14, SOLVER yielded a value of 2.17 years when salvage value decreases by $45,000 per year. DPBP and Salvage Value When using DPBP, salvage values should not be ignored. However, determining salvage values for periods of use ranging from 1 year to n years tends to be a very inexact process. The need to know salvage values for all possible periods of use is a limitation of the DPBP method. PBP A special case of the discounted payback period is the payback period (PBP). As the difference in the names imply, PBP does not incorporate the TVOM in its calculations. Hence, PBP ≤ DPBP. For an investment in which PW < $0, DPBP will not exist; however, because of the TVOM, PBP can exist when DPBP does not exist. Further, as with DPBP, PBP tends to ignore salvage values if the investment is terminated before the end of the planning horizon. In essence, PBP users assume a value of zero for the MARR and the salvage value. Because PBP violates the first rule laid down in the text, it should come as no surprise that we have little good to say about it. On the other hand, it can be used as a “tie breaker” when two or more alternatives have essentially the same PW. Yet, PBP is very, very popular. Why? The following reasons are given by its advocates: It is simple mathematically—does not require interest rate calculations. It does not require a decision concerning the MARR. It is much more easily explained and understood than are DCF methods. It reflects a manager’s attitudes when capital is limited. It provides a hedge against uncertainty of future cash flows. It provides a rough measure of the liquidity of an investment.

4.4.2 Discounted Payback Period for Multiple Alternatives We have emphasized the use of DPBP as a supplemental tool when comparing investment alternatives. We do not recommend it as the sole basis for choosing the preferred alternative. To understand why, consider the following example.

EXAMPLE 4.12 Using DPBP When Choosing Among Three Alternatives Recall Example 4.2 involving a theme park and two design alternatives. Now we add a third design alternative, which requires an initial investment of $150,000 and will produce annual after-tax revenue represented by a decreasing gradient series, with a revenue of $45,000 the first year and decreasing by $5,000 per year to a final value of 0 in the last year. Which alternative has the smallest DPBP? Key Data Given The cash flows shown in Figure 4.15; MARR = 10% Find DPBP for each alternative

FIGURE 4.15 CFDs for the Alternatives in Example 4.12 Solution The present worth values for the three alternatives (A, B, and C), for DPBP years are given by PWA = −$300,000 + $55,000(P |A 10%,DPBPA ) PWB = −$450,000 + $80,000(P |A 10%,DPBPB ) PWC = −$150,000 + $45,000(P |A 10%,DPBPC )−$5,000(P |G 10%,DPBPC )

Excel® Solution The Excel® NPER function can be used to solve for the values of DPBPA and DPBPB that equates the present worth to 0: DPBPA  =NPER(10%,−55000,300000) = 8.273 years DPBPB  =NPER(10%,−80000,450000) = 8.674 years

The Excel® SOLVER tool is used to obtain the value of DPBPC, as shown in Figure 4.16. DPBPC = 6.273 years

Thus, alternative C has the smallest DPBP.

FIGURE 4.16 Excel® SOLVER Solution for DPBPC in Example 4.12 Excel® Data File Exploring the Solution Based on the discounted payback period, the rank-ordered preference for the alternatives is C, B, A—exactly the reverse order of their present worth values. The present worth for C can be shown to be PWC

=

−$150,000 + $45,000(P |A

10%,10)−$5,000(P |G

=

−$150,000 + $45,000(6.14457)−$5,000(22.89134)

=

$12,048.95

10%,10)

or, using Excel®, PWc =1000*NPV(10%,45,40,35,30,25,20,15,10,5,0)−150000 = $12,048.81 whereas, using Excel®, PWA = $37,951.19 and PWB = $41,565.37. Figure 4.17 illustrates how the three present worths change over the duration of the planning horizon.

FIGURE 4.17 Discounted Payback Period Analysis for Example 4.12 Excel® Data File As Example 4.12 illustrates, the DPBP can lead to wrong conclusions regarding the investment alternative that maximizes economic worth.

Concept Check 04.04-CC001 The DPBP is ideal as a stand-alone measure of economic worth. True or False?

Concept Check 04.04-CC002 When comparing multiple alternatives, the DPBP can lead to wrong conclusions regarding the maximum economic worth. True or False?

4.5 Capitalized Worth LEARNING OBJECTIVE Calculate the capitalized worth (CW) of an investment for a given interest rate when the planning horizon is infinitely long. Video Lesson: Capitalized Worth A special type of cash flow series is a perpetuity, a uniform series that continues indefinitely. This is a special case, because an infinite series of cash flows would rarely be encountered in the business world. However, for such very long-term investment projects as bridges, tunnels, railways, highways, hydroelectric dams, nuclear power plants, forest harvesting, or the establishment of endowment funds where the estimated life is 50 years or more, assuming that an infinite cash flow series exists will generally be a good approximation. The present worth equivalent of an infinitely long series of cash flows is called the capitalized worth (CW). Because, in most applications, the capitalized worth is being computed for investments that have few if any positive returns, the capitalized worth is more generally referred to as the capitalized cost (CC) of an undertaking.

Capitalized Worth (CW) The value of all cash flows converted to a single sum equivalent at time zero using i = MARR when the planning horizon is infinitely long. The capitalized worth method is applicable only if there is reason to believe a series of cash flows will extend indefinitely into the future. Because it does not use the same planning horizon as PW, FW, and AW, there is no reason to assume that the results will be the same as those that would occur when using a finite planning horizon. If one wants to compute the present worth of an infinitely long uniform series of cash flows, which we call capitalized worth, then A CW = A(P |A i%,∞) =

(4.7)

i

Therefore, if the project in question involves an indefinite repetition of a life cycle, then one can convert the life cycle costs to an annual equivalent and divide the result by the MARR to obtain the CW.

4.5.1 Capitalized Worth for a Single Alternative As with the other measures of economic worth, capitalized worth can be used in the absence of alternatives. Because capitalized worth calculations are generally performed when the do-nothing alternative is not feasible, however, the requirement that CW > 0 in deciding whether or not to proceed with the investment is not appropriate. If only costs occur, then our objective is to minimize capitalized cost.

EXAMPLE 4.13 Capitalized Cost of Building Maintenance Every 10 years, the dome of the state capitol building has to be cleaned, sandblasted, and retouched. It costs $750,000 to complete the work. Using a 5% MARR, what is the capitalized cost for refurbishing the capitol dome? Key Data Given Cash flow profile in Figure 4.18 Find Capitalized Cost (CC)

FIGURE 4.18 CFD for Example 4.13 Solution CC

=

$750,000(A|P

=

$750,000(0.12950)/0.05

5%,10)/0.05

=

$1,942,500

Alternatively, CC

=

$750,000 + $750,000(A|F

=

$750,000 + $750,000(0.07950)/0.05

5%,10)/0.05

=

$1,942,500

The alternative approach yielded the same result because of the following relationship between the A|P and A|F factors: (A|P i%,n) = (A|F i%,n) + i. Exploring the Solution A useful interpretation of the CC is as follows: If $1,942,500 is deposited in a fund that pays 5% annual compound interest, then $750,000 can be paid out every 10 years, forever, to cover the cost of cleaning the capitol dome.

EXAMPLE 4.14 Capitalized Cost for Highway Construction A new highway is to be constructed, and asphalt paving will be used. The asphalt will cost $150 per foot, including the material and the paving operation. The asphalt is expected to last 5 years before requiring resurfacing. It is anticipated that the cost of resurfacing will remain the same per foot. Concrete drainage ditches will be installed on each side of the highway; they will each cost $7.75 per foot to install. Ditches will have to be replaced every 15 years; the cost of replacing them will also be $7.75 per foot. Four pipe culverts will be installed every mile; each culvert will cost $8,000 and will last 10 years; replacement culverts will cost $10,000 each, indefinitely. Annual maintenance of the highway will cost $9,000 per mile. Cleaning each culvert will cost $1,250 per year. Cleaning and maintaining each ditch will cost $3.75 per foot per year. Determine the capitalized cost (CC) per mile of highway using a MARR of 5%. Key Data Given The relevant data is organized in Table 4.3

TABLE 4.3 Data For Highway Construction Costs Initial Cost $150/ft

Paving Maintenance Ditches (install) (cleaning/maintenance)

Duration Replacement Cost 5 yrs $150/ft

$9,000/mi

1 yr

$7.75/ft $3.75/ft

15 yrs 1 yr

$7.75/ft

10 yrs

4 × $10,000/mi

1 yr

4 × $1,250/mi

Culverts (install + replace) 4 × $8,000/mi (cleaning)

4 × $1,250/mi Excel® Data File

Find   CC(highway/mile) = CC(paving/mile) + CC(maintenance/mile) + CC(ditches/mile) + CC

Solution First, let’s compute the capitalized cost of paving the road and the ditches. Because paving must occur every 5 years, there are two approaches that can be used: (1) convert the initial paving of the highway and ditches to an annual equivalent and then divide by the MARR to obtain the capitalized cost, or (2) treat the initial paving as a present cost and convert the resurfacing and replacing costs to an annual equivalent and then divide by the MARR to obtain the capitalized cost of resurfacing. The following results: 1.

CC(paving/mile)

= 5,280 ft/mi [$150/ft(A|P

5%,5) + $7.75/ft(A|P

5%,15)]/(0.05)

= $3,737,409 =PMT((5%,5,−5280*150)+PMT(5%,15,−5280*7.75))/0.05 = $3,737,487

2.

CC (paving/mile)

= 5,280 ft/mi{$150/ft[1 + (A|F

5%,5)] + $7.75/ft[1 + (A|F

5%,15)]}/(0.05)

= $3,737,409 =5280*(150+PMT (5%,5,,−150)/0.05+7.75+PMT (5%,15,,−7.75)/ = $3,737,487

Excel® Video Lesson: The PMT Function To complete the example, we add the capitalized cost for maintaining the highway and ditches, plus the installation and maintenance of the culverts: CC (maintenance/mile) CC(ditches/mile) CC(culverts/mile)

= $9,000/0.05 = $180,000 = [(2)(5,280 ft/mi)($3.75/ft)]/(0.05) = $792,000 = (4)[$8,000 + $1,250/0.05 + $10,000(A|F

5%,10)/0.05]

= $195,600 =4*(8000+1250/0.05+PMT(5%,10,,−10000)/0.05) = $195,604

Therefore, CC (highway/mile)

=

$3,737,487 + $180,000 + $792,000 + $195,604

=

$4,905,091

The capitalized cost for paving the highway and ditches is $3,737,487. The capitalized cost to maintain the highway and ditches is $972,000. The capitalized cost to install culverts every 10 years and maintain them annually is obtained by using the alternative approach in Example 4.13, because the replacement costs differ from the initial installation cost; the result is a capitalized cost of $195,604. Therefore, the overall capitalized cost to build and maintain the highway is $4,905,091 per highway mile. (Based on state highway estimates, the initial cost to construct a 4-lane divided highway comparable to interstates is currently approximately $25 million per highway mile. The initial cost to construct a 2-lane highway, as above, is approximately $1 million per highway mile, depending on terrain and a host of other factors. Hence, the cost estimates provided above are somewhat conservative.)

4.5.2 Capitalized Worth for Multiple Alternatives When multiple alternatives exist, the alternative having the greatest CW over the infinitely long planning horizon is recommended. Again, because capitalized worth alternatives usually involve costs, not revenues, and costs are designated with positive signs, the alternative with the smallest CC is recommended.

EXAMPLE 4.15 Capitalized Cost for Water Delivery Video Example In a developing country, two alternatives are under consideration for delivering water from a mountainous area to an arid area in the country’s southern region. A coated heavy-gauge plastic pipeline can be installed, with pumps spaced appropriately along the pipeline. Alternatively, a canal can be built; however, it will have greater water loss than the pipeline, due to evaporation and poaching along the canal route. To compensate for the water loss, the canal will have a greater carrying capacity than the pipeline. It is estimated it will cost $125 million to install the pipeline. Major replacements are planned every 15 years at a cost of $10 million. Pumping and other annual operating and maintenance costs are estimated to be $5 million. The canal will cost $200 million to construct; its annual operating and maintenance costs are anticipated to be $1 million. Major upgrades of the canal are anticipated every 10 years, at a cost of $5 million. Based on a five percent MARR and an infinitely long planning horizon, which alternative has the lowest capitalized cost? Key Data Given Initial Cost Annual O&M Major Pipeline $125 million $5 million $10 million every 15 years Canal

$200 million

$1 million

$5 million every 10 years

Find Capitalized Cost (CC) of each alternative Solution Pipeline CC

= $125,000,000 + [$10,000,000(A|F

5%,15) + $5,000,000]/0.05

= $234,268,000.00 =125000000+(PMT(5%,15,,−10000000)+5000000)/0.05 = $234,268,457.52

Canal CC

= $200,000,000 + [$5,000,000(A|F

5%,10) + $1,000,000]/0.05

= $227,950,000.00 =200000000+(PMT1(5%,10,,−5000000)+1000000)/0.05 = $227,950,457.50

The canal has the smaller capitalized cost and would be recommended.

Concept Check 04.05-CC001 The capitalized worth method a. Is applicable only if there is reason to believe a series of cash flows will extend indefinitely into the future b. Can be used in the absence of alternatives c. Is used for investments that have few, if any, positive returns d. All of the above

Concept Check 04.05-CC002 Capitalized worth alternatives generally involve revenues, not costs, and the alternative with the largest value is recommended. True or False?

Notes 1. Rather than make the statement “absent nonmonetary criteria” every time to indicate which investment is recommended, henceforth it will be assumed that the single criterion is monetary and the objective is to maximize economic worth. 2. Income taxes and after-tax analysis are presented in Chapter 9. 3. As discussed in a later section, CW > 0 is not generally a viable test, because the do-nothing alternative is not feasible for most of its applications.

CHAPTER 5 Annual Worth and Future Worth

Chapter 5 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced e-text and WileyPLUS

FE-Like Problems 05-FE001 Consider a palletizer at a bottling plant that has a first cost of $150,000, operating and maintenance costs of $17,500 per year, and an estimated net salvage value of $25,000 at the end of 30 years. Assume an interest rate of 8%. What is the annual equivalent cost of the investment if the planning horizon is 30 years? a. $29,760 b. $30,600 c. $31,980 d. $35,130 05-FE002 When using annual worth to evaluate the attractiveness of a single alternative, what value is the calculated AW compared to? a. PW b. FW c. 0.0 d. MARR Correct or Incorrect? Clear

  Check Answer

05-FE003 The annual worth of an alternative is 0. Which of the following is (are) also true? 1. PW = 0

2. PW = 0 a. (1) only b. (2) only c. Both (1) and (2) d. Neither (1) nor (2) 05-FE004 The overhead costs in a highly automated factory are expected to increase at an annual compound rate of 10% for the next 7 years. The overhead cost at the end of the first year is $200,000. What is the annual worth of the overhead costs for the 7-year period? The time value of money rate is 8%/year. a. $263,250 b. $231,520 c. $200,000 d. $187,020 Correct or Incorrect? Clear

  Check Answer

05-FE005 The operating and maintenance expenses for a mining machine are expected to be $11,000 in the first year and increase by $800 per year during the 15-year life of the machine. What uniform series of payments would cover these expenses over the life of the machine? Interest is 10%/year compounded annually. a. $11,000 b. $4,223 c. $13,423 d. $15,223 05-FE006 Scott wants to accumulate $2,500 over a period of 7 years so that a cash payment can be made for roof maintenance on his summer cottage. To have this amount when it is needed, he will make annual deposits at the

end of each year into a savings account that earns 8% annual interest per year. How much must each annual deposit be? a. $244 b. $259 c. $280 d. $357 Correct or Incorrect? Clear

  Check Answer

05-FE007 On the day your daughter is born, you deposit $1,000 in a college savings account that earns 8% compounded annually. On each of her birthdays thereafter, up to and including her 18th birthday, you deposit an additional $1,000. How much money is in the college account the day after her 18th birthday? a. $37,450 b. $38,950 c. $41,450 d. $46,800 05-FE008 A deposit of $800 is planned for the end of each year into an account paying 8%/year compounded annually. The deposits were not made for the 10th and 11th years. All other deposits were made as planned. What amount will be in the account after the deposit at the end of year 25? a. $55,397 b. $55,357 c. $59,537 d. $53,597 Correct or Incorrect? Clear

  Check Answer

05-FE009 If you invest $3,000 3 years from now, how much will be in the account 15 years from now if i = 8% compounded annually? a. $3,500 b. $7,555 c. $9,415 d. $9,516 05-FE010 Consider a palletizer at a bottling plant that has a first cost of $150,000, has operating and maintenance costs of $17,500 per year, and an estimated net salvage value of $25,000 at the end of 30 years. Assume an interest rate of 8%/year. What is the future equivalent cost of the investment if the planning horizon is 30 years? a. $3,371,000 b. $3,467,000 c. $3,623,000 d. $3,980,000 Correct or Incorrect? Clear

  Check Answer

Problems Note to Instructors and Students Many of the problems in this chapter are similar to problems in previous chapters. This similarity is intentional. It is designed to illustrate the use of different measures of merit on the same problem. Section 5.1 Annual Worth LEARNING OBJECTIVE 5.1 Calculate an annual worth (AW) converting all cash flows to an equivalent uniform annual series of cash flows over the planning horizon for a given interest rate.

Section 5.1.1 Annual Worth of a Single Alternative 05.01-PR001 Carlisle Company has been cited and must invest in equipment to reduce stack emissions or face EPA fines of $18,500 per year. An emission reduction filter will cost $75,000 and have an expected life of 5 years. Carlisle’s MARR is 10%/year. a. What is the annual worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on annual worth? c. Is the filter economically justified? 05.01-PR002 Video Solution DuraTech Manufacturing is evaluating a process improvement project. The estimated receipts and disbursements associated with the project are shown below. MARR is 6%/year. End of Year Receipts Disbursements 0

$0

$5,000

1

$0

$200

2

$2,000

$300

3

$4,000

$600

4 5

$3,000 $1,600

$1,000 $1,500

a. What is the annual worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on annual worth? c. Should DuraTech implement the proposed process improvement?

05.01-PR003 Video Solution Eddie’s Precision Machine Shop is insured for $700,000. The present yearly insurance premium is $1.00 per $100 of coverage. A sprinkler system with an estimated life of 20 years and no salvage value can be installed for $20,000. Annual maintenance costs for the sprinkler system are $400. If the sprinkler system is installed, the system must be included in the shop’s value for insurance purposes, but the insurance premium will reduce to $0.40 per $100 of coverage. Eddie uses a MARR of 15%/year. a. What is the annual worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on annual worth? c. Is the sprinkler system economically justified?

05.01-PR004 Fabco, Inc., is considering the purchase of flow valves that will reduce annual operating costs by $10,000 per year for the next 12 years. Fabco’s MARR is 7%/year. Using an annual worth approach, determine the maximum amount Fabco should be willing to pay for the valves. 05.01-PR005 Galvanized Products is considering the purchase of a new computer system for its enterprise data management system. The vendor has quoted a purchase price of $100,000. Galvanized Products is planning to borrow one-fourth of the purchase price from a bank at 15% compounded annually. The loan is to be repaid using equal annual payments over a 3-year period. The computer system is expected to last 5 years and has a salvage value of $5,000 at that time. Over the 5-year period, Galvanized Products expects to pay a technician $25,000 per year to maintain the system but will save $55,000 per year through increased efficiencies. Galvanized Products uses a MARR of 18%/year to evaluate investments. a. What is the annual worth of this investment?

b. What is the decision rule for judging the attractiveness of investments based on annual worth? c. Should the new computer system be purchased? 05.01-PR006 Quilts R Us (QRU) is considering an investment in a new patterning attachment with the cash flow profile shown in the table below. QRU’s MARR is 13.5%/year. EOY Cash Flow EOY Cash Flow 0 −$1,400 8 $600 1 2 3

$0 $500 $500

9 10 11

$700 $800 $900

4 5

$500 $500

12 13

−$1,000 −$2,000

6 7

$0 $500

14 15

−$3,000 $1,400

a. What is the annual worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on annual worth? c. Should QRU invest? 05.01-PR007 Imagineering, Inc., is considering an investment in CADCAM-compatible design software with the cash flow profile shown in the table below. Imagineering’s MARR is 18%/year. EOY Cash Flow (M$) EOY Cash Flow (M$) 0 1 2

−$12 −$1 $5

4 5 6

$5 $5 $2

3

$2

7

$5

a. What is the annual worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on annual worth? c. Should Imagineering invest? 05.01-PR008 Jupiter is considering investing time and administrative expense on an effort that promises one large payoff in the future, followed by additional expenses over a 10-year horizon. The cash flow profile is shown in the table below. Jupiter’s MARR is 12%/year. EOY Cash Flow (K$) EOY Cash Flow (K$) 0 −$2 6 $200 1 2 3

−$10 −$12 −$14

7 8 9

−$10 −$12 −$14

4 5

−$16 −$18

10

−$100

a. What is the annual worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on annual worth? c. Should Jupiter invest? 05.01-PR009 Video Solution Aerotron Electronics is considering the purchase of a water filtration system to assist in circuit board manufacturing. The system costs $40,000. It has an expected life of 7 years at which time its salvage value will be $7,500. Operating and maintenance expenses are estimated to be $2,000 per year. If the filtration system is not purchased, Aerotron Electronics will have to pay Bay City $12,000 per year for water purification. If the system is purchased, no water purification from Bay City will be needed. Aerotron Electronics must borrow half of the purchase price, but it cannot start repaying the loan for 2 years. The bank has agreed to three equal annual payments, with the first payment due at the end of year 2. The loan interest rate is 8% compounded annually. Aerotron Electronics’ MARR is 10% compounded annually.

a. What is the annual worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on annual worth? c. Should Aerotron Electronics buy the water filtration system?

05.01-PR010 Home Innovation is evaluating a new product design. The estimated receipts and disbursements associated with the new product are shown below. MARR is 10%/year. End of Year Receipts Disbursements 0 $0 $1,000 1 2 3

$600 $600 $700

$300 $300 $300

4 5

$700 $700

$300 $300

a. What is the annual worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on annual worth? c. Should Home Innovations pursue this new product? 05.01-PR011 Mayberry, Inc., is considering a design change that will cost $6,000 and will result in an annual savings of $1,000 per year for the 6-year life of the project. A cost of $2,000 will be avoided at the end of the project as a result of the change. MARR is 8%/year. a. What is the annual worth of this investment?

b. What is the decision rule for judging the attractiveness of investments based on annual worth? c. Should Mayberry implement the design change? 05.01-PR012 Nancy’s Notions pays a delivery firm to distribute its products in the metro area. Delivery costs are $30,000 per year. Nancy can buy a used truck for $10,000 that will be adequate for the next 3 years. Operating and maintenance costs are estimated to be $25,000 per year. At the end of 3 years, the used truck will have an estimated salvage value of $3,000. Nancy’s MARR is 24%/year. a. What is the annual worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on annual worth? c. Should Nancy buy the truck? Section 5.1.2 Annual Worth of Multiple Alternatives 05.01-PR013 The engineering team at Manuel’s Manufacturing, Inc., is planning to purchase an enterprise resource planning (ERP) system. The software and installation from Vendor A costs $380,000 initially and is expected to increase revenue $125,000 per year every year. The software and installation from Vendor B costs $280,000 and is expected to increase revenue $95,000 per year. Manuel’s uses a 4-year planning horizon and a 10%/year MARR. a. What is the annual worth of each investment? b. What is the decision rule for determining the preferred investment based on annual worth ranking? c. Which (if either) ERP system should Manuel purchase? 05.01-PR014 Video Solution Parker County Community College (PCCC) is trying to determine whether to use no insulation or to use insulation that is either 1 inch thick or 2 inches thick on its steam pipes. The heat loss from the pipes without insulation is expected to cost $1.50 per year per foot of pipe. A 1-inch thick insulated covering will eliminate 89% of the loss and will cost $0.40 per foot. A 2-inch thick insulated

covering will eliminate 92% of the loss and will cost $0.85 per foot. PCCC Physical Plant Services estimates that there are 250,000 feet of steam pipe on campus. The PCCC Accounting Office requires a 10%/year return to justify capital expenditures. The insulation has a life expectancy of 10 years. Determine which insulation (if any) should be purchased using annual worth analysis.

05.01-PR015 Final Finishing is considering three mutually exclusive alternatives for a new polisher. Each alternative has an expected life of 10 years and no salvage value. Polisher 1 requires an initial investment of $20,000 and provides annual benefits of $4,465. Polisher 2 requires an initial investment of $10,000 and provides annual benefits of $1,770. Polisher 3 requires an initial investment of $15,000 and provides annual benefits of $3,580. MARR is 15%/year. a. What is the annual worth of each polisher? b. Which polisher should be recommended? 05.01-PR016 Quantum Logistics, Inc., a wholesale distributor, is considering the construction of a new warehouse to serve the southeastern geographic region near the Alabama–Georgia border. There are three cities being considered. After site visits and a budget analysis, the expected income and costs associated with locating in each of the cities have been determined. The life of the warehouse is expected to be 12 years, it will have a negligible salvage value, and MARR is 15%/year. City

Initial Cost Net Annual Income

Lagrange $1,260,000 Auburn $1,000,000 Anniston $1,620,000

$480,000 $410,000 $520,000

a. What is the annual worth of each site? b. What is the decision rule for determining the preferred site based on annual worth ranking? c. Which city should be recommended?

05.01-PR017 Nadine Chelesvig has patented her invention. She is offering a potential manufacturer two contracts for the exclusive right to manufacture and market her product. Plan A calls for an immediate single lump sum payment to her of $30,000. Plan B calls for an annual payment of $1,000 plus a royalty of $0.50 per unit sold. The remaining life of the patent is 10 years. Nadine uses a MARR of 10%/year. What must be the uniform annual sales volume of the product for Nadine to be indifferent between the contracts, based on an annual worth analysis? 05.01-PR018 DelRay Foods must purchase a new gumdrop machine. Two machines are available. Machine 7745 has a first cost of $10,000, an estimated life of 10 years, a salvage value of $1,000, and annual operating costs estimated at $0.01 per 1,000 gumdrops. Machine A37Y has a first cost of $8,000, a life of 10 years, and no salvage value. Its annual operating costs will be $300 regardless of the number of gumdrops produced. MARR is 6%/year, and 30 million gumdrops are produced each year. a. What is the annual worth of each machine? b. What is the decision rule for determining the preferred machine based on annual worth ranking? c. Which machine should be recommended? 05.01-PR019 Tempura, Inc., is considering two projects. Project A requires an investment of $50,000. Estimated annual receipts for 20 years are $20,000; estimated annual costs are $12,500. An alternative project, B, requires an investment of $75,000, has annual receipts for 20 years of $28,000, and has annual costs of $18,000. Assume both projects have zero salvage value and that MARR is 12%/year. a. What is the annual worth of each project? b. Which project should be recommended? 05.01-PR020 Alpha Electronics can purchase a needed service for $90 per unit. The same service can be provided by equipment that costs $100,000 and that will have a salvage value of $0 at the end of 10 years. Annual operating costs for the equipment will be $7,000 per year plus $25 per unit produced. MARR is 12%/year.

a. Based on an annual worth analysis, should the equipment be purchased if the expected production is 200 units/year? b. Based on an annual worth analysis, should the equipment be purchased if the expected demand is 500 units/year? c. Determine the breakeven value for annual production that will return MARR on the investment in the new equipment. 05.01-PR021 Packaging equipment for Xi Cling Wrap costs $60,000 and is expected to result in end of year net savings of $23,000 per year for 3 years. The equipment will have a market value of $10,000 after 3 years. The equipment can be leased for $21,000 per year, payable at the beginning of each year. Xi Cling’s MARR is 10%/year. Based on an annual worth analysis, determine if the packaging equipment should be purchased or leased. 05.01-PR022 Video Solution Orpheum Productions in Nevada is considering three mutually exclusive alternatives for lighting enhancements to one of its recording studios. Each enhancement will increase revenues by attracting directors who prefer this lighting style. The following table shows the cash flow details, in thousands of dollars, for these enhancements. MARR is 10%/year. Based on an annual worth analysis, which alternative (if any) should be implemented? End of Year Light Bar Sliding Spots Reflected Beam 0 1

−$6,000 $2,000

−$14,000 $3,500

−$20,000 $0

2

$2,000

$3,500

$2,300

3

$2,000

$3,500

$4,600

4 5

$2,000 $2,000

$3,500 $3,500

$6,900 $9,200

6

$2,000

$3,500

$11,500

05.01-PR023 RealTurf is considering purchasing an automatic sprinkler system for its sod farm by borrowing the entire $30,000 purchase price. The loan would be repaid with four equal annual payments at an interest rate of

12%/year. It is anticipated that the sprinkler system would be used for 9 years and then sold for a salvage value of $2,000. Annual operating and maintenance expenses for the system over the 9-year life are estimated to be $9,000 per year. If the new system is purchased, cost savings of $15,000 per year will be realized over the present manual watering system. RealTurf uses a MARR of 15%/year for economic decision making. Based on an annual worth analysis, is the purchase of the new sprinkler system economically attractive? Section 5.2 Future Worth LEARNING OBJECTIVE 5.2 Calculate a future worth (FW) converting all cash flows to a single sum equivalent at the end of the planning horizon for a given interest rate. Section 5.2.1 Future Worth of a Single Alternative 05.02-PR001 An investment has the following cash flow profile. MARR is 12%/year. What is the minimum value of X such that the investment is attractive based on a future worth measure of merit? End of Year Cash Flow 0 1

−$30,000 $6,000

2

$13,500

3 4

$X $13,500

05.02-PR002 A 22-year-old engineering graduate wants to accumulate $2,000,000 to be available when she retires 40 years from today. She investigates several investment options and decides to invest in a stock market index fund after discovering that the long-term average return for the stock market is 10.4% per year. Because this will be a tax-sheltered account, she plans to ignore the impact of taxes. a. If she plans to make 40 uniform annual deposits starting 1 year from today, what is the dollar amount of the required deposits?

b. If she makes the first of the 40 deposits starting today rather than 1 year from today, what is the dollar amount of the required deposits? c. If she plans to make the first payment 1 year from today and each annual payment will be $200 greater than the previous year’s payment, i. What is the dollar amount of the first deposit? ii. What is the dollar amount of the last deposit? d. If she plans to make the first payment 1 year from today and each annual payment will be 5% greater than the previous year’s payment, i. What is the dollar amount of the first deposit? ii. What is the dollar amount of the last deposit? 05.02-PR003 Reconsider the situation described in Problem 26. Assume that rather than annual deposits, she makes monthly deposits. The first deposit will be 1 month from today, and the last deposit will be 40 years from today. Assume that the stock market return is 10.4% per year compounded monthly. a. If she plans to make uniform monthly deposits, what is the dollar amount of the monthly deposit? b. If she earns a 4% annual raise each year throughout her career (starting 1 year from today) and adjusts her monthly deposits by the same 4% each year, how much will be in the account immediately after the last deposit? 05.02-PR004 You decide to set up a college fund for your 10-year-old child and plan to make annual deposits into the account each year on your child’s birthday. Because “other things” consistently use more of your money than anticipated, your deposits are actually somewhat erratic. One year even resulted in a withdrawal. The account earns 5% per year. Birthday Deposit Birthday Deposit 10 $1,000 15 $3,000 11

$1,000

16

$2,500

12

$2,000

17

$2,000

13 14

$1,500 −$1,500

18

$2,000

a. How much is in the account immediately after the deposit on your child’s 18th birthday? b. How much would you have needed to deposit on each birthday to accumulate the same total if you had started on your child’s 10th birthday and made equal annual deposits with no withdrawals? c. How much would you have needed to deposit on each birthday to accumulate the same total if you had started on your child’s 1st birthday and made equal annual deposits with no withdrawals? 05.02-PR005 You decide to open an individual retirement account (IRA) at your local bank that pays 8%/year. At the end of each of the next 40 years, you will deposit $2,000 per year into the account (40 total deposits). Three years after the last deposit, you will begin making annual withdrawals. If you want the account to last 30 years (30 withdrawals), what amount will you be able to withdraw each year? 05.02-PR006 You decide to open an individual retirement account (IRA) at your local stockbroker that pays 10%/year for the life of the account. You deposit $2,000 today to open the account. For the next 41 years, you will deposit $2,000 per year into the account at the end of each year. There are a total of 42 $2,000 deposits. Exactly 1 year after the last deposit, you will start making withdrawals. a. What is the balance in the account immediately after the last deposit? b. What annual withdrawal can you make if you want the withdrawals to last 15 years? c. What annual withdrawal can you make if you want the withdrawals to last 20 years? d. What annual withdrawal can you make if you want the withdrawals to last 25 years? 05.02-PR007 On your child’s 1st birthday, you open an account to fund his college education. You deposit $300 to open the account. Each year, on his birthday, you make another deposit. Each subsequent deposit is 8% larger than the previous one. The account pays interest at 5%/year compounded annually. How much money is in the account immediately after the deposit on his 18th birthday?

05.02-PR008 Video Solution Financial planners (and engineering economists) unanimously encourage people to start early in planning for retirement. To illustrate this point, they frequently produce a table similar to the one below. Fill in the blank cells in this table assuming that your goal is to have $1,000,000 on your 65th birthday and that deposits start on the birthday shown and continue annually in the same amount on each birthday up to and including your 65th birthday. Assume that interest is compounded annually at 10%/year. Birthday of First Deposit Amount of Required Annual Deposit 25 30 35 40 45 50 55 60 65

$1,000,000

05.02-PR009 Video Solution Financial planners (and engineering economists) unanimously encourage people to seek out the highest rate of return possible within their personal level of risk tolerance. To illustrate this point, they frequently produce a table similar to the one below. Fill in the blank cells in this table assuming that your goal is to have $1,000,000 on your 65th birthday and that deposits start on your 26th birthday and continue annually in the same amount on each birthday up to and including your 65th birthday.

Interest Rate Earned Amount of Required Annual Deposit 4%/year 5%/year 6%/year 7%/year 8%/year 9%/year

05.02-PR010 Video Solution Bailey, Inc., is considering buying a new gang punch that would allow it to produce circuit boards more efficiently. The punch has a first cost of $100,000 and a useful life of 15 years. At the end of its useful life, the punch has no salvage value. Annual labor costs would increase $2,000 using the gang punch, but annual raw material costs would decrease $12,000. MARR is 5%/year. a. What is the future worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on future worth? c. Should Bailey buy the gang punch?

05.02-PR011 Carlisle Company has been cited and must invest in equipment to reduce stack emissions or face EPA fines of $18,500 per year. An emission reduction filter will cost $75,000 and have an expected life of 5 years. Carlisle’s MARR is 10%/year. a. What is the future worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on future worth?

c. Is the filter economically justified? 05.02-PR012 DuraTech Manufacturing is evaluating a process improvement project. The estimated receipts and disbursements associated with the project are shown below. MARR is 6%/year. End of Year Receipts Disbursements 0

$0

$5,000

1 2

$0 $2,000

$200 $300

3

$4,000

$600

4 5

$3,000 $1,600

$1,000 $1,500

a. What is the future worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on future worth? c. Should DuraTech implement the proposed process improvement? 05.02-PR013 Galvanized Products is considering the purchase of a new computer system for its enterprise data management system. The vendor has quoted a purchase price of $100,000. Galvanized Products is planning to borrow one-fourth of the purchase price from a bank at 15% compounded annually. The loan is to be repaid using equal annual payments over a 3-year period. The computer system is expected to last 5 years and have a salvage value of $5,000 at that time. Over the 5-year period, Galvanized Products expects to pay a technician $25,000 per year to maintain the system but will save $55,000 per year through increased efficiencies. Galvanized Products uses a MARR of 18%/year to evaluate investments. a. What is the future worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on future worth? c. Should the new computer system be purchased?

05.02-PR014 Quilts R Us (QRU) is considering an investment in a new patterning attachment with the cash flow profile shown in the table below. QRU’s MARR is 13.5%/year. EOY Cash Flow EOY Cash Flow 0

−$1,400

8

$600

1 2

$0 $500

9 10

$700 $800

3

$500

11

$900

4 5

$500 $500

12 13

−$1,000 −$2,000

6

$0

14

−$3,000

7

$500

15

$1,400

a. What is the future worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on future worth? c. Should QRU invest? 05.02-PR015 Video Solution Imagineering, Inc., is considering an investment in CAD-CAM compatible design software with the cash flow profile shown in the table below. Imagineering’s MARR is 18%/year. EOY Cash Flow (M$) EOY Cash Flow (M$) 0 1

−$12 −$1

4 5

$5 $5

2

$5

6

$2

3

$2

7

$5

a. What is the future worth of this investment?

b. What is the decision rule for judging the attractiveness of investments based on future worth? c. Should Imagineering invest?

05.02-PR016 Jupiter is considering investing time and administrative expense on an effort that promises one large payoff in the future, followed by additional expenses over a 10-year horizon. The cash flow profile is shown in the table below. Jupiter’s MARR is 12%/year. EOY Cash Flow (K$) EOY Cash Flow (K$) 0 1

−$2 −$10

6 7

$200 −$10

2

−$12

8

−$12

3

−$14

9

−$14

4 5

−$16 −$18

10

−$100

a. What is the future worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on future worth? c. Should Jupiter invest? Section 5.2.2 Future Worth of a Multiple Alternatives 05.02-PR017 Video Solution The following three investment opportunities are available. The returns for each investment for each year vary, but the first cost of each is $20,000. Based on a future worth analysis, which investment is preferred? MARR is 9%/year.

End of Year Investment 1 Investment 2 Investment 3 1 2

$8,000 $9,000

$11,000 $10,000

$9,500 $9,500

3

$10,000

$9,000

$9,500

4

$11,000

$8,000

$9,500

05.02-PR018 The engineering team at Manuel’s Manufacturing, Inc., is planning to purchase an enterprise resource planning (ERP) system. The software and installation from Vendor A costs $380,000 initially and is expected to increase revenue $125,000 per year every year. The software and installation from Vendor B costs $280,000 and is expected to increase revenue $95,000 per year. Manuel’s uses a 4-year planning horizon and a 10%/year MARR. a. What is the future worth of each investment? b. What is the decision rule for determining the preferred investment based on future worth ranking? c. Which (if either) ERP system should Manuel purchase? 05.02-PR019 Parker County Community College (PCCC) is trying to determine whether to use no insulation or to use insulation that is either 1 inch thick or 2 inches thick on its steam pipes. The heat loss from the pipes without insulation is expected to cost $1.50 per year per foot of pipe. A 1-inch thick insulated covering will eliminate 89% of the loss and will cost $0.40 per foot. A 2-inch thick insulated covering will eliminate 92% of the loss and will cost $0.85 per foot. PCCC Physical Plant Services estimates that there are 250,000 feet of steam pipe on campus. The PCCC Accounting Office requires a 10%/year return to justify capital expenditures. The insulation has a life expectancy of 10 years. Determine which insulation (if any) should be purchased using a ranking future worth analysis. 05.02-PR020 Nadine Chelesvig has patented her invention. She is offering a potential manufacturer two contracts for the exclusive right to manufacture and market her product. Plan A calls for an immediate single lump sum

payment to her of $30,000. Plan B calls for an annual payment of $1,000 plus a royalty of $0.50 per unit sold. The remaining life of the patent is 10 years. Nadine uses a MARR of 10%/year. What must be the uniform annual sales volume of the product for Nadine to be indifferent between the contracts based on a future worth analysis? 05.02-PR021 Quantum Logistics, Inc., a wholesale distributor, is considering the construction of a new warehouse to serve the southeastern geographic region near the Alabama–Georgia border. There are three cities being considered. After site visits and a budget analysis, the expected income and costs associated with locating in each of the cities have been determined. The life of the warehouse is expected to be 12 years, its salvage value is expected to be negligible, and MARR is 15%/year. City

Initial Cost Net Annual Income

Lagrange $1,260,000 Auburn $1,000,000

$480,000 $410,000

Anniston $1,620,000

$520,000

a. What is the future worth of each site? b. What is the decision rule for determining the preferred site based on future worth ranking? c. Which city should be recommended? 05.02-PR022 Video Solution Final Finishing is considering three mutually exclusive alternatives for a new polisher. Each alternative has an expected life of 10 years and no salvage value. Polisher 1 requires an initial investment of $20,000 and provides annual benefits of $4,465. Polisher 2 requires an initial investment of $10,000 and provides annual benefits of $1,770. Polisher 3 requires an initial investment of $15,000 and provides annual benefits of $3,580. MARR is 15%/year. a. What is the future worth of each polisher? b. Which polisher should be recommended?

05.02-PR023 Xanadu Mining is considering three mutually exclusive alternatives. Cash flows, measured in thousands of dollars, are shown in the table below. MARR is 10%/year. EOY A001 B002 C003 0 −$210 −$110 −$160 1

$80

$60

$80

2

$90

$60

$80

3 4

$100 $110

$60 $70

$80 $80

a. What is the future worth of each alternative? b. Which alternative should be recommended? 05.02-PR024 Yani has $12,000 for investment purposes. His bank has offered the following three choices. a. A special savings certificate that will pay $100 each month for 5 years and a lump sum payment at the end of 5 years of $13,000. b. Buy a share of a racehorse for $12,000 that will be worth $20,000 in 5 years. c. Put the money in a savings account that will have an interest rate of 12% per year compounded monthly. Use a future worth analysis to make a recommendation to Yani. 05.02-PR025 Two numerically controlled drill presses are being considered by the production department of Zunni’s Manufacturing; one must be selected. Comparison data is shown in the table below. MARR is 10%/year.

Drill Press T Drill Press M Initial Investment

$20,000

$30,000

Estimated Life

10 years

10 years

Estimated Salvage Value Annual Operating Cost

$5,000 $12,000

$7,000 $6,000

Annual Maintenance Cost

$2,000

$4,0007

a. What is the future worth of each drill press? b. Which drill press should be recommended? 05.02-PR026 Video Solution Alpha Electronics can purchase a needed service for $90 per unit. The same service can be provided by equipment that costs $100,000 and that will have a salvage value of $0 at the end of 10 years. Annual operating costs for the equipment will be $7,000 per year plus $25 per unit produced. MARR is 12%/year. a. Based on a future worth analysis, should the equipment be purchased if the expected production is 200 units/year? b. Based on a future worth analysis, should the equipment be purchased if the expected production is 500 units/year? c. Determine the breakeven value for annual production that will return MARR on the investment in the new equipment.

05.02-PR027 The management of Brawn Engineering is considering three alternatives to satisfy an OSHA requirement for safety gates in the machine shop. Each gate will completely satisfy the requirement, so no combinations need to be considered. The first costs, operating costs, and salvage values over a 5-year planning horizon are shown below. Using a future worth analysis with a MARR of 20%/year, determine the preferred gate.

End of Year 0 1 2 3 4 5

Gate 1 −$15,000 −$6,500 −$6,500

Gate 2 −$19,000 −$5,600 −$5,600

Gate 3 −$24,000 −$4,000 −$4,000

−$6,500 −$5,600 −$4,000 −$6,500 −$5,600 −$4,000 −$6,500 + $0 −$5,600 + $2,000 −$4,000 + $5,000

05.02-PR028 RealTurf is considering purchasing an automatic sprinkler system for its sod farm by borrowing the entire $30,000 purchase price. The loan would be repaid with four equal annual payments at an interest rate of 12%/year. It is anticipated that the sprinkler system would be used for 9 years and then sold for a salvage value of $2,000. Annual operating and maintenance expenses for the system over the 9-year life are estimated to be $9,000 per year. If the new system is purchased, cost savings of $15,000 per year will be realized over the present manual watering system. RealTurf uses a MARR of 15%/year for economic decision making. Based on a future worth analysis, is the purchase of the new sprinkler system economically attractive? 05.02-PR029 Orpheum Productions in Nevada is considering three mutually exclusive alternatives for lighting enhancements to one of its recording studios. Each enhancement will increase revenues by attracting directors who prefer this lighting style. The cash flow details, in thousands of dollars, for these enhancements are shown in the chart below. MARR is 10%/year. Based on a future worth analysis, which alternative (if any) should be implemented?

End of Year Light Bar Sliding Spots Reflected Beam 0 1 2

−$6,000 $2,000 $2,000

−$14,000 $3,500 $3,500

−$20,000 $0 $2,300

3 4

$2,000 $2,000

$3,500 $3,500

$4,600 $6,900

5 6

$2,000 $2,000

$3,500 $3,500

$9,200 $11,500

05.02-PR030 Deep Seas Submarine must implement a new engine in its submarines to meet the needs of clients who desire quieter operation. Two designs, both technologically feasible, have been created, and Deep Seas wishes to know which one to pursue. Design 1 would require an up-front manufacturing cost of $15,000,000 and will cost $2,500,000 per year for 3 years to swap out the engines in all its current submarines. Design 2 will cost $20,000,000 up front but due to a higher degree of compatibility will only require $1,500,000 per year to implement. MARR is 10%/year. Based on a future worth analysis, determine which design should be chosen. Section 5.2.3 Portfolio Analysis 05.02-PR031 An investor has $100,000 to invest in a business venture, or she can earn 10%/year with a $100,000 certificate of deposit for 4 years. Three possible business ventures have been identified. Any money not invested in the business venture can be put into a bank account that earns 7%/year. Based on a future worth analysis, what should be done with the $100,000? End of Year BV01 BV02 BV03 0 −$35,000 −$80,000 −$60,000 1 2

$0 $10,000 $0 $0 $10,000 $40,000

3 4

$0 $10,000 $0 $50,000 $90,000 $40,000

05.02-PR032 Reconsider data from Problem 53 (Orpheum Productions lighting enhancement). Assume that any money not invested in the lighting enhancements will be placed in an interest-bearing account earning MARR and will be used for future studio modernization projects. a. Use the total portfolio approach to examine the future worth of each alternative. b. Compare the future worth results from Problem 53 with your FW results for part (a). Explain your conclusions from this comparison.

Chapter 5 Summary and Study Guide Summary 5.1: Annual Worth

Learning Objective 5.1: Calculate an annual worth (AW) converting all cash flows to an equivalent uniform annual series of cash flows over the planning horizon for a given interest rate. (Section 5.1) If the AW is positive, then the investment is recommended. When comparing multiple investments, the alternative with the highest AW is preferred. Recall that in engineering economic analysis, AW is more frequently used than FW. Mathematically, the objective of AW analysis can be stated as follows: (5.2)

n

Maximize AWj = [∑ Atj (1 + MARR)

n−t

] (A|F MARR%,n)

∀j t=0

5.2: Future Worth

Learning Objective 5.2: Calculate a future worth (FW) converting all cash flows to a single sum equivalent at the end of the planning horizon for a given interest rate. (Section 5.2) The FW analysis is typically used as a supplement to the PW analysis but offers a unique forward-looking perspective to the investor to assist with goal setting. If the FW is positive, then the investment is recommended.

When comparing multiple investments, the alternative with the highest FW is preferred. Mathematically, the objective of FW analysis can be stated as follows: n

Maximize FWj = ∑ Ajt (1 + MARR)

n−t

(5.4)

∀j t=0

Important Terms and Concepts Annual Worth (AW) The value of all cash flows converted to an equivalent uniform annual series of cash flows over the planning horizon using i = MARR. Future Worth (FW) The value of all cash flows converted to a single sum equivalent at the end of the planning horizon using i = MARR.

Chapter 5 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 5.1: Annual Worth LO 5.1: Calculate an annual worth (AW) converting all cash flows to an equivalent uniform annual series of cash flows over the planning horizon for a given interest rate. Video Lesson: Annual Worth Video Lesson Notes: Annual Worth Excel Video Lesson: PMT Financial Function Excel Video Lesson Spreadsheet: PMT Financial Function Video Example 5.3: Using Annual Worth to Choose Between Two Alternatives Video Solution: 05.01-PR002 Video Solution: 05.01-PR003 Video Solution: 05.01-PR009 Video Solution: 05.01-PR014 Video Solution: 05.01-PR022 5.2: Future Worth

LO 5.2: Calculate a future worth (FW) converting all cash flows to a single sum equivalent at the end of the planning horizon for a given interest rate. Video Lesson: Future Worth Video Lesson Notes: Future Worth Excel Video Lesson: FV Financial Function Excel Video Lesson Spreadsheet: FV Financial Function Excel Video Lesson: SOLVER Tool Excel Video Lesson Spreadsheet: SOLVER Tool Video Example 5.5: Using Future Worth to Achieve a Financial Goal Video Example 5.6: Using Future Worth to Choose Between Two Alternatives Video Solution: 05.02-PR008 Video Solution: 05.02-PR009 Video Solution: 05.02-PR010 Video Solution: 05.02-PR015 Video Solution: 05.02-PR017 Video Solution: 05.02-PR022 Video Solution: 05.02-PR026 These chapter-level resources will help you with your overall understanding of the content in this chapter.

Appendix A: Time Value of Money Factors Appendix B: Engineering Economic Equations Flashcards: Chapter 05 Excel Utility: TVM Factors: Table Calculator Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values Excel Utility: Monthly Payment Sensitivity Excel Utility: TVM Factors: Discrete Compounding Excel Utility: TVM Factors: Geometric Series Future Worth Excel Utility: TVM Factors: Geometric Series Present Worth Excel Data Files: Chapter 05

CHAPTER 5 Annual Worth and Future Worth LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 5.1 Calculate an annual worth (AW) converting all cash flows to an equivalent uniform annual series of cash flows over the planning horizon for a given interest rate. (Section 5.1) 5.2 Calculate a future worth (FW) converting all cash flows to a single sum equivalent at the end of the planning horizon for a given interest rate. (Section 5.2)

Engineering Economics in Practice Josh Liu Josh Liu has been investing in the stock market for several years, beginning when he was a freshman in high school. His recent investments in Google and Akamai stock have been quite successful. Based on a stellar academic record as an undergraduate engineering student, Josh is contemplating graduate school, with plans to pursue an MBA. Due to his fascination with investments, Josh has a keen interest in financial engineering and possibly working on Wall Street following completion of his master’s degree. With an eye toward the future, Josh is interested in knowing what annual investments and annual returns on investments are required for him to amass an investment portfolio valued at $2 million within 30 years. He is also interested in knowing how long it will take for him to achieve a net worth in excess of $1 million with this same investment strategy. Not only do individual investors establish financial goals for the future but so do businesses. Although we are not aware of firms using future worth analysis as the principal tool to measure the economic worth of investment alternatives, future worth is used as a supplement to present worth in an engineering economic analysis. Also, some firms use future worth analysis to estimate terminal values in business acquisition analysis. The most popular use of future worth analysis is in retirement planning. Discussion Questions 1. Thus far, Josh’s annual investments have been in the stock market. What other investment options might you suggest to Josh? What factors should Josh consider in his stock investment portfolio? 2. Josh wants to know how long it will take him to become a millionaire. (Don’t we all!) What factors will affect this timeline? 3. To reach his $2 million goal within 30 years, Josh could determine the necessary uniform annual savings. However, one might expect Josh’s investment capacity to grow as his career advances. How would this impact an investment plan for Josh? 4. Think of a specific example of a business acquisition or a capital facilities project where a future analysis would be an important

consideration. Who might be most interested in this terminal value of the investment? Why?

Introduction In this chapter, we continue the discussion of using measures of economic worth to compare mutually exclusive investment alternatives. Specifically, we will look at annual worth analysis and future worth analysis.

Systematic Economic Analysis Technique 1. Identify the investment alternatives 2. Define the planning horizon 3. Specify the discount rate 4. Estimate the cash flows 5. Compare the alternatives 6. Perform supplementary analyses 7. Select the preferred investment Future worth analysis uses the MARR to express the economic worth of a set of cash flows, occurring over the planning horizon, as a single equivalent value at an ending time called “the future.” Future worth analysis typically is used as a supplement to present worth analysis, rather than a principal tool, in engineering economic analysis. It is also of great interest to individuals in determining what the value of their investment portfolios will be at some future time, such as for retirement planning. In engineering economic analysis, annual worth is more frequently used than future worth. As the name implies, annual worth is used to express economic equivalency in the form of a uniform annual series over the planning horizon. We begin this chapter by looking at annual worth analysis methods. Annual Worth (AW) The value of all cash flows converted to an equivalent uniform annual series of cash flows over the planning horizon using i = MARR.

5.1 Annual Worth LEARNING OBJECTIVE Calculate an annual worth (AW) converting all cash flows to an equivalent uniform annual series of cash flows over the planning horizon for a given interest rate. Video Lesson: Annual Worth In this section, we learn how to determine the uniform series equivalent for the cash flows that occur for an investment alternative during the planning horizon, and, when multiple mutually exclusive alternatives are available for investment, how to choose the one that maximizes economic worth.

5.1.1 Annual Worth of a Single Alternative Recalling our work in Chapter 2, the annual worth of an investment can be expressed mathematically as follows: (5.1)

n

AW = [∑ At (1 + MARR)

n−t

] (A|F  MARR%,n)

t=0

Of course, the annual worth can also be calculated using Equation 4.1 to compute the present worth and multiplying the result by (A|P MARR%,n). As with present worth analysis, the decision to pursue an investment opportunity depends on AW > 0, assuming the do-nothing alternative is feasible and has a negligible net cost. If the annual worth is positive, then the investment will be recommended.

EXAMPLE 5.1 Annual Worth of a Single Alternative We continue to use the acquisition of the surface mount placement (SMP) machine, introduced in Example 4.1, as our example of a single alternative. Recall, it involved a $500,000 investment, with annual returns of $92,500 and a $50,000 salvage value at the end of the 10-year planning horizon with a MARR of 10%. Key Data Given The cash flows outlined in Figure 5.1; MARR = 10%; planning horizon = 10 years Find AW of the investment. Is this investment recommended?

FIGURE 5.1 CFD for Example 5.1 Excel® Video Lesson: The PMT Function Solution The annual worth for the investment will be

AW

=

−$500,000(A|P  10%,10) + $92,500 + $50,000(A|F  10%,10)

=

−$500,000(0.16275) + $92,500 + $50,000(0.06275) = $14,262.50

Excel® Solutions or, using Excel®, AW =PMT(10%,10,500000,−50000)+92500 = $14,264.57

Because the choice is either do nothing or invest in the SMP machine and obtain an annual worth of $14,264.57 for the 10-year planning horizon, we choose to make the investment. Of course, the same choice would occur as long as AW > $0. In the previous chapter, we examined the impact of changes in the MARR on the economic worth of an investment. We did so, however, only when we were considering multiple alternatives. There was no reason for not performing a similar analysis for a single alternative. But we chose to wait until now to do so.

EXAMPLE 5.2 Using Excel® to Examine the Impact of Changes in the MARR Let’s examine the behavior of annual worth for the SMP machine when the MARR changes values, ranging from 0% to 20%. The plot of annual worth for the 10-year planning horizon is provided in Figure 5.2. Also shown in the figure is the result of using the Excel® SOLVER tool to determine the value of the MARR that makes the annual worth equal $0; the value obtained is 13.80%. (In the next chapter, we explore the internal rate of return, which is the MARR value that makes the annual worth, present worth, or future worth of an investment equal $0.)

FIGURE 5.2 Analyzing the Effect on AW of Changes in MARR for Example 5.2 Excel® Data File

5.1.2 Annual Worth of Multiple Alternatives

When attempting to determine the preferred alternative from among multiple mutually exclusive alternatives and when using annual worth as the measure of economic worth, choose the alternative that maximizes annual worth over the planning horizon of n years. Mathematically, the objective can be stated as (5.2)

n

Maximize AWj = [∑ Atj (1 + MARR) ∀j

t=0

n−t

] (A|F MARR%,n)

EXAMPLE 5.3 Using Annual Worth to Choose Between Two Alternatives Video Example In Chapter 4, we considered the installation of a new ride (the Scream Machine) at a theme park in Florida. Two alternative designs (A and B) were considered: Alternative A required a $300,000 investment, produced net annual after-tax revenues of $55,000, and had a negligible salvage value at the end of the 10-year planning horizon. Alternative B required a $450,000 investment, generated net annual after-tax revenues of $80,000, and also had a negligible salvage value at the end of the 10-year planning horizon. The do-nothing (DN) alternative was feasible and had an economic worth of $0. Based on a 10% MARR and using an annual worth measure, which design, if either, should be chosen? Key Data Given The cash flows outlined in Figure 5.3; MARR = 10%; planning horizon = 10 years Find AW of each investment alternative. Which investment is recommended?

FIGURE 5.3 CFDs for Example 5.3 Solution AWA

= −$300,000(A|P  10%,10) + $55,000 = −$300,000(0.16275) + $55,000 = $6,175.00 =PMT(10%,10,300000)+55000 = $6176.38

Likewise,

AWB  

= −$450,000(A|P  10%,10) + $80,000 = −$450,000(0.16275) + $80,000 = $6,762.50 =PMT(10%,10,450000)+80000 = $6764.57

Because AWB > AWA > AWDN, Design B is recommended.

Concept Check 05.01-CC001 Identify the one true statement about the annualized worth. a. The AW is the value of all cash flows converted to an equivalent uniform annual series of cash flows over a 5-year period b. The AW may provide a different decision than when present worth or future worth is calculated c. The AW will result in the same decision as the present worth or future worth analysis d. The AW is rarely used by individuals to determine the value of their investment portfolio in the future, such as for retirement

Concept Check 05.01-CC002 When comparing multiple mutually exclusive alternatives, select the alternative that __________ the annual worth. a. minimizes b. maximizes

5.2 Future Worth

LEARNING OBJECTIVE Calculate a future worth (FW) converting all cash flows to a single sum equivalent at the end of the planning horizon for a given interest rate. Video Lesson: Future Worth Now we focus on what is variously referred to as future worth, future value, and terminal value. Future worth analysis uses the MARR to express the economic worth of a set of cash flows, occurring over the planning horizon, as a single equivalent value at an ending or termination time called “the future.” Future Worth (FW) The value of all cash flows converted to a single sum equivalent at the end of the planning horizon using i = MARR. In the previous chapter, we noted that many investors prefer to express the economic worth of the set of cash flows as a single monetary sum at a point in time called “the present.” To obtain the single sum equivalent, we discount cash flows that occur at various points in time in the future. Hence, the name discounted cash flow analysis. As popular as discounting money is, we have found that people have far more difficulty understanding discounting than compounding. For many, it is easier to grasp the notion of money growing in value as one moves forward in time, rather than shrinking in value as one moves backward in time. For them, future worth analysis is more intuitively appealing than present worth analysis. Yet another reason for performing future worth analysis is goal setting. When performing financial planning, many individuals are interested in knowing what the value of their investment portfolio will be at some particular point in the future. For them, future worth is more relevant than present worth. In this section, we learn how to make decisions regarding the economic viability of a single investment using future worth analysis. In addition, we learn how to use future worth analysis to choose from among multiple investment alternatives the one having the greatest economic worth. Finally, we learn how to maximize the value of an investment portfolio by considering both the money invested in a particular alternative and the available capital that is not invested.

5.2.1 Future Worth of a Single Alternative

Recalling our work in Chapter 2 and letting i denote the MARR, the future worth of an investment can be expressed mathematically as follows: n

FW = ∑ At (1 + MARR)

n−t

(5.3)

t=0

As with present worth analysis, the decision to pursue an investment opportunity is dependent on FW > 0. If the future worth is positive, then the investment will be recommended.

EXAMPLE 5.4 Future Worth of a Single Investment Previously, we considered the acquisition of a new SMP machine having an initial cost of $500,000. It was anticipated that the investment would result in annual operating and maintenance costs being reduced by $92,500 per year, after taxes. The manufacturing engineer estimated the machine would be worth $50,000 at the end of the 10-year planning horizon. Using a 10% after-tax MARR and future worth analysis, should the investment be made? Key Data Given The cash flows outlined in Figure 5.1; MARR = 10%; planning horizon = 10 years Find FW of the investment. Is this investment recommended? Solution The future worth for the investment will be FW

=

−$500,000(F |P  10%,10) + $92,500(F |A 10%,10) + $50,000

=

−$500,000(2.59374) + $92,500(15.93742) + $50,000

=

$227,341.40

Excel® Solution or, using Excel®, FW =FV(10%,10,−92500,500000)+50000 = $227,340.55

Because FW > $0, the investment is recommended. We noted previously that future worth analysis is appropriate when planning to achieve a particular financial goal at some point in the future. The following example illustrates the use of future worth analysis in financial planning. Excel® Video Lesson: The FV Function

EXAMPLE 5.5 Using Future Worth to Achieve a Financial Goal Video Example A recent engineering graduate decided to begin an investment program at the age of 23, with the hope of achieving an investment goal of $5 million by age 58. If a gradient series describes the engineer’s investment pattern over the 35year period and if the annual return on the engineer’s investments is approximately 6.5%, what gradient step is required to achieve the goal if the first of the 36 investments equals $5,000? Key Data Given F = $5 million; i = 6.5%; A1 = $5,000; n = 36 Find G Solution Because the investment’s future worth is given and the unknown is the size of the gradient step (G), the following equation is to be solved for G: G = [$5,000,000(A|F  6.5%,36) − $5,000]/(A|G 6.5%,36)

where (A|F  6.5%,36) = 0.065/[(1.065)

36

− 1] = 0.0075133

and 36

(A|G 6.5%,36) =

{(1.065)

=

11.22339

− [1 + 36(0.065)]}/{0.065[(1.065)

36

Therefore, G = [$5,000,000(0.0075133) − $5,000]/11.22339 = $2,901.66

Excel® Solution

− 1]}

SOLVER can be used to solve for G. As shown in Figure 5.4, we let cell C20 contain the value of G. Then we generate the deposits by adding C20 to the preceding deposit. The balance in the investment account is computed by adding the most recent deposit to the product of the previous balance and 1.065. After 36 years, when the engineer is 58, the balance is to be $5 million. (If G = $3,000, the final balance is $5,146,882.19.) Excel® Video Lesson: SOLVER Tool

FIGURE 5.4 Set Up to Use the Excel® SOLVER Tool to Determine the Gradient Step Needed to Achieve a Financial Goal Excel® Data File Figure 5.5 contains the solution obtained using SOLVER. Notice, SOLVER is set up to make F19 equal 5000000 by changing C20. As shown, if G = $2,901.67, then a $5 million balance will occur in the investment account after 36 years.

FIGURE 5.5 Excel® SOLVER Solution to Example 5.5 Excel® Data File Exploring the Solution What if the investments do not earn 6.5%? To gain an understanding of the impact on the future worth of the investments, Figure 5.6 was developed, showing the growth in the investment portfolio over time for various annual returns on the investment, where the formula for (F/G i%,n) is given in Chapter 2. In anticipation of annual returns being between 6% and 8%, the engineer anticipates the size of the investment portfolio will be between $4.6 million and $6.4 million after the 36th deposit.

FIGURE 5.6 Impact of Annual Returns on an Investment Portfolio Excel® Data File

5.2.2 Future Worth of Multiple Alternatives When attempting to determine the preferred alternative from among multiple mutually exclusive alternatives and when using future worth as the measure of economic worth, choose the alternative that maximizes FW. Mathematically, letting MARR denote the interest rate used, the objective is to n

Maximize FWj = ∑ Ajt (1 + MARR) ∀j t=0

n−t

(5.4)

EXAMPLE 5.6 Using Future Worth to Choose Between Two Alternatives Video Example Recall previously, we compared economically two design alternatives (A and B) for a new ride (the Scream Machine) at a theme park. Alternative A required a $300,000 investment, produced after-tax net annual revenue of $55,000, and had a negligible salvage value at the end of the 10-year planning horizon. Alternative B required a $450,000 investment, generated after-tax net annual revenue of $80,000, and also had a negligible salvage value at the end of the 10-year planning horizon. The do-nothing alternative is feasible. Based on an after-tax MARR of 10% and using a future worth analysis, which alternative, if any, is the economic choice? (The do-nothing alternative is assumed to have a future worth of $0.) Key Data Given The cash flows outlined in Figure 5.3; MARR = 10%; planning horizon = 10 years Find FW of each investment alternative. Which investment is recommended? Solution FWA  =

−$300,000(F |P  10%,10) + $55,000(F |A 10%,10)

=

−$300,000(2.59374) + $55,000(15.93742)

=

$98,436.10

or, using Excel®, FWA  =FV(10%,10,−55000,300000) = $98,435.62

Likewise, FWB  =

−$450,000(F |P  10%,10) + $80,000(F |A 10%,10)

=

−$450,000(2.59374) + $80,000(15.93742)

=

$107,810.60

or, using Excel®, FWB  =FV(10%,10,−80000,450000) = $107,809.86

Because FWB > FWA > $0, Design Alternative B is recommended.

EXAMPLE 5.7 Using Future Worth to Choose a Retirement Plan During a meeting with a human resources representative of her new employer, a recent engineering graduate learns that she needs to choose between two retirement plans. One plan involves investments that are matched by the employer; the plan is managed by a committee of employees in the firm. Another option is to establish an investment plan that she will manage. In both cases, deposits are tax-deferred; also in both cases, withdrawal of funds before age 62 will result in a significant financial penalty. Her current age is 22. Under the first plan, up to 4% of an employee’s annual compensation is matched by the employer. Funds are invested in a mix of securities, including the company’s own stock; however, no more than 20% of the investment portfolio is allowed to be invested in the firm’s stock. The portfolio includes a mix of stocks, bonds, and U.S. Treasury notes. Over the past 15 years, the investment portfolio has increased in value at an annual compound rate of 6%. Under the second plan, individuals can manage their own investment portfolios. Although there are limits on the investments that can be made, a number of riskier choices are available. Individuals who choose to manage their own portfolio have to pay a management fee of 1.5% of their annual deposit. The company still matches up to 4% of the employee’s annual compensation. After considering the investments available in the second plan, the new employee narrows the choices to a set of investments that historically have earned between 2% and 12% annually. If the employee’s current salary is $55,000, she invests the maximum allowed, and her annual salary increases at an annual rate of 5%, which retirement plan should she choose? Key Data Given The employee and plan details summarized below.

First Plan

Second Plan

4% of annual salary invested + 4% company match Current annual salary = $55,000 Annual salary increase = 5% Historical portfolio annual growth = 6% No management fees

4% of annual salary invested + 4% company match Current annual salary = $55,000 Annual salary increase = 5% Historical portfolio annual growth = 2% to 12% Annual management fee = 1.5%

Find FW of each plan. Solution Under the first plan, her investment portfolio will have the following value after 40 years: FW1  =

2(0.04)($55,000)(F |A1 6%,5%,40) 40

=

$4,400[(1.06)

=

$1,428,120.90

− (1.05)

40

]/(0.06 − 0.05)

Under the second plan, in the pessimistic case (earns 2%/year), her investment portfolio will have the following value after 40 years: FW2p  =

2(0.04)($55,000)(0.985)(F |A1 2%,5%,40)

=

$4,334[(1.02)

=

$698,055.57

40

− (1.05)

40

]/(0.02 − 0.05)

Under the second plan, in the optimistic case (earns 12%/year), her investment portfolio will have the following value after 40 years: FW2o  =

2(0.04)($55,000)(0.985)(F |A1 12%,5%,40) 40

=

$4,334[(1.12)

=

$5,325,308.50

− (1.05)

40

]/(0.12 − 0.05)

After giving the matter considerable thought, she decided to choose the second plan and manage her own investment portfolio. What would you have done?

Given the turbulence of investment returns in 2008 and 2018, how reasonable are the annual increases in salary and the annual returns on her investment over a 40-year period?

5.2.3 Portfolio Analysis In Chapter 1 we noted the assumption that any money not invested in a candidate alternative remains in an investment pool and earns a return equal to the MARR. A portfolio analysis looks at the value of an entire portfolio of investments, including the money in the investment pool. Future worth is a convenient means of conducting a portfolio analysis. Portfolio Analysis The future worth of all cash flows, including uninvested money, using i = MARR.

EXAMPLE 5.8 Future Worth of the Portfolio In the case of the two design alternatives (A and B) for the Scream Machine, the most expensive one required a $450,000 investment. For the design alternative to be feasible, $450,000 must have been available for investment. If it is not invested in a new ride at the theme park, it could be invested and earn 10% annual compound returns. Compare the future worth of each alternative over a 10-year planning horizon using an investment portfolio approach, including residual capital in the investment pool. Key Data Given The cash flows outlined in Figure 5.3; MARR = 10%; planning horizon = 10 years Find FW of the investment portfolio for the do-nothing alternative, Alternative A, and Alternative B (including the investment and residual capital) Solution The future worth for the do-nothing (DN) alternative would be FWDN = $450,000(F |P  10%,10) = $450,000(2.59374) = $1,167,183.00

or, using Excel®, FWDN  =FV(10%,10%,−450,000) = $1,167,184.11

The $450,000 invested in Design Alternative B will result in annual revenue of $80,000. If the recovered funds are added to the investment pool, in 10 years they will be worth FWB = $80,000(F |A 10%,10) = $80,000(15.93742) = $1,274,993.60

or, using Excel®, FWB  =FV(10%,10,−80000) = $1,274,993.97

If Design Alternative A is chosen for investment, $55,000 will be recovered annually for 10 years. Placing the $55,000 in the investment pool will result in

a future value of FWA = $55,000(F |A 10%,10) = $55,000(15.93742) = $876,558.10

or, using Excel®, FWA  =FV(10%,10,−55000) = $876,558.35

In addition, because $450,000 was available and only $300,000 was required for Alternative A, $150,000 of residual capital (RC) remains in the investment pool and earns an annual return of 10%. Therefore, the $150,000 will grow at a rate of 10% compounded annually to achieve a value of FWRC = $150,000(F |P  10%,10) = $150,000(2.59374) = $389,061

or, using Excel®, FWRC  =FV(10%,10,−150000) = $389,061.37

Hence, if Design Alternative A is chosen, the investment portfolio will total $876,558.10 + $389,061.00, or $1,265,619.10. Or, using Excel®, the investment portfolio will total $876,558.35 + $389,061.37, or $1,265,619.72. In summary, if neither design is selected, the investment portfolio will have a value, based on results from Excel®, of $1,167,184.11. If Design B is installed, the investment portfolio will have a value of $1,274,993.97. If Design A is installed, the total value of the investment portfolio will be $1,265,619.72. Hence, to maximize the investment portfolio, Design B should be selected, and the value of the overall investment portfolio will be $1,274,993.97 − $1,265,619.72, or $9374.25 greater than if Design A is purchased.

Concept Check 05.02-CC001 The FW analysis: a. Is appropriate when one plans to achieve a particular financial goal at some point in the future b. Sometimes results in the same decision as PW and AW c. Does not utilize the concept of time value of money d. All of the above

Concept Check 05.02-CC002 When comparing multiple mutually exclusive alternatives, select the alternative that __________ the future worth. a. maximizes b. minimizes

Concept Check 05.02-CC003 A portfolio analysis a. Looks at the value of an entire portfolio of investments, not including the money in the investment pool b. Looks at the value of an entire portfolio of investments, including the money in the investment pool c. Is useful when considering multiple investments d. b and c

CHAPTER 6 Rate of Return

Chapter 6 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced e-text and WileyPLUS

FE-Like Problems 06-FE001 Consider the following cash flow diagram. What is the value of X if the internal rate of return is 15%?

a. $246 b. $255 c. $281 d. $290 06-FE002 If the internal rate of return (IRR) of a well-behaved investment alternative is equal to MARR, which of the following statements about the other measures of worth for this alternative must be true? i. PW = 0 ii. AW = 0 a. I only b. II only

c. Neither I nor II d. Both I and II Correct or Incorrect? Clear

  Check Answer

06-FE003 An investment is guaranteed to have a unique value of IRR if which of the following is true? a. Alternating positive and negative cash flows b. An initial negative cash flow followed by all positive cash flows and the sum of the positive cash flows is greater than the magnitude of the negative cash flow c. A unique value for ERR d. A positive PW at MARR 06-FE004 diagram?

What is the internal rate of return of the following cash flow

a. 20.0% b. 18.2% c. 17.5% d. 15.0% Correct or Incorrect? Clear

  Check Answer

06-FE005 A snow cone machine at an ice cream shop costs $15,000. The machine is expected to generate profits of $2,500 each year of its 10-year useful life. At the end of the 10 years the machine will have a salvage value of zero. Within what interest rate range does the IRR fall? a. Less than 10% b. 10% to 12% c. 12% to 14% d. Greater than 14% The next two questions are based on the following “present worth versus interest rate” graph for a well-behaved investment.

06-FE006 If the interest rate at B is 20%, then which of the following best describes the analysis of the investment? a. The IRR of the investment is less than 20% b. The IRR of the investment is equal to 20% c. The IRR of the investment is greater than 20% d. None of the above are true Correct or Incorrect? Clear

  Check Answer

06-FE007

The IRR of this investment is located at which point?

a. A b. C c. D d. E 06-FE008 A company is considering two alternatives, one of which must be implemented. Of the two projects, A has the higher maintenance cost, but B has the higher investment cost. The appropriate (and properly calculated) incremental IRR is 17.6%. Which alternative is preferred if the Minimum Attractive Rate of Return is 20%? a. A b. B c. The company is indifferent between A and B d. Cannot be determined from the information given Correct or Incorrect? Clear

  Check Answer

06-FE009 If the IRR of Alternative A is 18%, the IRR of Alternative B is 16%, and MARR is 12%, which of the following is correct?

a. Alternative B is preferred over alternative A b. Alternative A is preferred over alternative B c. Not enough information is given to determine which alternative is preferred d. Neither alternative A nor alternative B is acceptable 06-FE010 Consider the IRR and ERR measures of worth. If we define a “root” to mean a value for the measure that results in PW = 0, then which of the following statements is true? a. IRR can have multiple roots and ERR can have multiple roots b. IRR has only a single root but ERR can have multiple roots c. ERR has only a single root but IRR can have multiple roots d. IRR has only a single root and ERR has only a single root Correct or Incorrect? Clear

  Check Answer

06-FE011 Consider the calculation of an external rate of return (ERR). The positive cash flows in the cash flow profile are moved forward to t = n using what value of i in the (F|P,i,n–t) factors? a. 0 b. The unknown value of ERR (i′) c. MARR d. IRR

Problems Note to Instructors and Students Many of the problems in this chapter are similar to problems in previous chapters. This similarity is intentional. It is designed to illustrate the use of different measures of merit on the same problem.

Section 6.1 Internal Rate of Return Calculations LEARNING OBJECTIVE 6.1 Compute the Internal Rate of Return (IRR) for an individual investment and perform incremental comparisons of mutually exclusive investment alternatives to determine the one that maximizes economic worth. Sections 6.1.1 and 6.1.2 IRR Calculations—Single Alternative 06.01-PR001 Match the measures of worth in the first column with one (or more) of the analysis approaches that is (are) appropriate for that measure. Measure of Worth

Analysis Approach

(a) Annual worth (1) Ranking approach (b) External rate of return (2) Incremental approach (c) Future worth (d) Internal rate of return (e) Present worth 06.01-PR002 Match the measures of worth in the first column with the appropriate unit of measure that results from the analysis. Measure of Worth (a) Annual worth

Resulting Units of Measure (1) Dollars

(b) External rate of return (2) Percentage (c) Future worth (d) Internal rate of return (e) Present worth 06.01-PR003 Draw a cash flow diagram of any investment that has both of the following properties: 1. The investment has a 4-year life. 2. The investment has a 10%/yr internal rate of return.

06.01-PR004 An investment has the following cash flow profile. For each value of MARR below, what is the minimum value of X such that the investment is attractive based on an internal rate of return measure of merit? End of Year Cash Flow 0 −$30,000 1 2

$6,000 $13,500

3 4

$X $13,500

a. MARR is 12%/yr. b. MARR is 15%/yr. c. MARR is 24%/yr. d. MARR is 8%/yr. e. MARR is 0%/yr. 06.01-PR005 Nu Things, Inc., is considering an investment in a business venture with the following anticipated cash flow results: EOY Cash Flow EOY Cash Flow EOY Cash Flow 0 −$70,000 7 14,000 14 7,000 1 2

20,000 19,000

8 9

13,000 12,000

15 16

6,000 5,000

3 4

18,000 17,000

10 11

11,000 10,000

17 18

4,000 3,000

5 6

16,000 15,000

12 13

9,000 8,000

19 20

2,000 1,000

Assume MARR is 20% per year. Based on an internal rate of return analysis (1) determine the investment’s worth; (2) state whether or not your results indicate the investment should be undertaken; and (3) state the decision rule you used to arrive at this conclusion.

06.01-PR006 Carlisle Company has been cited and must invest in equipment to reduce stack emissions or face EPA fines of $18,500 per year. An emission reduction filter will cost $75,000 and have an expected life of 5 years. Carlisle’s MARR is 10%/yr. a. What is the internal rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? c. Is the filter economically justified? 06.01-PR007 Fabco, Inc., is considering the purchase of flow valves that will reduce annual operating costs by $10,000 per year for the next 12 years. Fabco’s MARR is 7%/yr. Using an internal rate of return approach, determine the maximum amount Fabco should be willing to pay for the valves. 06.01-PR008 Imagineering, Inc., is considering an investment in CADCAM-compatible design software with the cash flow profile shown in the table below. Imagineering’s MARR is 18%/yr. EOY Cash Flow (M$) 0 −$12 1 2

−$1 $5

3 4 5

$2 $5 $5

6 7

$2 $5

a. What is the internal rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? c. Should Imagineering invest?

06.01-PR009 DuraTech Manufacturing is evaluating a process improvement project. The estimated receipts and disbursements associated with the project are shown below. MARR is 6%/yr. End of Year Receipts Disbursements 0 $0 $5,000 1 2 3

$0 $2,000 $4,000

$200 $300 $600

4 5

$3,000 $1,600

$1,000 $1,500

a. What is the internal rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? c. Should DuraTech implement the proposed process improvement? 06.01-PR010 A design change being considered by Mayberry, Inc., will cost $6,000 and will result in an annual savings of $1,000 per year for the 6-year life of the project. A cost of $2,000 will be avoided at the end of the project as a result of the change. MARR is 8%/yr. a. What is the internal rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? c. Should Mayberry implement the design change? 06.01-PR011 Home Innovation is evaluating a new product design. The estimated receipts and disbursements associated with the new product are shown below. MARR is 10%/yr.

End of Year Receipts Disbursements 0 $0 $1,000 1 2 3

$600 $600 $700

$300 $300 $300

4 5

$700 $700

$300 $300

a. What is the internal rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? c. Should Home Innovations pursue this new product? 06.01-PR012 A project has been selected for implementation. The net cash flow (NCF) profile associated with the project is shown below. MARR is 10%/yr. EOY

NCF

0 1 2

−$70,000 $30,000 $30,000

3 4

$30,000 $30,000

5 6

$30,000 $30,000 + $2,000

a. What is the internal rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? c. Is the project economically justified?

06.01-PR013 Video Solution Bailey, Inc., is considering buying a new gang punch that would allow it to produce circuit boards more efficiently. The punch has a first cost of $100,000 and a useful life of 15 years. At the end of its useful life, the punch has no salvage value. Labor costs would increase $2,000 per year using the gang punch but raw material costs would decrease $12,000 per year. MARR is 5%/yr. a. What is the internal rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? c. Should Bailey buy the gang punch?

06.01-PR014 Eddie’s Precision Machine Shop is insured for $700,000. The present yearly insurance premium is $1.00 per $100 of coverage. A sprinkler system with an estimated life of 20 years and no salvage value can be installed for $20,000. Annual maintenance costs for the sprinkler system are $400. If the sprinkler system is installed, the system must be included in the shop’s value for insurance purposes but the insurance premium will reduce to $0.40 per $100 of coverage. Eddie uses a MARR of 15%/yr. a. What is the internal rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? c. Is the sprinkler system economically justified? 06.01-PR015 Nancy’s Notions pays a delivery firm to distribute its products in the metro area. Shipping costs are $30,000 per year. Nancy can buy a used truck for $10,000 that will be adequate for the next 3 years. Operating and maintenance costs are estimated to be $25,000 per year. At the end of 3 years, the used truck will have an estimated salvage value of $3,000. Nancy’s MARR is 24%/yr.

a. What is the internal rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? c. Should Nancy buy the truck? 06.01-PR016 Brock Associates invested $80,000 in a business venture with the following results: EOY

CF

EOY

CF

0 1

−$80,000 10,000

5 6

$28,000 22,000

2 3 4

16,000 22,000 28,000

7 8

16,000 10,000

MARR is 12%. Determine the internal rate of return and whether or not this is a desirable venture. 06.01-PR017 Shrewd Endeavors, Inc., invested $70,000 in a business venture with the following cash flow results: EOY

CF

EOY

CF

0 −$70,000 11 10,000 1 20,000 12 9,000 2 3

19,000 13 18,000 14

8,000 7,000

4

17,000 15

6,000

5

16,000 16

5,000

6 7

15,000 17 14,000 18

4,000 3,000

8

13,000 19

2,000

9 10

12,000 20 11,000

1,000

MARR is 10%. Determine the internal rate of return and whether or not this is a desirable venture. 06.01-PR018 An investment of $20,000 for a new condenser is being considered. Estimated salvage value of the condenser is $5,000 at the end of an estimated life of 6 years. Annual income each year for the 6 years is $8,500. Annual operating expenses are $2,300. Assume money is worth 15% compounded annually. Determine the internal rate of return and whether or not the condenser should be purchased. 06.01-PR019 Smith Investors places $50,000 in an investment fund. One year after making the investment, Smith receives $7,500 and continues to receive $7,500 annually until 10 such amounts are received. Smith receives nothing further until 15 years after the initial investment, at which time $50,000 is received. Over the 15-year period, determine the internal rate of return and whether or not this is a desirable investment if MARR = 10%. 06.01-PR020 Video Solution What do you know about the mathematical value of the internal rate of return of a project under each of the following conditions? a. The present worth of the project is greater than zero. b. The present worth of the project is equal to zero. c. The present worth of the project is less than zero. d. The future worth of the project is greater than zero. e. The future worth of the project is equal to zero. f. The future worth of the project is less than zero.

g. The annual worth of the project is greater than zero. h. The annual worth of the project is equal to zero. i. The annual worth of the project is less than zero.

06.01-PR021 Value Lodges is the owner of an economy motel chain. Value Lodges is considering building a new 200-unit motel. The cost to build the motel is estimated at $8,000,000; Value Lodges estimates furnishings for the motel will cost an additional $700,000 and will require replacement every 5 years. Annual operating and maintenance costs for the motel are estimated to be $800,000. The average rental rate for a unit is anticipated to be $40/day. Value Lodges expects the motel to have a life of 15 years and a salvage value of $900,000 at the end of 15 years. This estimated salvage value assumes that the furnishings are not new. Furnishings have no salvage value at the end of each 5-year replacement interval. Assuming average daily occupancy percentages of 50%, 60%, 70%, 80% for years 1 through 4, respectively, and 90% for the 5th through 15th years, MARR of 12%/yr, 365 operating days/year, and ignoring the cost of land, should the motel be built? Base your decision on an internal rate of return analysis. 06.01-PR022 Baon Chemicals Unlimited purchases a computer-controlled filter for $100,000. Half of the purchase price is borrowed from a bank at 15% compounded annually. The loan is to be paid back with equal annual payments over a 5-year period. The filter is expected to last 10 years, at which time it will have a salvage value of $10,000. Over the 10-year period, the operating and maintenance costs are expected to equal $20,000 in year 1 and increase by $1,500/yr each year thereafter. By making the investment, annual fines of $50,000 for pollution will be avoided. Baon expects to earn 12% compounded annually on its investments. Based on an internal rate of return analysis, determine whether the purchase of the filter is economically justified. 06.01-PR023 Video Solution RealTurf is considering purchasing an automatic sprinkler system for its sod farm by borrowing the entire $30,000 purchase price. The loan would be repaid with four equal annual payments at

an interest rate of12%/yr. It is anticipated that the sprinkler system would be used for 9 years and then sold for a salvage value of $2,000. Annual operating and maintenance expenses for the system over the 9-year life are estimated to be $9,000 per year. If the new system is purchased, cost savings of $15,000 per year will be realized over the present manual watering system. RealTurf uses a MARR of 15%/yr for economic decision making. Based on an internal rate of return analysis, is the purchase of the new sprinkler system economically attractive?

06.01-PR024 Galvanized Products is considering the purchase of a new computer system for its enterprise data management system. The vendor has quoted a purchase price of $100,000. Galvanized Products is planning to borrow one-fourth of the purchase price from a bank at 15% compounded annually. The loan is to be repaid using equal annual payments over a 3-year period. The computer system is expected to last 5 years and have a salvage value of $5,000 at that time. Over the 5-year period, Galvanized Products expects to pay a technician $25,000 per year to maintain the system but will save $55,000 per year through increased efficiencies. Galvanized Products uses a MARR of 18%/yr to evaluate investments. a. What is the internal rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? c. Should the new computer system be purchased? 06.01-PR025 Aerotron Electronics is considering the purchase of a water filtration system to assist in circuit board manufacturing. The system costs $40,000. It has an expected life of 7 years at which time its salvage value will be $7,500. Operating and maintenance expenses are estimated to be $2,000 per year. If the filtration system is not purchased, Aerotron Electronics will have to pay Bay City $12,000 per year for water purification. If the system is purchased, no water purification from Bay City will be needed. Aerotron Electronics must borrow half of the purchase price, but they cannot start repaying the loan for 2 years. The bank has agreed to three equal annual payments, with the first payment due at end of year 2. The loan interest rate is

8% compounded annually. Aerotron Electronics’ MARR is 10% compounded annually. a. What is the internal rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? c. Should Aerotron Electronics buy the water filtration system? 06.01-PR026 Delta Dawn’s Bakery is considering purchasing a new van to deliver bread. The van will cost $18,000. Two-thirds ($12,000) of this cost will be borrowed. The loan is to be repaid with 4 equal annual payments (first payment at t = 1) based on an interest rate of 4%/yr. It is anticipated that the van will be used for 6 years and then sold for a salvage value of $500. Annual operating and maintenance expenses for the van over the 6-year life are estimated to be $700 per year. If the van is purchased, Delta will realize a cost savings of $3,800 per year. Delta uses a MARR of 6%/yr. Based on an internal rate of return analysis, is the purchase of the van economically attractive? 06.01-PR027 Consider the following cash flow profile and assume MARR is 10%/yr. EOY NCF EOY NCF 0 −$100 4 −$30 1

$25

5

$60

2 3

$25 $60

6

$25

a. What does Descartes’ rule of signs tell us about the IRR(s) of this project? b. What does Norstrom’s criterion tell us about the IRR(s) of this project? c. Determine the IRR(s) for this project. d. Is this project economically attractive? 06.01-PR028 Video Solution Consider the following cash flow profile and assume MARR is 10%/yr.

EOY NCF EOY NCF 0 1

−$100 $25

4 5

$25 $25

2

$25

6

$25

3

$25

a. What does Descartes’ rule of signs tell us about the IRR(s) of this project? b. What does Norstrom’s criterion tell us about the IRR(s) of this project? c. Determine the IRR(s) for this project. d. Is this project economically attractive?

06.01-PR029 Consider the following cash flow profile and assume MARR is 10%/yr. EOY NCF EOY NCF 0

−$100

4

$15

1

$15

5

$15

2 3

$15 $15

6

$15

a. What does Descartes’ rule of signs tell us about the IRR(s) of this project? b. What does Norstrom’s criterion tell us about the IRR(s) of this project? c. Determine the IRR(s) for this project. d. Is this project economically attractive?

06.01-PR030 Consider the following cash flow profile and assume MARR is 10%/yr. EOY NCF EOY NCF 0 1

−$100 $25

4 5

$250 −$200

2

$200

6

−$100

3

−$100

a. What does Descartes’ rule of signs tell us about the IRR(s) of this project? b. What does Norstrom’s criterion tell us about the IRR(s) of this project? c. Determine the IRR(s) for this project. d. Is this project economically attractive? 06.01-PR031 Consider the following cash flow profile and assume MARR is 10%/yr. EOY NCF EOY NCF 0

−$100

4

−$950

1 2

$800 −$750

5 6

$700 −$800

3

$900

a. What does Descartes’ rule of signs tell us about the IRR(s) of this project? b. What does Norstrom’s criterion tell us about the IRR(s) of this project? c. Determine the IRR(s) for this project. d. Is this project economically attractive? 06.01-PR032 Consider the following cash flow profile and assume MARR is 10%/yr.

EOY NCF EOY NCF 0

−$101

4

$2

1 2

$411 −$558

5 6

$8 −$14

3

$253

a. Determine the IRR(s) for this project. b. Is this project economically attractive? 06.01-PR033 Quilts R Us (QRU) is considering an investment in a new patterning attachment with the cash flow profile shown in the table below. QRU’s MARR is 13.5%/yr. EOY Cash Flow EOY Cash Flow 0 1

−$1,400 $0

8 9

$600 $700

2

$500

10

$800

3 4

$500 $500

11 12

$900 −$1,000

5

$500

13

−$2,000

6 7

$0 $500

14 15

−$3,000 $1,400

a. What is the internal rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? c. Should QRU invest? Section 6.1.3 IRR Calculations—Multiple Alternatives 06.01-PR034 Quantum Logistics, Inc., a wholesale distributor, is considering the construction of a new warehouse to serve the southeastern geographic region near the Alabama–Georgia border. There are three cities being

considered. After site visits and a budget analysis, the expected income and costs associated with locating in each of the cities have been determined. The life of the warehouse is expected to be 12 years and MARR is 15%/yr. Based on an internal rate of return analysis, which city should be recommended? City

Initial Cost Net Annual Income

Lagrange $1,260,000 Auburn $1,000,000

$480,000 $410,000

Anniston $1,620,000

$520,000

06.01-PR035 Two mutually exclusive proposals, each with a life of 5 years, are under consideration. MARR is 12%. Each proposal has the following cash flow profile: EOY NCF(A) NCF(B) 0

−$30,000 −$42,000

1 2

$9,300 $12,625 $9,300 $12,625

3

$9,300 $12,625

4 5

$9,300 $12,625 $9,300 $12,625

Determine which (if either) alternative the decision maker should select using the internal rate of return method. 06.01-PR036 Chingos and Daughters Construction is considering three investment proposals: A, B, and C. Proposals A and B are mutually exclusive, and Proposal C is contingent on proposal B. The cash flow data for the investments over a 10-year planning horizon are given below. The company has a budget limit of $1 million for investments of the type being considered currently. MARR = 15%.

Initial investment

NCF(A) NCF(B) NCF(C) $600,000 $800,000 $470,000

Planning horizon

10 years 10 years 10 years

Salvage values Annual receipts

$70,000 $130,000 $65,000 $400,000 $600,000 $260,000

Annual disbursements $130,000 $270,000 $70,000 Determine which alternative should be selected using the internal rate of return method. 06.01-PR037 ZeeZee’s Construction Company has the opportunity to select one of four projects (A, B, C, or D) or the null (Do Nothing) alternative. Each project requires a single initial investment and has an internal rate of return as shown in the first table below. The second table shows the incremental IRR(s) for pairwise comparisons between each project and all other projects with a smaller initial investment. Investments and IRR(s) Project Initial Investment IRR Null

$0

0.0%

A

$600,000

44.0%

B C

$800,000 $470,000

40.0% 39.2%

D

$540,000

36.0%

Incremental IRR(s)

Increment Incremental IRR B-A B-D

28.3% 48.8%

B-C

41.4%

A-D A-C

116.5% 61.0%

D-C

18.4%

For each of the values of MARR below indicate which project is preferred based on an incremental IRR analysis. a. MARR = 50%. b. MARR = 41%. c. MARR = 25%. 06.01-PR038 A large company has the opportunity to select one of seven projects: A, B, C, D, E, F, G, or the null (Do Nothing) alternative. Each project requires a single initial investment as shown in the table below. Information on each alternative was fed into a computer program that calculated the IRR for each project as well as all the pertinent incremental IRR(s) as shown in the table below. Incremental Rate of Return of “Row”—“Column” Project

Null

A

Initial Investment $10,000

A

B

C

D

B

12,000

9

7%

C

13,000

8

2

0.1%

D E

15,000 16,000

7 6

9 5

F

18,000

5

G

23,000

7

E

5 1

9% 6

3%

8

2

5

5

5%

3

8

7

4

3

F

10%

2%

For example, the IRR for Project A is 10% and the incremental IRR of Project C minus Project B (C-B) is 0.1%. For each value of MARR below indicate which project is preferred and the evaluations you made to arrive at this conclusion. a. MARR = 12%. b. MARR = 9.5%. c. MARR = 8%. d. MARR = 3.5%. e. MARR = 1.5%. 06.01-PR039 Dark Skies Observatory is considering several options to purchase a new deep space telescope. Revenue would be generated from the telescope by selling “time and use” slots to various researchers around the world. Four possible telescopes have been identified in addition to the possibility of not buying a telescope if none are financially attractive. The table below details the characteristics of each telescope. An internal rate of return analysis is to be performed. T1 Useful life

T2

T3

T4

10 years 10 years 10 years 10 years

First cost $600,000 $800,000 $470,000 $540,000 Salvage value $70,000 $130,000 $65,000 $200,000 Annual revenue $400,000 $600,000 $260,000 $320,000 Annual expenses $130,000 $270,000 $70,000 $120,000 a. Determine the preferred telescope if MARR is 25%/yr. b. Determine the preferred telescope if MARR is 42%/yr. 06.01-PR040 Orpheum Productions in Nevada is considering three mutually exclusive alternatives for lighting enhancements to one of its recording studios. Each enhancement will increase revenues by attracting directors who prefer this lighting style. The cash flow details, in thousands of dollars, for these enhancements are shown in the chart below. MARR is 10%/yr. Based on

an internal rate of return analysis, which alternative (if any) should be implemented? End of Year Light Bar Sliding Spots Reflected Beam 0

−$6,000

−$14,000

−$20,000

1 2

$2,000 $2,000

$3,500 $3,500

$0 $2,300

3

$2,000

$3,500

$4,600

4 5

$2,000 $2,000

$3,500 $3,500

$6,900 $9,200

6

$2,000

$3,500

$11,500

06.01-PR041 Yani has $12,000 for investment purposes. His bank has offered the following three choices. 1. A special savings certificate that will pay $100 each month for 5 years and a lump sum payment at the end of 5 years of $13,000; 2. Buy a share of a racehorse for $12,000 that will be worth $20,000 in 5 years; 3. Put the money in a savings account that will have an interest rate of 12% per year compounded monthly. Use an internal rate of return analysis to make a recommendation to Yani. 06.01-PR042 On-Site Testing Service has received four investment proposals for consideration. Two of the proposals, X1 and X2, are mutually exclusive. The other two proposals, Y1 and Y2, are also mutually exclusive. Proposal Y1 is contingent on X1 and Y2 is contingent on X2. Other than these restrictions, any combination of proposals (including null) is feasible. MARR is 10%/yr. The expected cash flows for the proposals are shown below. An internal rate of return analysis is to be conducted. End of Year

X1

X2

Y1

Y2

0 −$10,000 −$15,000 −$6,000 −$9,000 1 through 8 $1,600 $2,600 $2,500 $3,500

a. List all the alternatives to be considered. b. Determine which (if any) proposals On-Site Testing should accept. 06.01-PR043 Arnold Engineering has available two mutually exclusive investment proposals, A and B. Their net cash flows are as shown in the table over a 10-year planning horizon. MARR is 12%. EOY NCF(A) NCF(B) 0 −$40,000 −$30,000 1

8,000

9,000

2 3 4

8,000 8,000 8,000

8,500 8,000 7,500

5 6

8,000 8,000

7,000 6,500

7 8 9

8,000 8,000 8,000

6,000 5,500 5,000

10

8,000

4,500

Determine which alternative the decision maker should select using an internal rate of return approach. 06.01-PR044 A firm is faced with four investment proposals, A, B, C, and D, having the cash flow profiles shown below. Proposals A and C are mutually exclusive, and Proposal D is contingent on Proposal B being chosen. Currently, $750,000 is available for investment and the firm has stipulated a MARR of 10%.

NCF(A) NCF(B) NCF(C) NCF(D) Initial investment Planning horizon Annual receipts

$400,000 $400,000 $600,000 $300,000 10 years 10 years 10 years 10 years $205,000 $215,000 $260,000 $230,000

Annual disbursements $110,000 $125,000 $120,000 $150,000 Salvage value $50,000 $50,000 $100,000 $50,000 Determine which alternative the decision maker should select. Use the internal rate of return method. 06.01-PR045 Three alternatives are being considered by the management of Brawn Engineering to satisfy an OSHA requirement for safety gates in the machine shop. Each of the gates will completely satisfy the requirement, so no combinations need to be considered. The first costs, operating costs, and salvage values over a 5-year planning horizon are shown below. Using an internal rate of return analysis with a MARR of 20%/yr, determine the preferred gate. End of Year 0 1 2 3 4 5

Gate 1 −$15,000

Gate 2 −$19,000

Gate 3 −$24,000

−$6,500 −$6,500

−$5,600 −$5,600

−$4,000 −$4,000

−$6,500 −$5,600 −$4,000 −$6,500 −$5,600 −$4,000 −$6,500 + $0 −$5,600 + $2,000 −$4,000 + $5,000

06.01-PR046 Five investment alternatives form the mutually exclusive, collectively exhaustive set under consideration. The cash flow profiles for the five projects are given in the table below.

Null

A

B

C

D

Life 10 years 10 years 10 years 10 years 10 years Initial investment 0 $600,000 $800,000 $470,000 $540,000 Salvage value 0 $70,000 $130,000 $65,000 $200,000 Annual revenues Annual expenses

0 $400,000 $600,000 $260,000 $320,000 0 $130,000 $270,000 $70,000 $120,000

Information on each investment alternative was fed into a computer program that calculated the IRR(s) and incremental IRR(s) as shown in the table below. Unfortunately, when the table was printed, one of the cells was overprinted with X’s and was unreadable. As the resident expert on incremental IRR analysis, you have been asked to assist. Incremental Rate of Return of “Row”—“Column” Alternative Null C D A C 39% D A B

36% 44% 40%

18% XXXXX 41%

117% 49%

28%

a. Specify the incremental cash flow profile that must be analyzed to determine the value in the overprinted incremental IRR cell. b. Determine the incremental IRR value that belongs in the overprinted cell. c. If MARR is 37%/yr, which investment alternative is preferred? d. Based on the data in the table, if MARR is 40%, specify whether the present worth of each investment alternative would be positive, negative, or zero when evaluated at MARR? 06.01-PR047 DelRay Foods must purchase a new gumdrop machine. Two machines are available. Machine 7745 has a first cost of $10,000, an estimated life of 10 years, a salvage value of $1,000, and annual operating costs estimated at $0.01 per 1,000 gumdrops. Machine A37Y has a first cost of $8,000, a life of 10 years, and no salvage value. Its annual operating costs will be $300 regardless of the number of gumdrops produced. MARR is 6%/yr,

and 30 million gumdrops are produced each year. Based on an internal rate of return analysis, which machine (if either) should be recommended? 06.01-PR048 Video Solution The engineering team at Manuel’s Manufacturing, Inc., is planning to purchase an Enterprise Resource Planning (ERP) system. The software and installation from Vendor A costs $380,000 initially and is expected to increase revenue $125,000 per year every year. The software and installation from Vendor B costs $280,000 and is expected to increase revenue $95,000 per year. Manuel’s uses a 4-year planning horizon and a 10% per year MARR. Based on an internal rate of return analysis, which ERP system should Manuel purchase?

06.01-PR049 Final Finishing is considering three mutually exclusive alternatives for a new polisher. Each alternative has an expected life of 10 years and no salvage value. Polisher I requires an initial investment of $20,000 and provides annual benefits of $4,465. Polisher II requires an initial investment of $10,000 and provides annual benefits of $1,770. Polisher III requires an initial investment of $15,000 and provides annual benefits of $3,580. MARR is 15%/yr. Based on an internal rate of return analysis, which polisher should be recommended? 06.01-PR050 Xanadu Mining is considering three mutually exclusive alternatives as shown in the table below. MARR is 10%/yr. Based on an internal rate of return analysis, which alternative should be recommended? EOY A001 B002 C003 0 −$210 −$110 −$160 1 $80 $60 $80 2 3 4

$90 $100 $110

$60 $60 $70

$80 $80 $80

06.01-PR051 Video Solution Parker County Community College (PCCC) is trying to determine whether to use no insulation or to use insulation 1-inch thick or 2-inches-thick on its steam pipes. The heat loss from the pipes

without insulation is expected to cost $1.50 per year per foot of pipe. A 1-inch thick insulated covering will eliminate 89% of the loss and will cost $0.40 per foot. A 2-inch thick insulated covering will eliminate 92% of the loss and will cost $0.85 per foot. PCCC Physical Plant Services estimates that there are 250,000 feet of steam pipe on campus. The PCCC Accounting Office requires a 10%/yr return to justify capital expenditures. The insulation has a life expectancy of 10 years. Determine which insulation (if any) should be purchased using an internal rate of return analysis.

06.01-PR052 Several years ago, a man won $27 million in the State Lottery. To pay off the winner, the State planned to make an initial $1 million payment today followed by equal annual payments of $1.3 million at the end of each year for the next 20 years. Just before receiving any money, the man offered to sell the winning ticket back to the State for a one-time immediate payment of $14.4 million. If the State uses a 6%/yr MARR, should the State accept the man’s offer? Use an internal rate of return analysis. 06.01-PR053 Calisto Launch Services is an independent space corporation and has been contracted to develop and launch one of two different satellites. Initial equipment will cost $750,000 for the first satellite and $850,000 for the second. Development will take 5 years at an expected cost of $150,000 per year for the first satellite; $120,000 per year for the second. The same launch vehicle can be used for either satellite and will cost $275,000 at the time of the launch 5 years from now. At the conclusion of the launch, the contracting company will pay Calisto $2,500,000 for either satellite. Calisto is also considering whether they should consider launching both satellites. Because Calisto would have to upgrade its facilities to handle two concurrent projects, the initial costs would rise by $150,000 in addition to the first costs of each satellite. Calisto would need to hire additional engineers and workers, raising the yearly costs to a total of $400,000. An additional compartment would be added to the launch vehicle at an additional cost of $75,000. As an incentive to do both, the contracting company will pay for both launches plus a bonus of $1,000,000. Using an internal rate of return analysis with a MARR of 10%/yr, what should Calisto Launch Services do?

06.01-PR054 Packaging equipment for Xi Cling Wrap costs $60,000 and is expected to result in end-of-year net savings of $23,000 per year for 3 years. The equipment will have a market value of $10,000 after 3 years. The equipment can be leased for $21,000 per year, payable at the beginning of each year. Xi Cling’s MARR is 10%/yr. Based on an internal rate of return analysis, determine if the packaging equipment should be purchased or leased. Section 6.2 External Rate of Return Calculations LEARNING OBJECTIVE 6.2 Compute the External Rate of Return (ERR) for an individual investment and perform incremental comparisons of mutually exclusive investment alternatives to determine the one that maximizes economic worth. Section 6.2.1 ERR Calculations—Single Alternative 06.02-PR001 Consider the following cash flow profile and assume MARR is 10%/yr. EOY NCF EOY NCF 0 −$100 4 −$950 1 $800 5 $700 2 3

−$750 $900

6

−$800

a. Determine the ERR for this project. b. Is this project economically attractive? 06.02-PR002 Video Solution Consider the following cash flow profile and assume MARR is 10%/yr.

EOY NCF EOY NCF 0 −$100 4 $25 1 2 3

$25 $25 $25

5 6

$25 $25

a. Determine the ERR for this project. b. Is this project economically attractive?

06.02-PR003 Consider the following cash flow profile and assume MARR is 10%/yr. EOY NCF EOY NCF 0 1 2

−$100 $25 $25

3

$60

4 5 6

−$30 $60 $25

a. Determine the ERR for this project. b. Is this project economically attractive? 06.02-PR004 Consider the following cash flow profile and assume MARR is 10%/yr. EOY NCF EOY NCF 0 −$101 4 $2 1 2 3

$411 −$558 $253

5 6

$8 −$14

a. Determine the ERR for this project. b. Is this project economically attractive? 06.02-PR005 Consider the following cash flow profile and assume MARR is 10%/yr. EOY NCF EOY NCF 0 −$100 4 $15 1 $15 5 $15 2 3

$15 $15

6

$15

a. Determine the ERR for this project. b. Is this project economically attractive? 06.02-PR006 Consider the following cash flow profile and assume MARR is 10%/yr. EOY NCF EOY NCF 0 1 2

−$100 $25 $200

3

−$100

4 5 6

$250 −$200 −$100

a. Determine the ERR for this project. b. Is this project economically attractive? 06.02-PR007 Nu Things, Inc., is considering an investment in a business venture with the following anticipated cash flow results:

EOY Cash Flow EOY Cash Flow EOY Cash Flow 0 −$70,000 7 14,000 14 7,000 1 2 3

20,000 19,000 18,000

8 9 10

13,000 12,000 11,000

15 16 17

6,000 5,000 4,000

4 5

17,000 16,000

11 12

10,000 9,000

18 19

3,000 2,000

6

15,000

13

8,000

20

1,000

Assume MARR is 20% per year. Based on an external rate of return analysis (1) determine the investment’s worth; (2) state whether or not your results indicate the investment should be undertaken; and (3) state the decision rule you used to arrive at this conclusion. 06.02-PR008 Solve problem 06.01-PR016, except use the external rate of return approach. 06.02-PR009 Solve problem 06.01-PR017, except use the external rate of return approach. 06.02-PR010 Solve problem 06.01-PR018, except use the external rate of return approach. 06.02-PR011 Solve problem 06.01-PR019, except use the external rate of return approach. 06.02-PR012 Quilts R Us (QRU) is considering an investment in a new patterning attachment with the cash flow profile shown in the table below. QRU’s MARR is 13.5%/yr.

EOY Cash Flow EOY Cash Flow 0 1 2

−$1,400 $0 $500

8 9 10

$600 $700 $800

3 4 5

$500 $500 $500

11 12 13

$900 −$1,000 −$2,000

6 7

$0 $500

14 15

−$3,000 $1,400

a. What is the external rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on external rate of return? c. Should QRU invest? 06.02-PR013 Aerotron Electronics is considering the purchase of a water filtration system to assist in circuit board manufacturing. The system costs $40,000. It has an expected life of 7 years at which time its salvage value will be $7,500. Operating and maintenance expenses are estimated to be $2,000 per year. If the filtration system is not purchased, Aerotron Electronics will have to pay Bay City $12,000 per year for water purification. If the system is purchased, no water purification from Bay City will be needed. Aerotron Electronics must borrow half of the purchase price, but they cannot start repaying the loan for 2 years. The bank has agreed to three equal annual payments, with the first payment due at end of year 2. The loan interest rate is 8% compounded annually. Aerotron Electronics’ MARR is 10% compounded annually. a. What is the external rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on external rate of return? c. Should Aerotron Electronics buy the water filtration system?

06.02-PR014 Home Innovations is evaluating a new product design. The estimated receipts and disbursements associated with the new product are shown below. MARR is 10%/yr. End of Year Receipts Disbursements 0 $0 $1,000 1 $600 $300 2 3 4

$600 $700 $700

$300 $300 $300

5

$700

$300

a. What is the external rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on external rate of return? c. Should Home Innovations pursue this new product? 06.02-PR015 Galvanized Products is considering the purchase of a new computer system for its enterprise data management system. The vendor has quoted a purchase price of $100,000. Galvanized Products is planning to borrow one-fourth of the purchase price from a bank at 15% compounded annually. The loan is to be repaid using equal annual payments over a 3-year period. The computer system is expected to last 5 years and has a salvage value of $5,000 at that time. Over the 5-year period, Galvanized Products expects to pay a technician $25,000 per year to maintain the system but will save $55,000 per year through increased efficiencies. Galvanized Products uses a MARR of 18%/yr to evaluate investments. a. What is the external rate of return of this investment? b. What is the decision rule for judging the attractiveness of investments based on external rate of return? c. Should the new computer system be purchased? Section 6.2.2 ERR Calculations—Multiple Alternatives

06.02-PR016 Three alternatives are being considered by the management of Brawn Engineering to satisfy an OSHA requirement for safety gates in the machine shop. Each of the gates will completely satisfy the requirement so no combinations need to be considered. The first costs, operating costs, and salvage values over a 5-year planning horizon are shown below. Using an external rate of return analysis with a MARR of 20%/yr, determine the preferred gate. End of Year

Gate 1

Gate 2

Gate 3

0 1 2

−$15,000 −$6,500 −$6,500

−$19,000 −$5,600 −$5,600

−$24,000 −$4,000 −$4,000

3 4 5

−$6,500 −$5,600 −$4,000 −$6,500 −$5,600 −$4,000 −$6,500 + $0 −$5,600 + $2,000 −$4,000 + $5,000

06.02-PR017 Video Solution Calisto Launch Services is an independent space corporation and has been contracted to develop and launch one of two different satellites. Initial equipment will cost $750,000 for the first satellite and $850,000 for the second. Development will take 5 years at an expected cost of $150,000 per year for the first satellite; $120,000 per year for the second. The same launch vehicle can be used for either satellite and will cost $275,000 at the time of the launch 5 years from now. At the conclusion of the launch, the contracting company will pay Calisto $2,500,000 for either satellite. Calisto is also considering whether they should consider launching both satellites. Because Calisto would have to upgrade its facilities to handle two concurrent projects, the initial costs would rise by $150,000 in addition to the first costs of each satellite. Calisto would need to hire additional engineers and workers, raising the yearly costs to a total of $400,000. An additional compartment would be added to the launch vehicle at an additional cost of $75,000. As an incentive to do both, the contracting company will pay for both launches plus a bonus of $1,000,000. Using an external rate of return analysis with a MARR of 10%/yr, what should Calisto Launch Services do?

06.02-PR018 Tempura, Inc., is considering two projects. Project A requires an investment of $50,000. Estimated annual receipts for 20 years are $20,000; estimated annual costs are $12,500. An alternative project, B, requires an investment of $75,000, has annual receipts for 20 years of $28,000, and annual costs of $18,000. Assume both projects have a zero salvage value and that MARR is 12%/yr. Based on an external rate of return analysis, which project should be recommended? 06.02-PR019 Solve problem 06.01-PR035, except use the external rate of return approach. 06.02-PR020 Solve problem 06.01-PR036, except use the external rate of return approach. 06.02-PR021 Solve problem 06.01-PR043, except use the external rate of return approach. 06.02-PR022 Solve problem 06.01-PR044, except use the external rate of return approach.

Chapter 6 Summary and Study Guide Summary 6.1: Internal Rate of Return Calculations

Learning Objective 6.1: Compute the Internal Rate of Return (IRR) for an individual investment and perform incremental comparisons of mutually exclusive investment alternatives to determine the one that maximizes economic worth. (Section 6.1) The IRR is a popular method to measure economic worth, especially for personal investment decision making. Unlike present, future, and annual worth methods, the IRR is not a ranking method, but rather incremental analysis must be performed to select from a set of mutually exclusive alternatives. The IRR method can return multiple solutions. Mathematically, investment j’s internal rate of return, denoted i , satisfies the following equality: ∗ j

n

n−t ∗

(6.1)

0 = ∑ Ajt  (1 + i ) j

t=0

In words, the internal rate of return is the interest rate that makes the future worth of an investment equal 0. 6.2: External Rate of Return Calculations

Learning Objective 6.2: Compute the External Rate of Return (ERR) for an individual investment and perform incremental comparisons of mutually exclusive investment alternatives to determine the one that maximizes economic worth. (Section 6.2) Like the IRR, the ERR is a popular method to measure economic worth. It is not a ranking method, but rather incremental analysis must be performed to select from a set of mutually exclusive alternatives. The ERR is “external” to make the point that the recovered funds are not reinvested at a rate equal to the IRR, and the return earned on recovered capital is external to the investment in question. The ERR method returns a single solution. The ERR is based on the following equation: n

∑ Rt  (1 + r)

n n−t

t=0

′

= ∑ Ct  (1 + i )

n−t

(6.2)

t=0

where Rt denotes the positive-valued cash flows in a cash flow series and Ct denotes the negative-valued cash flows in a cash flow series; r is the reinvestment rate, which we call the MARR; and i′ is the external rate of return.

Important Terms and Concepts Internal Rate of Return (IRR) The interest rate that makes the present worth, the future worth, and the annual worth equal to 0. Also referred to as the discounted cash flow rate of return, the cash flow rate of return, the rate of return (ROR), the return on investment (ROI), and the true rate of return. External rate of return (ERR) The interest rate that makes the absolute value of the future worth of negative-valued cash flows equal to the future worth of positive-valued cash flows that are reinvested at the MARR.

Chapter 6 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 6.1: Internal Rate of Return Calculations LO 6.1: Compute the Internal Rate of Return (IRR) for an individual investment and perform incremental comparisons of mutually exclusive investment alternatives to determine the one that maximizes economic worth. Video Lesson: Rate of Return Video Lesson Notes: Rate of Return Excel Video Lesson: RATE Financial Function Excel Video Lesson Spreadsheet: RATE Financial Function Excel Video Lesson: SOLVER Tool Excel Video Lesson Spreadsheet: SOLVER Tool Video Example 6.4: IRR Analysis with Mutually Exclusive Alternatives Video Solution: 06.01-PR013 Video Solution: 06.01-PR020 Video Solution: 06.01-PR023 Video Solution: 06.01-PR028 Video Solution: 06.01-PR048 Video Solution: 06.01-PR051 6.2: External Rate of Return Calculations LO 6.2: Compute the External Rate of Return (ERR) for an individual investment and perform incremental comparisons of mutually exclusive

investment alternatives to determine the one that maximizes economic worth. Excel Video Lesson: FV Financial Function Excel Video Lesson Spreadsheet: FV Financial Function Excel Video Lesson: MIRR Financial Function Excel Video Lesson Spreadsheet: MIRR Financial Function Excel Video Lesson: IRR Financial Function Excel Video Lesson Spreadsheet: IRR Financial Function Excel Video Lesson: GOAL SEEK Tool Excel Video Lesson Spreadsheet: GOAL SEEK Tool Video Example 6.6: Computing the External Rate of Return for a Single Alternative Video Solution: 06.02-PR002 Video Solution: 06.02-PR017 These chapter-level resources will help you with your overall understanding of the content in this chapter. Appendix A: Time Value of Money Factors Appendix B: Engineering Economic Equations Flashcards: Chapter 06 Excel Utility: TVM Factors: Table Calculator Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values Excel Utility: Monthly Payment Sensitivity Excel Utility: TVM Factors: Discrete Compounding Excel Utility: TVM Factors: Geometric Series Future Worth

Excel Utility: TVM Factors: Geometric Series Present Worth Excel Data Files: Chapter 06

CHAPTER 6 Rate of Return LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 6.1 Compute the Internal Rate of Return (IRR) for an individual investment and perform incremental comparisons of mutually exclusive investment alternatives to determine the one that maximizes economic worth. (Section 6.1) 6.2 Compute the External Rate of Return (ERR) for an individual investment and perform incremental comparisons of mutually exclusive investment alternatives to determine the one that maximizes economic worth. (Section 6.2)

Engineering Economics in Practice Motorola Solutions Motorola was founded in Illinois in 1928. From car radios, to televisions, to twoway radios, to automatic identification technologies, Motorola evolved into a communications company that today is designing, building, marketing, and selling products, services, and applications globally. On January 4, 2011, Motorola’s mobile devices and home businesses were spun off as a separate company, Motorola Mobility Holdings, Inc., subsequently were sold to Google in 2012, and acquired by Lenovo in 2014. The remaining business was renamed Motorola Solutions, Inc. Since spinning off its mobile phone in 2011, the company has become a leading global provider of mission-critical communication infrastructure, devices, accessories, software and services for use in the government, public safety, and commercial sectors to improve their operations through increased effectiveness, efficiency, and safety of their mobile workforces. Motorola Solutions offers key solutions in land mobile radio, public safety, managed & support services, video surveillance & analytics, and public safety command centers. If you see someone at an airport or police personnel using a two-way radio, it probably is a Motorola Solutions product. If you receive a delivery from FedEx, information regarding delivery very likely was communicated using Motorola Solutions products. In addition, there are numerous behind-the-scenes applications of Motorola’s products. Motorola Solutions, Inc., had annual sales in 2017 of $6.4 billion. In 2018 it had approximately 16,000 employees in 60 countries. During 2018, it had more than 100,000 customers in 100 countries. The company is committed to innovation and in 2017 it spent $568 million in Research & Development and has 5,500+ patents granted or pending. A technology company, Motorola Solutions is highly disciplined in managing its capital. Discounted cash flow methods are used throughout the company to ensure that shareholders receive attractive returns on their invested capital. The required rate of return on particular capital investment is based on weighted average cost of capital calculations, as well as considerations of the type of investment, its duration, and risks involved in the particular investment. The required return on the investment can change from year to year due to economy dynamics, but the need to perform present worth and internal rate of return calculations never stops. Discussion Questions

1. What factors would determine the rate of return values that Motorola Solutions would use for its investment decisions? 2. Motorola Solutions is a high-tech company with significant new product development. What unique considerations would such a company have with respect to expected returns on its investments as opposed to a “lower-tech” company? 3. Motorola Solutions is a global company with operations around the world. What complexity does this introduce into its capital investment decisions? 4. What financial gains might the parent company Motorola have sought as it spun off Motorola Mobility (sold to Google in 2012)?

Introduction Although this chapter does not have worth in its title, rates of return are also measures of economic worth. Instead of measuring economic worth in dollars, here we measure it in percentages. Among the DCF measures of economic worth, rates of return are probably the second most popular among corporations, ranked just behind present worth. For personal investment decision making, however, rates of return are used more frequently than present worth. No doubt the popularity of this analysis method is due to investors’ familiarity with interest rates and the ease with which investment returns can be compared with costs of capital for investment. Like the other DCF methods we have studied so far, rates of return can be used to compare mutually exclusive alternatives and choose the one having the greatest economic worth. However, they tend to be used more frequently in industry as supplements to one of the traditional “worth” methods—present, future, or annual worth.

Systematic Economic Analysis Technique 1. Identify the investment alternatives 2. Define the planning horizon 3. Specify the discount rate 4. Estimate the cash flows 5. Compare the alternatives 6. Perform supplementary analyses 7. Select the preferred investment Unlike present, future, and annual worths, rates of return are not ranking methods. When used to choose from among mutually exclusive alternatives on the basis of monetary considerations, incremental analysis is required when using rates of return. When performed correctly, however, the incremental analysis will result in the same investment alternative being recommended as when using one of the ranking methods. Although many different rates of return exist, here we consider only two, both of which are discounted cash flow methods: internal rate of return and external rate of return. In this chapter, we learn how to compare mutually exclusive investment alternatives using rate of return methods. In the case of the internal rate of return method, we learn that multiple solutions can occur, and we figure out how to deal with them.

6.1 Internal Rate of Return Calculations LEARNING OBJECTIVE Compute the Internal Rate of Return (IRR) for an individual investment and perform incremental comparisons of mutually exclusive investment alternatives to determine the one that maximizes economic worth. Video Lesson: Rate of Return The internal rate of return is also referred to as the discounted cash flow rate of return, the cash flow rate of return, the rate of return (ROR), the return on investment (ROI), and the true rate of return. The more common name, however, is internal rate

of return (IRR). Mathematically, investment j’s internal rate of return, denoted i , satisfies the following equality: ∗ j

n

n−t ∗

(6.1)

0 = ∑ Ajt  (1 + i ) j

t=0

Internal Rate of Return (IRR) The interest rate that makes the present worth, the future worth, and the annual worth equal to 0. Also referred to as the discounted cash flow rate of return, the cash flow rate of return, the rate of return (ROR), the return on investment (ROI), and the true rate of return. In words, the internal rate of return is the interest rate that makes the future worth of an investment equal 0. Likewise, it is the interest rate that equates the present worth and annual worth to 0. If the internal rate of return is at least equal to the MARR, the investment should be made.

6.1.1 IRR Calculations—Single Alternative As with other DCF methods, when applied to a single alternative, IRR is used to determine whether the investment opportunity is preferred to the do-nothing alternative.

EXAMPLE 6.1 The Surface Mount Placement Machine Investment Recall the manager of an electronics manufacturing plant who was asked to approve the purchase of a surface mount placement (SMP) machine having an initial cost of $500,000 in order to reduce annual operating and maintenance costs by $92,500 per year. At the end of the 10-year planning horizon, it was estimated that the SMP machine would be worth $50,000. Using a 10% MARR and internal rate of return analysis, should the investment be made? Key Data Given The cash flows shown in Figure 6.1; MARR = 10%; planning horizon = 10 years Find The IRR of the investment. Is this investment recommended? Solution Setting the future worth for the investment equal to 0 gives ∗

0 = −$500,000(F |P  i %,10) + $92,500(F |A i

∗

%, 10) + $50,000

Recalling the formulas used to compute capital recovery cost, the following annual worth formulation can be solved for the internal rate of return: ∗

0 = ($500,000 − $50,000)(A|P  i ,10) + $50,000i

∗

− $92,500

For i = 12%, $450,000(A|P  12%,10) + $50,000(0.12) − $92,500 = −$6,850

FIGURE 6.1 CFD for Example 6.1 For i = 15%, $450,000(A|P  15%,10) + $50,000(0.15) − $92,500 = $4,685

Interpolating for i* gives 13.78%. Excel® Solution Alternately, one can use the Excel® RATE worksheet function to solve for i* in this example: r* =RATE(10,92500,−500000,50000)      = 13.8%

Because i* > 10%, the investment is recommended. Excel® Video Lesson: RATE Financial Function

Example 6.1 illustrates one of the attractions of the IRR method. We can say “Invest $500,000 to obtain a 13.8% return on your investment.” This is often more informative or appealing than a dollar figure such as present or future worth.

6.1.2 IRR Calculations—Multiple Roots

Descartes’ Rule of Signs Descartes’ Rule of Signs, as applied to internal rate of return analysis, indicates there will be at most as many positive rates of return as there are sign changes in the cash flow profile. For example, one change of sign—usually one or more negative cash flows followed by one or more positive cash flows—will have at most one positive IRR value. Three sign changes, for example, will have at most three positive IRR values. Most cash flow profiles encountered in practice, however, will have a unique internal rate of return, despite multiple changes in sign. Example 6.2 illustrates a cash flow profile with multiple internal rates of return.

EXAMPLE 6.2 Multiple Roots To illustrate a cash flow profile having multiple roots, consider the data given in Table 6.1. The future worth of the cash flow series will be 0 using a 20%, 40%, or 50% interest rate. FW1 (20%) =−$4,000(1.2) FW2 (40%) =−$4,000(1.4) FW3 (50%) =−$4,000(1.5)

3

3

3

+$16,400(1.2) +$16,400(1.4) +$16,400(1.5)

2

2

2

−$22,320(1.2) +$10,080 = 0 −$22,320(1.4) +$10,080 = 0 −$22,320(1.5) +$10,080 = 0

TABLE 6.1 Cash Flow Profile EOY 0

CF − $4,000 

1

$16,400 

2 3

− $22,320  $10,080 

A plot of the future worth for this example is given in Figure 6.2. It is simply a plot of the future worth evaluated at values of i from 10% to 60% in increments of 2%.

FIGURE 6.2 Plot of Future Worth for Example 6.2

Norstrom’s Criterion In addition to Descartes’ rule of signs, Norstrom’s Criterion can be applied to determine if there is at most one real positive internal rate of return. If the cumulative cash flow series begins with a negative value and changes only once to a positive value, then there exists at most a single positive internal rate of return. Example 6.3 illustrates a cash flow profile having multiple sign changes, yet only a single positive IRR value. This example also illustrates Norstrom’s criterion that if the cumulative cash flow series begins with a negative number and changes only once to a positivevalued series, then there exists a unique positive rate of return. In the example, interestingly, if the additional investment is $200,000 instead of $150,000, Norstrom’s criterion is not met, but a single positive internal rate of return still exists. Norstrom’s criterion is a sufficient, not a necessary, condition for at most a single real positive rate of return to exist.

EXAMPLE 6.3 Rate of Return with Multiple Sign Changes and a Single Root Julian Stewart invested $250,000 in a limited partnership to drill for natural gas. His investment yielded annual returns of $45,000 the first year, followed by annual increases of $10,000 until the 6th year, at which time an additional $150,000 had to be invested for deeper drilling. Starting in the seventh year, following the supplemental investment, the annual returns decrease by $10,000 annually from $85,000 to $5,000. What is the IRR of Julian’s investment when future worth is maximized? Key Data Given The cash flows for Julian’s investment are shown in column B of the spreadsheet in Figure 6.3. Find IRR when FW is at a maximum. Solution The cash flow profile in column B of Figure 6.3 shows three sign changes, so according to Descartes’ rule of signs there will be at most three positive IRR values. The cumulative cash flow series in column C of Figure 6.3 begins with a negative value and changes only once to a positive value. Therefore, according to Norstrom’s criterion, there exists at most a single positive IRR. From the plot of future worth shown in Figure 6.3, it is evident that a single root exists, at i* = 19.12%, and using the Excel® SOLVER tool, FW is maximized at $52,700 when MARR equals 8.5469%. For the investment IRR = 19.12%. Excel® Video Lesson: SOLVER Tool

FIGURE 6.3 IRR for a Natural Gas Investment Excel® Data File

6.1.3 IRR Calculations—Multiple Alternatives In computing the internal rate of return for each of several investments, we continue to search for the interest rate that equates the economic worth to 0. Mathematically, from Equation 6.1, the internal rate of return for alternative j, denoted i , satisfies the ∗ j

following equality: n

n−t ∗

0 = ∑ Ajt  (1 + i ) j

t=0

Rate of return methods must be used incrementally when comparing mutually exclusive investment alternatives. That is, let the investment with the lowest initial cost be Alternative 1 (or the base alternative), let the investment with the next lowest cost be Alternative 2, and so on. Then, analyze each alternative in order of increasing initial cost. For alternatives 2 and beyond, rate of return is determined for the additional increment of investment (above the base alternative), rather than the entire cost. As long as each successive alternative’s rate of return exceeds the MARR, it is preferred to the previous (lower cost) investment. There is value in investing the

additional incremental capital for a subsequent alternative when its rate of return is equal to or exceeds the MARR. The preferred alternative will not necessarily have the greatest internal rate of return. Whereas the present, future, and annual worths are compared to 0, the comparison with rate of return analyses is with the MARR. The following example illustrates why incremental analysis is used when performing rate of return comparisons of investment alternatives.

EXAMPLE 6.4 IRR Analysis with Mutually Exclusive Alternatives Video Example Recall the example of the theme park in Florida that is considering two designs for a new ride called the Scream Machine. The first design alternative (A) requires a $300,000 investment and will produce net annual after-tax revenue of $55,000 over the 10-year planning horizon; the second alternative (B) requires a $450,000 investment and will produce net annual after-tax revenue of $80,000 annually. Both alternatives are expected to have negligible salvage values after the 10-year planning horizon. Based on a 10% MARR and an IRR comparison, which design (if either) should be chosen? Key Data Given The cash flows outlined in Figure 6.4; MARR = 10%; planning horizon = 10 years Find Recommended investment using incremental IRR analysis. Solution First, we examine each alternative looking for the smallest initial investment, in this case, Design A. Next, we compute the IRR for Design A. Recall that the computed IRR will be compared with the MARR. i

∗ A

=RATE (10,−55000,300000) = 12.87% > MARR = 10%

Because Design A is justified (because its return exceeds the MARR), the next step is to compute the internal rate of return on the $150,000 incremental investment in Design B (over and above the initial investment in Design A): i

∗ B−A

=RATE (10,−25000,150000) = 10.56% > MARR = 10%

FIGURE 6.4 CFDs for Example 6.4 Because the internal rate of return for the incremental investment in Design B is greater than the MARR (albeit only slightly), this incremental investment needed to acquire Design B is justified. Therefore, the overall internal rate of return for Design B can be computed as follows: i

∗ B

=RATE (10,−80000,450000) = 12.11%

In summary, Design B is the preferred alternative. The return on the first $300,000 is 12.87%, and the return on the last $150,000 is 10.56% for an overall return of 12.11%. Exploring the Solution Note that if the IRR(s) had been calculated for each design independently and then ranked, the wrong investment decision would have been made: Design A would have been ranked higher with an IRR of 12.87% as compared to Design B with a lower IRR of 12.11%. An incremental investment analysis is required when comparing mutually exclusive alternatives using IRR.

Example 6.4 illustrates Principle #6: Continue to invest as long as each additional increment of investment yields a return that is greater than the investor’s TVOM. When comparing mutually exclusive investment alternatives, each of which has well behaved cash flows, it is quite likely that multiple roots will occur. The reason for this is, except for the initial step, incremental analyses are performed. When one wellbehaved cash flow series is subtracted from another, there is no guarantee that the difference in the cash flow series will be well behaved, as the following example illustrates.

EXAMPLE 6.5 IRR Analysis with Regular Cash Flow Series for Alternative Investments, but Irregular Incremental Analysis Cash Flows Two mutually exclusive investment alternatives are being considered. The MARR is 12%. Alternative 1 requires an initial investment of $100,000; it returns $33,600 in year 1, $72,320 in year 2, and $39,920 in year 3. It has a regular cash flow, with a single change of sign and an IRR of 20.8122%. Alternative 2 requires an initial investment of $104,000 and has equal annual returns of $50,000 over the three years. It also has a regular cash flow, with a single change of sign. Which alternative is preferred? Key Data Given The cash flow profiles for the alternatives are given in Table 6.2. Find FW of the incremental investment.

TABLE 6.2 Data for Example 6.5 EOY CF(1) CF(2) 0 1 2

CF(2-1)

−$100,000.00 −$104,000.00 −$4,000.00  $33,600.00 $50,000.00 $16,400.00  $72,320.00 $50,000.00 −$22,320.00 

3 IRR = IRR = IRR =

$39,920.00 20.8122%

$50,000.00 $10,080.00  20.0000%  40.0000%  50.0000% 

Solution To determine which of the two alternatives is preferred, they are ordered by increasing investment. Because Alternative 1 has the smaller initial investment, it is considered first; it is justified, because IRR1 = 20.8122% > MARR > 12%. Next considered is Alternative 2. The incremental investment required to move from Alternative 1 to Alternative 2 is $4,000 initially, followed by incremental returns of $16,400, −$22,320, and $10,080. Note that this is an irregular

incremental cash flow, with three changes of sign. In fact, this incremental cash flow is identical to the cash flow of Example 6.2, analyzed previously, with multiple IRR values of 20%, 40%, and 50%. The future worth plot is seen in Figure 6.5; it is identical to Figure 6.2. Using Figure 6.5, we can conclude that if MARR < 20% or if 40% < MARR < 50%, the incremental investment yields a positive future worth and therefore Alternative 2 is preferred to Alternative 1. If 20% < MARR < 40% or if 50% < MARR, the future worth of the incremental investment is negative and Alternative 1 is preferred. Because MARR is 12%, Alternative 2 is selected.

FIGURE 6.5 Plot of Future Worth of Incremental Investment CF(2-1) for Example 6.5

Concept Check 06.01-CC001 Identify the one false statement about the internal rate of return. a. The IRR is also known as the discounted cash flow rate of return, the rate of return, and the return on investment b. The IRR is a ranking method c. The IRR will result in the same investment alternative being recommended as the PW, AW, and FW d. The IRR is the interest rate that makes the PW, AW, and FW equal to zero

Concept Check 06.01-CC002 Consider the following cash flows over a 5-year period: −$100,000 @ t = 0; $35,000 @ t = 1; $45,000 @ t = 2; −$40,000 @ t = 3; $50,000 @ t = 4; and $60,000 @ t = 5. Using Descartes’ rule of signs, how many positive rates of return are there? a. Exactly one b. Exactly two c. Exactly three d. At most two e. At most three

Concept Check 06.01-CC003 The IRR method must be used incrementally when comparing mutually exclusive investment alternatives. True or False?

6.2 External Rate of Return Calculations LEARNING OBJECTIVE Compute the External Rate of Return (ERR) for an individual investment and perform incremental comparisons of mutually exclusive investment alternatives to determine the one that maximizes economic worth. Because of the possibilities of multiple roots when using the internal rate of return method, an alternative approach called the external rate of return (ERR) method was developed.1 The approach is based on the following equation: n

∑ Rt  (1 + r) t=0

n n−t

′

= ∑ Ct  (1 + i ) t=0

n−t

(6.2)

External Rate of Return (ERR) The interest rate that makes the absolute value of the future worth of negative-valued cash flows equal to the future worth of positive-valued cash flows that are reinvested at the MARR. where Rt denotes the positive-valued cash flows in a cash flow series and Ct denotes the absolute value of the negative-valued cash flows in a cash flow series; r is the reinvestment rate, which we call the MARR; and i′ is the external rate of return. The reasoning in creating the ERR went like this: Any funds remaining in the investment pool are assumed to earn returns equal to the MARR. Therefore, money recovered from an investment should also earn returns equal to the MARR. The interest rate that makes the future worth of negative-valued cash flows equal to the future worth of positive-valued cash flows that are reinvested at the MARR is the external rate of return. The word external makes the point that recovered funds are not reinvested at a rate equal to the internal rate of return, and the return earned on recovered capital is external to the investment in question. Two features of the ERR are important: There exists a unique solution (no multiple rates of return), and the ERR is always between the IRR and the MARR. Hence, if IRR > MARR, then IRR > ERR > MARR; likewise, if IRR < MARR, then IRR < ERR < MARR; and if IRR = MARR, then IRR = ERR = MARR.

6.2.1 ERR Calculations—Single Alternative We begin by considering ERR for a single-alternative, the now-familiar SMP machine.

EXAMPLE 6.6 Computing the External Rate of Return for a Single Alternative Video Example The investment of $500,000 in an SMP machine is expected to reduce manufacturing costs by $92,500 per year. At the end of the 10-year planning horizon, it is expected the SMP machine will be worth $50,000. Using a 10% MARR and an external rate of return analysis, should the investment be made? Key Data Given The cash flows outlined in Figure 6.1; MARR = 10%; planning horizon = 10 years Find The ERR of the investment. Is this investment recommended? Solution The only negative-valued cash flow is the initial investment. Therefore, Equation 6.2 becomes $500,000(F |P  i', 10) $500,000(1 + i') (1 + i')

10

10

i'

=

$92,500(F |A 10%,10) +$50,000

=

$92,500(F |A 10%, 10) +$50,000

=

[$92,500(15.93742) +$50,000]/$500,000 = 3.048423

=

11.79117%

or Excel® Solution i' =RATE(10,,−500000,FV(10%,10,−92500)+50000)     = 11.79117%

Excel® Video Lesson: The FV Function Because i′> MARR, the machine is justified economically. The ERR can also be obtained using the Excel® MIRR worksheet function, only if the first cash flow is negative and all subsequent cash flows are positive. The modified internal rate of return [MIRR] as a measure of economic worth is not

covered in this text. The syntax for the Excel® MIRR worksheet function is =MIRR(values,finance_rate,reinvest_rate); for purposes of computing the ERR, the finance rate is ignored. As shown in Figure 6.6, using the MIRR worksheet function, we obtain the exact result for the ERR: 11.79118%. Recall, for this example, IRR = 13.8%. Excel® Video Lesson: The MIRR Financial Function The IRR function may also be used to determine the ERR, even if there are negative cash flows in other than year 0, as follows. 1. Define two new cash flow series: CF(+) and CF(−). Using the Excel® IF function, separate the original cash flow roots into two series: {Rt, t = 0, …, 10} and {Ct, t = 0, …, 10}. 2. Next, develop a new cash flow series {Ct, t = 0, …, 10}, which, except for t = 10, is the negative of {Ct}. 3. Then, compute the future worth of the positive-valued cash flows and add it to the negative cash flow (although in this example none exists) in the last year of the planning horizon. 4. The resulting series (E3:E13 in Figure 6.6) contains zeros, the negativevalued cash flow from the original cash flow series, and the future worth of all positive-valued cash flows; the latter (shown in cell E13) is obtained using the MARR of 10%. 5. Recalling the definition of the external rate of return (the interest rate that makes the absolute value of the future worth of the negative-valued cash flows equal to the future worth of the positive-valued cash flows, based on the MARR), use the Excel® IRR worksheet function to solve for the ERR, as indicated in cell E14. ERR = 11.79118%. Excel® Video Lesson: The IRR Financial Function

FIGURE 6.6 ERR Solution to Example 6.6 Excel® Data File

EXAMPLE 6.7 Using the ERR to Resolve a Multiple Root Problem Recall the data in Example 6.2 that led to multiple solutions to the IRR formulation. The following four cash flows occurred at t = 0, 1, 2, and 3: − $4,000, $16,400,−$22,320, and $10,080, respectively. Using a MARR of 12%, find the ERR. Solution From Equation 6.2, $4,000(F |P  i', 3) + $22,320(F |P  i', 1) = $16,400(F |P  12%,2) + $10,080

To determine the value of i′ that satisfies the equality, trial-and-error methods can be used, followed by interpolation. Due to the multiple negative signs, the Excel® MIRR function cannot be used to obtain the ERR value. However, the Excel® SOLVER and GOAL SEEK tools can be used. Figure 6.7 provides the SOLVER setup and solution. Both SOLVER and GOAL SEEK yield a value of 12.0911% for the ERR obtained. Excel® Video Lesson: GOAL SEEK Tool Shown in Table 6.3 are ERR values for various values of the MARR. Notice, for MARR < 20%, ERR > MARR; for 20% < MARR < 40%, ERR < MARR; for 40% < MARR < 50%, ERR > MARR; and for MARR > 50%, ERR < MARR.

FIGURE 6.7 Excel® SOLVER Setup and Solution to Example 6.7 Excel® Data File

TABLE 6.3 ERR Solutions for Various MARR Values in Example 6.7 MARR ERR MARR ERR MARR ERR  0%  0.4654% 20% 20.0000% 40% 40.0000%  2%  2.3768% 22% 21.9900% 42% 42.0030%  4%  6%  8%

 4.2999%  6.2338%  8.1775%

24% 26% 28%

23.9837% 25.9805% 27.9799%

44% 46% 48%

44.0049% 46.0052% 48.0037%

10% 12%

10.1302% 12.0911%

30% 32%

29.9812% 31.9840%

50% 52%

50.0000% 51.9939%

14% 16%

14.0592% 16.0399%

34% 36%

33.9877% 35.9919%

54% 56%

53.9850% 55.9732%

18%

18.0144%

38%

37.9962%

58%

57.9581%

6.2.2 ERR Calculations—Multiple Alternatives When mutually exclusive investment alternatives exist, the external rate of return method can be used to select the economically preferred one. However, like the internal rate of return method, it must be applied incrementally. Although there are multiple alternatives and each has its own external rate of return (i′j), there is a common reinvestment rate (r). Mathematically, for alternative j, the following equality must hold: n

∑ Rjt  (1 + r) t=0

n n−t

n−t ′

(6.3)

= ∑ Cjt  (1 + i ) j

t=0

where Rjt denotes the positive-valued cash flows and Cjt denotes the absolute value of the negative-valued cash flows in the cash flow series for alternative j.

EXAMPLE 6.8 Using the ERR to Compare Two Alternatives for the Scream Machine Recall the example involving two design alternatives for a new ride called the Scream Machine, the first of which requires an initial investment of $300,000 and yields net annual revenue of $55,000 over a 10-year planning period; the second requires an initial investment of $450,000 and yields a net annual revenue of $80,000 over a 10-year period. Using the ERR, and based on a 10% MARR, which alternative is preferred? Key Data Given The cash flows outlined in Figure 6.4; MARR = 10%; planning horizon = 10 years Find ERR for each alternative and the incremental investment. Solution The following steps are taken: 1. Compute the ERR for the smaller initial investment alternative, Design A: $300,000(1 + i (1 + i i

′ A

′ A

)

10

′ A

)

10

=

$55,000(F |A 10%,10)

=

$55,000(15.93742)/$300,000

=

11.31814% > MARR = 10%

or i i

′ A ′ A

 =RATE(10,,−300000,FV(10%,10,−55000))  = 11.31814% > MARR = 10%

Design Alternative A is justified economically. 2. Compute the ERR for the $150,000 incremental investment to move from Design A to Design B:

10

$150,000(1 + i

′ B−A

)

=

$25,000(F |A 10%,10)

=

$25,000(15.9374)/$150,000

=

10.26219% > MARR = 10%

10

(1 + i

i

′ B−A

)

′ B−A

or i

′ B−A

 =RATE(10,,−150000,FV(10%,10,−25000))

          = 10.26219% > MARR = 10%

The incremental investment to accept Design B is justified economically. 3. Design Alternative B will have an overall external rate of return of $450,000(1 + i (1 + i i

′ B

)

′ B

)

10

10

′ B

=

$80,000(F |A 10%,10)

=

$80,000(15.9374)/$450,000

=

10.97611%

or i

′ B

 =RATE(10,,−450000,FV(10%,10,−80000))

     = 10.97611%

EXAMPLE 6.9 ERR Analysis with Regular Cash Flow Series for Alternative Investments, but Irregular Incremental Analysis Cash Flows Recall Example 6.5 where two mutually exclusive alternative investments are considered. The MARR is 12%. The cash flow profiles for the alternatives are given in Table 6.2. Alternative 1 requires an initial investment of $100,000; it returns $33,600 in year 1, $72,320 in year 2, and $39,920 in year 3. It has a regular cash flow, with a single change of sign and an ERR of 17.7031%. Alternative 2 requires an initial investment of $104,000 and has equal annual returns of $50,000 over the 3 years. It also has a regular cash flow, with a single change of sign. Using ERR, which alternative is preferred? Solution To determine which of the two alternatives is preferred, they are ordered by increasing investment. Because Alternative 1 has the smaller initial investment, it is considered first. It is justified, because, as shown in Figure 6.8, ERR1 = 17.7031% > MARR > 12%. Next considered is Alternative 2. The incremental investment required to move from Alternative 1 to Alternative 2 is $4,000 initially, followed by $16,400, − $22,320, and $10,080. Note that this is an irregular incremental cash flow, with three changes of sign. Of course, it is identical to the incremental cash flow of Example 6.5, analyzed previously, which had multiple IRR values of 20%, 40%, and 50%. Likewise, it is identical to the incremental cash flow of Example 6.7, analyzed previously, which had a unique ERR of 12.0911%. Because ERR2−1 = 12.0911% > MARR = 12%, Alternative 2 is selected.

FIGURE 6.8 Data and ERR Analysis for Example 6.9 Excel® Data File

Concept Check 06.02-CC001 Differences between the ERR and the IRR include the following: a. The ERR will yield a unique solution (no multiple rates of return as in the IRR) b. In the ERR, funds recovered from the investment are assumed to earn returns equal to the MARR c. The ERR is always a value somewhere between the IRR and the MARR d. All of the above

Concept Check 06.02-CC002 The ERR can be used to compare multiple alternatives by ranking the ERR of each mutually exclusive alternative. True or False?

Note 1. White, J. A., K. E. Case, and M. H. Agee, “Rate of Return: An Explicit Reinvestment Rate Approach,” Proceedings of the 1976 AIIE Conference, American Institute of Industrial Engineers, Norcross, GA, 1976.

CHAPTER 7 Replacement Analysis

Chapter 7 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced e-text and WileyPLUS

FE-Like Problems 07-FE001 A company owns a 6-year-old gear hobber that has a book value of $60,000. The present market value of the hobber is $80,000. A new gear hobber can be purchased for $450,000. Using an insider’s point of view, what is the net first cost of purchasing the new gear hobber? a. $310,000 b. $370,000 c. $390,000 d. $450,000 07-FE002 A company owns a 6-year-old gear hobber that has a book value of $60,000. The present market value of the hobber is $80,000. A new gear hobber can be purchased for $450,000. Using an outsider’s point of view, what is the net first cost of purchasing the new gear hobber? a. $310,000 b. $370,000 c. $390,000 d. $450,000 Correct or Incorrect? Clear

  Check Answer

07-FE003 In evaluating a piece of equipment for its optimum replacement interval, the following table of equivalent uniform annual costs is obtained. What is the optimum replacement interval for the equipment?

n EUAC(n) 1 1582.00 2 1550.00 3 1575.00 4 1580.00 a. 1 year b. 2 years c. 3 years d. 4 years 07-FE004 A radiology clinic is considering buying a new $700,000 x-ray machine, which will have no salvage value after installation because the cost of removal will be approximately equal to its sales value. Maintenance is estimated at $24,000 per year as long as the machine is owned. After 10 years the x-ray source will be depleted and the machine must be scrapped. Which of the following represents the most economic life of this x-ray machine? a. One year, because it will have no salvage after installation b. Five years, because the maintenance costs are constant c. Ten years, because maintenance costs don't increase d. Cannot be determined from the information given Correct or Incorrect? Clear

  Check Answer

07-FE005 analysis?

Which of the following is not an approach to replacement

a. Cash flow approach b. Insider viewpoint c. Outsider viewpoint d. Supply chain approach

07-FE006 A company owns a 5-year-old turret lathe that has a book value of $20,000. The present market value of the lathe is $16,000. A new turret lathe can be purchased for $45,000. Using an outsider’s point of view, what is the first cost of keeping the old lathe? a. $29,000 b. $45,000 c. $20,000 d. $16,000 Correct or Incorrect? Clear

  Check Answer

07-FE007 A company owns a 5-year-old turret lathe that has a book value of $20,000. The present market value of the lathe is $16,000. A new turret lathe can be purchased for $45,000. Using an insider’s point of view, what is the first cost of the new lathe? a. $29,000 b. $45,000 c. $25,000 d. $16,000 07-FE008 What two cost categories form the trade off that leads to an optimum replacement interval? a. Direct costs and indirect costs b. Insider costs and outsider costs c. Operating & maintenance costs and capital recovery costs d. Sunk costs and opportunity costs Correct or Incorrect? Clear

  Check Answer

07-FE009 Increasing the magnitude of the initial investment tends to __________ the optimum replacement interval. a. decrease b. increase c. reverse d. not affect

Problems Section 7.1 Cash Flow and Opportunity Cost Approaches LEARNING OBJECTIVE 7.1 Perform a replacement analysis using a cash flow or opportunity cost approach. 07.01-PR001 Identify something you own, perhaps even something you still use regularly. a. Give a list of at least 6 potential reasons why you might consider replacing the identified item. b. Identify at least two possible “challengers” that you might consider to replace the item, and state for each challenger the main reason you would consider it. 07.01-PR002 Give two examples each of (i) functional obsolescence, (ii) technological obsolescence, and (iii) economic obsolescence for items that you or your family own. 07.01-PR003 Ten reasons why companies use equipment long after replacements would be justified economically are given in the text. In many cases, these reasons do not apply just to companies; rather, they apply to us as individuals. Give specific examples of at least 3 of the 10 reasons that apply to things owned by you or your family. 07.01-PR004 The Container Corporation of America is considering replacing an automatic painting machine purchased 9 years ago for $700,000. It has a

market value today of $40,000. The unit costs $350,000 annually to operate and maintain. A new unit can be purchased for $800,000 and will have annual O&M costs of $120,000. If the old unit is retained, it will have no salvage value at the end of its remaining life of 10 years. The new unit, if purchased, will have a salvage value of $100,000 in 10 years. Using an EUAC measure and a MARR of 20% should the automatic painting machine be replaced if the old automatic painting machine is taken as a trade-in for its market value of $40,000? a. Use the cash flow approach (insider’s viewpoint approach). b. Use the opportunity cost approach (outsider’s viewpoint approach). 07.01-PR005 A specialty concrete mixer used in construction was purchased for $300,000 7 years ago. Its annual O&M costs are $105,000. At the end of the 8-year planning horizon, the mixer will have a salvage value of $5,000. If the mixer is replaced, a new mixer will require an initial investment of $375,000 and at the end of the 8-year planning horizon, the new mixer will have a salvage value of $45,000. Its annual O&M cost will be only $40,000 due to newer technology. Use an EUAC measure and a MARR of 15% to see if the concrete mixer should be replaced if the old mixer is sold for its market value of $65,000. a. Use the cash flow approach (insider’s viewpoint approach). b. Use the opportunity cost approach (outsider’s viewpoint approach). 07.01-PR006 Online Educators (OE), a not-for-profit firm exempt from taxes, is considering replacement of some electronic equipment associated with its distance learning (DL) facility. It spent $100,000 on the equipment 3 years ago and has depreciated it to a current book value of $40,000. Its end-of-year O&M costs are $9,000 per year. Today’s newer technology, including high definition digital TV, has a price tag of $125,000 and, after some negotiation, OE negotiates a price of either (1) $108,000 in cash or (2) $91,000 in cash plus the current equipment as a trade-in. OE checked around and determined it could do no better by selling the soon-to-be-obsolete equipment to someone else. OE will use a 5-year planning horizon. If the newer equipment is purchased, it will have end-of-year O&M costs of $8,000 and a salvage value of $20,000 at that time. If the old equipment is retained, it will have to be supplemented in years 3, 4, and 5 by leasing a hi-def add-on unit costing

$30,000 per year payable at the beginning of the year with additional end-ofyear O&M costs of $7,000. The old equipment will also have no salvage value at the end of the planning horizon. MARR is 10%. a. What is the market value of the old equipment? b. Use the cash flow approach (insider’s viewpoint approach) to determine whether to keep or replace the current equipment. c. What is the market value of the old equipment? d. Use the opportunity cost approach (outsider’s viewpoint approach) to determine whether to keep or replace the current equipment. 07.01-PR007 A currently owned shredder for use in a refuse-powered electrical generating plant has a present net realizable value of $210,000 and is expected to have a market value of $10,000 after 4 years. Operating and maintenance disbursements are $100,000 per year. An equivalent shredder can be leased for $200 per day plus $80 per hour of actual use as determined by an hour meter, with both components assumed to be paid at the end of the year. Actual use is expected to be 1,500 hours and 250 days per year. Using a 4-year planning horizon and a MARR of 15%, determine the preferred alternative using the annual cost criterion. a. Consider only the above information and use the cash flow approach (insider’s viewpoint approach). b. Consider the additional information that the annual cost of operating without a shredder is $190,000 and use the cash flow approach (insider’s viewpoint approach). c. Consider only the above information and use the opportunity cost approach (outsider’s viewpoint approach). d. Consider the additional information that the annual cost of operating without a shredder is $190,000 and use the opportunity cost approach (outsider’s viewpoint approach). 07.01-PR008 Dell is considering replacing one of its material handling systems. It has an annual O&M cost of $48,000, a remaining operational life of 8 years, and an estimated salvage value of $6,000 at that time. A new system can be purchased for $175,000. It will be worth $50,000 in 8 years,

and it will have annual O&M costs of only $17,000 per year due to new technology. If the new system is purchased, the old system will be traded in for $55,000, even though the old system can be sold for only $45,000 on the open market. Leasing a new system will cost $31,000 per year, payable at the beginning of the year, plus operating costs of $15,000 per year payable at the end of the year. If the new system is leased, the existing material handling system will be sold for its market value of $45,000. Use a planning horizon of 8 years, an annual worth analysis, and MARR of 15% to decide which material handling system to recommend: (i) keep existing, (ii) trade in existing and purchase new, or (iii) sell existing and lease. a. Use the cash flow approach (insider’s viewpoint approach). 07.01-PR009 Video Solution Allen Construction purchased a crane 6 years ago for $130,000. It needs a crane of this capacity for the next 5 years. Normal operation costs $35,000 per year. The current crane will have no salvage value at the end of 5 more years. Allen can trade in the current crane for its market value of $40,000 toward the purchase of a new one that costs $150,000. The new crane will cost only $8,000 per year under normal operating conditions and will have a salvage value of $55,000 after 5 years. If MARR is 20%, determine which option is preferred. a. Use the cash flow approach (insider’s viewpoint approach). b. Use the opportunity cost approach (outsider’s viewpoint approach).

07.01-PR010 A division of Raytheon owns a 5-year-old turret lathe that has a book value of $24,000. It has a current market value of $18,000. The expected decline in market value is $3,000 per year from this point forward to a minimum of $3,000. O&M costs are $8,000 per year. Additional capability is needed. If the old lathe is kept, that new capability will be contracted out for $13,000, assumed payable at the end of each year. A new turret lathe has the increased capability to fulfill all needs, replacing the existing turret lathe and requiring no outside contracting. It can be purchased for $65,000 and will have an expected life of 8 years. Its market value is expected to be

$65,000(0.7t) at the end of year t. Annual O&M costs are expected to equal $10,000. MARR is 15% and the planning horizon is 8 years. a. Clearly show the cash flow profile for each alternative using a cash flow approach (insider’s viewpoint approach). b. Using an EUAC and a cash flow approach, decide which is the more favorable alternative. c. Clearly show the cash flow profile for each alternative using an opportunity cost approach (outsider’s viewpoint approach). d. Using an EUAC comparison and an opportunity cost approach, decide which is the more favorable alternative. 07.01-PR011 Five years ago, a multi-axis NC machine was purchased for the express purpose of machining large, complex parts used in commercial and military aircraft worldwide. It cost $350,000, had an estimated life of 15 years, and O&M costs of $50,000 per year. It was originally thought to have a salvage value of $20,000 at the end of 15 years but is now believed to have a remaining life of 5 years with no salvage value at that time. With business booming, the existing machine is no longer sufficient to meet production needs. It can be kept and supplemented by purchasing a new, smaller Machine S for $210,000 that will cost $37,000 per year for O&M, have a life of 10 years, and a salvage value of $210,000(0.8t) after t years. As an alternative, a larger, faster, and more capable Machine L can be used alone to replace the current machine. It has a cash price without trade-in of $450,000, O&M costs of $74,000 per year, a salvage value of $450,000(0.8t) after t years, and a 15-year life. The present machine can be sold on the open market for a maximum of $70,000. MARR is 20% and the planning horizon is 5 years. a. Clearly show the cash flow profile for each alternative using a cash flow approach (insider’s viewpoint approach). b. Using an EUAC comparison and a cash flow approach, decide which is the more favorable alternative. c. Clearly show the cash flow profile for each alternative using an opportunity cost approach (outsider’s viewpoint approach).

d. Using an EUAC comparison and an opportunity cost approach, decide which is the more favorable alternative. 07.01-PR012 Five years ago, ARCHON, a regional architecture/contractor firm, purchased an HVAC unit for $25,000, which was expected to last 15 years. It will have a salvage value of $0 in 10 more years. The annual operating cost of this unit started at $2,000 in the first year and has increased steadily at $250 per year ever since; last year the cost was $3,000. Its book value is now $13,000. ARCHON has been phenomenally successful due to its reputation for highly functional, high-quality, cost-effective designs and construction. It is building a new wing at its regional headquarters to accommodate a much larger computer design emphasis requiring larger, faster computers, architectural printers, e-storage for a construction repository of previous designs, and an increased human heat load. ARCHON can buy an additional unit to air-condition the new wing for $18,000. It will have a service life of 15 years, net salvage of $0 at that time, and a $3,000 market value after 10 years. It will have annual operating costs of $1,800 in the first year, increasing at $100 per year. As an alternative, ARCHON can buy a new unit to heat and cool the entire building for $35,000. It will last for 15 years, have a net salvage of $0 at that time; however, it will have a market value of $8,500 after 10 years. It will have first-year operating costs of $3,700 per year, increasing at $200 per year. The present unit can be sold now for $7,000. MARR is 20%, and the planning horizon is 10 years. a. Clearly show the cash flow profile for each alternative using a cash flow approach (insider’s viewpoint approach). b. Using a PW analysis and a cash flow approach, decide which is the more favorable alternative. c. Clearly show the cash flow profile for each alternative using an opportunity cost approach (outsider’s viewpoint approach). d. Using a PW analysis and an opportunity cost approach, decide which is the more favorable alternative. 07.01-PR013 Video Solution Clear Water Company has a downhole well auger that was purchased 3 years ago for $30,000. O&M costs are $13,000 per year. Alternative A is to keep the existing auger. It has a current market value of $12,000 and it will have a $0 salvage value after 7 more years.

Alternative B is to buy a new auger that will cost $54,000 and will have a $14,000 salvage value after 7 years. O&M costs are $6,000 for the new auger. Clear Water can trade in the existing auger on the new one for $15,000. Alternative C is to trade in the existing auger on a “treated auger” that requires vastly less O&M cost at only $3,000 per year. It costs $68,000, and the tradein allowance for the existing auger is $17,000. The “treated auger” will have an $18,000 salvage value after 7 years. Alternative D is to sell the existing auger on the open market and to contract with a current competitor to use its equipment and services to perform the drilling that would normally be done with the existing auger. The competitor requires a beginning-of-year retainer payment of $10,000. End-of-year O&M cost would be $6,000. MARR is 15% and the planning horizon is 7 years. a. Clearly show the cash flow profile for each alternative using a cash flow approach (insider’s viewpoint approach). b. Using an EUAC comparison and a cash flow approach, decide which is the more favorable alternative. c. Clearly show the cash flow profile for each alternative using an opportunity cost approach (outsider’s viewpoint approach). d. Using an EUAC comparison and an opportunity cost approach, decide which is the more favorable alternative.

07.01-PR014 Apricot Computers is considering replacing its material handling system and either purchasing or leasing a new system. The old system has an annual operating and maintenance cost of $32,000, a remaining life of 8 years, and an estimated salvage value of $5,000 at that time. A new system can be purchased for $250,000; it will be worth $25,000 in 8 years; and it will have annual operating and maintenance costs of $18,000/year. If the new system is purchased, the old system can be traded in for $20,000.

Leasing a new system will cost $26,000/year, payable at the beginning of the year, plus operating costs of $9,000/year, payable at the end of the year. If the new system is leased, the old system will be sold for $10,000. MARR is 15%. Compare the annual worths of keeping the old system, buying a new system, and leasing a new system based upon a planning horizon of 8 years. a. Use the cash flow approach. b. Use the opportunity cost approach. 07.01-PR015 Fluid Dynamics Company owns a pump that it is contemplating replacing. The old pump has annual operating and maintenance costs of $8,000/year: it can be kept for 4 years more and will have a zero salvage value at that time. The old pump can be traded in on a new pump. The trade-in value is $4,000. The new pump will cost $18,000 and have a value of $9,000 in 4 years and will have annual operating and maintenance costs of $4,500/year. Using a MARR of 10%, evaluate the investment alternative based upon the present worth method and a planning horizon of 4 years. a. Use the cash flow approach. b. Use the opportunity cost approach. 07.01-PR016 Video Solution A company owns a 5-year-old turret lathe that has a book value of $25,000. The present market value for the lathe is $16,000. The expected decline in market value is $2,000/year to a minimum market value of $4,000; maintenance plus operating costs for the lathe equal $4,200/year. A new turret lathe can be purchased for $45,000 and will have an expected life of 8 years. The market value for the turret lathe is expected to equal $45,000(0.70)k at the end of year k. Annual maintenance and operating cost is expected to equal $1,600. Based on a 12% MARR, should the old lathe be replaced now? Use an equivalent uniform annual cost comparison, a planning horizon of 7 years, and the cash flow approach.

07.01-PR017 Esteez Construction Company has an overhead crane that has an estimated remaining life of 7 years. The crane can be sold for $14,000. If the crane is kept in service it must be overhauled immediately at a cost of $6,000. Operating and maintenance costs will be $5,000/year after the crane is overhauled. After overhauling it, the crane will have a zero salvage value at the end of the 7-year period. A new crane will cost $36,000, will last for 7 years, and will have an $8,000 salvage value at that time. Operating and maintenance costs are $2,500 for the new crane. Esteez uses an interest rate of 15% in evaluating investment alternatives. Should the company buy the new crane based upon an annual cost analysis? a. Use the cash flow approach. b. Use the opportunity cost approach. 07.01-PR018 A small foundry is considering the replacement of a No. 1 Whiting cupola furnace that is capable of melting gray iron only with a reverberatory-type furnace that can melt gray iron and nonferrous metals. Both furnaces have approximately the same melting rates for gray iron in pounds per hour. The foundry company plans to use the reverberatory furnace, if purchased, primarily for melting gray iron, and the total quantity melted is estimated to be about the same with either furnace. Annual raw material costs would therefore be about the same for each furnace. Available information and cost estimates for each furnace are given below. Cupola furnace. Purchased used and installed 8 years ago for a cost of $20,000. The present market value is determined to be $8,000. Estimated remaining life is somewhat uncertain but, with repairs, the furnace should remain functional for 7 years more. If kept 7 years more, the salvage value is estimated as $2,000 and expected average annual expenses are: Fuel

$35,000

Labor (including maintenance) $40,000 Payroll taxes 10% of direct labor costs Taxes and insurance on furnace 1% of purchase price Other

$16,000

Reverberatory furnace. This furnace costs $32,000. Expenses to remove the cupola and install the reverberatory furnace are about $2,400. The new furnace

has an estimated salvage value of $3,000 after 7 years of use and annual expenses are estimated as: Fuel

$29,000

Labor (operating) Payroll taxes

$30,000 10% of direct labor costs

Taxes and insurance on furnace 1% of purchase price Other

$16,000

In addition, the furnace must be relined every 2 years at a cost of $4,000/occurrence. If the foundry presently earns an average of 15% on invested capital before income taxes, should the cupola furnace be replaced by the reverberatory furnace? a. Use the cash flow approach. b. Use the opportunity cost approach. 07.01-PR019 Metallic Peripherals, Inc., has received a production contract for a new product. The contract lasts for 5 years. To do the necessary machining operations, the firm can use one of its own lathes, which was purchased 3 years ago at a cost of $16,000. Today the lathe can be sold for $8,000. In 5 years the lathe will have a zero salvage value. Annual operating and maintenance costs for the lathe are $4,000/year. If the firm uses its own lathe, it must also purchase an additional lathe at a cost of $12,000; its value in 5 years will be $3,000. The new lathe will have annual operating and maintenance costs of $3,500/year. As an alternative, the presently owned lathe can be traded in for $10,000 and a new lathe of larger capacity purchased for a cost of $24,000; its value in 5 years is estimated to be $8,000, and its annual operating and maintenance costs will be $6,000/year. An additional alternative is to sell the presently owned lathe and subcontract the work to another firm. Company X has agreed to do the work for the 5-year period at an annual cost of $12,000/end-of-year. Using a 15% interest rate, determine the least-cost alternative for performing the required production operations.

a. Use the EUAC cash flow approach. b. Use the EUAC opportunity cost approach. 07.01-PR020 A machine was purchased 5 years ago for $12,000. At that time, its estimated life was 10 years with an estimated end-of-life salvage value of $1,200. The average annual operating and maintenance costs have been $14,000 and are expected to continue at this rate for the next 5 years. However, average annual revenues have been and are expected to be $20,000. Now, the firm can trade in the old machine for a new machine for $5,000. The new machine has a list price of $15,000, an estimated life of 10 years, annual operating and maintenance costs of $7,500, annual revenues of $13,000, and salvage values at the end of the jth year according to Sj = $15, 000 − $1, 500j,

for j = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Determine whether to replace or not by the annual worth method using a MARR equal to 15% compounded annually. Use a 5-year planning horizon and the cash flow approach. 07.01-PR021 A building supplies distributor purchased a gasoline-powered forklift truck 4 years ago for $8,000. At that time, the estimated useful life was 8 years with a salvage value of $800 at the end of this time. The truck can now be sold for $2,500. For this truck, average annual O&M expenses for year j have been Cj = $2, 000 + $400(j − 1)

Now the distributor is considering the purchase of a smaller battery-powered truck for $6,500. The estimated life is 10 years, with the salvage value decreasing by $600 each year. Average annual O&M expenses are expected to be $1,200. If a MARR of 10% is assumed and a 4-year planning horizon is adopted, based on an annual worth cash flow approach should the replacement be made now? 07.01-PR022 Kwik-Kleen Car Wash has been experiencing difficulties in keeping its equipment operational. The owner is faced with the alternative of overhauling the present equipment or replacing it with new equipment. The cost of overhauling the present equipment is $8,500. The present equipment

has annual operating and maintenance costs of $7,500. If it is overhauled, the present equipment will last for 5 years more and be scrapped at zero value. If it is not overhauled, it has a trade-in value of $3,200 toward the new equipment. New equipment can be purchased for $28,000. At the end of 5 years the new equipment will have a resale value of $12,000. Annual operating and maintenance costs for the new equipment will be $3,000. Using a MARR of 12%, what is your recommendation to the owner of the car wash? Base your recommendation on a present worth comparison and the cash flow approach. 07.01-PR023 A firm is contemplating replacing a computer it purchased 3 years ago for $400,000. It will have a salvage value of $20,000 in 4 more years. Operating and maintenance costs have been $75,000/year. Currently the computer has a trade-in value of $100,000 toward a new computer that costs $300,000 and has a life of 4 years, with a salvage value of $50,000 at that time. The new computer will have annual operating and maintenance costs of $80,000. If the current computer is retained, another small computer will have to be purchased in order to provide the required computing capacity. The smaller computer will cost $150,000, has a salvage value of $20,000 in 4 years, and has annual operating and maintenance costs of $30,000. Using an annual worth comparison before taxes, with a MARR of 15%, determine the preferred course of action using a cash flow approach. 07.01-PR024 National Chemicals has an automatic chemical mixer that it has been using for the past 4 years. The mixer originally cost $18,000. Today the mixer can be sold for $10,000. The mixer can be used for 10 years more and will have a $2,500 salvage value at that time. The annual operating and maintenance costs for the mixer equal $6,000/year. Because of an increase in business, a new mixer must be purchased. If the old mixer is retained, a new mixer will be purchased at a cost of $25,000 and have a $4,000 salvage value in 10 years. This new mixer will have annual operating and maintenance costs equal to $5,000/year. The old mixer can be sold and a new mixer of larger capacity purchased for $32,000. This mixer will have a $6,000 salvage value in 10 years and will

have annual operating and maintenance costs equal to $8,000/year. Based on a MARR of 15% and using an EUAC cash flow approach, what do you recommend? 07.01-PR025 The Ajax Specialty Items Corporation has received a 5-year contract to produce a new product. To do the necessary machining operations, the company is considering two alternatives. Alternative A involves continued use of the currently owned lathe. The lathe was purchased 5 years ago for $20,000. Today the lathe is worth $8,000 on the used machinery market. If this lathe is to be used, special attachments must be purchased at a cost of $3,500. At the end of the 5-year contract, the lathe (with attachments) can be sold for $2,000. Operating and maintenance costs will be $7,000/year if the old lathe is used. Alternative B is to sell the currently owned lathe and buy a new lathe at a cost of $25,000. At the end of the 5-year contract, the new lathe will have a salvage value of $13,000. Operating and maintenance costs will be $4,000/year for the new lathe. Using an annual worth analysis, should the firm use the currently owned lathe or buy a new lathe? Base your analysis on a minimum attractive rate of return of 15% and use a cash flow approach. 07.01-PR026 The Telephone Company of America purchased a numerically controlled production machine 5 years ago for $300,000. The machine currently has a trade-in value of $70,000. If the machine is continued in use, another machine, X, must be purchased to supplement the old machine. Machine X costs $200,000, has annual operating and maintenance costs of $40,000, and will have a salvage value of $30,000 in 10 years. If the old machine is retained, it will have annual operating and maintenance costs of $55,000 and will have a salvage value of $15,000 in 10 years. As an alternative to retaining the old machine, it can be replaced with Machine Y. Machine Y costs $400,000, has anticipated annual operating and maintenance costs of $70,000, and has a salvage value of $140,000 in 10 years. Using a MARR of 15%, a cash flow approach, and a present worth comparison, determine the preferred economic alternative.

07.01-PR027 A highway construction firm purchased a particular earthmoving machine 3 years ago for $125,000. The salvage value at the end of 8 years was estimated to be 35% of first cost. The firm earns an average annual gross revenue of $105,000 with the machine and the average annual operating costs have been and are expected to be $65,000. The firm now has the opportunity to sell the machine for $70,000 and subcontract the work normally done by the machine over the next 5 years. If the subcontracting is done, the average annual gross revenue will remain $105,000 but the subcontractor charges $85,000/end-of-year for these services. If a 15% rate of return before taxes is desired, use a cash flow approach to determine by the annual worth method whether or not the firm should subcontract. 07.01-PR028 Bumps Unlimited, a highway contractor, must decide whether to overhaul a tractor and scraper or replace it. The old equipment was purchased 5 years ago for $130,000; it had a 12-year projected life. If traded for a new tractor and scraper, it can be sold for $60,000. Overhauling the equipment will cost $20,000. If overhauled, O&M cost will be $25,000/year and salvage value will be negligible in 7 years. If replaced, a new tractor and scraper can be purchased for $150,000. O&M costs will be $12,000/year. Salvage value after 7 years will be $35,000. Using a 15% MARR and an annual worth analysis, should the equipment be replaced? Section 7.2 Optimum Replacement Interval LEARNING OBJECTIVE 7.2 Determine the optimum replacement interval in cases where an asset will be needed for an indefinite period of time. 07.02-PR001 You plan to purchase a car for $28,000. Its market value will decrease 20% per year. You have determined that the IRS-allowed mileage reimbursement rate for business travel is about right for fuel and maintenance at $0.505 per mile in the first year. You anticipate that it will increase at a rate of 10% each year, with the price of oil rising, influencing gasoline, oils, greases, tires, and so on. You normally drive 15,000 miles per

year. What is the optimum replacement interval for the car? Your MARR is 9%. 07.02-PR002 Video Solution Griffin Dewatering purchases a wellpoint pump connected to a skid-mounted diesel engine for $14,000. Its market value for salvage purposes decreases 30% each year. When installed on a construction job, a wellpoint system operates virtually 24/7, and operating and maintenance costs will be $3,500 the first year, increasing $600 each year thereafter. What is the optimum replacement interval if MARR is 15%?

07.02-PR003 Polaris Industries wishes to purchase a multiple-use in-plant “road test” simulator that can be used for ATVs, motorcycles, and snowmobiles. It takes digital data from relatively short drives on a desired surface—from smooth to exceptionally harsh—and simulates the ride over and over while the vehicle is mounted to a test stand under load. It can run continuously if desired and provides opportunities to redesign in areas of poor reliability. It costs $128,000 and its market value decreases 30% each year. Operating costs are modest; however, maintenance costs can be significant due to the rugged use. O&M in the first year is expected to be $10,000, increasing 25% each subsequent year. MARR is 15%. What is the optimum replacement interval? 07.02-PR004 Video Solution A granary purchases a conveyor used in the manufacture of grain for transporting, filling, or emptying. It is purchased and installed for $70,000 with a market value for salvage purposes that decreases at a rate of 20% per year with a minimum of value $3,000. Operation and maintenance is expected to cost $14,000 in the first year, increasing $1,000 per year thereafter. The granary uses a MARR of 15%. What is the optimum replacement interval for the conveyor?

07.02-PR005 Milliken uses a digitally controlled “dyer” for placing intricate and integrated patterns on manufactured carpet squares for home and commercial use. It is purchased for $400,000. Its market value will be $310,000 at the end of the first year and decrease $40,000 per year

thereafter to a minimum of $30,000. Operating costs are $20,000 the first year, increasing 8% per year. Maintenance costs are only $8,000 the first year but will increase 35% each year thereafter. Milliken’s MARR is 20%. Determine the optimum replacement interval for the dyer. 07.02-PR006 A firm is presently using a machine that has a market value of $11,000 to do a specialized production job. The requirement for this operation is expected to last only 6 years more, after which it will no longer be done. The predicted costs and salvage values for the present machine are: Year

1

2

3

4

5

6

Operating cost $1,500 $1,800 $2,100 $2,400 $2,700 $3,000 Salvage value

8,000 6,000 5,000 5,000 3,000 2,000

A new machine has been developed that can be purchased for $17,000 and has the following predicted cost performance: Year

1

2

3

4

5

6

Operating cost $1,000 $1,100 $1,200 $1,300 $1,400 $1,500 Salvage value 13,000 11,000 10,000 9,000 8,000 7,000 If interest is at 0%, when should the new machine be purchased? 07.02-PR007 A particular unit of production equipment has been used by a firm for a period of time sufficient to establish very accurate estimates of its operating and maintenance costs. Replacements can be expected to have identical cash flow profiles in successive life cycles. The appropriate discount rate is 15%. Operating and maintenance costs for a unit of equipment in its tth year of service, denoted by Ct, are as follows:

t

Ct

t

Ct

1 $6,000 6 $15,000 2 7,500 7 17,250 3 9,150 8 19,650 4 10,950 9 22,200 5 12,900 10 24,900 Each unit of equipment costs $45,000 initially. Because of its special design, the unit of equipment cannot be disposed of at a positive salvage value following its purchase; hence, a zero salvage value exists, regardless of the replacement interval used. a. Determine the optimum replacement interval assuming an infinite planning horizon. (Maximum feasible interval = 10 years.) b. Determine the optimum replacement interval assuming a finite planning horizon of 15 years, with Ct+1 = Ct + $1,500 + $150(t − 1) for t = 10, 11, …, 15. c. Solve parts (a) and (b) using a discount rate of 0%. d. Based on the results obtained, what can you conclude concerning the effect the discount rate has on the optimum replacement interval? 07.02-PR008 Given an infinite planning horizon, identical cash flow profiles for successive life cycles, and the following functional relationships for Ct, the operating and maintenance cost for the ith year of service for the unit of equipment in current use, and Fn, the salvage value at the end of n years of service: Ct = $4, 000(1.10)

t

Fn = $44, 000(0.50)

t = 1, 2, … , 12 n

n = 0, 1, 2, … , 12

Determine the optimum replacement interval assuming a MARR of (a) 0%, (b) 10%. (Maximum life is 12 years.)

07.02-PR009 $100,000 is invested in equipment having a negligible salvage value regardless of when the equipment is replaced. O&M costs equal $25,000 the first year and increase $5,000 per year. Based on a MARR of 10%, what are the optimum replacement interval and minimum EUAC? 07.02-PR010 $100,000 is invested in equipment having a negligible salvage value regardless of when the equipment is replaced. O&M costs equal $25,000 the first year and increase 15% per year. Based on a MARR of 10%, what are the optimum replacement interval and minimum EUAC? 07.02-PR011 $250,000 is invested in equipment having a negligible salvage value regardless of the number of years used. O&M costs equal $60,000 the first year and increase $12,000 per year. Based on a MARR of 10%, what are the optimum replacement interval and minimum EUAC? 07.02-PR012 $250,000 is invested in equipment having a salvage value equal to $250,000(0.80n) after n years of use. O&M costs equal $60,000 the first year and increase $8,000 per year. Based on a MARR of 10%, what are the optimum replacement interval and minimum EUAC? 07.02-PR013 $225,000 is invested in equipment having a salvage value equal to $200,000(0.75n) after n years of use. O&M costs equal $45,000 the first year and increase $15,000 per year. Based on a MARR of 10%, what is optimum replacement interval and what is the minimum EUAC? 07.02-PR014 $500,000 is invested in equipment having a negligible salvage value regardless of the number of years used. O&M costs equal $125,000 the first year and increase $25,000 per year. Based on a MARR of 10%, what are the optimum replacement interval and minimum EUAC? 07.02-PR015 $500,000 is invested in equipment having a salvage value equal to $500,000(0.80n) after n years of use. O&M costs equal $125,000 the first year and increase $15,000 per year. Based on a MARR of 10%, what are the optimum replacement interval and minimum EUAC? 07.02-PR016 $500,000 is invested in equipment having a salvage value equal to $500,000(0.75n) after n years of use. O&M costs equal $125,000 the first year and increase 15% per year. Based on a MARR of 10%, what are the optimum replacement interval and minimum EUAC?

07.02-PR017 $500,000 is invested in equipment having a salvage value equal to $500,000(0.65n) after n years of use. O&M costs equal $125,000 the first year and increase 10% per year. Based on a MARR of 10%, what are the optimum replacement interval and minimum EUAC? 07.02-PR018 True or False: Consider an optimum replacement interval problem of the type considered in this chapter. If O&M costs are a gradient annual series, the optimum replacement interval tends to increase as the magnitudes of the gradient step decreases. 07.02-PR019 True or False: Consider an optimum replacement interval problem of the type considered in this chapter. If the annual O&M cost is an increasing gradient series “sitting on top of” a base uniform series, then decreasing the magnitude of the base uniform series will tend to increase the optimum replacement interval. 07.02-PR020 True or False: Consider an optimum replacement interval problem of the type considered in this chapter. If salvage value is negligible regardless of how long the equipment is used and O&M cost is represented by a uniform annual series, then the equipment in question should not be replaced until it no longer can function. 07.02-PR021 True or False: Consider an optimum replacement interval problem of the type considered in this chapter. If O&M costs increase at a geometric rate, then decreasing the magnitude of the O&M cost in the first year will tend to decrease the optimum replacement interval. 07.02-PR022 True or False: Consider an optimum replacement interval problem of the type considered in this chapter. If the ORI equals 5 years based on a salvage value equal to 0.60n, where n is the years used before replacement, then, if the salvage value for a replacement becomes 0.80n, the new ORI will not be greater than 5 years.

Chapter 7 Summary and Study Guide Summary 7.1: Cash Flow and Opportunity Cost Approaches to Replacement Analysis

Learning Objective 7.1: Perform a replacement analysis utilizing a cash flow or opportunity cost approach. (Section 7.1) In a replacement analysis, the defender (existing asset) is compared to one or more challengers (potential replacement assets). Often, replacement analysis is undertaken because of the obsolescence of the defender—it lacks needed functional or technological capabilities, or its economic worth is less than that of the challenger(s). Sunk costs are past costs associated with the defender and have no bearing on current decisions. The cash flow approach, also called the insider’s viewpoint approach, “follows the money.” By this it is meant that for each replacement alternative, cash flows are shown each year for each alternative throughout the planning horizon. The cash flow and the opportunity cost approaches for replacement analysis are equivalent; however, the cash flow approach is preferred because it is more direct and simpler to evaluate when income taxes are considered. In the opportunity cost approach, also called the outsider’s viewpoint approach, the salvage value of the current asset is treated as its investment cost if it is retained; thus, by deciding to keep the asset, one gives up the opportunity to receive a monetary amount for it. 7.2: Optimum Replacement Interval

Learning Objective 7.2: Determine the optimum replacement interval in cases where an asset will be needed for an indefinite period of time. (Section 7.2) Often an asset is used for many years, for example in the case of over-theroad trailers and rental cars. In this situation, it is common to assume an identical asset that will be used over an indefinite period of time. We assume that the planning horizon equals an integer multiple of the ORI value obtained, future cash flow profiles are identical to their predecessors, and the present worth must be minimized over the planning horizon.

Important Terms and Concepts Replacement Analysis The comparison of investment alternatives that involve replacing an asset or service provider. Defender The current asset being compared in a replacement analysis. Challenger The potential replacement asset being compared in a replacement analysis. Functional Obsolescence Resulting from physical deterioration of the defender, increased demand that exceeds the defender’s capacity, or new requirements that the defender cannot meet. Technological Obsolescence Occurs through the introduction of new technology, such that challengers possess capabilities not present in the defender. Economic Obsolescence Occurs when the economic worth of one or more challengers exceeds the defender’s economic worth. Sunk Costs

Past costs that have no bearing on current decisions. Salvage Value The estimated value of an asset at the end of its useful life. Also referred to as the market value. Opportunity Cost The cost of a forgone alternative (or opportunity) that is incurred in order to pursue another alternative. Equivalent Uniform Annual Cost The EUAC is the equivalent uniform annual cost to keep or operate an asset. Optimum Replacement Interval (ORI) The interval at which an asset should be replaced to minimize cost (or maximize worth). Capital Recovery Cost The cost of ownership, a value that typically decreases with increased usage.

Chapter 7 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 7.1: Cash Flow and Opportunity Cost Approaches to Replacement Analysis LO 7.1: Perform a replacement analysis using a cash flow or opportunity cost approach. Excel Video Lesson: PMT Financial Function Excel Video Lesson Spreadsheet: PMT Financial Function Excel Video Lesson: FV Financial Function Excel Video Lesson Spreadsheet: FV Financial Function Video Example 7.1: Cash Flow Approach to Replacement Analysis Video Solution 07.01-PR009 Video Solution 07.01-PR013 Video Solution 07.01-PR016 7.2: Optimum Replacement Interval LO 7.2: Determine the optimum replacement interval in cases where an asset will be needed for an indefinite period of time. Video Lesson: Capital Recovery Cost Video Lesson Notes: Capital Recovery Cost Video Lesson: EUAC + ORI Video Lesson Notes: EUAC + ORI Video Example 7.3: Computing the Optimum Replacement Interval Video Solution 07.02-PR002

Video Solution 07.02-PR004 These chapter-level resources will help you with your overall understanding of the content in this chapter. Appendix A: Time Value of Money Factors Appendix B: Engineering Economic Equations Flashcards: Chapter 07 Excel Utility: TVM Factors: Table Calculator Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values Excel Utility: Monthly Payment Sensitivity Excel Utility: TVM Factors: Discrete Compounding Excel Utility: TVM Factors: Geometric Series Future Worth Excel Utility: TVM Factors: Geometric Series Present Worth Excel Data Files: Chapter 07

CHAPTER 7 Replacement Analysis LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 7.1 Perform a replacement analysis using a cash flow or opportunity cost approach. (Section 7.1) 7.2 Determine the optimum replacement interval in cases where an asset will be needed for an indefinite period of time. (Section 7.2)

Engineering Economics in Practice J.B. Hunt Transport Services J.B. Hunt Transport Services Inc. (JBHT) is one of the largest surface transportation companies in North America, operating in the United States, Canada, and Mexico with a workforce of about 24,000. In 2017, the company took steps to prepare for a more digital, self-service, and growth-oriented economy. They are harnessing their staff of transportation professionals to advance technology using real-time data to ensure effective transportation solutions for their customers. A full truckload transportation firm that delivers goods all over the United States, J.B. Hunt drivers make short- and long-haul deliveries. In 2017, JBHT’s consolidated revenue was $7.2 billion divided across three types of transportation services. Their largest business, Intermodal (JBI), generated revenues of $4.1 billion (57%) in 2017 and is also the largest privately owned fleet of domestic 53-foot containers (88,610 units). It is supported by a 77,946 company-owned chassis fleet, 4,776 class 8 tractors, operated by 5,782 drivers and 764 independent contractors. Their secondlargest business is Dedicated Contract Services (DCS), generating $1.7 billion (23.6%) in total revenue and representing North America’s largest private-fleet provider. DCS owns and operates 8,124 tractors as well as operates another 544 customer-owned power units. This segment of inventory management for their customers is usually close to the point of sales/purchase or consumption. Finally, their third business is referred to as “Highway Services.” This business unit generated $1.4 billion (19.4%) in revenue in 2017 utilizing 1,281 company-owned tractors and 7,120 trailers in virtually all classes of freight management, including truckload, temperature-control, flatbed, Less Than Truckload (LTL), and parcel. With more than 16,000 company-owned tractors and 123,000 trailers and containers used to ship nearly 5.3 million loads in 2018, JBHT is constantly faced with decisions regarding the acquisition and replacement of tractors and trailers. Also, using its own fleet versus contracting with independent truckers and purchasing versus leasing over-the-road equipment are examples of decisions the company must make on a regular basis. In a typical year, JBHT purchases hundreds of new over-the-road tractors, each of which has a retail price of approximately $121,000. In addition, JBHT annually purchases a large quantity of trailers and vans at an average cost of approximately $26,000. Also, JBHT annually purchases several hundred intermodal containers for approximately $13,000 each, plus the chassis to haul the containers (each chassis costs approximately $15,000). Tractors tend to be replaced after driving approximately 500,000 miles; during the life of a tractor, it incurs annual repair and maintenance expenses of approximately $4,000. Trade-in values on used tractors vary depending on use, but it is typical for JBHT to receive approximately $41,000 for a used tractor and $11,000 for a used trailer or van. Clearly, with hundreds of millions of dollars being spent annually on tractors and trailers, determining the most economic time to replace its moving equipment is a key to JBHT’s economic performance. Discussion Questions 1. Do you suppose that JBHT makes replacement decisions for its equipment on an item-by-item level or for a fleet of equipment? What might be some of the advantages and disadvantages to replacement at the fleet level? 2. What factors might influence the useful life of the JBHT equipment? 3. Why use miles instead of years (time) as a parameter for replacement? 4. How might safety factor into the replacement analysis decision?

Introduction One of the most common investment alternatives considered by businesses and individuals involves the replacement of an asset. No doubt, you have made many replacement decisions, such as replacing a car, a computer, a calculator, a television, a cell phone service provider, and so forth. Businesses make replacement

decisions on an ongoing basis. Replacement decisions are influenced by economics, capacity, quality of service provided, changing requirements, prestige, fads, and a host of other factors. In Chapters 4, 5, and 6, we examined various ways to compare investment alternatives. We considered two situations: a single alternative (where the decision was between investing and not investing) and a set of mutually exclusive investment alternatives (where we recommended the alternative having the greatest economic worth). In both situations, the alternatives in question could have been replacement alternatives. In this chapter, we consider two equivalent approaches that can be used to perform a replacement analysis: the cash flow approach and the opportunity cost approach. We conclude the chapter with a consideration of the optimum replacement frequency for equipment that will be used year in and year out; such analyses are called optimum replacement interval analyses.

7.1 Cash Flow and Opportunity Cost Approaches to Replacement Analysis LEARNING OBJECTIVE Perform a replacement analysis using a cash flow or opportunity cost approach.

7.1.1 Fundamentals of Replacement Analysis Because decisions to replace versus continuing to use an asset occur so frequently, a body of literature has evolved on this subject. We use the term replacement analysis when referring to the comparison of investment alternatives that involve replacing an asset or service provider. Replacement Analysis The comparison of investment alternatives that involve replacing an asset or service provider. Although replacement decisions occur for a variety of reasons, the following are typical: 1. The current asset, which we call the defender, has developed deficiencies, such as high setup cost, excessive maintenance expense, declining productivity, high energy cost, limited capability, or physical impairment. 2. Potential replacement assets, which we call the challengers, are available and have a number of advantages over the defender, such as new technology that is quicker to set up and easier to use, lower labor cost, lower maintenance expense, lower energy cost, higher productivity, or additional capabilities. 3. A changing external environment, including a. changing user and customer preferences and expectations; b. changing requirements; c. new, alternative ways of obtaining the functionality provided by the defender, including the availability of leased equipment and third-party suppliers; and d. increased demand that cannot be met with the current equipment—either supplementary or replacement equipment is required to meet demand. Defender The current asset being compared in a replacement analysis. Challenger The potential replacement asset being compared in a replacement analysis. Obsolescence is a frequently cited reason for replacing an asset. Various types of obsolescence can occur, including the following: 1. Functional obsolescence, which can result from physical deterioration of the defender, increased demand that exceeds the defender’s capacity, or new requirements that the defender cannot meet.

2. Technological obsolescence, which occurs through the introduction of new technology, such that challengers possess capabilities not present in the defender. 3. Economic obsolescence, which occurs when the economic worth of one or more challengers exceeds the defender’s economic worth. Functional Obsolescence Resulting from physical deterioration of the defender, increased demand that exceeds the defender’s capacity, or new requirements that the defender cannot meet. Technological Obsolescence Occurs through the introduction of new technology, such that challengers possess capabilities not present in the defender. Economic Obsolescence Occurs when the economic worth of one or more challengers exceeds the defender’s economic worth. Replacement analyses are basically just another type of alternative comparison. As such, the same systematic seven-step approach described in Chapter 1 can be used. Likewise, the consistent measures of economic worth described previously can be used. Yet, replacement decisions seem to pose unique difficulties. Why? Perhaps it is because an “emotional” attachment to present equipment can occur. Also, perhaps it is due to the presence of sunk costs. Sunk costs are past costs that have no bearing on current decisions. (A long history of maintenance and repair can cause owners to be reluctant to replace equipment—doing so might appear as an admission that mistakes were made in the past by keeping the equipment longer than it should have been.) Sunk Costs Past costs that have no bearing on current decisions.

Systematic Economic Analysis Technique 1. Identify the investment alternatives 2. Define the planning horizon 3. Specify the discount rate 4. Estimate the cash flows 5. Compare the alternatives 6. Perform supplementary analyses 7. Select the preferred investment Based on a DCF comparison, many companies use equipment long after replacements would be justified economically. Why? Several reasons come to mind: 1. The firm is currently making a profit, so there is no compelling reason to invest in new technology. 2. The current equipment still works and produces a product of acceptable quality—an “if it isn’t broken, don’t replace it” attitude prevails. 3. There are risks and uncertainties associated with change—replacing the “tried and true” proven defender with an unproven challenger is viewed as too risky. (Recall the cartoon character Pogo saying, “Change is good! You go first!”) 4. A decision to replace existing technology is a stronger commitment, for a period of time into the future, than continuing to use the defender—given the rapidly changing world and the growth of outsourcing and offshoring, some managers are reluctant to invest new capital in domestic operations. 5. Due to limitations on investment capital, replacements are given secondary consideration over investments that expand operations or add new capabilities (even though the replacement might have a greater economic

worth). 6. Uncertainty regarding the future—the defender has a track record insofar as annual costs are concerned, the challenger is unproven, estimates of future demand might not materialize, and annual costs estimates for the challenger could be incorrect. 7. The psychological impact of sunk costs—it is very hard for some people to ignore what was spent in the past; as a result, they continue to waste money by supporting equipment they should replace. 8. The technological improvement trap—because technological improvements occur so frequently, some are reluctant to replace the defender with this year’s challenger, because next year’s challenger will be less expensive and will have greater capability than this year’s. 9. Some companies prefer to be “technology followers” instead of “technology leaders”—they want others to debug the new generation of technology, and because new technology is being introduced so rapidly, they can never bring themselves to make the replacement decision. 10. Management is concerned about taking a hit on the quarterly financial statement by writing off an asset that is not fully depreciated, regardless of the DCF analysis.

7.1.2 Cash Flow and Opportunity Cost Approaches As noted, two equivalent approaches in a replacement analysis are the cash flow and the opportunity cost approaches. The two approaches differ in how the market value for the currently owned asset is treated if replacement occurs and if replacement does not occur. The cash flow approach, also called the insider’s viewpoint approach, “follows the money.” Specifically, with the cash flow approach, for each replacement alternative, cash flows are shown for each alternative for each year in the planning horizon. If replacement does not occur, the cash flow shown in year zero is often zero. If money must be expended in order to continue the use of the current asset, however, then a negative-valued cash flow is shown in year zero. On the other hand, if a replacement occurs, then the amount of money received in trade-in or the amount of money required to dispose of the current asset is shown in year zero. With the opportunity cost approach, the salvage value or market value of the current asset is treated as its investment cost if it is retained. Opportunity cost refers to the cost of a foregone alternative (or opportunity) that is incurred in order to pursue another alternative. In the opportunity cost approach to replacement analysis, by deciding to keep the asset, one gives up the opportunity to receive a monetary amount for it. The opportunity cost approach is also known as the outsider’s viewpoint approach. Salvage Value The estimated value of an asset at the end of its useful life. Also referred to as the market value. Opportunity Cost The cost of a forgone alternative (or opportunity) that is incurred in order to pursue another alternative. An example will illustrate how the cash flow and opportunity cost approaches are performed. This example introduces a new term, EUAC, or equivalent uniform annual cost. The EUAC is the annual cost to keep or operate the asset. It is similar to the annual worth, AW. Equivalent Uniform Annual Cost The EUAC is the equivalent uniform annual cost to keep or operate an asset.

EXAMPLE 7.1 Cash Flow Approach to Replacement Analysis Video Example A chemical mixer was purchased 8 years ago for $100,000. If retained, it will require an investment of $50,000 to upgrade it; if upgraded, it will cost $35,000/year to operate and maintain (O&M) and will have a negligible salvage value after 5 years. A new mixer can be purchased for $120,000; it will have an annual O&M cost of $15,000 and a salvage value of $40,000 after 5 years. Alternatively, a mixer can be leased with 5 beginning-of-year lease payments of $20,000; O&M costs will be $18,000/year. If the mixer is replaced, the old mixer can be sold on the used equipment market for $15,000. Using an insider’s approach, what are (a) the EUAC of keeping the current mixer, (b) the EUAC of replacing with a new mixer, and (c) the EUAC of replacing with a leased mixer? The MARR is 10%. Key Data Given MARR = 10%. Using a cash flow or insider’s approach, the cash flows are shown below for each replacement alternative. EOY CF(Keep)

CF(Replace)

CF(Lease)

0 1–4

−$50,000.00 −$105,000.00   −$5,000.00 −$35,000.00  −$15,000.00  −$38,000.00

5

−$35,000.00  $25,000.00  −$18,000.00

Find EUAC for each alternative. Solution The EUACs for each alternative are as follows: a.

EUAC(Keep)

=$35,000 +$50,000(A|P 10%, 5) =$35,000 +$50,000(0.26380)  =$48,190.00 =PMT(10%,5,−50000)+35000 =$48,189.87

b. EUAC(Replace)

=$15,000 +$105,000(A|P 10%, 5) −$40,000 (A|F 10%, 5) =$15,000 +$105,000(0 =PMT(10%,5,−105000,40000)+15000 =$36,146.84

c. EUAC(Lease)

=$18,000 +$20,000(F |P 10%, 1) −$15,000(A|P 10%, 5) =$18,000 +$20,000(1.10) − =PMT(10%,5,15000)+18000−FV(10%,1,,20000) =$36,043.04

Excel® Video Lesson: PMT Financial Function Excel® Video Lesson: FV Financial Function Notice, the $100,000 purchase price for the mixer is a sunk cost. It does not affect the cash flows over the planning horizon.

EXAMPLE 7.2 Opportunity Cost Approach to Replacement Analysis The replacement problem in Example 7.1 can also be addressed by the opportunity cost approach. Key Data Given MARR = 10%. Using an opportunity cost or outsider’s viewpoint approach, the cash flows are shown below for each alternative. EOY CF(Keep)

CF(Replace)

CF(Lease)

0

−$65,000.00 −$120,000.00  −$20,000.00

1–4

−$35,000.00  −$15,000.00  −$38,000.00

5

−$35,000.00  $25,000.00  −$18,000.00

Find EUAC for each alternative. Solution The EUACs for each alternative are as follows: a.

EUAC(Keep)

=$35,000 +$65,000(A|P  10%, 5) =$35,000 +$65,000(0.26380) =$52,147.00 =PMT(10%,5,−$65000)+$35000 =$52,146.84

b. EUAC(Replace)

=$15,000 +$120,000(A|P  10%, 5) −$40,000(A|F  10%, 5) =$15,000 +$120,000(0. =PMT(10%,5,−$120000,40000)+$15000 =$40,103.80

c.

EUAC(Lease)

=$18,000 +$20,000(F |P  10%, 1) =$18,000 +$20,000(1.10) =$40,000.00 =18000−FV(10%,1„20000) =$40,000.00

Exploring the Solution Notice, the same conclusion is reached with the two approaches: leasing has the lowest equivalent uniform annual cost. Furthermore, because the EUAC values obtained using the two approaches differ by $15,000(A|P 10%,5) = $3,957.00 or =PMT(10%,5,−15000) = $3,956.96, the difference in the EUAC(Keep) and either the EUAC(Replace) or the EUAC(Lease) will be the same for the cash flow and the opportunity cost approaches: EUAC(K − R)

EUAC(K − L)

=

$52,147.00 −$40,104.00 =$12,043.00 (opportunity cost approach)

=

$48,190.00 −$36,147.00 =$12,043.00 (cash flow approach)

=

$52,147.00 −$40,000.00 =$12,147.00 (opportunity cost approach)

=

$48,190.00 −$36,043.00 =$12,147.00 (cash flow approach).

Although the two approaches are equivalent, we prefer the cash flow approach. We believe it is more direct, especially when multiple replacement candidates are available and each offers a different trade-in value for the current asset. Also, when income taxes are being considered, the cash flow approach, in our view, is simpler to use than the opportunity cost approach. Consequently, unless we specify otherwise, in solving the problems at the end of the chapter, use the cash flow approach.

Concept Check 07.01-CC001 Which of the following statements is false with respect to replacement analysis? a. Functional obsolescence is the only type of obsolescence that should be considered in a replacement analysis b. A changing external environment can lead to the need for a replacement analysis c. Companies frequently use equipment long after a replacement would be economically justified d. In a replacement analysis, the current asset is referred to as the defender

Concept Check 07.01-CC002 In a replacement analysis, the market value of the currently owned asset a. does not impact the cash flows of the replacement asset(s) in the opportunity cost approach b. does not impact the cash flows of the current asset in the cash flow approach if the current asset is kept c. is treated as a salvage value for the current asset in the cash flow approach if the current asset is kept d. is added to the first cost of the replacement asset(s) in the opportunity cost approach

7.2 Optimum Replacement Interval LEARNING OBJECTIVE Determine the optimum replacement interval in cases where an asset will be needed for an indefinite period of time. In this section, we consider situations in which a particular asset will be needed for an indefinite period of time. When it wears out, it will be replaced by an identical asset, as many times as necessary. Examples would include over-the-road trailers, lift trucks, a basic machine tool, and rental cars, among others. For engineering economic analysis, we would like to know the optimum replacement interval for such an asset—the interval at which it should be replaced to minimize cost or maximize worth. Optimum Replacement Interval (ORI) The interval at which an asset should be replaced to minimize cost (or maximize worth).

7.2.1 Parameters for Determining ORI As equipment ages, operating and maintenance (O&M) costs tend to increase. At the same time, the cost of ownership tends to decrease with increased usage. The sum of the ownership and operating costs can often be represented by Figure 7.1. In this case, the equivalent uniform annual cost is a convex function of the life of the equipment.

FIGURE 7.1 EUAC Components Used to Determine the Optimum Replacement Interval Video Lesson: Capital Recovery Cost The cost of ownership is often called the capital recovery cost and is computed as follows: CR = P (A|P  i%, n) − F (A|F  i%, n)

(7.1)

where P is the initial investment in the equipment, F is its salvage value after n years of use, and the time value of money is i%. Capital Recovery Cost The cost of ownership, a value that typically decreases with increased usage. By converting the acquisition and disposal cost (salvage value) to a uniform annual series and by converting the O&M cost to a uniform annual series, one obtains the EUAC for a given value of n. By enumerating or searching over n, the optimum replacement interval (ORI) is obtained. Video Lesson: EUAC + ORI

EXAMPLE 7.3 Computing the Optimum Replacement Interval Video Example Five hundred thousand dollars is invested in a surface-mount placement machine. Experience shows that the salvage value for the SMP machine decreases by 25% per year. Hence, its salvage value equals $500,000(0.75n) after n years of use. O&M costs increase at a rate of 15% per year; the O&M cost equals $125,000 the first year of use. Based on a MARR of 10%, what is the ORI and what is the minimum EUAC (EUAC*)? Key Data Given Initial investment (P) = $500,000, Salvage value (F) = $500,000(0.75n), O&M = $125,000 in year one, increasing 15% per year Find ORI and EUAC* Solution For the example problem, EUAC = [$500,000 + $125,000(P|A1 10%,15%,n)] (A|P 10%,n) − $500,000(0.75n)(A|F 10%,n). As shown below, ORI = 5 years and EUAC* = $276,611. n (A|P 10%,n) (P|A1 10%,15%,n) (A|F 10%,n) EUAC 4 5

0.31547 0.26380

3.89190 4.97789

0.21547 0.16380

$277,119 $276,611

6

0.22961

6.11325

0.12691

$278,729

7.2.2 Behavior of the ORI as Parameters Change From Figure 7.1, will the ORI increase or decrease if the slope of the annual O&M cost curve increases? In general, the more expensive it becomes to operate and maintain equipment, the sooner it needs to be replaced. Hence, when the slope of the annual O&M cost curve is increased, the ORI will tend to decrease (due to integer values of n, the ORI might remain unchanged for some changes in the slope of the annual O&M cost curve). From Figure 7.1, will the ORI increase or decrease if the initial investment increases? In general, the more expensive an asset (all other parameters remaining constant), the longer it will be used. Hence, when the initial investment is increased, the ORI will tend to increase. By varying other parameters in Example 7.3, you will find that increasing the MARR tends to increase ORI and increasing the rate of decrease in salvage value tends to increase ORI. Three observations come to mind regarding ORI computations. 1. Implicit in the solution procedure we used to obtain the ORI is an assumption that the planning horizon equals an integer multiple of the ORI value obtained. 2. Subsequent replacements must have cash flow profiles that are identical to their predecessors. 3. For the ORI to be truly optimal, it must minimize present worth over the planning horizon. Suppose the planning horizon is fixed and greater than the ORI. How might one determine the optimum time to replace the asset? The following example addresses this scenario.

EXAMPLE 7.4 Optimum Replacement Timing with a Given Planning Horizon For the SMP machine in the previous example, suppose a planning horizon of 9 years is to be used. Recall, we obtained an ORI of 5 years. Because the planning horizon is not an integer multiple of 5, which of the following replacement sequences should we use: {5,4}; {6,3}; {7,2}; {8,1}; {9,0}? Key Data Given Initial investment (P) = $500,000, Salvage value (F) = $500,000(0.75n), O&M = $125,000 in year one, increasing 15% per year, Planning horizon = 9 years Find The replacement sequence that minimizes PW of costs given a 9-year planning horizon. Solution Shown below is the EUAC for values of n ranging from 1 to 9. n (A|P 10%,n) (P|A1 10%,15%,n) (A|F 10%,n) EUAC 1 1.10000 0.90909 1.00000 $300,000 2 3

0.57619 0.40211

1.85950 2.85312

0.47619 0.30211

$288,095 $280,737

4 5

0.31547 0.26380

3.89190 4.97789

0.21547 0.16380

$277,119 $276,611

6 7

0.22961 0.20541

6.11325 7.30022

0.12961 0.10541

$278,729 $283,112

8 9

0.18744 0.17364

8.54113 9.83846

0.08744 0.07364

$289,462 $297,599

For a replacement sequence {x, y}, the present worth over the 9-year planning horizon is given by PW{x, y} = EUAC(x)(P |A 10%, x) + EUAC(y)(P |A 10%, y)(P |F 10%, x)

Hence, the present worth for each of the sequences is as shown below. For a 9-year planning horizon, the first SMP machine should be replaced after 5 years and the successor should be used for the balance of the planning horizon. {x,y} PW{x,y} {5,4} $1,593,998.52 {6,3} $1,608,014.16 {7,2} $1,634,857.37 {8,1} $1,671,520.06 {9,0} $1,713,885.95 Perhaps you wondered why we did not consider the sequences {1,8}, {2,7}, {3,6}, and {4,5}. Based on the time value of money, the replacement interval having the smallest EUAC should be placed first in the sequence. An examination of the EUAC for various values of n shows that we always chose {x, y} sequences such that EUAC(x) < EUAC(y).

Concept Check 07.02-CC001 Optimal replacement interval is negatively correlated with which of the following? a. First cost b. Operating and maintenance cost c. MARR d. None of the above

CHAPTER 8 Depreciation

Chapter 8 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced e-text and WileyPLUS

FE-Like Problems 08-FE001 depreciable?

Which of the following is not a requirement for an asset to be

a. It must have a life longer than 1 year b. It must have a basis (initial purchase plus installation cost) greater than $1,000 c. It must be held with the intent to produce income d. It must wear out or get used up 08-FE002 A lumber company purchases and installs a wood chipper for $200,000. The chipper is classified as MACRS 7-year property. The chipper’s useful life is 10 years. The estimated salvage value at the end of 10 years is $25,000. Using straight-line depreciation, compute the first-year depreciation. a. $28,571.43 b. $20,000.00 c. $17,500.00 d. $25,000.00 Correct or Incorrect? Clear

  Check Answer

08-FE003

Which of the following is not true about depreciation?

a. Depreciation is not a cash flow b. To be depreciable, an asset must have a life longer than 1 year

c. A 5-year property will generate regular MACRS-GDS depreciation deductions in 6 fiscal years d. For MACRS-GDS an estimate of the salvage values is required 08-FE004 The depreciation deduction for year 11 of an asset with a 20year useful life is $4,000. If the salvage value of the asset was estimated to be zero and straight-line depreciation was used to calculate the depreciation deduction for year 11, what was the initial cost of the asset? a. $42,105 b. $67,682 c. $72,682 d. $80,000 Correct or Incorrect? Clear

  Check Answer

08-FE005 The depreciation deduction for year 11 of a 15-year property with a 20-year class life is $4,000. If the salvage value of the asset is estimated to be $5,000 and MACRS-GDS is used to calculate the depreciation deduction for year 11, what was the initial cost of the asset? a. $42,105 b. $67,682 c. $72,682 d. $80,000 08-FE006 MACRS-GDS deductions are a combination of which other methods of depreciation? a. Sum of Years Digits and Straight Line b. Sum of Years Digits and Declining Balance c. Double Declining Balance and 150% Declining Balance d. Double Declining Balance and Straight Line

Correct or Incorrect? Clear

  Check Answer

08-FE007 Production equipment used in the bottling of soft drinks (MACRS-GDS, 10-year property) is purchased and installed for $630,000. What is the depreciation deduction in the 4th year under MACRS-GDS? a. $90,720 b. $78,687 c. $72,576 d. $48,510 08-FE008 The depreciation allowance for a $100,000 MACRS-GDS asset was $8,550 after its 3rd year. What was the depreciation allowance after its 2nd year? a. $8,550 b. $9,500 c. $18,000 d. Cannot be determined with the information given Correct or Incorrect? Clear

  Check Answer

08-FE009 A lumber company purchases and installs a wood chipper for $200,000. The chipper is classified as MACRS 7-year property. The chipper’s useful life is 10 years. The estimated salvage value at the end of 10 years is $25,000. Using MACRS depreciation, compute the first-year depreciation. a. $17,500.00 b. $20,000.00 c. $25,007.50 d. $28,580.00

08-FE010 An x-ray machine at a dental office is MACRS 5-year property. The x-ray machine costs $6,000 and has an expected useful life of 8 years. The salvage value at the end of 8 years is expected to be $500. Assuming MACRS depreciation, what is the book value at the end of the third year? a. $1,584 b. $1,728 c. $3,916 d. $4,272 Correct or Incorrect? Clear

  Check Answer

08-FE011 Which of the following is (are) required to calculate MACRSGDS depreciation deductions? I. Property Class II. Salvage Value III. First Cost IV. Annual Maintenance Costs a. I and III only b. II and III only c. I, II, and III d. I, II, III, and IV 08-FE012 A new top-of-the-line semi-truck tractor (the tractor pulls the trailer, the combination making up an “18-wheeler”) costs $200,000 and is MACRS-GDS 3- year property. What is the depreciation charge and book value at the end of year 2 using MACRS-GDS with 50% bonus depreciation? a. $133,330 and $66,670 b. $44,450 and $155,550 c. $44,450 and $22,220 d. $14,810 and $14,820

Correct or Incorrect? Clear

  Check Answer

08-FE013 Computers and peripheral equipment will be purchased for $27,000. It is 5-year MACRS-GDS property and qualifies for 50% bonus depreciation. What are the depreciation values in years 1 and 2? a. $13,500 and $4,320 b. $16,200 and $4,320 c. $27,000 and $0 d. $5,400 and $8,640 08-FE014 A general-purpose truck is purchased for $36,000. It is a MACRS-GDS 5-year property and qualifies for 100% bonus depreciation and is expected to last for 8 years. What is the depreciation and book value at the end of year 3? a. $0 and $0 b. $6,912 and $10,368 c. $21,456 and $14,544 d. $21,456 and $0 Correct or Incorrect? Clear

  Check Answer

08-FE015 A robot for the manufacture of motor vehicles is purchased for $420,000. It has a class life of 10 years and is MACRS-GDS 7-year property. In evaluating the impact of this equipment on company finances, it is desired to compare the effect of using (1) no bonus depreciation, (2) 50% bonus depreciation, and (3) 100% bonus depreciation. Determine each of these three for depreciation taken in year 1. a. $60,018; $240,009; $420,000 b. $42,000; $210,000; $420,000

c. $60,018; $105,000; $210,000 d. $60,018; $315,000; $420,000 08-FE016 Vision-guided robotics equipment used to improve efficiency and purity when manufacturing food products is purchased for $1,200,000. It is expected to have a useful life of 10 years and is MACRS-GDS 7-year property. Determine the depreciation taken and book value at the end of year 8 if 50% bonus depreciation is used. a. $53,520 and $0 b. $26,760 and $0 c. $53,520 and $26,760 d. $26,760 and $26,760 Correct or Incorrect? Clear

  Check Answer

08-FE017 A 27.5-year residential rental property near a university is placed in service in the first month of a year. The property cost is $340,000. To improve the financials, it has been proposed that 50% bonus depreciation be used. Determine the depreciation the owner hopes to take in year 1. a. $170,000 b. $255,000 c. Bonus depreciation does not apply d. $181,849

Problems Section 8.1 Depreciation in Economic Analysis LEARNING OBJECTIVE 8.1 Demonstrate an understanding of depreciation concepts and terms.

08.01-PR001 Michelin is considering going “lights out” in the mixing area of the business that operates 24/7. Currently, personnel with a loaded cost of $600,000 per year are used to manually weigh real rubber, synthetic rubber, carbon black, oils, and other components prior to manual insertion in a Banbary mixer that provides a homogeneous blend of rubber for making tires. New technology is available that has the reliability and consistency desired to equal or exceed the quality of blend now achieved manually. It requires an investment of $3.75 million, with $80,000 per year operational costs and will replace all of the manual effort described above. a. How are the current manual expenditures handled for tax purposes? b. How would the new technology expenditures be handled for tax purposes? 08.01-PR002 Video Solution Taxes are paid each year on some measure of financial gain. We typically think of financial gain as “cash inflows minus cash outflows,” and yet simply subtracting outflows due to capital investments in plant and equipment that will be used over multiple years is not allowed. a. Why are these outflows not allowed to just be subtracted? b. Where do depreciation allowances fit into the tax picture, especially because they are not cash flows?

08.01-PR003 SteelTubes had sales of $300 million this year. Expenses were $250 million. Aside from these figures, the company also invested in new mills for carbon steel tubing, complete with peripheral loading, straightening, and coiling equipment plus facility reconfiguration totaling $14 million. SteelTubes believes the usable life of the mill will be only 7 years, owing to technological advances. There were no other financial considerations. a. Looking strictly at cash flows, what will be reported as the financial gain or loss? Is this a fair representation of financial performance?

b. If, for internal financial reporting, the manufacturer writes off equal amounts of the capital investment over the usable life, beginning this year, what will be the reported financial gain or loss? 08.01-PR004 Which of the following statements are TRUE? a. The reason for including a treatment of depreciation in this book is to allow you to develop a reasonably accurate report to the owners of a business regarding its value at any given point in time. b. Depreciation spreads investment costs over the useful life of equipment purchased. c. Depreciation allowances can be treated as expenses because they are cash flows. d. Depreciation affects income taxes, which are cash flows. Section 8.1.2 Language of Depreciation 08.01-PR005 Video Solution For each of the following assets, state whether the asset is tangible/intangible property, personal/real property, and depreciable/nondepreciable property. a. A cell phone tower. b. A plot of land for your personal use. c. A computer used in your job. d. A Mooney viscometer used in a polymers lab. e. An electric generator purchased by a public utility.

08.01-PR006 For each of the following assets, state whether the asset is tangible/intangible property, personal/real property, and depreciable/nondepreciable property. a. A melt-indexer used in a company research lab. b. A plot of land for the production of income. c. A restaurant franchise. d. An amateur radio tower attached to land with multiple antennas on it. e. Fencing and landscaping around an office complex. 08.01-PR007 For each asset given below, state whether the asset is tangible/intangible property, personal/real property, and depreciable/nondepreciable property. a. A tractor (part of tractor-trailer rig, an 18-wheeler). b. A copyright. c. An all-in-one copier, scanner, fax machine used in your business. d. A rental home for the purpose of generating rental income. 08.01-PR008 For each of the following assets, state whether the asset is tangible/intangible property, personal/real property, and depreciable/nondepreciable property. a. A computer used for personal e-mail, blogging, and hobbies. b. A file cabinet in your business office. c. A commercial delivery truck. d. An office complex for business. LEARNING OBJECTIVE 8.2 Calculate depreciation using the StraightLine (SLN), Declining Balance (DB), and Double Declining Balance (DDB) methods. Section 8.2.2 Straight-Line and Declining Balance Depreciation Methods

08.02-PR001 Video Solution A high-precision programmable router for shaping furniture components is purchased by Henredon for $190,000. It is expected to last 12 years and have a salvage value of $5,000. Calculate the depreciation deduction and book value for each year. a. Use straight-line depreciation. b. Use declining balance depreciation with a rate that ensures the book value equals the salvage value. c. Use double declining balance depreciation. d. Use double declining balance switching to straight-line depreciation.

08.02-PR002 A land grant university has upgraded its Course Management System (CMS), integrating the system throughout all of its main campus and branch campuses around the state. It has purchased a set of 15 servers and peripherals for needs associated with the CMS. The total cost basis is $120,000 and expected use will be 5 years, after which it will have no projected value. Calculate the depreciation deduction and book value for each year. a. Use straight-line depreciation. b. Use declining balance depreciation with a rate that ensures the book value equals the salvage value. c. Use double declining balance depreciation. d. Use double declining balance switching to straight-line depreciation. 08.02-PR003 A small truck is purchased for $17,000. The truck is expected to be of use to the company for 6 years, after which it will be sold for $3,500. a. Use straight-line depreciation.

b. Use declining balance depreciation with a rate that ensures the book value equals the salvage value. c. Use double declining balance depreciation. d. Use double declining balance switching to straight-line depreciation. 08.02-PR004 Video Solution A surface mount PCB placement/soldering line is to be installed for $1.6 million. It will have a salvage value of $100,000 after 5 years. Determine the depreciation deduction and the resulting unrecovered investment during each year of the asset’s life. a. Use straight-line depreciation. b. Use declining balance depreciation with a rate that ensures the book value equals the salvage value. c. Use double declining balance depreciation. d. Use double declining balance switching to straight-line depreciation.

08.02-PR005 A tractor for over-the-road hauling is purchased for $90,000. It is expected to be of use to the company for 6 years, after which it will be salvaged for $4,000. Calculate the depreciation deduction and the unrecovered investment during each year of the tractor’s life. a. Use straight-line depreciation. b. Use declining balance depreciation with a rate that ensures the book value equals the salvage value. c. Use double declining balance depreciation.

d. Use double declining balance switching to straight-line depreciation.

08.02-PR006 WindPower Inc. designs and commissions the manufacture of a wind-powered inverter-based constant voltage generator for research and experimentation with low-rated, highly variable, wind fields as a form of alternative energy. The unit cost $35,000 plus $3,000 for shipping and installation. After 3 years, WindPower had no further use for the experimental unit and was able to sell it for $2,000, less $500 for removal. WindPower had depreciated the generator cost over the 3 years with an estimated life of 5 years and a terminal book value of $1,000. All of the following parts relate to financial reporting, not computing income taxes. a. What is the total investment cost (basis) for this generator? b. What is the net market value actually received after 3 years? c. By what amount did the book value differ from the net market value at the end of 3 years if the following depreciation method was used? i. Straight-line depreciation. ii. Declining balance depreciation using a rate that ensures the book value equals the salvage value. iii. Double declining balance depreciation. iv. Double declining balance switching to straight-line depreciation. 08.02-PR007 A digitally controlled plane for manufacturing furniture is purchased on April 1 by a calendar-year taxpayer for $66,000. It is expected to last 12 years and have a salvage value of $5,000. Calculate the depreciation deduction during years 1, 4, and 8. a. Use straight-line depreciation. b. Use declining balance depreciation, with a rate that ensures the book value equals the salvage value. c. Use double declining balance depreciation.

d. Use declining balance depreciation, switching to straight-line depreciation. 08.02-PR008 A file server and peripherals are purchased in December by a calendar-year taxpayer for $8,000. The server will be used for 6 years and be worth $200 at that time. Calculate the depreciation deduction during years 1, 3, and 6. a. Use straight-line depreciation. b. Use declining balance depreciation, with a rate that ensures the book value equals the salvage value. c. Use double declining balance depreciation. d. Use declining balance depreciation, switching to straight-line depreciation. Section 8.3 Modified Accelerated Cost Recovery System (MACRS) LEARNING OBJECTIVE 8.3 Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS). 08.03-PR001 Which of the following statements are FALSE? a. MACRS is the only depreciation method approved by the IRS for computing income-tax liability and it is also the most commonly used method in the United States for financial reporting. b. MACRS stands for Modified Annuitized Cost Recovery System. c. MACRS-GDS is based on double declining balance switching to straightline depreciation. d. MACRS-ADS is based on straight-line depreciation. 08.03-PR002 Which of the following statements are TRUE? a. MACRS-GDS uses a half-year convention, whereas MACRS-ADS does not.

b. The half-year convention has the effect of depreciating over n − 1 full years (2, 3, . . . , n), and two half-years (1 and n + 1). c. The investment’s property class establishes the number of years over which the cost basis is to be recovered (depreciated). d. In general, MACRS-GDS has a longer recovery period (depreciation period) than MACRS-ADS. 08.03-PR003 A mold for manufacturing medical supplies (MACRS-GDS 5year property) is purchased at the beginning of the fiscal year for $30,000. The estimated salvage value after 10 years is $3,000. Calculate the depreciation deduction and the book value during each year of the asset’s life using MACRS-GDS allowances. 08.03-PR004 A panel truck (MACRS-GDS 5-year property) is purchased for $17,000. The truck is expected to be of use to the company for 6 years, after which it will be sold for $2,500. Calculate the depreciation deduction and the book value during each year of the asset’s life using MACRS-GDS allowances. 08.03-PR005 A digitally controlled plane for manufacturing furniture (MACRS-GDS 7-year property) is purchased on April 1 by a calendar-year taxpayer for $66,000. It is expected to last 12 years and have a salvage value of $5,000. Calculate the depreciation deduction during years 1, 4, and 8 using MACRS-GDS allowances. 08.03-PR006 Material-handling equipment used in the manufacture of grain products (MACRS-GDS 10-year property) is purchased and installed for $180,000. It is placed in service in the middle of the tax year. If it is removed just before the end of the tax year approximately 4.5 years from the date placed in service, determine the depreciation deduction during each of the tax years involved using MACRS-GDS allowances. 08.03-PR007 Repeat the previous problem (Problem 08.03-PR006) if the material-handling equipment is removed just after the tax year, again using MACRS-GDS allowances. 08.03-PR008 Electric utility transmission and distribution equipment (MACRS-GDS 20-year property) is placed in service at a cost of $300,000. It

is expected to last 30 years with a salvage value of $15,000. Determine the depreciation deduction and the book value during each year of the first 4 tax years using MACRS-GDS allowances. 08.03-PR009 A business building (MACRS-GDS 39-year property) is placed in service by a calendar-year taxpayer on January 4 for $300,000. Calculate the depreciation deduction for years 1 and 10, assuming the building is kept longer than 10 years, using MACRS-GDS allowances. 08.03-PR010 A building used for the overhaul of dewatering systems (MACRS-GDS 39-year property) is placed in service on October 10 by a calendar-year taxpayer for $140,000. It is sold almost 4 years later on August 15. Determine the depreciation deduction during years 1, 3, and 5 and the book value at the end of years 1, 3, and 5 using MACRS-GDS allowances. 08.03-PR011 A rental apartment complex (MACRS-GDS 27.5-year property) is placed in service by a calendar-year taxpayer on January 4 for $200,000. If the apartments are kept for 5 years and 2 months (sold on March 6), determine the depreciation deduction during each of the 6 tax years involved using MACRS-GDS allowances. 08.03-PR012 A rental house (MACRS-GDS 27.5-year property) is placed in service by a calendar-year taxpayer during July for $70,000. Determine the depreciation deduction and resulting book value for each applicable year using MACRS-GDS allowances. 08.03-PR013 For each of the following assets, state both the MACRS-GDS property class, if applicable, and the specific depreciation method to be used (e.g., 15-year property, 150% DBSLH). a. A melt-indexer used in a company research lab. b. A plot of land for the production of income. c. A restaurant franchise. d. An amateur radio tower attached to land with multiple antennas on it. e. Fencing and landscaping around an office complex. 08.03-PR014 For each of the following assets, state both the MACRS-GDS property class, if applicable, and the specific depreciation method to be used

(e.g., 15-year property, 150% DBSLH). a. A cell phone tower. b. A plot of land for your personal use. c. A computer used in your job. d. A Mooney viscometer used in a polymers lab. e. An electric generator purchased by a public utility. 08.03-PR015 For each of the following assets, state both the MACRS-GDS property class, if applicable, and the specific depreciation method to be used (e.g., 15-year property, 150% DBSLH). a. A computer used for personal e-mail, blogging, and hobbies. b. A file cabinet in your business office. c. A commercial delivery truck. d. An office complex for business. 08.03-PR016 For each of the following assets, state both the MACRS-GDS property class, if applicable, and the specific depreciation method to be used (e.g., 15-year property, 150% DBSLH). a. A tractor (part of tractor-trailer rig, an 18-wheeler). b. A copyright. c. An all-in-one copier, scanner, fax machine used in your business. d. A rental home for the purpose of generating rental income. 08.03-PR017 Electric generating and transmission equipment is placed in service at a cost of $3,000,000. It is expected to last 30 years with a salvage value of $250,000. a. What is the MACRS-GDS property class? b. Determine the depreciation deduction and the unrecovered investment during each of the first 4 tax years.

08.03-PR018 Bell’s Amusements purchased an expensive ride for their theme and amusement park situated within a city-owned Expo Center. Bell’s had a multi-year contract with Expo Center. The ride cost $1.2 million. Bell’s anticipated that the ride would have a useful life of 12 years, after which the net salvage value would be $0. After 4 years, the city and Bell’s were unable to come to an agreement regarding an extended contract. In order to expedite Bell’s departure, Expo Center agreed to purchase the ride and leave it in place. Right at the end of the 4th fiscal year, Expo Center paid to Bell’s the unrecovered investment (remember the half-year convention for MACRSGDS). Determine the amount paid, assuming: a. Straight-line depreciation used for valuation purposes was agreed upon. b. MACRS-GDS depreciation used for tax purposes (state the property class) was agreed upon. 08.03-PR019 A virtual mold apparatus for producing dental crowns permits an infinite number of shapes to be custom constructed based upon mold imprints taken by dentists. It costs $28,500 and is purchased at the beginning of the tax year. It is expected to last 9 years with no salvage value at that time. The dental supplier depreciates assets using MACRS and yet values assets of the company using straight-line depreciation. Determine the depreciation allowance and the unrecovered investment for each year: a. For tax purposes (be sure to identify the MACRS-GDS property class). b. For company valuation purposes. 08.03-PR020 Milliken uses a digitally controlled “dyer” for placing intricate and integrated patterns on manufactured carpet squares for home and commercial use. It is purchased on April 1 for $350,000. It is expected to last 8 years and have a salvage value of $30,000. a. What is the MACRS-GDS property class? b. Determine the depreciation deduction during each year of the asset’s 8year life. c. Determine the unrecovered investment at the end of each of the 8 years.

08.03-PR021 A high-precision programmable router for shaping furniture components is purchased by Henredon for $190,000. It is expected to last 12 years. Calculate the depreciation deduction and book value for each year using MACRS-GDS allowances. a. What is the MACRS-GDS property class? b. Assume the complete allowable depreciation schedule is used. c. Assume the asset is sold during the 5th year of use. 08.03-PR022 Material handling equipment used in the manufacture of sugar is purchased and installed for $250,000. It is placed in service in the middle of the tax year and removed from service 5.5 years later. Determine the MACRSGDS depreciation deduction during each of the tax years involved assuming: a. What is the MACRS-GDS property class? b. The equipment is removed 1 day before the end of the tax year. c. The equipment is removed 1 day after the end of the tax year. 08.03-PR023 Video Solution Englehard purchases a slurry-based separator for the mining of clay that costs $700,000 and has an estimated useful life of 10 years, a MACRS-GDS property class of 7 years, and an estimated salvage value after 10 years of $75,000. It was financed using a $200,000 down payment and a loan of $500,000 over a period of 5 years with interest at 10%. Loan payments are made in equal annual amounts (principal plus interest) over the 5 years. a. What is the amount of the MACRS-GDS depreciation taken in the 3rd year? b. What is the book value at the end of the 3rd year? c. Returning to the original situation, what is the amount of the MACRSGDS depreciation taken in the 3rd year if the separator is also sold during the 3rd year?

08.03-PR024 A portable concrete test instrument used in construction for evaluating and profiling concrete surfaces (MACRS-GDS 5-year property class) is purchased in December by a calendar-year taxpayer for $22,000. The instrument will be used for 6 years and be worth $2,000 at that time. a. Calculate the depreciation deduction for years 1, 3, and 6. b. If the instrument is sold in year 4, determine the depreciation deduction for years 1, 3, and 4. 08.03-PR025 A mold for manufacturing powdered metal firearm parts is purchased by Remington at the beginning of the fiscal year for $120,000. The estimated salvage value after 8 years is $10,000. Calculate the depreciation deduction and book value for each year using MACRS-GDS allowances. a. What is the MACRS-GDS property class? b. Assume the complete depreciation schedule is used. c. Assume the asset is sold during the 6th year of use. 08.03-PR026 A hydroprocessing reactor, an asset used in petroleum refining, is placed into service at a cost of $2.7 million. It is thought to have a useful life, with turnarounds and proper maintenance, of 18 years and will have a negligible salvage value at that time. a. What is the MACRS-GDS property class? b. Determine the depreciation deduction and the unrecovered investment over the depreciable life of the reactor. 08.03-PR027 A manufacturing system for fabricated metal products is purchased in the middle of the fiscal year for $800,000. The estimated salvage value after 10 years is $130,000. a. What is the MACRS-GDS property class? b. Determine the depreciation deduction during each year of the asset’s 10year life. c. Determine the unrecovered investment at the end of each of the 10 years.

08.03-PR028 Video Solution A surface mount PCB placement/soldering line for the manufacture of electronic components is to be installed for $1.6 million with an expected life of 6 years. Determine the depreciation deduction and the resulting unrecovered investment during each year of the asset’s life using MACRS-GDS allowances. a. What is the MACRS-GDS property class? b. Assume the line equipment will be sold shortly after the 6th year. c. Assume the line equipment is sold during the 3rd year of use.

08.03-PR029 A gas-powered electric generator is purchased by a public utility as part of an expansion program. It is expected to be useful, with proper maintenance, for an estimated 30 years. The cost is $17 million, installed. The salvage value at the end of 30 years is expected to be 10% of the original cost, not counting installation. a. What is the MACRS-GDS property class? b. Determine the depreciation deduction and the unrecovered investment for years 1, 5, and the last depreciable year of the generator. 08.03-PR030 A tractor for over-the-road hauling is purchased for $90,000. It is expected to be of use to the company for 6 years, after which it will be salvaged for $4,000. Calculate the depreciation deduction and the unrecovered investment during each year of the tractor’s life using MACRS-GDS allowances. a. What is the MACRS-GDS property class? b. Assume the tractor is used for the full 6 years. c. Assume the tractor is sold during the 4th year of use. d. Assume the tractor is sold during the 3rd year of use.

08.03-PR031 Pretend that you have misplaced your MACRS tables. Develop the tables for a property class of 3 years assuming 200% DB depreciation switching to straight-line; half-year convention; salvage value equal to $0. Your answers should match those for MACRS-GDS 3-year property. 08.03-PR032 Pretend that you have misplaced your MACRS tables. Develop the tables for a property class of 5 years assuming 200% DB depreciation switching to straight-line; half-year convention; salvage value equal to $0. Your answers should match those for MACRS-GDS 5-year property. 08.03-PR033 Suppose the IRS has decided to institute a new MACRSGDS property class of only 2 years. It will follow the usual depreciation conventions, determined in the same way as 3-, 5-, 7-, and 10-year property. Determine the yearly MACRS-GDS percentages for each year. 08.03-PR034 Suppose the IRS has decided to institute a new MACRS-GDS property class of 4 years. It will follow the usual depreciation conventions, determined in the same way as 3-, 5-, 7-, and 10-year property. Determine the yearly MACRS-GDS percentages for each year. 08.03-PR035 A nonresidential business building is placed in service by a calendar year taxpayer on March 3 for $300,000. a. What is the MACRS-GDS property class? b. Calculate the depreciation deduction for years 1, 4, and 8 if it is kept longer than 8 years. c. Calculate the depreciation deduction for years 1, 4, and 8 if it is sold on August 12 in the 8th calendar year. d. Why is it highly unlikely the building will ever be completely depreciated? 08.03-PR036 Now that you are making the big bucks, your spouse has decided to venture into the rental property business. Your spouse purchases a rental house and after making some improvements it has a basis of $85,000. Your spouse places it in service as a calendar-year taxpayer during May and sells it in September, just over 4 years later. a. What is the MACRS-GDS property class?

b. Determine the depreciation deduction during each of the years involved. c. Determine the unrecovered investment during each of the years involved. 08.03-PR037 A permanent steel building used for the overhaul of dewatering systems (engines, pumps, and wellpoints) is placed in service on July 10 by a calendar-year taxpayer for $240,000. It is sold almost 5 years later on May 15. a. What is the MACRS-GDS property class? b. Determine the depreciation deduction during each of the years involved. c. Determine the unrecovered investment during each of the years involved. 08.03-PR038 A residential rental apartment complex is placed in service by a calendar-year taxpayer on February 27 for $530,000. The apartments are kept for slightly more than 6 years and sold on March 6. a. What is the MACRS-GDS property class? b. Determine the depreciation deduction during each of the 7 years involved. c. Determine the unrecovered investment during each of the 7 years involved. 08.03-PR039 Video Solution Equipment for manufacturing vegetable oil products is purchased from Alfa. Items such as oil expellers, filter presses, and a steam generator are purchased for $1.2 million. These devices are expected to be used for 11 years with no salvage value at that time. Compare MACRS to traditional depreciation methods by calculating yearly depreciation allowances, present worth of the depreciation allowances, and book value for each year using each of the following. MARR is 9%. a. MACRS-GDS as is proper over its property class depreciation life. b. DDB taking a full deduction in the first year, with the last deduction in year 10.

c. DDB switching to straight-line, taking a full deduction in the 1st year, with the last deduction in year 10. d. SLN taking a full deduction in the 1st year, with the last deduction in year 10.

08.03-PR040 A building with business offices, a reception area, and numerous small diagnosis rooms is placed in service by a group of three orthopedic surgeons on January 4 for $650,000. a. What is the MACRS-GDS property class? b. Calculate the depreciation deduction for years 1, 4, and 7 if it is kept longer than 7 years. c. Calculate the depreciation deduction for years 1, 4, and 7 if it is sold on July 1 in the 7th calendar year. d. Why is it highly unlikely the building will ever be completely depreciated? 08.03-PR041 Ultra-clean special handling devices used in the filling process for the manufacture of baby food are placed into use at a cost of $850,000. These devices are expected to be useful for 4 years with a negligible salvage value at that time. Compare MACRS to traditional depreciation methods by calculating yearly depreciation allowances, present worth of the depreciation allowances, and book value for each year using each of the following. MARR is 11%. a. MACRS-GDS as is proper over its property class depreciation life. b. DDB taking a full deduction in the first year, with the last deduction in year 3. c. DDB switching to straight-line, taking a full deduction in the 1st year, with the last deduction in year 3. d. SLN taking a full deduction in the 1st year, with the last deduction in year 3.

Section 8.4 MACRS with Bonus Depreciation LEARNING OBJECTIVE 8.4 Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS) with bonus depreciation. 08.04-PR001 A mold for manufacturing medical supplies (MACRS-GDS 5year property) is purchased at the beginning of the fiscal year for $30,000. The estimated salvage value after 10 years is $3,000. Calculate the depreciation deduction and the book value during each year of the asset’s life using MACRS-GDS allowances. a. Use 50% bonus depreciation. b. Use 100% bonus depreciation. 08.04-PR002 A panel truck (MACRS-GDS 5-year property) is purchased for $17,000. The truck is expected to be of use to the company for 6 years, after which it will be sold for $2,500. Calculate the depreciation deduction and the book value during each year of the asset’s life using MACRS-GDS allowances. a. Use 50% bonus depreciation. b. Use 100% bonus depreciation. 08.04-PR003 A digitally controlled plane for manufacturing furniture (MACRS-GDS 7-year property) is purchased on April 1 by a calendar-year taxpayer for $66,000. It is expected to last 12 years and have a salvage value of $5,000. Calculate the depreciation deduction during years 1, 4, and 8 using MACRS-GDS allowances. a. Use 50% bonus depreciation. b. Use 100% bonus depreciation. 08.04-PR004 Material-handling equipment used in the manufacture of grain products (MACRS-GDS 10-year property) is purchased and installed for $180,000. It is placed in service in the middle of the tax year. If it is removed just before the end of the tax year approximately 4.5 years from the date

placed in service, determine the depreciation deduction during each of the tax years involved using MACRS-GDS allowances. a. Use 50% bonus depreciation. b. Use 100% bonus depreciation. 08.04-PR005 Repeat the previous problem (Problem 08.04-PR004) if the material-handling equipment is removed just after the tax year, again using MACRS-GDS allowances. a. Use 50% bonus depreciation. b. Use 100% bonus depreciation. 08.04-PR006 Electric utility transmission and distribution equipment (MACRS-GDS 20-year property) is placed in service at a cost of $300,000. It is expected to last 30 years with a salvage value of $15,000. Determine the depreciation deduction and the book value during each year of the first 4 tax years using MACRS-GDS allowances. a. Use 50% bonus depreciation. b. Use 100% bonus depreciation. 08.04-PR007 Bell’s Amusements purchased an expensive ride for their theme and amusement park situated within a city-owned Expo Center. Bell’s had a multi-year contract with Expo Center. The ride cost $1.2 million. Bell’s anticipated that the ride would have a useful life of 12 years, after which the net salvage value would be $0. After 4 years, the city and Bell’s were unable to come to an agreement regarding an extended contract. In order to expedite Bell’s departure, Expo Center agreed to purchase the ride and leave it in place. Right at the end of the 4th fiscal year, Expo Center paid to Bell’s the unrecovered investment (remember the half-year convention for MACRSGDS). Determine the property class and calculate the amount paid, assuming: a. 50% bonus depreciation. b. 100% bonus depreciation. 08.04-PR008 A virtual mold apparatus for producing dental crowns permits an infinite number of shapes to be custom constructed based upon mold

imprints taken by dentists. It costs $28,500 and is purchased at the beginning of the tax year. It is expected to last 9 years with no salvage value at that time. The dental supplier depreciates assets using MACRS. Determine the depreciation allowance for each year using: a. 50% bonus depreciation. b. 100% bonus depreciation.

Chapter 8 Summary and Study Guide Summary 8.1: Depreciation in Economic Analysis

Learning Objective 8.1: Demonstrate an understanding of depreciation concepts and terms. (Section 8.1) Most property deceases in value with use and time—it depreciates. U.S. income‐tax law permits depreciation allowances to be treated as expenses and deducted from taxable income. This is an important concept to consider, as it can affect the way investments and annual operating costs are treated from an after‐tax (as opposed to before‐tax) perspective. The depreciation method used can significantly impact the present worth of depreciation allowances and thereby can significantly impact the income taxes a business pays. The Internal Revenue Service has the following requirements for depreciable property: It must be used in business or held for the production of income. It must have a life that can be determined, and that life must be longer than a year. It must be something that wears out, decays, gets used up, becomes obsolete, or loses value from natural causes. Depreciable property may be tangible or intangible. Tangible property can be seen or touched and can be categorized as personal or real. Personal property is goods such as cars, trucks, machinery, furniture, equipment, and anything that is tangible except real property. Real property is land and generally anything that is erected on, growing on, or attached to it.

Intangible property cannot be seen or touched but has value to the owner; it includes copyrights, brands, software, goodwill, formulas, designs, patents, trademarks, licenses, information bases, and franchises. 8.2: Straight‐Line and Declining Balance Depreciation Methods

Learning Objective 8.2: Calculate depreciation using the Straight‐Line (SLN), Declining Balance (DB), and Double Declining Balance (DDB) methods. (Section 8.2) SLN, DB, and DDB represent the first category of depreciation methods typically used for financial reporting purposes. In SLN, depreciation charges are uniform across the useful life of the property. DB and DDB are extensions of SLN with intent to accelerate the depreciation schedule. The following table summarizes the key equations for depreciation and book values using these methods. Depreciation Method Straight Line Declining Balance

Depreciation Value Bt

dt = (P − F )/n

dt

= =

Book Value =

P − tdt

=

P − t(P − F )/n

pBt−1 pP (1 − p)

t−1

Bt = P (1 − p)

t

It is often desirable to switch from using a specified declining balance rate to a straight‐line rate at the point where the straight‐line depreciation charges exceed those of the declining balance rate. The equation that defines this point is the smallest value of t for which: Bt−1 − F n − (t − 1)

> pBt−1

8.3: Modified Accelerated Cost Recovery System (MACRS)

(8.6)

Learning Objective 8.3: Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS). (Section 8.3) The second category of depreciation methods is for income‐tax purposes. In the United States, MACRS is the only depreciation method approved for use by the IRS. MACRS is an accelerated cost recovery system using a half‐year convention for various types of properties. Accelerated depreciation methods are preferred to nonaccelerated depreciation methods because of the desire to maximize the present worth of depreciation allowances. The recovery period for depreciable property is specified by the IRS and is usually shorter than the useful life of the property. 8.4: MACRS with Bonus Depreciation

Learning Objective 8.4: Calculate depreciation using MACRS with Bonus Depreciation. (Section 8.4) In recent years, the IRS has allowed bonus depreciation to be included in the allowance for the first year for qualifying investments. With the exception of buildings, all capital expenditures considered in chapters involving depreciation in performing economic justifications are considered to qualify for bonus depreciation. To date, the amount of the bonus has been either 50% or 100% of the cost basis for the asset. With 50% bonus depreciation, for the first year, one‐half of the initial cost basis is added to the standard allowance based on an adjusted cost basis; thereafter, the standard allowance is applied to the adjusted cost basis. With 50% bonus depreciation, the adjusted cost basis equals one‐half of the initial cost basis. With 100% bonus depreciation, the allowance the first year is, in fact, the initial cost basis. Therefore, the entire investment is recovered with the first year depreciation allowance.

Important Terms and Concepts Depreciation The concept that takes into account the fact that most property decreases in value with use and time. Depreciation Allowances The periodic amounts that may be deducted from taxable income—these amounts are treated as expenses even though they are not cash flows. Depreciable Property Property meeting the specific requirements as defined by the U.S. Internal Revenue Service whereby it is used in business or held for the production of income; has a life that can be determined, and that life must be longer than one year; and is something that wears out, decays, gets used up, becomes obsolete, or loses value from natural causes. Tangible Property Property that can be seen or touched and can be categorized as personal or real. Personal Property Goods such as cars, trucks, machinery, furniture, equipment, and anything that is tangible except real property. Real Property Land and generally anything that is erected on, growing on, or attached to it. Intangible Property Property that cannot be seen or touched, but has value to the owner such as copyrights, brands, software, goodwill, formulas, designs, patents, trademarks, licenses, information bases, and franchises. Amortization As defined by the U.S. income tax code “the recovery of certain capital expenditures, that are not ordinarily deductible, in a manner that is similar to straight-line depreciation.” Straight-Line Depreciation (SLN)

The depreciation method that treats annual depreciation charges as a uniform annual series. Cost Basis The amount paid for acquisition and installation of a depreciable asset. Book Value The undepreciated portion of an asset determined by subtracting the cumulative depreciation charge from the cost basis. Also referred to as the unrecovered investment, the adjusted cost basis, and the adjusted basis. Declining Balance (DB) A type of accelerated depreciation method that produces a negative geometric series of depreciation charges providing larger depreciation charges in the early years and smaller depreciation charges in the later years. Double Declining Balance (DDB) A type of accelerated depreciation method that is twice, or 200%, the straight-line rate. Modified Accelerated Cost Recovery System (MACRS) The only depreciation method approved by the U.S. IRS for computing income-tax liability. Modified Accelerated Cost Recovery System– General Depreciation System (MACRSGDS) The MACRS depreciation method based on double declining balance switching to straight-line depreciation. Modified Accelerated Cost Recovery System– Alternative Depreciation System (MACRSADS) The MACRS depreciation method based on straightline depreciation, with a longer recovery than GDS. Half-Year Convention A feature used in both MACRS-GDS and MACRSADS assuming that on average, a property is used for half of the first year of service and half of the last year of service.

Economic Stimulus Act In 2008, to stimulate capital investments, the U.S. Congress created provisions in U.S. tax law allowing businesses to rapidly recover capital expenditures by providing additional first-year depreciation allowances, called bonus depreciation. Each year since 2008, such provisions have been in place, with the magnitude of the bonus depreciation being either 50% or 100%. Bonus Depreciation To stimulate capital expenditures, the U.S. Congress frequently creates provisions for additional first-year depreciation allowances, called bonus depreciation. Tax Relief Act In 2008, the U.S. Congress named its income tax provision the Economic Stimulus Act to stimulate capital expenditures by corporations. In 2010, its provision was named the Tax Relief Act. The 2008 Economic Stimulus Act allowed an additional first-year depreciation allowance equal to 50% of the original cost basis. The 2010 Tax Relief Act increased the allowance to 100% of the original cost basis.

Chapter 8 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 8.1: Depreciation in Economic Analysis LO 8.1: Demonstrate an understanding of depreciation concepts and terms. Video Lesson: Depreciation Video Lesson Notes: Depreciation Video Solution: 08.01-PR002 Video Solution: 08.01-PR005 8.2: Straight-Line and Declining Balance Depreciation Methods LO 8.2: Calculate depreciation using the Straight-Line (SLN), Declining Balance (DB), and Double Declining Balance (DDB) methods. Excel Video Lesson: SLN Financial Function Excel Video Lesson Spreadsheet: SLN Financial Function Excel Video Lesson: RATE Financial Function Excel Video Lesson Spreadsheet: RATE Financial Function Excel Video Lesson: DB Financial Function Excel Video Lesson Spreadsheet: DB Financial Function Excel Video Lesson: DDB Financial Function Excel Video Lesson Spreadsheet: DDB Financial Function Excel Video Lesson: VDB Financial Function Excel Video Lesson Spreadsheet: VDB Financial Function Video Example 8.1: Straight-Line Depreciation Applied to the SMP Machine

Video Example 8.2: Declining Balance Depreciation Applied to the SMP Machine Video Solution: 08.02-PR001 Video Solution: 08.02-PR004 8.3: Modified Accelerated Cost Recovery System (MACRS) LO 8.3: Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS). Video Example 8.4: Applying MACRS with the SMP Machine Investment Video Solution: 08.03-PR023 Video Solution: 08.03.PR028 Video Solution: 08.03.PR039 8.4: MACRS with Bonus Depreciation LO 8.4: Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS) with bonus depreciation. These chapter-level resources will help you with your overall understanding of the content in this chapter. Appendix A: Time Value of Money Factors Appendix B: Engineering Economic Equations Flashcards: Chapter 08 Excel Utility: TVM Factors: Table Calculator Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values Excel Utility: Monthly Payment Sensitivity Excel Utility: TVM Factors: Discrete Compounding

Excel Utility: TVM Factors: Geometric Series Future Worth Excel Utility: TVM Factors: Geometric Series Present Worth Excel Data Files: Chapter 08

CHAPTER 8 Depreciation LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 8.1 Demonstrate an understanding of depreciation concepts and terms. (Section 8.1) 8.2 Calculate depreciation using the Straight-Line (SLN), Declining Balance (DB), and Double Declining Balance (DDB) methods. (Section 8.2) 8.3 Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS). (Section 8.3) 8.4 Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS) with bonus depreciation. (Section 8.4)

Engineering Economics in Practice Amazon Amazon.com, or Amazon, incorporated in 1994, is a Seattle, Washington based corporation. Its Nasdaq symbol is “AMZN,” and it is known as a multinational technology company with key competencies in e-commerce, cloud computing and artificial intelligence. Amazon is the largest internet company by revenue in the world and the second largest employer (behind Walmart) in the United States. At its helm is Jeffrey Bezos, serving as President, Chief Executive Officer and Chairman of the Board. On March 6, 2018, Forbes declared Bezos as the wealthiest person in the world with a reported net worth of $112 billion. Amazon utilizes a multi-level e-commerce strategy. In its early inception it focused on business-toconsumer relationships between itself and its customers and business-to-business relationships between itself and its suppliers. Later it expanded into customer-to-customer relationships acting as an intermediary to facilitate transactions. In 2017, Amazon acquired Whole Foods Market for $13.4 billion, increasing its presence as a brick-and-mortar retailer. This strategy was believed by many analysts as a direct challenge to Walmart’s traditional retail stores. In 2018 the company employed 647,500 employees (compared to 566,000 in 2017, a 14% increase) and reported revenue of $232.9 billion (compared to $177.9 billion in 2017, a 31% increase), net income of $10.1 billion (compared to $3.0 billion in 2017, a 332% increase), and total assets of $162.6 billion (compared to $131.3 billion in 2017, a 14% increase). For an annual fee, shoppers can become Amazon Prime members entitling them to free fast shipping, streaming of movies, exclusive shopping deals and more. It is estimated that in June 2018, Amazon had 95 million Prime members being served by more than 75 fulfillment centers and 25 sortation centers across North America. Clearly this rapid growth both domestically and globally increases the complexity of Amazon’s business placing a strain on its management, personnel, operations, systems, technical performance, and finances. The company’s expansion into new products, services technologies and geographic regions opens them to additional business, legal, financial and competitive risks. They acknowledge that they have relatively little operating experience in international market segments exposing themselves further to a number of risks putting their profitability in this market in jeopardy. Amazon’s business has seasonality and they have invested heavily on a massive amount of server capacity for their website, especially to handle the excessive traffic during the December holiday season. Amazon’s state sales tax collection policy has changed over the years as it did not collect any state sales taxes in its early years. This prompted states to create online sales tax laws thereby compelling Amazon to collect sales taxes from customers to satisfy a complaint that the company had a competitive advantage over companies with a physical store presence. When a firm’s capital assets are distributed around the world, when they are expanding rapidly, when income-tax laws vary from country to country, and when capital equipment is purchased, replaced, and modified frequently, it is a challenge to maintain accurate records regarding the value of assets owned by the shareholders. Discussion Questions 1. What impact does depreciation have on capital investment decision making at Amazon? 2. What will be the impact of these capital investment decisions on income for Amazon and its shareholders? 3. Amazon is experiencing rapid growth both domestically and globally. What complexities does this create for the company?

4. Amazon’s business has seasonality with peak periods during the December holiday season. How does this impact their business?

Introduction Most property decreases in value with use and time—that is, it depreciates. Depreciation is the subject of this chapter. We present it to ensure that you understand the implications of making investments in things that must (by U.S. tax law) be depreciated versus those that do not. Also, we want to ensure that you understand how a depreciation method can impact the profitability of an investment alternative. Depreciation The concept that takes into account the fact that most property decreases in value with use and time. In addition to learning the mechanics of calculating depreciation charges, we want you to gain an understanding of why depreciation and the accompanying income-tax treatments are important in performing engineering economic analyses. Engineers make decisions in the process of designing products and processes that can have significant after-tax effects. So, as you learn the material in the chapter, don’t lose sight of WHY you are learning about depreciation. Remember, you are learning about depreciation because your design decisions can affect the way investments and annual operating costs are treated from an income-tax perspective. In the next chapter you will further explore the topic of income taxes as they relate to engineering economic analyses.

8.1 Depreciation in Economic Analysis LEARNING OBJECTIVE Demonstrate an understanding of depreciation concepts and terms. Video Lesson: Depreciation

8.1.1 The Role of Depreciation in Economic Analysis We use a cash flow approach throughout this book. For that reason, we seldom use the word depreciation, because depreciation is not a cash flow. Typically, there are two cash flows associated with a capital asset: what you pay for it and what you get when you sell or trade it for a replacement. Depreciation is an artifice that reflects the decrease in the asset’s value over time or with usage. A principal reason for developing the concept of depreciation is to allow a reasonably accurate report to the owners of a business regarding its value at any given point in time. Another reason is to allow reasonable estimates to be made concerning the cost of doing business; prices need to be set at levels that will recapture investments made in depreciable property—property that is used to produce the goods and services sold by the business. As such, when the time comes to replace an asset, funds will be available to do so (unless prices have increased over the asset’s life). A report to the owners of the business regarding its profitability, based strictly on cash flows, can be misleading. For example, suppose you own a business with annual revenues that exceed annual expenses by, say, $1 million. This year, however, you spend $1.5 million on revenue-producing equipment that will be used for 5 years. If all you count is cash, then a cash flow calculation will indicate that you lost $0.5 million this year. On the other hand, if you spread the $1.5 million investment over the 5-year period of its use, then your business will be accurately portrayed as being profitable.

To spread the investment costs over the useful life of the equipment purchased, there are various approaches you can use. We consider a number of them in this chapter. As good as the foregoing arguments are for considering depreciation, they are not the reason for including a treatment of depreciation in this book. We include it because U.S. income-tax law permits depreciation allowances to be deducted from taxable income; in other words, depreciation allowances can be treated as expenses even though they are not cash flows. As a result, depreciation affects income taxes, which are cash flows. Therefore, when comparing investment alternatives using an after-tax analysis, an accurate consideration of deprecation becomes important. Depreciation Allowances The periodic amounts that may be deducted from taxable income—these amounts are treated as expenses even though they are not cash flows. For example, whether to design a production process that will be performed by people versus one that will be performed by robots is a typical choice made by mechanical and manufacturing engineers. But the aftertax consequences are quite different, depending on the choice made. If people perform the work, labor costs are treated as expenses and are fully deducted from taxable income in the year in which they occur. If robots perform the work, however, only the robots’ operating costs are treated as expenses. The acquisition cost for the robots must be depreciated. As such, the investment will be recovered over multiple years.

8.1.2 The Language of Depreciation Business expenditures are either expensed or depreciated. Expensing an expenditure is akin to depreciating it fully in the year in which it occurs. Expenditures for labor, materials, and energy are examples of items that are fully deducted from taxable income. On the other hand, expenditures for production equipment, vehicles, and buildings cannot be fully deducted from taxable income in the year in which they occur; instead, they must be spread out or distributed over some allowable recovery period. Technically, U.S. tax law permits deduction from taxable income of a reasonable allowance for wear and tear, natural decay or decline, exhaustion, or obsolescence of property used in a trade or business or of property held for producing income. Specifically, the Internal Revenue Service requires that the following requirements be met for depreciable property: 1. It must be used in business or held for the production of income. 2. It must have a life that can be determined, and that life must be longer than a year. 3. It must be something that wears out, decays, gets used up, becomes obsolete, or loses value from natural causes. Depreciable Property Property meeting the specific requirements as defined by the U.S. Internal Revenue Service whereby it is used in business or held for the production of income; has a life that can be determined, and that life must be longer than one year; and is something that wears out, decays, gets used up, becomes obsolete, or loses value from natural causes. Depreciable property may be tangible or intangible. Tangible property can be seen or touched and can be categorized as personal or real. Personal property is goods such as cars, trucks, machinery, furniture, equipment, and anything that is tangible except real property. Real property is land and generally anything that is erected on, growing on, or attached to it. Land, by itself, does not qualify for depreciation, but the buildings, structures, and equipment on it do qualify; land does not qualify because it does not pass the third requirement, given above. In contrast to tangible property, intangible property cannot be seen or touched

but has value to the owner; it includes copyrights, brands, software, goodwill, formulas, designs, patents, trademarks, licenses, information bases, and franchises. Tangible Property Property that can be seen or touched and can be categorized as personal or real. Personal Property Goods such as cars, trucks, machinery, furniture, equipment, and anything that is tangible except real property. Real Property Land and generally anything that is erected on, growing on, or attached to it. Intangible Property Property that cannot be seen or touched, but has value to the owner such as copyrights, brands, software, goodwill, formulas, designs, patents, trademarks, licenses, information bases, and franchises. A term used synonymously with depreciation is amortization. Generally speaking, depreciation is associated with personal property and real property (other than land), whereas amortization is associated with intangible property. The U.S. income tax code defines amortization as “the recovery of certain capital expenditures, that are not ordinarily deductible, in a manner that is similar to straight-line depreciation.” Amortization As defined by the U.S. income tax code “the recovery of certain capital expenditures, that are not ordinarily deductible, in a manner that is similar to straight-line depreciation.” (While intangible property is amortized (depreciated) for financial reporting purposes, it is expensed for income-tax purposes. Business expenditures for software, for example, can be deducted from taxable income, although they are typically spread out uniformly over the amortization period for financial reports.)

Concept Check 08.01-CC001 Which of the following statements about depreciation is not true? a. Depreciation helps companies estimate the cost of doing business b. U.S. tax law permits deductions from taxable income for depreciation c. Depreciable property may be tangible or intangible d. Depreciation is a cash flow that is recognized as an expense to a business

Concept Check 08.01-CC002 Which of the following items would not be depreciable in computing income taxes? a. A company’s corporate headquarter building b. Land on which a company’s factory sits c. A machine tool used for production d. A company’s office furniture

8.2 Straight-Line and Declining Balance Depreciation Methods LEARNING OBJECTIVE Calculate depreciation using the Straight-Line (SLN), Declining Balance (DB), and Double Declining Balance (DDB) methods. There are several methods of calculating depreciation. Firms may use any of the methods described in this section (SLN, DB, DDB, or DDB with switch to SLN) for accounting and financial reporting purposes, though use of the straight-line (SLN) method is most common. For income-tax purposes, however, firms must use one of the Modified Accelerated Cost Recovery System (MACRS) methods described later in this chapter.

8.2.1 Straight-Line Depreciation (SLN) Straight-line depreciation is the oldest of the depreciation methods in use today. As the name implies, annual depreciation charges form a uniform annual series. If P denotes the cost basis, i.e., the amount paid for acquisition and installation of a depreciable asset, and F denotes the asset’s salvage value at the end of the useful life, recovery period, or depreciable life of n years, then the annual SLN depreciation charge dt for year t is dt

= (P − F )/n

(8.1)

=SLN(P,F,n)

Straight-Line Depreciation (SLN) The depreciation method that treats annual depreciation charges as a uniform annual series. Cost Basis The amount paid for acquisition and installation of a depreciable asset. where SLN is the Excel® worksheet function for calculating the annual depreciation charge. The syntax for the SLN function, SLN(cost,salvage,life), is self-explanatory. Excel® Video Lesson: SLN Financial Function The undepreciated portion of an asset is called the book value. Also referred to as the unrecovered investment, the adjusted cost basis, and the adjusted basis, the book value at the end of year t, Bt, based on straight-line depreciation, is the cost basis less the cumulative depreciation charge, or Bt

=

P − tdt

=

P − t(P − F )/n

(8.2)

Book Value The undepreciated portion of an asset determined by subtracting the cumulative depreciation charge from the cost basis. Also referred to as the unrecovered investment, the adjusted cost basis, and the adjusted basis. Hence, book value is a straight line. Further, because the salvage value is incorporated in the depreciation calculation, book value at the end of the recovery period is equal to the salvage value. While corporate income taxes, and consequently engineering project cash flows, are based on MACRS depreciation methods, we know of companies that require the use of straight-line depreciation in engineering economic justifications. Why? We were given the following reasons:

1. Management does not want an investment’s economic viability to be decided on the basis of the depreciation method used. Recognizing that cash flow estimates are just that, estimates, and they are likely to prove to be incorrect, management does not feel that it is justified to be so precise regarding depreciation when errors in the cash flow estimates can easily offset any gains made by using accelerated depreciation. 2. Management wishes to use a standardized approach and, because of its simplicity, chooses to use SLN in all economic justifications.

EXAMPLE 8.1 Straight-Line Depreciation Applied to the SMP Machine Video Example An SMP machine costs $500,000 and has a $50,000 salvage value after 10 years of use. Using straight-line depreciation, what are the depreciation charge and book value for the machine after year 5? Key Data Given Cost Basis P = $500,000; Salvage Value F = $50,000; Useful Life n = 10 yrs Find d5, B5 Solution The following values occur for t = 5, with SLN: d5

= ($500,000 − $50,000)/10 = $45,000 =SLN(500000,50000,10) = $45,000

B5

= $500,000 − 5($45,000) = $275,000

8.2.2 Declining Balance and Double Declining Balance Depreciation (DB and DDB) Declining balance depreciation is an accelerated depreciation method. As such, it provides larger depreciation charges in the early years and smaller depreciation charges in the later years. Declining Balance (DB) A type of accelerated depreciation method that produces a negative geometric series of depreciation charges providing larger depreciation charges in the early years and smaller depreciation charges in the later years. Accelerated depreciation methods are desirable because they take into account the time value of money. Assuming a TVOM greater than 0, because depreciation charges can be deducted from taxable income, after-tax present worth is maximized when the present worth of depreciation charges is maximized. Just as one prefers to receive money sooner rather than later, due to the TVOM, so should one prefer to depreciate an asset sooner rather than later.

Where SLN produced a uniform annual series of depreciation charges, DB produces a negative geometric series of depreciation charges. With DB, the depreciation charge in a given year is a fixed percentage of the book value at the beginning of the year. Letting p denote the DB percentage or depreciation rate used, dt

= =

(8.3)

pBt−1 pP (1 − p)

t−1

and Bt = P (1 − p)

(8.4)

t

Notice, salvage value is not incorporated in the calculations for DB. As a result, the book value at the end of the cost recovery period is unlikely to equal the salvage value obtained for the asset. (We will address the income-tax implications of book values not equaling salvage values in Chapter 9.) What depreciation rate or value of p should be used? Among the most often used values are 125%, 150%, and 200% (the maximum amount allowed by the IRS) of the straight-line rate, which is 1/n. If it is desired to use a percentage such that the book value equals the salvage value after n years, then, from Equation 8.4, F = P (1 − p)

n

Solving for p gives p

= 1 − (F |P )

1/n

(8.5)

=−RATE(n,,−P ,F )

Excel Video Lesson Spreadsheet: RATE Financial Function Notice the minus sign between the equal sign and the Excel® RATE worksheet function. Also, when p = 2/n, which is twice, or 200%, the straight-line rate, the DB plan is called double declining balance (DDB) depreciation. Double Declining Balance (DDB) A type of accelerated depreciation method that is twice, or 200%, the straight-line rate. Excel® has three declining balance worksheet functions: DB, DDB, and VDB. The VDB, or variable declining balance function, returns the depreciation of an asset for any period specified, including partial periods, using the DDB method or another specified method. The answers obtained from the Excel® DB function sometimes differ from answers obtained using Equations 8.3 and 8.4. Therefore, if Excel® functions are to be used, DDB and VDB are recommended. The syntaxes for DDB and VDB are DDB(cost,salvage,life,period,factor); and VDB(cost,salvage,life,start_period,end_period,factor,no_switch) Excel® Video Lesson: DB Financial Function where

cost = the cost basis of the asset salvage = the salvage value of the asset life = the cost recovery period of the asset period = the time period for which the depreciation charge is desired factor = the depreciation rate (p) to use; in the case of DDB, if omitted, it is assumed to equal 2, the double-declining rate start_period = the starting period for which depreciation is calculated; must use the same units as life end_period = the ending period for which depreciation is calculated; must use the same units as life no_switch = a logical value specifying whether to switch to straight-line depreciation when the straightline depreciation is greater than the declining balance calculation Excel® Video Lesson: DDB Financial Function The Excel® DDB function uses a depreciation rate equal to 2/n. When using the Excel® VDB function, switching from declining balance to straight-line depreciation is an option; if the declining balance depreciation rate is not specified, it is assumed to equal 2/n. Excel® Video Lesson: VDB Financial Function We address the concept of switching depreciation methods in the next section. Because we have introduced the syntax for the Excel® VDB function, note that when no_switch is TRUE, Excel® does not switch to straight-line depreciation even when the depreciation is greater than the declining balance calculation; if no_switch is FALSE or omitted, Excel® switches to straight-line depreciation when it is greater than the declining balance depreciation. Finally, all arguments except no_switch must be positive numbers; therefore, the cost basis is not shown as a negative-valued cash flow.

EXAMPLE 8.2 Declining Balance Depreciation Applied to the SMP Machine Video Example For the SMP machine in Example 8.1, what are the depreciation charge and book value after year 5 using declining balance depreciation? Key Data Given Cost Basis P = $500,000; Salvage Value F = $50,000; Useful Life n = 10 years Find d5, B5 Formula Solution Before computing d5, B5, we use Equation 8.5 to determine the depreciation rate that equates the book value at the end of the recovery period to the salvage value. Recall, a $500,000 investment is made in the SMP machine; it has an estimated $50,000 salvage value after 10 years of usage. Therefore, p

= 1 − ($50,000/$500,000)

0.1

= 20.5672% =−RATE (10,,500000,50000) = 20.5672%

Unfortunately, this rate exceeds 2/n and is not permitted by the IRS. So, suppose the declining balance rate is to be twice the straight-line rate. Letting p = 2/10, from Equations 8.3 and 8.4, d5

B5

=

0.20($500,000)(0.80)

=

$40,960.00

=

$500,000(0.80)

=

$163,840.00

4

5

Excel® Solution Using the Excel® DDB worksheet function gives d5

=DDB(500000,50000,10,5) = $40,960.00

When using the Excel® VDB worksheet function, it is necessary to specify a beginning and ending year for the calculation of the depreciation charge. To compute the depreciation charge for t = 5, d5

=VDB(500000,50000,10,4,5,2) = $40,960.00

To obtain the book value using the Excel® DDB, and VDB worksheet functions, it is necessary to compute the depreciation charges for all previous years and then subtract the cumulative depreciation charges from the cost basis. Table 8.1 provides the depreciation charges obtained and the associated book values for the $500,000 SMP investment for t = 1, . . . , 10 using the Excel® worksheet functions SLN, DDB, and VBD. Notice that the final book value with DDB does not equal $50,000. Also, notice that the depreciation charges are identical for VDB and DDB until year 9, at which time VDB switches from a declining balance rate to a straight-line rate. TABLE 8.1 Depreciation Allowances and Book Values for the SMP Investment SLN Bt DDB dt DDB Bt VDB dt VDB Bt t SLN dt 0 1 2 3 4 5 6 7 8 9 10

$45,000.00 $45,000.00 $45,000.00 $45,000.00 $45,000.00

$500,000.00 $455,000.00 $410,000.00 $365,000.00 $320,000.00 $275,000.00

$45,000.00 $45,000.00 $45,000.00 $45,000.00 $45,000.00

$230,000.00 $185,000.00 $140,000.00 $95,000.00 $50,000.00

$100,000.00 $80,000.00 $64,000.00 $51,200.00 $40,960.00

$500,000.00 $400,000.00 $320,000.00 $256,000.00 $204,800.00 $163,840.00

$100,000.00 $80,000.00 $64,000.00 $51,200.00 $40,960.00

$500,000.00 $400,000.00 $320,000.00 $256,000.00 $204,800.00 $163,840.00

$32,768.00 $26,214.40 $20,971.52 $16,777.22 $13,421.77

$131,072.00 $104,857.60 $83,886.08 $67,108.86 $53,687.09

$32,768.00 $26,214.40 $20,971.52 $16,943.04 $16,943.04

$131,072.00 $104,857.60 $83,886.08 $66,943.04 $50,000.00

Excel® Data File

8.2.3 Switching from DDB to SLN with the Excel® VDB Function LEARNING OBJECTIVE Determine at what point it is optimal to switch from a declining balance rate to a straight-line rate. As noted, the Excel® VDB worksheet function includes an optional feature: It can switch from using a specified declining balance rate to a straight-line rate when it is optimum to do so. But what criterion is used to define optimum? Accelerated depreciation methods are preferred to straight-line depreciation but only up to a point, and VDB determines that point. The declining balance methods front-end-load depreciation charges. Because declining balance methods compute depreciation charges using a constant percentage of the book value, however, toward the recovery period’s latter stages, the amount of depreciation charged drops off precipitously with declining balance methods. And, as was the case with DDB in the previous example, the book value calculated with declining balance methods might not reach the salvage value during the recovery period. The Excel® VDB function computes the depreciation charge that would result if declining balance continued to be used and compares it with the depreciation charge that would result if straight-line depreciation were used for the balance of the recovery period. Hence, to maximize the present worth of

depreciation charges, the optimum time to switch is the first time the depreciation charge with straight-line depreciation is greater than would result if declining balance were continued. Hence, a switch to straightline depreciation occurs at the first year for which Bt−1 − F n − (t − 1)

> pBt−1

(8.6)

Notice, the estimated salvage value is used in determining the straight-line depreciation component, even though it is neglected in the DDB calculations. Switching to straight-line depreciation is never desirable if the estimated salvage value, F, exceeds Bn, the declining balance unrecovered investment for the last year of the recovery period.

EXAMPLE 8.3 Switching from DDB to SLN with the SMP Machine For the SMP machine in previous examples, what is the optimal point for the depreciation method to switch from DDB to SLN? Key Data Given P = $500,000, F = $50,000, n = 10, and p = 0.2 for DDB Find The t for which

Bt−1 − F n − (t − 1)

> pBt−1

Solution When t = 1, the left-hand side (LHS) of Equation 8.6 is ($500,000 − $50,000)/10 = $45,000; the righthand side (RHS) of Equation 8.6 is 0.20 ($500,000) = $100,000. Because LHS < RHS, try t = 2. When t = 2, the LHS of Equation 8.6 is ($400,000 − $50,000)/9 = $38,888.89; the RHS of Equation 8.6 is 0.20($400,000) = $80,000. Because LHS < RHS, try t = 3. This iterative approach continues until the LHS of Equation 8.6 exceeds the RHS as it does in year t = 9. That is the point at which depreciation switches from DDB to SLN, and the depreciation charges dt remain the same from that point until t = n. The calculations performed and the process used to determine the optimum time to switch from DDB to SLN are shown in Figure 8.1.

FIGURE 8.1 Depreciation and Book Value Using DDB Switching to SLN Excel® Data File

Concept Check 08.02-CC001 Of the following, what information is required to calculate depreciation allowances under straight-line depreciation? I. Useful life II. First cost III. Salvage value IV. Property class a. I and II only b. II and IV only c. I, II, and III only d. I, II, III, and IV Correct or Incorrect? Clear

  Check Answer

Concept Check 08.02-CC002 Which of the following statements is true when comparing double declining balance depreciation to straight-line depreciation? a. Double declining balance deductions will be greater than straight-line deductions in all years of the asset’s depreciable life b. Double declining balance deductions will be greater than straight-line deductions in early years of the asset’s depreciable life but less in later years c. Straight-line deductions will be greater than double declining balance deductions in all years of the asset’s depreciable life d. Double declining balance deductions will be less than straight-line deductions in early years of the asset’s depreciable life but greater in later years Correct or Incorrect? Clear

  Check Answer

Concept Check 08.02-CC003 The optimal switch from declining balance depreciation to straight-line depreciation occurs when what condition is met? a. The first time the depreciation deduction from straight-line depreciation is greater than the salvage value of the asset b. The first time the depreciation deduction from straight-line depreciation is less than would result if declining balance were continued c. The first time the depreciation deduction from straight-line depreciation is greater than would result if declining balance were continued d. The asset is no longer in service Correct or Incorrect? Clear

  Check Answer

8.3 Modified Accelerated Cost Recovery System (MACRS) LEARNING OBJECTIVE Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS). Although SLN is the most commonly used depreciation method in the United States for purposes of financial reporting, Modified Accelerated Cost Recovery System (MACRS) is the only depreciation method approved by the IRS for computing income-tax liability. In general, MACRS depreciation should be used for engineering economic analysis because we are interested in analyzing the actual cash flows (including taxes) involved in an investment decision. Modified Accelerated Cost Recovery System (MACRS) The only depreciation method approved by the U.S. IRS for computing income-tax liability. Most depreciable property placed in service after 1986 qualifies for MACRS. There are two variations of MACRS: the General Depreciation System (GDS) and the Alternative Depreciation System (ADS). For most applications of interest to us, GDS is based on double declining balance, switching to straight-line depreciation. ADS is based on straight-line depreciation, with a longer recovery period than GDS. Both MACRS-GDS and MACRS-ADS have pre-established recovery periods for most property; in the case of MACRS-GDS, most property is assigned to eight property classes, which establish the number of years over which the cost basis is to be recovered.

Modified Accelerated Cost Recovery System–General Depreciation System (MACRS-GDS) The MACRS depreciation method based on double declining balance switching to straight-line depreciation. Modified Accelerated Cost Recovery System–Alternative Depreciation System (MACRS-ADS) The MACRS depreciation method based on straight-line depreciation, with a longer recovery than GDS. Both MACRS-GDS and MACRS-ADS include a feature we have not used previously: a half-year convention. It is assumed that, on average, a property is used for half of the first year of service. Similarly, it is assumed the property is used for half of the last year of service. In addition, if depreciable property is sold or taken out of service before being fully depreciated, in the year of disposal, a half-year depreciation charge is allowed. The letter H is used to designate the half-year convention. Hence, 200% DBSLH-GDS denotes double declining balance, switching to straight-line depreciation with the half-year convention applied to the first and last years of service over the GDS recovery period. Exceptions to the half-year convention are 27.5-year and 39-year property. Half-Year Convention A feature used in both MACRS-GDS and MACRS-ADS assuming that on average, a property is used for half of the first year of service and half of the last year of service.

8.3.1 MACRS-GDS Depreciable tangible property may be assigned to one of the eight MACRS (MACRS-GDS unless otherwise specified) classes, shown in Table 8.2. From the property class descriptions, it is obvious that determining the class to which a given property belongs is a complex task. For that reason, if there is any uncertainty regarding the appropriate class to use, we recommend you consult tax professionals or review publications available from the IRS, Department of the Treasury, U.S. Government.

TABLE 8.2 MACRS-GDS Property Classes Property Class Personal Property 3-Year Property Qualifying property with a class life of 4 years or less, including: tractor units for over-the-road use; special tools for manufacturing motor vehicles; special handling devices for manufacture of food and beverages; and special tools for manufacturing rubber, finished plastic, glass, and fabricated metal products 5-Year Property Qualifying property with a class life of more than 4 years but less than 10 years, including automobiles and light, general-purpose trucks; certain research and experimentation equipment; alternative energy and biomass property; computers and peripheral equipment; satellite space segment property; data-handling equipment (typewriters, calculators, copiers, printers, facsimile machines, etc.); heavy, generalpurpose trucks; timber cutting assets; offshore oil and gas well drilling assets; certain construction assets; computer-based telephone central office switching equipment; and many assets used for the manufacture of knitted goods, carpets, apparel, medical and dental supplies, chemicals, and electronic components 7-Year Property Qualifying property with a class life of 10 or more years but less than 16 years, property without any class life and not included in the 27.5- and 39-year categories, including office furniture, fixtures, and equipment; theme and amusement park assets; assets used in the exploration for and production of petroleum and natural gas deposits; most assets used for manufacturing such things as food products, spun yarn, wood products and furniture, pulp and paper, rubber products, finished plastic products, leather products, glass products, foundry products, fabricated metal products, motor vehicles, aerospace products, athletic goods, and jewelry 10-Year Property Qualifying property with a class life of 16 or more years but less than 20 years, including vessels, tugs, and similar water transportation equipment; petroleum refining assets; assets used in the manufacture of grain, sugar, vegetable oil products, and substitute natural gas-coal gasification 15-Year Property Qualifying property with a class life of 20 or more years but less than 25 years, including land improvements, such as sidewalks, roads, drainage facilities, sewers, bridges, fencing, and landscaping; cement manufacturing assets; some water and pipeline transportation assets; municipal wastewater treatment plants; telephone distribution plant assets; certain electric and gas utility property; and some liquefied natural gas assets 20-Year Property

Qualifying property with a class life of 25 years or more, other than real property with a class life of 27.5 years or more, plus municipal sewers, including such assets as farm buildings; some railroad structures; some electric generating equipment, such as certain transmission lines, pole lines, buried cable, repeaters, and much other utility property

Property Class 27.5-Year Property

Real Property Residential rental property, including a rental home or structure for which 80% or more of the gross rental income for the tax year is rental income from dwelling units. A dwelling unit is a house or an apartment used to provide living accommodations in a building or structure, but not a unit in a hotel, motel, inn, or other establishment in which more than one-half of the units are used on a transient basis Nonresidential real property: depreciable property with a class life of 27.5 years or more and that is not residential rental property

39-Year Property

Excel® Data File The Excel® VDB function can be used easily to calculate the depreciation allowance for an asset, given its MACRS property class. For example, letting P = 1, F = 0, n = 5, start_period = 0.0, end_period = 0.5, and factor = 2, VDB will determine the first year’s depreciation rate for 5-year property to be 20%. Repeating this for five more periods (0.5, 1.5; 1.5, 2.5; 2.5, 3.5; 3.5, 4.5; and 4.5, 5.0) will determine the remaining five rates. However, it is much easier to use the values provided in Table 8.3 for 3-, 5-, 7-, 10-, 15-, and 20year property. For 27.5- and 39-year property, use the percentages given in Table 8.4, which are based on the month in which the property was placed in service. For 3-, 5-, 7-, and 10-year property, the annual depreciation charge is based on 200% DBSLH-GDS; for 15- and 20-year property, the annual depreciation charge is based on 150% DBSLH-GDS; for 27.5- and 39-year property, the annual depreciation charge is based on SLM, or straight-line depreciation, mid-month convention. TABLE 8.3 MACRS-GDS percentages for 3-, 5-, 7-, and 10-year property are 200% DBSLH and 15- and 20year property are 150% DBSLH 3-Year 5-Year 7-Year 10-Year 15-Year 20-Year EOY Property Property Property Property Property Property  1 33.33 20.00 14.29 10.00 5.00 3.750  2 44.45 32.00 24.49 18.00 9.50 7.219  3 14.81 19.20 17.49 14.40 8.55 6.677  4  7.41 11.52 12.49 11.52 7.70 6.177  5  6  7  8  9 10 11 12 13 14 15

11.52  5.76

16 17 18 19 20 21 Excel® Data File

 8.93  8.92  8.93  4.46

 9.22  7.37  6.55  6.55  6.56  6.55  3.28

6.93 6.23 5.90 5.90 5.91 5.90 5.91 5.90 5.91 5.90 5.91

5.713 5.285 4.888 4.522 4.462 4.461 4.462 4.461 4.462 4.461 4.462

2.95

4.461 4.462 4.461 4.462 4.461 2.231

TABLE 8.4 a. MACRS-GDS percentages for 27.5-year residential rental property using mid-month convention Month in Tax Year Property Placed in Service Year 1 2 3 4 5 6 7 8 9 10 11 1 3.485% 3.182% 2.879% 2.576% 2.273% 1.970% 1.667% 1.364% 1.061% 0.758% 0.455% 2–9 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 10– 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 26* 11– 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 27** 28 1.970% 2.273% 2.258% 2.879% 3.182% 3.485% 3.636% 3.636% 3.636% 3.636% 3.636% 29 0.152% 0.455% 0.758% 1.061% 1.364% b. MACRS-GDS percentages for 39-year nonresidential real property using mid-month convention Month in Tax Year Property Placed in Service Year 1 2 3 4 5 6 7 8 9 10 11 1 2.461% 2.247% 2.033% 1.819% 1.605% 1.391% 1.177% 0.963% 0.749% 0.535% 0.321% 2–39 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 40

12 0.152% 3.636% 3.637% 3.636% 3.636% 1.667%

12 0.107% 2.564%

0.107% 0.321% 0.535% 0.749% 0.963% 1.177% 1.391% 1.605% 1.819% 2.033% 2.247% 2.461%

*Even-numbered year. **Odd-numbered year.

Excel® Data File For a given property class, the sum of the depreciation rates equals 100%. Therefore, the asset is fully depreciated over the recovery period. The depreciation allowance for a given year is the product of the cost basis and the depreciation rate taken from Table 8.3 or 8.4, as appropriate. Letting pt denote the MACRS depreciation rate for year t, the depreciation charge for year t and the book value at the end of year t are given by (8.7)

dt = pt P

and t

t

Bt = P − ∑ dj = P (1 − ∑ pj ) j=1

j=1

(8.8)

EXAMPLE 8.4 Applying MACRS with the SMP Machine Investment Video Example What are the depreciation allowances and book values for the SMP machine using MACRS depreciation? Key Data Use Tables 8.2 and 8.3 to determine the property class and MACRS percentages for the SMP machine. Given P = $500,000 Find dt and Bt Solution The IRS allows the $500,000 investment in a surface mount placement machine to qualify as 5-year property. Using the MACRS percentages for 5-year properties and applying Equations 8.7 and 8.8, the depreciation charges and the book values for the SMP machine are: d1 = 0.20($500,000) = $100,000

B1 = $500,000 − $100,000 = $400,000

d2 = 0.32($500,000) = $160,000

B2 = $400,000 − $160,000 = $240,000

d3 = 0.192($500,000) = $96,000

B3 = $240,000 − $96,000 = $144,000

d4 = 0.1152($500,000) = $57,600

B4 = $144,000 − $57,600 = $86,400

d5 = 0.1152($500,000) = $57,600

B5 = $86,400 − $57,600 = $28,800

d6 = 0.0576($500,000) = $28,800

B6 = $28,800 − $28,800 = $0

EXAMPLE 8.5 Applying MACRS with the SMP Machine Disposed of Early What are the depreciation allowances and book values for the SMP machine using MACRS-GDS depreciation if the machine is sold in the 4th year? Key Data From Example 8.4, based on a 5-year property class, the depreciation allowances book values for years 1, 2, 3, and 4 are available. Given P = $500,000 Find dt and Bt for t = 1, 2, 3, and 4 Solution In the year of disposal, with MACRS-GDS depreciation a half-year depreciation allowance occurs in the first year and in the year of disposal. For 5-year property, MACRS-GDS percentages include the half-year allowance in the first year and in the 6th year. In this case, because the equipment is sold in the 4th year, a half-year allowance occurs. Therefore, for years 1, 2, 3, and 4, from Example 8.4, d1 = $100,000

B1 = $400,000

d2 = $160,000

B2 = $240,000

d3 = $96,000

B3 = $144,000

d4 = $57,600/2 = $28,800

B4 = $144,000 − $28,800 = $115,200

Many variations exist in how depreciation charges are computed for the wide range of depreciable properties that exist. As noted at the chapter’s beginning, our objective is to ensure that you understand the after-tax consequences of the design decisions that engineers make. Because depreciation affects income taxes, engineers must understand how depreciation allowances are determined.

8.3.2 MACRS-ADS Although MACRS-GDS is far more popular, taxpayers may elect to claim MACRS-ADS deductions. The MACRS-ADS method is required for use on some property, including property (1) used predominantly outside the United States; (2) having any tax-exempt use; (3) financed by tax-exempt bonds; or (4) imported and covered by executive order of the U.S. president. The MACRS-ADS method is simply straight-line depreciation with either a half-year (SLH) or a mid-month (SLM) convention, as appropriate. Other than for public sector applications, it is unlikely that a taxpayer would choose to use MACRS-ADS depreciation. Hence, we have limited detailed consideration to MACRS-GDS.

Concept Check 08.03-CC001 Of the following, what information is required to calculate depreciation allowances under MACRSGDS? I. Useful life II. First cost III. Salvage value IV. Property class a. II and IV only b. II, III, and IV only c. I, II, and IV only d. I, II, III, and IV Correct or Incorrect? Clear

  Check Answer

Concept Check 08.03-CC002 MACRS-ADS is based on which of the following depreciation methods? a. Declining balance depreciation b. Straight-line depreciation c. Double declining balance depreciation d. Declining balance depreciation with optimal switch to straight-line depreciation Correct or Incorrect? Clear

  Check Answer

8.4 MACRS with Bonus Depreciation LEARNING OBJECTIVE Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS) with bonus depreciation.

The 2008 Economic Stimulus Act approved by the U.S. Congress created a provision to spur capital investments by allowing businesses to rapidly recover capital expenditures by providing an additional firstyear depreciation, called bonus depreciation. Specifically, for assets placed in service after December 31, 2007 and before January 1, 2011, for most property, and before January 1, 2012, for certain property, taxpayers were allowed to claim 50% of an asset’s basis in the year the asset was placed in service; the remaining 50% of the asset’s basis was claimed using the standard MACRS depreciation allowances. Economic Stimulus Act In 2008, to stimulate capital investments, the U.S. Congress created provisions in U.S. tax law allowing businesses to rapidly recover capital expenditures by providing additional first-year depreciation allowances, called bonus depreciation. Each year since 2008, such provisions have been in place, with the magnitude of the bonus depreciation being either 50% or 100%. Bonus Depreciation To stimulate capital expenditures, the U.S. Congress frequently creates provisions for additional first-year depreciation allowances, called bonus depreciation. Thereafter, for a portion of 2010 and all of 2011, the 2010 Tax Relief Act increased bonus depreciation to 100%. For 2012 thru 2017, 50% bonus depreciation was to be in effect, but it was to be reduced to 40% in 2018 and 30% in 2019. However, the tax reform actions of the U.S. Congress in 2017 resulted in 100% bonus depreciation being restored retroactively for purchases made after September 26, 2017 and continuing through 2022. Tax Relief Act In 2008, the U.S. Congress named its income tax provision the Economic Stimulus Act to stimulate capital expenditures by corporations. In 2010, its provision was named the Tax Relief Act. The 2008 Economic Stimulus Act allowed an additional first-year depreciation allowance equal to 50% of the original cost basis. The 2010 Tax Relief Act increased the allowance to 100% of the original cost basis. Property having a MACRS recovery period greater than 20 years is not eligible for bonus depreciation. Therefore, most buildings do not qualify. In addition, residential rental property does not qualify for bonus depreciation. We are unable to predict what the U.S. Congress will do in the future regarding bonus depreciation. Therefore, in this and subsequent chapters in which depreciation is included in economic justifications, with the exceptions noted above, we consider three possibilities, all based on MACRS-GDS depreciation: no bonus depreciation, 50% bonus depreciation, and 100% bonus depreciation.

EXAMPLE 8.6 Applying MACRS with Bonus Depreciation to the SMP Machine Investment What are the depreciation allowances and book values for the SMP machine using MACRS depreciation with 50% and 100% bonus depreciation? Key Data From Example 8.4, based on a 5-year property class, the MACRS percentages are 20%, 32%, 19.2%, 11.52%, 11.52%, and 5.76%. Given P = $500,000 Find dt and Bt with 50% bonus depreciation and 100% bonus depreciation Solution With 50% bonus depreciation, the IRS allows 50% of the $500,000 investment (or $250,000) to be included in the depreciation allowance the first year. The remaining 50% of the investment ($250,000) becomes an adjusted basis for purposes of calculating standard depreciation allowances for years 1 through 6. Therefore, d1 = $250,000 + 0.20($250,000) = $3000,000

B1 = $5000,000 − $300,000 = $200,000

d2 = 0.32($250,000) = $80,000

B2 = $2000,000 − $80,000 = $120,000

d3 = 0.192($250,000) = $48,000

B3 = $120,000 − $48,000 = $72,000

d4 = 0.1152($250,000) = $28,800

B4 = $7200 − $28,800 = $43,200

d5 = 0.1152($250,000) = $28,800

B5 = $43,200 − $28,800 = $14,400

d6 = 0.0576($250,000) = $14,400

B6 = $14,400 − $14,400 = $0

With 100% bonus depreciation, the IRS allows 100% of the $500,000 investment to be included in the depreciation allowance the first year. Therefore, d1 = $500,000 d2 = d3 = d4 = d5 = d6 = $0

B1 = $500,000 − $500,000 = $0    and   

B2 = B3 = B4 = B5 = B6 = $0

Concept Check 08.04-CC001 With 100% bonus depreciation, the entire investment is recovered in the first year. True or False?

CHAPTER 9 Income Taxes

Chapter 9 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced e-text and WileyPLUS

FE-Like Problems 09-FE001 A company has depreciation of $250,000 for the year. Interest is $75,000 on an outstanding loan of $1,000,000. Employee pay, outside services, repairs, utilities, transportation, legal fees, and similar expenses are $2,150,000. Taxable income is $225,000. What is the gross income for the year? a. $3,700,000 b. $2,700,000 c. $2,625,000 d. $1,700,000 09-FE002 A professional engineer who teaches at a university also does consulting as time permits. Her efforts do not conflict with her professorial job and she does this work through her own (one-person) company. Gross income for the year is $207,000; depreciation of equipment is $23,000; expenses such as travel, liability insurance, fees to specialty contractors, rent, and similar expenses are $34,500; and interest on a small loan is $3,500. What is her taxable income? a. $207,000 b. $149,500 c. $192,000 d. $146,000 Correct or Incorrect? Clear

  Check Answer

09-FE003 The correctly calculated taxes due on a corporate taxable income of $13,000,000 are closest to which of the following? a. $2,730,000 b. $3,250,000 c. $3,750,000 d. $4,200,000

09-FE004 When a business calculates taxable income from gross income, which of the following is true? a. Depreciation, interest, and principal are all subtracted b. Depreciation and interest are subtracted, principal is not c. Depreciation is subtracted, interest and principal are not d. Interest and principal are subtracted, depreciation is not Correct or Incorrect? Clear

  Check Answer

09-FE005 When considering the use of debt capital to finance a project, the upper limit for the interest rate on an attractive loan can be determined by which of the following? a. MARR b. MARR*(1 + tax rate) c. MARR/(1 − tax rate) d. MARR*(1 − tax rate) 09-FE006 calculation.

Consider the following data extracted from an after-tax cash flow

Before-Tax-and-Loan = $22,500 Loan Principal Payment = $7,434 Loan Interest Payment = $892 MACRS Deduction = $7,405 Taxes Due = $3,550.75 Which of the following is closest to the after-tax cash flow? a. $1,372 b. $8,777 c. $10,623 d. $16,211 Correct or Incorrect? Clear

  Check Answer

09-FE007 calculation.

Consider the following data extracted from an after-tax cash flow

Before-Tax-and-Loan = $22,500 Loan Principal Payment = $5,926 Loan Interest Payment = $2,400 MACRS Depreciation Deduction = $16,665 Which of the following is closest to the Taxable Income? a. −$2,491 b. −$91 c. $3,435 d. $14,174 09-FE008 Consider the following data for 2020 from an after-tax cash flow analysis. What is the loan interest payment for 2020? Before-Tax-and-Loan = $20,000 Loan Principal Payment = $4,018 Depreciation Deduction = $8,920 Taxable Income = $8,018 Taxes Due = $2,004.50 After-Tax Cash Flow = $10,915.50 a. $1,274 b. $3,062 c. $7,062 d. $11,080 Correct or Incorrect? Clear

  Check Answer

09-FE009 Consider the following data for 2021 from an after tax cash flow analysis. What is the taxable income for 2021?

Before-Tax-and-Loan = $23,000 Loan Principal Payment = $3,203 Loan Interest Payment = $3,877 Depreciation Deduction = $12,490 Taxes Due = $1,658 After-Tax Cash Flow = $14,262 a. $40,000 b. $35,540 c. $6,633 d. $28,460 09-FE010 Consider the following data for 2019 from an after tax cash flow analysis. What is the after tax cash flow for 2019? Before-Tax-and-Loan = $23,000 Loan Principal Payment = $3,203 Loan Interest Payment = $3,878 Depreciation Deduction = $12,490 Taxable Income = $6,633 Taxes Due = $1,658 a. $20,744 b. $14,262 c. $3,430 d. $1,175 Correct or Incorrect? Clear

  Check Answer

09-FE011 The following data are from an after-tax cash flow analysis in year 1 for a new MACRS 5-year property. How much money would be saved in year 1 if 50% bonus depreciation is used?

Initial Investment = $180,000 Regular MACRS Depreciation Deduction in Year 1 = $36,000 Before-Tax-and-Loan Cash Flow = $280,000 Loan Principal Payment = $17,500 Interest on Loan = $5,650 a. $22,500 b. $18,000 c. $13,500 d. $23,000 09-FE012 The following data are from an after-tax cash flow analysis in year 1 for a new MACRS 5-year property. How much money would be saved in year 1 if 100% bonus depreciation is used? Initial Investment = $180,000 Regular MACRS Depreciation Deduction in Year 1 = $36,000 Before-Tax-and-Loan Cash Flow = $280,000 Loan Principal Payment = $17,500 Interest on Loan = $5,650 a. $30,240 b. $75,600 c. $37,800 d. $36,000 Correct or Incorrect? Clear

  Check Answer

09-FE013 A MACRS 7-year oil field exploration property is depreciated using 50 bonus depreciation. This will reduce the taxes due by $21,430 in year 1 compared to not using any bonus depreciation at all. Which of the following is closest to the initial investment being depreciated? a. $126,960 b. $188,943 c. $120,000

d. $200,023 09-FE014 A company has total asset purchases of $2,800,000 during the year. It has borrowed $1,400,000 toward these purchases at a rate of 5.25%. What is the amount it can expense using the Section 179 expense deduction? a. $800,000 b. $300,000 c. $700,000 d. $0 Correct or Incorrect? Clear

  Check Answer

09-FE015 A small company has purchased Section 179-qualifying assets for $750,000 during the year. How much money will be saved in taxes during year 1 if the following data apply: Depreciation in year 1 if Section 179 is NOT used = $187,500 Loan repayment = $70,000 Interest payment on loan = $20,000 a. $118,125 b. $140,625 c. $187,500 d. $280,000 09-FE016 A company has purchased $2,200,000 in Section 179-qualifying property during the year. In addition to using the Section 179 expense deduction, it wishes to use 50% bonus depreciation. Being in the offshore oil and gas business, all such assets are MACRS 5-year property. What will be the tax savings in year 1 as compared to simple MACRS depreciation without Section 179 and 50% bonus depreciation? a. $170,000 b. $290,000 c. $320,000 d. $268,800 Correct or Incorrect? Clear

  Check Answer

Problems Section 9.1 Corporate Income-Tax Rates LEARNING OBJECTIVE 9.1 Calculate corporate income taxes. 09.01-PR001 What are the various taxes paid in the USA by corporations? Which of those has the most significant impact upon economic analyses? 09.01-PR002 Income taxes are calculated based on gross income less certain allowable deductions. They are also assessed on gains resulting from the disposal of property. What is a 10-word or less definition appropriate for a corporation, based on Wikipedia, for each of the following factors? a. Gross income. b. Expenses. c. Depreciation. d. Interest. e. Property (e.g., equipment) disposition. 09.01-PR003 For each statement in parts a to d, give a short answer or indicate True or False. a. Which of the following is a cash flow: (1) depreciation, (2) loan interest paid, and/or (3) income tax. b. We know that depreciation law has changed dramatically since 1950; however, tax law has not changed. (T or F). c. Depreciation method affects taxes owed. (T or F). d. In an alternative evaluation, the inclusion of taxes will change the amount of the measures of merit (e.g., PW or AW), but will not change which alternative is selected as most desirable. (T or F). 09.01-PR004 Explain the primary effect of (a) the 2008 Economic Stimulus Act, (b) the 2010 Tax Relief Act, and (c) the American Tax Cuts and Jobs Act of 2018. 09.01-PR005 Explain why it is generally preferred to do economic analyses on an aftertax basis, rather than on a before-tax basis. 09.01-PR006 What is the federal income tax for each of the following corporate taxable incomes?

a. $25,000 b. $70,000 c. $95,000 d. $200,000 e. $1,000,000 f. $12,000,000 g. $17,000,000 h. $25,000,000 09.01-PR007 Calculate the corporate income tax for each of the following corporate taxable incomes. a. $15,000 b. $88,000 c. $180,000 d. $400,000 e. $16,700,000 09.01-PR008 Video Solution Calculate the corporate income tax for each of the following corporate taxable incomes. a. $12,000 b. $65,000 c. $220,000 d. $1,000,000 e. $19,300,000

Section 9.2.1 After-Tax Analysis Using Retained Earnings—Single Alternative (No Borrowing)

LEARNING OBJECTIVE 9.2 Analyze the after-tax implications for investment alternatives using retained earnings. 09.02-PR001 An air purifier for use in manufacturing semiconductors is placed in service with a first cost of $50,000. It will be used for 8 years, have an annual gross income less operating expenses of $14,000 and will have no salvage value. Corporate income-tax rate is 25%. a. Determine the after-tax cash flows for years 0–8 if depreciation allowances are $10,000, $16,000, $9,600, $5,760, $5,760, $2,880, $0, and $0 during the 8 years (MACRS-GDS 5-year property class). b. Determine the after-tax cash flows for years 0–8 if depreciation allowances are $10,000 for years 1–5 and $0 in years 6–8 (SLN depreciation over only the first 5 years). 09.02-PR002 A special handling device for the manufacture of food is placed in service. It costs $30,000 and has a salvage value of $2,000 after a useful life of 5 years. The device generates a savings of $14,000 per year. Corporate income-tax rate is 25%. Find the after-tax cash flow for each year if: a. Depreciation allowances are $9,999, $13,335, $4,443, $2,223, and $0 for years 1–5 (MACRS-GDS 3-year property class). b. Depreciation allowances are $7,500 for each of years 1–4 and $0 in year 5 (SLN depreciation only over the first 4 years). 09.02-PR003 A specially coated mold for manufacturing tires (MACRS-GDS 3-year property) costs $35,000 and has a salvage value of $1,750 after a useful life of 5 years. The mold generates a net savings of $14,000 per year. The corporate tax rate is 25%. Find the after-tax cash flow for each year using MACRS-GDS allowances. 09.02-PR004 Raytheon wishes to use an automated environmental chamber in the manufacture of electronic components. The chamber is to be used for rigorous reliability testing and burn-in. It is installed for $1.4 million and will have a salvage value of $200,000 after 8 years. Its use will create an opportunity to increase sales by $650,000 per year and will have operating expenses of $250,000 per year. Corporate income-tax rate is 25%. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR if the chamber is kept for 8 years. After-tax MARR is 10%. a. Use straight-line depreciation (no half-year convention). b. Use MACRS-GDS and state the appropriate property class. c. Use double declining balance depreciation (no half-year convention, no switching).

09.02-PR005 Milliken uses a digitally controlled “dyer” for placing intricate and integrated patterns on manufactured carpet squares for home and commercial use. It is purchased for $400,000. It is expected to last 8 years and have a salvage value of $30,000. Increased net income due to this dyer is $95,000 per year. Milliken’s tax rate is 25% and the after-tax MARR is 12%. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR after 8 years. a. Use straight-line depreciation (no half-year convention). b. Use MACRS-GDS and state the appropriate property class. c. Use double declining balance depreciation (no half-year convention, no switching). 09.02-PR006 Suppose Milliken has an opportunity with similar cash flows to those for a digitally controlled “dyer,” although there are no depreciable items. They can invest in a marketing study by a blue-ribbon consultancy costing $400,000. Expected net returns are $95,000 per year over 7 years and $125,000 during the 8th year. Milliken’s tax rate is 25% and the after-tax MARR is 12%. a. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR after 8 years. b. Compare the results of part a with those of Problem 09.02-PR005b where MACRSGDS is used. Explain the differences. 09.02-PR007 A subsidiary of AEP places in service electric generating and transmission equipment at a cost of $3,000,000. It is expected to last 30 years with a salvage value of $250,000. The equipment will increase net income by $500,000 in the first year, increasing by 2.4% each year thereafter. The subsidiary’s tax rate is 25% and the after-tax MARR is 9%. There is some concern that the need for this equipment will last only 10 years and will need to be sold off for $550,000 at that time. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR after only 10 years to see if the venture would be worthwhile economically. a. Use straight-line depreciation (no half-year convention). b. Use MACRS-GDS and state the appropriate property class. c. Use double declining balance depreciation (no half-year convention, no switching). 09.02-PR008 Bell’s Amusements purchased an expensive ride for their theme and amusement park situated within a city-owned Expo Center. Bell’s had a multiyear contract with Expo Center. The ride cost $1.35 million, installed. Gross income from the ride was $420,000 per year, with operating expenses of $120,000. Bell’s anticipated that the ride would have a useful life of 12 years, after which the net salvage value would be $0. After 4 years, the city and Bell’s were unable to come to an agreement regarding an extended contract. In order to expedite Bell’s departure, Expo Center agreed to purchase the ride and leave it in place. Right at the end of the 4th fiscal year, Expo Center paid to Bell’s the

$900,000 unrecovered investment based on using straight-line depreciation. Corporate income-tax rate is 25% and the after-tax MARR is 9%. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR after 4 years. a. Use straight-line depreciation (no half-year convention). b. Use MACRS-GDS and state the appropriate property class. c. Use double declining balance depreciation (no half-year convention, no switching). 09.02-PR009 Video Solution A high-precision programmable router for shaping furniture components is purchased by Henredon for $190,000. It is expected to last 12 years and have a salvage value of $5,000. It will produce $45,000 in net revenue each year during its life. Corporate income-tax rate is 25% and the after-tax MARR is 10%. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR if the router is kept for 12 years. a. Use straight-line depreciation (no half-year convention). b. Use MACRS-GDS and state the appropriate property class. c. Use double declining balance depreciation (no half-year convention, no switching).

09.02-PR010 Henredon can spend $190,000 now for a design portfolio with a different furniture look inspired by some of the ultra-modern culture in the metropolitan areas of Dubai. While some consider this a gamble, it is generally conceded that such a new line can result in increased net revenues of $45,000 per year for 11 years plus $50,000 in the 12th year. Henredon’s income-tax rate is 25% and MARR is 10% on the after-tax cash flows. a. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR after 12 years. b. Compare the results of part a with those of Problem 09.02-PR009b where MACRSGDS is used. Explain the differences. 09.02-PR011 A tractor for over-the-road hauling is to be purchased by AgriGrow for $90,000. It is expected to be of use to the company for 6 years, after which it will be salvaged for $4,000. Transportation cost savings are expected to be $170,000 per year; however, the cost of drivers is expected to be $70,000 per year and operating expenses are expected to be $63,000 per year, including fuel, maintenance, insurance and the like. The

company’s income-tax rate is 25% and MARR is 10% on after-tax cash flows. Suppose that, to AgriGrow’s surprise, they actually dispose of the tractor at the end of the 4th tax year for $6,000. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR after only 4 years. a. Use straight-line depreciation (no half-year convention). b. Use MACRS-GDS and state the appropriate property class. c. Use double declining balance depreciation (no half-year convention, no switching). 09.02-PR012 AgriGrow can invest in a “100-day” short-term project costing $90,000 to improve customer service. They believe the return on the project will be a net increase in sales of $37,000 per year over 3 years and $43,000 in the 4th year. AgriGrow’s income-tax rate is 25% and MARR is 10% on the after-tax cash flows. a. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR after 4 years. b. Compare the results of part a with those of Problem 09.02-PR011b where MACRSGDS is used. Explain the differences. 09.02-PR013 Specialized production equipment is purchased for $125,000. The equipment qualifies as 5-year equipment for MACRS-GDS depreciation. The BTCF profile for the acquisition, shown below, includes a $30,000 salvage value at the end of the 5-year planning horizon. A 25% tax rate applies. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR values if the after-tax MARR is 10%. EOY

BTCF

0

−$125,000

1 2 3 4 5

50,000 60,000 70,000 80,000 120,000

09.02-PR014 A company purchases a machine for $800,000. The equipment qualifies as 5-year property for MACRS-GDS depreciation. Before-tax cash flows are as shown below, including a $200,000 salvage value after 5 years. Using a 25% income-tax rate, determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR using an 8% MARRAT.

EOY BTCF 0 −$800,000 1

100,000

2

200,000

3

300,000

4

400,000

5

700,000

09.02-PR015 An investment of $800,000 is made in equipment that qualifies as 3-year equipment for MACRS-GDS depreciation. The BTCF profile for the investment is given below, including a $200,000 salvage value at the end of the 5-year planning horizon. A 25% tax rate applies and the after-tax MARR is 8%. Determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR. EOY

BTCF

0

−$800,000

1

100,000

2

200,000

3

300,000

4

400,000

5

700,000

09.02-PR016 In Problem 09.02-PR015, suppose the equipment still qualifies as MACRSGDS 3-year property but is sold for $300,000 after 3 years of use. A 25% tax rate applies and the after-tax MARR is 8%. Determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR. Section 9.2.2 After-Tax Analysis Using Retained Earnings—Multiple Alternatives (No Borrowing) 09.02-PR017 Two investments involving a virtual mold apparatus for producing dental crowns qualify for different property classes. Investment A has a cost of $58,500, lasts 9 years with no salvage value, and costs $150,000 per year in operating expenses. It is in the 3-year property class. Investment B has a cost of $87,500, lasts 9 years with no salvage value, and costs $125,000 per year. Investment B, however, is in the 7-year property class. The company income-tax rate is 25% and MARR is an after-tax 10%. a. Based upon the use of MACRS-GDS depreciation, compare the AW of each alternative to determine which should be selected.

b. What must be Investment B’s cost of operating expenses for these two investments to be equivalent? 09.02-PR018 A virtual mold apparatus for producing dental crowns permits an infinite number of shapes to be custom constructed based upon mold imprints taken by dentists. Two models are available. One costs $58,500 and is expected to last 9 years with no salvage value at that time. Costs of use are $30 per crown and 5,000 crowns per year are produced. The other mold apparatus costs $87,500, lasts 9 years, has no salvage value, and is less costly to use at $25 per crown. The dental supplier depreciates assets using MACRS and yet values assets of the company using straight-line depreciation. The income-tax rate is 25% and MARR is an after-tax 10%. a. Based upon the use of MACRS-GDS depreciation (be sure to state the property class), compare the AW of each alternative to determine which should be selected. b. Based upon the use of straight-line (no half-year) depreciation, compare the AW of each alternative to determine which should be selected. 09.02-PR019 Video Solution A portable concrete test instrument used in construction for evaluating and profiling concrete surfaces (MACRS-GDS 5-year property class) is under consideration by a construction firm for $22,000. The instrument will be used for 6 years and be worth $2,000 at that time. The annual cost of use and maintenance will be $9,500. Alternatively, a more automated instrument (same property class) available from the manufacturer costs $29,000, with use and maintenance costs of only $7,500 and salvage value after 6 years of $3,000. The income-tax rate is 25% and MARR is an after-tax 12%. Determine which alternative is less costly, based upon comparison of after-tax annual worth.

09.02-PR020 Two investments involving a granary qualify for different property classes. Investment A costs $70,000 with $3,000 salvage value after 16 years and is depreciated as MACRS-GDS in the 10-year property class. Investment B costs $110,000 with a $4,000 salvage value after 16 years and is in the MACRS-GDS 5-year property class. Operation and maintenance for each is expected to be $18,000 and $14,000 per year, respectively. The income-tax rate is 25%, and MARR is 9% after taxes. a. Determine which alternative is less costly, based upon comparison of after-tax annual worth. b. What must the cost of the second (more expensive) investment be for there to be no economic advantage between the two? 09.02-PR021 A granary has two options for a conveyor used in the manufacture of grain for transporting, filling, or emptying. One conveyor can be purchased and installed for $70,000 with $3,000 salvage value after 16 years. The other can be purchased and

installed for $110,000 with $4,000 salvage value after 16 years. Operation and maintenance for each is expected to be $18,000 and $14,000 per year, respectively. The granary uses MACRS-GDS depreciation, has a income-tax rate of 25%, and a MARR of 9% after taxes. a. Determine which alternative is less costly, based upon comparison of after-tax annual worth. b. What must the cost of the second (more expensive) conveyor be for there to be no economic advantage between the two? 09.02-PR022 A firm may invest $30,000 in a numerically controlled lathe for use in furniture manufacturing (MACRS-GDS 7-year property). It would last for 11 years and have zero salvage value at that time. Alternatively, the firm could invest $X in a methods improvement study (this is non depreciable). Each of the investment alternatives will yield an increase in income of $15,000 for 11 years. If the tax rate for the firm is 25%, for what value of $X will the firm be indifferent between the two investment alternatives? The after-tax MARR is 15%. (Hint: Excel® SOLVER would be an excellent way to determine the answer.) 09.02-PR023 Two mutually exclusive alternatives, A and B (both MACRS-GDS 5-year property), are available. Alternative A requires an original investment of $100,000, has a useful life of 6 years, annual operating costs of $2,500, and a salvage value at the end of year k given by $100,000(0.70)k. Alternative B requires an original investment of $150,000, has a life of 8 years, zero annual operating costs, and a salvage value at the end of year k given by $150,000(0.80)k. The after-tax MARR is 15%, and a 25% tax rate is applicable. Perform an annual worth comparison and recommend the least-cost alternative. a. Use a planning horizon of 6 years. b. Use a planning horizon of 8 years and assume that an identical A will be purchased for the final two years of use. Section 9.3 After-Tax Analysis Using Borrowed Capital—Single Alternative LEARNING OBJECTIVE 9.3 Analyze the after-tax implications for investment alternatives using borrowed capital. 09.03-PR001 What is the difference or distinction being made when we speak of beforetax cash flows and before-tax and loan cash flows. Be precise in your answer. 09.03-PR002 Video Solution Chevron Phillips has put into place new laboratory equipment for the production of chemicals; the cost is $1,800,000 installed. CP borrows 45% of all capital needed and the borrowing rate is 12.5%. In the first year, 25%

of the principal borrowed will be paid back. The throughput rate for in-process test samples has increased the capacity of the lab, saving a net of $X per year. In this first year, depreciation is $360,000 and taxable income is $328,000. a. What is the “gross income” or annual savings $X? b. Determine the income tax for the first year assuming a income-tax rate of 25%. c. What is the after-tax cash flow for the first year?

09.03-PR003 Hyundai USA has numerous robotic welders as well as robotic checkers with vision. One underbody robotic welder was installed and is increasing productivity by 2.5% in one area. The result is a savings of $500,000 per year. Deductible expenses other than depreciation and interest associated with the installed robotic welder are only $120,000. Depreciation is $171,480 this year. Interest on borrowed money is $60,500 and no principal is paid back this year. a. Determine the taxable income for the year. b. Determine the income tax for the year assuming a income-tax rate of 25%. c. Determine the after-tax cash flow for the year. 09.03-PR004 Abbott placed into service a flexible manufacturing cell costing $850,000 early this year for production of their analytical testing equipment. Gross income due to the cell is expected to be $750,000 with deductible expenses of $475,000. Depreciation is based on MACRS-GDS and the cell is in the 7-year property class calling for a depreciation percentage of 14.29% or $121,465 in the first year. Half of the cell cost is financed at 11% with principal paid back in equal amounts over 5 years. The first year’s interest is therefore $46,750 while the principal payment is $85,000. a. Determine the taxable income for the first year. b. Determine the tax paid due to the cell during the first year using a 25% income-tax rate. c. Determine the after-tax cash flow for the first year. 09.03-PR005 A Boeing contractor responsible for producing a portion of the landing gear for huge airliners experienced a storm-related power glitch during the multi-axis milling, to tolerances less than 0.001 inch, of a large and complex part. The value already in the part, plus the equipment damage, was $300,000. Risk analysis indicates that a similar event might occur once per year on average if nothing is done. PolyPhaser, a leader in

lightning and surge protection, was commissioned to do a turnkey installation to protect this critical portion of the process. The first cost is $480,000 installed. A total of $275,000 is borrowed at a rate of 12% per year and no principal is repaid in the first year. Deductible annual costs are $Y, and depreciation is MACRS-GDS in the 7-year property class, or 14.29% in the first year. The taxable income is $15,000. a. What is the value of the “deductions” or $Y in the first year? b. What is the income tax paid in the first year assuming a income-tax rate of 25%? c. What is the after-tax cash flow for the first year? Note: Problems 09.03-PR006–09.03-PR014 use one or more of the following loan plans: Plan 1—pay the accumulated interest at the end of each interest period and pay the principal at the end of the loan period. Plan 2—make equal principal payments plus interest on the unpaid balance at the end of the period. Plan 3—make equal principal-plus-interest end-of-period payments. Plan 4—make a single payment of principal and interest at the end of the loan period. 09.03-PR006 An investment of $250,000 is made in equipment that qualifies as MACRSGDS 7-year property. The before-tax cash flow profile for the investment is given below, including a $100,000 salvage value at the end of the 5-year planning horizon. A loan is taken for 80% of the investment capital at an annual compound interest rate of 18% and the loan is repaid over a 5-year period. A 25% tax rate and an MARRAT of 7% apply. EOY

BTCF

0

−$250,000

1 2 3 4 5

40,000 40,000 40,000 40,000 140,000

Determine the PW of the ATCFs using: a. Loan payment Plan 1. b. Loan payment Plan 2. c. State which of the two loan payment plans is preferred, and explain why it is preferred in terms of the relationship between the loan rate and the MARRAT.

09.03-PR007 A company purchases a machine for $800,000. The equipment qualifies as 5-year property for MACRS-GDS depreciation. The machine is paid for by borrowing $500,000, to be repaid over a 5-year period at an annual compound interest rate of 12%. Before-tax cash flows are as shown on the next page, including a $200,000 salvage value after 5 years. The income-tax rate is 25%. An MARRAT of 10% applies. EOY

BTCF

0

−$800,000

1

100,000

2

200,000

3

300,000

4

400,000

5

700,000

Determine the PW of the ATCFs using: a. Loan payment Plan 2. b. Loan payment Plan 4. c. State which of the two loan payment plans is preferred, and explain why it is preferred in terms of the relationship between the loan rate and the MARRAT. 09.03-PR008 Specialized production equipment is purchased for $125,000. The equipment qualifies as 5-year equipment for MACRS-GDS depreciation. Suppose 40% of the investment capital is borrowed at an annual compound rate of 18% and the loan is repaid over a 4-year period.The BTCF profile for the acquisition, shown below, includes a $30,000 salvage value at the end of the 5-year planning horizon. A 25% tax rate applies. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR values if the after-tax MARR is 10%. EOY BTCF 0 −$125,000 1 50,000 2 60,000 3 70,000 4 80,000 5 120,000 Determine the PW of the ATCFs using: a. Loan payment Plan 1.

b. Loan payment Plan 3. c. State which of the two loan payment plans is preferred, and explain why it is preferred in terms of the relationship between the loan rate and the MARRAT. 09.03-PR009 Hyundai USA has numerous robotic welders as well as robotic checkers with vision. One underbody robotic welder costing $1,200,000 (7-year property class) was installed and is increasing productivity by 2.5% in one area. The result is a savings of $500,000 per year. Deductible expenses other than depreciation and interest associated with the installed robotic welder are only $120,000. Hyundai borrowed $550,000 at 11% for 5 years. They plan to keep the welder for 8 years. Hyundai’s income-tax rate is 25% and their MARR is 9% after taxes. Determine the PW, FW, AW, IRR, and ERR for the investment if: a. The loan is paid back using Plan 1. b. The loan is paid back using Plan 2. c. The loan is paid back using Plan 3. d. The loan is paid back using Plan 4. 09.03-PR010 Abbott placed into service a flexible manufacturing cell costing $850,000 early this year. They financed $425,000 of it at 11% per year over 5 years. Gross income due to the cell is expected to be $750,000 with deductible expenses of $475,000. Depreciation is based on MACRS-GDS and the cell is in the 7 year property class. Abbott’s income-tax rate is 25%, MARR is 10% after taxes, and they expect to keep the cell for 8 years. Determine the PW, FW, AW, IRR, and ERR for the investment if: a. The loan is paid back using Plan 1. b. The loan is paid back using Plan 2. c. The loan is paid back using Plan 3. d. The loan is paid back using Plan 4. 09.03-PR011 A Boeing contractor responsible for producing a portion of the landing gear for huge airliners experienced a storm-related power glitch during the multi-axis milling, to tolerances less than 0.001 inch, of a large and complex part. The value already in the part, plus the equipment damage, was $300,000. Risk analysis indicated that a similar cost might occur once per year on average if nothing is done. PolyPhaser, a leader in lightning and surge protection, was commissioned to do a turnkey installation to protect the process from similar yearly losses. The first cost is $480,000 installed. A total of $275,000 is borrowed at a rate of 12% per year for the entire 10-year planning horizon. Deductible annual operating and maintenance costs are $Y, and depreciation is MACRS-GDS in the 7-year property class. The income-tax rate is 25%, MARR is 10%, and the expected life of the PolyPhaser equipment is 10 years. There is no salvage value. Use Goal Seek or Solver

in Excel® to determine the value of Y such that MARR is exactly achieved, no more and no less, if: a. The loan is paid back using Plan 1. b. The loan is paid back using Plan 2. c. The loan is paid back using Plan 3. d. The loan is paid back using Plan 4. 09.03-PR012 Video Solution Raytheon wishes to use an automated environmental chamber in the manufacture of electronic components. The chamber is to be used for rigorous reliability testing and burn-in. It is installed for $1.4 million, $600,000 of which is borrowed at 11% for 5 years, and will have a salvage value of $200,000 after 8 years. Its use will create an opportunity to increase sales by $650,000 per year and will have operating expenses of $250,000 per year. Corporate income-tax rate is 25%. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR if the chamber is kept for 8 years. After-tax MARR is 10%. Determine for each year the ATCF and the PW, FW, AW, IRR, and ERR for the investment if (for parts a, b, and c, use straight-line depreciation (over 8 years with no half-year convention), and for parts d, e, and f, use MACRS-GDS depreciation with the appropriate property class: a. The loan is paid back using Plan 1. b. The loan is paid back using Plan 2. c. The loan is paid back using Plan 3. d. The loan is paid back using Plan 1. e. The loan is paid back using Plan 2. f. The loan is paid back using Plan 3.

09.03-PR013 A subsidiary of AEP places in service electric generating and transmission line equipment at a cost of $3,000,000 with half of it borrowed at 11% over 8 years. It is expected to last 30 years with a salvage value of $250,000. The equipment will increase

net income by $500,000 in the first year, increasing by 2.4% each year thereafter. The subsidiary’s tax rate is 25% and the after-tax MARR is 9%. There is some concern that the need for this equipment will last only 10 years and will need to be sold off for $550,000 at that time. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR after only 10 years to see if the venture would be worthwhile economically if (for parts a, b, and c, use straight-line depreciation; no halfyear convention, and for parts d, e, and f, use MACRS-GDS depreciation): a. The loan is paid back using Plan 1. b. The loan is paid back using Plan 2. c. The loan is paid back using Plan 3. d. The loan is paid back using Plan 1. e. The loan is paid back using Plan 2. f. The loan is paid back using Plan 3. 09.03-PR014 Chevron Phillips has put into place new laboratory equipment for the production of chemicals; the first cost is $1,800,000 installed. Chevron Phillips borrows 45% of all capital needed and the borrowing rate is 1.5% over 4 years. The throughput rate for in-process test samples has increased the capacity of the lab, with a net savings of $X per year. Depreciation follows MACRS-GDS, MARR is 11%, the income-tax rate is 25%, and the planning horizon is 6 years with a salvage value of $250,000 at that time. Use Goal Seek or Solver in Excel® to determine the value of X such that MARR is exactly achieved, no more and no less, if: a. The loan is paid back using Plan 1. b. The loan is paid back using Plan 2. c. The loan is paid back using Plan 3. d. The loan is paid back using Plan 4. 09.03-PR015 An asset is purchased for $90,000 with the intention of keeping it for 10 years, but is sold at the end of year 3. A total of $30,000 was borrowed money that was to be repaid over three years in equal annual payments, including principal and interest. The depreciation is correct for the appropriate MACRS Recovery Period. MARRAT = 12%.

EOY BT&LCF

PPMT

0

− − $90,000.00 $30,000.00

1

$35,000.00

2

$35,00.00

3

IPMT

DWO

TI

TAX

ATCF

$8,639.31 $4,500.00 $12,861.00 $17,639.00 $4,409.75 $3,204.10

$40,000.00 $11,425.49

$9,754.90 $2,438.72 $19,421.97 $7,870.50

$31,063.65 PWAT =

a. What is the MACRS property class of the asset? b. What is the salvage value received at the end of year 3? c. What is the loan interest rate? d. What is the book value at the end of year 3? e. Determine the values of the entries in the empty cells. 09.03-PR016 A project has a first cost of $180,000, an estimated salvage value of$20,000 after 6 years, and other economic attributes as detailed in the table below. Unfortunately, as the end of year 4 neared, the project had to be abandoned and the market value of the asset at that time was different from the original estimated salvage value. There was a loan to help finance the project. It was being paid back in equal annual installments (the yearly principal plus interest payment was the same). The remaining principal had to be paid at the end of year 4 when the project was stopped. Following is the table: EOY BT&LCF 0 1 2 3 4

PPMT

IPMT

DWO

TI

TAX

ATCF

− − $180,000.00 $70,000.00 $75,000.00 $9,785.71 $4,900.00 $36,000.00 $8,525.00 $51,789.29 $75,000.00 $4,215.00 $13,185.00 $57.018.04 $75,000.00 $11,203.65 $34,560.00 $9,239.49 $51.074.81 $105,000.00 $2,697.80 $50,452.20 $12,615.55 MARRAT = PWAT $44.059.74

a. What is the MACRS property class? b. What is the salvage value received at the end of year 4? c. What is the loan interest rate?

d. What is the book value at the end of year 4? e. Determine the values of the entries in the empty cells. Section 9.3 After-Tax Analysis Using Borrowed Capital—Multiple Alternatives Note: Problems 09.03-PR017–09.03-PR022 use one or more of the following loan plans: Plan 1—pay the accumulated interest at the end of each interest period and pay the principal at the end of the loan period. Plan 2—make equal principal payments, plus interest on the unpaid balance at the end of the period. Plan 3—make equal principal-plus-interest end-of-period payments. Plan 4—make a single payment of principal and interest at the end of the loan period. 09.03-PR017 Recall Problem 09.02-PR018. Suppose both models will require a loan of $40,000 at an interest rate of 17%, with payment using Plan 1 over a period of 4 years. Rework the problem. 09.03-PR018 Recall Problem 09.02-PR019. Suppose both alternatives will require a loan of 50% of the investment cost, an annual interest rate of 18%, with payment using Plan 3 over the 6-year planning horizon. Rework the problem. 09.03-PR019 Recall Problem 09.02-PR020. Suppose both investments will require a loan of $14,000 at an interest rate of 10%, with payment using Plan 4 over a period of 3 years. Rework the problem. 09.03-PR020 Recall Problem 09.02-PR021. Suppose both conveyors will require a loan of $14,000 at an interest rate of 10%, with payment using Plan 2 over a period of 3 years. Rework the problem. 09.03-PR021 Recall Problem 09.02-PR022. Suppose both alternatives will involve a loan of 40% of the investment cost, with payment over 10 years using Plan 4 and a sweet Federal government loan rate of 2%. Rework the problem. 09.03-PR022 Recall Problem 09.02-PR023. Suppose both options will involve a loan of 40% of the investment cost, with payment over (a) 6 years and (b) 8 years using Plan 1 and a sweet Federal government loan rate of 2%. Rework the problem. Section 9.4 Purchasing Versus Leasing Equipment LEARNING OBJECTIVE 9.4 Illustrate the different tax consequences for the common example of a lease vs. purchase alternative.

09.04-PR001 Video Solution Griffin Dewatering is considering three alternatives. The first is the purchase of a permanent steel building to house their existing equipment for the overhaul of dewatering systems (engines, pumps, and well points). The building can be put into service for $240,000 in early January of this year. The planning horizon for this is 10 years, at which time the building can be sold for $120,000 in late December. Maintenance and upkeep of the equipment, plus labor and materials for overhaul, costs $130,000 per year. A second alternative is to lease a building for $15,000 per year at the beginning of each year in which case all operating costs are identical except for an additional $4,000 per year cost due to the inconvenient location of the lease property. Third, they could simply contract out the overhaul work for $170,000 per endof-year, with an immediate credit through salvage of their present equipment for $45,000. The income-tax rate is 25% and the after-tax MARR is 12%. a. Determine the annual worth associated with buying the building. Be sure to give the appropriate MACRS-GDS property class. b. Determine the annual worth of leasing. c. Determine the annual worth of contracting out the work. d. Determine the annual contract price that makes contracting and leasing economically equivalent.

09.04-PR002 Michelin is considering going “lights out” in the mixing area of the business that operates 24/7. Currently, personnel with a loaded cost of $600,000 per year are used to manually weigh real rubber, synthetic rubber, carbon black, oils, and other components prior to manual insertion in a Banbary mixer that provides a homogeneous blend of rubber for making tires (rubber products). New technology is available that has the reliability and consistency desired to equal or exceed the quality of blend now achieved manually. It requires an investment of $2.5 million, with $110,000 per year operational costs and will replace all of the manual effort described above. The planning horizon is 8 years and there will be a $300,000 salvage value at that time for the new technology. The income-tax rate is 25% and the after-tax MARR is 10%. a. Determine the annual cost of purchasing the new technology. b. Determine the annual cost of continuing with the manual mixing. c. Determine the amount of the investment in new technology that would make the two alternatives equivalent.

Section 9.5 After-Tax Analysis with Bonus Depreciation LEARNING OBJECTIVE 9.5 Analyze the after-tax implications of bonus depreciation for investment alternatives. 09.05-PR001 Video Solution Raytronics wishes to use an automated environmental chamber in the manufacture of electronic components. The chamber is to be used for rigorous reliability testing and burn-in. It is installed for $1.4 million and will have a salvage value of $200,000 after 8 years. Its use will create an opportunity to increase sales by $650,000 per year and will have operating expenses of $250,000 per year. Corporate income-tax rate is 25%. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR if the chamber is kept for 8 years. After-tax MARR is 10%. a. Use MACRS-GDS(5) alone. b. Use MACRS-GDS(5) with 50% bonus depreciation. c. Use MACRS-GDS(5) with 100% bonus depreciation. d. Comment on the percentage differences in PWAT for these three analyses.

09.05-PR002 A subsidiary of AEP places in service electric generating and transmission equipment at a cost of $3,000,000. It is expected to last 30 years with a salvage value of $250,000. The equipment will increase net income by $500,000 in the first year, increasing by 2.4% each year thereafter. The subsidiary’s tax rate is 25% and the after-tax MARR is 9%. There is some concern that the need for this equipment will last only 10 years and will need to be sold off for $550,000 at that time. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR after only 10 years to see if the venture would be worthwhile economically. a. Use MACRS-GDS(20) alone. b. Use MACRS-GDS(20) with 50% bonus depreciation. c. Use MACRS-GDS(20) with 100% bonus depreciation. d. Comment on the percentage differences in PWAT for these three analyses.

09.05-PR003 A virtual mold apparatus for producing dental crowns permits an infinite number of shapes to be custom constructed based upon mold imprints taken by dentists. Two models are available. One costs $58,500 and is expected to last 9 years with no salvage value at that time. Costs of use are $30 per crown and 5,000 crowns per year are produced. The other mold apparatus costs $87,500, lasts 9 years, has no salvage value, and is less costly to use at $25 per crown. The dental supplier depreciates assets using MACRS(5). The income tax rate is 25% and MARR is an after-tax 10%. a. Use MACRS-GDS(5) alone. b. Use MACRS-GDS(5) with 50% bonus depreciation. c. Use MACRS-GDS(5) with 100% bonus depreciation. d. Comment on the differences in AWAT for each of these three analyses. 09.05-PR004 A granary has two options for a conveyor used in the manufacture of grain for transporting, filling, or emptying. One conveyor can be purchased and installed for $70,000 with $3,000 salvage value after 16 years. The other can be purchased and installed for $110,000 with $4,000 salvage value after 16 years. Operation and maintenance for each is expected to be $18,000 and $14,000 per year, respectively. The granary uses MACRS-GDS depreciation, has a income-tax rate of 25%, and a MARR of 9% after taxes. a. Use MACRS-GDS(10) alone. b. Use MACRS-GDS(10) with 50% bonus depreciation. c. Use MACRS-GDS(10) with 100% bonus depreciation. d. Comment on the differences in AWAT for each of these three analyses. Section 9.6 After-Tax Analysis with a Section 179 Expense Deduction LEARNING OBJECTIVE 9.6 Analyze the after-tax implications for small businesses of a Section 179 expense deduction. 09.06-PR001 Video Solution Raytronics wishes to use an automated environmental chamber in the manufacture of electronic components. The chamber is to be used for rigorous reliability testing and burn-in. It is installed for $1.4 million and will have a salvage value of $200,000 after 8 years. Its use will create an opportunity to increase sales by $650,000 per year and will have operating expenses of $250,000 per year. Corporate income-tax rate is 25%. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR if the chamber is kept for 8 years. After-tax MARR is 10%. a. Use MACRS-GDS(5) alone.

b. Use MACRS-GDS(5) with the Section 179 expense deduction. c. Comment on the percentage differences in PWAT for these two analyses.

09.06-PR002 A subsidiary of AEP places in service electric generating and transmission equipment at a cost of $3,000,000. It is expected to last 30 years with a salvage value of $250,000. The equipment will increase net income by $500,000 in the first year, increasing by 2.4% each year thereafter. The subsidiary’s tax rate is 25% and the after-tax MARR is 9%. There is some concern that the need for this equipment will last only 10 years and will need to be sold off for $550,000 at that time. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR after only 10 years to see if the venture would be worthwhile economically. a. Use MACRS-GDS(20) alone. b. Use MACRS-GDS(20) with the Section 179 expense deduction. c. Comment on the percentage differences in PWAT for these two analyses. 09.06-PR003 Raytronics wishes to use an automated environmental chamber in the manufacture of electronic components. The chamber is to be used for rigorous reliability testing and burn-in. It is installed for $1.4 million and will have a salvage value of $200,000 after 8 years. Its use will create an opportunity to increase sales by $650,000 per year and will have operating expenses of $250,000 per year. Corporate income-tax rate is 25%. Develop tables using a spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR if the chamber is kept for 8 years. After-tax MARR is 10%. a. Use MACRS-GDS(5) alone. b. Use MACRS-GDS (5) with the Section 179 expense deduction and 50% bonus depreciation. c. Comment on the percentage differences in PWAT for these two analyses. 09.06-PR004 A subsidiary of AEP places in service electric generating and transmission equipment at a cost of $3,000,000. It is expected to last 30 years with a salvage value of $250,000. The equipment will increase net income by $500,000 in the first year, increasing by 2.4% each year thereafter. The subsidiary’s tax rate is 25% and the after-tax MARR is 9%. There is some concern that the need for this equipment will last only 10 years and will need to be sold off for $550,000 at that time. Develop tables using a

spreadsheet to determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR after only 10 years to see if the venture would be worthwhile economically. a. Use MACRS-GDS(20) alone. b. Use MACRS-GDS(20) with the Section 179 expense deduction and 50% bonus depreciation. c. Comment on the percentage differences in PWAT for these two analyses. 09.06-PR005 Specialty Valves has robotic micro-welders as well as robotic checkers with vision for making high-precision valves in a niche market for production of petroleum and natural gas. One welder costing $1,200,000 (MACRS-7-year property class) was installed and is increasing productivity and providing a savings of $500,000 per year. This welder is the only capital investment for the year. Deductible expenses other than depreciation associated with the installed micro-welder are only $120,000. Specialty Valves plans to keep the welder for 8 years, after which there will be no salvage value. Their income tax rate is 25% and their MARR is 9% after taxes. Determine the after-tax PW, FW, AW, IRR, and ERR for the investment. a. Use MACRS-GDS(7) alone. b. Use MACRS-GDS(7) with the Section 179 expense deduction and 50% bonus depreciation. c. Comment on the percentage differences in PWAT for these two analyses. 09.06-PR006 Fluid Power has put into place new laboratory equipment for the production of chemicals. The first cost is $1,800,000 installed. This will be their only equipment capital investment for the year. The throughput rate for in-process test samples has increased the capacity of the lab, with a net annual savings of $X. Depreciation follows MACRS-GDS(5), the after-tax MARR is 11%, and the planning horizon is 6 years with a salvage value of $250,000 at that time. Use Goal Seek or Solver in Excel® to determine the value of X such that MARR is exactly achieved, no more and no less. a. Use MACRS-GDS(5) alone. b. Use MACRS-GDS(5) with the Section 179 expense deduction and 50% bonus depreciation. c. Comment on the percentage differences in X for these two analyses.

Chapter 9 Summary and Study Guide Summary 9.1: Corporate Income-Tax Rates

Learning Objective 9.1: Calculate corporate income taxes. (Section 9.1) The taxes a corporation pays are a true cost of doing business and may impact selection of the best investment alternative. Therefore, it is best to perform economic analyses on an after-tax basis so that the tax effect can be considered. Income-tax rates and regulations change rapidly, so consulting an expert in income taxes is advisable when income taxes will play a major role in determining the economic viability of an investment. The overall income-tax rate can be expressed as itr = str + ftr(1 − str)

(9.1)

where str = state income-tax rate, and ftr = federal income-tax rate. 9.2: After-Tax Analysis Using Retained Earnings (No Borrowing)

Learning Objective 9.2: Analyze the after-tax implications for investment alternatives using retained earnings. (Section 9.2) The viability of or preference for an investment alternative may change when tax implications are considered. In general, the faster an investment is depreciated, the greater its after-tax present worth, as illustrated in several chapter examples.

The after-tax cash flow of an investment alternative without borrowed capital can be expressed as: ATCF = BTCF(1 − itr) + itr(DWO)

(9.5)

where BTCF = before-tax cash flow DWO = depreciation write-off or allowance itr = income tax rate ATCF = after-tax cash flow 9.3: After-Tax Analysis Using Borrowed Capital

Learning Objective 9.3: Analyze the after-tax implications for investment alternatives using borrowed capital. (Section 9.3) For borrowed capital, interest paid by businesses is deductible from taxable income. This necessitates performing an after-tax analysis in order to determine the attractiveness of the investment. We learned that using someone else’s money can possibly make even more money for ourselves. The after-tax cash flow of an investment alternative with borrowed capital can be expressed as: ATCF = BT&LCF(1 − itr) − LCF + itr(DWO + IPMT)

where BT&LCF = before-tax-and-loan cash flow LCF = loan cash flow IPMT = interest payment In the year of the depreciable property’s disposal,

(9.9)

TIn = BT&LCFn (including Fn ) − IPMTn − DWOn − Bn

(9.10)

where TIn = taxable income in the year of property’s disposal Bn = book value at time of disposal Fn = salvage value at time of disposal Because interest on a loan can be deducted from taxable income, the effective after-tax interest rate paid on borrowed funds is ieff = i(1 − itr)

(9.11)

9.4: Leasing Versus Purchasing Equipment

Learning Objective 9.4: Illustrate the different tax consequences for the common example of a lease vs. purchase alternative. (Section 9.4) The decision to lease vs. purchase is a common investment alternative that should be analyzed using an after-tax analysis, as income-tax considerations can change the recommendation regarding the investment to be made. For example, in the lease versus purchase case, the DWO allowance can make the purchase alternative quite attractive. 9.5: After-Tax Analysis with Bonus Depreciation

Learning Objective 9.5: Analyze the after-tax implications of bonus depreciation for investment alternatives. (Section 9.5) The 2008 Economic Stimulus Act approved by the U.S. Congress created a provision to spur capital investments by allowing businesses to rapidly

recover capital expenditures by providing an additional first-year depreciation, called bonus depreciation. Since 2008, the magnitude of the bonus has varied from 50% of the initial cost basis to 100% of the initial cost basis. When bonus depreciation is allowed, after-tax present worth will increase over that which occurs in the absence of bonus depreciation. The depreciation allowance in the first year is the sum of bonus depreciation and the standard allowance based on the adjusted cost basis, which is the initial cost basis less the bonus depreciation. Taxable income equals before-tax cash flow less bonus depreciation, less depreciation allowance and, in the cases of borrowed funds, less interest on borrowed capital. 9.6: After-Tax Analysis with a Section 179 Expense Deduction

Learning Objective 9.6: Analyze the after-tax implications for small businesses of a Section 179 expense deduction. (Section 9.6) To stimulate capital investments by small businesses, a one-time expense deduction is allowed in the first year following an investment in qualifying property. For purposes of calculating depreciation allowances, the cost basis is adjusted by subtracting the Section 179 expense deduction from the initial cost basis. The magnitude of the expense deduction depends on the magnitude of the aggregate cost of qualifying equipment and the limits approved for a given tax year. (Because changes occur frequently in the limits, check the IRS Web site to see what the limits are for a given tax year.) When bonus depreciation is allowed, small businesses can take a Section 179 expense deduction, as well as a bonus depreciation deduction in the first year following the capital investment.

Important Terms and Concepts Taxable Income Gross income less allowable deductions.

Income Tax Rate (itr) The income-tax rate reflecting the combined federal and state incometax rates. This rate is typically not additive, as state income taxes are generally deductible expenses when computing federal income tax. After-Tax Cash Flow (ATCF) The amount remaining after income taxes and deductions (including interest, but excluding depreciation allowances) are subtracted from gross income. Before-Tax Cash Flow (BTCF) A term used when no borrowed money is involved, equal to the gross income less deductions (excluding depreciation allowances). Before-Tax-and-Loan Cash Flow (BT&LCF) A term used when borrowed money is involved, equal to the gross income less deductions (not including either depreciation or principal or interest on the loan). Loan Cash Flow (LCF) The cash flow reflecting the sum of the principal payment and the interest payment. Bonus Depreciation To stimulate capital expenditures, the U.S. Congress frequently creates provisions for additional first-year depreciation allowances, called bonus depreciation. American Tax Cuts and Jobs Act Among other things, the American Tax Cuts and Jobs Act adjusts the parameters of Section 179 of the United States Internal Revenue Code. In 2018, it increased Section 179 limits to $1 million for businesses with total assets purchases of no more than $2.5 million. Section 179 Expense Deduction Section 179 of the United States Internal Revenue Code was introduced in 1981 by the U.S. Congress to stimulate capital investments by small businesses. Over the years, the magnitude of the additional first-year depreciation allowance has been increased. Because the added allowance

reduces taxable income in the same way normal business expenses do, the allowance is called a Section 179 expense deduction.

Chapter 9 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 9.1: Corporate Income-Tax Rates LO 9.1: Calculate corporate income taxes. Video Lesson: Corporate Taxes Video Lesson Notes: Corporate Taxes Video Solution: 09.01-PR008 9.2: After-Tax Analysis Using Retained Earnings (No Borrowing) LO 9.2: Analyze the after-tax implications for investment alternatives using retained earnings. Video Example 9.3: After-Tax Analysis with MACRS Depreciation Video Example 9.6: After-Tax Comparison of Manual Versus Automated Solutions Video Solution: 09.02-PR009 Video Solution: 09.02-PR019 9.3: After-Tax Analysis Using Borrowed Capital LO 9.3: Analyze the after-tax implications for investment alternatives using borrowed capital. Excel Video Lesson: FV Financial Function Excel Video Lesson Spreadsheet: FV Financial Function Excel Video Lesson: MIRR Financial Function Excel Video Lesson Spreadsheet: MIRR Financial Function

Excel Video Lesson: IRR Financial Function Excel Video Lesson Spreadsheet: IRR Financial Function Excel Video Lesson: SOLVER Tool Excel Video Lesson Spreadsheet: SOLVER Tool Video Solution: 09.03-PR002 Video Solution: 09.03-PR012 9.4: Leasing Versus Purchasing Equipment LO 9.4: Illustrate the different tax consequences for the common example of a lease vs. purchase alternative. Video Solution: 09.04-PR001 9.5: After-Tax Analysis with Bonus Depreciation LO 9.5: Analyze the after-tax implications of bonus depreciation for investment alternatives. Video Solution: 09.05-PR001 9.6: After-Tax Analysis with a Section 179 Expense Deduction LO 9.6: Analyze the after-tax implications for small businesses of a Section 179 expense deduction. Video Solution: 09.06-PR001 These chapter-level resources will help you with your overall understanding of the content in this chapter. Appendix A: Time Value of Money Factors Appendix B: Engineering Economic Equations Flashcards: Chapter 09 Excel Utility: TVM Factors: Table Calculator

Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values Excel Utility: Monthly Payment Sensitivity Excel Utility: TVM Factors: Discrete Compounding Excel Utility: TVM Factors: Geometric Series Future Worth Excel Utility: TVM Factors: Geometric Series Present Worth Excel Data Files: Chapter 09

CHAPTER 9 Income Taxes LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 9.1 Calculate corporate income taxes. (Section 9.1) 9.2 Analyze the after-tax implications for investment alternatives using retained earnings. (Section 9.2) 9.3 Analyze the after-tax implications for investment alternatives using borrowed capital. (Section 9.3) 9.4 Illustrate the different tax consequences for the common example of a lease vs. purchase alternative. (Section 9.4) 9.5 Analyze the after-tax implications of bonus depreciation for investment alternatives. (Section 9.5) 9.6 Analyze the after-tax implications for small businesses of a Section 179 expense deduction. (Section 9.6)

Engineering Economics in Practice Intel With revenue of $62.8 billion in 2017, Intel is the world’s largest chip maker and also a leading manufacturer of computer, networking, and communications products. Major products include platforms designed for notebooks and desktops, wireless and wired connectivity devices, workload-optimized platforms, and related products designed for the enterprise, cloud and communication infrastructure market segments. Intel’s global workforce in 2017 totaled 102,700 worldwide, with approximately 50% located in the United States. Intel employees are highly educated, with approximately 87% working in technical roles. In 2017, Intel invested $24.9 billion in R&D and capital spending to strengthen its competitive position. The company is in the midst of a corporate transformation as it expands beyond the traditional PC and server businesses into markets addressing the growing demands to process, analyze, store, and transform data. Intel has therefore shifted its R&D focus while maintaining investment at approximately 20% of revenue. Its capital investment in silicon wafer manufacturing of platform products has remained flat, while capital investment in memory has increased significantly. Intel looks for acquisitions that further leverage its R&D investments and in 2017 acquired Mobileye to strengthen its leadership in autonomous vehicles. Semiconductor manufacturing is very expensive. Many of Intel’s competitors do not own manufacturing, assembly, and test facilities. Instead, they contract with third parties to perform those functions. Intel considers its ownership of fabrication facilities to be an important strategic advantage. In 2017, Intel made significant progress in strengthening its businesses, including introducing new products, pursuing emerging opportunities, and making strategic acquisitions. Discussion Questions 1. Intel is a high-tech company with significant R&D. What income-tax considerations would such a company have as compared to a “lower-tech” company? 2. What tax implications might there be for a global company such as Intel? 3. Do you suspect that Intel is making large capital investments? If so, what impact will depreciation have on these investments? 4. What barriers of entry would competitors of Intel have entering into this market?

Introduction From our discussion of depreciation methods, we know that income taxes can significantly impact the economic viability of a capital investment. Tax dollars are cash flows, and therefore it is necessary to consider them explicitly, just like costs of wages, equipment, materials, and energy. One of the major factors affecting income taxes is depreciation. Although depreciation allowances are not cash flows, their magnitudes and timing affect income taxes. Therefore, proper knowledge and application of tax laws can make the economic difference between accepting or rejecting an investment alternative, as well as between profit or loss on the corporate bottom line. The chapter begins with basic concepts of income taxes, focusing on corporate, not personal, investments. Next, we consider the impact of depreciation allowances on after-tax cash flows and demonstrate the after-tax effects of accelerated depreciation methods. Then we expand our

consideration of after-tax consequences to the use of borrowed capital, because interest paid by businesses is deductible from taxable income. After considering the after-tax implications of leasing versus purchasing, we analyze the effects of bonus depreciation and Section 179 expense deductions on the after-tax present worth of a capital investment. The taxes a corporation pays represent a real cost of doing business and, consequently, affect the cash flow profile of projects. For this reason, it is wise to perform economic analyses on an after-tax basis. After-tax analysis procedures are identical to before-tax procedures; however, cash flows are adjusted for taxes paid or saved. Corporations typically pay a variety of taxes, including ad valorem (property), sales, excise (a tax on the manufacture, sale, or consumption of a commodity), and income taxes. Among the various types of taxes paid, corporate income taxes tend to be the most significant when performing an economic analysis, because most of the other taxes are not affected by the kinds of investments generally included in an engineering economic analysis. Income taxes are assessed on gross income less certain allowable deductions and on gains resulting from the disposal of property. Federal and state income-tax regulations are detailed, intricate, and subject to change over time. Hence, only general concepts and procedures for calculating after-tax cash flow profiles and performing aftertax analyses are emphasized here.

Systematic Economic Analysis Technique 1. Identify the investment alternatives 2. Define the planning horizon 3. Specify the discount rate 4. Estimate the cash flows 5. Compare the alternatives 6. Perform supplementary analyses 7. Select the preferred investment

9.1 Corporate Income-Tax Rates LEARNING OBJECTIVE Calculate corporate income taxes. Video Lesson: Corporate Taxes For income-tax purposes, a corporation includes associations, business trusts, joint stock companies, insurance firms, and trusts and partnerships that actually operate as associations or corporations. Corporate income tax, however, is not limited to traditional business organizations. Engineers, doctors, lawyers, and other professional people may be treated as corporations if they have formally organized under state professional association acts. Corporate income taxes are computed by multiplying taxable income by the appropriate income-tax rate.

9.1.1 Determining Taxable Income Taxable income must first be determined before any tax rate can be applied. Taxable income is gross income less allowable deductions. Gross income is income in a general sense less any monies specifically exempt from tax liability. Corporate deductions are subtracted from gross income and commonly include items such as salaries, wages, repairs, rent, bad debts, taxes (other than income), charitable contributions, casualty losses, interest, and depreciation. Interest and depreciation are of particular importance, because we can control them to some extent through financing arrangements and accounting procedures. Taxable income is represented pictorially in Figure 9.1. These components are not all cash flows, because the depreciation allowance is treated simply as an expense in determining taxable income. Taxable Income Gross income less allowable deductions.

FIGURE 9.1 Pictorial Representation of Taxable Income For many years, the federal corporate income-tax rate varied, depending on the taxable income. For tax years 1993 through 2017, as shown in Table 9.1, there were 8 income tax brackets. However, the Tax Cuts and Jobs Act introduced on January 1, 2018 reduced the U.S. federal corporate income-tax rate from a maximum 35% to a flat rate of 21%, regardless of the magnitude of taxable income. TABLE 9.1 Corporate Income-Tax Rates for Tax Years January 1, 1993 to December 31, 2017 Taxable Income (TI), in $ Tax Rate (itr) Income Tax (T) 0 < TI ≤ 50,000 0.15 0.15(TI) 50,000 < TI ≤ 75,000 0.25 7,500 + 0.25(TI − 50,000) 75,000 < TI ≤ 100,000 0.34 13,750 + 0.34(TI − 75,000) 100,000 < TI ≤ 335,000 0.39(0.34 + 0.05) 22,250 + 0.39(TI − 100,000) 335,000 < TI ≤ 10,000,000 0.34 113,900 + 0.34(TI − 335,000) 10,000,000 < TI ≤ 15,000,000 0.35 3,400,000 + 0.35(TI − 10,000,000) 15,000,000 < TI ≤ 18,333,333 0.38(0.35 + 0.03) 5,150,000 + 0.38(TI − 15,000,000) 18,333,333 < TI 0.35 0.35(TI) For personal income tax calculations, income tax brackets remain. Different tax rates apply for different taxable incomes and for the category of the income tax filer. Regarding the latter, different thresholds apply for unmarried individuals, married individuals filing joint returns, and heads of households. A number of additional things must be considered in computing income taxes for

individuals, including, but not limited to, standard deduction and personal exemption, alternative minimum tax, and earned income tax credit. The text includes a consideration of income taxes from a business perspective. Except for the purchase of a principal residence, few opportunities exist for individuals to perform after-tax economic justifications. Principally, for this reason, we do not explore after-tax analyses for individuals. (Furthermore, income tax calculations for individuals can become quite complex. Except for fairly straightforward situations, individuals should consider retaining the services of income tax professionals.) As illustrated in Figure 9.2, corporate income-tax rates have changed dramatically over the past century. Therefore, it is likely further changes will occur in the future. Because the rate used in the text might differ significantly from the actual rate at the time of your economic justification, we recommend you consult the U.S. Internal Revenue Service Web site (http://www.irs.gov) for up-to-date information on U.S. tax laws.

FIGURE 9.2 Highest U.S. Federal Income-Tax Rate for Corporations (1909 thru 2018) Source: www.irs.gov/pub/irs-soi/histabb.xls supplemented for years following 2010.

In addition to paying federal income taxes, corporations pay state income taxes. In 2018, state incometax rates for businesses ranged from zero percent in Ohio, South Dakota, Texas, Washington, and Wyoming to rates greater than 9% in Alaska, Illinois, Iowa, Minnesota, and Pennsylvania. Weighted by population, the Tax Foundation calculated an average rate of 6% for states. (Source: https://taxfoundation.org/us-corporate-income-tax-more-competitive/). Because corporations can deduct state income taxes from taxable income when calculating federal income taxes, the combination of state and federal income taxes will be less than the sum of the state rate (str) and federal rate (ftr). Specifically, the combined income tax rate (itr) should equal itr = str + ftr(1 − str)

(9.1)

Arbitrarily, letting str equal 5% and ftr equal 21%, applying Equation 9.1, the combined income-tax rate equals 24.95%. Therefore, in the text, a value of 25% will be used for the combined income-tax rate in most examples and end-of-chapter problems. Depending on the particular state (and country) in which the investment is made, a different combined income-tax rate might be more appropriate.

Income Tax Rate (itr) The income-tax rate reflecting the combined federal and state income-tax rates. This rate is typically not additive, as state income taxes are generally deductible expenses when computing federal income tax.

EXAMPLE 9.1 Calculating Taxable Income and Income Tax for a Small Business In 2020, a small business has gross income of $350,000. In the same year, its overall depreciation totals $80,000 and the interest on borrowed funds totals $15,000. The business has no other deductions to claim. What will be its taxable income? Based on a corporate income-tax rate of 25%, what will be its income tax liability in 2020? Key Data Given Gross income = $350,000; Depreciation allowance = $80,000; Interest on borrowed money = $15,000; Income tax rate (itr) = 25% Find Taxable income (TI) and Income Tax (T) Solution From Figure 9.1, TI = Gross income − Depreciation allowance − Interest on borrowed money. Also, T = itr(TI). Therefore, TI 

=

$350,000 − $80,000 − $15,000 = $255,000 and

T

=

$255,000(0.25) = $63,750

Because they are commonly used in business, let’s consider briefly the differences in and tax treatments of four types of financial instruments: capital leases, operating leases, rental agreements, and financial loans. The principal difference between a capital lease and an operating lease is duration. For example, in a capital lease agreement the lessor agrees to transfer the ownership rights to the lessee after completion of the lease period. Capital leases are typically long-term, noncancellable, and involve major equipment that will be sold at a bargain price or automatically transferred to the lessor at the end of the lease period. Operating leases are used for short-term leasing of assets and are like a rental agreement to pay the owner periodically for using the asset, as they do not involve any transfer of ownership. Periodic operating lease payments are treated as operating expenses and are expensed on the income statement, impacting both the operating and net income. A financial loan is the lending of money by one or more individuals, organizations (often banks), or other entities to other individuals or organizations. The recipient (i.e., the borrower) incurs a debt, is usually liable to pay interest on that debt until it is repaid, and must repay the principal amount borrowed. In Chapter 8, we noted different depreciation allowances are used for financial reporting versus income-tax calculations. Straight-line depreciation is typically used in financial reports and modified accelerated cost recovery system depreciation is required for computing income-tax liability. Similarly, there are differences in the way leases, loans, and rental agreements are treated. Currently, both capital leases and operating leases must be amortized for purposes of financial reporting; the same applies to long-term loans and long-term rental agreements.

Our interest in leases, loans, and rental agreements pertains to their tax treatment. As indicated in Figure 9.1, interest paid on borrowed money is treated as an operating expense and deducted from taxable income; rental payments and lease payments are also treated as operating expenses.

Concept Check 09.01-CC001 When performing economic justifications, income taxes are important because a. income tax dollars are cash flows b. depreciation does not impact income taxes c. federal taxes are important, but state taxes are negligible d. income-tax laws remain constant over time

9.2 After-Tax Analysis Using Retained Earnings (No Borrowing) LEARNING OBJECTIVE Analyze the after-tax implications for investment alternatives using retained earnings. Before proceeding with after-tax economic analyses, it will be useful to introduce the concept of a before-tax MARR (MARRBT) and an after-tax MARR (MARRAT). An approximation to account for income taxes when they are not explicitly considered in our calculations is to set MARRBT greater than MARRAT to account for the effective income-tax rate. Hence, an approximation of MARRAT will include the effect of income taxes as follows: MARRAT ≈ MARRBT (1 − effective income-tax rate)

The base elements needed to calculate after-tax cash flow (ATCF) for an alternative are summarized in Figure 9.3, which shows that the ATCF is the amount remaining after income taxes and deductions, including interest but excluding depreciation allowances, are subtracted from gross income. In many of the following tables and spreadsheets, we simplify our terminology by speaking of before-tax cash flow (BTCF). This term is used when no borrowed money is involved; it equals gross income less deductions, not including depreciation. The term before-tax-and-loan cash flow (BT&LCF) is used when borrowed money is involved; it equals gross income less deductions, not including either depreciation or principal or interest on the loan.

After-Tax Cash Flow (ATCF) The amount remaining after income taxes and deductions (including interest, but excluding depreciation allowances) are subtracted from gross income. Before-Tax Cash Flow (BTCF) A term used when no borrowed money is involved, equal to the gross income less deductions (excluding depreciation allowances). Before-Tax-and-Loan Cash Flow (BT&LCF) A term used when borrowed money is involved, equal to the gross income less deductions (not including either depreciation or principal or interest on the loan).

FIGURE 9.3 Pictorial Representation of After-Tax Cash Flow (ATCF) The following notation will be used for after-tax analysis of an investment alternative’s economic worth, without borrowed capital: BTCF =  before-tax cash flow DWO  =  depreciation write-off or allowance TI  =  taxable income itr  =  income tax rate T = income tax ATCF  =  after-tax cash flow

The following equation holds for investments in depreciable property: ATCF = BTCF − T

(9.2)

T = itr(TI)

(9.3)

TI = BTCF − DWO

(9.4)

ATCF = BTCF(1 − itr) + itr (DWO)

(9.5)

where

where

Therefore,

9.2.1 Single Alternative In Examples 9.1 through 9.3 we focus on after-tax analysis, using retained earnings, for a single investment alternative. Later, we will look at a similar analysis for multiple alternatives.

EXAMPLE 9.2 After-Tax Analysis with SLN Depreciation Recall the $500,000 investment in the surface-mount placement (SMP) machine in an electronics manufacturing plant. It produced $92,500 in net revenue (after taxes) for 10 years, plus a $50,000 salvage value at the end of the 10-year period. Now, to determine the after-tax economic worth of the investment, we will use a 25% income-tax rate and will perform an after-tax analysis using a 10% after-tax MARR. Key Data Given P = $500,000, itr = 25%, MARRAT = 10%, ATCF (years 1–10) = $92,500, S = $50,000 Find PWAT, AWAT, FWAT, IRRAT, ERRAT Solution In the previous chapters, the $92,500 net revenue was given to be an after-tax figure. Although we did not specify the depreciation method used to generate the ATCF amount of $92,500, it was based on straight-line depreciation over a 10-year recovery period. As shown in Table 9.2, a BTCF of $108,333.33 results in an ATCF of $124,166.67 when straight-line depreciation is used with a $50,000 salvage value.

TABLE 9.2 Before-Tax and After-Tax Analysis of the SMP Investment with SLN Depreciations EOY BTCF DWO TI T ATCF  0  1

−$500,000.00 $108,333.33 $45,000.00 $63,333.33 $15,833.33

−$500,000.00  $92,500.00 

 2  3

$108,333.33 $45,000.00 $63,333.33 $15,833.33 $108,333.33 $45,000.00 $63,333.33 $15,833.33

$92,500.00  $92,500.00 

 4  5

$108,333.33 $45,000.00 $63,333.33 $15,833.33 $108,333.33 $45,000.00 $63,333.33 $15,833.33

$92,500.00  $92,500.00 

 6  7  8

$108,333.33 $45,000.00 $63,333.33 $15,833.33 $108,333.33 $45,000.00 $63,333.33 $15,833.33 $108,333.33 $45,000.00 $63,333.33 $15,833.33

$92,500.00  $92,500.00  $92,500.00 

 9 10

$108,333.33 $45,000.00 $63,333.33 $15,833.33 $158,333.33 $45,000.00 $63,333.33 $15,833.33

$92,500.00  $142,500.00 

MARRBT =

13.333%

MARRAT =

10% 

PWBT =

$94,396.20

 PWAT =

$87,649.62 

FWBT =

$330,013.08

FWAT =

$227,340.55 

AWBT =

$17,628.61

AWAT =

$14,264.57 

IRRBT =

17.80%

IRRAT =

13.80% 

ERRBT =

15.31%

ERRAT =

11.79% 

Excel® Data File The computation of ATCF is as shown in Table 9.2. In the 10th year, notice how the salvage value is handled. The $500,000 investment is not fully recovered through depreciation allowances. Straight-line depreciation allowances of $45,000 over a 10-year period reduce the book value to $50,000 at the end of the 10-year planning horizon. Because the book value exactly equals the salvage value, there is no income tax associated with the sale of the SMP machine for $50,000. Therefore, the taxable income in year 10 is identical to that in each of the 9 previous years. Exploring the Solution Table 9.2 also provides values for the various DCF-based measures of economic worth treated in previous chapters. As expected, the BTCF values and the after-tax measures of economic worth shown in Table 9.2 are identical to those obtained in Chapters 4 through 6. If, in the previous chapters, BTCF was used and a before-tax analysis was performed, then a before-tax minimum attractive rate of return (MARRBT) must be used. An estimate of MARRBT is obtained by dividing the after-tax minimum attractive rate of return (MARRAT) by 1 minus the income tax rate. Hence, with a 25% tax rate and a 10% MARRAT, a MARRBT of 13.333% would be used.

EXAMPLE 9.3 After-Tax Analysis with MACRS Depreciation Video Example As noted in Chapter 8, the SMP machine purchased for $500,000 qualifies as 5-year property for MACRS depreciation. What is the after-tax effect on the measures of economic worth if MACRS is used with a 25% tax rate and a 10% MARRAT? Key Data Given P = $500,000, itr = 25%, MARRAT = 10%, BTCF (years 1–9) = $108,333.33, BTCF (year 10) = $158,333.33 Find PWAT, AWAT, FWAT, IRRAT, ERRAT Solution Table 9.3 provides the results of the after-tax analysis. Notice, because MACRS fully recovers the investment, the book value at the end of year 10 is 0. If the equipment is sold for $50,000, then the full amount of the salvage value is taxable income. (When a depreciable asset is sold for more than its book value, the gain is called depreciation recapture and is taxed as ordinary income under current tax law.)

TABLE 9.3 After-Tax Analysis of the SMP Investment with MACRS Depreciation EOY BTCF  0 −$500,000.00

DWO

TI

T

ATCF −$500,000.00 

 1  2

$108,333.33 $100,000.00 $8,333.33 $2,083.33 $106,250.00  $108,333.33 $160,000.00 −$51,666.67 −$12,916.67 $121,250.00 

 3  4

$108,333.33 $96,000.00 $12,333.33 $3,083.33 $105,250.00  $108,333.33 $57,600.00 $50,733.33 $12,683.33 $95,650.00 

 5  6

$108,333.33 $57,600.00 $50,733.33 $12,683.33 $108,333.33 $28,800.00 $79,533.33 $19,883.33

$95,650.00  $88,450.00 

 7  8  9

$108,333.33 $108,333.33 $108,333.33

$0.00 $108,333.33 $27,083.33 $0.00 $108,333.33 $27,083.33 $0.00 $108,333.33 $27,083.33

$81,250.00  $81,250.00  $81,250.00 

10

$158,333.33

$0.00 $158,333.33 $39,583.33 $118,750.00  PWAT = $110,361.50  FWAT = $286,249.32  AWAT =

$17,960.83 

IRRAT =

15.18% 

ERRAT =

12.22% 

Excel® Data File

Effects of Different Depreciation Methods and Salvage Values Examples 9.1 and 9.2 bring out some important observations. Note that the after-tax PW, FW, AW, IRR, and ERR values at the bottom of Table 9.3 using MACRS depreciation are all better than the after-tax PW, FW, AW, IRR, and ERR values at the bottom of Table 9.2 using SLN depreciation. As pointed out in Chapter 8, because depreciation is deducted from taxable income, accelerated depreciation yields a greater economic worth than occurs using straight-line depreciation. If the salvage value exactly matches the book value at the planning horizon’s end as in Example 9.1 using SLN depreciation, there are no tax implications. If the salvage value exceeds the book value, as it does in Example 9.2 where the SMP machine was sold for $50,000 more than the book value at the horizon’s end, the excess amount is taxed at the itr (25%). If the salvage value is less than the book value at the horizon’s end, the difference is termed a book loss and is deducted from taxable income. In general, we recommend that the following relationship be used to compute taxable income in the year of property disposal. Specifically, if disposal occurs at the end of year n at a salvage value of Fn, TIn = BTCFn [including Fn ] − DWOn − Bn

(9.6)

In words, taxable income in the year of a depreciable asset’s disposal equals the before-tax cash flow, including the salvage value, less the depreciation allowance, less the book value at the time of disposal.

As occurred in Example 9.4, taxable income in one or more years can be negative. In this case, the SMP machine investment reduced taxable income in year 2 and produced a tax savings of $14,333.33. Also, if depreciable property is disposed of before the end of the recovery period, when using MACRS depreciation, a half-year allowance (in the case of personal property) or midmonth allowance (in the case of real property) is permitted in the year of disposal. For example, if an asset is sold during the 6th tax year and it is 7-year property, then the MACRS depreciation allowance percentages are 14.29%, 24.49%, 17.49%, 12.49%, 8.93%, and 4.46%; normally, the depreciation percentage for the 6th year would be 8.92%. For our purposes, when a depreciable asset is sold for less than its book value, the difference is termed a book loss and is deducted from taxable income. Although there are carry-forward and carry-back rules, for purposes of the text, we do not present the details; instead, we encourage you to consult income-tax guides or tax experts if you encounter such a situation. Nondepreciable Expenditures Not all expenditures are used to acquire depreciable property. Consider, for example, an expenditure on a consulting study to identify cost-saving opportunities. Another example of alternatives with different tax consequences is leasing versus purchasing equipment. Likewise, let’s not forget expenditures on advertising to generate increased revenue. For tax purposes, these types of expenditures are “expensed” or “written off” in the year in which they occur. Other expenditures that can be expensed include software and most R&D expenses. The following example shows the after-tax effects of expenditures that can be expensed versus those that must be capitalized and depreciated.

EXAMPLE 9.4 After-Tax Analysis with Nondepreciable Expenditures In Example 9.2, $500,000 was invested in depreciable property to obtain before-tax annual revenues of $108,333.33 each of the following 9 years and $158,333.33 the 10 year. Now suppose the expenditure is on a consulting study that identifies cost-saving opportunities that will produce the same BTCF. With a 25% income-tax rate and an after-tax MARR of 10%, what will be the after-tax measures of economic worth for the investment? Key Data Given P = $500,000, itr = 25%, MARRAT = 10%, BTCF (years 1–9) = $108,333.33, BTCF (year 10) = $158,333.33 Find PWAT, AWAT, FWAT, IRRAT, ERRAT Solution Table 9.4 provides the results of the after-tax analysis. TABLE 9.4 After-Tax Analysis of a $500,000 Investment in a Consulting Study EOY BTCF TI T ATCF  0

−$500,000.00 −$500,000.00 −$125,000.00 −$375,000.00 

 1  2

$108,333.33 $108,333.33

$108,333.33 $108,333.33

$27,083.33 $27,083.33

$81,250.00  $81,250.00 

 3

$108,333.33

$108,333.33

$27,083.33

$81,250.00 

 4

$108,333.33

$108,333.33

$27,083.33

$81,250.00 

 5  6

$108,333.33 $108,333.33

$108,333.33 $108,333.33

$27,083.33 $27,083.33

$81,250.00  $81,250.00 

 7

$108,333.33

$108,333.33

$27,083.33

$81,250.00 

 8

$108,333.33

$108,333.33

$27,083.33

$81,250.00 

 9 10

$108,333.33 $158,333.33

$108,333.33 $158,333.33

$27,083.33 $39,583.33

$81,250.00  $118,750.00 

PWAT =

$138,703.95 

FWAT =

$359,762.33 

AWAT =

$22,573.43 

IRRAT =

17.80% 

ERRAT =

13.52% 

Excel® Data File

Now, Examples 9.2 and 9.3 show some interesting observations. Note that the values at the bottom of Table 9.4 when the investment is expensed are all better than the values at the bottom of Table 9.3 when the investment is capitalized and depreciated using MACRS. One way to think about this is that expensing is the extreme of accelerated depreciation. Rather than spreading out the asset’s value evenly over time (as in SLN), or in a relatively accelerated fashion (with DDB), we are able to achieve the most tax benefit by expensing the total amount in year 0. Expensing would be preferred to depreciating, but tax law dictates when we can and cannot expense, as well as how we must depreciate.

9.2.2 Multiple Alternatives An after-tax comparison of investment alternatives follows the procedures described in previous chapters. The only changes are that we are now explicitly calculating ATCF and using MARRAT. Depending on the alternatives being considered, different depreciation classes might apply to the alternatives or some may even be expensed.

EXAMPLE 9.5 After-Tax Comparison of Alternatives with Different Property Classes A manufacturing firm is considering different investments that qualify for different property classes. Alternative A is for specialized tools that qualify as 3-year property, require an investment of $300,000, and provide before-tax annual savings of $63,333.33 per year over the 10-year planning horizon. Alternative B is for production equipment that qualifies as 7-year property, requires an investment of $450,000, and provides before-tax annual savings of $91,666.67 per year over the 10-year horizon. An after-tax MARR of 10% applies, and with a 25% income-tax rate this is equivalent to a before-tax MARR of 13.33%. Which alternative is preferred? Key Data Given Alternative A: P = $300,000; 3-year property; itr = 25%; MARRAT = 10%; BTCF (years 1−10) = $63,333.33 Alternative B: P = $450,000; 7-year property; itr = 25%; MARRAT = 10%; BTCF (years 1−10) = $91,666.67 Find PWBT and PWAT of Alternatives A and B Solution As shown in Table 9.5, Alternative A is the preferred investment, with an after-tax present worth of $54,284.67 versus $53,602.08 for Alternative B. On a before-tax basis, however, Alternative B is preferred to Alternative A.

TABLE 9.5 After-Tax Comparison of Investment Alternatives with Different Property Classes EOY  0  1

BTCF(A)

TI(A)

T(A)

ATCF(A)

−$9,164.17

−$300,000.00  $72,497.50 

 2

$63,333.33 $133,350.00 −$70,016.67 −$17,504.17

$80,837.50 

 3

$63,333.33

$44,430.00

$18,903.33

$4,725.83

$58,607.50 

 4  5

$63,333.33 $63,333.33

$22,230.00 $0.00

$41,103.33 $63,333.33

$10,275.83 $15,833.33

$53,057.50  $47,500.00 

 6

$63,333.33

$0.00

$63,333.33

$15,833.33

$47,500.00 

 7

$63,333.33

$0.00

$63,333.33

$15,833.33

$47,500.00 

 8  9

$63,333.33 $63,333.33

$0.00 $0.00

$63,333.33 $63,333.33

$15,833.33 $15,833.33

$47,500.00  $47,500.00 

10

$63,333.33

$0.00

$63,333.33

$15,833.33

$47,500.00 

PWBT =

$39,132.06

PWAT =

$54,284.67 

EOY

−$300,000.00  $63,333.33

DWO(A)

BTCF(B)

$99,990.00 −$36,656.67

DWO(B)

TI(B)

T(B)

ATCF(B)

 0

−$450,000.00

−$450,000.00 

 1

$91,666.67

$27,361.67

$6,840.42

$84,826.25 

 2  3

$91,666.67 $110,205.00 −$18,538.33 $91,666.67 $78,705.00 $12,961.67

−$4,634.58 $3,240.42

$96,301.25  $88,426.25 

 4

$91,666.67

$56,205.00

$35,461.67

$8,865.42

$82,801.25 

 5

$91,666.67

$40,185.00

$51,481.67

$12,870.42

$78,796.25 

 6  7

$91,666.67 $91,666.67

$40,140.00 $40,185.00

$51,526.67 $51,481.67

$12,881.67 $12,870.42

$78,785.00  $78,796.25 

 8

$91,666.67

$20,070.00

$71,596.67

$17,899.17

$73,767.50 

 9

$91,666.67

$0.00

$91,666.67

$22,916.67

$68,750.00 

10 PWBT =

$91,666.67 $40,849.04

$0.00

$91,666.67

$22,916.67 PWAT =

$68,750.00  $53,602.08 

$64,305.00

Excel® Data File

EXAMPLE 9.6 After-Tax Comparison of Manual Versus Automated Solutions Video Example In a distribution center, loads to be shipped have been palletized manually. A proposal has been made to use a robot to perform the palletizing operation. Because cartons to be palletized are not dimensionally uniform, a vision system coupled with optimization software will be required with the robot. Currently, two people perform palletizing. The labor cost for this is $50,000 per year. A fully equipped robot to perform the task will cost $125,000 and will have annual operating costs of $500. The robot qualifies as 3-year property. If a tax rate of 25% and a MARR of 10% are used, should the robot be purchased? Use a 5-year planning horizon and perform an after-tax annual worth analysis; assume a salvage value of $25,000 for the robot after 5 years of use. Key Data Given Manual Alternative: A = $50,000/year; itr = 25%; MARRAT = 10%; N = 5 Robotic Alternative: P = $125,000; A = $500; 3-year property; itr = 25%; MARRAT = 10%; N = 5; SV = $25,000 Find EUACAT Solution Manual palletizing: Because the labor cost can be expensed in the year in which it occurs, the after-tax equivalent uniform annual cost of manually palletizing equals $50,000(0.75), or $37,500. Robotic palletizing: As shown in Table 9.6, the equivalent uniform annual cost of robotic palletizing equals $23,417.80. Therefore, the robot is justified economically. TABLE 9.6 After-Tax Comparison of Manual Versus Robotic Palletizing EOY BTCF DWO TI T ATCF 0

−$125,000.00

−$125,000.00 

1

−$500.00 $41,662.50 −$42,162.50 −$10,540.63

$10,040.63 

2 3

−$500.00 $55,562.50 −$56,062.50 −$14,015.63 −$500.00 $18,512.50 −$19,012.50 −$4,753.13

$13,515.63  $4,253.13 

4

−$500.00 $9,262.50 −$9,762.50 −$2,440.63

$1,940.63 

5

$24,500.00

$0.00 $24,500.00

Excel® Data File

$6,125.00

$18,375.00 

EUACAT =

$23,417.80 

Concept Check 09.02-CC001 The present worth of the after-tax cash flows of an investment will be maximized when the cost of the investment is a. expensed b. depreciated using straight-line depreciation c. depreciated using double declining balance depreciation d. depreciated using MACRS depreciation

Concept Check 09.02-CC002 When comparing multiple alternatives in an after-tax cash flow analysis, which of the following are true? I. All alternatives should be depreciated using the same depreciation method II. Some alternatives may be expensed III. Some alternatives may use different depreciation methods IV. All alternatives should be expensed a. I only b. IV only c. III only d. II and III only

9.3 After-Tax Analysis Using Borrowed Capital LEARNING OBJECTIVE Analyze the after-tax implications for investment alternatives using borrowed capital. The previous after-tax analyses assumed the investments were paid for using retained earnings. However, depreciable property is frequently purchased using borrowed funds. Because interest is deductible from taxable income, it is useful to consider how an after-tax analysis is performed when using borrowed funds. When borrowed funds are used, additional notation is needed. Let PPMT = principal payment

IPMT = interest payment LCF = loan cash flow = PPMT + IPMT Loan Cash Flow (LCF) The cash flow reflecting the sum of the principal payment and the interest payment. The following formulas apply when borrowed funds are used: ATCF = BT&LCF − LCF − T

(9.7)

After-tax cash flow equals before-tax-and-loan cash flow, less the loan payment, less the income tax. As before, T = itr(TI)

but, now, TI = BT&LCF − IPMT − DWO

(9.8)

Taxable income equals before-tax-and-loan cash flow, less the interest paid, less the depreciation allowance. Therefore, ATCF = BT&LCF(1 − itr) − LCF + itr (DWO  +  IPMT)

(9.9)

Finally, in the year of the depreciable property’s disposal, TIn = BT&LCFn (including Fn ) − IPMTn − DWOn − Bn

(9.10)

Because interest on a loan can be deducted from taxable income, the effective after-tax interest rate paid on borrowed funds is ieff = i (1 − itr)

(9.11)

MARRBT is therefore given by MARRAT/(1 − itr). Hence, if the income-tax rate is 25%, and MARRAT is 10%, then MARRBT is 0.10/0.75, or 13.333%. When the interest rate, for example i = 12%, is less than MARRBT, one should borrow as much as possible and delay repaying the loan as long as possible; when the interest rate is greater than MARRBT, one should borrow as little as possible and repay the principal as quickly as possible. After-tax analysis of a single alternative when money is borrowed is very similar to the analysis of a single alternative financed using retained earnings. When borrowed money is used, however, the interest payment on the loan (IPMT) must be subtracted from BT&LCF to determine taxable income (TI). Also, remember that both the principal and interest payments are cash flows. This is demonstrated in Example 9.7.

EXAMPLE 9.7 Using Excel® to Analyze After-Tax Effects of Borrowed Funds—Single Alternative Now, let’s examine four payment plans as they are used to secure $300,000 financing for the $500,000 SMP machine acquisition considered in earlier examples. The plans are: (a) Plan 1— pay the accumulated interest at the end of each interest period and pay the principal at the end of the loan period; (b) Plan 2—make equal principal payments, plus interest on the unpaid balance at the end of the period; (c) Plan 3—make equal end-of-period payments; and (d) Plan 4—make a single payment of principal and interest at the end of the loan period. The loan interest rate = 12%. Key Data Given Loan amount = $300,000; P = $500,000; itr = 25%; MARRAT = 10%; BT&LCF (years 1–9) = $108,333.33; BT&LCF (year 10) = $156,333.33 Find PWAT, AWAT, FWAT, IRRAT, ERRAT for Plans 1, 2, 3, and 4 Solution a. Figure 9.4 depicts the after-tax analysis for Plan 1, in which only the interest is paid each year; at the end of year 10, the $300,000 principal is paid.

FIGURE 9.4 After-Tax Analysis of the SMP Investment with $300,000 of Borrowed Capital Repaid Using Plan 1 Excel® Data File

As shown, the after-tax present worth (PWAT), based on a 12% interest rate on the $300,000 loan and a 10% MARRAT, is $128,795.20. When no borrowed funds were used, PWAT was $110,361.50 (see Example 9.3). Borrowing the $300,000 at 12% compound annual interest increased PWAT, because interest is deducted from taxable income. Excel® Video Lesson: The FV Function Notice, because of the single payment of principal in year 10, ATCF is negative. Hence, there are two negative values in the ATCF column. Descartes’ rule of signs indicates that either two or zero roots exist; the two roots occur at −18.50% and at 33.19%. Because multiple negative values exist, the Excel® MIRR worksheet function cannot be used to compute ERR. However, by following the process first outlined in Example 6.6, the IRR function can be used to calculate ERRAT. As shown in Figure 9.4, ERRAT equals 15.61%. Excel® Video Lesson: MIRR Financial Function Excel® Video Lesson: IRR Financial Function b. Table 9.7 provides the after-tax analysis for Plan 2, in which equal annual principal payments are made, plus interest on the unpaid loan balance. As shown, PWAT is $121,927.80. Plan 2 is not as attractive as Plan 1. Why? Because the after-tax interest rate is less than the MARR, it is better to delay paying principal. c. Table 9.8 contains the results for Plan 3, in which equal annual payments are made over the 10-year period. For Plan 3, PWAT is $123,485.16. As expected, because principal payments increase over time, Plan 3 does not perform as well as Plan 1, where no principal payments are made until the end of the loan period. d. Figure 9.5 provides the results for Plan 4, in which no payment is made until the end of the loan period. Notice, with an after-tax PW worth of $112,022.00. Plan 4 ranks last among the four plans.

TABLE 9.7 After-Tax Analysis of the SMP Investment with $300,000 of Borrowed Capital Repaid Using Plan 2 MARRAT = 10% income tax rate = 25% EOY BT&LCF PPMT IPMT

DWO

 0

− − $500,000.00 $300,000.00

 1

$108,333.33 $30,000.00 $36,000.00 $100,000.00

 2  3

interest rate = 12% TI Tax ATCF − $200,000.00 

− − $49,250.00  $27,666.67 $6,916.67 $108,333.33 $30,000.00 $32,400.00 $160,000.00 − − $66,950.00  $84,066.67 $21,016.67 $1,08,333.33 $30,000.00 $28,800.00 $96,000.00 − − $53,650.00  $16,466.67 $4,116.67

 4  5

$1,08,333.33 $30,000.00 $25,200.00 $57,600.00 $25,533.33 $6,383.33 $46,750.00  $1,08,333.33 $30,000.00 $21,600.00 $57,600.00 $29,133.33 $7,283.33 $49,450.00 

 6

$1,08,333.33 $30,000.00 $18,000.00 $28,800.00 $61,533.33 $15,383.33 $44,950.00 

 7

$1,08,333.33 $30,000.00 $14,400.00

$0.00 $93,933.33 $23,483.33 $40,450.00 

 8  9

$108,333.33 $30,000.00 $10,800.00 $108,333.33 $30,000.00 $7,200.00

$0.00 $97,533.33 $24,383.33 $43,150.00  $0.00 $101,133.33 $25,283.33 $45,850.00 

10

$158,333.33 $30,000.00 $3,600.00

$0.00 $154,733.33 $38,683.33 $86,050.00  PWAT = $121,927.80  FWAT = $316,249.32  AWAT = $19,843.19 

Excel® Data File

IRRAT =

22.97% 

ERRAT =

15.36% 

TABLE 9.8 After-Tax Analysis of the SMP Investment with $300,000 of Borrowed Capital Repaid Using Plan 3 MARRAT = 10% income tax rate = 25% EOY BT&LCF

PPMT

IPMT

interest rate = 12% DWO

 0

− − $500,000.00 $300,000.00

 1

$108,333.33 $17,095.25 $36,000.00 $100,000.00

 2

$108,333.33 $19,146.68 $33,948.57 $160,000.00

TI

Tax

ATCF − $200,000.00 

− − $62,154.75  $27,666.67 $6,916.67

 4

− − $76,641.89  $85,615.24 $21,403.81 $108,333.33 $21,444.28 $31,650.97 $96,000.00 − − $60,067.49  $19,317.64 $4,829.41 $108,333.33 $24,017.59 $29,077.65 $57,600.00 $21,655.68 $5,413.92 $49,824.16 

 5

$108,333.33 $26,899.71 $26,195.54 $57,600.00 $24,537.79 $6,134.45 $49,103.64 

 6

$108,333.33 $30,127.67 $22,967.58 $28,800.00 $56,565.75 $14,141.44 $41,096.65 

 7  8

$108,333.33 $33,742.99 $19,352.26 $108,333.33 $37,792.15 $15,303.10

$0.00 $88,981.07 $22,245.27 $32,992.82  $0.00 $93,030.23 $23,257.56 $31,980.53 

 9

$108,333.33 $42,327.21 $10,768.04

$0.00 $97,565.29 $24,391.32 $30,846.76 

10

$158,333.33 $47,406.47 $5,688.78

$0.00 $152,644.56 $38,161.14 $67,076.94 

 3

PWAT = $123,485.16  FWAT = $320,288.71  AWAT = $20,096.64 

Excel® Data File

IRRAT =

25.08% 

ERRAT =

15.42% 

FIGURE 9.5 After-Tax Analysis of SMP Investment with $300,000 of Borrowed Capital Repaid Using Plan 4 Excel® Data File With Plan 4, no payment is made until year 10. Hence, the most negative cash flow occurs at that time. Descartes’ rule of signs indicates either two or zero roots exist; as shown in Figure 9.5, the two roots occur at −0.046% and at 48.858%. Because multiple negative values exist, the Excel® MIRR worksheet function cannot be used to compute ERR; however, as shown, ERR can be calculated using the Excel® IRR worksheet function with the modified ATCF column and equals 15.00%. Notice, PWAT is maximized when MARRAT equals 12.35%. (We used the Excel® SOLVER tool to obtain the MARRAT value that maximized PWAT.) Excel® Video Lesson: SOLVER Tool Exploring the Solution Given the results obtained for the four payment plans, it is anticipated that after-tax PW will increase as the amount borrowed increases. Figure 9.6 presents the results for Plan 1 if the investment is entirely paid for using borrowed funds. As expected, the PWAT of $141,084.34 is greater than the PWAT of $128,795.20 that occurred when only $300,000 was borrowed.

FIGURE 9.6 After-Tax Analysis of $500,000 Investment with 100% Borrowed Capital Repaid Using Plan 1 Excel® Data File When 100% of investment capital is obtained by borrowing, no initial investment occurs at the beginning of the ATCF series. As shown in Figure 9.6, the only negative-valued cash flow occurs at the end of year 10. As such, increases in MARR will increase PWAT. This is just the opposite of what we observed previously. Because there are no negative-valued cash flows before the planning horizon’s end, IRR and ERR are not defined. (Notice, PWAT is maximized when MARR = 18.106%.) Example 9.7 concerns a single investment alternative. After-tax analysis of multiple alternatives with borrowed capital is similar. Again, comparison of alternatives is based on ATCF instead of BTCF, the use and repayment of borrowed capital, and the use of MARRAT instead of MARRBT. Also, depending on the alternatives being considered, different depreciation allowances might apply to the alternatives. Example 9.7 presents ample guidance for after-tax analysis of investment alternatives using borrowed capital. Among the four payment plans, if the lender’s interest rate is less than the borrower’s MARRAT, Plan 4 yields the greatest PWAT. If the lender’s interest rate is greater than the borrower’s MARRAT, but less than the borrower’s MARRAT divided by one minus the borrower’s itr, Plan 1 has the greatest PWAT. If the lender’s interest rate is greater than the borrower’s MARRAT divided by one minus the borrower’s itr, it is best to use your own investment capital; however, if borrowing is necessary, then Plan 2 yields the greatest PWAT. For Example 9.7, with the lender’s interest rate equal to 12%, the borrower’s MARRAT equal to 10%, and the income-tax rate equal to 25%, MARRAT/(1 − itr) is equal to 13.33%. Therefore, because the

lender’s interest rate (12%) is between MARRAT = 10% and MARRAT/(1 − itr) = 13.33%, Plan 1 is preferred, as demonstrated.

Concept Check 09.03-CC001 When determining the present worth of the after-tax cash flows of an investment purchased using borrowed funds, which of the following are required? I. Principal and interest component for each loan payment II. Tax rate applied to the taxable income generated by the investment III. Depreciation deductions for the investment IV. Before-tax and loan cash flows for the investment V. MARR a. I, III, IV, and V only b. II, III, IV, and V only c. IV and V only d. All items (I, II, III, IV, and V)

9.4 Leasing Versus Purchasing Equipment LEARNING OBJECTIVE Illustrate the different tax consequences for the common example of a lease vs. purchase alternative. As mentioned previously, another example of alternatives with different tax consequences is leasing versus purchasing.

EXAMPLE 9.8 After-Tax Comparison of Leasing Versus Purchasing The Acme Brick Company is considering adding five lift trucks to its fleet. Both purchasing and leasing were discussed with the fork-truck supplier. If the lift trucks are purchased, they will have a first cost of $18,000; annual operating and maintenance costs are estimated to be $3,750 per truck. At the end of the 5-year planning horizon, the lift trucks are estimated to have salvage values of $3,000 each. The lift trucks qualify as MACRS 3-year property. If the lift trucks are leased, beginning-of-year lease payments will be $5,900 per truck. The supplier includes maintenance in the lease price. However, the company must pay annual operating costs of $1,800 per truck. Lease payments can be expensed for tax purposes. Using an after-tax MARR of 12% and an income-tax rate of 25%, should the company lease the lift trucks? Key Data Given Buy P = $18,000; A = $3,750 per truck; SV = $3,000 per truck at end of Year 5; itr = 25%; MARRAT = 12% Lease A = $5,900 per truck plus $1,800 per truck for annual operating cost Find Should the trucks be leased or bought assuming that the trucks qualify as a MACRS 3year property? Solution As shown in Table 9.9, an after-tax analysis indicates the lift trucks should be leased. The present worth cost of purchasing is $113,735.42, whereas the present worth cost of leasing is $106,136.23. The $7,599.19 difference in present worth costs is probably great enough for the company to lease the lift trucks.

TABLE 9.9 After-Tax Comparison of Purchasing Versus Leasing Lift Trucks EOY

BTCF(P)

DWO(P)

TI(P)

T(P)

ATCF(P)

0

−$90,000.00

1

−$18,750.00 $29,997.00 −$48,747.00 −$12,186.75

2 3

−$18,750.00 $40,005.00 −$58,755.00 −$14,688.75 −$4,061.25  −$18,750.00 $13,329.00 −$32,079.00 −$8,019.75 −$10,730.25 

4

−$18,750.00 $6,669.00 −$25,419.00 −$6,354.75 −$12,395.25 

5

−$90,000.00 

−$3,750.00

$0.00 −$3,750.00

PWBT(P) = −$144,251.31 EOY

BTCF(L)

−$937.50

−$6,563.25 

−$2,812.50 

PWAT(P) = −$113,735.42  DWO(L)

TI(L)

T(L)

ATCF(L)

0

−$29,500.00

−$29,500.00 −$7,375.00 −$22,125.00 

1

−$38,500.00

$0.00

−$38,500.00 −$9,625.00 −$28,875.00 

2

−$38,500.00

$0.00

−$38,500.00 −$9,625.00 −$28,875.00 

3

−$38,500.00

$0.00

−$38,500.00 −$9,625.00 −$28,875.00 

4

−$38,500.00

$0.00

−$38,500.00 −$9,625.00 −$28,875.00 

5 −$9,000.00 PWBT(L) = −$141,514.97

$0.00

−$9,000.00 −$2,250.00 −$6,750.00  PWAT(L) = −$106,136.23 

Excel® Data File Interestingly, as shown in Table 9.9, if the decision had been made using a before-tax MARR of 0.12/(1 − 0.25), or 16%, the before-tax present worth cost of purchasing is $144,251.31, and the present worth cost of leasing is $141,514.97. Again, the $7,599.19 difference in the present worths is probably great enough for the company to decide to lease the lift trucks.

Concept Check 09.04-CC001 When comparing purchasing equipment versus an operating leasing of equipment, the purchase cost of equipment is generally __________ and the lease cost of equipment is generally __________. a. depreciated; depreciated b. depreciated; expensed c. expensed; expensed d. expensed; depreciated

9.5 After-Tax Analysis with Bonus Depreciation LEARNING OBJECTIVE Analyze the after-tax implications of bonus depreciation for investment alternatives. Thus far, our analyses have been based on no bonus depreciation. From the previous analyses, we learned the faster an asset is depreciated, the greater will be its after-tax present worth. Therefore, bonus depreciation will yield increased after-tax present worth. Drawing on the material in Chapter 8, we will explore the impact of bonus depreciation on the acquisition of the SMP machine. Because the use of bonus depreciation does not depend on the source of investment capital, we assume the acquisition occurs with retained earnings. Bonus Depreciation To stimulate capital expenditures, the U.S. Congress frequently creates provisions for additional first-year depreciation allowances, called bonus depreciation.

EXAMPLE 9.9 Applying Bonus Depreciation to the Purchase of the SMP Machine A surface-mount placement (SMP) machine is purchased for $500,000. Over its 10-year life, it produces before-tax cash flows of $108,333.33 and has a salvage value of $50,000. Based on an income-tax rate of 25%, an after-tax required return on investment of 10%, MACRS-GDS depreciation, and bonus depreciation of 50% and 100%, compare the after-tax present worth of the acquisition with that obtained in the absence of bonus depreciation. Key Data Given P = $500,000, itr = 25%, MARRAT = 10%, BTCF (years 1 − 10) = $108,333.33, S = $50,000, bonus depreciation of 50% and 100% Find Compare PWAT without bonus depreciation to PWAT with 50% bonus depreciation and PWAT with 100% bonus depreciation Solution The computation of PWAT with 50% bonus depreciation and with 100% bonus depreciation is given in Table 9.10. The depreciation allowances were calculated in Example 8.5. From Example 9.3, the after-tax present worth without bonus depreciation is $110,361.50; with 50% bonus depreciation, the after-tax present worth is $118,850.91; and with 100% bonus depreciation, the after-tax present worth is $127,340.31.

TABLE 9.10 After-Tax Analysis of the SMP Machine Investment with Bonus Depreciation Bonus Depreciation = 50% EOY  0

BTCF

DWO

TI

T

−$500,000.00

ATCF −$500,000.00 

 1

$108,333.33 $300,000.00 −$191,666.67 −$47,916.67

$156,250.00 

 2

$108,333.33

$80,000.00

$28,333.33

$7,083.33

$101,250.00 

 3

$108,333.33

$48,000.00

$60,333.33

$15,083.33

$93,250.00 

 4  5

$108,333.33 $108,333.33

$28,800.00 $28,800.00

$79,533.33 $79,533.33

$19,883.33 $19,883.33

$88,450.00  $88,450.00 

 6

$108,333.33

$14,400.00

$93,933.33

$23,483.33

$84,850.00 

 7

$108,333.33

$0.00

$108,333.33

$27,083.33

$81,250.00 

 8

$108,333.33

$0.00

$108,333.33

$27,083.33

$81,250.00 

 9

$108,333.33

$0.00

$108,333.33

$27,083.33

$81,250.00 

10

$158,333.33

$0.00

$158,333.33

$39,583.33

$118,750.00 

PWAT =

$118,850.91 

Bonus Depreciation = 100% EOY 0

BTCF

DWO

TI

T

−$500,000.00

ATCF −$500,000.00 

1 2

$108,333.33 $500,000.00 −$391,666.67 −$97,916.67 $108,333.33 $0.00 $108,333.33 $27,083.33

$206,250.00  $81,250.00 

3

$108,333.33

$0.00

$108,333.33

$27,083.33

$81,250.00 

4

$108,333.33

$0.00

$108,333.33

$27,083.33

$81,250.00 

5

$108,333.33

$0.00

$108,333.33

$27,083.33

$81,250.00 

6

$108,333.33

$0.00

$108,333.33

$27,083.33

$81,250.00 

7

$108,333.33

$0.00

$108,333.33

$27,083.33

$81,250.00 

8

$108,333.33

$0.00

$108,333.33

$27,083.33

$81,250.00 

9 10

$108,333.33 $158,333.33

$0.00 $0.00

$108,333.33 $158,333.33

$27,083.33 $39,583.33

$81,250.00  $118,750.00 

PWAT =

$127,340.31 

Excel® Data File Recall, from Example 9.4, if the $500,000 had been expensed rather than capitalized, the $500,000 expense would have reduced taxable income by $500,000 in the year in which it occurred; as a result, the after-tax present worth was $138,703.95. With 100% bonus depreciation, we reduce taxable income by $500,000 in the first year. In practice, companies typically pay income taxes quarterly. Therefore, with 100% bonus

depreciation, a company would take a full allowance at the end of the first quarter, not at the end of the first year.

Concept Check 09.05-CC001 Utilizing bonus depreciation will result in an increase in after-tax present worth. a. True b. False

9.6 After-Tax Analysis with a Section 179 Expense Deduction LEARNING OBJECTIVE Analyze the after-tax implications for small businesses of a Section 179 expense deduction. The American Tax Cuts and Jobs Act signed by President Donald Trump in December of 2017 increased Section 179 deduction limits to $1,000,000 for businesses with total asset purchases of $2,500,000 or less for the tax year beginning in 2018. The deduction limits apply to each taxpayer, not each investment. The amount eligible to be expensed for any tax year is reduced dollar-for-dollar by the amount the aggregate cost of qualifying property exceeds $2,500,000. If the aggregate cost of qualifying equipment is less than or equal to $1,000,000, the full amount can be expensed; if the aggregate cost of qualifying equipment is greater than $1,000,000 but less than or equal to $2,500,000, then $1,000,000 can be expensed; if the aggregate cost of qualifying equipment is greater than $2,500,000 but less than or equal to $3,500,000, then the amount by which the aggregate cost of qualifying equipment exceeds $2,500,000 can be expensed; and if the aggregate cost of qualifying equipment exceeds $3,500,000, no expense deduction can occur. American Tax Cuts and Jobs Act Among other things, the American Tax Cuts and Jobs Act adjusts the parameters of Section 179 of the United States Internal Revenue Code. In 2018, it increased Section 179 limits to $1 million for businesses with total assets purchases of no more than $2.5 million. To illustrate the calculation of the expense deduction, suppose the aggregate cost of qualifying equipment equals $800,000; in this case, $800,000 can be expensed. If the aggregate cost of qualifying equipment equals $2,000,000, then $1,000,000 can be expensed. If the aggregate cost of qualifying equipment equals $3,000,000, then $3,000,000 minus $2,500,000, or $500,000, can be expensed. If the aggregate cost of qualifying equipment equals $4,000,000, then no expense deduction can occur. After taking the expense deduction, the cost basis for qualifying equipment is reduced by the expense deduction taken. Furthermore, small businesses can take both a Section 179 expense deduction and bonus depreciation, if it is available. (Of course, if 100% bonus depreciation is allowed, the full amount of the cost basis is deducted in the first year.) To illustrate the application of Section 179 expense deduction, we consider it without bonus depreciation and with 50% bonus depreciation. Section 179 Expense Deduction Section 179 of the United States Internal Revenue Code was introduced in 1981 by the U.S. Congress to stimulate capital investments by small businesses. Over the years, the magnitude of the additional first-year depreciation allowance has been increased. Because the added allowance reduces taxable income in the same way normal business expenses do, the allowance is called a Section 179 expense deduction.

EXAMPLE 9.10 Applying the Section 179 Expense Deduction To illustrate the application of a Section 179 expense deduction, assume three identical SMP machines are purchased for $1,500,000 and placed in service during the 2018 tax year. We also assume no other qualifying equipment is purchased during the year. The annual BTCF equals $325,000 and the salvage value equals $150,000. Because the $1,500,000 investment does not exceed $2,500,000, a $1,000,000 expense deduction is allowed and the cost basis is reduced to $1,500,000 minus $1,000,000, or $500,000. Based on an income-tax rate of 25%, an after-tax required return on investment of 10%, and MACRS-GDS depreciation as shown in Table 9.11, the after-tax present worth is $365,042. Therefore, the aftertax present worth for each SMP machine is $121,681, compared to $110,362 without Section 179 expense deduction. TABLE 9.11 Applying the Section 179 Expense Deduction

Excel® Data File

EXAMPLE 9.11 Applying the Section 179 Expense Deduction with Bonus Depreciation To illustrate the application of a Section 179 expense deduction with bonus depreciation, assume 50% bonus depreciation applies for the expenditures in Example 9.10. First, the small business takes the Section 179 expense deduction in the first year. Then, a 50% bonus depreciation allowance is taken in the first year. Finally, MACRS 5-year allowances are taken for years 1 thru 6. As shown in Table 9.12, the after-tax present worth is $373,531.54 for the three SMP machines. Therefore, the after-tax present worth for each SMP machine is $124,510.51, compared to $118,850.91 with 50% bonus depreciation and no Section 179 expense deduction, versus $121,680.71 with a Section 179 expense deduction and no bonus depreciation. TABLE 9.12 Applying the Section 179 Expense Deduction and 50% Bonus Depreciation

Excel® Data File

Concept Check 09.06-CC001 Small businesses can take both a Section 179 expense deduction and bonus depreciation. a. True b. False

CHAPTER 10 Inflation

Chapter 10 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced e-text and WileyPLUS

FE-Like Problems 10-FE001 Logan is conducting an economic evaluation under inflation using the then-current approach. If the inflation rate is j and the real time value of money rate is d, which of the following is the interest rate he should use for discounting the cash flows? a. j b. d c. j + d d. j + d + dj 10-FE002 Mike’s Veneer Shop owns a vacuum press that requires annual maintenance. Mike has a contract to cover the maintenance expenses for the next 5 years. The contract calls for an annual payment of $600 with adjustment each year for inflation. Inflation is expected to hold constant at 6%/yr over this period. The then-current cash flow pattern for this expense is best described by which of the following? a. Uniform series b. Gradient series c. Geometric series d. Continuous series Correct or Incorrect? Clear

  Check Answer

10-FE003 Mike’s Veneer Shop owns a vacuum press that requires annual maintenance. Mike has a contract to cover the maintenance expenses for the next 5 years. The contract calls for an annual payment of $600 with adjustment each year for inflation. Inflation is expected to hold constant at 6%/yr over this period. The constant-dollar cash flow pattern for this expense is best described by which of the following? a. Uniform series b. Gradient series c. Geometric series d. Continuous series 10-FE004 An economist has predicted that there will be a 7% per year inflation of prices during the next 10 years. If this prediction proves to be correct, an item that presently sells for $10 would sell for what price in 10 years? a. $5.08 b. $10.70 c. $17.00 d. $19.67 Correct or Incorrect? Clear

  Check Answer

10-FE005 If the real discount rate is 7% and the inflation rate is 10%, which of the following interest rates will be used to find the present worth of a series of cash flows that are in then-current dollars? a. 10.0% b. 17.7% c. 7.0% d. 10.7% 10-FE006 If the real discount rate is 7% and the inflation rate is 10%, hich of the follo ing interest rates ill be sed to find the present orth of a

which of the following interest rates will be used to find the present worth of a series of cash flows that are in constant-worth dollars? a. 10.0% b. 17.7% c. 7.0% d. 10.7% Correct or Incorrect? Clear

  Check Answer

10-FE007 When done correctly, what is the relationship between the present worth of an alternative calculated using a then-current approach and the present worth of the alternative calculated using a constant-worth approach? a. They are equal b. Then-current PW is higher because it uses inflated dollars c. Constant-worth PW is higher because it uses a lower discount rate d. Cannot be determined without knowing the cash flows and inflation rate 10-FE008 Ten years ago Jennifer bought an investment property for $100,000. Over the 10-year period inflation has held consistently at 3% annually. If Jennifer expects a 13%/yr real rate of return, what would she sell the property for today? a. $116,000 b. $134,400 c. $339,500 d. $456,200 Correct or Incorrect? Clear

  Check Answer

10-FE009 As reported by the Bureau of Labor Statistics, the CPI for 2005 was 585.0 (using a Base Year of 1967 = 100). The CPI for 2006 was 603.9. Based on this data, what was the inflation rate for 2006? a. 3.23% b. 5.85% c. 6.04% d. 18.9% 10-FE010 Your small business has purchased only two new assets totaling $2,800,000 during the tax year and you want to take advantage of the Section 179 expense deduction. Which would you legally use for the deduction? a. $2,000,000 b. $700,000 c. $1,000,000 d. $2,800,000 Correct or Incorrect? Clear

  Check Answer

10-FE011 You have just added a purchase of a high-capacity weld mill for $4,000,000. You want to maximize the deduction you can take when calculating taxable income. Which would you use? a. Section 179 and MACRS depreciation b. 150% bonus depreciation c. 50% bonus depreciation alone d. 100% bonus depreciation 10-FE012 Given an inflation rate of 10% and a combined discount rate of 21%, what is the real (or constant) discount rate? a. 11% b. 10%

c. 33.1% d. 2.1% Correct or Incorrect? Clear

  Check Answer

Problems Section 10.1 The Meaning and Measure of Inflation LEARNING OBJECTIVE 10.1 Quantify the effects of inflation on purchasing power with the inflation rate. 10.01-PR001 What is a good working definition of “inflation” in 10 words or less? 10.01-PR002 Give four examples of goods or services that have exhibited inflation in recent years. 10.01-PR003 What is the relationship between “inflation” and “deflation”? Give an example of deflation experienced in your everyday life. 10.01-PR004 Give two examples of goods or services that you have seen inflate dramatically and also deflate dramatically over the past few years. 10.01-PR005 What is meant by a “market basket rate”? 10.01-PR006 What are the differences between the Consumer Price Index and the Producer Price Index? 10.01-PR007 What is the Higher Education Price Index (HEPI), where does it fit in with the CPI and PPI, and how is the HEPI related to the CPI? 10.01-PR008 Suppose a friend argues that the CPI does not represent them because they do not purchase some of the things, including big-ticket items, in the market basket. Can they conclude that the CPI is irrelevant to them? Explain your reasoning.

10.01-PR009 Video Solution The CPI-U (U.S. city average, all items) has the following annual averages: Year Index 2013 233.0 2014 236.7 2015 237.0 2016 240.0 2017 245.1 a. For each year from 2014 to 2017 determine the annual inflation rate in percent to two decimal places. b. Because inflation, like interest, is compounded from period to period (e.g., year to year), estimate the overall annual inflation rate per year from 2013 to 2017. Suggestion! Do not simply average the rates of part (a).

10.01-PR010 The CPI-U for Americans 62 years of age and older (some of your professors and some of your authors are interested in this!) present the following annual inflation rates in percent: Year Rate % 2012 1.6 2013 2014 2015

2.4 1.9 3.3

2016

3.4

a. Assuming the index value in year 2011 was 100.0, determine the index for each year from 2012 to 2016 to one place after the decimal.

b. Because inflation, like interest, is compounded from period to period (e.g., year to year), estimate the overall annual inflation rate per year from 2013 to 2017. Suggestion! Do not simply average the rates given above. 10.01-PR011 The Korean manufacturing output per hour index is given for years 2014–2018 as follows: Year Index 2014 214.8 2015 235.8 2016 252.2 2017 281.2 2018 305.1 a. For each year from 2015 to 2018 determine the rate of increase in Korean manufacturing output per hour to two decimal places. b. Because this index, like inflation, is compounded from period to period (e.g., year to year), estimate the overall annual rate of increase in Korean manufacturing output per hour from 2015 to 2018. Suggestion! Do not simply average the rates of part (a). 10.01-PR012 The Korean hourly compensation rates of increase are presented as follows: Year Rate % 2014 −14.84 2015 13.75 2016 10.09 2017 18.87 2018 18.79 a. Assuming the index value in year 2013 was 165.9, determine the index for each year from 2014 to 2018 to one place after the decimal.

b. Because this index, like inflation, is compounded from period to period (e.g., year to year), estimate the overall annual rate of increase per year from 2014 to 2018. Suggestion! Do not simply average the rates given above.

Section 10.2 Before-Tax Analysis LEARNING OBJECTIVE 10.2 Conduct a before-tax analysis with inflation. 10.02-PR001 Video Solution You are earning 5.2% on a certificate of deposit. Inflation is running 3.5%. What is the real rate of return on your investment?

10.02-PR002 You are considering a bond that pays annually at 6.2%. Inflation is projected to be running 4.0%. What will be your real interest rate? 10.02-PR003 Array Solutions requires a 14.0% return on their projects. Analysis shows that even though they have been earning the desired 14.0%, their real return appears to be only 10.0% when they look at what they can buy with their returns. a. Explain why there is this discrepancy. b. Determine the inflation rate. 10.02-PR004 Chevron-Phillips requires a real return of 14.2%. If inflation is running 3.8%, what must be their MARR or “hurdle rate” on capital investments when using then-current dollars in analyses? 10.02-PR005 Suppose you want to earn a real interest rate of 5%. For inflation rates of 0.0, 1.0, 2.0, …, 9.0, 10.0, 15.0, 20.0, and 50.0%, determine the combined rate of interest you must earn. 10.02-PR006 How much must you invest exactly 5 years from now to have $500,000 in today’s buying power 20 years from now? You can invest your money at 10% per year and inflation runs 4%. 10.02-PR007 Video Solution A software company’s labor requirements currently cost $350,000/year. The labor hour requirements are expected to increase by 10% per year over the next five years. If inflation is 4%, determine the labor costs after 5 years using:

a. Then-current dollars. b. Constant-worth dollars.

10.02-PR008 You invested $10,000 on January 1, 2018, at 7% interest compounded annually. You have not touched the investment since that date. You are planning to take your money and close out the investment on January 1, 2028. a. If average inflation is 3.7%, what has been your “real” annual interest rate? b. At the time you originally invested, there was a boat you admired costing $8,000. Over the years, boats are inflating at a rate of 7%. You were also interested in a marine navigation system costing $4,000; similar systems are dropping in price at a 2% rate. If you decide to buy one of each when you close out the account, how much will the purchases cost you? c. How much money will you have left over (or be short) after your purchases? 10.02-PR009 Your department is budgeting miscellaneous expenses for the next 5 years. Your best guess at the annual inflation rate is 3.9% and the combined MARR is 15%. Expenses currently run $14,500 per year. Assume that expenses are end-of-year payments. a. Determine the then-current dollar amounts for years 1, 2, 3, 4, and 5. b. Determine the constant-dollar amount for years 1, 2, 3, 4, and 5. c. Determine the PW of the then-current-dollar amounts. d. Determine the PW of the constant dollar amounts. 10.02-PR010 Shea is pricing materials (wood, wire, pipe, etc.) for new home construction on a “per unit” basis. Inflation on materials has been running at 16.0% for the past 3 years and is expected to remain at that rate for the next 10 years. The actual dollars paid for expenses at the end of year 3 for a “unit” are $55,000.

a. What did the same set of materials cost 3 years ago? b. If the trend continues, what will they cost 10 years from now? 10.02-PR011 Video Solution Global steel prices have a yearover-year inflationary rate increase of 12.4%. Tube Fab purchased $700,000 of a particular carbon steel during the year just ended right now, and they intend to purchase the same quantity at the end of each of the next 5 years. Tube Fab earns a real rate of 16.0% on their money. a. Determine the then-current amounts they will pay for steel at the end of each of the next 5 years. b. Determine the constant-value amounts they will pay for steel at the end of each of the next 5 years. c. Determine Tube Fab’s PW of expenditures over the next 5 years using then-current dollars. d. Determine Tube Fab’s PW of expenditures over the next 5 years using constant-value dollars.

10.02-PR012 Global steel prices have a year-over-year inflationary rate increase of 12.4%. Tube Fab purchased $700,000 of a particular carbon steel during the year just ended right now. Their business has been increasing and they intend to purchase 20% more steel each year, over the previous year’s purchase, for the next 5 years. Tube Fab earns a real rate of 9.0% on their money. a. Determine the then-current amounts they will pay for steel at the end of each of the next 5 years. b. Determine the constant-value amounts they will pay for steel at the end of each of the next 5 years.

c. Determine Tube Fab’s PW of expenditures over the next 5 years using then-current dollars. d. Determine Tube Fab’s PW of expenditures over the next 5 years using constant-value dollars. 10.02-PR013 Padayappa has now retired after 40 years of employment. He just made an annual deposit to his investment portfolio and realized he has $2 million (not counting home, cars, furniture, etc.). His money has been earning 7% per year and inflation has been running 4% per year over the past 40 years. a. What equal amount of money did he put into his investment at the end of each year? b. What is the buying power of his $2 million in terms of a base 40 years ago? c. If he could buy a TV 40 years ago for $400, what would a comparable one cost today if the consumer electronics inflation rate is −3%? 10.02-PR014 A 24-year-old December 2019 graduate has decided he wants the equivalent of $2,500,000 in January 1, 2020 buying power to be available exactly 40 years later on January 1, 2060. He plans to make his first investment of $Z on January 1, 2021 and every year thereafter, with the last payment of $Z on January 1, 2060. He can earn 8% on his money and expects inflation to run 5%. a. How many actual dollars will there be in his account immediately after his last deposit? b. What is $Z? 10.02-PR015 An investment of $8,000 is made at time 0 with returns of $3,500 at the end of each of years 1–4, with all monetary amounts being in real dollars. Inflation is running 7% per year over that time. Also, the real rate of return is 15% per year. Determine the present worth of the investment using both real dollars and then-current dollars. 10.02-PR016 The winner of a lottery is given a choice of $1,000,000 cash today or $2,000,000 paid out as follows: $100,000 cash per year for 20 years

with the first payment today and 19 subsequent annual payments thereafter. The inflation rate is expected to be constant at 4%/yr over the award period and the winner’s TVOM (real interest rate) is 3.5%/yr. a. Which choice is better for the winner? Neglect the effect of taxes, life span, and uncertainty. b. At what value of inflation are the two choices economically equivalent? c. What would you do if you do NOT neglect the effect of life span and uncertainty? 10.02-PR017 Parents wish to provide for their child’s college education. Being a bit risk averse, they plan to invest in stable, yet unspectacular, opportunities yielding a 6.0% return. Their best guess at inflation is 4.0% for the foreseeable future. They plan to make investments on the child’s birthday (USA-style—first birthday is 1 year after date of birth), every year from ages 1 through 18. They envision their child needing $100,000 at the beginning of the first year of college, with inflated amounts to follow for 3 more years. The first $100,000 will be needed right at the end of the 18th birthday’s investment —right at the beginning of the 19th year. a. What equal amount of money must they invest at the end of each year? b. If the parents decide that their earning power will increase, and each year they will invest 10% more, how much must they invest in the first year? c. If a grandparent offers to put up the entire sum of money needed, on the date of the child’s birth, what sum must the grandparent put up? 10.02-PR018 Labor costs over a 4-year period have been forecast in thencurrent dollars as follows: $10,000, $12,000, $15,000, and $17,500. The general inflation rate for the 4 years is forecast to be 5%. Determine the constant-dollar labor costs for each of the 4 years. 10.02-PR019 Yearly labor costs of a highway maintenance group are currently $420,000/year. If labor costs increase at a 12% rate and general inflation increases at 9%, determine for each of the next 6 years the labor cost in thencurrent and constant dollars.

10.02-PR020 If you desire a real return of 8% on your money and inflation is running at 3%, what combined rate of return should you require on your investments? 10.02-PR021 Mellin Transformers, Inc., uses a required return of 15% in all economic justifications. Inflation is anticipated to be 5% over the foreseeable future. What real discount rate are they implicitly using? 10.02-PR022 The following material costs are anticipated over a 5-year period: $9,000, $11,000, $14,000, $18,000, and $23,000. It is estimated that a 4% inflation rate will apply over the time period in question. The material costs given above are expressed in then-current dollars. The time value of money, excluding inflation, is estimated to be 7%. Determine the present worth equivalent for material cost using the following: a. Then-current costs. b. Constant-worth costs. 10.02-PR023 The inflation rates for 4 years are forecast to be 3%, 3%, 4%, and 5%. The interest rate exclusive of inflation is anticipated to be 6%, 5%, 4%, and 5% over this same period. If labor is projected to be $1,000, $1,500, $2,000, and $1,000 in then-current dollars, during those years, determine the present worth equivalent for labor cost. 10.02-PR024 A landfill has a first cost of $270,000. Annual operating and maintenance costs for the first year will be $40,000. These costs will increase at 11% annually. Income for dumping rights at the landfill will be held fixed at $120,000 per year. The landfill will be operating for 10 years. Inflation will average 4.5%, and a real return of 3.6% is desired. What is the present worth of this project using? a. A then-current analysis? b. A constant-worth analysis? 10.02-PR025 The unit price of personal computers, measured in constant dollars, is expected to decrease at an annual rate of 10%. However, the number of microcomputers purchased by the company is expected to increase over a 5-year period at an annual rate of 30%. The unit price of a personal computer

is currently $2,000. This year the company will purchase 100 microcomputers at the current price. Using a real interest rate of 8% and an inflation rate of 6%, determine the following: a. The yearly expenditures for personal computers over the 5-year period, measured in constant-worth dollars. b. The yearly expenditures for personal computers over the 5-year period, measured in then-current dollars. c. The single sum equivalent at the end of the current year of the expenditures on microcomputers over the 5-year period, measured in current dollars. d. The future worth equivalent of the expenditures on microcomputers at the end of the 5-year period, measured in then-current dollars. 10.02-PR026 Ninety-thousand dollars is invested in a program to reduce the material requirements in a production process. As a result of the investment, the annual material requirement is reduced by 10,000 pounds. The present unit cost of the material is $2 per pound. The price of a pound of the material is expected to increase at an annual rate of 8%, due to inflation. Determine the combined interest rate that equals the present worth of the savings to the present worth of the investment over a 5-year period; based on the result obtained, determine the real interest rate that equates the two present worths. 10.02-PR027 A rental car agency rents cars for an average price of $45/day. The number of cars owned in 2020 totals 150. The number of days rented per car in 2020 equals 200. Hence, the rental revenue in 2020 totals $45(150) (200), or $1,350,000. If the agency increases the number of cars at an annual rate of 10%, the number of days rented per car each year remains constant, and the average rental rate increases 5%/year due to inflation, what is the present worth of rental income in 2025 if the agency’s real MARR is 10%? (Allow for a fractional valued number of cars.) 10.02-PR028 In 2020, Samsung introduced a new smartphone on the market at a retail price of $250; the cost of producing the phone was $50/unit. During 2020, 5 million smartphones were sold. The number of smartphones sold increased at an annual rate of 10% through 2024; in 2025, Apple introduced a new product that caused the demand for the Samsung smartphone to equal one-half the number sold in 2024. Over the 5-year period (2021, 2022, 2023,

2024, and 2025), the retail price for the Samsung smartphone (measured in then-current dollars) decreased at an annual rate of 20% and the cost of producing the smartphone (measured in constant or 2020 dollars) decreased at an annual rate of 25%. Over the 5-year period, annual inflation was 2%. Samsung has a real required return on investment of 10%. What was the cumulative present worth of the profits for the Samsung smartphone for the years 2020 through 2025? 10.02-PR029 A ham radio operator wishes to borrow $160,000 to construct a world-class antenna system, transceiver, and amplifier at an electrically quiet location that can be accessed remotely and controlled via the Internet. Microphone, Morse code, radio teletype, slow-scan TV, and a host of other modes may be used for contesting, amassing DX awards, and chatting from anywhere in the world. She borrows the money at 8.5%. Annual inflation is 3.8%. Her combined MARR is 9%. The loan is to be paid back over 5 years. What is the amount to be paid at each year-end and the PW (using both thencurrent and constant-dollar approaches) if repayment follows: a. Plan 1 (pay accumulated interest each year and principal at the end of the last year)? b. Plan 2 (make equal annual principal payments each year, plus interest on the unpaid balance)? c. Plan 3 (make equal annual payments)? d. Plan 4 (make a single payment of principal and interest at the end of the last year)? 10.02-PR030 For the situation stated in Problem 10.02-PR029, let the interest on borrowed money go from 5% to 15% in 1% increments. For each borrowing rate, which payment plan is preferred? Do your answers match what is predicted in the text? 10.02-PR031 Steinway R&D is pursuing the development of an attachment that can easily clean the inside of grand pianos. This innovation will require a loan of $500,000 for fabrication and testing of several units. Inflation is 3.9%, and the loan is available at a rate of 10%. The combined MARR is 17%. The loan is to be paid back over 4 years. What is the amount to be paid at the end of each year and the PW (using both then-current and constant-dollar approaches) if payment follows:

a. Plan 1 (pay accumulated interest each year and principal at the end of the last year)? b. Plan 2 (make equal annual principal payments each year, plus interest on the unpaid balance)? c. Plan 3 (make equal annual payments)? d. Plan 4 (make a single payment of principal and interest at the end of the last year)? 10.02-PR032 For the situation stated in Problem 10.02-PR031, let the interest on borrowed money go from 7% to 20% in 1% increments. For each borrowing rate, which payment plan is preferred? Do your answers match what is predicted in the text? 10.02-PR033 True or False: When the real return on investment is 6% and the combined minimum attractive rate of return is 10%, then inflation must be less than 4%. 10.02-PR034 True or False: If the combined minimum attractive rate of return is 12% and inflation is 3%, then an investment that yields a real external rate of return of 9% is not justified economically. 10.02-PR035 True or False: If the real MARR is 10% and inflation is 4%, then an investment that yields a combined external rate of return of 13.5% is not justified economically.

Section 10.3 After-Tax Analysis LEARNING OBJECTIVE 10.3 Conduct an after-tax analysis with inflation. 10.03-PR001 An economic analysis is being performed in real (not actual) dollars. The company’s combined MARR is 10%, and the inflation rate is 4%. The asset has a first cost of $10,000. It will be depreciated as MACRS 3-year property using rates of 33.33%, 44.45%, 14.81%, and 7.41%. What depreciation amount will be shown in year 3 of the analysis? 10.03-PR002 Electronic Games is moving very quickly to introduce a new interrelated set of video games. The initial investment for equipment to produce the necessary electronic components is $9 million. The salvage value after 6 years is $700,000. Anticipated net contribution to income is $6 million the first year, decreasing by $1 million each year for 6 years, with all dollar amounts expressed in real dollars. Depreciation follows MACRS 5-year property, taxes are 25%, the real MARR is 18%, and inflation is 4%. a. Determine the actual after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows. c. Determine the AW of the after-tax cash flows. d. Determine the FW of the after-tax cash flows. e. Determine the combined IRR of the after-tax cash flows. f. Determine the combined ERR of the after-tax cash flows. g. Determine the real IRR of the after-tax cash flows. h. Determine the real ERR of the after-tax cash flows. 10.03-PR003 Reconsider Problem 10.03-PR002 exactly as written. a. Determine the real after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows. c. Determine the AW of the after-tax cash flows.

d. Determine the FW of the after-tax cash flows. e. Determine the real IRR of the after-tax cash flows. f. Determine the real ERR of the after-tax cash flows. g. Determine the combined IRR of the after-tax cash flows. h. Determine the combined ERR of the after-tax cash flows. 10.03-PR004 Video Solution Henredon purchases a highprecision programmable router for shaping furniture components for $190,000. It is expected to last 12 years and have a salvage value of $5,000. It will produce $45,000 in net revenue each year during its life. All dollar amounts are expressed in real dollars. Depreciation follows MACRS 7-year property, taxes are 25%, the real after-tax MARR is 10%, and inflation is 3.9%. a. Determine the actual after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows. c. Determine the AW of the after-tax cash flows. d. Determine the FW of the after-tax cash flows. e. Determine the combined IRR of the after-tax cash flows. f. Determine the combined ERR of the after-tax cash flows. g. Determine the real IRR of the after-tax cash flows. h. Determine the real ERR of the after-tax cash flows.

10.03-PR005 Reconsider Problem 10.03-PR004 exactly as written. a. Determine the real after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows. c. Determine the AW of the after-tax cash flows. d. Determine the FW of the after-tax cash flows. e. Determine the real IRR of the after-tax cash flows. f. Determine the real ERR of the after-tax cash flows. g. Determine the combined IRR of the after-tax cash flows. h. Determine the combined ERR of the after-tax cash flows. 10.03-PR006 Video Solution Raytheon wishes to use an automated environmental chamber in the manufacture of electronic components. The chamber is to be used for rigorous reliability testing and burn-in. It is installed for $1.4 million and will have a salvage value of $200,000 after 8 years. Its use will create an opportunity to increase sales by $650,000 per year and will have operating expenses of $250,000 per year. All dollar amounts are expressed in real dollars. Depreciation follows MACRS 5year property, taxes are 25%, the real after-tax MARR is 10%, and inflation is 4.2%. a. Determine the actual after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows. c. Determine the AW of the after-tax cash flows. d. Determine the FW of the after-tax cash flows.

e. Determine the combined IRR of the after-tax cash flows. f. Determine the combined ERR of the after-tax cash flows. g. Determine the real IRR of the after-tax cash flows. h. Determine the real ERR of the after-tax cash flows.

10.03-PR007 Reconsider Problem 10.03-PR006 exactly as written. a. Determine the real after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows. c. Determine the AW of the after-tax cash flows. d. Determine the FW of the after-tax cash flows. e. Determine the real IRR of the after-tax cash flows. f. Determine the real ERR of the after-tax cash flows. g. Determine the combined IRR of the after-tax cash flows. h. Determine the combined ERR of the after-tax cash flows. 10.03-PR008 Henredon purchases a high-precision programmable router for shaping furniture components for $190,000. It is expected to last 12 years and have a salvage value of $5,000. It will produce $45,000 in net revenue each year during its life. All dollar amounts are expressed in actual dollars. Depreciation follows MACRS 7-year property, taxes are 25%, the actual aftertax MARR is 14.62%, and inflation is 4.2%. a. Determine the real after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows. c. Determine the AW of the after-tax cash flows.

d. Determine the FW of the after-tax cash flows. e. Determine the real IRR of the after-tax cash flows. f. Determine the real ERR of the after-tax cash flows. g. Determine the combined IRR of the after-tax cash flows. h. Determine the combined ERR of the after-tax cash flows. 10.03-PR009 Reconsider Problem 10.03-PR008 exactly as written. a. Determine the actual after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows. c. Determine the AW of the after-tax cash flows. d. Determine the FW of the after-tax cash flows. e. Determine the combined IRR of the after-tax cash flows. f. Determine the combined ERR of the after-tax cash flows. g. Determine the real IRR of the after-tax cash flows. h. Determine the real ERR of the after-tax cash flows. 10.03-PR010 A surface-mount placement machine is purchased for $500,000. The SMP machine qualifies as 5-year equipment for MACRS-GDS depreciation. The before-tax cash flows, in constant dollars, include an annual uniform series of $120,000 plus a $200,000 salvage value at the end of the 4year planning horizon. A 25% tax rate applies. Inflation is 3%/yr. The real ATMARR is 8%. a. Determine the after-tax cash flows, in constant dollars, for each year. b. Determine the present worth for the investment. c. Determine the real internal rate of return for the investment. 10.03-PR011 An investment of $600,000 is made in equipment that qualifies as 3-year equipment for MACRS-GDS depreciation. The before-tax cash flows, measured in constant dollars, for the investment consist of a uniform annual series of $200,000 plus a $200,000 salvage value at the end of the 5-

year planning horizon. A 25% tax rate and 3% inflation rate apply. The real ATMARR is 10%. a. Determine the after-tax cash flows, in constant dollars, for each year. b. Determine the present worth for the investment. c. Determine the real internal rate of return for the investment. 10.03-PR012 Specialized production equipment is purchased for $125,000. The equipment qualifies as 5-year equipment for MACRS-GDS depreciation. The BTCF profile for the acquisition, in then-current dollars, is an increasing $10,000 gradient series, beginning with $50,000 the first year. In addition, a $30,000 salvage value occurs at the end of the 5-year planning horizon. A 25% tax rate applies. Inflation is 3%/yr. The real ATMARR is 9%. a. Determine the after-tax cash flows, in constant dollars, for each year. b. Determine the present worth for the investment. c. Determine the real internal rate of return for the investment. 10.03-PR013 A manufacturing company decides to purchase a computer for $800,000. The equipment qualifies as 5-year equipment for MACRS-GDS depreciation. The constant-dollar before-tax cash flows can be represented by a $25,000 increasing gradient series; the BTCF the first year is $125,000; and a $100,000 salvage value occurs at the end of the 7-year planning horizon. A 25% tax rate applies. Inflation is 5%/yr. The real ATMARR is 10%. a. Determine the after-tax cash flows, in constant dollars, for each year. b. Determine the present worth for the investment. c. Determine the real internal rate of return for the investment. 10.03-PR014 A construction company purchases a crane for $250,000. The crane qualifies as 5-year equipment for MACRS-GDS depreciation. The BTCF profile for the acquisition, expressed in constant dollars, consists of an annual uniform series of $50,000, plus a $50,000 salvage value at the end of the 7-year planning horizon. A 25% tax rate applies. Inflation is 4%/yr. The real ATMARR is 8%. a. Determine the after-tax cash flows, in constant dollars, for each year.

b. Determine the present worth for the investment. c. Determine the real internal rate of return for the investment. 10.03-PR015 An investment of $450,000 is made in equipment that qualifies as 5-year equipment for MACRS-GDS depreciation. The then-current dollar before tax cash flows are given by a $50,000 increasing gradient series, with the cash flow the first year equaling $100,000. In addition, a $50,000 thencurrent salvage value occurs at the end of the 5-year planning horizon. A 25% tax rate and 4% inflation rate apply. The real ATMARR is 8%. a. Determine the after-tax cash flows, in constant dollars, for each year. b. Determine the present worth for the investment. c. Determine the real internal rate of return for the investment. 10.03-PR016 Solve Problem 10.03-PR002 with (a) 50% bonus depreciation and (b) 100% bonus depreciation. 10.03-PR017 Solve Problem 10.03-PR004 with (a) 50% bonus depreciation and (b) 100% bonus depreciation. 10.03-PR018 Assets used for the manufacture of medical supplies are purchased by an Abbott supplier for $1,500,000. They qualify as 5-year equipment for MACRS-GDS depreciation. The before-tax cash flows, in constant dollars, include an annual uniform series of $480,000 plus a $400,000 salvage value at the end of the 4-year planning horizon. A 25% tax rate and a 3% inflation rate apply. The real ATMARR is 8%. Determine the after-tax cash flow in constant dollars for each year, the real present worth for the investment, and the real internal rate of return. Use the Section 179 expense deduction with (a) 50% bonus depreciation and (b) 100% bonus depreciation. 10.03-PR019 Assets used for special handling in the manufacture of food and beverages are purchased for $2,800,000. These qualify as 3-year equipment for MACRS-GDS depreciation. The before-tax cash flows, measured in constant dollars, consist of a uniform annual series of $925,000 plus an $800,000 salvage value at the end of the 5-year planning horizon. A 25% tax rate and a 3%/yr inflation rate apply. The real ATMARR is 10%. Determine the after-tax cash flow in constant dollars for each year, the real present worth for the investment, and the real internal rate of return. Use the Section 179

expense deduction with (a) 50% bonus depreciation and (b) a 100% bonus depreciation. 10.03-PR020 WI, Inc., is replacing a large portion of its computer and data handling equipment for $2,000,000. This investment qualifies as 5-year property for MACRS-GDS depreciation. The BTCF profile of savings due to the acquisition, in then-current dollars, is an increasing gradient of $100,000 beginning with $500,000 the first year. In addition, a $200,000 salvage value occurs at the end of the 5-year planning horizon. A 25% tax rate and a 3%/yr inflation rate apply. The real ATMARR is 9%.Determine the after-tax cash flow in constant dollars for each year, the real present worth for the investment, and the real internal rate of return. Use the Section 179 expense deduction with (a) 50% bonus depreciation and (b) 100% bonus depreciation.

Section 10.4 After-Tax Analysis with Borrowed Capital LEARNING OBJECTIVE 10.4 Conduct an after-tax analysis with inflation using borrowed capital. 10.04-PR001 Electronic Games is moving very quickly to introduce a new interrelated set of video games. The initial investment for equipment to produce the necessary electronic components is $9 million, with $4 million borrowed at 12% over 6 years and paying only the interest each year and the entire principal in the last year. The salvage value after 6 years is $700,000. Anticipated net contribution to income is $6 million the first year, decreasing by $1 million each year for 6 years, with all dollar amounts expressed in real or constant dollars. Depreciation follows MACRS 5-year property, taxes are 25%, the real MARR is 18%, and inflation is 4%. a. Determine the actual or then-current after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows. c. Determine the AW of the after-tax cash flows. d. Determine the FW of the after-tax cash flows. e. Determine the combined IRR of the after-tax cash flows. f. Determine the combined ERR of the after-tax cash flows. g. Determine the real IRR of the after-tax cash flows. h. Determine the real ERR of the after-tax cash flows. 10.04-PR002 Reconsider Problem 10.04-PR001 exactly as written. a. Determine the real after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows. c. Determine the AW of the after-tax cash flows. d. Determine the FW of the after-tax cash flows. e. Determine the real IRR of the after-tax cash flows.

f. Determine the real ERR of the after-tax cash flows. g. Determine the combined IRR of the after-tax cash flows. h. Determine the combined ERR of the after-tax cash flows. 10.04-PR003 Video Solution Henredon purchases a highprecision programmable router for shaping furniture components for $190,000. It is expected to last 12 years and have a salvage value of $5,000. Henredon will borrow $100,000 at 13% over 6 years, paying only interest each year and paying all the principal in the 6th year. It will produce $45,000 in net revenue each year during its life. All dollar amounts are expressed in real or constant dollars. Depreciation follows MACRS 7-year property, taxes are 25%, the real after-tax MARR is 10%, and inflation is 3.9%. a. Determine the then-current or actual after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows. c. Determine the AW of the after-tax cash flows. d. Determine the FW of the after-tax cash flows. e. Determine the combined IRR of the after-tax cash flows. f. Determine the combined ERR of the after-tax cash flows. g. Determine the real IRR of the after-tax cash flows. h. Determine the real ERR of the after-tax cash flows.

10.04-PR004 Reconsider Problem 10.04-PR003 exactly as written. a. Determine the real or constant dollar after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows. c. Determine the AW of the after-tax cash flows. d. Determine the FW of the after-tax cash flows. e. Determine the real IRR of the after-tax cash flows. f. Determine the real ERR of the after-tax cash flows. g. Determine the combined IRR of the after-tax cash flows. h. Determine the combined ERR of the after-tax cash flows. 10.04-PR005 Video Solution Raytheon wishes to use an automated environmental chamber in the manufacture of electronic components. The chamber is to be used for rigorous reliability testing and burn-in. It is installed for $1.4 million and will have a salvage value of $200,000 after 8 years. Raytheon will borrow $800,000 at 12% to be paid back over 8 years. The environmental chamber will create an opportunity to increase sales by $650,000 per year and will have operating expenses of $250,000 per year. All dollar amounts are expressed in real dollars. Depreciation follows MACRS 5-year property, taxes are 25%, the real aftertax MARR is 10%, and inflation is 4.2%. Determine the actual after-tax cash flows for each year and the PW, FW, AW, IRRc, ERRc, IRRr, and ERRr for each of the four loan payment plans considered in the chapter: a. Plan 1. b. Plan 2. c. Plan 3. d. Plan 4.

e. Which is the preferred plan for Raytheon?

10.04-PR006 Electronic Games is moving very quickly to introduce a new interrelated set of video games. The initial investment for equipment to produce the necessary electronic components is $9 million, with $4 million borrowed at 12% over 6 years. The salvage value after 6 years is $700,000. Anticipated net contribution to income is $6 million the first year, decreasing by $1 million each year for 6 years, with all dollar amounts expressed in real dollars. Depreciation follows MACRS 5-year property, taxes are 25%, the real MARR is 18%, and inflation is 4%. Determine the actual after-tax cash flows for each year and the PW, FW, AW, IRRc, ERRc, IRRr, and ERRr for each of the four loan payment plans considered in the chapter: a. Plan 1. b. Plan 2. c. Plan 3. d. Plan 4. e. Which is the preferred plan for Electronic Games? 10.04-PR007 A construction company decides to purchase a crane for $250,000. The crane qualifies as 5-year equipment for MACRS-GDS depreciation. The BTCF profile for the acquisition, expressed in constant dollars, consists of an annual uniform series of $50,000, plus a $50,000 salvage value at the end of the 7-year planning horizon. The crane is paid for by borrowing $100,000. The loan is to be repaid over a 5-year period at an annual compound interest rate of 12%. A 25% tax rate applies. Inflation is 4%/yr. The real ATMARR is 8%. Among the four payment plans considered in the chapter, use the one that maximizes ATPW for the borrower. a. Determine the present worth for the investment. b. Determine the real internal rate of return for the investment.

10.04-PR008 An investment of $250,000 is made in equipment that qualifies as 7-year equipment for MACRS-GDS depreciation. Measured in constant dollars, the investment yields annual returns of $40,000, plus a salvage value of $100,000 at the end of the 5-year planning horizon. The investment capital of $200,000 is obtained by borrowing money at an annual compound interest rate of 18% and the loan is repaid over a 5-year period. A 25% tax rate and 3% inflation rate apply. The ATMARR, is 7%. Among the four payment plans considered in the chapter, use the one that maximizes the after-tax present worth and determine the after-tax present worth for the investment. 10.04-PR009 An investment of $800,000 is made in equipment that qualifies as 3-year equipment for MACRS-GDS depreciation. The before-tax cash flows, measured in constant dollars, for the investment consist of a uniform annual series of $200,000 plus a $200,000 salvage value at the end of the 5year planning horizon. The investment capital of $400,000 is obtained by borrowing at an annual compound interest rate of 10% and repaying the loan over a 5-year period. A 25% tax rate and 3% inflation rate apply. The real ATMARR is 10%. Among the four payment plans considered in the chapter, use the one that maximizes ATPW. a. Determine the present worth for the investment. b. Determine the real internal rate of return for the investment. 10.04-PR010 In Problem 10.04-PR001, change a., b., c., …, h. to be 1., 2., 3., … 8. a. Use 50 % bonus depreciation. b. Use 100% bonus depreciation. 10.04-PR011 A construction company decides to purchase two related items of heavy equipment for $1,500,000. The purchases qualify as 5-year equipment for MACRS-GDS depreciation. The BTCF profile for the acquisition, expressed in constant dollars, consists of an annual uniform series of $250,000, plus a $250,000 salvage value at the end of a 7-year planning horizon. The equipment is paid for by borrowing $500,000. The-loan is to be repaid over a 5-year period at an annual compound interest rate of 12%. A 25% tax rate and a 4% inflation rate apply. The real ATMARR is 8%. Among the four loan payment plans provided in the chapter, use the one that

maximizes the ATPW for the borrower. Determine (1) the real present worth, (2) the real internal rate of return, (3) the actual present worth, and (4) the actual internal rate of return for the investment. a. Use the Section 179 expense deduction and regular MACRS-GDS depreciation. b. Use 50% bonus depreciation. c. Use both the Section 179 expense deduction and 50% bonus depreciation. 10.04-PR012 An investment of $1,500,000 is made in assets for manufacturing precision glass products. The equipment qualifies as 7-year property for MACRS-GDS depreciation. Measured in constant dollars, the investment yields annual returns of $350,000 plus a salvage value of $600,000 at the end of a 5-year planning horizon. Investment capital is obtained by borrowing $1,200,000 at an annual compound rate of 18% and the loan is repaid over a 5-year period. A 25% tax rate and 3% inflation rate apply. The real ATMARR is 7%. Among the four loan payment plans considered in the chapter, use the one that maximizes the ATPW for the borrower. Determine (1) the after-tax real present worth, (2) the after-tax actual present worth, (3) the after-tax real internal rate of return, and (4) the after-tax actual rate of return for the investment. a. Use the Section 179 expense deduction and regular MACRS-GDS depreciation. b. Use 50% bonus depreciation. c. Use both the Section 179 expense deduction and 50% bonus depreciation. 10.04-PR013 Special tools for manufacturing fabricated metal products are purchased for $1,200,000. They qualify for 3-year MACRS-GDS depreciation. The before-tax cash flows, measured in constant dollars, consist of a uniform annual series of $250,000 plus a $300,000 salvage value at the end of the 5-year planning horizon. Investment capital of $800,000 is obtained by borrowing at an annual compound rate of 12% and repaying the loan over a 5-year period. A 25% tax rate and 3% inflation rate apply. The real ATMARR is 8%. Among the four loan repayment plans considered in the chapter, use the one that maximizes the ATPW for the borrower. Determine (1) the after-tax real present worth, (2) the after-tax actual present worth, (3) the after-tax real

internal rate of return, and (4) the after-tax actual rate of return for the investment. a. Use the Section 179 expense deduction and regular MACRS-GDS depreciation. b. Use 50% bonus depreciation when it applies. c. Use both the Section 179 expense deduction and 50% bonus depreciation.

Chapter 10 Summary and Study Guide Summary 10.1: The Meaning and Measure of Inflation

Learning Objective 10.1: Quantify the effects of inflation on purchasing power with the inflation rate. (Section 10.1) Inflation reduces the purchasing power of money and is an important factor to consider in any type of engineering economic decision, especially one where the decision spans multiple years. Inflation rates can be estimated in many ways, including: CPI: Within the United States, the consumer price index (CPI) is often used to estimate the inflation rate. The CPI measures price changes that occur from one month to the next for a specified bundle of products. Even if one purchases products outside this selected set, they will tend to see similar impacts of inflation, as we live in a global society where the economy is so interconnected. For example, increases in the price of oil in the Middle East may increase the price of consumer products in the United States, which uses the Middle East oil to make gasoline to fuel the trucks that transport the products to the stores where the consumers shop. PPI: Because not all businesses experience inflation as reflected by the market basket used to measure CPI, the producer price index (PPI) is an alternative method to be used that is comprised of a family of over 10,000 indexes. 10.2: Before-Tax Analysis

Learning Objective 10.2: Conduct a before-tax analysis with inflation. (Section 10.2) If inflation is to be considered in a before-tax analysis, then both the cash flows and the discount rate must be adjusted to include inflation or exclude inflation; otherwise, the analysis will be flawed. One must carefully consider whether constant dollars or then-current dollars should be used for the analysis. When performed correctly, the constant-dollars approach and the then-current dollars approach will yield identical present worths. The relationship between the real interest rate (ir), inflation rate (f), and combined rate (ic) can be expressed as: ic = ir + f + ir (f )

(10.2)

The relationships between the real minimum attractive rate of return (MARRr), the inflation rate, and the combined minimum attractive rate of return (MARRc) can be expressed as: MARRc = MARRr + f + MARRr (f )

(10.3)

Two key equations relating constant-dollars ($Ck) and then-current dollars ($Tk) at the end of year k are: k

(10.6)

−k

(10.8)

$T k = $Ck (1 + f )

and $C k = $T k (1 + f )

10.3: After-Tax Analysis

Learning Objective 10.3: Conduct an after-tax analysis with inflation. (Section 10.3) After-tax analyses can be a bit trickier than before-tax analyses when inflation is considered because in the United States, depreciation allowances are not permitted to increase with inflation. When performing after-tax analyses, there are two options: Remove the effects of inflation from the depreciation allowances in order to express them in constant dollars; Or, express other cash flows in then-current dollars. This is often the simplest option. Inflation reduces the after-tax present worth of an investment in capital equipment unless borrowed capital is used. 10.4: After-Tax Analysis with Borrowed Capital

Learning Objective 10.4: Learning Objective: Conduct an after-tax analysis with inflation using borrowed capital. (Section 10.4) When money is borrowed at fixed rates, the loan payments and depreciation allowances are then-current cash flows. It is easiest to perform after-tax analyses using a then-current approach when both inflation and borrowed funds are present because the taxable income will be reduced by interest payments and depreciation allowances. Depending on the fraction of investment capital that is borrowed, inflation can increase the after-tax present worth of an investment in capital equipment when using borrowed capital.

Important Terms and Concepts Inflation

A decrease in the purchasing power of money that is caused by an increase in general price levels of goods and services without an accompanying increase in the value of the goods and services. Consumer Price Index (CPI) A measure by the U.S. Department of Labor’s Bureau of Labor Statistics used to determine inflation rates, measuring the price changes that occur from one month to the next for a specified set of products. Also known as the retail price index. Producer Price Index (PPI) A family of over 10,000 indexes published monthly by the Bureau of Labor Statistics. PPI measures price changes from the seller’s or producer’s perspective. Higher Education Price Index (HEPI) A value used to determine inflation rates specifically within the higher education sector. Constant Dollars Dollars considered as being free from inflation. Also known as real dollars, today’s dollars, inflationfree dollars, constant purchasing power dollars, constant-value dollars, and constant-worth dollars. Then-Current Dollars Dollars that include inflation. Also known as future dollars, nominal dollars, actual dollars, and inflated dollars. Real Interest Rate An interest rate that does not include inflation. It is a pure discount rate, one that expresses the real desired return on an investment. Inflation Rate A rate representing the loss of a dollar’s purchasing power in 1 year. Combined Rate An inflation rate that includes both the real required return on investment and the inflation rate. It is also known as the market rate and the inflation-

adjusted interest rate.

Chapter 10 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 10.1: The Meaning and Measure of Inflation LO 10.1: Quantify the effects of inflation on purchasing power with the inflation rate. Video Lesson: Inflation Video Lesson Notes: Inflation Video Solution: 10.01-PR009 10.2: Before-Tax Analysis LO 10.2: Conduct a before-tax analysis with inflation. Excel Video Lesson: NPV Financial Function Excel Video Lesson Spreadsheet: NPV Financial Function Video Example 10.2: Before-Tax Impact of Inflation on the Present Worth of Raw Material Cost Video Solution: 10.02-PR001 Video Solution: 10.02-PR007 Video Solution: 10.02-PR011 10.3: After-Tax Analysis LO 10.3: Conduct an after-tax analysis with inflation. Excel Video Lesson: FV Financial Function Excel Video Lesson Spreadsheet: FV Financial Function

Excel Video Lesson: PMT Financial Function Excel Video Lesson Spreadsheet: PMT Financial Function Excel Video Lesson: IRR Financial Function Excel Video Lesson Spreadsheet: IRR Financial Function Excel Video Lesson: MIRR Financial Function Excel Video Lesson Spreadsheet: MIRR Financial Function Video Example 10.3: Using Excel® to Analyze the After-Tax Impact of Inflation on the SMP Investment Video Solution: 10.03-PR004 Video Solution: 10.03-PR006 10.4: After-Tax Analysis with Borrowed Capital LO 10.4: Conduct an after-tax analysis with inflation using borrowed capital. Video Solution: 10.04-PR003 Video Solution: 10.04-PR005 These chapter-level resources will help you with your overall understanding of the content in this chapter. Appendix A: Time Value of Money Factors Appendix B: Engineering Economic Equations Flashcards: Chapter 10 Excel Utility: TVM Factors: Table Calculator Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values

Excel Utility: Monthly Payment Sensitivity Excel Utility: TVM Factors: Discrete Compounding Excel Utility: TVM Factors: Geometric Series Future Worth Excel Utility: TVM Factors: Geometric Series Present Worth Excel Data Files: Chapter 10

CHAPTER 10 Inflation LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 10.1 Quantify the effects of inflation on purchasing power with the inflation rate. (Section 10.1) 10.2 Conduct a before-tax analysis with inflation. (Section 10.2) 10.3 Conduct an after-tax analysis with inflation. (Section 10.3) 10.4 Conduct an after-tax analysis with inflation using borrowed capital. (Section 10.4)

Engineering Economics in Practice Caterpillar In 1890, Benjamin Hold and Daniel Best competed to see who could produce the best steam-powered tractor for use in farming. By 1925, they had merged their companies, forming Caterpillar Tractor Company. Headquartered in Peoria, Illinois, Caterpillar today is the world’s leading manufacturer of construction and mining equipment, diesel and natural gas engines, industrial gas turbines, and diesel-electric locomotives. Caterpillar’s Jim Umpleby stated in the 2017 Annual Report, “One of my first initiatives after becoming CEO in January 2017 was to appoint a diverse, cross-functional team of Caterpillar leaders to develop a strategy for long-term profitable growth. We never lost sight of the fact that our customers have been using our equipment to help improve living standards around the world for more than 90 years.” In 2017, the company had sales and revenues totaling $45.5 billion, an increase of 18% from 2016. Caterpillar is a global company with 96,000 employees worldwide with 41% of sales and revenues in 2017 from inside the United States and 59% from outside the United States. Caterpillar’s profits were $754 million, compared with a loss of $67 million in 2016, and capital expenditures totaled $2.3 billion, down from $2.9 billion in 2016. R&D expenditures in 2017 were $1.9 billion, representing 4.2% of sales and revenues. The company acknowledges that its business is highly sensitive to global economic conditions in the industries and markets that it serves. Inflation is an important economic factor, especially when one considers that Caterpillar is making significant investments to increase their production of construction equipment not only within the United States but also in Brazil and Asia/Pacific, regions that have experienced high inflation over the past decade and weak economies. Global companies such as Caterpillar must pay close attention to inflation rates in the countries in which they operate and serve. Inflation rates can differ markedly from one economy to another, as do income-tax laws. Discussion Questions 1. In 2017, Caterpillar generated 59% of its sales and revenues from outside the United States. Why is this relevant to our study of inflation? 2. Given high inflation rates in regions in which Caterpillar has production facilities, what decisions might management consider making? 3. Inflation rates in the United States have been fairly steady in recent years. Does Caterpillar need to develop different business strategies for its ventures within and outside the United States? 4. The inflation rate in different regions of the world is outside Caterpillar’s control. What can the company do to protect itself from the adverse effects of inflation, such as rising commodity or component prices and rising labor rates?

Introduction Previously, we have said little about inflation and its effect on engineering economic justification. In Chapter 1, we stated, “Money has time value in the absence of inflation.” In Chapter 2, we indicated that annual increases in expenditures represented by a geometric series of cash flows could be due to inflation. In this chapter, we examine the before-tax and after-tax effects of inflation on the economic worth of an investment when capital is borrowed and when it comes from retained earnings. We show that inflation can significantly impact the economic viability of a capital investment. For that reason, when comparing investment alternatives in an inflationary economy, it is important to give explicit consideration to inflation.

Systematic Economic Analysis Technique 1. Identify the investment alternatives 2. Define the planning horizon 3. Specify the discount rate 4. Estimate the cash flows 5. Compare the alternatives 6. Perform supplementary analyses 7. Select the preferred investment

10.1 The Meaning and Measure of Inflation LEARNING OBJECTIVE Quantify the effects of inflation on purchasing power with the inflation rate. Video Lesson: Inflation Consumers understand the impact of inflation on their ability to purchase goods and services. As the value of a dollar diminishes over time, the effects of inflation are manifested. More precisely, we can say that inflation is a decrease in the purchasing power of money that is caused by an increase in general price levels of goods and services without an accompanying increase in the value of the goods and services. Inflationary pressure is created when more dollars are put into an economy without an accompanying increase in goods and services. In other words, printing more money without an increase in economic output generates inflation. Inflation A decrease in the purchasing power of money that is caused by an increase in general price levels of goods and services without an accompanying increase in the value of the goods and services. Inflation can have such a significant impact on an investment’s economic worth that it should be considered when comparing investment alternatives. Although it is important to consider the impact of inflation on investments made within one country, it is especially important to do so in multinational investment situations. The inflation rates in different countries can be dramatically different. Hence, a firm that is faced with making decisions concerning investments of capital in various nations must give strong consideration to the inflationary conditions in the countries in question. As noted, inflation adversely affects the purchasing power of money. To illustrate what we mean by the purchasing power of money, suppose a firm purchases 1 million pounds of a particular material each year, and the price of the material increases by 3% per year. The quantity of material the firm can purchase with a fixed amount of money— that is, the purchasing power of the firm’s money—decreases over time. The only way the firm can afford to continue purchasing the material is to decrease its usage rate or to increase its source of funds. In the latter case, the firm might increase the price of the products it sells; if so, then the purchasing power of its customers’ money will be decreased. In these situations, the continuing spiral of price increases does not contribute real increases to the firm’s profits; instead, it results in an inflated representation of the firm’s profits. The overall process of price increases without accompanying increases in the quality or value of the goods or services is referred to as inflation.

10.1.1 Measuring Inflation: The Consumer Price Index (CPI) How is the inflation rate determined? The approach varies, depending on what rate is desired. For example, the most commonly used way of estimating the inflation rate for consumers in the United States is the consumer price index, developed by the U.S. Department of Labor’s Bureau of Labor Statistics. Published monthly, the consumer

price index (CPI) measures the price changes that occur from one month to the next for a specified set of products. Also called the retail price index, it is the index most often referred to by the U.S. media when referring to inflation rates. Consumer Price Index (CPI) A meausre by the U.S. Department of Labor’s Bureau of Labor Statistics used to determine inflation rates, measuring the price changes that occur from one month to the next for a specified set of products. Also known as the retail price index. The CPI is a market basket rate, in that it is based on 80,000 prices that are recorded in 87 urban areas. The socalled market basket is purchased each month and reflects what consumers buy for their day-to-day living. The market basket includes more than 200 categories of expenditures, arranged in eight major groups: food and beverages; housing; apparel; transportation; medical care; recreation; education and communication; and other goods and services. The Bureau of Labor Statistics has collected data since 1913. Major changes in definitions and methodology have occurred over the years. The CPI reflects spending patterns, not costs of living, for two population groups in the United States: all urban consumers, and urban wage earners and clerical workers. Table 10.1 shows CPI values from 1913 through 2018. The table and Figure 10.1 also show the year-to-year percentage changes in the CPI. These percentages are the estimates of inflation used by many businesses and financial institutions. As shown, double-digit annual inflation has occurred in the past. From 2013 through 2018, annual inflation rates have been 1.5%, 1.6%, 0.1%, 1.3%, 2.1%, and 2.4% respectively.

TABLE 10.1 CPI Data 1913–2018 Year CPI % Change Year CPI % Change Year CPI % Change 1913  9.9

1948 24.1

 8.1%

1983  99.6

3.2%

1914 10.0

  1.0%

1949 23.8

−1.2%

1984 103.9

4.3%

1915 10.1

  1.0%

1950 24.1

 1.3%

1985 107.6

3.6%

1916 10.9

  7.9%

1951 26.0

 7.9%

1986 109.6

1.9%

1917 12.8

 17.4%

1952 26.5

 1.9%

1987 113.6

3.6%

1918 15.1

 18.0%

1953 26.7

 0.8%

1988 118.3

4.1%

1919 17.3

 14.6%

1954 26.9

 0.7%

1989 124.0

4.8%

1920 20.0 1921 17.9

 15.6% −10.5%

1955 26.8 1956 27.2

−0.4%  1.5%

1990 130.7 1991 136.2

5.4% 4.2%

1922 16.8

 −6.1%

1957 28.1

 3.3%

1992 140.3

3.0%

1923 17.1

  1.8%

1958 28.9

 2.8%

1993 144.5

3.0%

1924 17.1

  0.0%

1959 29.1

 0.7%

1994 148.2

2.6%

1925 17.5

  2.3%

1960 29.6

 1.7%

1995 152.4

2.8%

1926 17.7

  1.1%

1961 29.9

 1.0%

1996 156.9

3.0%

1927 17.4

 −1.7%

1962 30.2

 1.0%

1997 160.5

2.3%

1928 17.1

 −1.7%

1963 30.6

 1.3%

1998 163.0

1.6%

1929 17.1

  0.0%

1964 31.0

 1.3%

1999 166.6

2.2%

1930 16.7

 −2.3%

1965 31.5

 1.6%

2000 172.2

3.4%

1931 15.2

 −9.0%

1966 32.4

 2.9%

2001 177.1

2.8%

1932 13.7

 −9.9%

1967 33.4

 3.1%

2002 179.9

1.6%

1933 13.0 1934 13.4

 −5.1%   3.1%

1968 34.8 1969 36.7

 4.2%  5.5%

2003 184.0 2004 188.9

2.3% 2.7%

1935 13.7

  2.2%

1970 38.8

 5.7%

2005 195.3

3.4%

1936 13.9

  1.5%

1971 40.5

 4.4%

2006 201.6

3.2%

1937 1938 1939 1940 1941 1942 1943 1944 1945 1946 1947

  3.6%  −2.1%  −1.4%   0.7%   5.0%  10.9%   6.1%   1.7%   2.3%   8.3%  14.4%

1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982

 3.2%  6.2%  11.0%  9.1%  5.8%  6.5%  7.6%  11.3%  13.5%  10.3%  6.2%

2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018

2.8% 3.8% −0.4% 1.6% 3.2% 2.1% 1.5% 1.6% 0.1% 1.3% 2.1% 2.4%

14.4 14.1 13.9 14.0 14.7 16.3 17.3 17.6 18.0 19.5 22.3

41.8 44.4 49.3 53.8 56.9 60.6 65.2 72.6 82.4 90.9 96.5

Excel® Data File

207.3 215.3 214.5 218.1 224.9 229.6 233.0 236.7 237.0 240.0 245.1 251.1

FIGURE 10.1 CPI Annual Percent Changes 1913–2018 Excel® Data File Some people contend that the CPI does not measure correctly the inflation they experience, because they do not purchase everything included in the CPI market basket calculation. As an example, if they do not intend to buy or sell a house, they argue that the changes in house prices have no influence on them. Further, if they are vegetarians, they argue that the change of prices for beef, pork, and chicken does not affect them. Similar arguments are made for a large number of the items in the market basket the Bureau of Labor Statistics uses to calculate the CPI. While it is true that few, if any, people purchase all of the items included in the CPI market basket, it is incorrect to conclude that such price increases do not impact everyone. Today’s economy is interconnected. Consider vegetarians. How can their household budgets be impacted by increases in meat prices? If owners of stores where vegetarians shop eat meat and increase the prices in their stores because of price increases for meat, then vegetarians are impacted indirectly by increases in meat prices. In general, if store owners increase prices in reaction to price increases for items in the CPI market basket they purchase, then all of their customers are indirectly affected by the price increases in the CPI market basket. This phenomenon causes all of us to be impacted by price increases in things we do not purchase. The same occurs in business. A soft-drink manufacturer that purchases large quantities of sugar but very little oil is still impacted when oil prices increase. Oil is consumed in manufacturing the plastic bottles and aluminum cans the soft-drink manufacturer uses; also, price increases for oil impact the cost of fuel to heat and air-condition the offices and production areas. Price increases in one segment of the economy are generally felt in all segments of the economy.

10.1.2 Other Inflation Indexes Because most businesses do not experience inflation as reflected by the market basket used to measure the CPI, a different index is used for producers of goods and services. The producer price index (PPI) is a family of over 10,000 indexes published monthly by the Bureau of Labor Statistics. It measures price changes from the seller’s or producer’s perspective. Formerly called the wholesale price index, the PPI generally provides a better measurement of the inflation effects for a particular business or for the purchase of particular equipment than the CPI. Producer Price Index (PPI) A family of over 10,000 indexes published monthly by the Bureau of Labor Statistics. PPI measures price changes from the seller’s or producer’s perspective. Another index, the Higher Education Price Index (HEPI), has been developed to measure the year-to-year changes in higher education costs. Due to rapid increases in expenditures for library holdings, Internet upgrades, wireless connectivity, computing and laboratory equipment, health insurance, utilities, and faculty salaries in some key academic disciplines, in general, HEPI increases faster than CPI.1

Higher Education Price Index (HEPI) A value used to determine inflation rates specifically within the higher education sector.

Concept Check 10.01-CC001 The __________ is commonly used to measure the impact of inflation on individuals and the __________ is commonly used to measure the impact of inflation on businesses. a. Producer price index; wholesale price index b. Producer price index; consumer price index c. Consumer price index; producer price index d. Wholesale price index; market basket rate Correct or Incorrect? Clear

  Check Answer

10.2 Before-Tax Analysis LEARNING OBJECTIVE Conduct a before-tax analysis with inflation. When incorporating inflation effects in economic analyses, it is essential for the discount rate and cash flows to be consistent. Specifically, if inflation is to be incorporated explicitly, then inflationary effects must be incorporated in both the cash flows and in the discount rate. Likewise, if inflation is not to be incorporated in the analysis, then both the cash flows and the discount rate must exclude inflation or be inflation-free. Otherwise, the analysis will be flawed. To facilitate a consideration of inflation in economic analyses, it is useful to define two terms: constant dollars and then-current dollars. Constant dollars are free of inflation; they are also called real dollars, today’s dollars, inflation-free dollars, constant purchasing power dollars, constant-value dollars, and constant-worth dollars. Then-current dollars include inflation; they are also called future dollars, nominal dollars, actual dollars, and inflated dollars. Constant Dollars Dollars considered as being free from inflation. Also known as real dollars, today’s dollars, inflation-free dollars, constant purchasing power dollars, constant-value dollars, and constant-worth dollars. Then-Current Dollars Dollars that include inflation. Also known as future dollars, nominal dollars, actual dollars, and inflated dollars. It is also useful to define three rates: the real interest rate, the inflation rate, and the combined rate. The real interest rate does not include inflation; it is a pure discount rate—one that expresses the real desired return on investment. The inflation rate is expressed as a percent and represents the loss of a dollar’s purchasing power in 1 year. The combined rate, also called the market rate and the inflation-adjusted interest rate, includes both the real required return on investment and the inflation rate.

Real Interest Rate An interest rate that does not include inflation. It is a pure discount rate, one that expresses the real desired return on an investment. Inflation Rate A rate representing the loss of a dollar’s purchasing power in 1 year. Combined Rate An inflation rate that includes both the real required return on investment and the inflation rate. It is also known as the market rate and the inflation-adjusted interest rate. Letting ir denote the real interest rate, f denote the inflation rate, and ic denote the combined rate, the following relationship exists among the three rates: (1 + ic ) = (1 + ir )(1 + f )

(10.1)

ic = ir + f + ir (f )

(10.2)

which reduces to

The same relationships exist among the real minimum attractive rate of return (MARRr), the inflation rate, and the combined minimum attractive rate of return (MARRc): MARRc = MARRr + f + MARRr (f )

(10.3)

When given ic and f, the value of ir can be obtained from Equation 10.2. Specifically, ir = (ic − f )/(1 + f )

(10.4)

MARRr = (MARRc − f )/(1 + f )

(10.5)

Likewise,

Letting $Tk denote the magnitude of a then-current cash flow at the end of year k and $Ck denote the magnitude of a constant-dollar cash flow at the end of year k, the following relationship holds: $Tk = $Ck (1 + f )

k

(10.6)

or $Tk = $Ck (F |P  f %,k)

(10.7)

Likewise, $Ck = $Tk (1 + f )

−k

(10.8)

or $Ck = $Tk (P |F  f %,k)

(10.9)

EXAMPLE 10.1 Calculating the Real Interest Rate If inflation averages 4% per year and your return on an investment, based on then-current dollars, is 10%, what is your real return on investment? Key Data Given f = 4%, ic = 10% Find ir Solution From Equation 10.4, ir

=

(ic − f )/(1 + f ) = (0.10 − 0.04)/(1 + 0.04)

=

0.057692 or 5.7692%

EXAMPLE 10.2 Before-Tax Impact of Inflation on the Present Worth of Raw Material Cost Video Example A small business spent $125,000 for raw material during 2020. It anticipates the cost of raw material will increase, due to inflation, at a rate of 3% per year. It also anticipates that the amount of raw material required will increase at a rate of 8% per year due to increased demand for the firm’s products. If the business has a 10% real required return on its investments, what is the present worth of the cost of raw material over the 5year period (2021–2025)? Key Data Given Cash flows shown in Table 10.2, f = 3%, MARRr = 10% Find PW of raw material cost TABLE 10.2 Cash Flows and Present Worth for Example 10.2 k $Ck (P|F, 10%,k) PWk ($C) $Tk

(P|F, 13.3%,k) PWk ($T)

2021 $135,000.00 2022 $145,800.00

0.90909 0.82645

$122,727.27 $139,050.00 $120,495.87 $154,679.22

0.88261 0.77900

$122,727.27 $120,495.87

2023 $157,464.00 2024 $170,061.12

0.75131 0.68301

$118,305.03 $172,065.16 $116,154.03 $191,405.29

0.68756 0.60685

$118,305.03 $116,154.03

2025 $183,666.01

0.62092 $114.042.14 $212,919.24 PWtotal = $591,724.35

0.53561 $114.042.14 PWtotal = $591,724.35

Excel® Data File Solution Table 10.2 shows the yearly cash flows for this example in constant and then-current dollars. The constantdollar cash flows equal $125,000(1.08)k − 2020 and the then-current cash flows equal $125,000[1.08(1.03)]k − 2020. The present worth of $591,724.35 for the constant-dollar cash flows can be obtained using the Excel® NPV worksheet function with a 10% MARRr. Similarly, the present worth of $591,724.35 for the then-current cash flows is obtained using a 10% + 3% + 10%(3%) or 13.3% MARRc. Alternatively, the present worths can be calculated by hand using calculated or tabulated values, as shown in Table 10.2. Excel® Video Lesson: NPV Function

Concept Check 10.02-CC001 A before-tax economic evaluation with inflation effects is being conducted. When comparing the constantdollar present worth with the then-current present worth, which of the following best represents their relationship? Assume both PWs are correctly calculated with appropriate values of MARR. a. PW in constant dollars < PW in then-current dollars b. PW in constant dollars = PW in then-current dollars c. PW in constant dollars > PW in then-current dollars d. Cannot be determined without knowing the real interest rate, the inflation rate, and the cash flows Correct or Incorrect? Clear

  Check Answer

10.3 After-Tax Analysis LEARNING OBJECTIVE Conduct an after-tax analysis with inflation. In previous chapters, one of two assumptions was made: Our before-tax analysis was based on cash flows and interest rates that were inflation-free or that included inflation effects. When cash flows are expressed in constant dollars and the minimum attractive rate of return is a real required return, nothing special needs to be done to incorporate inflation when performing before-tax analyses. That is not true for after-tax analyses! Why the difference? Because, in the United States, depreciation allowances are not permitted to increase with inflation. Depreciation allowances are expressed in then-current dollars, not constant dollars. Therefore, either the effects of inflation must be “stripped” out of the depreciation allowances in order to express them in constant dollars, or all other cash flows must be expressed in then-current dollars.

EXAMPLE 10.3 Using Excel® to Analyze the After-Tax Impact of Inflation on the SMP Investment Video Example Recall the $500,000 investment in a surface-mount placement machine that produces net annual savings, before taxes, of $108,333.33 for 10 years, plus a $50,000 salvage value at the end of the 10-year period. The estimates were in constant dollars. The SMP machine qualified as 5-year property for MACRS-GDS depreciation. A 25% income-tax rate and a MARRr of 10% were used to perform an after-tax analysis. To illustrate the effect of inflation in after-tax analyses, suppose annual inflation equals 4%. Assuming no bonus depreciation applies and with a combined after-tax minimum attractive rate of return equal to 10% plus 4% plus 10% times 4%, or 14.4%, what are the PW and IRR values of the investment in then-current dollars? Excel® Video Lesson: FV Financial Function Key Data Given Cash flows and depreciation as outlined in Figure 10.2, MARRr = 10%, f = 4%, income tax = 25%. MARRcAT = 14.4% Find PW of the SMP investment with 4% inflation, IRRc, IRRr

FIGURE 10.2 After-Tax, Then-Current Dollar Analysis of the SMP Investment with 4% Inflation Excel® Data File Solution As shown in Figure 10.2, then-current, before-tax cash flows ($TBTCF) are obtained by increasing constantdollar, before-tax cash flows ($CBTCF) at an annual compound rate of 4%; depreciation allowances are in then-current dollars. The resulting then-current, after-tax present worth is $101,119.32, compared to $110,361.50 if inflation is negligible. Likewise, when inflation is negligible, the after-tax internal rate of return is 15.18%. With 4% inflation, the combined after-tax internal rate of return is 19.33%; therefore, the real after-tax internal rate of return is IRRr =

(0.1933 − 0.04)/(1.04) = 0.1474, or 14.74%.

Excel® Video Lesson: PMT Financial Function

Excel® Video Lesson: IRR Financial Function

Excel® Video Lesson: MIRR Financial Function Exploring the Solution An alternative approach is to convert the depreciation allowances to constant dollars using Equation 10.8 or 10.9. As shown in Figure 10.3, the constant-dollar, after-tax present worth is $101,119.32, which is identical to the then-current result. The real after-tax internal rate of return is 14.74%.

FIGURE 10.3 After-Tax, Constant Dollar Analysis of the SMP Investment with 4% Inflation Excel® Data File

Concept Check 10.03-CC001 The cash flows for an economic evaluation are expressed in constant dollars. What must be done to the MACRS depreciation allowances in order to correctly conduct an after-tax present worth analysis? a. Divide each depreciation allowance by (1 + f)k, where f is the inflation rate and k is the year index b. Multiply each depreciation allowance by (1 + f)k, where f is the inflation rate and k is the year index c. Multiply each depreciation allowance by (F|P f%, k), where f is the inflation rate and k is the year index d. Nothing must be done, and the MACRS allowances can be used as is Correct or Incorrect? Clear

  Check Answer

10.4 After-Tax Analysis with Borrowed Capital

LEARNING OBJECTIVE Conduct an after-tax analysis with inflation using borrowed capital. When money is borrowed at fixed rates, the loan payments (principal and interest) are then-current cash flows. Depreciation allowances are also then-current cash flows. Because taxable income is reduced by interest payments and depreciation allowances, it is more convenient to perform after-tax analyses using a then-current approach when both inflation and borrowed funds are present. If one wishes to use a constant-dollar approach, however, then the principal payments, interest payments, and depreciation allowances must be converted to constant-dollar cash flows. Equation 10.8 or 10.9 can be used to perform the conversion.

EXAMPLE 10.4 Using Excel® to Analyze the Impact of Inflation on the SMP Investment When Money Is Borrowed Suppose $300,000 is borrowed at a 12% annual compound interest rate and is repaid over a 10-year period using one of four payment plans. This is the same scenario we considered in Example 9.7, but we now include inflation of 4%/per year in our analysis. To review, the four payment plans are: (a) Plan 1—pay the accumulated interest at the end of each interest period and repay the principal at the end of the loan period; (b) Plan 2—make equal principal payments, plus interest on the unpaid balance at the end of the period; (c) Plan 3—make equal end-of-period payments; and (d) Plan 4—make a single payment of principal and interest at the end of the loan period. Assuming bonus depreciation is not available, which of the payment plans should be used if the MARRr equals 10%? (As before, a 25% income-tax rate is used.) Key Data Given i = 12%, f = 4%, MARRc = 14.4%, tax rate = 25%, $300,000 borrowed with each of the four payment plans Find PWAT for each payment plan Solution Figure 10.4 presents the results for Plan 1. Tables 10.3 through 10.5 provide the after-tax results for Plans 2, 3, and 4, respectively. Plan 1 yields an after-tax present worth of $184,317.66; Plan 2 yields an after-tax present worth of $155,842.70; Plan 3 yields an after-tax present worth of $162,586.49; Plan 4 yields an after-tax present worth of $199,571.84. On this basis, Plan 4 should be used to repay the $300,000 loan over a 10-year period. (Similar conclusions are reached using either FW or AW then-current comparisons.) Notice that for Plans 1 and 4 multiple negative values exist for ATCF, so the Excel® MIRR worksheet function cannot be used to compute the ERR. Instead, as was done in Example 9.7, the ERR is calculated using the Excel® IRR worksheet function with the modified ATCF column.

FIGURE 10.4 After-Tax Analysis of the SMP Investment with $300,000 Borrowed, Repaid with Plan 1, and 4% Inflation Excel® Data File

TABLE 10.3 After-Tax Analysis of the SMP Investment; $300,000 Borrowed and Repaid with Plan 2; 4% Inflation MARRr = 10% interest rate = 12% income-tax rate = 25% EOY $TBTCF $TPPMT $TIPMT $TDWO 0 − − $500,000.00 $300,000.00 1 $112,666.66 $30,000.00 $36,000.00 $100,000.00

inflation rate = 4% $TTI $TTax $TATCF −$200,000.00

2

$117,173.33

$30,000.00 $32,400.00 $160,000.00

3 4 5 6

$121,860.26 $126,734.67 $131,804.06 $137,076.22

$30,000.00 $30,000.00 $30,000.00 $30,000.00

$28,800.00 $25,200.00 $21,600.00 $18,000.00

$96,000.00 $57,600.00 $57,600.00 $28,800.00

− $23,333.34 − $75,226.67 −$2,939.74 $43,934.67 $52,604.06 $90,276.22

7 8 9 10

$142,559.27 $148,261.64 $154,192.11 $234,372.01

$30,000.00 $14,400.00 $30,000.00 $10,800.00 $30,000.00 $7,200.00 $30,000.00 $3,600.00

$0.00 $0.00 $0.00 $0.00

$128,159.27 $137,461.64 $146,992.11 $230,772.01

Excel® Data File

−$5,833.33

$52,500.00

− $18,806.67 −$734.93 $10,983.67 $13,151.02 $22,569.06

$73,580.00

$32,039.82 $34,365.41 $36,748.03 $57,693.00

$66,119.45 $73,096.23 $80,244.08 $143,079.01

PW $T =

$155,842.70

FW $T =

$598,338.17

AW $T =

$30,344.98

IRRc =

30.58%

ERRc =

21.18%

IRRr =

25.55%

ERRr =

16.52%

$63,795.20 $60,551.01 $67,053.05 $66,507.17

TABLE 10.4 After-Tax Analysis of the SMP Investment; $300,000 Borrowed and Repaid with Plan 3; 4% Inflation MARRr = 10% interest rate = 12% income-tax rate = 25% EOY $TBTCF $TPPMT $TIPMT $TDWO 0 − − $500,000.00 $300,000.00 1 $112,666.66 $17,095.25 $36,000.00 $100,000.00

inflation rate = 4% $TTI $TTax $TATCF −$200,000.00 − $23,333.34 − $76,775.24 −$5,790.71 $40,057.02 $48,008.52

−$5,833.33

$65,404.75

− $19,193.81 −$1,447.68 $10,014.25 $12,002.13

$83,271.89

2

$117,173.33

$19,146.68 $33,948.57 $160,000.00

3 4 5

$121,860.26 $126,734.67 $131,804.06

$21,444.28 $31,650.97 $24,017.59 $29,077.65 $26,899.71 $26,195.54

$96,000.00 $57,600.00 $57,600.00

6 7 8 9

$137,076.22 $142,559.27 $148,261.64 $154,192.11

$30,127.67 $33,742.99 $37,792.15 $42,327.21

$22,967.58 $19,352.26 $15,303.10 $10,768.04

$28,800.00 $85,308.64 $0.00 $123,207.01 $0.00 $132,958.54 $0.00 $143,424.07

$21,327.16 $30,801.75 $33,239.64 $35,856.02

$62,653.81 $58,662.27 $61,926.76 $65,240.84

10

$234,372.01

$47,406.47

$5,688.78

$0.00 $228,683.23

$57,170.81 PW $T =

$124,105.95 $162,586.49

FW $T =

$624,230.08

AW $T =

$31,658.10

IRRc =

33.05%

ERRc =

21.41%

IRRr =

27.93%

ERRr =

16.74%

Excel® Data File

$70,212.69 $63,625.17 $66,706.68

TABLE 10.5 After-Tax Analysis of the SMP Investment; $300,000 Borrowed and Repaid with Plan 4; 4% Inflation MARRr = 10% interest rate = 12% income-tax rate = 25% EOY $TBTCF $TPPMT $TIPMT $TDWO 0 − − $500,000.00 $300,000.00 1 $112,666.66 $0.00 $0.00 $100,000.00 2 $117,173.33 $0.00 $0.00 $160,000.00 3 4 5 6 7

$121,860.26 $126,734.67 $131,804.06 $137,076.22 $142,559.27

$0.00 $0.00 $0.00 $0.00 $0.00

$0.00 $0.00 $0.00 $0.00 $0.00

8 $148,261.64 $0.00 $0.00 9 $154,192.11 $0.00 $0.00 10 $234,372.01 $300,000.00 $631,754.46

inflation rate = 4% $TTI $TTax $TATCF $TA − −$200, $200,000.00 $12,666.66 $3,166.67 $109,500.00 − − $127,880.00 $42,826.67 $10,706.67

$96,000.00 $25,860.26 $6,465.07 $57,600.00 $69,134.67 $17,283.67 $57,600.00 $74,204.06 $18,551.02 $28,800.00 $108,276.22 $27,069.06 $0.00 $142,559.27 $35,639.82

$115,395.20 $109,451.01 $113,253.05 $110,007.17 $106,919.45

$0.00 $148,261.64 $37,065.41 $111,196.23 $0.00 $154,192.11 $38,548.03 $115,644.08 $0.00 − − − $1,534, $397,382.46 $99,345.61 $598,036.84 PW $T = $199,571.84 FW $T = $766,230.63 AW $T =

$38,859.72

IRRc =

54.17%

ERRc =

22.60%

IRRr =

48.24%

ERRr =

17.88%

Excel® Data File Exploring the Solution In Chapter 9, PWAT was $128,795.20 for Plan 1; $121,927.80 for Plan 2; $123,485.16 for Plan 3; and $112,022.00 for Plan 4. A comparison of the results indicates PWAT differences varied significantly among the payment plans. With borrowed funds, however, inflation produced a greater PWAT for each plan, as summarized in Table 10.6.

TABLE 10.6 Comparison of PWAT for Four Payment Plans, with and Without Inflation Plan 1 2 3 4

PWAT(0% Inflation)

PWAT(4% Inflation)

Difference

$128,795.20 $121,927.80 $123,485.16 $112,022.00

$184,317.66 $155,842.70 $162,586.49 $199,571.84

$55,522.46 $33,914.90 $39,101.33 $87,549.84

Excel® Data File Notice the impact of inflation on the preferred plan. With no inflation, Plan 1 is preferred; with 4% inflation, Plan 4 is preferred. Among the four payment plans, if the lender’s interest rate is less than the borrower’s combined after-tax minimum attractive rate of return (MARRcAT), Plan 4 yields the greatest PWAT. If the lender’s interest rate is greater than MARRcAT, but less than MARRcAT/(1 − itr), Plan 1 has the greatest PWAT. If the lender’s interest rate is greater than MARRcAT/(1 − itr), it is best to not borrow investment capital; however, if borrowing is necessary, then Plan 2 yields the greatest PWAT. For Example 10.4, the lender’s interest rate = 12% and MARRcAT = 14.4%. Therefore, because the lender’s interest rate (12%) is less than MARRcAT (14.4%), Plan 4 is preferred, as demonstrated. In bringing our treatment of inflation to a close, we remind the reader that, even though we did not include examples of such, after-tax analyses can be performed under conditions of inflation when bonus depreciation and Section 179 expense deduction are available. End-of-chapter problems provide opportunities to perform such analyses. Keep in mind, both bonus depreciation and Section 179 expense deduction are in then-current dollars, as are principal and interest payments and depreciation allowances.

Concept Check 10.04-CC001 The before-tax and loan cash flows for an economic evaluation are expressed in constant dollars. What must be done to loan and depreciation cash flows in order to correctly conduct an after-tax present worth analysis with borrowed capital? I. Convert the loan principal payments to constant dollar II. Convert the loan interest payments to constant dollar III. Convert the MACRS depreciation allowances to constant dollar a. I and II only b. II and III only c. I, II, and III d. No conversions are required Correct or Incorrect? Clear

Note

  Check Answer

1. For a year-to-year comparison of CPI and HEPI, see White, J. A., K. E. Case, D. B. Pratt, Principles of Engineering Economic Analysis, 6th edition, John Wiley & Sons, NY, 2012; for more information on CPI, see the Web site for the Bureau of Labor Statistics (www.bls.gov/cpi/cpifact2htm); and for more information on HEPI, see the Web site for the Commonfund Institute (www.commonfund.org/Commonfund).

CHAPTER 11 Break-Even, Sensitivity, and Risk Analysis

Chapter 11 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced e-text and WileyPLUS

FE-Like Problems 11-FE001 If the total cost for producing widgets can be represented by TC = 8,000 + 0.75*X, where X is the number of widgets produced and total revenue can be represented by TR = 4.00*X, what is the break-even value for number of widgets produced? a. 1,684 b. 2,000 c. 2,462 d. 3,763 The next five questions refer to the following sensitivity graph:

11-FE002

The analysis is most sensitive to changes in which component?

a. Annual Revenue b. Initial Investment c. Salvage Value d. Cannot be determined from the information given Correct or Incorrect? Clear

  Check Answer

11-FE003

The analysis is least sensitive to changes in which component?

a. Annual Revenue b. Initial Investment c. Salvage Value d. Cannot be determined from the information given 11-FE004 What is the numeric value of the present worth of the original project (i.e., no changes)? a. −10 b. 20 c. 1,000 d. Cannot be determined from the information given Correct or Incorrect? Clear

  Check Answer

11-FE005 What percentage change in initial investment would cause the project to become unattractive? a. −10 b. +20 c. +1,000

d. Cannot be determined from the information given 11-FE006 If a line for “annual expenses” were added to the graph, what slope would you expect the line to have? a. Positive slope (line rises as it goes left to right) b. Negative slope (line falls as it goes left to right) c. Zero slope (horizontal line) d. Infinite slope (vertical line) Correct or Incorrect? Clear

  Check Answer

11-FE007 Which of the following is not a method typically used for supplementary analysis of engineering economy problems? a. Break-even analysis b. Depreciation analysis c. Risk analysis d. Sensitivity analysis 11-FE008 The probability of weather-related crop damage during the growing season in a typical year is given by the following table. If the interest rate is 8%, what is the expected present worth of crop damage over the next 5 years? Value of Crop Damage Probability $0

60%

$100,000

25%

$200,000

13%

$300,000

2%

a. $57,000 b. $167,000

c. $228,000 d. $285,000 Correct or Incorrect? Clear

  Check Answer

11-FE009 Gooey Bites sells snack packs for $3 per pack. Variable expenses involved in producing snack packs are estimated to be $1 per pack and fixed costs for operating the production line are estimated to be $14,000. How many snack packs must Gooey Bites sell to break even? a. 14,000 b. 3,500 c. 4,667 d. 7,000 11-FE010 Reconsider the previous problem. After making changes to the production line, Gooey Bites made a profit of $36,000 by selling 20,000 snack packs. Variable costs were modified by the line changes but fixed costs were unaffected. What is the new variable cost per pack? a. $0.33 b. $0.50 c. $1.00 d. $1.50 Correct or Incorrect? Clear

  Check Answer

Problems Section 11.1 Break-Even Analysis

LEARNING OBJECTIVE 11.1 Calculate the break-even value using a break-even analysis approach. 11.01-PR001 Uncertainty can impact many elements of an engineering economic analysis. Given the list of factors below, rank them from most to least uncertain and briefly justify why you ranked them in the order you did. Factor List (alphabetic) First cost MARR Operating and maintenance costs Planning horizon Salvage value 11.01-PR002 Match the terms in the first column with an appropriate definition from the second column. Terms Breakeven (a) analysis

Definitions Determining how the worth of a investment changes with (1) changes in one or more parameters

Determining probabilistic statements about the worth of Sensitivity an investment based on probabilistic values assigned to (b) analysis (2) one or more parameters Risk (c) analysis

Determining the indifference value of a particular (3) parameter

11.01-PR003 Cecil’s Manufacturing is considering production of a new product. The sales price would be $10.25 per unit. The cost of the equipment is $100,000. Operating and Maintenance costs are expected to be $3,500 annually. Based on a 7-year planning horizon and a MARR of 12%, determine the number of units that must be sold annually to achieve break-even.

11.01-PR004 Reconsider Problem 11.01-PR003. Determine whether each of the following statements is true or false by determining the new break-even for each case. Each case is independent of the other cases. a. If the cost of the equipment doubles, the break-even volume will double. b. If the revenue per unit doubles, the break-even volume will halve. c. If the O&M costs double, the break-even volume will double. 11.01-PR005 The Fence Company is setting up a new production line to produce top rails. The relevant data for two alternatives are shown below. Installed Cost Expected Life Salvage Value Variable Cost per top rail

Flow Line Manufacturing Cell $15,000 $10,000 5 years $0 $6.00

5 years $0 $7.00

a. Based on MARR of 8%, determine the annual rate of production for which the alternatives are equally economical. b. If it is estimated that production will be 300 top rails per year, which alternative is preferred and what will be the total annual cost? 11.01-PR006 A manufacturer offers an inventor the choice of two contracts for the exclusive right to manufacture and market the inventor’s patented design. Plan I calls for an immediate single payment of $50,000. Plan II calls for an annual payment of $2,000 plus a royalty of $1.00 for each unit sold. The remaining life of the patent is 10 years. MARR is 10% per year. a. What must be the uniform annual sales to make Plan I and Plan II equally attractive? b. If fewer than the number in (a) are scheduled for production and sales, which plan is more attractive? 11.01-PR007 Video Solution A pipeline contractor can purchase a needed truck for $40,000. Its estimated life is 6 years and it has no salvage value. Maintenance is estimated to be $2,400 per year. Operating expense is

$60 per day. The contractor can hire a similar unit for $150 per day. MARR is 7%. a. How many days per year must the services of the truck be needed such that the two alternatives are equally costly? b. If the truck is needed for 180 days/year, should the contractor buy the truck or hire the similar unit? Determine the dollar amount of annual savings generated by using the preferred alternative rather than the non preferred.

11.01-PR008 Video Solution A firm has the capacity to produce 650,000 units of product per year. At present, it is operating at 64% of capacity. The firm’s income per unit is $1.00, annual fixed costs are $192,000, and variable costs are $0.376 per unit of product. a. What is the firm’s current annual profit or loss? b. At what volume of product does the firm break even? c. What would be the profit or loss at 80% of capacity?

11.01-PR009 Spending $1,500 more today for a hybrid engine rather than a gasoline engine will result in annual fuel savings of $300. How many years must this savings continue in order to justify the extra investment if money is worth 10% per year, compounded annually? 11.01-PR010 The Cooper Company is considering investing in a recuperator. The recuperator will have an initial cost of $20,000 and a service life of 10 years. Operating and maintenance costs for the first year are estimated to be $1,500, increasing by $100 every year thereafter. It is estimated that the salvage value of the recuperator will be 20% of its initial cost. The recuperator

will result in equal annual fuel savings throughout its service life. Assuming MARR is 12%, what is the minimum value of fuel savings for which the recuperator is attractive? 11.01-PR011 Snow Valley Ski Resort has been contracting snow removal from its parking lots at a cost of $400/day. A snow removal machine can be purchased for $25,000. The machine is estimated to have a useful life of 6 years with a zero salvage value at that time. Annual costs for operating and maintaining the equipment are estimated to be $5,000. Determine the break even value for the number of days per year that snow removal is required in order to justify purchasing the snow removal machine. MARR is 12%/yr. 11.01-PR012 A small utility company is considering the purchase of a powerdriven post hole digger. The equipment will cost $8,000 and have an estimated life of 8 years and a salvage value of $1,000 at that time. Annual maintenance costs for the digger are estimated to be 15% of the first cost regardless of its level of usage. Operations costs are $40 per day with an output rate of 25 holes per day. At present, holes are manually dug at a rate of 1.5 per day by a laborer whose marginal cost is $11.20 per day. Determine the break-even value in holes per year. MARR is 8%. 11.01-PR013 To make a batch of 1,000 units, it is estimated that 120 direct labor hours are required at a cost of $10 per hour. Direct material costs are estimated at $1,500 per batch. The overhead costs are calculated based on an overhead rate of $7.50 per direct labor hour. The item can be readily purchased from a local vendor for $4 per unit. a. Should the item be manufactured or purchased? b. What is the break-even value for the overhead rate (dollars per direct labor hour)? Assume that the material costs, labor hours, and labor costs do not change. 11.01-PR014 A manufacturer of precision castings has the capacity to produce 1,000,000 castings per year. Each sells for $15. The variable cost per unit to produce the casting is $9 per casting. Annual fixed costs for the plant are $3,500,000. a. If the plant is currently operating at 50% of capacity, how much profit (loss) is being earned?

b. What percent of production capacity is required for break-even? 11.01-PR015 Video Solution A new manufacturing plant costs $5 million to build. Operating and maintenance costs are estimated to be $45,000 per year and a salvage value of 25% of the initial cost is expected. The units the plant produces are sold for $35 each. Sales and production are designed to run 365 days per year. The planning horizon is 10 years. Find the break-even value for the number of units sold per day for each of the following values of MARR. a. MARR is 5%. b. MARR is 10%. c. MARR is 15%.

11.01-PR016 A subsidiary of a major furniture company manufactures wooden pallets. The plant has the capacity to produce 300,000 pallets per year. Presently the plant is operating at 70% of capacity. The selling price of the pallets is $18.25 per pallet and the variable cost per pallet is $15.75. At zero output, the subsidiary plant’s annual fixed costs are $550,000. This amount remains constant for any production rate between zero and plant capacity. a. With the present 70% of capacity production, what is the expected annual profit or loss for the subsidiary plant. b. What annual volume of sales (units) is required in order for the plant to break even? c. What would be the annual profit or loss if the plant were operating at 90% of capacity? d. If fixed costs could be reduced by 40%, what would be the new breakeven sales volume? 11.01-PR017 Cowboy Metal Cutting produces a laser cut part based on customer orders. The number of units requested on a customer’s order for the

laser cut part can vary from 1 unit to 150 units. Cowboy has determined that three different cutting machines can be used to produce this part. An economic analysis of production costs has produced the data in the table below. Cutting Tool ID Fixed Cost per Order Variable Cost per Unit CT 1 CT 2 CT 3

$300 $750 $500

$9.00 $3.00 $5.00

a. For all order sizes between 1 and 150, determine the preferred (most economical) cutting machine for an order of that size. b. For an order of size 75, what is the minimum cost production? 11.01-PR018 Two 100-horsepower motors are under consideration by the Mighty Machinery Company. Motor Q costs $5,000 and operates at 90% efficiency. Motor R costs $3,500 and is 88% efficient. Annual operating and maintenance costs are estimated to be 15% of the initial purchase price. Power costs 3.2¢/kilowatt-hour. How many hours of full-load operation are necessary each year in order to justify the purchase of motor Q? Use a 15-year planning horizon; assume that salvage values will equal 20% of the initial purchase price; and let the MARR be 15%. (Note: 0.746 kilowatts = 1 horsepower.) 11.01-PR019 An aluminum extrusion plant manufactures a particular product at a variable cost of $0.04 per unit, including material cost. The fixed costs associated with manufacturing the product equal $30,000/year. Determine the break-even value for annual sales if the selling price per unit is (a) $0.40, (b) $0.30, and (c) $0.20. 11.01-PR020 Owners of a nationwide motel chain are considering locating a new motel in Snyder, Arkansas. The complete cost of building a 150-unit motel (excluding furnishings) is $5 million; the firm estimates that the furnishings in the motel must be replaced at a cost of $1,875,000 every 5 years. Annual operating and maintenance cost for the facility is estimated to be $125,000. The average rate for a unit is anticipated to be $55/day. A 15year planning horizon is used by the firm in evaluating new ventures of this type; a terminal salvage value of 20% of the original building cost is

anticipated; furnishings are estimated to have no salvage value at the end of each 5-year replacement interval; land cost is not to be included. Determine the break-even value for the daily occupancy percentage based on a MARR of (a) 0%, (b) 10%, (c) 15%, and (d) 20%. (Assume that the motel will operate 365 days/year.) 11.01-PR021 A consulting engineer is considering two pumps to meet a demand of 15,000 gallons/minute at 12 feet total dynamic head. The specific gravity of the liquid being pumped is 1.50. Pump A operates at 70% efficiency and costs $12,000; pump B operates at 75% efficiency and costs $18,000. Power costs $0.015/kilowatt-hour. Continuous pumping for 365 days/year is required (i.e., 24 hours/day). Using a MARR of 10% and assuming equal salvage values for both pumps, how many years of service are required for pump B to be justified economically? (Note: Dynamic head times gallon/minute times specific gravity divided by 3960 equals horsepower required. Horsepower times 0.746 equals kilowatts required.) 11.01-PR022 A business firm is contemplating the installation of an improved material-handling system between the packaging department and the finished goods warehouse. Two designs are being considered. The first consists of an automated guided vehicle system (AGVS) involving three vehicles on the loop. The second design consists of a pallet conveyor installed between packaging and the warehouse. The AGVS will have an initial equipment cost of $280,000 and annual operating and maintenance costs of $50,000. The pallet conveyor has an initial cost of $360,000 and annual operating and maintenance costs of $35,000. The firm is not sure what planning horizon to use in the analysis; however, the salvage value estimates given in the following table have been developed for various planning horizons. Using a MARR of 10%, determine the break-even value for n, the planning horizon.

Salvage Value Estimates n AGVS Pallet Conveyor 1 $230,000 $300,000 2 3

185,000 145,000

245,000 200,000

4 5 6

110,000 80,000 55,000

160,000 125,000 95,000

11.01-PR023 A manufacturing plant in Michigan has been contracting snow removal at a cost of $400/day. The past 3 years have produced heavy snowfalls, resulting in the cost of snow removal being of concern to the plant manager. The plant engineer has found that a snow-removal machine can be purchased for $25,000; it is estimated to have a useful life of 6 years and a zero salvage value at that time. Annual costs for operating and maintaining the equipment are estimated to be $5,000. Determine the break-even value for the number of days per year that snow removal is required in order to justify the equipment, based on a MARR of (a) 0%, (b) 10%, and (c) 15%. 11.01-PR024 The motor on a gas-fired furnace in a small foundry is to be replaced. Three different 15-horsepower electric motors are being considered. Motor X sells for $2,500 and has an efficiency rating of 90%; motor Y sells for $1,750 and has a rating of 85%; motor Z sells for $1,000 and is rated to be 80% efficient. The cost of electricity is $0.065/kilowatt-hour. An 8-year planning horizon is used, and zero salvage values are assumed for all three motors. A MARR of 12% is to be used. Assume that the motor selected will be loaded to capacity. Determine the range of values for annual usage of the motor (in hours) that will lead to the preference of each motor. (Note: 0.746 kilowatts = 1 horsepower.) 11.01-PR025 A machine can be purchased at t = 0 for $20,000. The estimated life is 5 years, with an estimated salvage value of zero at that time. The average annual operating and maintenance expenses are expected to be $5,500. If MARR = 10%, what must the average annual revenues be in order to be indifferent between (a) purchasing the machine or (b) doing nothing?

11.01-PR026 Two condensers are being considered by the Ajax Company. A copper condenser can be purchased for $5,000; annual operating and maintenance costs are estimated to be $500. Alternatively, a ferrous condenser can be purchased for $3,500; since the Ajax company has not had previous experience with ferrous condensers, they are not sure what annual operating and maintenance cost estimate is appropriate. A 5-year planning horizon is to be used, salvage values are estimated to be 15% of the original purchase price, and a MARR of 20% is to be used. Determine the break-even value for the annual operating and maintenance cost for the ferrous condenser. Section 11.2 Sensitivity Analysis LEARNING OBJECTIVE 11.2 Perform a sensitivity analysis to examine the impact on the measure of economic worth and the robustness of the economic decision when values of one or more parameters vary over specified ranges. 11.02-PR001 Video Solution A pork processing facility is considering the installation of either a storage facility or a holding pond. A biosystems engineer has been hired to evaluate the economic trade-offs for the two alternatives. The engineer estimates the cost of the storage facility to be $213,000, with annual costs for maintenance to be $3,200 per year. She estimates the cost of the pond to be $90,000, plus $45,000 for pumps and piping, and annual operating and maintenance costs for the holding pond are estimated to be $8,500. The engineer estimates the life of the storage facility and the pond to be around 20 years but is concerned about the accuracy of this estimate. She decides to do a sensitivity analysis. a. Develop the equation that she would use to determine how sensitive the economic decision is to changes in life. Use a MARR of 15%. b. Determine which alternative is preferred for lives ranging from 15 to 25 years.

11.02-PR002 You have been asked to perform a sensitivity analysis on a company’s plan to modernize its facilities to determine the impact of possible errors in estimating the net annual savings. The initial investment in the modernization is $30,000. The expected net annual savings are $13,000. The salvage value is $7,000 after a planning horizon of 7 years. MARR is 12% per year. a. Determine if the modernization is economically attractive based on the initial estimates and an annual worth (AW) analysis. b. Determine the AW if the net annual savings change by the following percentages from the initial estimate: −80%, −60%, −40%, −20%, +20%, +40%. c. Determine the percentage change in net annual savings that causes a reversal in the decision regarding the attractiveness of the project. 11.02-PR003 Plot a sensitivity graph for annual worth versus initial cost, annual revenue, and salvage value for the data in the table shown below. Vary only one parameter at a time, each within the range of −20% to +20%. MARR is 3%/yr. Project life is 4 years. Initial Cost $120,000 Annual Revenue $25,000 Salvage Value

$35,000

Based on your graph, which parameter shows the most sensitivity? the least? 11.02-PR004 Video Solution A new project will cost $80,000 initially and will last for 7 years, at which time its salvage value will be $2,500. Annual revenues are anticipated to be $15,000 per year. For a MARR of 12%/yr, plot a sensitivity graph for annual worth versus initial cost, annual revenue, and salvage value, varying only one parameter at a time, each within the range of +/− 50%.

11.02-PR005 In Problem 11.01-PR020 suppose the following pessimistic, most likely, and optimistic estimates are given for building cost, furnishings

cost, annual operating and maintenance costs, and the average rate per occupied unit. Most Pessimistic Likely Building cost

Optimistic

$7,500,000 $5,000,000 $4,000,000

Furnishings cost Annual operating and maintenance costs Average rate

3,000,000

1,875,000

1,000,000

200,000

125,000

75,000

35/day

55/day

75/day

Determine the pessimistic and optimistic limits on the break-even value for the daily occupancy percentage based on a MARR of 12%. Assume the motel will operate 365 days/year. 11.02-PR006 Plot a sensitivity graph for annual worth versus initial cost, annual revenue, and salvage value for the data in the table below. Vary only one parameter at a time, each within the range of −20% to +20%. MARR is 20%/yr. Project life is 10 years. Initial Cost

$800,000

Annual Revenue $330,000 Salvage Value

$130,000

Based on your graph, which parameter shows the most sensitivity? the least? 11.02-PR007 A warehouse modernization plan requires an investment of $3 million in equipment. At the end of the 10-year planning horizon, it is anticipated the equipment will have a salvage value of $600,000. Annual savings in operating and maintenance costs due to the modernization are anticipated to total $1,500,000/year. A MARR of 10% is used by the firm. Perform a sensitivity analysis to determine the effects on the economic feasibility of the plan due to errors in estimating the initial investment required and the annual savings. 11.02-PR008 The owners of a discount motel chain are considering building a new motel. Optimistic, pessimistic, and most likely estimates of

several key parameters have been obtained from the local builders and the chamber of commerce. These estimates are shown in the table below. The life of the motel is estimated to be 15 years and MARR is 20%. Parameter

Pessimistic Most Likely Optimistic

Initial Cost

$10,500,000 $ 8,875,000 $6,000,000

Annual Operating $350,000 $175,000 $150,000 Annual Revenue $1,500,000 $2,500,000 $3,500,000 a. Based only on the pessimistic estimates, is the new motel economically attractive? b. Based only on the most likely estimates, is the new motel economically attractive? c. Based only on the optimistic estimates, is the new motel economically attractive? d. Approximating the expected value for each parameter by dividing by 6 the sum of the pessimistic estimate, the optimistic estimate, and 4 times the most likely estimate, based on an expected value analysis, is the new motel economically attractive? 11.02-PR009 Initial estimates of the parameters for an investment are given below. Parameter

Initial Estimate Sensitivity

Initial Investment

$15,000

None

Net Annual Receipts

$2,500

−30%, 0%, +30%

10 years $500

−20%, 0%, +20% −50%, 0%, +50%

15% per year

none

Project Life Salvage Value MARR

You wish to do a multiparameter sensitivity analysis based on the sensitivities shown. AW is the preferred measure of worth. a. How many values of AW need to be calculated?

b. Determine the AW values. 11.02-PR010 Reconsider the data in Problem 11.02-PR009. Based on a present worth measure of worth, complete a multi-parameter sensitivity analysis that examines all possible combinations of the estimates. Section 11.3 Risk Analysis LEARNING OBJECTIVE 11.3 Perform risk analysis using analytical or simulation solutions to derive exact values or estimates for the expected value and standard deviation of the measure of economic worth. 11.03-PR001 An investment of $15,000 is to be made into a savings account. The interest rate to be paid each year is uncertain; however, it is estimated that it is twice as likely to be 6% as it is to be 4%, and it is equally likely to be either 6% or 8%. Determine the probability distribution for the amount in the fund after 3 years, assuming the interest rate is statistically independent over time. 11.03-PR002 Assume an initial investment of $12,000, annual receipts of $4,500, and an uncertain life for the investment. Use a 15% MARR. Let the probability distribution for the life of the investment be given as follows: n p(n) 1 0.10 2 0.15 3 0.20 4 0.25 5 0.15 6 0.10 7 0.05 Analytically determine the probability of the investment being profitable. 11.03-PR003 In Problem 11.03-PR002 suppose the MARR is not known with certainty and the following probability distribution is anticipated to hold:

i

p(i)

0.10 0.20 0.12 0.60 0.15 0.20 Use analytical methods to determine the probability of the investment being profitable. 11.03-PR004 In Problem 11.03-PR002 suppose the magnitude of the annual receipts (R) is subject to random variation. Assume that each annual receipt will be identical in value and the annual receipt has the following probability distribution: R

p(R)

$5,000 0.20 $4,500 0.60 $4,000 0.20 Use analytical methods to determine the probability of the investment’s being profitable. 11.03-PR005 In Problem 11.03-PR002 suppose the minimum attractive rate of return is distributed as given in Problem 11.03-PR003 and suppose the annual receipts are distributed as given in Problem 11.03-PR004. Analytically determine the probability that the investment will be profitable. 11.03-PR006 In Problem 11.03-PR005 suppose the initial investment is equally likely to be either $9,000 or $11,000. Analytically determine the probability that the investment will be profitable. 11.03-PR007 Suppose n = 4, i = 0%, a $11,000 investment is made, and the receipt in year j, j = 1, …, 4, is statistically independent and distributed as in Problem 11.03-PR004. What is the probability distribution for present worth? 11.03-PR008 Consider an investment alternative having a 6-year planning horizon and expected values and variances for statistically independent cash flows as given below:

j

E(Aj)

V(Aj)

0 −$22,500 625 × 104 1

4,000 16 × 104

2

5,000 25 × 104

3

6,000 36 × 104

4

7,000 49 × 104

5

8,000 64 × 104

6

9,000 81 × 104

Using a discount rate of 10%, determine the expected values and variances for both present worth and annual worth. Based on the central limited theorem, compute the probability of a positive present worth; compute the probability of a positive annual worth. 11.03-PR009 Solve Problem 11.03-PR008 using a discount rate of (a) 0%, (b) 15%. 11.03-PR010 Two investment alternatives are being considered. Alternative A requires an initial investment of $15,000 in equipment; annual operating and maintenance costs are anticipated to be normally distributed, with a mean of $5,000 and a standard deviation of $500; the terminal salvage value at the end of the 8-year planning horizon is anticipated to be normally distributed, with a mean of $2,000 and a standard deviation of $800. Alternative B requires endof-year annual expenditures over the planning horizon. The annual expenditure will be normally distributed, with a mean of $8,000 and a standard deviation of $750. Using a MARR of 15%, what is the probability that Alternative A is the most economic alternative? 11.03-PR011 In Problem 11.03-PR010, suppose the MARR is 10% with probability 0.25, 12% with probability 0.50, and 15% with probability 0.25; what is the probability that Alternative A is the most economic alternative? 11.03-PR012 Company W is considering investing $12,500 in a machine. The machine will last n years, at which time it will be sold for L. Maintenance costs for this machine are estimated to increase by 10%/year over its life. The

maintenance cost for the first year is estimated to be $1,500. The company has 10% MARR. Based on the probability distributions given below for n and L, what is the expected equivalent uniform annual cost for the machine? n

L

p(n)

6 $5,000 0.2 8 3,000 0.4 10 1,000 0.4 11.03-PR013 An initial investment of $22,500 results in independent annual receipts of $6,250 until the end of the project life. The probability distribution for the life of the project is shown in the table below. MARR is 15%/yr. Life (years) Probability 4

0.10

5 6

0.25 0.45

7

0.15

8

0.05

For the following questions, employ an analytical solution: a. Determine the probability that the present worth of the project is greater than zero. b. Determine the probability that the present worth of the project is greater than $1,000. 11.03-PR014 Main Electric is deciding whether to invest $10,600,000 in a new plant. An analyst forecasts that the plant will generate the independent random cash flows shown in the table below at the end of each year. MARR is 15%/yr.

End of Year

Expected Net Cash Flow

Standard Deviation of Net Cash Flow

0

− $19,700,000

$0

1

$1,200,000

$120,000

2 3

$3,600,000 $6,000,000

$240,000 $650,000

4

$9,600,000

$750,000

5

$11,000,000

$1,080,000

For the following questions, employ an analytical solution: a. Determine the mean and standard deviation of the present worth. b. If the present worth is normally distributed, what is the probability that present worth is greater than zero? Assume each end-of-year cash flow is normally distributed with the mean and standard deviation shown in the table above. c. Using a Monte Carlo simulation with 10,000 iterations, estimate the mean and standard deviation of present worth and the probability of a positive present worth. 11.03-PR015 A new CNC mill is expected to cost $263,000 and have a useful life of 6 years. The net annual savings generated by the mill are independent from year to year and are estimated to follow a uniform distribution with a lower bound of $45,000 and an upper bound of $55,000. MARR is 12%/yr. For the following questions, employ an analytical solution: a. Determine the probability that the present worth of the CNC mill is positive. b. Using a Monte Carlo simulation with 10,000 iterations, estimate the mean and standard deviation of present worth and the probability of a positive present worth. 11.03-PR016 One of two mutually exclusive alternatives must be selected for implementation. Alternative A is an equipment purchase; Alternative B is a lease arrangement with annual payments. The characteristics of the two

investments are shown in the table below. Use an 8 year planning horizon and a MARR of 15%/yr. Alt.

Parameter

Mean Std. Dev. Distribution

A Initial Cost A Annual Maintenance

$13,000 $5,000

None $500

None Normal

A Salvage Value

$2,000

$800

Normal

B End-of-Year Lease Payment

$7,500

$750

Normal

For the following questions, employ an analytical solution: a. Determine the probability that Alternative A is the preferred alternative. b. Using a Monte Carlo simulation with 10,000 iterations, estimate the mean and standard deviation of present worth and the probability of a positive present worth. 11.03-PR017 An investment of $5,000 is expected to generate the probabilistic returns shown in the table below over its 3-year life. Assume the annual cash flows are independent and that the distribution of present worth is normal. EOY

Annual Return

1

$2,500 with probability 0.4 or $1,800 with probability 0.6

2

$3,000 with probability 0.5 or $2,000 with probability 0.5

3

$3,500 with probability 0.7 or $2,500 with probability 0.3

For the following questions, employ an analytical solution: a. Determine the probability that the present worth is negative. b. Using a Monte Carlo simulation with 10,000 iterations, estimate the probability that the present worth is negative. 11.03-PR018 A project under consideration has a 10-year projected life. The initial investment for the project is estimated to have a mean of $10,000 and a standard deviation of $1,000. The annual receipts are independent with each

year’s expected return having a mean of $1,800 and a standard deviation of $200. MARR is 12%. For the following questions, employ an analytical solution: a. Determine the probability that the present worth is negative. Assume the initial investment and annual receipts are independent and normally distributed. b. Using a Monte Carlo simulation with 10,000 iterations, estimate the probability that the present worth is negative. 11.03-PR019 Video Solution A proposed project has the following cash flow estimates. End of Year Mean Net Cash Flow Standard Deviation of Cash Flow 0 −$32,000 $1,000 1

$4,000

$2,000

2 3

$8,000 $12,000

$3,000 $5,000

4

$12,000

$6,000

5

$12,000

$7,000

Assuming independent cash flows, a normally distributed net present value, and a minimum attractive rate of return of 18%, determine the following. For the following questions, employ an analytical solution: a. The mean and standard deviation of net present value. b. The probability that the net present value is positive. c. The probability that the net present value is greater than $5,000. Assume the initial investment and annual receipts are normally distributed.

d. Using a Monte Carlo simulation with 10,000 iterations, estimate the probability that the present worth is positive and estimate the probability that the present worth is greater than $5,000.

11.03-PR020 A proposed project has the following cash flow estimates. End of Year Mean Net Cash Flow Standard Deviation of Cash Flow 0

−$800,000

$250,000

1 2

$1,000,000 $1,000,000

$450,000 $600,000

Assuming statistically independent cash flows, a normally distributed net present value, and a minimum attractive rate of return of 15%, determine the following. For the following questions, employ an analytical solution: a. the mean and standard deviation of net present value. b. the probability that the net present value is negative. c. the probability that the net present value is greater than $1,000,000. Assume the initial investment and annual receipts are normally distributed. d. Using a Monte Carlo simulation with 10,000 iterations, estimate the probability that the present worth is negative. 11.03-PR021 A $5,000 process improvement project is expected to increase annual expenses for the next 3 years by an average of $20,000 with a standard deviation of $3,000. The annual savings generated over the 3 years will average $24,000 with a standard deviation of $4,000. MARR is 20%. Assume independent cash flows. For the following questions, employ an analytical solution:

a. Assuming that present worth is normally distributed, determine the probability that the process improvement will result in a loss. b. Assuming that present worth is normally distributed, determine the probability that the process improvement will result in a present worth of $10,000 or greater. c. Using a Monte Carlo simulation with 10,000 iterations, estimate the probability that the process improvement will result in a loss and the probability that the present worth is $10,000 or greater.

Chapter 11 Summary and Study Guide Summary 11.1: Break-Even Analysis

Learning Objective 11.1: Calculate the break-even value using a break-even analysis approach. (Section 11.1) The break-even value is the value at which we are indifferent between two alternatives, that is, they are equivalent. The internal rate of return itself is a break-even value because it is the interest rate at which the economic worth of an investment equals zero. In an engineering economic analysis, breakeven analysis is often used to determine the level of annual savings required to justify a particular capital investment. 11.2: Sensitivity Analysis

Learning Objective 11.2: Perform a sensitivity analysis to examine the impact on the measure of economic worth and the robustness of the economic decision when values of one or more parameters vary over specified ranges. (Section 11.2) Sensitivity analysis is performed when we want to gauge the impact on the economic worth if one or more parameters take on values over some specified range; typically, we are interested in knowing what percent change in a parameter’s value will result in a different recommendation regarding an investment. A popular approach of parameter estimation is to consider three estimates for the value of the parameter typically depicted as an optimistic, pessimistic, and most likely value for a parameter. Compared to sensitivity analysis, risk analysis attempts to reflect the imprecision inherent in

assigning values to parameters in an engineering economic analysis. This imprecision is represented in the form of a probability distribution, and the parameters themselves are treated as random variables. 11.3: Risk Analysis

Learning Objective 11.3: Perform risk analysis using analytical or simulation solutions to derive exact values or estimates for the expected value and standard deviation of the measure of economic worth. (Section 11.3) Risk analysis incorporates probabilistic estimates with the values of the parameters; typically we want to know the probability of an investment being profitable or the probability of the measure of economic worth having at least a particular value. In some cases analytical approaches can be used, and the probabilities of certain outcomes can be obtained mathematically. However, analytical solutions can be difficult to obtain except for the simplest problems, thus simulation solutions are often preferred. Simulation is a popular approach for performing risk analysis for more complex situations where an analytical solution would be difficult to obtain. Numerous computer software packages are available to support simulation analyses. Simulation approaches have certain advantages and disadvantages. Advantages are that simulation is useful in selling a system modification to management; they can be used to verify an analytical solution; they are versatile; and less background in mathematical analysis and probability theory is generally required. Some of the major disadvantages of simulation are that simulations can be quite time-consuming; they introduce a source of randomness not present in analytical solutions in the form of sampling error; Monte Carlo simulations do not reproduce the input distribution exactly; validation of the simulation model is often overlooked; and it is so easily applied that it is often used when analytical solutions can be easily obtained at considerably less cost.

Important Terms and Concepts

Break-Even Analysis A method used when an accurate estimate of a parameter’s value cannot be provided, but intelligent judgments can be made as to whether or not the value is less than or greater than some break-even value. Break-Even Value The value of a parameter at which the measure of economic worth equates to zero. The term breakeven point is sometimes used instead of break-even value; however, within this text, we will use the latter term. Sensitivity Analysis A method used to determine the impact on the measure of economic worth when values of one or more parameters vary over specified ranges. Risk Analysis The process of incorporating explicitly random variation in one or more parameters. Autocorrelated Correlated with itself over time.

Chapter 11 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 11.1: Break-Even Analysis LO 11.1: Calculate the break-even value using a break-even analysis approach. Video Lesson: Break-Even Analysis Video Lesson Notes: Break-Even Analysis Excel Video Lesson: NPER Financial Function Excel Video Lesson Spreadsheet: NPER Financial Function Excel Video Lesson: PMT Financial Function Excel Video Lesson Spreadsheet: PMT Financial Function Excel Video Lesson: SOLVER Tool Excel Video Lesson Spreadsheet: SOLVER Tool Excel Video Lesson: RATE Financial Function Excel Video Lesson Spreadsheet: RATE Financial Function Video Example 11.2: Break-Even Analysis with Excel® for the SMP Investment Video Solution: 11.01-PR007 Video Solution: 11.01-PR008 Video Solution: 11.01-PR015 11.2: Sensitivity Analysis LO 11.2: Perform a sensitivity analysis to examine the impact on the measure of economic worth and the robustness of the economic decision when values of one or more parameters vary over specified ranges.

Video Lesson: Sensitivity Analysis Video Lesson Notes: Sensitivity Analysis Video Example 11.3: Sensitivity Analysis for a Single Alternative Video Solution: 11.02-PR001 Video Solution: 11.02-PR004 11.3: Risk Analysis LO 11.3: Perform risk analysis using analytical or simulation solutions to derive exact values or estimates for the expected value and standard deviation of the measure of economic worth. Excel Video Lesson: NORM.S.DIST Function Excel Video Lesson Spreadsheet: NORM.S.DIST Function Video Example 11.6: Risk Analysis for the SMP Investment Video Solution: 11.03-PR019 These chapter-level resources will help you with your overall understanding of the content in this chapter. Appendix A: Time Value of Money Factors Appendix B: Engineering Economic Equations Flashcards: Chapter 11 Excel Utility: TVM Factors: Table Calculator Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values Excel Utility: Monthly Payment Sensitivity Excel Utility: TVM Factors: Discrete Compounding Excel Utility: TVM Factors: Geometric Series Future Worth Excel Utility: TVM Factors: Geometric Series Present Worth

Excel Data Files: Chapter 11

CHAPTER 11 Break-Even, Sensitivity, and Risk Analysis LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 11.1 Calculate the break-even value using a break-even analysis approach. (Section 11.1) 11.2 Perform a sensitivity analysis to examine the impact on the measure of economic worth and the robustness of the economic decision when values of one or more parameters vary over specified ranges. (Section 11.2) 11.3 Perform risk analysis using analytical or simulation solutions to derive exact values or estimates for the expected value and standard deviation of the measure of economic worth. (Section 11.3)

Engineering Economics in Practice Uber Uber, a transportation network company headquartered in San Francisco, California, is perhaps best known for its peer-to-peer ridesharing services. The company was founded in March 2009, and in May 2019 held an initial public offering (IPO) at $45 a share, raising more than $8 billion. During the past decade, the company has had well-publicized problems, including a toxic culture, criminal investigations, and a decline in its valuation from a peak of $69 billion to $30 billion as of August 5, 2019. The ousting of founder and CEO Travis Kalanick was an attempt to set the right cultural tone of humility and determination and to move the company forward in the right way. Although Uber is a global company, a major issue in the United States is that it is trying to hold onto its dominant share of the online ride-hailing market, estimated at 65–70%. Uber’s approach is to subsidize drivers and riders to a significant level. Amir Efrati stated in The Information, “Both Uber and Lyft are playing a game of chicken as to which one will cut spending first in order to reduce losses and stomach a much slower growth rate.” Nocera (Bloomberg, January 23, 2018) reports that Uber recently spent 90% of the company’s $9.7 billion in quarterly revenue on driver payouts and bonuses and rider discounts and insurance costs, and questions whether this level of spending is sustainable. In order to combat the high driver cost, estimated at about $7 billion per quarter, Uber has been working on building a self-driving car. Sure, a fleet of self-driving cars would revolutionize the business, and significantly reduce Uber’s driver cost, but the company must ask itself whether this is one of its core competencies. Might it not make more sense to let someone else build a self-driving car and then purchase the technology? By all reports, Google is far ahead of Uber with this new technology and has invested billions of dollars in R&D on autonomous automobiles, leading us to conclude that a better strategy for Uber might be to focus on its core business of ridesharing. Furthermore, experts estimate that it might be decades before this technology could be helpful to Uber, considering the technology development and regulatory issues that need to be resolved. Uber might be better served by increasing its R&D on artificial intelligence and machine learning to help manage its fleet more effectively and provide better customer service. Discussion Questions 1. Is the risk experienced by Uber limited to financial risk? If not, what other risks are present? 2. Is the exposure to risk at Uber limited to their investors and employees? If not, who else is at risk? 3. What are some specific steps that Uber can take to limit its risks? 4. How might sensitivity analysis be used to mitigate the risk faced by Uber? 5. Given the complex investment decisions made by a company such as Uber, is an economic break-even analysis sufficient to make decisions? If not, what other factors might need to be considered and how might this be done?

Introduction To date, we have assumed complete certainty when specifying the values of all parameters involved in an economic justification. We have assigned quantities such as the magnitude of the initial investment, the magnitudes of the annual cash flows, the length of the planning horizon, the value of the discount rate, the terminal salvage value for the investment, and inflation rates. However, unexpected things happen, such as Hurricane Matthew, which devastated parts of the Caribbean and southeastern and mid-Atlantic portions of the United States in late September and early October 2016; BP’s oil spill in the Gulf of Mexico in 2010; the global economic crisis in 2009 and 2010; and the United States terrorist

attacks on September 11, 2001, to name just a recent few. Each example is a major event that impacted many lives and companies. There are also thousands of smaller events that do not produce headlines but that affect the financial outcomes of capital investments. And, you don’t have to read the newspaper to know how rapidly the price of a barrel of oil changes; all you have to do is fill your car or truck with gasoline or diesel fuel. In this chapter, we examine several techniques that are used to increase the confidence of managers making capital investment decisions in the face of uncertainty. As noted in Chapter 1, this chapter addresses the 6th step in the systematic economic analysis technique: perform supplementary analyses.

Systematic Economic Analysis Technique 1. Identify the investment alternatives 2. Define the planning horizon 3. Specify the discount rate 4. Estimate the cash flows 5. Compare the alternatives 6. Perform supplementary analyses 7. Select the preferred investment We begin the chapter by addressing situations where we want to know what single value for a particular parameter will make someone indifferent as to whether or not to make an investment. Such analyses are termed break-even analyses. The second type of supplementary analysis we consider is sensitivity analysis. In performing such analyses, we are interested in learning how sensitive the economic worth for one or more investments is to various values of one or more parameters. The third type of supplementary analysis considered is risk analysis. In contrast to sensitivity analysis, probabilities are assigned to various values of one or more parameters, and a probabilistic statement is made regarding the economic worth for one or more investments. Typically, the probabilistic statement takes the form of “the probability of the investment having a positive-valued present worth is …” or “the probability of the internal rate of return for the investment being greater than the minimum attractive rate of return is …” or “the investment alternative having the greatest probability of having a positive-valued present worth is …”.

11.1 Break-Even Analysis LEARNING OBJECTIVE Calculate the break-even value using a break-even analysis approach. Video Lesson: Break-Even Analysis Break-even analysis is normally used when an accurate estimate of a parameter’s value cannot be provided, but intelligent judgments can be made as to whether or not the value is less than or greater than some break-even value. The break-even value is the value at which someone is indifferent as to whether or not to make an investment. More precisely, it is the value of a parameter at which the measure of economic worth being used (PW, FW, AW) is equal to zero. In performing a break-even analysis, we want to determine the break-even value.

Break-Even Analysis A method used when an accurate estimate of a parameter’s value cannot be provided, but intelligent judgments can be made as to whether or not the value is less than or greater than some break-even value. Break-Even Value The value of a parameter at which the measure of economic worth equates to zero. The term break-even point is sometimes used instead of break-even value; however, within this text, we will use the latter term. Although we did not label it as such, in Chapter 3 we performed several break-even analyses and referred to the process as equivalence. In particular, we posed a situation in which we determined the value of a particular parameter in order for two cash flow profiles to be equivalent. Another way of stating the problem could have been, “Determine the value of X that will yield a break-even situation between two alternatives.” In this case, X denotes the parameter whose value is to be determined. Additionally, when the cash flow profiles for two alternatives are equivalent, a break-even situation can be said to exist between two alternatives. The internal rate of return itself is a break-even value because it is the interest rate that equates to 0 the economic worth of an investment. In other words, it is the interest rate that equates the present worth of the positive-valued cash flows to the present worth of the negative-valued cash flows. Stated another way, the internal rate of return is the break-even value for the reinvestment rate because such a reinvestment rate will yield a future worth of 0 for either an individual alternative or the differences in cash flows for two alternatives. (Similarly, the external rate of return is a break-even value because it is the interest rate that equates the future worth of negative-valued cash flows to the future worth of positive-valued cash flows when the positive-valued cash flows are reinvested and earn interest equal to the minimum attractive rate of return.) Recall in Chapter 4 we introduced another example of break-even analysis. Specifically, we determined how long it would take to recover the initial investment, which is termed the discounted payback period, when the TVOM is greater than 0. (Also, recall that the Excel® NPER worksheet function can be used to determine how long uniform annual savings must occur in order to fully recover an initial investment, given a desired return on investment.) Excel® Video Lesson: NPER Financial Function In an engineering economic analysis, break-even analysis is often used to determine the level of annual savings required to justify a particular capital investment. In a replacement study, break-even analysis is often performed on the purchase price or salvage value of the replacement alternatives. Likewise, break-even analysis might be used to determine how long annual savings must occur in order to economically justify a particular investment.

EXAMPLE 11.1 Determining the Break-Even Value for Units Sold The Gizmo Manufacturing Company is considering making and selling a new product. The following data have been provided to management: Sales price Equipment cost

$17.50/unit $250,000

Incremental overhead cost

$50,000/year

Sales and marketing cost

$150,000/year

Operating and maintenance cost $25/operating hour Production time/1,000 units 100 hours Packaging and shipping cost

$0.50/unit

Planning horizon

5 years

Minimum attractive rate of return 15% Some managers are reluctant to launch a new product because of the uncertainty of future sales. To provide management with information that might make it easier to draw the correct conclusion, a break-even analysis will be performed for annual sales required to economically justify introducing the new product. What is the break-even value of units sold annually? Key Data Given Product cost parameters as defined in the chart above Find Break-even value of units sold annually (X units/year) Solution Assuming a negligible salvage value for all equipment at the end of the 5-year planning horizon and letting X denote the number of units sold annually for the product, the annual worth for the investment alternative can be determined as follows: AW(15%)

=

−$250,000(A|P 15%,5) − $50,000 −$150,000 − 0.1($25)X − $0.50X + $17.50X

=

−$274,578.89 + $14.50X

Setting the annual worth equal to 0 and solving for X gives a break-even value of 18,936.475 units per year. Exploring the Solution How should management interpret this break-even value? Suppose they are confident that annual sales will be between 20,000 and 30,000 units. Because this projection is greater than the break-even value, the product should be launched. Clearly, it is not necessary to know precisely what the annual sales volume will be; instead, we need to know if it will be greater than or less than the break-even value. The break-even value will not always fall outside the range of what are judged to be realistic values of a parameter. For example, if annual sales will be between 15,000 and 20,000 units per year, it is not obvious what decision should be made. A graphical representation of the example is given in Figure 11.1. The chart is referred to as a break-even chart because one can determine graphically the break-even point by observing the value of X when annual revenue equals annual cost.

FIGURE 11.1 Break-Even Analysis for the Annual Sales of a New Product Excel® Data File

EXAMPLE 11.2 Break-Even Analysis with Excel® for the SMP Investment Video Example Recall the $500,000 investment in a surface-mount placement machine, treated in previous chapters. (a) Based on a 10-year planning horizon, a $50,000 salvage value, and a 10% MARR, what annual savings are required for the investment to break even? (b) Assume $92,500 is an accurate estimate of the annual savings that will result from the SMP investment, but it is not clear how long the machine will be used. What is the break-even value for the investment’s duration? Solution a. Letting X denote the break-even value for annual savings, the following annual worth relationship must hold: $500,000(A |P 10%,10) = X + $50,000(A| F 10%,10)

or X = $500,000(A |P 10%,10) − $50,000(A| F 10%,10)

Using the Excel® PMT function, X

=PMT(10%,10,−500000,50000) = $78,235.43

Excel® Video Lesson: The PMT Function b. To determine the break-even value for the investment’s duration when we do not know how long the SMP machine will be used, an assumption is needed regarding its salvage value. First, suppose salvage value decreases linearly from $500,000 to $50,000 over a 10-year period. With salvage value decreasing at a rate of $45,000 per year, setting the EUAC of purchasing the SMP machine equal to the annual worth of the annual savings and salvage value gives $500,000(A |P 10%,n) = $92,500 + ($50,000 − $45,000n)(A| F 10%,n)

What value of n is required for the equality to hold? As shown in Figure 11.2, using the Excel® SOLVER tool yields a value of 2.17 years for the break-even value of n. Excel® Video Lesson: SOLVER Tool

FIGURE 11.2 Using the Excel® SOLVER Tool to Determine the Break-Even Value for the Planning Horizon Excel® Data File Second, suppose salvage value decreases geometrically from $500,000 to $50,000 over a 10-year period. Using the Excel® RATE function, the geometric rate is j

=RATE(10,,−500000,50000) = −20.5672%

Therefore, n

$500,000(A |P 10%,n) = $92,500 + $500,000(0.794328 )(A| F 10%,n)

Excel® Video Lesson: RATE Financial Function As shown in Figure 11.2, using the Excel® SOLVER tool yields a value of 6.95 years for the breakeven value of n. If we assume the salvage value is not a function of n and remains $50,000 regardless of the investment’s duration, then the Excel® NPER function can be used to determine the break-even value of n: n

=NPER(92500,−500000,50000) = 7.57668 years

Likewise, if the salvage value is negligible, regardless of investment duration, the break-even value of n is given by n

=NPER(92500,−500000) = 8.15972 years

Concept Check 11.01-CC001 A company is using break-even analysis to determine how many units of a new product must be sold for the product to be profitable. Which of the following actions will cause the break-even point of a product to increase? a. Reduction in the purchase price of the equipment needed to produce the product b. Reduction in the per unit production cost of the product c. Reduction in the sales price of the product d. All of the answer choices are correct

11.2 Sensitivity Analysis LEARNING OBJECTIVE Perform a sensitivity analysis to examine the impact on the measure of economic worth and the robustness of the economic decision when values of one or more parameters vary over specified ranges. Video Lesson: Sensitivity Analysis Sensitivity analyses are performed to determine the impact on the measure of economic worth when values of one or more parameters vary over specified ranges. If modest changes of parameter values adversely affect an investment’s economic worth, then it is said to be sensitive to changes in the particular parameters. On the other hand, if the economic choice is not affected by significant changes in the values of one or more parameters, then the decision is said to be insensitive to changes in the parameter values. Sensitivity Analysis A method used to determine the impact on the measure of economic worth when values of one or more parameters vary over specified ranges. We perform sensitivity analyses in several chapters, although we do not always refer to them as such. For example: in Chapters 4 and 5, we examine the impact on economic worth of the minimum attractive rate of return; in Chapter 7, we examine the sensitivity of the optimum replacement interval to changes in the initial investment, the salvage value, the rate of increase in annual operating and management costs, and the minimum attractive rate of return; in Chapter 9, we examine the sensitivity of after-tax present worth to changes in the depreciation method and to changes in the amount of money borrowed; in Chapter 10, we examine the sensitivity of after-tax present worth to changes in the inflation rate; we also examine the impact of inflation on the preferred loan repayment method; and in Chapter 12, we examine the sensitivity of the optimum investment portfolio to changes in the level of investment capital available and the minimum attractive rate of return. A derisive term occasionally used to describe analytical models is GIGO—“garbage-in, garbage-out.” In other words, what you get out of the model is no better than what you put into it. While one cannot always trust the

results obtained from models of reality, perfect information is not often required to produce correct decisions. Sensitivity analysis can be used to determine if, in fact, less-than-perfect estimates of the parameter values will result in the best decision being made.

EXAMPLE 11.3 Sensitivity Analysis for a Single Alternative Video Example To illustrate how a sensitivity analysis might be performed, we consider once more the SMP investment: a $500,000 initial investment, annual savings of $92,500 for a 10-year period, and a salvage value of $50,000. As before, a 10% MARR applies. Let’s consider how sensitive the annual worth for the investment is to errors in estimating the initial investment, the annual savings, the salvage value, the investment’s duration, and the MARR. Specifically, for an error range of ±50% for each parameter, what is the impact on AW? Solution If it is assumed that all estimates are correct except that for annual savings, the alternative’s annual worth can be given as AW(10%) =  

−$500,000(A|P 10%, 10) +$50,000(A|F 10%, 10) + $92,500(1 + X)

where X denotes the percent error (decimal equivalent) in estimating the value for annual savings. Plotting annual worth as a function of the percent error in estimating the value of annual savings yields the straight line having positive slope in Figure 11.3. Performing similar analyses for the initial investment required, the salvage value, the investment’s duration (planning horizon), and the MARR yields the remaining results given in Figure 11.3.

FIGURE 11.3 Sensitivity Analysis for Example 11.3 Excel® Data File (Note: For the example, we assumed the salvage value was $50,000, regardless of the investment’s duration. In reality, salvage value depends on investment life. Also, we used annual worth instead of present worth in the sensitivity analysis because of the variability of the planning horizon.) As shown in Figure 11.3, the investment’s net annual worth is affected differently by errors in estimating the values of the various parameters. Based on the slopes of the sensitivity curves, annual worth is most sensitive to errors in estimating the values of the initial investment, the annual savings, and the planning horizon. It is insensitive to changes in salvage value and is moderately sensitive to changes in the MARR. Exploring the Solution One may also explore the impact to the break-even value during the sensitivity analysis. The break-even values for the individual parameters are $587,649.62 for the initial investment, $78,235.43 for the annual savings, −$177,340.55 for the salvage value, 7.5767 years for the investment’s duration, and 13.8% for the MARR. These values are determined by equating the alternative’s annual worth to zero and solving for the break-even value in question. (As noted previously, the discounted payback period (DPBP) is the breakeven value for the investment duration, and the internal rate of return (IRR) is the break-even value for the MARR.)

EXAMPLE 11.4 Sensitivity Analysis for Multiple Alternatives Recall, in previous chapters we considered two alternative designs for a new ride called the Scream Machine at a theme park in Florida. Design A had an initial cost of $300,000 and net annual after-tax revenues of $55,000; Design B had an initial investment of $450,000 and net annual after-tax revenues of $80,000. A 10% MARR was used over the 10-year planning horizon. Using sensitivity analysis, determine under what circumstances Design A will be preferred over Design B, and vice versa. Key Data Given The cash flows outlined in Figure 11.4; MARR = 10%; planning horizon = 10 years Find Sensitivity of design choice to estimation errors in annual revenue

FIGURE 11.4 CFDs for Example 11.4 Solution Letting x denote the error in estimating the annual revenue for Design A, and letting y denote the error in estimating the annual revenue for Design B, the present worth for each is as follows: PWA = −$300,000 + $55,000(1 + x)(P|A  10%,10)

and PWB = −$450,000 + $80,000(1 + y)(P|A  10%,10)

For Design A to be preferred over Design B, PW(A) > PW(B). Therefore, assuming either Design A or Design B must be chosen, Design A will be preferred so long as −$300,000 + $55,000(1 + x)(P|A 10%,10) > −$450,000 + $80,000(1 + y)(P|A 10%,10)

or $3,614.18 − $337,951.20x + $491,565.00y < $0

Solving for y, y < −0.007352385 + 0.6875x

As shown in Figure 11.5, Design A is preferred for combinations of estimation errors that fall below the indifference diagonal plotted. Design B is preferred for error combinations above the indifference diagonal. If the estimates are correct for Design B, an increase of 1.07% in annual revenue for Design A will cause it to be the preferred choice. Likewise, assuming the estimates for Design A are correct, a decrease of 0.735% in annual revenue for Design B will make it no longer be the preferred choice. Therefore, the preferred design is sensitive to errors in estimating annual revenue.

FIGURE 11.5 Sensitivity Analysis for Example 11.4 Excel® Data File

The sensitivity analyses we’ve considered so far involve just one parameter at a time. In practice, estimation errors occur for multiple parameters simultaneously. A popular method of performing multiparameter sensitivity analysis is to provide three estimates for each parameter subject to estimation error. The estimates, typically, represent optimistic, pessimistic, and most likely values for each parameter. The following example illustrates this approach.

EXAMPLE 11.5 Multiparameter Sensitivity Analysis For Example 11.5, suppose (for both designs) there is uncertainty concerning the values for the initial investments required and the annual revenue that will result. Specifically, suppose the estimates shown in Table 11.1 are available for the four parameters. Under what circumstances is Design A preferred over Design B, and vice versa? TABLE 11.1 Optimistic, Pessimistic, and Most Likely Estimates for Four Scream Machine Parameters Parameter Optimistic Pessimistic Most Likely Initial Investment (A)

$285,000

$310,000

$300,000

Initial Investment (B) Annual Revenue (A)

$400,000 $65,000

$510,000 $40,000

$450,000 $55,000

Annual Revenue (B)

$85,000

$70,000

$80,000

PWA

$114,396.86

−$64,217.32

$37,951.19

PWB

$122,288.20

−$64,217.32

$41,565.37

Solution From Table 11.1, assuming the pessimistic scenario occurs for each, Design A is preferred. If, however, the optimistic or the most likely scenario occurs, then Design B is preferred. It seems unlikely that pessimistic estimates will occur for all 4 parameters and for both designs. Likewise, it is unlikely that the optimistic estimates will occur for all 4 parameters for both designs. And, unfortunately, neither is it likely that the most likely estimates will be realized for each of the 8 parameters (4 for each design). Instead, combinations of pessimistic, optimistic, and most likely values generally will occur. There are 81 possible combinations of the 4 parameters and 3 estimates (34 = 81) for each design. After computing the present worth for each design and for each possible combination, we show in Figure 11.6 the preferred design. In 41 cases of the 81 combinations considered, Design A has the greatest present worth.

FIGURE 11.6 Considering 81 Possible Combinations of 3 Estimates and 4 Parameters

Excel® Data File Some might conclude that Design A is best because more than 50% of the combinations favor it. However, there is no reason to believe each combination is equally likely to occur. In order to explicitly account for the likelihood of outcomes, we may assign probabilities to one or more parameters and conduct a risk analysis. This process is demonstrated in Example 11.7 of the next section.

Concept Check 11.02-CC001 Consider the sensitivity analysis of a single alternative where the sensitivity curves of all parameters are linear. To determine the parameter whose errors would most significantly impact the annual worth, look for the sensitivity curve (line) with the numerically __________. a. Highest slope b. Highest absolute value of slope c. Lowest slope d. Lowest absolute value of slope Correct or Incorrect? Clear

  Check Answer

11.3 Risk Analysis1 LEARNING OBJECTIVE Perform risk analysis using analytical or simulation solutions to derive exact values or estimates for the expected value and standard deviation of the measure of economic worth. Risk analysis is performed when probabilities can be assigned to various values of one or more parameters. We define risk analysis as the process of incorporating explicitly random variation in one or more parameters. For example, estimates of probabilities for possible values of annual savings and estimates of probabilities for possible salvage values might be used to analyze the economic worth of investing in a new machine tool. Risk Analysis The process of incorporating explicitly random variation in one or more parameters. In comparison with sensitivity analyses, risk analyses attempt to reflect the imprecision inherent in assigning values to parameters in an engineering economic analysis. The imprecision is represented in the form of a probability distribution. The parameters are treated as random variables. Probability distributions for the random variables in question are often based on subjective probabilities. Occasionally, there might be historical data on which the probability distributions are based. Typically, the more distant in the future an event is, the less precise is our estimate of the value of the event’s outcome. Hence, letting the variance reflect our degree of precision, we expect the variance of the probability distributions to increase with time. Among the probability distributions commonly used in risk analysis are the normal distribution and the beta distribution. Examples of these are depicted in Figures 11.7 and 11.8. For discussions of several probability distributions and their process generators in the context of simulation, see any number of simulation texts.

FIGURE 11.7 Normal Distribution

FIGURE 11.8 Sample Beta Distributions Using either analytic or simulation approaches, risk analysis develops exact values or estimates for the expected value and standard deviation of the measure of economic worth. In addition, the probability of present worth, future worth, or annual worth being greater than 0 and the probability of internal rate of return or external rate of return being greater than the MARR are typically determined in risk analyses, either analytically or with simulation. The magnitudes of cash flows, the planning horizon’s duration, and the value of the MARR frequently can be considered to be random variables. For example, the cash flows occurring in a given year are often functions of several other factors, such as selling prices, market size and share, market growth rate, investment required,

inflation rate, tax rates, operating costs, fixed costs, and salvage values of all assets. The values of a number of these random variables can be correlated with each other and can be autocorrelated. Consequently, an analytical development of the probability distribution for the measure of economic worth is not easily achieved in most real-world situations. Thus, simulation is widely used in performing risk analyses. Autocorrelated Correlated with itself over time.

11.3.1 Analytical Solutions To illustrate the use of an analytical approach to develop the probability distribution for present worth, consider the following present worth formulation: (11.1)

n

PW = ∑ At (1 + i)

−t

t=0

Suppose the cash flows, At, are random variables with expected values E(At) and variances V(At). Because the expected value of a sum of random variables is given by the sum of the expected values of the random variables, the expected present worth is given by (11.2)

n

E(PW)= ∑ E[At (t + i)

−t

]

t=0

Further, because the expected value of the product of a constant and a random variable is given by the product of the constant and the expected value of the random variable, Equation 11.2 reduces to n

E(PW)= ∑ E(At )(1 + i)

−t

(11.3)

t=0

Hence, the expected present worth of a series of cash flows is found by summing the present worths of the expected values of the individual cash flows. To determine the variance of present worth, we recall that the variance of the sum of statistically independent random variables is the sum of the variances of the random variables; also, we recall that the variance of the product of a constant and a random variable equals the product of the square of the constant and the variance of the random variable. Hence, from Equation 11.1, when the random cash flows are statistically independent, the variance of present worth is given by n

V (PW)= ∑ V (At )(1 + i)

−2t

(11.4)

t=0

Excel® Video Lesson: NORM.S.DIST Function When the annual cash flows are statistically independent, and the value of n is large,based on the Central Limit Theorem, we can expect the distribution of present worth to approximate a normal distribution. The Excel® NORM.S.DIST function can be used to approximate the probability of present worth being greater than 0. The syntax for the NORM.S.DIST function provides the probability of a normally distributed random variable with mean m and standard deviation s being less than or equal to z by entering =NORM.S.DIST((z-m)/s,TRUE) in any cell. Due to the symmetry of the normal distribution, the probability of a normally distributed random variable with mean m and standard deviation s being greater than or equal to z can be obtained by entering =NORM.S.DIST((m-z)/s,TRUE) in any cell. Hence, for z = 0,

Pr(PW > 0) =NORM.S.DIST(E(PW)/SD(PW),TRUE)

where E(PW) and SD(PW) denote the expected value and standard deviation of present worth, respectively.

(11.5)

EXAMPLE 11.6 Risk Analysis for the SMP Investment Video Example To illustrate the calculation of the expected value and the variance of present worth, recall the SMP investment: $500,000 initial investment; $92,500 annual savings; $50,000 salvage value; 10-year planning horizon; and 10% MARR. Suppose the annual savings and the salvage value are random variables, distributed as shown in Table 11.2. TABLE 11.2 Means, Variances, Standard Deviations, and Probability Distributions for Annual Savings and Salvage Value for the SMP Investment A p(A) Ap(A) SV p(SV) SVp(SV) A2p(A) SV2p(SV)  $75,000

0.070

 $5,250

  393,750,000

$40,000

0.10

 $4,000

  160,000,000

 $80,000  $85,000

0.095 0.131

 $7,600 $11,135

 608,000,000  946,475,000

$45,000 $50,000

0.20 0.40

 $9,000 $20,000

  405,000,000 1,000,000,000

 $90,000

0.178

$16,020

1,441,800,000

$55,000

0.20

$11,000

  605,000,000

 $95,000 $100,000

0.183 0.181

$17,385 $18,100

1,651,575,000 1,809,999,998

$60,000 Sum

0.10 1.00

$6,000 $50,000

  360,000,000 2,530,000,000

$105,000

0.162

$17,010

1,786,050,000

Sum 1.000 E(A) = $92,500

$92,500

8,637,649,998 E(SV) = $50,000

V(A) = 81,400,001.7

V(SV) = 30,000,000

SD(A) = $9,022.19

SD(SV) = $5,477.23

Excel® Data File Computing the expected value for the annual cash flow gives $92,500. Similarly, the expected value for the salvage value is $50,000. The variance for annual savings is V (A)

2

2

=

E(A ) − [E(A) ]

=

8,637,649,998 − (92,500)

=

81,400,001.7

2

and the variance for salvage value is V (SV)

2

) − [E(SV)]

2

=

E(SV

=

2,530,000,000 − (50,000)

=

30,000,000

2

Table 11.3 provides the results of applying the expected values and variances for the annual savings and salvage value to obtain the statistical parameters for the SMP investment’s present worth. As noted, the expected present worth is $87,649.62, and the variance of present worth is 334,461,261.82. Notice, the

expected cash flow in the 10th year equals the sum of the expected annual savings and the expected salvage value; likewise, the variance of the cash flow in the 10th year equals the sum of the annual savings variance and the salvage value variance. TABLE 11.3 Computing the Expected Value and Variance for the Present Worth of the SMP Investment, Plus the Probability of the Present Worth Being Greater Than Zero E(A) = $92,500 E(SV) = $50,000 V(A) = 81,400,001.7 SD(A) $9,022.19 = EOY(t)

E(CF)

V(SV) = SD(SV) = (1.10)−t E(CF) (1.10)−t

30,000,000 $5,477.23

V(CF)

(1.10)−2t

V(CF) (1.10)−2t

 0

−$500,000

1.0000

− $500,000.00

0.0

1.0000

0.00

 1  2

$92,500 $92,500

0.9091 0.8264

$84,090.91 $76,446.28

81,400,001.7 81,400,001.7

0.8264 0.6830

67,272,728.68 55,597,296.43

 3

$92,500

0.7513

$69,496.62

81,400,001.7

0.5645

45,948,178.87

 4

$92,500

0.6830

$63,178.74

81,400,001.7

0.4665

37,973,701.54

 5  6

$92,500 $92,500

0.6209 0.5645

$57,435.22 $52,213.84

81,400,001.7 81,400,001.7

0.3855 0.3186

31,383,224.41 25,936,549.10

 7

$92,500

0.5132

$47,467.13

81,400,001.7

0.2633

21,435,164.55

 8  9

$92,500 $92,500

0.4665 0.4241

$43,151.93 $39,229.03

81,400,001.7 81,400,001.7

0.2176 0.1799

17,715,012.02 14,640,505.80

10

$142,500

0.3855

$54,939.92

111,400,001.7

0.1486

16,558,900.41

E(PW) = $87,649.62

V(PW) = 334,461,261.82 SD(PW) = $18,288.28 Pr(PW > 0)* = 0.999999178

*Central Limit Theorem approximation Excel® Data File

EXAMPLE 11.7 Risk Analysis for the Scream Machine Investment Recall the example involving two design alternatives (A and B) for a new ride (the Scream Machine) in a Florida theme park. Design A has an initial cost of $300,000 and net annual after-tax revenue of $55,000; Design B has an initial investment of $450,000 and net annual after-tax revenue of $80,000. A 10% MARR was used over the 10-year planning horizon. Now, suppose annual revenue for each machine is a statistically independent random variable. Further, suppose annual revenue for Design A is normally distributed with a mean of $55,000 and a standard deviation of $5,000; annual revenue for Design B is normally distributed with a mean of $80,000 and a standard deviation of $7,500. What is the probability of each design having a positive present worth? What is the probability of Design B having a greater present worth than Design A? Key Data Given MARR = 10%; planning horizon = 10 years Design A: Initial cost = $300,000; Annual Revenue is normally distributed with mean = $55,000 & standard deviation = $5,000 Design B: Initial cost = $450,000; Annual Revenue is normally distributed with mean = $80,000 & standard deviation = $7,500 Find Pr[PW(A) > 0], Pr[PW(B) > 0], Pr[PW(B) > PW(A)] Solution Because the annual revenues are statistically independent, normally distributed random variables, present worth will be normally distributed. The expected present worth for each design will be the same as in Chapter 4: $37,951.19 for Design A and $41,565.37 for Design B. Calculations for the variance and standard deviation for present worth for each design are summarized in Table 11.4. Also shown is the variance for the difference in cash flows, (B − A). Notice, to determine which of the two designs is more profitable, we take advantage of Pr[PW(B) > PW(A)] being the same as Pr[PW(B − A) > 0].

TABLE 11.4 Risk Analysis for the Selection of the Scream Machine Design EOY(t) (1.10)−2t V(At)

V(At) (1.10)−2t V(Bt)

V(Bt)(1.10)−2t V(Bt − At) V(Bt − At) (1.10)−2t

 0  1

1.0000 0 0.0000 0 0.0000 0.8264 25,000,000 20661157.0248 56,250,000 46487603.3058 81,250,000

67148760.3306

 2

0.6830 25,000,000 17075336.3841 56,250,000 38419506.8643 81,250,000

55494843.2484

 3  4

0.5645 25,000,000 14111848.2513 56,250,000 31751658.5655 81,250,000 0.4665 25,000,000 11662684.5052 56,250,000 26241040.1368 81,250,000

45863506.8169 37903724.6420

 5

0.3855 25,000,000 9638582.2357 56,250,000 21686810.0304 81,250,000

31325392.2661

 6  7

0.3186 25,000,000 7965770.4428 56,250,000 17922983.4962 81,250,000 0.2633 25,000,000 6583281.3577 56,250,000 14812383.0547 81,250,000

25888753.9390 21395664.4124

 8

0.2176 25,000,000 5440728.3948 56,250,000 12241638.8882 81,250,000

17682367.2829

 9

0.1799 25,000,000 4496469.7477 56,250,000 10117056.9324 81,250,000

14613526.6801

10

0.1486 25,000,000 3716090.7006 56,250,000 8361204.0764 81,250,000 12077294.7770 V[PW(A)] = 101351949.0447 V[PW(B − A)] = 329393834.3954 SD[PW(A)] = $10,067.37

SD[PW(B − A)] = $18,149.21

Pr[PW(A) > 0]* = 0.999918

Pr[PW(B − A) > 0]* = 0.578922 V[PW(B)] = 228041885.3507 SD[PW(B)] = $15,101.06 Pr[PW(B) > 0]* = 0.997043

*Central Limit Theorem approximations Excel® Data File Recall, the variance of the difference in two statistically independent random variables is the sum of the variances of the two random variables. Therefore, the variance of the difference in annual revenues equals the sum of (5,000)2 for Design A and (7,500)2 for Design B, or 81,250,000. Notice, the probability of Design A having a positive present worth is 0.999918, the probability of Design B having a positive present worth is 0.997043, and the probability of Design B having a present worth greater than the present worth for Design A is 0.578922. Hence, the probability of Design A being the best choice economically is 0.421078, or there is a 58% chance that Design B is best and a 42% chance that Design A is best. The probabilities were calculated using the Excel® NORM.S.DIST function as follows:

Pr[PW(A) > 0]

=NORM.S.DIST(37951.19/10067.37,TRUE) = 0.999918

Pr[PW(B) > 0]

=NORM.S.DIST(41565.37/15101.06,TRUE) = 0.997043

Pr[PW(B − A) > 0]

=NORM.S.DIST(3614.18/18149.21,TRUE) = 0.578922

For Example 11.6 and Example 11.7, we solved for the variance of present worth using a solution procedure that applies for general distributions of statistically independent annual cash flows. However, because the annual cash flows in both examples were also identically distributed, we could have obtained the variance of present worth using the following equations (1 + i)

2n

− 1)

V (PW) = V (A) [ (1 + i)

2n

(11.6)

1 ][ (1 + i)

2

]

where V(A) is the variance of the annual cash flows. Alternately, V (PW) = V (A) (P|A i%,2n) (A|Fi%,2)

(11.7)

Using Excel® financial functions, V (PW) =PV (i%,2n,V (A)) *PMT (i%,2,,1)

(11.8)

For Example 11.6, because salvage value and annual cash flows are statistically independent, V(PW) is equal to the sum of the variance of present worth for the annual cash flows and the variance of present worth for the salvage value. Because V(A) equals 81,400,001.7, from Equation 11.8, V[PW(A)] for the annual cash flows is V (PW (A)) =PV (10%,20,81400001.7) *PMT (10%,2,,1) = 330,001,952.98.

For salvage value, V[PW(SV)] = V(SV)(P|Fi%,2n,1) = 30,000,000/1.1020 = 4,459,308.84. Therefore, V(PW) = 330,001,952.98 + 4,459,308.84 = 334,461,261.82, as shown in Table 11.3. For Example 11.7, the annual cash flows for designs A and B are statistically independent and identically distributed, with the variance of the cash flows for Design A equal to 25,000,000 and the variance of the cash flows for Design B equal to 56,250,000. Therefore, the variance for the difference in cash flows for designs A and B equal the sum of the variances, or 81,250,000. Therefore, from Equation 11.8, the variances of present worth for Design A, Design B, and the difference in their cash flows is V [PW (A)] =PV (10%,20,25000000) *PMT (10%,2,1) = 101,351,949.0447, V [PW (B)] =PV (10%,20,56250000) *PMT (10%,2,1) = 228,031,885.3507,

and V [PW (B − A)] =PV (10%,20,81250000) *PMT (10%,2,1) = 329,393,834.3954,

as shown in Table 11.4. For some problems at the end of the chapter, using Equations 6, 7, or 8 can simplify calculating the variance for present worth.

11.3.2 Simulation Solutions2

Some of the major reasons for using simulation in risk analysis include the following: 1. Except for the simplest problems, analytical solutions are difficult to obtain. 2. Simulation is useful in selling a system modification to management. 3. Simulation can be used as a verification of analytical solutions. 4. Simulation is very versatile. 5. Less background in mathematical analysis and probability theory is generally required. Some of the major disadvantages of simulation are the following: 1. Simulation models can be quite time-consuming to formulate. 2. Simulations introduce a source of randomness not present in analytical solutions (sampling error). 3. Monte Carlo simulations do not reproduce the input distribution exactly (especially the tails of the distribution).3 4. Validation of the simulation model is easily overlooked. 5. Simulation is so easily applied that it is often used when analytical solutions can be more easily obtained at considerably less cost. Computer software is available to support simulation analyses. Among the vast array of options, we have found Pallisade Corporation’s Excel®-based software, @RISK, to be easy to use. In addition, Microsoft’s VBA (Visual Basic for Applications) software language can be used with Excel® to perform simulations of engineering economic investments. @RISK software includes the option of using either Monte Carlo simulation or Latin Hypercube simulation. Practically every known probability distribution is included in @RISK’s menu of input distributions. Among the discrete probability mass functions available are the binomial, discrete, discrete uniform (rectangular), hypergeometric, negative binomial, and Poisson distributions; the continuous probability density functions include, among others, the beta, chi square, Erlang, gamma, geometric, lognormal, normal, Pareto, PERT, triangular, uniform, and Weibull distributions. The following examples illustrate the use of @RISK 7.0 simulation software in performing engineering economic analyses.

EXAMPLE 11.8 Monte Carlo Simulation with Excel® and @Risk for the SMP Investment Recall, in Example 11.6, we performed an analytical analysis of the SMP investment, with annual savings and salvage value assumed to be statistically independent random variables. The probability distributions used are given in Table 11.5. The analytical solution yielded an expected present worth of $87,649.62, a variance for present worth of 334,461,261.82, and a 0.99999918 probability of a positive-valued present worth. TABLE 11.5 Probability Distributions for Example 11.8 Annual Savings Probability Salvage Value Probability  $75,000 0.070 $40,000 0.1  $80,000

0.095

$45,000

0.2

 $85,000

0.131

$50,000

0.4

 $90,000

0.178

$55,000

0.2

 $95,000

0.183

$60,000

0.1

$100,000

0.181

$105,000

0.162 Excel® Data File

We now perform a Monte Carlo simulation using @RISK; 100,000 simulated investments yielded an average present worth of $87,677.55, a variance of 336,070,498, and an estimate of 0.99999914 (based on the Central Limit Theorem) for the probability of a positive-valued present worth. The number of present worths that were greater than 0 totaled 100,000; not once was the present worth less than 0. Based on our analytical solution, the expected number of instances of present worth being less than 0 is (1.0 − 0.999999178)(100,000), or 0.0822. The histogram obtained from the Monte Carlo simulation is provided in Figure 11.9.4

FIGURE 11.9 PW Histogram from 100,000 Simulation Trials in Example 11.8. (The Figure was Generated with the help of @RISK, a Product of Palisade Corporation, Ithaca, NY; www.palisade.com.) Excel® Data File Because we can only determine the internal rate of return numerically, we did not consider the distribution of internal rate of return in Example 11.6. However, with @Risk software and the Excel® IRR function, we can obtain an approximation of the distribution of IRR, as well as an estimate of the probability of IRR being less than the MARR. (Because Pr(IRR < MARR) = Pr(PW < 0), we already know how many times IRR is less than 10% during the 100,000 simulation trials—zero!) The Monte Carlo simulation yielded an estimate of 13.80126% for the expected internal rate of return for the SMP investment when annual savings and salvage value were statistically independent random variables. The sample standard deviation obtained for IRR was 0.7926308%. As noted, all simulated investments yielded an IRR value greater than or equal to the MARR. The histogram obtained from the Monte Carlo simulation is provided in Figure 11.10.

FIGURE 11.10 IRR Histogram from 100,000 Simulation Trials in Example 11.8. (The Figure was Generated with the Help of @RISK, a Product of Palisade Corporation, Ithaca, NY; www.palisade.com.)

Excel® Data File

EXAMPLE 11.9 Monte Carlo Simulation with Excel® and @Risk for the Scream Machine Investment In Example 11.7, we presented analytical results for the two design alternatives being considered for the Scream Machine. Specifically, we estimated the probabilities of Design A being profitable, of Design B being profitable, and of Design B having a greater present worth than Design A. Here, we duplicate the analysis but with Monte Carlo simulation, and we assume net annual revenue produced by the new ride is a statistically independent random variable. Because the revenue in any given year is independent of the revenue in any other year, as shown in Figure 11.11, the Excel® NPV function is used to calculate present worth. The histograms for PW(A), PW(B), and PW(B − A) resulting from the 100,000 simulated investments are given in Figure 11.12. Of the 100,000 simulation trials, all but 12 resulted in a positive-valued present worth for Design A, all but 280 resulted in a positive-valued present worth for Design B, and 57,877 trials resulted in Design B having a greater present worth than Design A. Hence, the probability of Design B being the most economic is estimated to be equal to 0.57877.

FIGURE 11.11 Setup for @RISK Monte Carlo Simulation of Example 11.9. (The Figure was Generated with the Help of @RISK, a Product of Palisade Corporation, Ithaca, NY; www.palisade.com.) Excel® Data File

FIGURE 11.12a PW Histogram from 100,000 Simulated Investments in Design A Excel® Data File

FIGURE 11.12b PW Histogram from 100,000 Simulated Investments in Design B Excel® Data File

FIGURE 11.12c PW Histogram from 100,000 Simulated Incremental Investments Between Designs B and A. (The figures were generated with the help of @RISK, a product of Palisade Corporation, Ithaca, NY; www.palisade.com.) Excel® Data File Recall, solving analytically for the mean and variance for PW(A), PW(B), and PW(B − A) yielded probability estimates of positive-valued present worths equal to 0.999918, 0.997043, and 0.578922, respectively, compared with the simulated results of 0.99988, 0.99720, and 0.57877, respectively. Also, the Monte Carlo simulation produced average present worths of $37,963.06 for Design A, $41,564.11 for Design B, and $3,601.05 for the difference in Designs B and A. From our analytical solution, the exact expected values are $37,951.19 for Design A, $41,565.37 for Design B, and $3,614.18 for the difference in Designs B and A. The Monte Carlo simulation results produced slightly smaller estimates for the probabilities of a positivevalued present worth for Design A and of Design B being more profitable than Design A, as well as the expected values for present worth of Design B and the difference in Designs B and A. However, the differences in the results obtained analytically and with simulation are, for all practical purposes, negligible.

Concept Check 11.03-CC001 The fact that we can expect the present worth of statistically independent annual cash flows to be distributed approximately according to a normal distribution is based on which of the following? a. The Central Limit Theorem b. The Law of Large Numbers c. Descartes’ rule of signs d. Chebyshev’s inequality Correct or Incorrect? Clear

  Check Answer

Concept Check 11.03-CC002 When comparing simulation solutions to analytical solutions, which of the following are considered advantages of simulation solutions? I. Less randomness II. Useful in selling solutions to management III. Very versatile a. I and II only b. I and III only c. II and III only d. I, II, and III Correct or Incorrect? Clear

  Check Answer

Notes 1. A basic understanding of probability theory and Monte Carlo simulation is assumed for this section. 2. A number of other analytical solutions and a more extensive discussion of simulation solutions are presented in White, J. A., K. E. Case, and D. B. Pratt, Principles of Engineering Economic Analysis, 6th edition, John Wiley & Sons, Inc., NY, 2012. 3. Latin hypercube simulation, included in Pallisade Decision Tools 7.5.2 @RISK software, uses stratified sampling of the input distributions to force sampling across the entire range of values of the random variables. It incorporates “sampling without replacement,” because only one sample is drawn randomly from a stratification or stratum. 4. Many of the figures and tables in this section were generated with the help of @RISK 7.0, a software product of Palisade Corporation, Ithaca, NY: www.palisade.com.

Chapter 12 Capital Budgeting

Chapter 12 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced e-text and WileyPLUS

FE-Like Problems 12-FE001 Sarah is considering two investment proposals. Proposal A is to purchase a new computer. Proposal B is to purchase a new printer. She will not buy the printer unless she buys the computer. The relationship between Proposals A and B is best described by which of the following? a. B and A are mutually exclusive b. B is contingent on A c. A is contingent on B d. Not enough information is given to determine a relationship 12-FE002 Consider a capital budgeting formulation where the binary variables x1 and x2 are used to represent the acceptance (xi = 1) or rejection (xi = 0) of each alternative. A mutual exclusivity constraint between the two alternatives can be represented by which of the following? a. x1 + x2 ≤ 1 b. x2 ≤ x1 c. x1 + x2 ≥ 1 d. x1 ≤ x2 Correct or Incorrect? Clear

  Check Answer

12-FE003 Consider a capital budgeting formulation where the binary variables x1 and x2 are used to represent the acceptance (xi = 1) or rejection (xi

= 0) of each alternative. The requirement that x2 is contingent upon x1 can be represented by which of the following? a. x1 + x2 ≤ 1 b. x2 ≤ x1 c. x1 + x2 ≥ 1 d. x1 ≤ x2 12-FE004 Consider a capital budgeting formulation where the binary variables x1 and x2 are used to represent the acceptance (xi = 1) or rejection (xi = 0) of each alternative. The requirement that the null alternative is not feasible can be represented by which of the following? a. x1 + x2 ≤ 1 b. x2 ≤ x1 c. x1 + x2 ≥ 1 d. x1 ≤ x2 Correct or Incorrect? Clear

  Check Answer

12-FE005 Sebastian is about to compare a set of mutually exclusive and indivisible alternatives using a ranking approach. Which of the following is not an appropriate measure of worth? a. Present worth b. Annual worth c. Future worth d. Internal rate of return

12-FE006 If 6 investment proposals are under consideration, how many investment combinations must be evaluated if a complete enumeration approach is being used? a. 6 b. 2*6 = 12 c. 62 = 36 d. 26 = 64 Correct or Incorrect? Clear

  Check Answer

12-FE007 Which of the following is not an approach that can be used to perform a capital budgeting economic analysis? a. Box-Jenkins algorithm b. Excel® SOLVER c. Exhaustive enumeration d. Lorie-Savage formulation 12-FE008 To determine an optimal portfolio of investments when the available choices are divisible, the investment choices should first be ranked in increasing order based on which of the following? a. FW b. Initial investment c. IRR d. PW Correct or Incorrect? Clear

  Check Answer

12-FE009 Consider the following binary linear programming formulation of a capital budgeting problem.

Max s.t.

1, 200x1 + 600x2 + 1, 650x4 15, 000x1 + 20, 000x2 + 25, 000x3 + 30, 000x4 ≤ 70, 000 x1 + x2 ≤ 1 x4 ≤ x3 x1 , x2 , x3 , x4 = (0, 1)

The first cost of project x3 is a. $70,000 b. $25,000 c. $950 d. $x4 12-FE010 Consider the following binary linear programming formulation of a capital budgeting problem. Max s.t.

1, 200x1 + 600x2 + 1, 650x4 15, 000x1 + 20, 000x2 + 25, 000x3 + 30, 000x4 ≤ 70, 000 x1 + x2 ≤ 1 x4 ≤ x3 x1 , x2 , x3 , x4 = (0, 1)

Projects x3 and x4 are a. Mutually exclusive b. x3 is contingent on x4 c. x4 is contingent on x3 d. Not related Correct or Incorrect? Clear

  Check Answer

12-FE011 Consider the following binary linear programming formulation of a capital budgeting problem.

Max s.t.

1, 200x1 + 600x2 + 1, 650x4 15, 000x1 + 20, 000x2 + 25, 000x3 + 30, 000x4 ≤ 70, 000 x1 + x2 ≤ 1 x4 ≤ x3 x1 , x2 , x3 , x4 = (0, 1)

The capital budget limit is a. $90,000 b. $70,000 c. $30,000 d. $4,400

Problems Section 12.2 Capital Budgeting Problem with Indivisible Investments LEARNING OBJECTIVE 12.2 Solve the capital budgeting problem with independent, indivisible investments as a binary linear programming problem. 12.02-PR001 True or False: In solving a classical capital budgeting problem using binary linear programming (BLP), the objective function can be either the sum of present worths or the sum of annual worths without affecting the optimum investment portfolio. 12.02-PR002 Aerotron Radio Inc. has $250,000 available and its engineering staff has proposed the following indivisible investments. With each, Aerotron can exit at the end of its planning horizon of 5 years and have its initial investment returned. In addition, each year Aerotron will receive the annual return shown below. MARR is 12%. Investment Initial Investment Annual Return 1

 $75,000

$10,800

2

 $65,000

$12,000

Investment Initial Investment Annual Return 3

 $50,000

 $7,500

4

 $80,000

$13,750

5

$100,000

$15,750

For the original problem: a. Which investments should Aerotron select for the optimum portfolio? b. What is the present worth for the optimum investment portfolio? c. What is the IRR for the optimum investment portfolio? In addition to the original problem statement, let investments 1 and 4 be mutually exclusive and investment 3 be contingent on investment 2: d. Now, which alternatives should Aerotron select? e. What is the present worth for the optimum investment portfolio? f. What is the IRR for the optimum investment portfolio? Consider the original problem: g. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) a MARR of 14.4%, and (3) a MARR of 9.6%. 12.02-PR003 Polaris Industries has $1,250,000 available for additional innovations on the Victory Vision motorcycle. These include the 5 indivisible, equal-lived alternatives, each of which guarantees the investment can be exited after 6 years with the initial investment returned. In addition, each year Polaris will receive an annual return as noted below. MARR is 15%. Investment Initial Investment Annual Return 1 $350,000  $90,000 2 $300,000  $85,000

Investment Initial Investment Annual Return 3 $250,000  $75,000 4 $500,000 $130,000 5

$400,000

$115,000

For the original problem: a. Which alternatives should Polaris select for the optimum portfolio? b. What is the present worth for the optimum investment portfolio? c. What is the IRR for the portfolio? In addition to the original problem statement, Polaris has noted that investments 1, 2, and 4 are mutually exclusive, and marketing believes at least 3 investments must be made. d. Which alternatives should now be selected? e. What is the present worth for the optimum investment portfolio? f. What is the IRR for the optimum investment portfolio? Return to the original problem statement: g. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) a MARR of 18%, and (3) a MARR of 12%. 12.02-PR004 CustomMetalworks is considering the expansion of their cable fabrication business for towers, rigging, winches, and many other uses. They have available $250,000 for investment and have identified the following indivisible alternatives, each of which will provide an exit with full return of the investment at the end of a 5-year planning horizon. Each year, CustomMetalworks will receive an annual return as noted below. MARR is 12%.

Investment Initial Investment Annual Return 1  $25,000  $7,500 2 3

 $40,000  $85,000

$12,000 $20,000

4 5

$100,000  $65,000

$22,000 $17,000

For the original problem: a. Which alternatives should be selected by CustomMetalworks? b. What is the present worth for the optimum investment portfolio? c. What is the IRR for the optimum investment portfolio? In addition to the original opportunity statement, CustomMetalworks has determined that investments 3 and 4 are mutually exclusive and investment 5 is contingent on either investment 1 or 2 being funded. d. Now, which alternatives should be selected? e. What is the present worth for the optimum investment portfolio? f. What is the IRR for the optimum investment portfolio? Reconsider the original problem: g. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) a MARR of 14.4%, and (3) a MARR of 9.6%. 12.02-PR005 Video Solution Gymnastics4Life is a high-end facility for beginning, intermediate, and elite gymnasts. The latter are drawn from the nearby region for exclusive and dedicated training. In order to maintain their edge, G4L trustees wish to invest up to $350,000 in new methods for critical evaluation and training and are considering the following

independent, indivisible, investments, each of which guarantees return of the initial investment at the end of a planning horizon of 7 years. In addition, G4L will receive annual returns as noted below. MARR is 12%. Investment Initial Investment Annual Return 1 2 3

$150,000 $130,000 $100,000

$24,000 $22,000 $15,000

4 5

$160,000 $200,000

$25,000 $30,000

For the original problem: a. Which alternatives should G4L select to form the optimum portfolio? b. What is the present worth for the optimum portfolio? c. What is the IRR for the optimum investment portfolio?

During review, the G4L trustees judge investments 2 and 5 to be considered as mutually exclusive. d. Which alternatives should now be selected? e. What is the present worth for G4L’s new optimum investment portfolio? f. What is the IRR for the portfolio?

Consider the original problem:

g. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) a MARR of 14.4%, and (3) a MARR of 9.6%.

12.02-PR006 Yaesu America wishes to enhance their already fine line of electronic equipment for commercial and individual use. Their engineering staff has proposed 5 independent, indivisible, equal-lived investments, cutting across different product lines, with each estimated to return the initial investment if it is exited after a planning horizon of 5 years. In addition, each year, Yaesu is projected to receive an annual return as noted below. They have available $1,250,000 to invest and their MARR is 10%. Investment Initial Investment Annual Return 1 $400,000 $50,000 2 3

$300,000 $200,000

$36,000 $25,000

4 5

$600,000 $500,000

$69,000 $55,000

For the original problem: a. Which alternatives should Yaesu America select as optimal? b. What is the present worth for the selected portfolio? c. What is the IRR for the optimum set of investments? In addition to the original problem statement, Yaesu America has noted that investment 4 is contingent on investment 2. d. Now, which alternatives should be selected? e. What is the present worth for the portfolio?

f. What is the IRR for the portfolio? Consider the original opportunity statement: g. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) a MARR of 12%, and (3) a MARR of 8%. 12.02-PR007 Suppose your own consulting firm has been doing well and you believe it is time to make a move to add a new, related area of engineering services. To do so, you have identified the following 5 independent, indivisible, equal-lived investments, each of which guarantees you can exit it after 4 years and have your initial investment returned to you. Each year, you receive an annual return as noted below. Your MARR is 10% and you have $250,000 to invest. Investment Initial Investment Annual Return 1 2 3

 $45,000  $60,000  $85,000

 $4,000  $7,000  $9,000

4 5

$100,000  $75,000

$12,000 $11,000

For the original problem: a. Which alternatives should you select to form the optimum portfolio? b. What is the present worth of your selected portfolio? c. What is the IRR for the optimum portfolio? In addition to the original problem statement, you now believe that investments 4 and 5 should be considered mutually exclusive. d. Which alternatives should you now select?

e. What is the present worth for this portfolio? f. What is the IRR now? Reconsider the original problem: g. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) a MARR of 12%, and (3) a MARR of 8%. 12.02-PR008 A laboratory within Bayer is considering the five indivisible investment proposals below to further upgrade their diagnostic capabilities to ensure continued leadership and state-of-the-art performance. The laboratory uses a 10-year planning horizon, has a MARR of 10%, and a capital limit of $1,000,000. Investment 1

Initial Investment $300,000

Annual Receipts $205,000

Annual Disbursements $125,000

Salvage Value $50,000

2 3

$400,000 $450,000

$230,000 $245,000

$130,000 $140,000

$50,000 $60,000

4

$500,000

$260,000

$135,000

$75,000

5

$600,000

$290,000

$150,000

$75,000

For the original opportunity statement: a. Which alternatives should be selected to form the optimum portfolio for the lab? b. What is the present worth for the optimum investment portfolio? c. What is the IRR for the portfolio? In addition to the original opportunity statement, Bayer declares that investments 2 and 4 are mutually exclusive, investment 5 is contingent on 2 being funded, and at least two investments must be made.

d. Now, which alternatives should be selected by Bayer? e. What is the present worth for the resulting investment portfolio? f. What is the resulting IRR? Again consider the original opportunity statement: g. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) a MARR of 12%, and (3) a MARR of 8%. 12.02-PR009 A division of Conoco-Phillips is involved in their periodic capital budgeting activity and the engineering and operations staffs have identified 10 indivisible investments with cash flow parameters shown below. Conoco-Phillips uses a 10-year planning horizon and a MARR of 10% in evaluating such investments. The division’s capital limit for this budgeting cycle is $2,500,000. Investment Initial Investment Annual Return Salvage Value  1

$150,000

 $35,000

 $25,000

 2  3

$200,000 $225,000

 $38,000  $45,000

 $50,000  $22,500

 4

$275,000

 $60,000

 $27,500

 5  6

$350,000 $400,000

 $75,000  $95,000

 $55,000  $75,000

 7

$475,000

$110,000

 $50,000

 8

$500,000

 $85,000

$100,000

 9 10

$550,000 $600,000

$120,000 $125,000

 $75,000  $75,000

For the original problem statement: a. Which alternatives should Conoco-Phillips select?

b. What is the present worth of the optimum portfolio? c. What is the IRR for the portfolio? In addition to the original problem statement, Conoco-Phillips has noted that investments 1 and 3 are mutually exclusive, investment 4 is contingent on either investment 2 or investment 5 being funded, and at least 5 investments must be made. d. Which alternatives should now be selected? e. What is the present worth for the new portfolio? f. What is the IRR for the investment portfolio? Consider the original problem: g. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) a MARR of 12%, and (3) a MARR of 8%. 12.02-PR010 A lending firm is considering six independent and indivisible investment alternatives which, at any time the firm chooses, can be exited with a full refund of the initial investment. A total of $200,000 is available for investment, and the MARR is 10% (Note! There is no planning horizon specified, so the firm can choose any number of years it wishes—the optimum portfolio and the IRR will remain the same because the initial investment and the salvage value are the same, and the annual returns are constant each year.). Alternative Initial Investment Annual Return 1 $25,000 $2,600 2

$35,000

$3,750

3

$30,000

$3,050

4 5

$40,000 $60,000

$4,775 $6,750

6

$50,000

$5,850

For the original problem: a. Which alternatives should the lending firm select as optimal? b. What is the present worth for the optimum portfolio? c. What is the IRR for the portfolio? Several possible constraints have been identified for additional analysis by the lending firm. Determine (1) the optimum investment portfolio, (2) the present worth, and (3) the IRR when: d. Investments 4 and 5 are mutually exclusive. e. Investment 1 is contingent on investment 2 being pursued. f. Exactly four investments must be pursued. g. All of the constraints d, e, and f are considered simultaneously. Reconsider the original problem: h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. i. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) a MARR of 12%, and (3) a MARR of 8%. 12.02-PR011 Rex Electric has decided to move into low-rise (2–8 floors) commercial building electrical wiring. After great success in upscale residential and small commercial wiring, they have identified four independent and indivisible investments, any or all of which will help make the move to the next level. Rex Electric’s MARR is 10%, and $500,000 is available for investment immediately, with $175,000 available for follow-up investment the next year. The cash flows are shown below, in thousands of dollars. EOY CF(1) CF(2) CF(3) CF(4) 0 1

−$50 −$125 −$200 −$250 −$100 −$75

$50

$75

EOY CF(1) CF(2) CF(3) CF(4) 2 $50 $70 $50 $75 3

$50

$70

$50

$75

4 5

$50 $50

$70 $70

$50 $75

$75 $75

6

$50

$70

$75

$85

7

$50

$70

$75

$85

8 9

$50 $50

$70 $70

$75 $75

$85 $85

10

$75 $100

$75 $100

a. Which alternatives should be selected by Rex Electric? b. What is the present worth for the selected investment portfolio? c. What is the IRR for the optimum portfolio? Using SOLVER for sensitivity analysis, d. determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current limit on investment capital, (2) plus 20% of the year 0 budget, and (3) minus 20% of the year 0 budget. e. determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) a MARR of 12%, and (3) a MARR of 8%. 12.02-PR012 Consider the following five indivisible investment alternatives which, at the end of 5 years, fully refund the initial investment. Given the annual returns shown below, $150,000 of investment capital available, and a MARR of 10%, determine the optimum investment portfolio. What are the PW and IRR for the investment portfolio? Alternative Initial Investment Annual Return A B

$75,000 $60,000

 $8,750  $7,250

C

$80,000

 $9,750

Alternative Initial Investment Annual Return D E

$55,000 $90,000

 $6,500 $10,750

12.02-PR013 Consider the six indivisible investment alternatives shown below. The planning horizon is 5 years. The MARR is 12%. $50,000 is available for investment. Alternative Initial Investment Annual Return Salvage Value A

 $8,000

$3,200

$1,000

B C

$15,000 $10,000

$4,750 $3,070

$1,750 $1,100

D

$20,000

$5,950

$2,000

E

$19,000

$5,150

$2,100

F

$12,000

$4,250

$1,200

a. Which investments should be made in order to maximize present worth? b. Solve part a when investments B and D are mutually exclusive and F is contingent on E. 12.02-PR014 Consider the six indivisible investment alternatives shown below. The planning horizon is 8 years. The MARR is 15%. $60,000 is available for investment. Alternative Initial Investment Annual Return Salvage Value M  $8,000 $3,200 $1,000 N

$15,000

$4,750

$1,750

O P

$10,000 $20,000

$3,070 $5,950

$1,100 $2,000

Q

$19,000

$5,150

$2,100

R

$12,000

$4,250

$1,200

a. Which investments should be made in order to maximize present worth?

b. Solve part a when investments N and P are mutually exclusive and R is contingent on Q. 12.02-PR015 The City of Clyde, Ohio, is using the binary linear programming (BLP) formulation of a capital budgeting problem shown below. Assume $175,000 is available for investment and all investments are indivisible; MARR = 12%; n = 10 yrs. Given the incomplete SOLVER parameter box shown, respond to parts a–f below if it is desired to obtain an optimum investment portfolio.

a. Specify the contents of Set Target Cell. b. Specify the contents for By Changing Cells. c. Specify all constraints to be added. d. Now, suppose the original problem is modified so that investments 1 and 2 are mutually exclusive. Show how you would incorporate that constraint in SOLVER.

e. Now, suppose the original problem is modified so investment 2 is contingent on investment 3 being pursued. Show how you would incorporate that constraint in SOLVER. f. Now, suppose the original problem is modified to specify that at least three and no more than four investments can be pursued. Show how you would incorporate that constraint in SOLVER. 12.02-PR016 The American Radio Relay League is using a binary linear programming (BLP) formulation to select from among several possible investments. Each investment continues for 5 years. Given the incomplete SOLVER parameter box shown, respond to parts a–f below in order to obtain the optimum investment portfolio. Assume $180,000 is available for investment and all investments are indivisible.

a. Specify the contents of Set Target Cell. b. Specify the contents for By Changing Cells. c. Specify all constraints to be added.

d. Now, suppose the original problem is modified so that investments 4 and 5 are mutually exclusive. Show how you would incorporate that constraint in SOLVER. e. Now, suppose the original problem is modified so that investment 6 is contingent on investment 5 being pursued. Show how you would incorporate that constraint in SOLVER. f. Now, suppose the original problem is modified so that exactly three investments must be pursued. Show how you would incorporate that constraint in SOLVER. Section 12.3 Capital Budgeting Problem with Divisible Investments LEARNING OBJECTIVE 12.3 Solve the capital budgeting problem with independent, divisible investments. 12.03-PR001 True or False: In solving a capital budgeting problem involving investment opportunities that are divisible (i.e., you can invest in portions of the opportunities instead of “all or nothing”), you rank the opportunities on the basis of present worth and add to the portfolio opportunities and fractions of opportunities, beginning with the largest present worth, until “the investment bucket is filled.” 12.03-PR002 True or False: In solving a capital budgeting problem involving investment opportunities that are divisible (i.e., you can invest in portions of the opportunities instead of “all or nothing”), you rank the opportunities on the basis of internal rate of return and add to the portfolio opportunities and fractions of opportunities, beginning with the largest internal rate of return, until “the investment bucket is filled.” 12.03-PR003 Aerotron Radio Inc. has $250,000 available and its engineering staff has proposed the following divisible investments. With each, Aerotron can exit at the end of its planning horizon of 5 years and have its initial investment returned. In addition, each year Aerotron will receive the annual return shown below. MARR is 12%. Investment Initial Investment Annual Return

Investment Initial Investment Annual Return 1  $75,000 $10,800 2

 $65,000

$12,000

3 4

 $50,000  $80,000

 $7,500 $13,750

5

$100,000

$15,750

a. Determine the optimum portfolio, including which investments are fully or partially (if partial, give percentage) selected. You may use Excel®; do not use SOLVER. b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER. c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current MARR, (2) a MARR of 14.4%, and (3) a MARR of 9.6%. Use Excel® and SOLVER. d. Determine the optimum investment portfolio and its PW when investments 1, 2, and 3 are divisible and investments 4 and 5 are indivisible. Use Excel® and SOLVER. 12.03-PR004 Polaris Industries has $1,250,000 available for additional innovations on the Victory Vision motorcycle. These include the five divisible, equal-lived alternatives, each of which guarantees the investment can be exited after 6 years with the initial investment returned. In addition, each year Polaris will receive an annual return as noted below. MARR is 15%. Investment Initial Investment Annual Return 1 2

$350,000 $300,000

 $90,000  $85,000

3

$250,000

 $75,000

4

$500,000

$130,000

Investment Initial Investment Annual Return 5

$400,000

$115,000

a. Determine the optimum portfolio, including which investments are fully or partially (if partial, give percentage) selected. You may use Excel®; do not use SOLVER. b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER. c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current MARR, (2) a MARR of 18%, and (3) a MARR of 12%. Use Excel® and SOLVER. d. Determine the optimum investment portfolio and its PW when investments 1, 2, and 3 are indivisible and investments 4 and 5 are divisible. Use Excel® and SOLVER. 12.03-PR005 Video Solution CustomMetalworks is considering the expansion of their cable fabrication business for towers, rigging, winches, and many other uses. They have available $250,000 for investment and have identified the following divisible alternatives, each of which will provide an exit with full return of the investment at the end of a 5-year planning horizon. Each year, CustomMetalworks will receive an annual return as noted below. MARR is 12%. Investment Initial Investment Annual Return 1 2

 $25,000  $40,000

 $7,500 $12,000

3

 $85,000

$20,000

4 5

$100,000 $65,000

$22,000 $17,000

a. Determine the optimum portfolio, including which investments are fully or partially (if partial, give percentage) selected. You may use Excel®; do

not use SOLVER. b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER. c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current MARR, (2) a MARR of 14.4%, and (3) a MARR of 9.6%. Use Excel® and SOLVER. d. Determine the optimum investment portfolio and its PW when investments 1, 2, and 5 are divisible and investments 3 and 4 are indivisible. Use Excel® and SOLVER.

12.03-PR006 Gymnastics4Life is a high-end facility for beginning, intermediate, and elite gymnasts. The latter are drawn from the nearby region for exclusive and dedicated training. In order to maintain their edge, G4L trustees wish to invest up to $350,000 in new methods for critical evaluation and training and are considering the following independent, divisible, investments, each of which guarantees return of the initial investment at the end of a planning horizon of 7 years. In addition, G4L will receive annual returns as noted below. MARR is 12%. Investment Initial Investment Annual Return 1

$150,000

$24,000

2

$130,000

$22,000

3 4

$100,000 $160,000

$15,000 $25,000

5

$200,000

$30,000

a. Determine the optimum portfolio, including which investments are fully or partially (if partial, give percentage) selected. You may use Excel®; do not use SOLVER. b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER. c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current MARR, (2) a MARR of 14.4%, and (3) a MARR of 9.6%. Use Excel® and SOLVER. d. Determine the optimum investment portfolio and its PW when investments 1 and 2 are divisible and investments 3, 4, and 5 are indivisible. Use Excel® and SOLVER. 12.03-PR007 Yaesu America wishes to enhance their already fine line of electronic equipment for commercial and individual use. Their engineering staff has proposed 5 independent, divisible, equal-lived investments, cutting across different product lines, with each estimated to return the initial investment if it is exited after a planning horizon of 5 years. In addition, each year, Yaesu is projected to receive an annual return as noted below. They have available $1,250,000 to invest and their MARR is 10%. Investment Initial Investment Annual Return 1 2

$400,000 $300,000

$50,000 $36,000

3

$200,000

$25,000

4

$600,000

$69,000

5

$500,000

$55,000

a. Determine the optimum portfolio, including which investments are fully or partially (if partial, give percentage) selected. You may use Excel®; do not use SOLVER. b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the

current limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER. c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current MARR, (2) a MARR of 12%, and (3) a MARR of 8%. Use Excel® and SOLVER. d. Determine the optimum investment portfolio and its PW when investments 2 and 4 are divisible and investments 1, 3, and 5 are indivisible. Use Excel® and SOLVER. 12.03-PR008 Suppose your own consulting firm has been doing well and you believe it is time to make a move to add a new, related area of engineering services. To do so, you have identified the following 5 independent, divisible, equal-lived investments, each of which guarantees you can exit it after 4 years and have your initial investment returned to you. Each year, you receive an annual return as noted below. Your MARR is 10% and you have $250,000 to invest. Investment Initial Investment Annual Return 1 2 3

 $45,000  $60,000  $85,000

 $4,000  $7,000  $9,000

4 5

$100,000  $75,000

$12,000 $11,000

a. Determine the optimum portfolio, including which investments are fully or partially (if partial, give percentage) selected. You may use Excel®; do not use SOLVER. b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER. c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the

current MARR, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER. d. Determine the optimum investment portfolio and its PW when investments 2 and 4 are indivisible and investments 1, 3, and 5 are divisible. Use Excel® and SOLVER. 12.03-PR009 A laboratory within Bayer is considering the five divisible investment proposals below to further upgrade their diagnostic capabilities to ensure continued leadership and state-of-the-art performance. The laboratory uses a 10-year planning horizon, has a MARR of 10%, and a capital limit of $1,000,000. Investment 1

Initial Investment $300,000

Annual Receipts $205,000

Annual Disbursements $125,000

Salvage Value $50,000

2 3 4

$400,000 $450,000 $500,000

$230,000 $245,000 $260,000

$130,000 $140,000 $135,000

$50,000 $60,000 $75,000

5

$600,000

$290,000

$150,000

$75,000

a. Determine the optimum portfolio, including which investments are fully or partially (if partial, give percentage) selected. You may use Excel®; do not use SOLVER. b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER. c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current MARR, (2) a MARR of 12%, and (3) a MARR of 8%. Use Excel® and SOLVER. d. Determine the optimum investment portfolio and its PW when all investments except investment 3 are divisible and at least 2 investments must be pursued fully or partially and no more than 3 can be pursued fully or partially. Use Excel® and SOLVER.

12.03-PR010 A division of Conoco-Phillips is involved in their periodic capital budgeting activity and the engineering and operations staffs have identified 10 divisible investments with cash flow parameters shown below. Conoco-Phillips uses a 10-year planning horizon and a MARR of 10% in evaluating such investments. The division’s capital limit for this budgeting cycle is $2,500,000. Investment Initial Investment Annual Return Salvage Value  1 $150,000  $35,000  $25,000  2 $200,000  $38,000  $50,000  3  4

$225,000 $275,000

 $45,000  $60,000

 $22,500  $27,500

 5  6  7

$350,000 $400,000 $475,000

 $75,000  $95,000 $110,000

 $55,000  $75,000  $50,000

 8  9 10

$500,000 $550,000 $600,000

 $85,000 $120,000 $125,000

$100,000  $75,000  $75,000

a. Determine the optimum portfolio, including which investments are fully or partially (if partial, give percentage) selected. You may use Excel®; do not use SOLVER. b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER. c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current MARR, (2) a MARR of 12%, and (3) a MARR of 8%. Use Excel® and SOLVER. d. Determine the optimum investment portfolio and its PW when investments 1 through 5 are divisible and investments 6 through 10 are indivisible. Use Excel® and SOLVER.

e. Determine the optimum investment portfolio and its PW when investments 1 through 5 are indivisible and investments 6 through 10 are divisible. Use Excel® and SOLVER. f. Determine the optimum investment portfolio and its PW when: investments 1, 3, and 5 are mutually exclusive; making any investment in 4 is contingent on either investment 2 or investment 5 being fully or partially funded; at least 5 investments must be made, albeit partially; investments 1 through 5 are indivisible and investments 6 through 10 are divisible. Use Excel® and SOLVER. 12.03-PR011 A lending firm is considering 6 independent and divisible investment alternatives which, at any time the firm chooses, can be exited with a full refund of the initial investment. A total of $200,000 is available for investment, and the MARR is 10% (Note! There is no planning horizon specified, so the firm can choose any number of years they wish—the optimum portfolio and the IRR will remain the same because the initial investment and the salvage value are the same, and the annual returns are constant each year.) Investment Initial Investment Annual Return 1 $25,000 $2,600 2 $35,000 $3,750 3 4 5

$30,000 $40,000 $60,000

$3,050 $4,775 $6,750

6

$50,000

$5,850

a. Determine the optimum portfolio, including which investments are fully or partially (if partial, give percentage) selected. You may use Excel®; do not use SOLVER. b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER.

c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current MARR, (2) a MARR of 12%, and (3) a MARR of 8%. Use Excel® and SOLVER. d. Determine the optimum investment portfolio and its PW when investments 1 through 3 are indivisible and investments 4 through 6 are divisible. Use Excel® and SOLVER. e. Determine the optimum investment portfolio when all of the investments are divisible, but fractional investments are limited to 0%, 25%, 50%, 75%, or 100%. Use Excel® and SOLVER. 12.03-PR012 Rex Electric has decided to move into low-rise (2–8 floors) commercial building electrical wiring. After great success in upscale residential and small commercial wiring, they have identified 4 independent and divisible investments, any or all of which will help make the move to the next level. Rex Electric’s MARR is 10%, and $500,000 is available for investment immediately, with $175,000 available for follow-up investment the next year. The cash flows, in thousands of dollars, are shown below. EOY CF(1) CF(2) CF(3) CF(4)  0 −$50 −$125 −$200 −$250  1 −$100 −$75 $50 $75  2  3  4

$50 $50 $50

$70 $70 $70

$50 $50 $50

$75 $75 $75

 5  6  7

$50 $50 $50

$70 $70 $70

$75 $75 $75

$75 $85 $85

 8  9 10

$50 $70 $50 $70 $75 $100

$75 $85 $75 $85 $75 $100

a. Determine the optimum portfolio, including which investments are fully or partially (if partial, give percentage) selected. You may use Excel®; do

not use SOLVER. b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current limit on investment capital at the end of year 0, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER. c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current MARR, (2) a MARR of 12%, and (3) a MARR of 8%. Use Excel® and SOLVER. d. Determine the optimum investment portfolio and its PW when investments 2 and 4 are indivisible, investments 1 and 3 are divisible, and only $125,000 in investment capital will be available for follow-up investment at the end of year 1. Use Excel® and SOLVER. 12.03-PR013 Consider the 6 divisible investment alternatives shown below. The planning horizon is 5 years. The MARR is 12%. $50,000 is available for investment. Alternative Initial Investment Annual Return Salvage Value A $8,000 $3,200 $1,000 B C D

$15,000 $10,000 $20,000

$4,750 $3,070 $5,950

$1,750 $1,100 $2,000

E F

$19,000 $12,000

$5,150 $4,250

$2,100 $1,200

a. What proportion of each investment is to be included in the optimum investment portfolio? b. Solve part a when full or partial investment cannot be made in more than 3 of the investments. c. Solve part a when investments D and F are mutually exclusive. 12.03-PR014 Consider the 6 divisible investment alternatives shown below. The planning horizon is 8 years. The MARR is 15%. $60,000 is available for

investment. Alternative Initial Investment Annual Return Salvage Value M  $8,000 $3,200 $1,000 N O P

$15,000 $10,000 $20,000

$4,750 $3,070 $5,950

$1,750 $1,100 $2,000

Q R

$19,000 $12,000

$5,150 $4,250

$2,100 $1,200

a. What proportion of each investment is to be included in the optimum investment portfolio? b. Solve part a when full or partial investment cannot be made in more than 3 of the investments. c. Solve part a when investments P and R are mutually exclusive.

Chapter 12 Summary and Study Guide Summary 12.1: The Classical Capital Budgeting Problem

Learning Objective 12.1: Explain the importance of capital budgeting. (Section 12.1) Capital budgeting (or capital rationing) is an important concept whereby an investment portfolio is formed from a set of economically attractive investment opportunities. The capital budgeting problem assumes an abundant number of independent investment opportunities and limited capital. The objective is then to maximize the worth of the investment given various constraints. Mathematical programming is an appropriate tool to solve the capital budgeting problem. The use of a heuristic approach to solve the problem provides the opportunity to obtain optimal or near-optimal solutions. We saw how the Excel® SOLVER tool can be used to solve such problems. There are several practical considerations relative to capital budgeting. A common one involves the timing of decision, specifically, that all capital budgeting investment decisions are not always made at the beginning of the year. New investment opportunities become available throughout the year, thus the investment decisions are made on an ongoing basis. Because of this, a company may wish to hold a portion of its capital in reserve for future investment opportunities. 12.2: Capital Budgeting Problem with Indivisible Investments

Learning Objective 12.2: Solve the capital budgeting problem with independent, indivisible investments as a binary linear programming problem. (Section 12.2) An indivisible investment must be pursued entirely or not at all. Mathematically, this investment type is considered binary or 0–1, and can be formulated as a binary linear programming (BLP) problem. The use of an optimization algorithm to solve this problem can be complicated and is beyond the scope of this textbook; however, powerful heuristic routines exist to provide optimal or near-optimal solutions. Excel® SOLVER utilizes a heuristic approach and can be used to solve this type of problem, although it should be noted that SOLVER is not guaranteed to yield an optimum solution. Constraints can be added to the capital budgeting problem when not all investment opportunities are independent. Excel® is also useful to perform sensitivity analysis, including analyzing the sensitivity of the amount of investment capital available and the MARR used on the optimum investment portfolio. 12.3: Capital Budgeting Problem with Divisible Investments

Learning Objective 12.3: Solve the capital budgeting problem with independent, divisible investments. (Section 12.3) A divisible investment may be pursued in a fractional portion. A common case of this is for certain oil and gas exploration investments involving the pooling together of resources by multiple investors. Compared to the indivisible investment, the divisible investment allows for a partial investment, thus eliminating the binary variable constraint. Additional constraints can be added when not all investment opportunities are independent and sensitivity analysis can be performed.

Important Terms and Concepts

Capital Budgeting Problem The need to choose from among a set of investments those that will be pursued, subject to a limitation on capital available for investment. Also referred to as a capital rationing problem. CAP EX The amount of money a company budgets for capital expenditures. Also referred to as Cap-X. Indivisible Investment An investment that must be pursued entirely or not at all. Divisible Investment An investment that may be pursed in a fractional portion.

Chapter 12 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 12.1: The Classical Capital Budgeting Problem LO 12.1: Explain the importance of capital budgeting. Video Lesson: Capital Budgeting Video Lesson Notes: Capital Budgeting Excel Video Lesson: SOLVER Tool Excel Video Lesson Spreadsheet: SOLVER Tool 12.2: Capital Budgeting Problem with Indivisible Investments LO 12.2: Solve the capital budgeting problem with independent, indivisible investments as a binary linear programming problem. Excel Video Lesson: SUMPRODUCT Function Excel Video Lesson Spreadsheet: SUMPRODUCT Function Excel Video Lesson: NPV Financial Function Excel Video Lesson Spreadsheet: NPV Financial Function Excel Video Lesson: IRR Financial Function Excel Video Lesson Spreadsheet: IRR Financial Function Video Example 12.1: Solving a Capital Budgeting Problem with the Excel® SOLVER Tool Video Solution: 12.02-PR005 12.3: Capital Budgeting Problem with Divisible Investments LO 12.3: Solve the capital budgeting problem with independent, divisible investments.

Video Example 12.3: Optimizing the Investment Portfolio with the Excel® SOLVER Tool When Investments Are Divisible Video Solution: 12.03-PR005 These chapter-level resources will help you with your overall understanding of the content in this chapter. Appendix A: Time Value of Money Factors Appendix B: Engineering Economic Equations Flashcards: Chapter 12 Excel Utility: TVM Factors: Table Calculator Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values Excel Utility: Monthly Payment Sensitivity Excel Utility: TVM Factors: Discrete Compounding Excel Utility: TVM Factors: Geometric Series Future Worth Excel Utility: TVM Factors: Geometric Series Present Worth Excel Data Files: Chapter 12

CHAPTER 12 Capital Budgeting LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 12.1 Explain the importance of capital budgeting. (Section 12.1) 12.2 Solve the capital budgeting problem with independent, indivisible investments as a binary linear programming problem. (Section 12.2) 12.3 Solve the capital budgeting problem with independent, divisible investments. (Section 12.3)

Engineering Economics in Practice Abbott Laboratories Founded by Dr. Wallace C. Abbott in 1888, incorporated in 1900, and headquartered in Lake Bluff, Illinois, Abbott Laboratories is a global, broad-based health care company with branded generic pharmaceuticals, medical devices, diagnostics, and nutrition products. Its chairman and CEO, Miles D. White, a mechanical engineering graduate from Stanford, noted, “Abbott is here to create value. We do so by helping the people who use our products achieve better health. And we do so by helping our shareholders achieve financial growth. 2017 was an outstanding year for our company in both respects.” 2017 marked an extraordinary year for the company, with two major acquisitions. The first was St. Jude Medical (for approximately $23.6 billion), which makes Abbott a leading medical-device innovator across the spectrum of cardiac care, also allowing them to enter the new field of neuromodulation to treat chronic pain and movement disorders. The second was Alere Inc. (for approximately $5.3 billion), a leader in rapid testing technologies. This acquisition strengthens Abbott in the area of diagnostics. Abbott has a strong commercial, manufacturing, and R&D infrastructure throughout the worlds’ largest and fastest growing markets, bringing Abbott’s corporate identity to more than three billion people worldwide. They employ approximately 90,000 people around the world. The impact of foreign operations is primarily derived from operations in Puerto Rico, Switzerland, Ireland, the Netherlands, Costa Rica, and Singapore. Abbott’s net sales in 2017 totaled $27.4 billion, up from $20.9 billion in 2016. Capital expenditures remained flat from 2016 to 2017 at $1.1 billion in support of upgrading and expanding manufacturing and research and development facilities and equipment in various segments, investments in information technology, and laboratory instruments placed with customers. As is true of health care innovators, Abbott invests heavily in R&D: $2.2 billion in 2017, up from $1.4 billion in 2016. It had more than 25 key product launches in 2017.

Health care technology is an R&D-intensive industry, it is highly competitive and it is very dynamic. Changes occur rapidly. Consequently, firms in the industry must be highly adaptable. Their very survival might depend on the speed with which an acquisition or divestiture is made, or how quickly a new product is brought to market. Leveraging R&D investments, every effort is made at Abbott to launch new products faster than the competition. Therefore, Abbott must manage its R&D portfolio carefully. Many opportunities exist for investment, but wise choices must be made. Discussion Questions 1. The acquisition of St. Jude Medical and Alere is aimed towards positioning Abbott for the future. What other benefits might be realized? 2. Abbott invests heavily in capital. How might this be affected with the two new acquisitions and what changes do you predict? 3. R&D increased significantly from 2016 to 2017 while capital expenditures remained flat during this period. Are these two occurrences random or linked? 4. Despite the fact that a capital investment limit is established for a fiscal year, capital investment considerations can expand well beyond one year. What is driving this lengthier period of time?

Introduction Thus far in the text we have addressed how to determine if an investment is fiscally attractive and which one of multiple mutually exclusive investment alternatives is the most attractive economically. In this chapter we address a different task: determining the investment portfolio when capital is limited and many economically viable investments are available that are independent, rather than mutually exclusive. This is the task of capital budgeting. Our study of capital budgeting in this chapter will focus on comparing alternatives and selecting the preferred investments (where the portfolio is developed). Due to the complexity of these analyses, we will use the Excel®

SOLVER tool to help organize the necessary data and automate the calculations for all of the Example problems.

Systematic Economic Analysis Technique 1. Identify the investment alternatives 2. Define the planning horizon 3. Specify the discount rate 4. Estimate the cash flows 5. Compare the alternatives 6. Perform supplementary analyses 7. Select the preferred investment

12.1 The Classical Capital Budgeting Problem LEARNING OBJECTIVE Explain the importance of capital budgeting. Video Lesson: Capital Budgeting Our focus in this chapter is on quantitative approaches to optimizing financial returns when capital is limited. Choosing from among a set of investments those that will be pursued, subject to a limitation on capital available for investment, is generally referred to as the capital rationing problem or the capital budgeting problem. We chose the latter because it is more commonly used in the engineering economics literature. Capital Budgeting Problem The need to choose from among a set of investments those that will be pursued, subject to a limitation on capital available for investment. Also referred to as a capital rationing problem.

What do you do when you have far more economically attractive investments than can be funded with the available investment capital? Nice problem to have, isn’t it? It is certainly better than the reverse—having more investment capital available than fiscally attractive investments. But, still, the question must be answered: How do you choose from among a set of really, really good investments? Capital budgeting provides a methodology for solving this problem. The “abundance of riches” scenario occurs far more frequently than you might imagine. Pharmaceutical, chemical, and semiconductor companies, among others, typically must make choices among investments. They must forgo making some investments that will generate returns significantly greater than their cost of capital. Indeed, companies that employ large numbers of engineers are frequently faced with deciding how to ration scarce investment capital. Otherwise, the engineers are not as effective as they should be. The amount of money a company budgets for capital expenditures (often called CAP EX or Cap-X) generally varies from year to year. It ranges from being significantly higher than annual depreciation to a level substantially below annual depreciation, depending on market conditions; and it depends on a combination of recent history and near-term future expansion plans, the condition of the overall economy, and other similar factors. However, it is not a positive sign regarding a company’s fiscal health if its capital expenditures are substantially less than its annual depreciation over a prolonged period of time. CAP EX The amount of money a company budgets for capital expenditures. Also referred to as Cap-X. Why not borrow the money necessary to make investments when the after-tax present worth will be positive after including the cost of debt service? For publicly traded companies, the stock market usually reacts negatively when a firm’s ratio of debt-to-equity capital increases significantly. Likewise, issuing additional stock to obtain equity capital is viewed negatively by shareholders, because it dilutes the fraction of the firm’s assets represented by a share of stock. In addition, rating agencies will downgrade a company when its debtto-equity ratio increases dramatically, causing the company’s cost of capital to increase and making the investment community nervous. So, the reality is, a

firm will not always be able to invest in projects that have positive after-tax present worths; choices will have to be made. Typically, companies create a hierarchy of approval levels for capital expenditures. Such practices generate what are called size gates. For example, in a corporation with multibillion-dollar sales, one might allow capital expenditures requiring less than a million dollars to be approved by the head of a division, those requiring more than a million dollars but less than $10 million to be approved by the head of a business unit within the corporation, those requiring more than $10 million but less than $30 million to be approved by the corporation’s chief financial officer, and those requiring more than $30 million to be approved by the board of directors. Although the numbers vary from company to company, size gates are frequently used in large firms. As examples, Eastman Chemical Company and Motorola Solutions use them. Where size gates occur, choices must be made, and the selection process does not occur only at the organization’s highest levels. Division heads, business heads, chief financial officers, and boards of directors frequently must choose from among attractive investment alternatives. How do they do it? That is the subject of this chapter. Excel® Video Lesson: SOLVER Tool In this chapter, we show 1. how to formulate a capital budgeting problem with independent, indivisible investments as a binary linear programming problem and how to solve (using Excel®’s SOLVER tool) reasonably sized problems by maximizing the investment portfolio’s present worth; 2. how to add mutually exclusive, contingent, “either/or” and other constraints to a formulation of a capital budgeting problem involving indivisible investments; and how to determine (using SOLVER) the investment portfolio that maximizes its present worth; 3. how to formulate a capital budgeting problem involving independent, divisible investments as a linear programming problem and how to solve (using SOLVER) reasonably sized problems by maximizing the investment portfolio’s present worth or by “filling the investment

portfolio bucket” with investments ranked in order of their internal rates of return. There are several practical considerations we should mention relative to capital budgeting. For example, even though a capital investment limit is established for a fiscal year, it is seldom the case that all prospective investments will be known or available for analysis at a particular point in time during the year. Instead of determining at the beginning of a fiscal year which of a known set of investment opportunities will be funded, it is more common to make capital investment decisions throughout the year. To ensure that the firm can fund a highly attractive investment that materializes toward the latter part of the year, some portion of the capital investment funds might be held in reserve for this purpose. Likewise, because of the uncertainty regarding new investments that might materialize, some investment decisions are postponed, but with the proviso that investments will be made if no others materialize that are more attractive financially. Despite these practicalities, the material presented on capital budgeting is valuable. At various times during a fiscal year, capital investment decisions are made. Furthermore, when they are made, they are done so in the face of multiple competing alternative uses for the investment capital. During a fiscal year, you must often answer multiple times the question, How do you choose from among a set of really, really good investments? Hopefully, you have a better idea of the considerations that should go into answering this question as well as some tools you can use to determine the optimum investment portfolio. We should note that many criticisms have been made regarding the various mathematical programming formulations of the capital budgeting problem that have been presented in research journals. Some want to incorporate in the formulation of the capital budgeting problem the ability to borrow money at stated interest rates, others want to do multiyear capital budgeting (as though they are prescient regarding future investment opportunities that will arise), still others want to incorporate probabilistic considerations, some argue for optimizing a firm’s annual cash flow, and a significant number of scholars advocate incorporating multiple criteria. While each of these preferences might be appropriate for specific applications, such formulations are beyond the scope of an introductory text. For those who wish to pursue capital

budgeting beyond the brief introduction offered in this chapter, we direct you to the vast capital budgeting and capital rationing literature.

Concept Check 12.01-CC001 Reasons why a company facing a capital rationing problem may choose not to raise additional capital to invest in all projects having a positive after-tax present worth include which of the following? I. Borrowing money increases a company’s debt-to-equity ratio II. Issuing stock dilutes shareholder value III. Cost of capital can increase as more money is borrowed a. I and II only b. I and III only c. II and III only d. I, II, and III Correct or Incorrect? Clear

  Check Answer

Concept Check 12.01-CC002 Because new investment opportunities can become available throughout a company’s fiscal year, companies frequently a. Wait until the end of the year to make capital budgeting decisions b. Reserve some capital for use later in the year c. Ignore opportunities that occur after the initial capital budgeting decisions d. Require that the alternatives considered during capital budget decision making at the beginning of the year include every possibility that may occur during the year Correct or Incorrect? Clear

  Check Answer

12.2 Capital Budgeting Problem with Indivisible Investments LEARNING OBJECTIVE Solve the capital budgeting problem with independent, indivisible investments as a binary linear programming problem. Investments are called indivisible when they must be pursued entirely or not at all. Hence, a binary decision variable (xj = 0 or 1) is used in a mathematical formulation of the capital budgeting problem to denote if investment j is to be included in the investment portfolio. Our objective is to maximize the present worth of the investment portfolio. Letting cj denote the capital investment

required for investment j and letting C denote the total amount of investment capital available, the capital budgeting problem can be formulated as follows: Indivisible Investment An investment that must be pursued entirely or not at all. Maximize PW1 x1 + PW2 x2 + … + PWn−1 xn−1 + PWn xn

(12.1)

c1 x1 + c2 x2 + … + cn−1 xn−1 + cn xn ≤ C

(12.2)

subject to

xj = (0, 1)

j = 1, …, n

(12.3)

Because xj equals 1 when the investment is included in the portfolio and equals 0 otherwise, Equation 12.1 is the present worth for the investment portfolio. The first constraint (Equation 12.2) assures that the total investment required for the portfolio is no greater than the amount of capital available. The second constraint (Equation 12.3) affirms that the decision variables are binary. The mathematical optimization problem is a binary linear programming (BLP) problem. An early heuristic solution procedure, the Lorie-Savage procedure (named for its developers, J. H. Lorie and L. J. Savage), was used to obtain, hopefully, good, if not optimal, solutions. For small-sized problems, enumeration can be used. However, we do not recommend forming all possible 2n combinations of the n investments and, for those not exceeding the capital limit, choosing the one having the greatest present worth. For problems with few investments, that might be feasible, but for a relatively small example with n equal to 10, there are 1,024 possible solutions.

12.2.1 Solving the Two-Constraint Capital Budgeting Problem For our purposes, we will solve the BLP formulation of the capital budgeting problem using the Excel® SOLVER tool. It is well suited for small-sized

problems of this type. However, we must caution that SOLVER is not guaranteed to yield an optimum solution. The search algorithm embedded in SOLVER can terminate prematurely and not produce an optimum solution.

EXAMPLE 12.1 Solving a Capital Budgeting Problem with the Excel® SOLVER Tool Video Example To illustrate using SOLVER in solving a BLP formulation of the capital budgeting problem with indivisible investments, suppose you are presented with six different “one-shot” investment opportunities with the parameters given in Figure 12.1. Your MARR is 10% and you have a 5year planning horizon. Two investments (5 and 6) terminate in fewer than 5 years. You have a total of $120,000 to invest. The investment decision for each opportunity is a binary decision. Hence, you can only invest in an opportunity once; hence, you cannot pursue investment 3 six times for a present worth of $12,149.69. The objective is to choose investment opportunities from among the six available in order to maximize present worth and not exceed the $120,000 limitation on capital.

FIGURE 12.1 SOLVER Set Up for Example 12.1 Excel® Data File Solution In Figure 12.1, for row k, k = 2, …, 7, the entry in cell Hk is obtained using the Excel® worksheet function SUMPRODUCT(Bk:Gk,B9:G9); for row k, k = 9, 10, and 11, the entry in cell Hk is obtained using the Excel® worksheet function SUM(Bk:Gk); PW in row 8 is obtained using

the Excel® NPV worksheet function; IRR in cell H13 is obtained using the Excel® IRR worksheet function; and for rows 10 and 11 multiply the values in rows 2 and 9 and in rows 8 and 9, respectively. Excel® Video Lesson: SUMPRODUCT Function Excel® Video Lesson: NPV Financial Function Excel® Video Lesson: IRR Financial Function To use the Excel® SOLVER tool, the parameter values are as shown in Figure 12.2. Two constraints are needed: one that requires the values of the decision variables to be binary valued and one that requires that the total amount invested be less than or equal to the amount of capital available. The solution obtained is shown in Figure 12.3. Specifically, $118,000 is to be invested in opportunities 1, 2, 3, 4, and 6; investment opportunity 5 is not pursued. The resulting PW is $15,930.12 and the resulting IRR for the investment portfolio is 15.21%.

FIGURE 12.2 SOLVER Parameters for Example 12.1 Excel® Data File

FIGURE 12.3 SOLVER Solution to Example 12.1 Excel® Data File

12.2.2 Addressing Additional Constraints

A nice feature of the Excel® SOLVER tool is its ability to handle additional constraints. Consider, for example, a type of constraint that is considered throughout the text—the mutually exclusive or “either, neither, but not both” constraint. This takes the form of requiring that the sum of the mutually exclusive decision variables be less than or equal to 1. In particular, suppose investment opportunities g and h are mutually exclusive. In such a case, the following constraint can be incorporated in the spreadsheet for solution using SOLVER: xg + xh ≤ 1. The constraint ensures that both xg and xh cannot equal 1, because x is a binary decision variable; hence, investment in both g and h cannot occur. Another type of constraint arises when not all investment opportunities are independent. In particular, suppose investment opportunity k cannot be pursued unless investment opportunity j is pursued; in other words, suppose k is contingent on j. In such a case, the following constraint can be added to the SOLVER parameters: xk ≤ xj. Again, because the decision variables are binary, the constraint ensures that xk cannot equal 1 if xj equals 0. In fact, the only way xk can equal 1 is for xj to equal 1. A third type of limitation that can be incorporated in the BLP formulation is the “either/or” contingent constraint. To illustrate the concept, suppose opportunity r cannot be pursued unless either opportunity s or opportunity t is pursued. (If both opportunities s and t are pursued, then opportunity r can also be pursued.) In such a case, the following constraint can be incorporated in the spreadsheet for solution using the Excel® SOLVER tool: xr ≤ xs + xt. The constraint ensures that xr cannot equal 1 unless at least one of the two opportunities (s and t) is pursued. Finally, a fourth type of limitation that the Excel® SOLVER tool can accommodate is the “at least, but not more than” constraint. For instance, suppose at least u but not more than v investments can be funded. In this case, the sum of the decision variables must be greater than or equal to u and less than or equal to v. Thus, two constraints must be added to the set of SOLVER parameters, both keying on a cell that contains the sum of the decision variables.

EXAMPLE 12.2 Incorporating Additional Constraints in the Capital Budgeting Problem with the Excel® SOLVER Tool To illustrate the addition of constraints in the capital budgeting problem, suppose in the previous example that no more than 4 investments can be pursued. Also, suppose investment opportunities 2 and 4 are mutually exclusive and investment opportunity 2 is contingent on either investment opportunity 1 or investment opportunity 3 being pursued. What is the impact on PW and IRR of these additional constraints? Solution From Figure 12.4, we find the solution obtained using the Excel® SOLVER tool. Now, investment opportunities 3 and 4 are not pursued, PW is reduced to $12,693.07, and IRR is 15.05%. The amount of investment capital required is $103,000, with the $17,000 balance earning the MARR of 10%. The Excel® SOLVER tool parameters for the constrained problem are shown in Figure 12.4. Three additional constraints are added to those used to solve Example 12.1. Specifically, a constraint is added that the value of cell D13 has to be less than or equal to 1; this satisfies the mutually exclusive constraint involving investments 2 and 4, because the value of cell D13 is the sum of the decision variables for investments 2 (C9) and 4 (E9). Another constraint requires that the value of cell D14 is greater than or equal to the value of cell C9; this satisfies the contingent requirement that investment 2 (C9) cannot be pursued if neither investment 1 (B9) nor investment 3 (D9) is pursued, because cell D14 contains the sum of the decision variables for investments 1 and 3. The third new constraint is that the sum of the decision variables (H9) is less than or equal to 4; this satisfies the constraint that no more than 4 investments be pursued.

FIGURE 12.4 SOLVER Solution for Example 12.2, with SOLVER Parameters Shown Excel® Data File Because small-sized BLP problems can be solved using the Excel® SOLVER tool, it is a relatively simple matter to perform sensitivity analysis such as described in Chapter 11. For example, one can easily determine how sensitive the optimum investment portfolio is to changes in, say, the limit on investment or changes to the MARR. These factors can simply be adjusted in small increments +/− the original value such as +/− 10%, +/− 20%, etc., so that the decision maker can gain insight into the robustness of the solution.

Concept Check 12.02-CC001 Four indivisible projects, A, B, C, and D, are being considered in a capital rationing problem. Which of the following constraints could be incorporated in an Excel® Solver formulation of the problem? I. Either project A or project B can be done but not both II. Project C cannot be done unless project A is done III. A maximum of 2 projects can be selected a. II only b. I and III only c. I, II, and III d. None of these constraints could be accommodated in Solver; these constraints would necessitate an enumeration solution Correct or Incorrect? Clear

  Check Answer

12.3 Capital Budgeting Problem with Divisible Investments LEARNING OBJECTIVE Solve the capital budgeting problem with independent, divisible investments. In the previous section, the investment opportunities were considered to be indivisible. You either invested in all of an opportunity or you did not invest in it at all. Investing in a fractional portion was not permitted. In this section, we

examine capital budgeting when, in fact, investment opportunities are divisible. Divisible Investment An investment that may be pursed in a fractional portion. Certain oil and gas exploration investments are often divisible, because it is not unusual for several individuals to get together to invest in drilling for oil or gas. Individual investors do not have to be equal. They share proportionately in the investment and in the returns. Similar divisible investment opportunities can occur when a block of stock is offered for sale; the same situation can exist for bonds, land, hotels, shopping centers, and so forth. With divisible investments, the decision variable is changed from investing fully (xj = 1) versus not investing at all (xj = 0) in investment j to investing wholly, partially, or not at all (0 ≤ pj ≤ 1) in investment j, where pj is the decision variable representing the percentage of investment j to be pursued. The capital budgeting problem with divisible investments can be formulated mathematically as follows: Maximize PW1 p1 + PW2 p2 + … + PWn−1 pn−1 + PWn pn

(12.4)

c1 p1 + c2 p2 + … + cn−1 pn−1 + cn pn ≤ C

(12.5)

subject to

0 ≤ pj ≤ 1

j = 1, …, n

(12.6)

EXAMPLE 12.3 Optimizing the Investment Portfolio with the Excel® SOLVER Tool When Investments Are Divisible Video Example To illustrate using the Excel® SOLVER tool to determine the optimum investment portfolio when investments are divisible, recall Example 12.1. The setup for a SOLVER solution is provided in Figure 12.5. The binary decision variables are replaced by fractional valued decision variables having values ranging from 0% to 100%. The SOLVER parameters also are shown in the figure. Notice, the objective function (Equation 12.4) and constraints (Equations 12.5 and 12.6) are incorporated in the SOLVER parameters. Also, notice that the constraint 0 ≤ pj ≤ 1 is implemented by using two constraints: B9:G9 ≤ 1 and B9:G9 ≥ 0. The investment portfolio obtained is to fully invest in opportunities 1, 2, 3, 4, and 6 and to partially invest (6.67%) in opportunity 5. The PW for the portfolio is $16,073.69; the IRR for the portfolio is 15.17%. Not surprisingly, all of the investment capital is used. (Why is this not surprising? Can you think of a situation in which not all of the investment capital will be used? What if not all investment opportunities have positive-valued present worths?)

FIGURE 12.5 SOLVER Parameters and Solution for Example 12.3 Excel® Data File When partial funding of investments is allowed, salvage value equals the initial investment, and annual returns are a uniform annual series, and so it is quite easy to obtain the optimum investment portfolio. Taking advantage of the special structure imposed on the investments, the optimum investment portfolio is obtained by (a) ranking the investment opportunities on their internal rates of return, and (b) forming the portfolio by “filling the investment bucket,” starting with the opportunity having the greatest internal rate of return and proceeding sequentially until the “bucket” is full. The following example illustrates the solution procedure.

EXAMPLE 12.4 Optimizing the Investment Portfolio with the Excel® SOLVER Tool When Investments Are Divisible and Have a Special Structure To illustrate using the “bucket filling” solution procedure when divisible investments have the special structure of (a) salvage value equaling the initial investment and (b) annual returns being a uniform series, consider the data for 5 investments shown in Table 12.1. Solution If $150,000 is available for investment, the order in which investments will be placed in the “investment bucket” is 1, 3, 4, 5, and 2. Because the sum of the investments is $200,000, not all 5 investments will “fit” in the “bucket.” After selecting investment 1, $135,000 remains available for investment; after adding investment 3 to the investment portfolio, $95,000 remains available; after selecting investment 4, $45,000 remains available. The next investment to be selected is investment 5. However, only $45,000 of the $70,000 investment will be pursued, filling the “bucket.” Hence, the optimum values of the decision variables are: p1 = p3 = p4 = 100%, p5 = 64.29% (45/70), and p2 = 0%. The IRR for the portfolio is $3,750 + $9,250 + $11,250 + $14,250 (45/70) = $33,410.71 divided by $150,000, or 22.27%. Exploring the Solution As an exercise, solve the example using the solution procedure from Example 12.3 and verify that using an IRR ranking procedure actually results in maximizing present worth, not IRR. To do so, you will need to select a planning horizon, such as 5 years or 10 years, and calculate the present worth for each investment.

TABLE 12.1 Characteristics of Five Investment Opportunities Investment Opportunity 1 2 3

4

5

Initial Investment

$15,000 $25,000 $40,000 $50,000 $70,000

Annual Return Salvage Value

$3,750 $5,000 $9,250 $11,250 $14,250 $15,000 $25,000 $40,000 $50,000 $70,000

Internal Rate of Return

25.00% 20.00% 23.13% 22.50% 20.36%

With divisible investments, consideration of mutually exclusive, contingency, and other constraints requires thought about what each means. For example, if investment opportunity 1 is contingent on investment opportunity 3, does that mean p1 cannot be greater than p3 or that p1 cannot be greater than zero unless p3 is greater than zero? These (and other) considerations are incorporated in end-of-chapter problems.

Concept Check 12.03-CC001 Consider the following statement: In a capital rationing problem with divisible investments, if all projects have positive present worths, all of the investment capital will be used. a. This statement is true b. This statement is false c. Not enough information is given; the numeric values of the present worths of the projects are required to know whether this statement is true or false d. Not enough information is given; the numeric values of the IRRs of the projects and the value of MARR are required to know whether this statement is true or false Correct or Incorrect? Clear

  Check Answer

CHAPTER 13 Obtaining and Estimating Cash Flows

Chapter 13 FE-Like Problems and Problems Problem available in WileyPLUS Tutoring Problem available in WileyPLUS Video Solution available in enhanced e-text and WileyPLUS

FE-Like Problems 13-FE001 On a balance sheet, a corporation’s economic obligations to nonowners are called a. Owners’ equity b. Liabilities c. Assets d. Retained earnings 13-FE002 The information below has been extracted from the books of the Shelley Company. Which of the following represents Shelley’s current ratio? Current Assets Current Liabilities Cash $86 Accounts payable $78 Accounts receivable $130 Wages payable $68 Inventory

$140 Taxes payable

Total

$356 Total

a. 0.76 b. 1.48 c. 2.05 d. 2.51 Correct or Incorrect? Clear

  Check Answer

$28 $174

13-FE003 sheet is

The fundamental equation used within an accounting balance

a. Net profit = gross profit − expenses − taxes b. Assets + liabilities = net worth c. Assets + liabilities + net worth = 0 d. Assets = liabilities + net worth 13-FE004

Marginal cost is

a. Any cost occurring after “time now” b. The ratio of total cost to total quantity of output c. The market value of an asset at the end of its life less its disposal costs d. The incremental cost of producing one more unit of output Correct or Incorrect? Clear

  Check Answer

13-FE005 The three major categories that comprise cost of goods manufactured are a. Material, labor, and overhead b. Average, marginal, and instantaneous c. Past, present, and future d. Initial, operating, and salvage 13-FE006 The total cost equation for producing X widgets is given by TC = $1,000 + $6X. The average cost per widget for producing 500 widgets is closest to which of the following? a. $1,000 b. $6 c. $8 d. $4,000

Correct or Incorrect? Clear

  Check Answer

13-FE007 The total cost equation for producing X widgets is given by TC = $1,000 + $6X. The variable cost per widget is closest to which of the following? a. $1,000 b. $6 c. $8 d. $4,000 13-FE008 The total cost equation for producing X widgets is given by TC = $1,000 + $6X. The marginal cost per widget at a production level of 300 units is closest to which of the following? a. $2,800 b. $6 c. $8 d. $4,000 Correct or Incorrect? Clear

  Check Answer

13-FE009

Fixed cost is

a. Any cost that does not vary with the quantity of output b. The ratio of total cost to total quantity of output c. The market value of an asset at the end of its life less its disposal costs d. The incremental cost of producing one more unit of output

13-FE010 When an organization considers its work-in-process, how should it be classified on a balance sheet? a. Asset b. Liability c. Net worth d. Expense Correct or Incorrect? Clear

  Check Answer

Problems Section 13.1 Cost Terminology LEARNING OBJECTIVE 13.1 Identify five unique cost viewpoints used for economic decision making. 13.01-PR001 Reconsider the scenario presented in Example 13.1, which details a firm considering purchasing a small turret lathe to manufacture part number 163H for the B&K Corporation. a. What additional assumptions/data would be required to complete an economic analysis to determine if the turret lathe should be purchased? b. What source(s) would you pursue to obtain the data you identified in part a? c. Given the facts and data presented, as well as any additional assumptions you made and data you acquired, should the turret lathe be purchased? 13.01-PR002 Match the terms in the first column with the appropriate definition in the second column. Terms (a) First cost

Definitions (1) Any cost occurring after “time now”

Terms

Definitions

(b) Average cost

(2) The total initial investment required to get an asset ready for service

(c) Fixed cost

(3) The ratio of total cost to total quantity of output

(d) Overhead cost

(4) The market value of an asset at the end of its life less its disposal costs

(e) Marginal cost

(5) The cost of obtaining funds through debt obligations

(f) Opportunity cost

(6) The incremental cost of producing one more unit of output

(g) Salvage value

(7) Any past cost or portion of past cost that is not recovered

(h) Sunk cost

(8) The recurring costs that are necessary to operate and maintain an asset (9) Any cost that varies with the quantity of output

(i) Variable cost (j) Past cost

(10) Any cost occurring prior to “time now”

(k) Maintenance costs

(11) Any cost that does not vary with the quantity of output

(l) Future cost

(12) The cost of forgoing an opportunity to earn interest on funds

(m) Cost of debt capital

(13) All costs in manufacturing other than direct material and direct labor

13.01-PR003 A firm has the capacity to produce 1,000,000 units of a product each year. At present, it is operating at 70% of capacity. The firm’s annual revenue is $700,000. Annual fixed costs are $300,000, and the variable costs are $0.50 per unit. a. What is the firm’s annual profit or loss? b. At what volume of sales does the firm break even? c. What will be the profit or loss if the plant runs at 90% of capacity assuming a constant income per unit and constant variable cost per unit? d. At what percent of capacity would the firm have to run to earn a profit of $90,000?

13.01-PR004 Two production methods are being considered for making thermal dryers. Method 1 has a fixed cost of $1,000 and a variable cost of $20 per unit. Method 2 has a fixed cost of $400 and a variable cost of $100 per unit. For what range of production volumes would you prefer each method? 13.01-PR005 The maker of Winglow is purchasing a new stamping machine. Two options are being considered, Rooney and Blair. The sales forecast for Winglow is 8,000 units for next year. If purchased, the Rooney will increase plant fixed costs by $20,000 and reduce variable costs by $5.60 per unit. The Blair will increase fixed costs by $5,000 and reduce variable costs by $4.00 per unit. If variable costs are now $20 per unit, which machine should be purchased? 13.01-PR006 Product X is sold for $500 per unit. The total cost of production per year, including capital recovery and a return, is given by the expression TC = 0.04n

3

− 700n + 50, 000

where n is the number of units sold. If TC represents the total of all fixed and variable costs, determine the following: a. The value of n that maximizes profit. b. The maximum profit for a year. c. The fixed cost per year. 13.01-PR007 Video Solution The cost curve for producing widgets passes through the following points and is piecewise linear in between. Units Produced Costs   0 200

 $600 $1,200

400 600

$1,600 $1,800

a. What is the fixed cost of producing 600 widgets?

b. What is the variable cost of producing 600 widgets? c. What is the cost per unit if only 400 widgets are produced? d. What is the incremental cost of producing the 100th widget? e. What is the incremental cost of producing the 500th widget? f. What is the fixed cost per unit for producing 1,000 widgets? g. What is the variable cost per unit for producing 1,000 widgets?

13.01-PR008 A manufacturer of precision cutting tools has the capacity to make 1,000,000 cutting tools per year. Each sells for $15. The variable cost per unit to produce the cutting tools is $9. Annual fixed costs for the manufacturer are $3,500,000. a. If the plant is operating at 50% of design capacity, how much profit (loss) is being earned? b. At what percent of capacity must the plant operate to break even? c. What is the cost per tool when operating at the level you determined in part b? 13.01-PR009 A new engineering building at a state university is to contain 10,000,000 square feet. The total cost of the building (TC) is given by the following expression: 2

TC = (200 + 80X + 2X )A

where X = number of floors and A = floor area in ft2/floor.

a. Create a table that shows the total building cost, average cost per floor, and marginal cost per floor (using the difference equation approach) for configurations ranging from 1 floor to 12 floors, inclusive. (Hint: You may want to create the table using your favorite spreadsheet program.) b. Based on your table, what is the optimal number of floors for the building? Justify your answer based on the “total cost” column. c. Justify your answer in part b based on the “marginal cost per floor” column. d. Demonstrate, using differential calculus, that your answer in parts b and c is correct. (Note: For this part, assume that X is a continuous variable.) 13.01-PR010 Video Solution The KMP Metal Machining Company produces widgets according to customer order. The company has determined that widgets can be produced on three different machine tools: M1, M2, or M3. An analysis of widget production cost reveals the following data: Machine Tool Fixed Cost/Order Variable Cost/Unit M1 $300 $9 M2 $750 $3 M3

$500

$5

a. Using a graphical approach, determine the most economical machine tool to use for all order sizes between 1 and 200 units. In other words, determine the subranges within the overall range of 1 to 200 for which each machine tool is preferred. b. Using an algebraic approach, determine the most economical machine tool to use for all order sizes between 1 and 200 units. (Hint: Determine the subranges within the overall range of 1 to 200 for which each machine tool is preferred.) c. For an order of size 75, which machine tool should be used to produce the order, and what is the total production cost?

d. For an order of size 160, assume that the preferred (most economical) machine is unavailable. What penalty (expressed in dollars of additional production cost) must be paid if the second most economical machine is used? The third?

13.01-PR011 Two processes are put in place for production. Neither will be removed. Process R is designed to produce 10,000 units per year and has a fixed cost of $90,000 per year. Process T has the same design capacity and has a fixed cost of $80,000 per year. Process R produces the initial 4,000 units at a variable cost of $8 per unit and the next 6,000 units at a variable cost of $17 per unit. Process T produces the first 5,000 units at a variable cost of $9 each and produces the next 5,000 at $5 each. Assume that the fixed costs are incurred even if no production is assigned to the process. a. What should be the loads assigned to Processes R and T if demand for the product is 5,500 units? b. What is the total cost and average cost per unit for part a? c. What should be the loads assigned to Processes R and T if demand for the product is 9,500 units? d. What is the total cost and average cost per unit for part c? 13.01-PR012 The variable cost of producing specialty value-added stainlesssteel tubing varies according to the quantity produced per year. The variable cost is $500 per ton from 0 to 700 tons. Above this level, the variable cost is different. At this time, we are producing 600 tons per year, and the average cost per ton is $800. Next year, we anticipate ramping up production to 1,000 tons, and the projected average cost for all 1,000 tons will be $710 per ton. a. What is the current fixed cost? b. If the fixed cost holds constant next year, what is the variable cost of production for tons above 700 tons? Section 13.2 Cost Estimation

LEARNING OBJECTIVE 13.2 Apply cost estimating principles to determine costs for given scenarios. 13.02-PR001 Find three Web sites where “cost estimating” is a topic. a. State the URL for each of the sites. b. Identify the one you would likely find most useful if you wanted to learn more about cost estimating. c. In 25 words or less, what does the site you identified in part b have that is most useful? 13.02-PR002 Video Solution A distributor has learned that an estimate for the packaging and processing costs can be predicted based on the weight of an order. To this end, a regression analysis is performed resulting in the mixed cost function: cost = 63.63 + 0.279 weight. a. What is the fixed cost within this cost function? b. What is the marginal cost within this cost function? c. What was the independent variable within the regression analysis? d. What was the dependent variable within the regression analysis? e. What is the estimated cost for a package weighing 600 pounds? f. The regression analysis reports an r2 of 0.9824. Interpret the meaning of this value.

13.02-PR003 For each of the situations below, identify a potential source of analogy data for the costs under each of the following cases: (1) this project/activity is a first of its kind for the company, and (2) this project/activity is similar to projects that the company has completed before. a. Construction of a high-rise in New York City. b. Construction of a two-lane bridge in a rural Oklahoma county. c. Acquisition of a punch press to be used by a manufacturing company in California. d. Hiring an information technology manager to oversee new network services. 13.02-PR004 A circuit board manufacturer is trying to determine the appropriate cost driver for estimating the cost of circuit board assembly support. The list of potential cost drivers has been narrowed to three: direct labor hours, number of circuit boards completed, and average cycle time. In preparation for a simple linear regression analysis, a data collection activity has been completed. The results of the data collection are shown in the table below. Use the data to complete the following: a. Determine a regression line for each of the three possible cost drivers. For each possible driver: i. plot a graph of the cost driver versus the support cost; ii. determine and clearly state the regression equation; and iii. determine and clearly state the associated r2. b. Which driver do you recommend? c. What percentage of the data does your driver and regression equation explain? Week Circuit Board Assembly Support Cost 1 2

$66,402 $56,943

Direct Labor Hours 7,619 7,678

Number of Boards Completed 2,983 2,830

Average Cycle Time 186.44 139.14

Week Circuit Board Assembly Support Cost 3 $60,337 4 $50,096

Direct Labor Hours 7,816 7,659

Number of Boards Completed 2,413 2,221

Average Cycle Time 151.13 138.30

5 6

$64,241 $60,846

7,646 7,765

2,701 2,656

158.63 148.71

7 8 9

$43,119 $63,412 $59,283

7,685 7,962 7,793

2,495 2,128 2,127

105.85 174.02 155.30

10 11

$60,070 $53,345

7,732 7,771

2,127 2,338

162.20 142.97

12 13 14

$65,027 $58,220 $65,406

7,842 7,940 7,750

2,685 2,602 2,029

176.08 150.19 194.06

15 16

$35,268 $46,394

7,954 7,768

2,136 2,046

100.51 137.47

17 18 19

$71,877 $61,903 $50,009

7,764 7,635 7,849

2,786 2,822 2,178

197.44 164.69 141.95

20

$49,327

7,869

2,244

123.37

21 22

$44,703 $45,582

7,576 7,557

2,195 2,370

128.25 106.16

23

$43,818

7,569

2,016

131.41

24

$62,122

7,672

2,515

154.88

25

$52,403

7,653

2,942

140.07

Section 13.3 General Accounting Principles

LEARNING OBJECTIVE 13.3 Demonstrate proficiency with balance sheets, income statements, and ratio analysis. 13.03-PR001 balance sheet.

Determine the missing values below to produce a valid BALANCE SHEET Year Ending December 31, 2020

ASSETS Cash

(a)

Accounts receivable Raw materials inventory

$8,000 $10,000

Work-in-process inventory

 $15,000

Finished goods inventory Land

$18,500 $30,000

Building

$80,000

Less: Depreciation reserve

$8,000

(b)

Equipment Less: Depreciation reserve

(c) $4,000

$36,000

TOTAL ASSETS LIABILITIES Notes payable Accounts payable Declared dividends

$283,727 $25,000 (d) $20,000

TOTAL LIABILITIES NET WORTH

 $51,000

Common stock

$200,000

Retained earnings TOTAL NET WORTH

(e) (f)

BALANCE SHEET Year Ending December 31, 2020 TOTAL LIABILITIES & NET WORTH

(g)

13.03-PR002 Following is a list of balance sheet accounts for Branman Co. as of December 31, 2018. Account Cash

Balance $45,000

Accounts receivable

$8,000

Notes payable Raw materials inventory Common stock

$25,000 $10,000 $100,000

Work-in-process inventory $15,000 Accounts payable

$6,000

Account

Balance

Declared dividends payable Finished goods inventory

$20,000 $18,500

Land

$30,000

Net building (less depreciation)

$72,000

Net equipment (less depreciation) $36,000 Retained earnings ?????? a. Classify each of the following accounts as an asset, liability, or net worth. b. What is the balance in the Retained Earnings account if the data forms a valid balance sheet? 13.03-PR003 Video Solution The Scott Company shows the following alphabetical list of account balances as of December 31, 2018. Account Accounts payable

Balance $50,000

Account

Balance

Accounts receivable

$80,000

Accrued taxes

$20,000

Buildings (net value) $205,000 Cash $40,000 Dividends payable

$20,000

Equipment (net value) $180,000 Account

Balance

Finished goods Land value

$30,000 $115,000

Long-term mortgages

$390,000

Notes payable

$60,000

Raw material inventory $40,000 Stockholder’s equity $150,000 a. Construct a balance sheet for Scott Company as of December 31, 2018. b. Demonstrate that the fundamental accounting equation holds by stating the dollar values of the terms in the equation.

13.03-PR004 In addition to the account balances in the previous problem, Scott Company recorded the following summary transactions for the year ending December 31, 2018. Category

Amount

Administrative expenses

$30,000

Factory depreciation Direct labor charges

$30,000 $70,000

Factory overhead

$35,000

General interest payments $40,000

Category Raw material expenses

Amount $90,000

Sales

$400,000

Taxes paid

$20,000

Construct an income statement for Scott Company for the period 1/1/2018 to 12/31/2018. 13.03-PR005 Betty Smith is the owner of Accurate Tax Service. For the year ending April 30, 2019, the following information is available for this service business. At the beginning of this accounting period, the balance in the Betty Smith Capital account was $32,000. Following is a summary of activities during this accounting period: Betty Smith, capital withdrawn: $18,600 Revenue from income tax preparation: $71,300 Revenue from monthly clients: $43,800 Salaries expense: $12,500 Advertising expense: $900 Rent expense: $6,000 Automobile expense: $1,300 Office supplies expense: $7,500 Note: You should assume that all profit (or loss) accrued during this accounting period is absorbed into the Betty Smith Capital account at the end of the period. Following are the account balances at the end of this accounting period (except Betty Smith Capital): Cash: $62,500 Accounts receivable: $3,700 Office furniture and fixtures: $11,300 Office machines and computers: $15,000 Automobile: $9,500 Accounts payable: $1,700

a. Prepare an income statement for the year ending April 30, 2019. b. Prepare a balance sheet as of April 30, 2019. 13.03-PR006 As published in its annual report, the balance sheet and income statements of JanCo include the items shown below (listed alphabetically). All balances are as shown on the statements; no additional transaction processing or adjustments are required. Accounts payable

$12,000 Notes payable

Accounts receivable $21,000 Prepaid rent

$6,000 $1,050

Advertising expense $600 Property, plant, and equipment $15,300 Cash $22,950 Rent expense $2,100 Common stock

$54,000 Retained earnings

Cost of goods sold $15,000 Sales revenue Depreciation expense $300 Wages expense Inventory

$14,400 Wages payable

$1,500 $27,000 $7,500 $1,200

a. Prepare a balance sheet for JanCo. b. Prepare an income statement for JanCo. 13.03-PR007 As published in its annual report, the consolidated balance sheet of Packers includes the items shown below (listed alphabetically). Accounts payable Accounts receivable Capital in excess of par value Cash Common stock at par value Inventory Long-term liabilities Notes payable

? Other assets $8,739 Other liabilities

$30,350 $57,786

$187,506 Property, plant, and equipment

$103,684

$92,827 Retained earnings $402 Total assets

$35,097 $378,045

? Total liabilities $9,474 Total liabilities and net worth $11,195 Total net worth

? ? ?

a. Determine the values for the missing items. b. Prepare a balance sheet for Packers. 13.03-PR008 The annual report from a major energy company contains the following: Net income represented 9% return on average total assets of approximately $27 billion and a 19% return on average stockholders’ equity. Net income per gallon on products sold worldwide averaged $0.035. Net income was $0.04 on each dollar of revenue. Based on this information, determine the following: a. Net income. b. Total revenues. c. Average stockholders’ equity. d. Gallons of products sold. 13.03-PR009 Video Solution Consider the comparative balance sheet and income statement for Starbucks provided in Figures 13.8 and 13.9. Based on these financial statements, determine the following for the year September 30, 2018: a. Return on assets employed. b. Return on owner’s equity. c. Current ratio. d. Acid test ratio. e. Accounts receivable turnover. f. Inventory turnover.

g. Debt to equity ratio. h. Operating income to total assets. i. EBIT. j. EBITDA.

13.03-PR010 Find three Web sites where “financial ratios” is a topic. a. State the URL for each of the sites. b. Identify the one you would likely find most useful if you wanted to learn more about financial ratios. c. In 25 words or less, what does the site you identified in part b have that is most useful? d. List the names and formulas for at least three additional financial ratios not stated in the chapter. Section 13.4 Cost Accounting Principles LEARNING OBJECTIVE 13.4 Apply cost accounting principles to determine relevant cost values. 13.04-PR001 Video Solution The Bryant Company manufactures a custom-designed part for the aerospace industry. Within the production facility, there are five operating departments. The sequence of departments through which the part passes during production is shown in the table below. The table also contains other production information. Two units of product can be obtained from one 4′ × 4′ sheet of aluminum that costs $3.50 per square

foot. Overhead costs to produce a lot of size 10,000 units are estimated to be $304,000. Sequence

Dept

Avg. Labor ($/hr)

Avg. Prod. Time (min/part)

FloorSpace (sq. ft.)

1 2

Cutting Stamping

$20.00 $18.20

3.0 1.0

1,300 1,600

3

Trimming

$17.20

2.0

1,200

4 5

Assemble Package

$16.00 $14.40

5.0 0.5

2,000 1,500

a. Assume that no other production costs exist. Determine the total cost of producing a 10,000-unit lot. b. Determine the cost per piece of producing a 10,000-unit lot. c. Assume that overhead is distributed on the basis of floor space. What is the overhead assigned to each department?

13.04-PR002 EmKay Company is divided into two departments for accounting purposes. The direct labor hours and overhead costs for the last year are as follows: Department Direct Labor Hours Overhead Costs A   900 $13,500 B

1,350

$27,000

What selling price should the company quote on Job Order A54 if the following conditions apply? Raw material costs are estimated as $750. Estimated direct labor hours required in Departments A and B are 25 hours and 50 hours, respectively.

Workers in both departments earn $15 per hour. Overhead costs in each department are allocated based on an overhead rate per direct labor hour. Profit is to be 25% of total cost. 13.04-PR003 For a small manufacturing firm, a current-period job that requires 35 hours of direct labor is to be allocated $437.50 of overhead cost based on a rate developed from previous-period data. The following additional information is available: Overhead is allocated based on direct labor hours. One thousand direct labor hours were expended in the previous period. The previous period’s average labor rate was $10/hour. The current period’s average labor rate is $10/hour. a. What were the overhead costs for the previous period? b. If overhead for the job in question had been allocated based on direct labor dollars, how much overhead cost would have been allocated? (If you did not get an answer to part a, you may assume a value of $15,000 for previous-period overhead costs.) 13.04-PR004 To make a batch of 1,000 units of a certain item, it is estimated that 120 direct labor hours are required at a cost of $10/hour. Direct material costs are estimated at $1,500 per batch. The overhead costs are calculated based on an overhead rate of $7.50 per direct labor hour. The item can be readily purchased from a local vendor for $4 per unit. a. Should the item be made or purchased? b. Over what range of overhead rate is your answer in part a valid? 13.04-PR005 Given the following information for Live Wire Electronics, construct an income statement for the month ending June 30. Sales Cost of goods sold

$500,000 $170,000

Administration cost

$75,000

Advertising expense $25,000 Taxes

$70,000

13.04-PR006 Cost Center D within the Welding Department of the Mizer Corporation collected the following data during 2019. Cost Center D Overhead Expenses: $4,800 Cost Center D Direct Labor Hours: 2,800 Cost Center D Direct Labor Cost: $11,200 Cost Center D Direct Material Cost: $3,000 Compute the 2020 cost center overhead rate by the following methods: a. Percentage of direct labor cost. b. Percentage of prime cost. c. Rate per direct labor hour. d. Mizer Corp. has adopted the direct labor hours method of overhead allocation. The New Orders Department has just received an order that requires an estimated 125 hours of direct labor in Cost Center D. What is the estimated overhead charge for this job in Cost Center D? 13.04-PR007 Find three Web sites where “activity-based costing” (ABC) is a topic. a. State the URL for each of the sites. b. Identify the one you would likely find most useful if you wanted to learn more about ABC. c. In 25 words or less, what does the site you identified in part b have that is most useful? 13.04-PR008 Video Solution Skrunchy Company produces two products, Lower and Upper. The following two tables give pertinent information about these two products. Category Direct labor

Lower Upper $700,000 $1,000,000

Category

Lower

Upper

Direct material $2,000,000 $1,500,000 Direct labor hours     28,000     25,000 Quantity produced     20,000     10,000 Total overhead

$2,500,000

ABC Activity

Driver

Rework

Hours $1,800,000

Material handling Moves

Cost

Lower

Upper

4,000 hours 42,500 hours

$700,000 40,000 moves 20,000 moves

a. What is the cost per unit of Upper if Skrunchy uses traditional overhead allocation based on direct labor hours? b. What is the cost per unit of Upper if Skrunchy uses activity-based costing to allocate overhead?

13.04-PR009 Reconsider Problem 13.04-PR008. a. What is the cost per unit of Lower if Skrunchy uses traditional overhead allocation based on direct labor hours? b. What is the cost per unit of Lower if Skrunchy uses activity-based costing to allocate overhead? 13.04-PR010 LockTite Company produces two products, Pretty Safe (PS) and Virtually Impenetrable (VI). The following two tables give pertinent information about these products. Category Number produced

PS 3,000

VI 2,000

Direct material cost $350,000 $150,000 Direct labor hours 5,000 hours 10,000 hours Direct labor cost

$100,000

$200,000

Category

PS

Total overhead

VI

ABC Activity

Driver

$750,000 Cost

PS

VI

Production

Machine hours

$500,000 1,000 hours 4,000 hours

Engineering

Engineering hours $250,000 2,000 hours 3,000 hours

a. What is the cost per unit of Pretty Safe if LockTite uses traditional overhead allocation based on number of units produced? b. What is the cost per unit of Pretty Safe if LockTite uses activity-based costing to allocate overhead? 13.04-PR011 Reconsider Problem 13.04-PR010. a. What is the cost per unit of Virtually Impenetrable if LockTite uses traditional overhead allocation based on number of units produced? b. What is the cost per unit of Virtually Impenetrable if LockTite uses activity-based costing to allocate overhead? Section 13.5 Economic Value Added LEARNING OBJECTIVE 13.5 Calculate the economic value added for an equity-based capital investment. 13.05-PR001 EOY Revenue

Consider the following table (shown in two parts): Operating Expenses

Depreciation Net Operating Profit

Taxes

 8 $2,765,000  9 $2,765,000

$2,379,200 $2,379,200

$38,400 $23,040

$347,400 $362,760

$86,850 $90,690

10 $2,765,000

$2,379,200

$13,824

$371,976

$92,994

EOY NOPAT MARR BOYBV EVA (ECON Profit)  8 $260,550

0.1

$93,080

 9 $272,070

0.1

XXX

XXX

EOY NOPAT MARR BOYBV EVA (ECON Profit) 10 $278,982

0.1

a. What is the EVA in year 8? b. What is the beginning-of-year book value (BOYBV) in year 9? 13.05-PR002 Consider the following table (shown in two parts): a. What is the EVA in year 5? b. What is the capital charge (Cost of Capital) in year 6? EOY Revenue 5 6

$2,765,000 $2,765,000

Operating Expenses $2,379,200 $2,379,200

7

$2,765,000

$2,379,200

Depreciation Net Operating Taxes Profit $49,536 $336,264 $84,066 $49,536 $336,264 $84,066 $49,536

$336,264

$84,066

EOY NOPAT MARR BOYBV EVA (ECON Profit) 6 7 8

$252,198 $252,198 $252,198

0.1 0.1 0.1

$247,680 XXX

XXX

13.05-PR003 A company is considering purchasing a new piece of machinery at a cost $60,000. It is expected to generate revenues of $20,000 per year for 4 years against $3,000 of annual operating expenses. The machinery is MACRS 3-year property. The income tax rate is 25%. MARR is 10%. What is the EVA for year 3? 13.05-PR004 A company is considering purchasing a new piece of machinery at a cost $50,000. It is expected to generate revenues of $25,000 per year for 4 years against $1,500 of annual operating expenses. The machinery is MACRS 3-year property. The income tax rate is 25%. MARR is 10%. What is the EVA for year 3? 13.05-PR005 Find three Web sites where “economic value added” (EVA) is a topic.

a. State the URL for each of the sites. b. Identify the one you would likely find most useful if you wanted to learn more about EVA. c. In 25 words or less, what does the site you identified in part b have that is most useful? 13.05-PR006 An investment of $1,250,000 is made in 7-year MACRS-GDS equipment. The investment yields annual before-tax returns of $200,000, plus a salvage value of $500,000 at the end of the 10-year planning horizon. The MARRAT is 7%, the income tax rate is 25%, the maximum Section 179 expense deduction is taken and 50% bonus depreciation applies. For the investment, calculate a. After-tax present worth, b. After-tax annual worth, and c. EVA. 13.05-PR007 For the investment described in Problem 13.05-PR006, let 50% bonus depreciation apply. For the investment, calculate a. After-tax present worth, b. After-tax annual worth, and c. EVA. 13.05-PR008 For the investment described in Problem 13.05-PR006, let the maximum Section 179 expense deduction be taken. For the investment, calculate a. After-tax present worth, b. After-tax annual worth, and c. EVA. 13.05-PR009 An investment of $1,400,000 is made in 5-year MACRS-GDS equipment. Measured in constant dollars, the investment yields annual returns of $400,000 and a salvage value of $500,000 at the end of the 7-year planning

horizon. A 25% tax rate and a 3% inflation rate apply. The real after-tax minimum attractive rate of return is 8%. For the investment, calculate a. After-tax present worth, b. After-tax annual worth measured in then-current dollars, c. Real internal rate of return, and d. EVA measured in then-current dollars. 13.05-PR010 For the investment described in Problem 13.05-PR009, let 50% bonus depreciation apply. For the investment, calculate a. After-tax present worth, b. After-tax annual worth measured in then-current dollars, c. Real internal rate of return, and d. EVA measured in then-current dollars. 13.05-PR011 For the investment described in Problem 13.05-PR009, let the maximum Section 179 expense deduction be taken and let 50% bonus depreciation apply. For the investment, calculate a. After-tax present worth, b. After-tax annual worth measured in then-current dollars, c. Real internal rate of return, and d. EVA measured in then-current dollars. 13.05-PR012 An investment of $4,500,000 is made in equipment qualifying as 5-year equipment for MACRS-GDS depreciation. Measured in constant dollars, the investment yields annual returns of $1,200,000, plus a salvage value of $300,000 at the end of the 6-year planning horizon. Based on a 25% income tax rate, a 4% inflation rate, and a real after-tax minimum attractive rate of return of 10%, calculate a. After-tax present worth, b. After-tax annual worth measured in then-current dollars,

c. Real internal rate of return, and d. EVA measured in then-current dollars. 13.05-PR013 For the investment described in Problem 13.05-PR0012, let 50% bonus depreciation apply. For the investment, calculate a. After-tax present worth, b. After-tax annual worth measured in then-current dollars, c. Real internal rate of return, and d. EVA measured in then-current dollars. 13.05-PR014 For the investment described in Problem 13.05-PR0012, let the maximum Section 179 expense deduction be taken and let 50% bonus depreciation apply. For the investment, calculate a. After-tax present worth, b. After-tax annual worth measured in then-current dollars, c. Real internal rate of return, and d. EVA measured in then-current dollars. 13.05-PR015 A company invests $5,000,000 in MACRS GDS 5-year property. Measured in constant dollars, the investment yields savings of $1,500,000 the first year; thereafter, annual savings decrease $100,000 per year for 10 years, at which time the equipment has a constant dollar salvage value of $900,000. Inflation is anticipated to equal 3% per year over the planning horizon. Section 179 expense deduction and 50% bonus depreciation are available. A real required return on investment of 10% is used by the company in performing economic justifications. Based on an income tax rate of 25%, calculate a. After-tax present worth, b. After-tax annual worth measured in then-current dollars, and c. EVA measured in then-current dollars.

Chapter 13 Summary and Study Guide Summary 13.1: Cost Terminology

Learning Objective 13.1: Identify 5 unique cost viewpoints used for economic decision making. (Section 13.1) The 5 unique cost viewpoints for economic decision making include: Life Cycle Viewpoint deals with costs and revenues based upon when they occur within an asset’s service life. Past/Future Viewpoint deals with costs and revenues based upon when they occur relative to “time now.” Manufacturing Cost Structure Viewpoint looks at a product’s selling price and breaks this price into its constituent pieces such as cost of goods sold and profit. Fixed and Variable Viewpoint acknowledges that some costs vary in proportion to quantity of output while other costs do not vary in proportion to the quantity of output. TC(x) = FC + VC(x)

(13.1)

Average and Marginal Viewpoint deals with costs expressed in terms of units of output. TC(x) AC(x) =

(13.4)

x

13.2: Cost Estimation

Learning Objective 13.2: Apply cost estimating principles to determine costs for given scenarios. (Section 13.2) Cost estimation is one of the most difficult challenges an engineering economist faces. There are 3 cost estimation principles discussed, including: Project Estimation allows the decision maker to compare alternative engineering investment projects and make a selection from these projects. It typically includes the following 6-step process: (1) plan the estimate; (2) research, collect, and analyze data; (3) develop the estimate structure; (4) determine the estimating methodologies; (5) compute the cost estimate; and (6) document and present the estimate. Estimating Methodologies, including (1) parametric estimating; (2) estimating by analogy; and (3) engineering estimates. General Sources of Data to include both internal and external sources. Internal sources include things like sales records, production control records, inventory records, quality control records, purchasing department records, work measurement and other industrial engineering studies, maintenance records, and personnel records. External sources include published information generally available and information available on request. 13.3: General Accounting Principles

Learning Objective 13.3: Demonstrate proficiency with balance sheets, income statements, and ratio analysis. (Section 13.3) Balance Sheet is a statement of the financial condition of a firm at a point in time divided into three main groups: assets, liabilities, and net worth items. Income Statement is a statement of the revenues and expenses incurred by a firm during a period of time. Ratio Analysis is used as a comparison to a company's historic values to spot trends and/or changes in performance; as a comparison to expected or desired benchmarks; and as a comparison to competitors’ performance. Ratios examine earning power, short-term liability obligations (or liquidity), and long-term liability obligations (or solvency). One may also examine the ability of a company to earn a profit through its operations. Common earning power ratios include: Net Income Return on Assets Employed =

(13.7)

Average Total Assets

Net Income Return on Owner's Equity =

(13.8)

Average Owner's Equity

Short-term liability obligation ratios include: Current Assets

Current Ratio =

(13.9)

Current Liabilities

Cash  + Accounts Receivable + Short Term Investments Acid Test Ratio =

(13.10)

Current Liabilities

Net Sales Accounts Receivable Turnover =

(13.11)

Average Accounts Receivable

Cost of Goods Sold Inventory Turnover =

(13.12)

Average Inventory

Long-term liability obligation ratios include: Total Liabilities Debt to Equity Ratio =

Net Income Before Taxesand Interest Times Interest Earned Ratio =

(13.13)

Total Capital Worth

(13.14)

Interest Charges

Net Operating Income Operating Income to Total Assets Ratio =

(13.15)

Total Assets

Earning a profit through its operations include: Earnings Before Interest and Taxes  (EBIT) = Net Income + Income Taxes + Interest Expense

(13.16) (13.17)

Earnings Before Interest, Taxes, Depreciation, and Amortization (EBITDA) = Net Income + Income Taxe

Interest Expense + Depreciation + Amortization

13.4: Cost Accounting Principles

Learning Objective 13.4: Apply cost accounting principles to determine relevant cost values. (Section 13.4) Four cost accounting principles are presented: Traditional Cost Allocation Methods include allocation based on direct labor hours, allocation based on direct labor dollars, and allocation based on prime dollars. Activity-Based Costing (ABC) is a costing method designed to associate costs with the activities that drive them. Standard Costs include the estimated material, labor, and overhead costs for a unit. Standard costs serve as a basis for measuring production efficiency and performance over time. 13.5: Economic Value Added

Learning Objective 13.5: Calculate the economic value added for an equity-based capital investment. (Section 13.5) Economic Value Added (EVA) focuses management attention on adding value for shareholders. EVA = NOPAT − CoC

(13.21)

For equity-based capital investments, EVA equals after-tax annual worth. Annual worth, economic value added, present worth, and other discounted cash flow measures of economic worth will yield the same recommendations regarding equity-based capital investments.

Important Terms and Concepts Accounting (as defined by the American Institute of Certified Public Accountants) “The art of recording, classifying and summarizing in a significant manner and in terms of money, transactions and events which are, in the past at least, of a financial character, and interpreting the results thereof.” Operating and Maintenance Costs and Revenues Recurring costs that are necessary to operate and maintain an item during its useful life. Salvage Value The estimated value of an asset at the end of its useful life. Also referred to as the market value. First Cost The total initial investment required to get an asset ready for service. Terminal Value Present worth of all cash flows occurring at and after the end of the planning horizon, including final disposal costs. Sunk Costs Past costs that have no bearing on current decisions. Past Costs Historical costs that have occurred for an item under consideration.

Future Costs All costs that may occur in the future. Opportunity Cost The cost of a forgone alternative (or opportunity) that is incurred in order to pursue another alternative. Cost of Capital The cost of obtaining funds for financing projects through debt obligations and/or equity sources. Cost of Goods Sold The total cost of manufacturing and marketing a product. Overhead Costs All costs of manufacturing other than direct material and direct labor, including indirect labor and indirect material. Direct Costs Generally labor and material costs that can be easily measured and conveniently allocated to a specific operation, product, or project. Indirect Costs Generally labor and material costs that are difficult or impossible to assign directly to a specific operation, product, or project. Fixed Costs Costs that do not vary in proportion to the quantity of output. Variable Costs Costs that vary in proportion to quantity of output. Break-Even Point Point where total revenues and total costs are equal. Average Cost The ratio of total costs incurred divided by the number of units produced. Marginal Cost The increase or decrease in the total cost of a production run for making one additional unit of an item. Economies of Scale As production increases, the cost per unit of output decreases because fixed costs are spread over a larger output. Economies of Scope Concept that suggests fixed costs be driven as low as possible, ideally to zero. Cost Estimating (as defined by the Association for the Advancement of Cost Engineering International) “A predictive process used to quantify, cost, and price the resources required by the scope of an asset investment option, activity or project. As a predictive process, estimating must address risks and uncertainties. The outputs of estimating are used primarily as inputs for budgeting, cost or value analysis, decision making in business, asset and project planning, or for project cost and schedule control processes.” Balance Sheet A statement of the financial condition of a firm at a point in time divided into three main groups: assets, liabilities, and net worth items. Income Statement A statement of the revenues and expenses incurred by a firm during a period of time.

Activity Based Costing (ABC) A costing method designed to associate costs with the activities that drive them. Standard Costs The estimated material, labor and overhead costs for a unit. Standard costs serve as a basis for measuring production efficiency and performance over time. Economic Value Added (EVA) A management tool that examines the difference between the net operating profit after taxes and the cost of capital.

Chapter 13 Study Resources Chapter Study Resources These multimedia resources will help you study the topics in this chapter. 13.1: Cost Terminology LO 13.1: Identify 5 unique cost viewpoints used for economic decision making. Video Example 13.6: Break-Even Point for Production Video Example 13.8: Marginal Cost Video Solution: 13.01-PR007 Video Solution: 13.01-PR010 13.2: Cost Estimation LO 13.2: Apply cost estimating principles to determine costs for given scenarios. Video Lesson: Cost Estimation Video Lesson Notes: Cost Estimation Video Solution: 13.02-PR002 13.3: General Accounting Principles LO 13.3: Demonstrate proficiency with balance sheets, income statements, and ratio analysis. Video Lesson: General Accounting Principles Video Lesson Notes: General Accounting Principles Video Solution: 13.03-PR003 Video Solution: 13.03-PR009

13.4: Cost Accounting Principles LO 13.4: Apply cost accounting principles to determine relevant cost values. Video Lesson: Cost Accounting Principles Video Lesson Notes: Cost Accounting Principles Video Solution: 13.04-PR001 Video Solution: 13.04-PR008 13.5: Economic Value Added LO 13.5: Calculate the economic value added for an equity-based capital investment. Excel Video Lesson: PMT Financial Function Excel Video Lesson Spreadsheet: PMT Financial Function Excel Video Lesson: NPV Financial Function Excel Video Lesson Spreadsheet: NPV Financial Function Video Example 13.3: Evaluating the Economic Performance of Two Firms These chapter-level resources will help you with your overall understanding of the content in this chapter. Appendix A: Time Value of Money Factors Appendix B: Engineering Economic Equations Flashcards: Chapter 13 Excel Utility: TVM Factors: Table Calculator Excel Utility: Amortization Schedule Excel Utility: Cash Flow Diagram Excel Utility: Factor Values Excel Utility: Monthly Payment Sensitivity

Excel Utility: TVM Factors: Discrete Compounding Excel Utility: TVM Factors: Geometric Series Future Worth Excel Utility: TVM Factors: Geometric Series Present Worth Excel Data Files: Chapter 13

CHAPTER 13 Obtaining and Estimating Cash Flows LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to: 13.1 Identify 5 unique cost viewpoints used for economic decision making. (Section 13.1) 13.2 Apply cost estimating principles to determine costs for given scenarios. (Section 13.2) 13.3 Demonstrate proficiency with balance sheets, income statements, and ratio analysis. (Section 13.3) 13.4 Apply cost accounting principles to determine relevant cost values. (Section 13.4) 13.5 Calculate the economic value added for an equity‐based capital investment. (Section 13.5)

Engineering Economics in Practice Starbucks In 1971 in Seattle’s Pike Place Market, a small coffee shop opened its doors for the first time. Like most new businesses, it had its share of successes and false starts, but slowly and surely it began to grow. Over the next 20 years the little coffee shop expanded both its menu and its presence in the U.S. Northwest, growing to 116 outlets. In 1991, success led to the stock market and the company “went public” by offering stock on the Nasdaq National Market under the trading symbol “SBUX.” Today, we know the Starbucks company as the leading retailer and roaster of specialty coffees in the world with an aim to serve the world’s most sustainable coffee. Starbucks employed approximately 238,000 workers in 2018 and welcomed customers into more than 29,000 retail stores in 78 countries. In 2018, Starbucks reported revenue of $24.7 billion (up by 10% from $22.4 billion in 2017), net income of $4.5 billion (up by 56% from $2.9 billion in 2017), and total assets of $24.2 billion (up by 68% from $14.4 billion in 2017). Without question, Starbucks has achieved remarkable success in its nearly 50‐year history. In addition to these accounting measures of success, Starbucks succeeds in other ways too. Accolades include being named one of the world’s most admired companies (Fortune), one of the most ethical companies (Ethisphere), and one of the most valuable brands (Fortune). How does Starbucks achieve these remarkable results? Undoubtedly, there are many explanations, including the way Starbucks supports key decision makers with appropriate and timely information. This support includes collecting, summarizing, and reporting relevant financial information in a timely manner to internal managers and external decision makers. This is the world of accounting. Discussion Questions 1. As Starbucks expands its growth both domestically and globally, what challenges does it face related to cost estimating? 2. What are examples of fixed and variable costs for Starbucks? 3. In the northwestern corner of the United States, you seemingly find a Starbucks retail store on every street corner. How do you suppose these retailers can stay in business with such close proximity? 4. How might Starbucks overhead costs be allocated? Do you expect one method to be better than the other?

Introduction One purpose of this chapter is to briefly explore, from an engineering economy perspective, the world of accounting. Accounting is a broad and complex discipline. Our purpose in Chapter 13 is not to cover accounting topics to the same depth and breadth that an accountant would, but instead to enable the engineering economist to be conversant with accountants and to have an understanding of the sources, development, and uses of typical accounting data. Accounting (as defined by the American Institute of Certified Public Accountants) “The art of recording, classifying and summarizing in a significant manner and in terms of money, transactions and events which are, in the past at least, of a financial character, and interpreting the results thereof.” A second purpose of this chapter is to address the topic of estimating cash flows. Frequently, cash flow estimation is based upon discovering the patterns underlying past cash flows and extrapolating them into the future, thus implying a reliance on accounting data from prior periods. In still other cases, the cash flows associated with engineering economic analysis cannot be extrapolated from history. Other types of estimating techniques are required for these situations. Engineering economic analysis is primarily concerned with comparing alternative projects on the basis of an economic measure of effectiveness. The comparison process utilizes a variety of cost terminologies and cost concepts. To begin our discussion of cost terminology, we exemplify a typical production situation.

EXAMPLE 13.1 Cost Terminology Introductory Example Let us assume that the primary business of a small manufacturing firm is job‐shop machining—that is, the firm produces a variety of products and component parts according to customer orders. Any given order may be for only five parts or as many a several hundred. The firm has periodically received orders to manufacture a part, which we will identify as Part Number 163H, for the B&K Corporation. The part has been manufactured in a four‐step production sequence consisting of (1) sawing bar stock to length, (2) machining on an engine lathe, (3) machining on an upright drill press, and (4) packaging. The unit cost to produce Part Number 163H by this sequence has been $25, where the unit cost is comprised of the major cost elements of direct labor, direct materials, and overhead (prorated costs for insurance, taxes, electric power, marketing expenses, etc.). The firm is now in the process of negotiations with the B&K Corporation to obtain a contract for producing 10,000 of these parts over a 4‐year period, or an average of 2,500 units/year. A contract for this volume of parts is highly desirable, but in order to obtain the contract, the firm must lower the unit cost. An engineer for the firm has been assigned to determine production methods to lower the unit cost. After study, the engineer recommends the purchase of a small turret lathe. With the turret lathe, the processing sequence of Part Number 163H would consist essentially of (1) machining bar stock on the turret lathe and (2) packaging. The estimated unit cost for Part Number 163H by this production method would be $15. Furthermore, the production rate with the new method would be increased over the old method, because the turret lathe would replace the sawing, engine lathe, and drill press operations. If the turret lathe is purchased, the saw, engine lathe, and drill press would not be sold but would be kept for other jobs for which the firm may receive orders. The turret lathe would be reserved for the production of Part Number 163H, but about 25% excess production capacity could be devoted to other jobs. The incremental investment required to purchase the turret lathe and the new tooling required, as well as installing the machine, is $50,000. The turret lathe’s physical life is judged to be about 15 years, but federal tax laws permit the investment capital to be recovered through annual depreciation charges in 5 years. At the end of 5 years, the firm estimates the turret lathe’s salvage value will be $25,000. If the maximum unit price the B&K Corporation will pay for Part Number 163H is $22, should the firm accept the contract for 10,000 parts and then purchase the turret lathe in order to execute the contract?

Using the techniques and methods presented in the text, we can answer this question. However, that is not our purpose here. The example situation has been cited to illustrate that considerable research and investigation is required to determine or estimate the cost figures used in economic decision making. This cost information is typically obtained from a variety of sources, such as company production records, accounting records, manufacturers’ catalogs, and technical specifications, publications from the U.S. Government Printing Office, and others. To make effective comparisons and intelligent recommendations, an engineering economist must be familiar with cost terminology, cost factors, and estimating techniques used by different specialists.

13.1 Cost Terminology LEARNING OBJECTIVE Identify 5 unique cost viewpoints used for economic decision making. In this section, cost definitions and concepts are considered from 5 different viewpoints: (1) life cycle, (2) past/future, (3) manufacturing cost structure, (4) fixed/variable, and (5) average/marginal. Each viewpoint offers unique advantages with respect to economic decision making. The viewpoint to be used in analyzing a particular decision depends upon the purpose of the analysis. Frequently, multiple viewpoints are employed in a single analysis. Examining a decision making situation from multiple viewpoints can help ensure that all relevant costs (and revenues) have been considered prior to conducting the analysis. Further, it helps to ensure that the identified costs and revenues are correctly documented and incorporated within the analysis.

13.1.1 Life Cycle Viewpoint The life cycle viewpoint deals with costs and revenues based upon when they occur within an asset’s service life. The term asset should be interpreted in the general sense as a machine, a unit of equipment, a product line, a project, a building, a system, and so forth. The life cycle viewpoint includes design and development costs, fabrication and testing costs, operating and maintenance costs, operating revenues, and salvage value. Operating and Maintenance Costs and Revenues Recurring costs that are necessary to operate and maintain an item during its useful life. Salvage Value The estimated value of an asset at the end of its useful life. Also referred to as the market value. This text is primarily concerned with economic justification of engineering projects, the replacement of existing projects or assets, and the economic comparison of alternative projects. For the purposes of these types of analysis, we will define the life cycle viewpoint to consist of (1) first cost, (2) operating and maintenance costs and revenues, and (3) salvage value. First Cost The first cost of an asset is the total initial investment required to get the asset ready for service; such costs are usually nonrecurring during the asset’s life. For the purchase of a machine tool, its first cost may consist of the following major elements: (1) the basic machine cost, (2) costs for training personnel, (3) shipping and installation costs, (4) initial tooling costs, and (5) supporting equipment costs. The installation costs may include, for example, preparing a foundation; vibration and noise insulation; providing heat, light, and power supply; and testing. Supporting equipment costs may include computer‐control hardware and software and spare‐parts inventory. First Cost The total initial investment required to get an asset ready for service. For other projects, a different set of first cost elements may be appropriate. Some projects may include working capital for inventories; accounts receivable; and cash for wages, materials, and so forth. In any case, the emphasis here is that an item’s first cost normally involves more cost elements than just the basic purchase price. Whether the first cost elements are aggregated or maintained separately depends on income‐tax considerations and whether

or not a before‐tax or after‐tax economic analysis is desired. Certain income‐tax laws and depreciation methods were presented in the text, so we won’t elaborate on this particular point here. Operating and Maintenance Costs and Revenues Operating and maintenance costs are recurring costs that are necessary to operate and maintain an item during its useful life. Operating costs usually consist of labor, material, and overhead items. Depending upon the accounting system a firm uses, a wide range of cost factors may be included in the major cost classification of overhead. Typical overhead items are fuel or electric power, insurance premiums, inventory charges, indirect labor (as opposed to direct labor), administrative and management expenses, and so forth. It is usually assumed that operating and maintenance costs are annual costs, but maintenance costs may not be on a recurring, annual basis— that is, a regular annual schedule of minor or preventive maintenance may be followed, or it could be policy that maintenance is performed only when necessary, such as when a major overhaul is required. In most cases, the maintenance policy consists of both preventive maintenance and maintenance on an as‐needed basis. Regardless, repair and upkeep result in costs that must be recognized in the economic analysis of engineering projects. Operating revenues are revenues that result from having and using the asset. In many cases, these revenues are estimated as recurring annual cash flows. They may have a consistent pattern (such as a uniform, gradient, or geometric series), or they may exhibit no consistent pattern and need to be estimated for each year individually. In the case of manufacturing assets and projects, the revenues are typically based on estimates of the volume and revenue generated by the parts that utilize the asset or result from the project. Salvage Value When an item’s life cycle ends, disposal costs usually result. These may include labor and material costs to remove the item; shipping costs; or special costs, such as those for disposing of hazardous materials. Although disposal costs may be incurred at the end of the life cycle, most items have some monetary value at the time of their disposal. This value is the market or trade‐in value (i.e., the actual dollar worth for which the item may be sold at the time of disposal). After deducting the cost of disposal from the market value at disposal, the resulting net dollar worth is termed the salvage value. Depending on the economic justification being performed, salvage value might be called terminal value. For example, when a planning horizon of 10 years is being used, but cash flows from the investment will continue for a longer period of time, salvage value is equal to the present worth of future cash flows resulting from the investment, including the final disposal of all assets associated with the investment. Terminal Value Present worth of all cash flows occurring at and after the end of the planning horizon, including final disposal costs. The market value, the disposal costs, and the salvage value are usually not known with certainty and therefore must be estimated. For an item that satisfies the IRS definition of a capital asset and that decreases in value over time through physical deterioration, the IRS has approved depreciation methods that can serve to estimate the rate of deterioration and consequent decrease in the asset’s value. The capital asset’s value at the end of a given accounting period during the asset’s life is termed the book value. Scrap value, on the other hand, refers only to the value of the material of which the item is made. For example, a 4‐year‐old automobile may have a scrap value of $500 but a market value of $5,000. A distinction between these terms is generally not important for evaluating potential investment projects; therefore, we use salvage value to denote the end‐of‐life value. For example, a trade‐ in value of $3,000 minus disposal costs of $500 equals a net salvage value of $2,500. The life cycle viewpoint obviously involves a time horizon, and the end of an item’s life may be judged from either a functional or an economic point of view. An item’s economic life is generally shorter than its functional life. For example, an engine lathe may remain functional for 15 years or more, but because of periodic advancements in machine design technology, newer engine lathes have higher production rates; therefore, an engine lathe’s economically useful life may be only 10 years. An item’s economic life is usually a matter of company policy, which is greatly influenced by income‐tax considerations.

13.1.2 Past/Future Viewpoint

The past/future viewpoint deals with costs and revenues based upon when they occur relative to “time now.” This viewpoint incorporates more than just a timeline; it also includes concepts related to past and future cash flows such as sunk costs, opportunity costs, and cost of capital. Each of these is explained in more detail in the examples and paragraphs that follow. Sunk Costs Past costs that have no bearing on current decisions.

EXAMPLE 13.2 Past Costs and Sunk Costs Past costs are historical costs that have occurred for the item under consideration. Sunk costs are past costs that are unrecoverable. The distinction is perhaps best made through examples. Past Costs Historical costs that have occurred for an item under consideration. Assume that an investor purchases 100 shares of common stock in the JHP Corporation through a broker at $25/share. In addition, the investor pays $85 in brokerage fees and other charges. Just 2 months later, and before receiving any dividend payments, the purchaser resells the 100 shares of common stock through the same broker at $35/share minus $105 for selling expenses. The purchaser realizes a net profit of $810 ($3,500 − $2,500 − $85 − $105) on these transactions. At the time of sale, the $2,500 and $85 are past costs, but because these are recovered after the sales transaction, sunk costs are not incurred. If, on the other hand, the investor sold the 100 shares 2 months after purchase and the market price was $20/share, with a $70 charge for selling fees, the investor would incur a capital loss of $655 ($2,000 − $2,500 − $85 − $70). In this instance, some of the past costs would be recovered, but the $655 capital loss would be a sunk cost. If the investor reasons that the market price will decline further or if he simply needs the money, the $655 sunk cost should be ignored if the shares are to be sold for $20 each. However, sunk costs are not totally irrelevant to a present decision. They may qualify as capital losses and serve to offset capital gains or other taxable income and thus reduce income taxes paid. Past costs and sunk costs provide information that can improve the accuracy of estimating future costs for similar items.

EXAMPLE 13.3 Another Example of Sunk Costs Another example of sunk costs is the purchase and sale of a piece of equipment. Assume the equipment is purchased for $10,000 and the salvage value at the end of 5 years of service is estimated to be $5,000. For illustrative purposes, we will further assume that the annual decrease in the equipment’s value through depreciation is $1,000 per year. The $1,000 annual cost of depreciation is a cost of production that, in theory, should be allocated to the equipment’s output. After allocating this and other manufacturing costs, general and administrative costs, and marketing costs to each unit of production, the total unit cost is determined. A profit is then added to each unit of production in order to arrive at the unit selling price. Thus, when a unit is sold, a portion of each sales dollar returns a portion of the depreciation expense. In this illustration, it is assumed that sales will return, or recover, the total estimated depreciation expense of $5,000 (first cost minus estimated salvage value) for the 5‐year period. However, if the equipment has a market value of $2,000 at the end of 5 years, there is a $3,000 ($5,000 − $2,000) sunk cost. The $3,000 capital loss represents an error in estimating the rate of depreciation, and the owner cannot insist that the equipment is worth $5,000 when the market value for the 5‐year‐old equipment is, in fact, only $2,000. If the equipment is kept, it is argued that the true value being kept is thus only $2,000.

All costs that may occur in the future are termed future costs. These may include operating costs for labor and materials, maintenance costs, overhaul costs, and disposal costs. In any case, by virtue of them occurring in the future, these costs are rarely known with certainty and must therefore be estimated. This is also true for future revenues or savings if these are involved in a given project. Estimates of future costs or revenues are uncertain and subject to error. Future Costs All costs that may occur in the future. The cost of forgoing an opportunity to earn interest, or a return, on investment funds is termed an opportunity cost. This concept is best explained through examples. If a person has $1,000 and stores this cash in a home safe, she is forgoing the opportunity to earn interest on the money by establishing a savings account in a local bank that pays, for example, 5% annual compound interest. (Of course, investments other than savings accounts are possible.) For a 1‐year period, the person is forgoing the opportunity to earn (0.05)($1,000) = $50. The $50 amount is thus termed the opportunity cost associated with storing the $1,000 in the home safe. Opportunity Cost The cost of a forgone alternative (or opportunity) that is incurred in order to pursue another alternative.

EXAMPLE 13.4 Opportunity Cost Assume that a person has $5,000 cash on hand. This amount is considered equity capital if the $5,000 was not borrowed (i.e., there is no debt obligation involved). The person has available secure investment opportunities such as establishing a personal savings account in a commercial bank or purchasing other financial instruments. From the available investment opportunities, suppose the optimum combination of risk (security level) and interest yield on the investment results in a 10% annual interest. Thus, the investment of $5,000 would yield (0.10)($5,000) = $500 each year. If, instead of investing the $5,000, the person purchases jewelry for the same amount, he will forgo the opportunity to earn $500 interest/yr. The $500 amount is again termed an annual opportunity cost associated with purchasing the jewelry. The same logic applies in defining an annual opportunity cost for investments in business and engineering projects. For example, purchasing production machinery with $20,000 of equity funds prevents this money from being invested elsewhere with greater security and/or higher profit potential. This concept of opportunity cost is fundamental to the study of engineering economy and is a cost element that is included in virtually all methodologies for comparing alternative projects. In manufacturing and retailing, one of the most common uses of the term opportunity cost is in conjunction with inventory. Holding inventory (raw material, work‐in‐process (WIP), or finished goods) is said to carry a high opportunity cost. For example, in the last fiscal year, Starbucks reported inventories of over $1,400 million in their financial reports. Most of these inventory items were likely planned and needed to provide the availability of products and high levels of customer service that patrons have come to expect from Starbucks. However, these inventories do not come without cost. If we assume that Starbucks could identify improvement projects that would return 15% if funded, then the $1,400 million tied up in inventories carries a $210 million opportunity cost. These types of realizations have companies like Starbucks constantly striving to find ways to maintain high levels of customer service while simultaneously lowering investments in inventories. In the text, we addressed the concept of opportunity costs under the heading of MARR. Some individuals define MARR based on the cost of capital. As used in this text, the term cost of capital refers to the cost of obtaining funds for financing projects through debt obligations and/or equity sources. Debt funds are usually obtained from external sources by (1) borrowing money from banks or other financial organizations (e.g., insurance companies and pension funds) and (2) issuing bonds. These debt obligations are normally long‐term and result in interest payments on, say, a monthly, quarterly, semiannual, or annual basis. The interest payments are thus a cost of borrowed capital. Financing projects through issuing bonds is a method of obtaining capital funds that may be less

familiar to the reader than borrowing money from a bank. Although we provided a brief introduction to bonds in Chapter 3, some elaboration of bonds is appropriate. Cost of Capital The cost of obtaining funds for financing projects through debt obligations and/or equity sources. Bonds are issued by various organizational units, including partnerships, corporations (profit or nonprofit), governmental units (municipal, state, and federal), or other legal entities. The sale of bonds represents a legal debt of the issuing organization; as such, bonds are generally secured (or guaranteed) by the organization’s assets (examples are mortgage bonds or collateral bonds). Debenture bonds, on the other hand, are promissory notes or just a promise to pay. In any case, the purchaser of a bond has legal claim to the assets of the issuing organization but has no ownership privileges in it. In the sense that bonds are debt obligations and not ownership shares, they are considered a more secure investment than either common or preferred stock. This statement should not be taken as a universal truth, however, because the security level for a bond or a stock depends on many factors, economic and otherwise; the principal factor is the financial soundness of the issuing unit. Additional details on interest payments for bank loans and interest payments for bonds were presented in the text. Another method of financing engineering projects is through the use of equity funds, which are generally obtained from one or both of the following sources: (1) common or preferred stock authorized by the company and sold through brokers to investors, or (2) earnings accumulated from prior years and retained by the company. Both of these sources of funds incur an opportunity cost. The text presented details on calculating these costs. Additional information related to a company’s equity funding is presented later in the “General Accounting Principles” section.

13.1.3 Manufacturing Cost Structure Viewpoint The manufacturing cost structure viewpoint looks at a product’s selling price and breaks this price into its constituent pieces. A typical cost structure for a manufactured item, adapted from Ostwald,1 is shown in Figure 13.1. Before examining the details of this figure, two general comments are appropriate. First, while this figure specifically targets a manufacturing cost structure, the major elements (the boxes in the figure) would remain essentially unchanged in other environments (retail, service, etc.). The changes for these other environments are the labels shown for the various combinations. For example, at Starbucks, Cost of Goods Manufactured would likely be referred to as Cost of Sales, while Factory Overhead would be referred to as Store Operating Expenses.

FIGURE 13.1 Manufacturing Cost Structure The second general comment about Figure 13.1 deals with the size of the boxes in the figure. Just because the boxes appear to be approximately equal in height does not mean the costs are approximately equal. The relative sizes of these component pieces can and do vary dramatically among sectors (manufacturing, retail, etc.), among industries within sectors, and among companies within industries. The purpose of the figure is to show the structure of selling price, not the relative size of the constituent pieces. The cost of goods sold, as shown in Figure 13.1, is the total cost of manufacturing and marketing a product. An amount of profit is added to this total cost to arrive at a selling price. Such a cost structure is helpful in arriving at a unit cost, which is a primary objective of cost accounting. The term cost of goods sold, as used here, has a different meaning than when used in general accounting practice, particularly for retail businesses. General accounting defines this term to be beginning‐of‐the period inventory plus purchases minus end‐of‐period inventory. Different meanings for the same terminology are unfortunate, but they do occur in the literature and in practice, and the reader is cautioned on this point. To simplify the treatment of the total cost of goods sold (as defined by Figure 13.1), the major cost elements can be defined as direct material, direct labor, and overhead costs. Cost of Goods Sold The total cost of manufacturing and marketing a product. Overhead Costs All costs of manufacturing other than direct material and direct labor, including indirect labor and indirect material. Direct material costs and direct labor costs are the costs of material and labor that are easily measured and can be conveniently allocated to a specific operation, product, or project. Indirect costs for both labor and material, on the other hand, are either very difficult or impossible to assign directly to a specific operation, product, or project.

The expense of directly assigning such costs is prohibitive, and costs are therefore considered to be indirect for accounting purposes. Direct Costs Generally labor and material costs that can be easily measured and conveniently allocated to a specific operation, product, or project. Indirect Costs Generally labor and material costs that are difficult or impossible to assign directly to a specific operation, product, or project

EXAMPLE 13.5 Direct and Indirect Costs As an example of these different cost elements, suppose the raw material for a given part is a rectangular gray iron casting. The casting is milled on 5 sides, the unmachined surface is painted and air‐dried, and then 4 through holes are drilled and tapped. The finished parts are stacked in wooden boxes, 30 per box, and are delivered to a customer. In this example, the direct labor required per part to machine, paint, and package is probably readily determined. The labor required to receive the raw materials, handle parts between workstations, load boxes onto a truck, and deliver material to the customer is less easily identified and assigned to each part. This labor would be classified as indirect, especially if the labor in receiving, handling, shipping, and delivery is responsible for dealing with many different parts during the normal workday. The unit purchase price of the gray iron casting is an identifiable direct material cost. The cost of paint used per part may or may not be easily determined; if it is not, it is an example of indirect material cost. Also, any lubricating oils used during the machining processes would be an indirect material cost, not readily assigned on a cost‐per‐part basis. Overhead costs consist of all costs of manufacturing other than direct material and direct labor. A given firm may identify different overhead categories, such as factory overhead, general and administrative overhead, and marketing expenses. Furthermore, overhead amounts may be allocated to a total plant, to departments within a plant, or even to a given piece of equipment. Typical specific items of cost included in the general category of overhead are indirect materials, indirect labor, taxes, insurance premiums, rent, maintenance and repairs, supervisory and administrative personnel (technical, sales, and management), and utilities (water, electric power, etc.). Depreciation expenses are also usually included in the general overhead but may occasionally be considered a part of direct costs. It is the task of cost accounting to assign a proportionate amount of these costs to various products manufactured or to services provided by a business organization. This topic is addressed in more detail later in this chapter.

13.1.4 Fixed and Variable Viewpoint Fixed costs do not vary in proportion to the quantity of output. General administrative expenses, taxes and insurance, rent, building and equipment depreciation, and utilities are examples of cost items that are usually invariant with production volume and hence are termed fixed costs. Such costs may be fixed only over a given range of production; they may then change and be fixed for another range of production. Hence the concept of a relevant range is frequently associated with fixed costs. For example, lighting costs in a manufacturing plant may be considered fixed for single‐shift operation, regardless of how many units are produced. However, if demand warrants adding a second shift, lighting costs will increase to a new, higher‐valued, fixed cost. In this case, we would define lighting costs to be fixed with two relevant ranges. This suggests that fixed costs can, in some cases, be represented graphically as a step function. The reader is cautioned to recognize that if the width of the relevant ranges becomes small, the cost may be more appropriately treated as a variable cost rather than a fixed cost. Fixed Costs Costs that do not vary in proportion to the quantity of output. Variable costs vary in proportion to quantity of output. These costs are usually for direct material and direct labor. For instance, in Example 13.5, gray iron casting was used to produce a part. This cost was presented to illustrate direct material cost, because it can be directly traced to the product produced. The cost of the castings is also an

example of a variable cost. This is the case because the cost of castings is directly proportional to the number of parts produced. In other words, if the factory produces and sells 1,000 parts, there will be an associated cost of 1,000 times the unit cost of the castings. Variable Costs Costs that vary in proportion to quantity of output. This example also illustrates another point—the notion of multiple viewpoints within a single analysis. If we are using the manufacturing cost structure viewpoint, the cost of castings is direct material. If we are using the fixed variable viewpoint, the cost of castings is a variable cost. The fact that this cost can be classified in two ways does not constitute a flaw in our representation scheme; it simply means the way we interpret a particular cost is dependent upon the viewpoint we have adopted. The intent of having and using multiple viewpoints is to facilitate communication with different audiences and to help the engineering economist ensure that all relevant costs have been incorporated in the analysis. Embracing the second of these two points is best assured by questioning each alternative’s cash flows from each of the various viewpoints. For example, the engineering economist might ask, “Have we considered all costs within the manufacturing cost structure?” and “Have we considered all fixed or variable costs?” and “Which (if any) of these costs we are considering are sunk costs?” and so on. Many cost items have both fixed and variable components. For example, a plant maintenance department may have a constant number of maintenance personnel at fixed salaries over a wide range of production output. However, the amount of maintenance work done and replacement parts required on equipment may vary in proportion to production output. Thus, total annual maintenance costs for a plant over several years would consist of both fixed and variable components. Indirect labor, equipment depreciation, and electrical power are other cost items that may consist of fixed and variable components. Determining the fixed and variable portion of such a cost item may not be possible; if it is possible, the expense of establishing detailed measurement techniques and accounting records may be prohibitive. A comprehensive discussion on this issue is beyond the scope of this book, and the reader is referred to books on general cost accounting for further information. Certain total costs (TC) can be expressed as the sum of fixed costs (FC) and variable costs (VC). As an example, the total annual cost for operating a personal automobile for a given year might be expressed as TC(x) = FC + VC(x)

(13.1)

where x = miles per year. Costs for insurance, license tags, depreciation, certain maintenance, and interest on borrowed money if the automobile was financed are essentially fixed costs, independent of the miles traveled per year. Expenses for gasoline, oil, tire replacements, and certain maintenance are proportional to, or functional with, the mileage per year. One could argue that depreciation expenses are comprised of both fixed and variable components, because wear and tear on an automobile increases as the number of miles driven increases. We will not pursue that argument here. Arbitrarily assigning numerical values to the total cost function, assume that TC(x) = $950 + $0.15x

is a valid relationship for a given year in question (the relevant range is restricted to a given year, because actual depreciation expenses, and hence the fixed expenses, may vary from year to year). This relationship is linear in terms of x, miles driven. In general, the variable cost component of total cost is not always a linear function. However, in many real situations, the relationship is sufficiently linear as to make the assumption of a linear relationship reasonable. The reader is cautioned not to make this assumption without first examining the data. Figure 13.2 graphically illustrates the total cost function.

FIGURE 13.2 Total Annual Cost as a Function of Mileage Now let us consider Figure 13.2 as a total cost function for a production line in a manufacturing firm where the output from the line is a single product. Furthermore, let it be assumed that each unit of production can be sold for $R and that the total revenue (TR) is a linear function of the production quantity: (13.2)

TR(x) = $Rx

Adding this functional relationship to Figure 13.2 and modifying the terminology for this example yields Figure 13.3.

FIGURE 13.3 Revenue and Cost as a Function of Production Volume In Figure 13.3 as noted in Chapter 11, the total annual revenue equals the total annual cost at the point labeled “break‐even point.” The break‐even production volume, referred to as x*, is the point at which the total revenue equals the total cost, as shown in Equation 13.3. ∗

TR(x )

(13.3)

∗

=

TC(x )

=

FC + VC(x )

∗

Break‐Even Point Point where total revenues and total costs are equal. Certain important observations can now be made. If the production volume is less than x*, an annual net loss will occur, the amount of which is equal to TC(x) − TR(x), evaluated for a particular value of x. By the same token, if the production volume is greater than x*, then an annual net revenue or profit will result. The amount of annual profit is equal to TR(x) − TC(x), evaluated for a particular value of x.

It is generally desirable to have a small break‐even value. For the general example of Figure 13.3, this can be accomplished in three ways: (1) increasing the slope of the total revenue line, (2) decreasing the slope of the variable cost line, and (3) decreasing the magnitude of the fixed cost line. Increasing the slope of the total revenue line means increasing the product’s selling price, which may be a poor marketing strategy in a competitive market environment where sales would be lost. Fixed costs, although not literally fixed in all cases, can be difficult to reduce. Thus, reducing variable costs for direct material and labor usually offers the first point of attack for the engineer or analyst for profit improvement. The concept of break‐even analysis is general. Assuming that a break‐even point exists for two relationships—y = g(·) and y = h(·)—that are functions of a single variable x, the value of x for break‐even, say x*, may be determined from equating g(x) = h(x) and solving for x*. The concept can be extended to more than two functions of a single variable, say y = h(x), y = g(x), and y = t(x). If these are all linear functions, then Figure 13.4 depicts two of the possible results.

FIGURE 13.4 Break‐Even Points (a) Single, (b) Multiple In Figure 13.4a, all three functions intersect at a single point; thus, a single value of x* can be determined (Point A). In Figure 13.4b, there is no unique break‐even value of x involving all three functional relationships. The linear equations y = h(x) and y = t(x) intersect at Point B, which is the break‐even value for these two relationships. Point C is the break‐even value for y = h(x) and y = g(x). Point D is the break‐even value for y = g(x) and y = t(x). The concept of break‐even analysis also extends to nonlinear functions, with one or more break‐even values and functions of more than a single variable, which may be of linear or nonlinear form. However, examples and problems dealing only with functions of a single variable will be presented in this chapter.

EXAMPLE 13.6 Break‐Even Point for Production Video Example The cost of tooling and direct labor required to set up for a machining job on a turret lathe is $300. Once set up, the variable cost to produce one finished unit consists of $2.50 for material and $1.00 for labor to operate the lathe. For simplicity, it is assumed these are the only relevant fixed and variable costs. If each finished unit can be sold for $5, determine (1) the production quantity required to break even and (2) the net profit (or loss) if the lot size is 1,000 units. Solution Letting x = the production volume in units, then R(x) = TC(x) = FC + VC(x)

and $5.00x = $300 + ($2.50 + $1.00)x

Simplifying and solving for x yields x* (the break‐even value): x

∗

= $300/$1.50 = 200 units

For a production output of 1,000 units, the net profit, P, is calculated to be P = $5(1,000 units) − ($2.50 + $1.00)(1,000 units) − $300 = $1,200

EXAMPLE 13.7 Alternative Analysis Using Break‐Even2 This example concerns selecting between two alternative methods of processing crude oil in a producing oil field, where the basis for the decision is the number of barrels of crude oil processed per year. The two methods of processing the crude oil are (1) a manually operated tank battery or (2) an automated tank battery. The tank batteries consist of heaters, treaters, storage tanks, and so forth, that remove saltwater and sediment from crude oil prior to its entrance to pipelines for transport to an oil refinery. For each alternative, fixed costs and variable costs are involved. Fixed costs include items such as pumper labor; maintenance (fixed over the production quantity of interest); taxes; certain energy costs (power to operate control panels and motors in continuous operation); and, for manual tank batteries, a cost for oil “shrinkage.” Variable costs for chemical additives, heating, and noncontinuous operating motors are proportionate to the volume of oil being processed. This relationship is assumed to be linear over the production quantity of interest. The necessary data are given in Tables 13.1 and 13.2 and are considered valid for production quantities up to 1,000 barrels/day (365,000 barrels/yr). TABLE 13.1 Cost Data for Automatic Tank Battery Operations Cost $/day Fixed cost Control panel power 0.15 Circulating pump power (3 hp) 0.82 Maintenance Meter Calibration Chemical pump power (1/4 hp) Total Fixed Cost Fixed Cost per year = $2.69 × 365 = $982 Variable Cost Pipeline pump (5 hp @ 50% utilization) Chemical additives (7:5 qt/day) Inhibitor (2 qt/day)

1.00 0.40 0.32 $2.69

0.63 3.75 1.00

Gas (10.8 mcf/day × $0.0275/mcf) 0.30 Total Variable Cost $5.68 Variable cost per barrel = 0.01136/bbl @ 500 bbl/day

TABLE 13.2 Cost Data for Manual Tank Battery Operation Cost

$/day

Fixed cost Chemical pump power Circulating pump power Total Fixed Cost

0.16 0.82 $0.98

Fixed Cost per year = $0.98 × 365 = $358 Variable Cost Chemical additives (7.5 qt/day) (10.8 mcf/day × $0.0275/mcf)

3.75 0.30

Total Variable Cost $4.05 Variable cost per barrel = 0.00810/bbl @ 500 bbl/day In addition to the fixed and variable costs given in Table 13.1 for the automatic tank battery operation, other annual fixed costs are D1

=

annual cost of depreciation and interest = $3,082

M1

=

annual cost of maintenance, taxes, and labor = $5,485

Solution Letting x = the number of barrels of oil processed per year, the total annual cost, TC1(x), for the automatic tank battery operations is given by TC1 (x)

=

FC1 + VC1 (x)

=

($982 + $3,082 + $5,485) + $0.01136x

=

$9,549 + $0.01136x

In addition to the fixed and variable costs given in Table 13.2 for the manual tank battery operations, other annual fixed costs are D2

=

annual cost of depreciation and interest = $2,017

M2

=

annual cost of maintenance, taxes, and labor = $7,921

Then, the total annual cost, TC2(x), for the manual tank battery operation is given by TC2 (x)

=

FC2 + VC2 (x)

=

($358 + $2,017 + $7,921) + $0.00810x

=

$10,296 + $0.00810x

By equating the two total cost functions, the break‐even production volume can be determined as TC1 (x)

=

TC2 (x)

$9,549 + $0.01136x

=

$10,296 + $0.00810x

=

229,141 barrels/year

x

∗

Exploring the Solution The interpretation of the break‐even point in this example is that x* = 229,141 barrels/year is the point of indifference between the choice of the two alternatives. If production volume is less than x*, then the first

alternative, or the automatic tank battery operation, would be preferred. For instance, if x = 200,000 barrels/year, then TC1(x) = $11,821 and TC2(x) = $11,916. Similarly, if production volume is greater than x*, the manual tank battery operation would be preferred. In our particular case, it would now be necessary to obtain a production estimate from the field engineers. Once determined, we could recommend which processing option is preferred.

13.1.5 Average and Marginal Viewpoint The average/marginal viewpoint deals with costs expressed in terms of units of output. While this sounds similar to variable cost, the concept here is significantly different. The average cost is the ratio of total costs incurred divided by the number of units produced while incurring those costs. The marginal cost is the incremental cost associated with increasing the output by 1 unit. The definition of marginal cost changes slightly if the output is continuous (e.g., gallons of paint) rather than discrete (e.g., number of automobiles). This difference is explored more fully in the material that follows. Average Cost The ratio of total costs incurred divided by the number of units produced. Marginal Cost The increase or decrease in the total cost of a production run for making one additional unit of an item. The average cost of 1 unit of output (unit cost) is the ratio of total cost to the quantity of output (miles traveled, production volume, etc.); that is, (13.4)

TC(x) AC(x) =

x

where AC(x)

=

average cost per unit of x

TC(x)

=

total cost for x units of output

x

=

output quantity

The average cost is usually a variable function of the output quantity and normally decreases with an increasing output quantity. Using the automobile example from Section 13.1.4, which had a total cost function of $(950 + 0.15x), the average cost, in dollars per mile, is given by 950 + 0.15x AC(x)

=

x

950 =

x

+ 0.15

If the automobile travels 10,000 miles/year, then the average operating cost is $(950/10,000 + 0.15) = $0.245/mile. For a total annual travel distance of 20,000 miles, the average operating cost decreases to $0.1975/mile. This can be explained by noting that as the output quantity x increases, the proportion of the fixed cost allocated to each unit of output decreases. This relationship is a fundamental economic principle often referred to as the economies of scale, which underlies the economic benefits of mass production. Such a relationship assumes that the variable cost coefficient remains constant over the range of the output variable x. In a production environment, it is possible that the variable cost coefficient will increase as the production volume increases due to increased maintenance expenses, defective product, and so on. Economies of Scale As production increases, the cost per unit of output decreases because fixed costs are spread over a larger output. For a total cost function that is continuous in the variable x, marginal cost is defined as the derivative of the total cost function (dependent variable) with respect to x, or dTC(x)/dx. This is true for continuous functions that are linear or nonlinear in the variable x. In the special case of a continuous total cost function that is linear in x, such as

TC(x) = $950 + 0.15x, then dTC(x)/dx = $0.15. In this case, marginal cost is the constant value $0.15 and is the cost required to increase the output quantity x by 1 unit. If the total cost function is discontinuous and defined only for discrete values of x (for example, x = 1,2,3, …), then difference equations must be used to determine marginal costs. For example, TC(6) − TC(5) is the marginal cost of increasing the output quantity from x = 5 to x = 6. Thus, in the discrete case, marginal cost is always the cost required to increase the output quantity x by 1 unit at a specified level of output. This is true for discrete total cost functions regardless of whether they are linear or nonlinear. Two additional observations are noteworthy with respect to average and marginal cost. First, examine the average cost equation above and consider what happens as x, the number produced, grows ever larger. As x increases, the proportion of the fixed cost assigned to each unit grows increasingly smaller, approaching 0 in the limit. This implies that, under the assumptions above, the average cost converges to the marginal cost. In the limit (when x is infinity) the average cost equals the marginal cost. A second observation is related to the emerging concept of economies of scope. In contrast to economies of scale, which encourages large production runs to drive down average cost, economies of scope suggests that fixed costs be driven as low as possible, ideally to 0. Note that in terms of our average cost equation above, this implies that the point of attack is to reduce the $950 numerator of the fixed cost term. Similar to the economies of scale approach, the limiting case for average cost is the $0.15 marginal cost per unit. However, unlike economies of scale, economies of scope strives to disconnect the ideal average cost from the production volume. Thus, costs are reduced due to the elimination (or at least minimization) of the need to create large inventories to drive down average cost. This approach has the added advantage of enhancing both cost performance and production flexibility. Economies of Scope Concept that suggests fixed costs be driven as low as possible, ideally to zero.

EXAMPLE 13.8 Marginal Cost For a certain production process, fixed costs are $60,000. Variable costs are $30 per unit of production. Therefore, the total cost function is given by TC(x) = 60,000 + 30x. What is the marginal cost at x = 10? x = 20? Video Example 13.8 Solution There are two ways to solve this problem: (1) difference equations, and (2) differentiation. For purposes of illustration, both approaches will be demonstrated. Difference Equations Marginal Cost at 10

=

Total Cost at 11 − Total Cost at 10

MC(10)

=

TC(11) − TC(10)

MC(10)

=

[$60,000 + $30(11)] − [$60,000 + $30(10)]

MC(10)

=

$60,330 − $60,300

MC(10)

=

$30

Similarly, MC(20) = TC(21) − TC(20) MC(20) = [$60,000 + $30(21)] − [$60,000 + $30(20)] MC(20) = $60,630 − $60,600 MC(20) = $30

Differentiation Marginal Cost at 10 = first derivative of the total cost function evaluated at 10 MC(x)

=

d/dx TC(x)

MC(x)

=

d/dx [$60,000 + $30x]

MC(x)

=

$30

therefore MC(10)

=

$30 and MC(20)

=

$30

The concept of marginalism is general and applies to other mathematical functions as well. For example, marginal revenues can be determined from total revenue functions, marginal profit values can be determined from total profit functions, and so forth. Marginal revenue (profit) is the additional revenue (profit) received from selling 1 more unit of the output quantity x at a specified level of output. Marginal and average values corresponding to a specified output quantity are generally different. If the marginal cost is smaller than the average cost, an increase in output will result in a reduction of average cost. This can be seen by recalling the now familiar automobile problem, when TC(x) = $950 + $0.15x. The average cost is AC(x) = $950/x + $0.15 and the marginal cost is MC(x) = $0.15. Thus, for all nonnegative finite values of x, marginal cost is always smaller than the average cost, and the unit cost will continue to decrease as x is increased. Such a relationship is not true in general for nonlinear total cost functions. Tables 13.3 and 13.4 summarize the relationships between marginal cost and total cost (Table 13.3) and between marginal cost and average cost (Table 13.4). The following example illustrates some of these cost, revenue, and profit relationships.

TABLE 13.3 Relationship Between Marginal Cost and Total Cost If MC(x) > 0 then TC(x + 1) > TC(x) If MC(x) = 0 then TC(x + 1) = TC(x) If MC(x) < 0 then TC(x + 1) < TC(x) TABLE 13.4 Relationship Between Marginal Cost and Average Cost If MC(x) < AC(x) then AC(x + 1) < AC(x) If MC(x) = AC(x) then AC(x + 1) = AC(x) If MC(x) > AC(x) then AC(x + 1) > AC(x)

EXAMPLE 13.9 Cost, Revenue, and Profit Relationships A small firm blends and bags chemicals, primarily for home gardening purposes. The market area for the firm is local, and all sales are to wholesale distributors. For one pesticide dust product, sales and production cost records over the past 10 seasons have been reviewed and analyzed. The following equations approximate the relationships among selling price, sales volume, production costs, and profit before income taxes. (The functional form for selling price implicitly reflects a fundamental relationship between price and demand. Namely, as the selling price is decreased, demand for the item increases. Alternatively, in order to increase the demand, the selling price must be reduced.) Given Let t

=

number of tons per season

SP(t)

=

selling price in order to sell t tons

=

$(800 − 0.8t)

=

total revenue when t tons are sold at a particular selling price

=

selling price X demand

=

$(800 − 0.8t)t

=

$(800t − 0.8t )

=

the marginal revenue at a sales volume of t tons

=

dTR(t)/dt =$(800 − 1.6t)

=

the total production cost for t tons

=

$(10,000 + 400t)

=

total profit when t tons are sold

=

TR(t) − TC(t)

=

$(800t − 0.8t ) −$(10,000 + 400t)

=

$(−0.8t

=

average profit per ton when t tons are sold

=

TP(t)/t

=

$(−0.8t + 400 − 10,000/t)

TR(t)

MR(t)

TC(t)

TP(t)

AP(t)

2

2

2

+ 400t − 10,000)

The equations apply for the range 0 ≤ t < 1,000. Figure 13.5 on the following page illustrates the total revenue, total cost, and total profit curves for this example.

FIGURE 13.5 Total Revenue, Total Cost, and Total Profit as a Function of Tons Excel® Data File Solution We first determine that the total revenue will be maximized when 500 tons are produced and sold per season —that is, by calculus, we take the first derivative of the revenue function, set it equal to 0, and solve for t. dTR(t)/dt = 800 − 2(0.8)t = 0

resulting in t

∗

= 500 tons

The total revenue with a sales volume of 500 tons is TR(500)

=

$800(500) − $0.8(500)

=

$200,000

2

The marginal revenue at the output level of 500 tons is MR(500) = $800 − $1.6(500) = 0

Thus, the rate of change in the TR(t) function with respect to t is 0 when TR(t) is evaluated at t = 500. The TR(t) function is strictly concave with a unique maximum value at TR(500). For sales from t = 1 to 500, the total revenue function is increasing at a decreasing rate. For sales volumes from t = 500 to 1,000, total revenues are decreasing at an increasing rate. Maximizing total revenues is not the issue in this example, however. We wish to maximize profits. Again, by calculus, we take the first derivative of the profit function, set it equal to 0, and solve for t. dTP(t)/dt = 2(−0.8)t + 400 = 0

resulting in t

∗

= 250 tons

Thus, total profit will be maximized for a sales volume of 250 tons, and the maximum profit per season would be TP(250) = −$0.8(250)

2

+ $400(250) − $10,000 = $40,000

The average profit per ton when 250 tons are sold is AP(250) = −0.8(250) + 400 − 10,000/250 = $160/ton

Finally, we note that there are two break‐even points in this example. By equating TR(t) = TC(t), or 800t − 0.8t

2

= 10,000 + 400t

we obtain −0.8t

2

+ 400t − 10,000 = 0

Solving for the positive roots of this quadratic equation yields t = 26.39 and t = 473.61

For a sales volume in the range 26.39 ≤ t ≤ 473.61, the firm will make a profit. Sales volumes outside this range will result in total costs exceeding total revenues and a net loss to this firm.

Concept Check 13.01‐CC001 The life cycle costs of an asset include which of the following? I. Purchase cost of the asset II. Shipping and installation cost of the asset III. Operating and maintenance cost of the asset IV. Salvage value of the asset a. I and IV only b. I, III, and IV only c. I and III only d. I, II, III, and IV Correct or Incorrect? Clear

  Check Answer

Concept Check 13.01‐CC002 Which of the following terms is associated with giving up the chance to earn a return on investment funds? a. Opportunity cost b. Cost of capital c. Sunk cost d. Past cost Correct or Incorrect? Clear

  Check Answer

Concept Check 13.01‐CC003 The three major components of cost of goods sold are a. Direct labor, direct material, and profit b. Direct labor, overhead, and profit c. Direct labor, direct material, and overhead d. Direct material, indirect material, and marketing Correct or Incorrect? Clear

  Check Answer

Concept Check 13.01‐CC004 The concept of a break‐even point a. Applies only to two functions b. Applies only to linear functions c. Applies only to two linear functions d. Is general and can be applied to multiple functions, including both linear and nonlinear functions Correct or Incorrect? Clear

  Check Answer

Concept Check 13.01‐CC005 The ratio of total cost to number of units produced defines a. Incremental cost b. Average cost c. Marginal cost d. Opportunity cost Correct or Incorrect? Clear

  Check Answer

13.2 Cost Estimation LEARNING OBJECTIVE Apply cost estimating principles to determine costs for given scenarios. The Association for the Advancement of Cost Engineering International (www.aacei.org) defines cost estimating as Video Lesson: Cost Estimation Cost Estimating (as defined by the Association for the Advancement of Cost Engineering International) “A predictive process used to quantify, cost, and price the resources required by the scope of an asset investment option, activity or project. As a predictive process, estimating must address risks and uncertainties. The outputs of estimating are used primarily as inputs for budgeting, cost or value analysis, decision making in business, asset and project planning, or for project cost and schedule control processes.” a predictive process used to quantify, cost, and price the resources required by the scope of an asset investment option, activity or project. As a predictive process, estimating must address risks and uncertainties. The outputs of estimating are used primarily as inputs for budgeting, cost or value analysis, decision making in business, asset and project planning, or for project cost and schedule control processes. Webster’s New Collegiate Dictionary states for estimate, “the comprehensive term, implies personal judgment the significance of which can only be made clear by the context.” These definitions make it clear that estimating, in particular cost estimating, is not an exact science. Rather, it is an approximation that involves the availability and relevancy of appropriate historical data, personal judgments based on the estimator’s experience, and the time frame available for completing the estimating activity. Cost estimation is one of the most difficult challenges an engineering economist faces. It is difficult because it involves future events that are not (cannot be) known with certainty. Nonetheless, estimation of the amounts and timing of future cash flows is a necessary part of engineering economic analysis. Many different terms pertain to the general subject of estimation. This text will not attempt to enumerate and explain all the terms exhaustively. Selected terminology will be given as needed to explain the topics. Furthermore, an in‐depth study of estimation procedures and the accuracy of estimated values is the study of mathematical statistics and probability theory, about which a vast literature exists.

It is difficult to state precisely in quantitative terms the relationship between the accuracy of an estimate and the cost of making the estimate. Intuitively, as more detailed information is obtained to provide the basis for an estimate and as more mathematical precision is used in calculating the estimate, the more accurate the estimate should be. However, as the level of detail increases, the cost involved in making the estimate increases. Ostwald and McLaren3 have conceptualized this notion by the function CT = C(M) + C(E)

(13.5)

where CT

=

the total cost of making the estimate in dollars

C(M)

=

the functional cost of making the estimate in dollars

C(E)

=

the functional cost of errors in the estimate in dollars.

As depicted in Figure 13.6, the optimal amount of detail is that which minimizes the total cost of estimating. Because the curves for C(M) and C(E) are in general nonlinear, the minimum cost will not necessarily occur at the point of the two curves’ intersection. Quantitatively determining the optimal amount of detail is at best difficult and may be a practical impossibility. However, this concept of the total cost of an estimate varying with the amount of detail involved in making the estimate is realistic and is important to the general subject of estimation. In the abbreviated discussion on cost estimation techniques that follows, we note that the individual techniques are based on varying amounts of detail with implied differences in the cost of making the estimate.

FIGURE 13.6 Cost of Detail in Estimating

13.2.1 Project Estimation In this textbook, we are concerned with estimation in the specific context of comparing alternative engineering investment projects and making a selection from among these projects. The annual revenues or savings, the initial and annual recurring costs, the life of a project, and the future salvage value of capital assets such as buildings and equipment that may be associated with a given project are rarely, if ever, known with certainty. For all categories of estimated items previously mentioned, four classes of estimates, based on accuracy and degree of detail, can be defined: (1) order‐of‐magnitude estimates, (2) preliminary estimates for feasibility studies, (3) semidetailed estimates for budget authorization, and (4) detailed estimates for execution and control. Order‐of‐magnitude estimates are useful for concept screening and are usually gross estimates based on experience and judgment and made without formal examination of the details involved. Preliminary estimates are useful for feasibility studies and are also gross estimates, but more consideration is given to detail in making the estimate than for order‐of‐magnitude estimates. Key subelements of the overall task are individually estimated; engineering specifications are considered, and so forth. Semidetailed estimates are useful for budget authorization and are made by expanding the list of individually estimated items from key subelements to all major subelements. Finally, detailed estimates, which are used for execution and control, consider each subelement individually. Detailed

estimates are expected to result in the most accurate estimate of actual cost. In preparing the estimate, each subelement of the overall task is considered, and an attempt is made to assign a realistic cost to it. Pricing a product or contract bidding usually involves detailed estimates of the costs involved. Given the uncertainty inherent in cost estimating, a systematic approach is called for. The Federal Aviation Administration’s (FAA) Life Cycle Cost Estimating Handbook4 suggests the following 6‐step process: 1. Plan the estimate 2. Research, collect, and analyze data 3. Develop the estimate structure 4. Determine the estimating methodologies 5. Compute the cost estimate 6. Document and present the estimate Planning the Estimate Planning the estimate focuses on determining its intended use and the initial identification of an anticipated estimating methodology. The four‐category classification scheme outlined earlier in this section (order‐of‐ magnitude, etc.) or a more comprehensive five‐level classification scheme available to members of the Association for the Advancement of Cost Engineering International (www.aacei.org) are useful in defining the estimate’s intended use. Appropriate time and attention are warranted at this step of the estimating process, because this step provides the framework for the remainder of the estimating process. Data Research, Collection, and Analysis The second step—data research, collection, and analysis—includes (1) determining the availability of the data required by the initial methodology, (2) collecting the data, and (3) assessing the applicability of the collected data. Part 1 of this effort focuses on an initial determination of the availability of data sources to support the categories of data and the anticipated methodology of the estimating process. If data sources are available, then initial collection can proceed; if not, a different approach must be considered. As data collection begins, the types of data collected generally fall within two broad categories: cost estimating relationship and historical cost. Relationship data are used in conjunction with mathematical functions to estimate a factor of interest (dependent variable) based on one or more related factors (independent variables). Historical data are typically time‐series data that represent or are directly related to the factor of interest. The final part of this step involves assessing the applicability of the collected data. It is not uncommon at this stage that iterations to earlier parts of this step, and potentially back to the planning step, are required. Developing the Estimate Structure During the initial data collection, broad categories of cost were identified, assessed for data availability, and initial data collected. At the current stage (developing structure), the initial data requirements are refined with additional detail, and additional structure is added. This involves breaking down the broad categories into discrete cost elements. Frequently the discrete cost elements are derived and validated by considering the various cost viewpoints presented in Section 13.2. This process minimizes the chances that a cost element will be overlooked or omitted. Determining the Estimating Methodology The choice of a good estimating methodology is an important factor in developing a good estimate. Three primary methodologies are (1) parametric estimating, (2) estimating by analogy, and (3) engineering estimates. Each of these methods is discussed in more detail in Section 13.3.2. Computing the Cost Estimate At this step, the data, the estimate structure, and the methodology come together to facilitate the calculation of the estimate itself. Frequently, a spreadsheet model is used to support the calculations and documentation of the process. In addition to the quality of the estimate, the engineering economist frequently incorporates two additional issues at this step: time phasing of the estimate and consideration of the impact of inflation.

Documenting and Presenting the Estimate The estimating task is not finished when numbers are determined. The estimate must be carefully documented and presented to the decision maker in a style and format that communicates effectively. Four key concerns are important. First, the documentation process should be completed concurrently with the estimate’s development. Postponing documentation until after the estimating process is complete inherently leads to incomplete documentation. Second, documentation should be complete and step‐by‐step. The estimator should not arbitrarily assume that the decision maker has in‐depth knowledge of the items being estimated or the estimating process. Third, the documentation should contain sufficient information to allow replication or enhancement of the estimate. Frequently, at this stage, the estimate will transfer into the decision maker’s hands. As the decision making process proceeds, refinements to the estimate are likely. The documentation should support this likelihood. Fourth, in many cases, the estimate documentation will be the primary means through which the credibility of the estimate and the estimator will be judged. Poor documentation leads to poor credibility, which leads to the decision maker lacking confidence in the estimate.

13.2.2 Estimating Methodologies Three primary methodologies for estimating are (1) parametric estimating, (2) estimating by analogy, and (3) engineering estimates. Each of these methods is discussed in more detail below. Parametric Estimating Parametric estimating develops estimates based on characteristics or features of the item being estimated. This process relies on a proven or assumed causal relationship between the characteristic and the cost. The relationships are usually expressed mathematically, frequently resulting from a regression analysis. (Detailed presentation of regression analysis is beyond the scope of this text, but note that Excel® includes a significant suite of regression tools. Additional regression capabilities for Excel® are available as add‐ins.) The primary advantage of the parametric method is that it can quickly produce good‐quality estimates with limited project detail. The primary disadvantage is that parametric estimates do not produce low‐level detail, and they assume that the relationships represented in regressed data are truly cause and effect and that they will continue similarly into the future. Estimating by Analogy Forecasting by analogy is based on the premise that no new project is totally new. Many, if not most, new projects and systems evolve from their predecessors. As such, many of the cost elements and cost relationships can be approximated from the predecessor. The idea is that costs for similar elements are best estimated from the predecessor. For components that have evolved, adjustments are made for complexity and technical or physical differences. The primary advantage of the analogy method is that, if a good analogy can be found, it facilitates development of a low‐level forecast relatively quickly. The primary disadvantage is identifying and verifying that the selected analogy is, in fact, appropriate. Engineering Estimating The engineering estimating method is also referred to as bottom‐up estimating. The process starts at the lowest level of detail available for the project, considers each cost element, and builds up to the total cost estimate. Detailed engineering estimating takes relatively large amounts of time and requires detailed information about the project. The primary advantage of this method is that the level of detail gives high credibility to the estimate developed. The primary disadvantage is that the time and information requirements are high.

13.2.3 General Sources of Data There are many sources for providing data to make the various estimates required in comparing alternative investment projects. Sources may be either internal or external to the firm. Examples of sources within a firm are sales records, production control records, inventory records, quality control records, purchasing department records, work measurement and other industrial engineering studies, maintenance records, and personnel records.

The accounting system can and usually does serve as an important, if not primary, internal source of detailed estimates on operating costs, maintenance costs, and material costs, among others. Sources of data external to the firm may be grouped into two general classes: (1) published information that is generally available and (2) information (published or otherwise) available on request. Available published information includes the vast literature of trade journals, professional society journals, U.S. government publications, reference handbooks, other books, and technical directories. Information not generally available except by request includes many sources listed in the previous category. For instance, many professional societies and trade associations publish handbooks, other books, special reports, and research bulletins that are available on request. Manufacturers of equipment and distributors of equipment are excellent sources of technical data, and most will readily supply this information without charge. Additionally, various government agencies, commercial banks (particularly holding companies involved in leasing buildings and equipment), and research organizations (commercial, governmental, industrial, and educational) may be sources of data to aid the estimating process. Estimates of the functionally useful physical life of a piece of equipment may be obtained from manufacturers and suppliers. Alternatively, if a company repeatedly buys a particular piece of equipment and keeps accurate maintenance records, these records may be used to obtain an estimate of the item’s functional life. For example, suppose records reveal that 100% of the items survive the first 3 years of service. Then, 10% of the items fail in the fourth year, 20% fail in the fifth year, 50% in the sixth year, 15% in the seventh year, and the remaining 5% fail in the eighth year. A weighted‐average time to failure for this equipment is [(0.10)(4) + (0.20)(5) + (0.50)(6) + (0.15)(7) + (0.05)(8)], or 5.85 years. This is an estimate of this particular item’s functional life. The texts by Ostwald and McLaren (footnote 1) and by Stewart, Johannes, and Wyskida5 provide detailed presentations of well‐founded general methodologies for cost estimating. The interested reader is referred to these texts for additional detail.

Concept Check 13.02‐CC001 As the level of detail in a cost estimate increases, the cost of making the estimate usually ____ and the cost of errors resulting from the estimate usually ____. a. increases; decreases b. increases; increases c. decreases; decreases d. decreases; increases Correct or Incorrect? Clear

  Check Answer

13.3 General Accounting Principles LEARNING OBJECTIVE Demonstrate proficiency with balance sheets, income statements, and ratio analysis. As mentioned, the engineer should have some understanding of basic accounting practice and cost accounting techniques in order to obtain data from the firm’s accounting system. The study of accounting is commonly divided into financial accounting and managerial accounting. Managerial accounting (particularly the subcategory of cost accounting) is more important to the engineer as a source of data for making cost estimates pertinent to engineering projects. Cost accounting will therefore receive the greater emphasis in this text. Our treatment of

accounting is general and high level and is directed toward fundamental accounting concepts rather than a comprehensive treatment of accounting detail. Video Lesson: General Accounting Principles Accounting is the language of business. Without an understanding of this language, it is virtually impossible for an engineer to acquire and correctly interpret the data needed for economic analysis or to communicate the results of an analysis in meaningful and significant terms to managers. Learning this language is also crucial for engineers who aspire to progress through a technical career track to higher levels of authority and responsibility within a company. The American Institute of Certified Public Accountants defines accounting as “the art of recording, classifying and summarizing in a significant manner and in terms of money, transactions and events which are, in part at least, of a financial character, and interpreting the results thereof.” This definition embodies the four key elements of an accounting system: recording, classifying, summarizing, and interpreting the financial data of an organization, whether profit or nonprofit. General accounting information is summarized in two basic financial reports: the balance sheet and the income statement. These two financial statements will be the focus of our treatment of accounting. The balance sheet provides a statement of a firm’s financial condition at a point in time and lists the values of the assets, liabilities, and net worth of the firm. The income statement details the revenues and expenses incurred by a firm during a period of time, usually a month, quarter, or year. The two statements are closely related. The income statement summarizes the financial activities that occur between two balance sheets. The balance sheet reflects the financial condition as a result of the activities reported in the income statement. This relationship is illustrated in Figure 13.7.

FIGURE 13.7 Relationship Between Balance Sheets and Income Statements

13.3.1 Balance Sheet The items listed on a balance sheet are usually classified into three main groups: assets, liabilities, and net worth items. Subgroups may also be identified, such as current and fixed assets or current and fixed liabilities. Assets are properties owned by the firm, and liabilities are debts owed by the firm against these assets. The dollar difference between assets and liabilities is the net worth of the business, which measures the investment made by the owners or stockholders of the business plus any accumulated profits left in the business by the owners or stockholders. Another term for net worth is owners’ equity. Balance Sheet A statement of the financial condition of a firm at a point in time divided into three main groups: assets, liabilities, and net worth items. The fundamental accounting equation is defined as Assets  −  Liabilities = Net Worth

Rewriting, we have the more common form of the fundamental equation of accounting: Assets = Liabilities + Net Worth

(13.6)

The usual format of a balance sheet follows the equation in this form. This can be observed in the condensed version of Starbucks’s balance sheet for fiscal year 2018, shown in Figure 13.8. Note in the figure that total assets equals the sum of total liabilities plus total net worth (called “shareholders’ equity” on Starbucks’s balance sheet). The double underline under the balancing totals is the accountant’s verification that the balance sheet is indeed balanced according to the fundamental equation of accounting.

FIGURE 13.8 Starbucks’ Condensed Consolidated Balance Sheet STARBUCKS CORPORATION CONSOLIDATED BALANCE SHEETS (in millions, except per share data) Sep 30, 2018

Oct 1, 2017

ASSETS Current assets: Cash and cash equivalents

$ $ 2,462.3 8,756.3

Short‐term investments Accounts receivable, net

181.5 693.1

Inventories Prepaid expenses and other current assets Total current assets Long‐term investments Equity and cost investments Property, plant and equipment, net Deferred income taxes, net Other long‐term assets Other intangible assets Goodwill

228.6 870.4

1,400.5 1,364.0 1,462.8    358.1 12,494.2

5,283.4

267.7 334.7

542.3 481.6

5,929.1 134.7

4,919.5 795.4

412.2

362.8

1,042.2 441.4 3,541,6   1,539.2

TOTAL ASSETS

$24,156.4 $14,365.6 LIABILITIES AND EQUITY

Current liabilities: Accounts payable

$ $  782.5 1,179.3

Accrued liabilities Insurance reserves

2,298.4 213.7

Stored value card liability and current portion of deferred revenue Current portion of long‐term debt

1,642.9 1,288.5 349.9       —

Total current liabilities Long‐term debt Deferred revenue Other long‐term liabilities Total liabilities

1,934.5 215.2

5,684.2

4,220.7

9,090.2 6,775.7

3,932.6 4.4

1,430.5    750.9 22,980.6 8,908.6

Shareholders’ equity: Common stock ($0.001 par value)—authorized, 2,400.0 shares; issued and outstanding, 1,309.1 and 1,431.6 shares, respectively Additional paid‐in capital Retained earnings Accumulated other comprehensive loss

1.3

1.4

41.1

41.1

1,457.4

5,563.2

(330.3)   (155.6)

Total shareholders’ equity Noncontrolling interests Total equity TOTAL LIABILITIES AND EQUITY

1,169.5

5,450.1

   6.3      6.9 1,175.8 5,457.0 $24,156.4 $14.365.6

Excel® Data File Usually each of the broad categories on the balance sheet (assets, liabilities, and net worth) are divided into subcategories. While reading the following material, the reader is encouraged to review Starbucks’s balance sheet (Figure 13.8) to reinforce and confirm the points made. Current assets include cash and other assets that can be readily converted into cash; a period of 1 year or 1 business cycle is usually assumed as a criterion for conversion. Similarly, current liabilities are the debts that are due and payable within 1 year (or 1 business cycle) from the date of the balance sheet in question. Typical current‐asset items are cash, accounts receivable, notes receivable, raw material inventory, work in process, finished goods inventory, and prepaid expenses. By contrast, fixed assets are the properties the firm owns that are not readily converted into cash within 1 year, and fixed liabilities are long‐term debts due and payable after 1 year from the balance sheet’s date. Fixed‐asset items are land, buildings, equipment, furniture, and fixtures. Items that are typically listed under current liabilities are accounts payable, notes payable, interest payable, taxes payable, prepaid income, and dividends payable. Fixed liabilities include notes payable, bonds payable, mortgages payable, and so forth. Net worth items appearing on a balance sheet are less standard and, to a degree, depend on whether the business is a sole proprietorship, a partnership, or a corporation. The corporation’s size is also an influencing factor on item designation. However, items such as capital stock, retained earnings, capital surplus, or earned surplus appear under the net worth group. To better understand balance sheets, we will examine several parts of a small tool manufacturing company’s balance sheet, shown in Table 13.5. Of first importance, note that this sample balance sheet is “balanced” because assets = liabilities + net worth (i.e., the fundamental equation of accounting holds). Note the fixed‐asset portion of the balance sheet. The building originally cost $200,000, and depreciation expenses have been charged annually, so the total depreciation charges (as of the balance sheet’s date) have been $50,000, the amount entered as depreciation reserve for the building. In theory, the depreciation reserve is an accumulated amount of funds held to repurchase the asset when its functionally useful life terminates. The first cost of the depreciable asset (building, in this case) minus the amount in the depreciation reserve equals the book value. If the book value were a true estimate of the salvage, or market, value, then the sum of the amount in the depreciation reserve plus the book value provides the firm with an amount of funds equal to the original purchase price. This sum can be applied toward the purchase of a replacement asset. Typically, however, the market value of the asset when sold differs from the book value. The difference results in either a capital gain or loss and affects the firm’s income taxes.

TABLE 13.5 Example Balance Sheet BuiltRite Tool and Engineering Company Balance Sheet as of December 31, 2018 ASSETS Current Assets Cash

$ 25,000

Accounts Receivable (net) Raw Materials

115,000 8,500

Work In Process

7,000

Finished Goods Small Tools

3,000   12,500

Total Current Assets Fixed Assets Land

$171,000 $ 30,000

Building Less depreciation reserve

$200,000   50,000 150,000

Equipment Less depreciation reserve

$750,000  150,000 600,000

Office Equipment

  10,000

Total Fixed Assets Total Assets

$790,000 $961,000

LIABILITIES AND CAPITAL Current Liabilities Accounts Payable Taxes Payable Total Current Liabilities

32,000   15,000 47,000

Fixed Liabilities Mortgage loan payable Equipment loan payable

$130,000  350,000

Total Fixed Liabilities Total Liabilities

$480,000 $527,000

Capital Common Stock Retained Earnings

$325,000 80,000

Earned Surplus (current year)

  29,000

Total Capital Total Liabilities and Capital

$434,000 $961,000

Excel® Data File A similar explanation applies to the fixed‐asset account equipment. In this particular balance sheet, the equipment account is an aggregate for all the equipment owned by the company instead of an individual listing of each

equipment item. There could be separate equipment accounts, grouped according to equipment class. In any case, a company normally keeps individual records on equipment items, which are then summarized on the balance sheet. The net worth section of the balance sheet in Table 13.5 contains three entries, which summarize the company’s ownership (or equity) accounts. The common stock account reflects the ownership position of stockholders, frequently referred to as contributed capital. The reporting of contributed capital accounts on a balance sheet is subject to several legal requirements, depending on the nature of the contributed capital (e.g., common stock, preferred stock, par value, no par value). Discussion of these issues is beyond the scope of this text. The next two entries in the net worth section of Table 13.5 summarize the equity generated through the operation of the business rather than through investor contributions. These accounts are frequently referred to as earned capital. The retained earnings account summarizes the surplus (or deficit) of earnings over expenses that have been held by the company (i.e., retained) for accounting periods prior to the current period. The earned surplus account highlights this same information for the current period and should match the net profit (or loss) reflected on the current income statement.

13.3.2 Income Statement The second basic financial report compiled by the accounting system is the income statement, or profit and loss statement. For the current accounting period, the income statement provides management with (1) a summary of the revenues received, (2) a summary of the expenses incurred to obtain the revenues, and (3) the profit or loss resulting from business operations. The income statement’s format varies widely (more so than for balance sheets), and the revenue and expense items depend on the type of business involved. Figure 13.9 shows a condensed version of Starbucks’s income statement (referred to as a statement of earnings). Income Statement A statement of the revenues and expenses incurred by a firm during a period of time.

FIGURE 13.9 Starbucks’ Condensed Consolidated Statement of Earnings STARBUCKS CORPORATION CONSOLIDATED STATEMENTS OF EARNINGS (in millions, except per share data) Fiscal Year Ended Net revenues:

Sep 30, 2018 Oct 1, 2017 Oct 2, 2016

Company‐operated stores Licensed stores

$19,690.3 2,652.2

$17,650.7 2,355.0

$16,844.1 2,154.2

Other

  2,377.0

  2,381.1

  2,317.6

24,719.5 10,174.5

22,386.8 9,034.3

21,315.9 8,509.0

Store operating expenses

7,193.2

6,493.3

6,064.3

Other operating expenses Depreciation and amortization expenses

539.3 1,247.0

500.3 1,011.4

499.2 980.8

1,759.0    224.4

1,450.7    153.5

1,408.9       —

Total net revenues Cost of sales including occupancy costs

General and administrative expenses Restructuring and impairments Total operating expenses

21,137.4

18,643.5

17,462.2

   301.2 3,883.3

   391.4 4,134.7

   318.2 4,171.9

1,376.4 499.2

— 93.5

— 5.4

191.4

181.8

102.6

  (170.3) 5,780.0

   (92.5) 4,317.5

   (81.3) 4,198.6

 1,262.0

 1,432.6

 1,379.7

4,518.0     (0.3)

2,884.9      0.2

2,818.9      1.2

Net earnings attributable to Starbucks Earnings per share—basic

$ 4,518.3 $   3.27

$ 2,884.7 $   1.99

$ 2,817.7 $   1.91

Earnings per share—diluted

$   3.24

$   1.97

$   1.90

1,382.7

1,449.5

1,471.6

1,394.6

1,461.5

1,486.7

Income from equity investees Operating income Gain resulting from acquisition of joint venture Net gain resulting from divestiture of certain operations Interest income and other, net Interest expense Earnings before income taxes Income tax expense Net earnings including noncontrolling interests Net earnings/(loss) attributable to noncontrolling interests

Weighted average shares outstanding: Basic Diluted Excel® Data File

The income statement usually begins with a revenues section. Revenues are generated from the sale of products or services marketed by the firm. Next, the income statement reflects the direct costs associated with generating the sales revenue. This section is generally referred to as the cost of goods. For a manufacturing company (Table 13.6), the cost of goods section usually includes the costs incurred in producing the product. For a retail company (Table 13.7 and Figure 13.9), this section includes the costs incurred in acquiring the retail goods to be sold. In either case, the result of subtracting the cost of goods from the sales is gross profit. Gross profit is not always

shown as a separate line on an income statement. For example, Tables 13.6 and 13.7 show a gross profit line, but Starbucks (Figure 13.9) does not. After gross profit is determined, other operating expenses of the business are incorporated. This section includes all expenses that are not incorporated in the cost of goods. Generally this consists of general and administrative expenses as well as the marketing expenses. The subtraction of these costs from gross profit results in net profit before taxes. Subtracting taxes results in the net income (or loss) for the period. This final value is transferred to the earned surplus line on the balance sheet. The format for the income statement in Table 13.6 is oversimplified, even for a small manufacturing firm. For example, the “cost of goods sold” entry may be considerably more detailed to reflect multiple product lines, the depreciation items may be detailed to a greater extent to reflect multiple classes of assets, and several common expense items—such as employee benefits contributions, insurance premiums, and advertising—are not included. Similarly, the income statement in Table 13.7 is highly simplified. This format is typical for a retail business, and the major difference in form concerns the method of determining the “cost of goods sold” item. TABLE 13.6 Example Income Statement—Manufacturing Format BuiltRite Tool and Engineering Company Income Statement For Year Ended December 31, 2018 Net Sales

$1,200,000

Less Cost of Goods Manufactured Direct Labor Direct Materials Indirect Labor Depreciation Repairs and Maintenance Utilities Gross Profit

$420,000 302,000 112,000 98,000 41,500 11,500

 985,000 $215,000

Other Expenses Marketing General and Administrative Interest payments Net Income before Taxes

$49,000 76,000   35,000   160,000 $55,000

Less Income Taxes

   26,000

Net Income (posted to Earned Surplus)

 $29,000

Excel® Data File

TABLE 13.7 Example Income Statement—Retail Format BuiltRite Tool and Engineering Company Income Statement For Year Ended December 31, 2018 Net Sales Less Cost of Goods Sold Inventory, December 31, 2017 Plus purchases Total

$1,200,000 $26,000  432,000 $458,000

Less Inventory, December 31, 2018   44,000   414,000 Gross Profit $786,000 Less Expenses Direct Labor Depreciation—Building

$420,000 10,000

Depreciation—Equipment

30,000

Repairs and Maintenance Indirect Labor

41,500 218,000

Utilities Supplies

9,800    1,700   731,000

Net Income before Taxes Less Income Taxes Net Income (posted to Earned Surplus)

$55,000    26,000  $29,000

Excel® Data File Actual balance sheets and income statements use the same general formats as the ones illustrated above but usually contain more detail, as was seen in the Starbucks examples. The Starbucks financial statements were extracted from their November 16, 2018, Form 10‐K, which is a required disclosure statement that publicly held companies file with the SEC (Securities and Exchange Commission). The Starbucks balance sheet (Figure 13.8) and income statement (Figure 13.9) are both referred to as comparative statements, because they show more than 1 fiscal year and therefore provide an opportunity for the reader to make year‐to‐year comparisons.

13.3.3 Ratio Analysis An important topic related to the interpretation of balance sheets and income statements is ratio analysis, a common practice that examines relationships between the values found on these statements. Ratio analysis is not critical to an engineer focused solely on the accounting function as a source of data for economic analysis. However, to interpret a company’s accounting statements to determine the attractiveness of a company’s stock as a potential investment or to determine the economic health of a company as a potential employer, ratio analysis is of fundamental importance. If the reader’s interest is focused solely on data sourcing, this section can be skipped with no loss of continuity for the remainder of the chapter. Before proceeding to a presentation of several popular ratio calculations, it is important to bear in mind that the values of the ratios themselves are neither good nor bad. They can only be interpreted in comparison to the ratios of peer group companies or to an individual’s personal expectations of a company’s performance. They can also be used over time to gain an understanding of trend lines for and changes within a company. Ratios are only rough

guides to interpreting financial statements; they are not mathematical conclusions. As noted, ratios are generally used in one of three ways to draw conclusions about a company’s financial health: 1. Comparison with a company’s historic values of the same ratio to spot trends or changes in performance, 2. Comparison with expected or desired performance benchmarks available from financial analysts, and 3. Comparison with competitors within the same industry to measure competitive performance or position. Generally, the analysis of financial statements focuses on three primary areas: 1. The company’s earning power, 2. The short‐term liability obligations, and 3. The long‐term liability obligations. We will use the financial statements of Carson’s Cutlery Company to illustrate the calculation of several common ratios in each of these areas. Table 13.8 contains comparative balance sheets (two balance sheets displayed side‐ by‐side) for the years 2017 and 2018. Table 13.9 contains comparative income statements for the years 2017 and 2018.

TABLE 13.8 Carson’s Cutlery Company Comparative Balance Sheet Carson’s Cutlery Comparative Balance Sheet as of December 31, 2017 and 2018 ASSETS

2018

Current Assets Cash Accounts Receivable (net) Inventory Prepaid Expenses Total Current Assets Fixed Assets Machinery Furniture Other Total Assets

2017

$61,750

$83,520

195,000

130,500

65,000   22,750

50,000   31,900

$344,500

$295,920

$208,000

$187,830

74,750 22,750

72,500 23,750

$650,000

$580,000

$92,950 147,212

$87,000 109,653

  69,438

  64,920

$309,600

$261,573

LIABILITIES AND CAPITAL Current Liabilities Notes Payable Accounts Payable Taxes Payable Total Current Liabilities Fixed Liabilities Loans Total Liabilities Capital Stock Retained Earnings Earned Surplus Total Capital Total Liabilities and Capital Excel® Data File

$100,000

$90,000

$409,600

$351,573

$100,000 88,427

$100,000 77,397

   51,973

   51,030

$240,400 $650,000

$228,427 $580,000

TABLE 13.9 Carson’s Cutlery Company Comparative Income Statement Carson’s Cutlery Comparative Income Statement For Years Ended December 31, 2017 and 2018 2018 2017 $1,625,450 $1,450,000

Net Sales Cost of Goods Sold Beginning Inventory

$50,000

$40,000

Direct Materials

406,000

350,000

Direct Labor Factory Overhead

801,500 94,603

700,000 90,000

Total Less: Ending Inventory Cost of Goods Sold

$1,352,103 $1,180,000 65,000 50,000 $1,287,103 $1,130,000

Gross Profit Other Operating Expenses Selling Expenses Depreciation & Amortization General and Administrative Total Other Operating Expenses Net Operating Income Less: Interest Expenses Less: Income Taxes Net Income

 $338,347

 $320,000

$43,980 58,122

$37,200 53,791

  122,484

  120,580

$224,586  $113,761

$211,570  $108,430

$21,600

$18,000

40,188   $51,973

39,400   $51,030

Excel® Data File The viability of maintaining a company’s financial health depends upon its earning power. Firms must earn and sustain profit over the long term to survive. The following ratios are commonly used to assess earning power: Return on Assets Employed =

(13.7)

Net Income Average Total Assets

(13.8)

Net Income Return on Owner’s Equity = Average Owner’s Equity

The calculation of earning power ratios for Carson’s Cutlery for 2018 are shown below: $51,973 Return on Assets Employed =

= 0.0845 or 8.45% ($580,000 + 650,000)/2 $51,973

Return on Owner’s Equity =

= 0.2217, or 22.17% ($228,427 + $240,400)/2

The second major area of focus in ratio analysis is short‐term liability obligations. This is referred to as liquidity. These ratios measure a company’s ability to meet its current obligations. The following ratios are commonly used

to measure liquidity: (13.9)

Current Assets Current Ratio = Current Liabilities

Acid Test Ratio =

Cash + Accounts Receivable + Short Term Investments

(13.10)

Current Liabilities

(13.11)

Net Sales Accounts Receivable Turnover =

Average Accounts Receivable

(13.12)

Cost of Goods Sold Inventory Turnover = Average Inventory

The calculation of liquidity ratios for Carson’s Cutlery for 2018 is shown below. $344,500 Current Ratio

=

= 1.112 $309,600 $61,750 + $195,000 + $0

Acid Test Ratio

=

= 0.829 $309,600 $1,625,450

Accounts Receivable Turnover

=

= 9.98 ($130,500 + $195,000)/2 $1,287,103

Inventory Turnover

=

= 22.38 ($50,000 + $65,000)/2

The final major area of focus in ratio analysis is long‐term liability obligations. This is referred to as solvency. These ratios measure a company’s ability to meet its long‐term obligations based on its current debt structure. The following ratios are commonly used to measure solvency: (13.13)

Total Liabilities Debt to Equity Ratio =

Total Capital Worth

Net Income Before Taxes and Interest Times Interest Earned Ratio =

(13.14)

Interest Charges

Net Operating Income Operating Income to Total Assets Ratio =

(13.15)

Total Assets

The calculation of solvency ratios for Carson’s Cutlery for 2018 is shown below: $409,600 Debt to Equity Ratio

=

= 1.70 $240,400 $113,761

Times Interest Earned Ratio

=

= 5.27 $21,600 $113,761

Operating Income to Total Assets Ratio

=

= 0.1750, or 17.50% $650,000

Two other measures may be encountered when the financial analysis is focused on a company’s ability to earn a profit through its operations. This focus is achieved by excluding certain nonoperating expenses (or revenues) from earnings. Earnings before interest and taxes (EBIT) considers a company’s revenues less its operating expenses. Interest charges (or revenues) and taxes are excluded from EBIT. EBIT may also be referred to as operating earnings, operating profit, or operating income. Earnings before interest, taxes, depreciation, and

amortization (EBITDA) takes this one step further by excluding noncash expenses (depreciation and amortization) from the earnings consideration. The following calculations define these measures: EBIT

=

Net Income  +  Income Taxes + Interest Expense

EBITDA = Net Income  +  Income Taxes + Interest Expense

(13.16) (13.17)

+Depreciation + Amortization

The calculation of these measures for Carson’s Cutlery for 2018 is shown below: EBIT

=

$51,973 + $40,188 + $21,600 = $113,761

EBITDA

=

$51,973 + $40,188 + $21,600 + $58,122 = $171,883

It is worth noting that EBIT was already shown on the income statement under net operating income. As stated earlier, ratios provide a useful and powerful means to express the relationships found in balance sheets and income statements. These ratios are generally only useful when compared to a meaningful set of standards or expectations, which typically take the form of previous year’s results, peer group comparisons, or industry averages. Many variations to the names and calculation formulas presented above can be found in financial literature. The reader is cautioned to ensure that ratios are calculated in consistent ways before making comparisons.

Concept Check 13.03‐CC001 A balance sheet classifies items into one of three main groups. These groups are a. Income, expenses, and profit b. Assets, liabilities, and profit c. Assets, liabilities, and net worth d. Income, expenses, and net worth Correct or Incorrect? Clear

  Check Answer

Concept Check 13.03‐CC002 An income statement documents revenues and expenses incurred during a period in order to determine a. Change in the value of assets for the period b. Profit (or loss) for the period c. Depreciation for the period d. Value of the company’s net worth at the end of the period Correct or Incorrect? Clear

  Check Answer

Concept Check 13.03‐CC003 Which of the following is not one of the typical uses of financial ratios? a. Comparisons to a company’s historic values b. Comparison to benchmarks c. Comparison to competitors’ values d. Comparison to zero with positive values indicating desirable ratios and negative values indicating undesirable ratios Correct or Incorrect? Clear

  Check Answer

13.4 Cost Accounting Principles LEARNING OBJECTIVE Apply cost accounting principles to determine relevant cost values. Video Lesson: Cost Accounting Principles The balance sheet and the income statement in Section 13.3 are sometimes considerably removed, both in time and in detail, from data required for engineering project‐level decision making. In these cases, more important to the engineer as a source of cost information is the cost accounting system within a particular firm. The firm may be involved in manufacturing or providing services, and if it is involved in manufacturing, production may be on a job‐shop or process basis. There are some fundamental differences in cost accounting procedures for determining manufacturing costs versus determining the cost of providing a service; also, there are differences in accounting procedures if manufacturing is on a job‐shop or process basis. In order to concentrate on basic principles instead of

details, the cost accounting system assumed will be that of a job‐shop manufacturing firm. Thus, the emphasis will be on determining the per‐order costs for a job order.

13.4.1 Traditional Cost Allocation Methods The total cost of producing any job order consists of direct material, direct labor, and overhead costs. Our approach here will not split the overhead cost into factory overhead, general overhead, and marketing expenses in order to simplify the presentation. Direct materials for a given job order may include purchased parts and in‐house fabricated parts. The cost for purchased materials is determined primarily from purchase invoices. The cost for fabricated parts is determined primarily from the job cost system. Direct labor expended on a job order is normally recorded by operators or captured electronically via information technology, and the direct labor cost is determined by applying the appropriate labor cost rates. The labor rates, as determined by the accounting system, will normally include the cost of employee fringe benefits in addition to the basic hourly rate. Determining direct material and direct labor costs can be problematic in some situations, but both are generally more readily determined than the overhead cost. Overhead costs typically cannot be allocated as direct charges to any single job order and therefore must be prorated among all the job orders on some rational basis. Three popular methods of allocating overhead costs to manufacturing jobs are in wide use today: 1. Allocation based on direct labor hours, 2. Allocation based on direct labor dollars, and 3. Allocation based on direct labor dollars plus direct material dollars (prime costs). These methods can be applied at any desired manufacturing unit level (i.e., an entire plant, specific departments, work centers, machines, or job orders). Step‐by‐step procedures for using each of these methods to (1) determine an appropriate overhead rate and (2) use this rate to estimate overhead on a specific job are outlined below. Example 13.10, which follows the step‐by‐step procedures, illustrates their application. Allocate Overhead Based on Direct Labor Hours 1. Determine (or estimate) values for previous period direct labor hours and overhead cost for the manufacturing unit 2. Calculate the rate per direct labor hour: (13.18)

Overhead Cost Rate = Direct Labor Hours

3. Determine (or estimate) the number of direct labor hours required by the particular job for which overhead cost is being estimated 4. Calculate the overhead cost for the job: Estimated Overhead = Rate × Estimated Direct Labor Hours

Allocate Overhead Based on Direct Labor Dollars 1. Determine (or estimate) values for previous‐period direct labor dollars and overhead cost for the manufacturing unit 2. Calculate the percentage ratio of overhead cost to direct labor dollars: Overhead Cost Ratio =

× 100%

(13.19)

Direct Labor Dollars

3. Determine (or estimate) the direct labor dollars required by the particular job for which overhead cost is being estimated 4. Calculate the overhead cost for the job:

Estimated Overhead = Ratio × Estimated Direct Labor Dollars

Allocate Overhead Based on Direct Labor Dollars and Direct Material Dollars 1. Determine (or estimate) values for previous‐period direct labor dollars, direct material dollars, and overhead cost for the manufacturing unit 2. Calculate the percentage ratio of overhead cost to direct labor dollars plus direct material dollars: Overhead Cost Ratio =

Direct Labor Dollars + Direct Material Dollars

× 100%

(13.20)

3. Determine (or estimate) the direct labor dollars and direct material dollars required by the particular job for which overhead cost is being estimated 4. Calculate the overhead cost for the job: Estimated Overhead

=

Ratio × (Estimated Direct Labor Dollars + Estimated Direct Material Dollars

EXAMPLE 13.10 Traditional Methods of Overhead Allocation The overhead allocation for a job is to be estimated. Assume the direct labor hours for the job are estimated to be 40 hours at a rate of $12.50 per hour. Direct material costs are estimated at $850. The overhead calculations are to be based on the following cost totals collected during the previous accounting period. Total direct labor hours Total direct labor dollars

48,000 $480,000

Total direct material dollars $600,000 Total overhead costs

$360,000

Using the step‐by‐step procedures above, we will calculate the overhead allocation based on (a) direct labor hours, (b) direct labor dollars, and (c) direct labor dollars plus direct material dollars. Solution a. Direct labor hours Step 1: Previous‐period direct labor hours = 48,000 Previous‐period overhead cost = $360,000 Step 2: Rate per direct labor hour = $360,000/48,000 = $7.50/hour Step 3: Estimated direct labor hours for job = 40 Step 4: Estimate overhead = $7.50/hour × 40 hours = $300 b. Direct labor dollars Step 1: Previous‐period direct labor dollars = $480,000 Previous‐period overhead cost = $360,000 Step 2: Percentage ratio per direct labor dollar = ($360,000/$480,000) × 100% = 75% Step 3: Estimated direct labor dollars for job = 40 hours × $12.50/hour = $500 Step 4: Estimate overhead = 75% × $500 = $375 c. Direct labor dollars + direct material dollars Step Previous‐period direct labor dollars = $480,000 1: Previous‐period direct material dollars = $600,000 Previous‐period overhead cost = $360,000 Step Percentage ratio per (direct labor dollar + direct material dollar) = [$360,000/($480,000 + 2: $600,000)] × 100% = 33.33% Step Estimated direct labor dollars + direct material dollars for job = (40 hours × $12.50/hour) + $850 3: = $1,350 Step Estimate overhead = 33.33% × $1,350 = $450 4: Determining the overhead cost for a job order by the rate per direct labor hour method will yield the same result as the percentage of direct labor cost method, provided that the rate per direct labor hour used on the job in question is equal to the average factory labor rate. The “percentage of prime cost” method will necessarily yield a different

assignment of overhead to a job order than the other two methods. The choice among these three methods varies by company; indeed, cost accountants use other methods in distributing overhead costs in some cases. The rate per direct labor hour method is perhaps the most commonly used. Whatever method is chosen for distributing overhead costs to job orders in a current year, the rates or percentages are based on the previous year’s cost figures. Thus, overhead rates may change from year to year within a particular firm. Because an average overhead rate for the entire factory may very well be too gross an estimate when actual overhead costs differ among departments within the factory, cost accounting may determine individual overhead rates for departments or cost centers. In addition, the hourly rates for direct labor may vary among these cost centers. A further refinement is to determine overhead rates for individual machines within cost centers. Then, as particular job orders progress through cost centers (departments or machines), the direct labor time (or machine time) spent on the job order in the various cost centers is recorded, the appropriate labor or machine rates and overhead rates are applied, and the total cost for the job is calculated. The following example illustrates the variety of methods used to distribute overhead to a given cost center; the total overhead for the cost center will then be distributed to particular products by yet another method.

EXAMPLE 13.11 Overhead Allocation to Departments The information in Table 13.10 has been accumulated for the Deetco Company’s two departments during the past year. Deetco distributes depreciation overhead based on (1) the first cost of equipment in each department, (2) a zero salvage value of the equipment in 10 years, and (3) a straight‐line rate of depreciation. All overhead other than depreciation is first distributed to each department according to the number of employees in each department, and then an overhead rate per direct labor hour is computed for each department. TABLE 13.10 Data for Example 13.11 Department A Department B Direct Material Cost Direct Labor Cost Direct Labor Hours

$720,000  $260,000 

Total

$240,000  $960,000  $140,000  $400,000 

25,200 

16,200 

41,400 

Number of Employees First Cost of Equipment

14  $250,000 

9  23  $200,000  $450,000 

Annual Depreciation Other Factory Overhead

$25,000 

$20,000  $45,000  $150,000 

General Overhead

$350,000 

What price should the company quote on Job Order D if raw material costs are estimated as $900; estimated direct labor hours required in Departments A and B are 30 hours and 100 hours, respectively; and profit is to be set at 25% of selling price? Solution For Department A, the total overhead allocation is determined as follows: Annual depreciation

$25,000

Other factory overhead (14/23)($150,000) $91,000 General overhead (14/23)($350,000) $213,043 TOTAL

$329,347

Thus, the overhead rate for Department A per direct labor hour is $329,347 Rate =

25,200

= $13.07 per direct labor hour

For Department B, the total overhead allocation is calculated as follows: Annual depreciation

$20,000

Other factory overhead (9/23)($150,000) $58,696 General overhead (9/23)($350,000) $136,957 TOTAL Thus, the overhead rate for Department B per direct labor hour is

$215,653

$215,653 Rate =

= $13.31 per direct labor hour 16,200

The estimated total cost for Job Order D is then computed as Direct material cost

$900

Direct labor cost A ($260,000/25,200)(30 hours) Overhead cost A ($13.07)(30 hours)

$309.52 $392.10

Direct labor B ($140,000/16,200)(100 hours)

$864.20

Overhead cost B ($13.31)(100 hours) TOTAL

$1,331 $3,796.82

If x = the selling price of Job Order D, then x = total cost + profit x = $3,796.82 + (0.25)x

solving for x yields $3,796.82 x = 0.75

thus, x = $5,062.43

13.4.2 Activity Based Costing Activity-based costing (ABC) emerged in the field of cost accounting due to dramatic changes in the nature and characteristics of manufacturing costs. Historically, direct labor and direct material constituted the most significant elements of the cost of goods. Overhead was the smallest element and hence was allocated based on direct labor or prime costs (see Section 13.5.1). In many cases, it is no longer accurate to assume that direct labor is the largest element in the cost pool and overhead the smallest. With the introduction and implementation of computer‐ controlled and automated manufacturing systems, it is not unusual for overhead costs to dominate the cost of producing items. Frequently, in fact, direct labor is the least significant in the cost pool and overhead the largest. Activity Based Costing (ABC) A costing method designed to associate costs with the activities that drive them. ABC is designed to meet the challenge of a changing cost mix by associating manufacturing costs with activities that drive them. First, costs must be identified by categories. These need not (and probably should not) be associated with products or organizational units; rather, they are associated with clearly defined cost categories or cost pools. Typical examples of cost pools include material handling costs, energy costs, tooling costs, or maintenance costs. Next, the activities that drive the significant cost pools must be identified. These are to be monitored and controlled under ABC. This task is difficult and complex. Many companies have never considered their processes from the cost driver or value‐added point of view. The newness of this approach and the implicit challenge of considering activities from a new perspective make this task a significant undertaking. Examples of cost‐driving activities include machine hours for energy costs, material moves or truck hours for material handling costs, and machine hours or production volume for tooling and maintenance costs. Next, the expected (or actual) rate of activity for each of the cost drivers is used to predict (or monitor) the costs associated with each cost pool. Such activity‐based accounting of costs can be used as a basis to eliminate high‐ cost activities, particularly if they generate low value added. Similarly, ABC analysis can be used to focus the

attention of process improvement activities toward those activities that drive high costs. Better still, product and process redesign can focus on changes that ultimately eliminate the activation of high‐cost driver activities. Many companies are employing ABC to make activity‐based decisions that result from a more realistic allocation of costs than was previously possible. In many cases, ABC‐generated process and product redesigns have been impressive when measured in terms of cost reduction and profit improvement. Activity‐based costing does, however, require information sharing and a cross‐functional perspective that is new to many companies. A vast literature exists for ABC and its derivative management philosophy, activity‐based management. The interested reader is encouraged to explore this literature, particularly the text by Kaplan and Anderson6 and the text by Cokins.7

13.4.3 Standard Costs Although the first task of cost accounting is to determine per‐item or per‐order costs, another major purpose is to interpret financial data so that management can (1) measure changes in production efficiency and (2) judge the adequacy of production performance. Establishing cost standards can be of great assistance in achieving these objectives. A standard‐cost system involves, in advance of manufacture, (1) the preparation of standard rates for material, labor, and overhead, and (2) the application of these rates to the standard quantities of material and labor required for a job order, or for each production operation required to complete the job order. Because a process‐type manufacturing firm, such as an oil refinery, outputs the same product (or a few products) over a long time period, cost standards are more readily determined for process firms than for job‐shop firms, where the variety of output is large and varies with customer order. However, the number and type of production operations required to complete various job orders are finite for a given manufacturing firm. Each job order is, of course, made up of single units. Thus, a standard amount of material, standard labor times, and machine times can be determined for each unit. It is usually the responsibility of the firm’s work measurement function to determine these standard quantities. Then, by applying standard unit material costs and standard labor rates, standard unit costs for material, labor, and overhead can be determined. The standard costs then serve as a basis for measuring production efficiency and performance over time. Deviations from standard costs may be caused by several factors, especially (1) raw material price variations and (2) actual quantities of material and labor used versus the standard amounts of these items. This latter factor is of primary concern in determining production efficiency and performance, measures of which provide information to management to aid in cost control. Standard Costs The estimated material, labor and overhead costs for a unit. Standard costs serve as a basis for measuring production efficiency and performance over time.

Concept Check 13.04‐CC001 The sum of overhead costs is the numerator for calculating the overhead application ratio in traditional overhead allocation methods. Possible denominators include all of the following except a. Direct labor hours b. Profit c. Direct labor dollars d. The sum of direct labor dollars and material dollars

Concept Check 13.04‐CC002 Activity‐based costing achieves more realistic allocation of overhead costs by associating these costs with a. Activities that drive them b. Activities under the direct control of the Board of Directors c. Profit d. Management activities Correct or Incorrect? Clear

  Check Answer

Concept Check 13.04‐CC003 Deviations from standard costs can be caused by I. Variations in the per unit cost of labor and/or material II. Variations in the quantity used of labor and/or material a. I only b. II only c. I and II d. neither I nor II Correct or Incorrect? Clear

  Check Answer

13.5 Economic Value Added LEARNING OBJECTIVE Calculate the economic value added for an equity‐based capital investment. Since the mid‐1980s, a management tool called economic value added (EVA) has been used by an impressive set of firms to make investment decisions. It focuses management’s attention on an important objective: adding value for the shareholders. In fact, it has been the principal tool used by upper management within the Coca‐Cola Company. EVA is used to facilitate decisions regarding major capital investment, as well as acquisitions and divestitures. It has also been used to analyze products and operating units to determine the economic winners and losers within the firm’s portfolio of products and businesses.

Economic Value Added (EVA) A management tool that examines the difference between the net operating profit after taxes and the cost of capital. Basically, EVA is a management tool that examines the difference between the net operating profit after taxes and the cost of capital, which includes the cost of both debt and equity capital.8 Hence, the interest charges and bond rates that contribute to the cost of debt capital are combined with the cost to the shareholders of providing the firm with equity capital (by purchasing its stock). Many firms that adopted EVA found that few of their managers knew how much capital was tied up in their business units. Moreover, the managers did not have an accurate understanding of the true cost of capital. EVA seeks to remedy this by focusing attention on adding value for the shareholder through more effective use of capital. The result of an increased emphasis on effective use of capital should be lower inventories, fewer warehouses, and so on. Stern Stewart, a corporate financial advisory service, is generally credited with the development of EVA. Stern Stewart argues that there are four ways to create value for the shareholder: 1. Increase profitability without using additional capital (e.g., increase profit margins and increase sales without using additional capital). 2. Invest in projects that earn more than the cost of capital. 3. Free up capital that earns less than the cost of capital. 4. Use debt to reduce the cost of capital. Real estate, equipment, facilities, working capital, and inventories are examples of capital being used within the firm. Other, not so obvious, examples of capital are investments in training and in research and development. The investments in training result in increased value of the firm’s human capital. While it is not easy to quantify the value of capital in research and development (R&D) and in training, they should not be overlooked in the quest for value. The following three examples illustrate how EVA is calculated and how it can be used to improve economic decision making.

EXAMPLE 13.12 Using EVA to Select a Firm for Acquisition Two firms (A and B) are being considered for acquisition. The assets of the firms are $100 million and $200 million, respectively. Both are debt free; hence, the equity equals the assets. The annual operating profits for the firms are $40 million and $72 million, respectively. Taxes equal 25% of operating profit. The after‐tax weighted average cost of capital (WACC) is 12%. Which firm is most attractive for acquisition from a return on equity perspective and from an EVA perspective? Given EquityA = $100M, EquityB = $200M, NOPA = $40M, NOPB = $72M, itr = 25%, CoC = 12% Find ROEA, ROEB, EVAA, EVAB, recommendation NOPATA

=

$40M (1 − 0.25) = $30M and NOPATB = $72M (1 − 0.25) = $54M

ROEA

=

$30M/$100M = 30% and ROEB = $54M/$200M = 25%

Based on ROE, Firm A is more attractive. CoCA

=

0.12($100M) = $12M and CoCB = 0.12($200M) = $24M

EVAA

=

$30M − $12M = $18M and EVAB = $54M − $24M = $30M

Based on EVA, Firm B is more attractive Example 13.12. As was the case with the IRR method, one cannot choose the alternative having the greatest return on equity; instead, incremental net operating profits should be compared with the cost of capital. Recall the 6th of 10 principles of engineering economic analysis presented in Chapter 1 and the 1924 quote from Donald Brown, the chief financial officer of General Motors: “The object of management is not necessarily the highest rate of return on capital, but … to assure profit with each increment of volume that will at least equal the economic cost of additional capital required.” Maximizing return on investment, return on equity, or return on assets is not the right objective.

EXAMPLE 13.13 Evaluating the Economic Performance of Two Firms Video Example An investor is considering investing in one of two firms (C and D). The WACC is 12% and the corporate income‐tax rate is 25% for both firms. Firm C’s equity totals $100 million and its net annual operating profit before taxes is $24 million. Firm D’s equity totals $200 million and its net annual operating profit before taxes is $36 million. Based on an EVA analysis, which firm is stronger economically? Given: EquityC = $100M, EquityD = $200M, NOPC = $24M, NOPD = $36M, itr = 25%, CoC = 12% Find: EVAC, EVAD, recommendation NOPATC

=

$24M (1 − 0.25) = $18M and NOPATD = $36M (1 − 0.25) = $27M

CoCC

=

0.12($100M) = $12M and CoCD = 0.12($200M) = $24M

EVAC

=

$18M − $12M = $6M and EVAD = $27M − $24M = $3M

Based on EVA, Firm C is more attractive, even though Firm D had a greater net operating profit after taxes. Alternatively, we could have performed the following analysis. The return on equity for Firm C is $18M/$100M, or 18%; for Firm D, the return on equity is $27M/$200M, or 13.5%. Therefore, the difference in return on equity and the cost of capital is 18% − 12%, or 6%, for Firm C and 13.5% − 12%, or 1.5%, for Firm D. Multiplying the difference by the firm’s equity gives the following EVA values: 6%($100M), or $6M, for Firm C, and 1.5%($200M), or $3M, for Firm D. In addition to using EVA for firm‐level decisions, it can be used in evaluating engineering economic analysis investment alternatives similar to those considered in Chapters 9 and 10. To demonstrate its use in such evaluations, recall EVA equals the difference in net operating profit after taxes (NOPAT) and after‐tax cost of capital (CoC), EVA = NOPAT − CoC

(13.21)

Drawing on our work in Chapter 9, net operating profit after taxes is equal to the product of taxable income (TI) and 1 minus the income tax rate (itr), NOPAT = (1 − itr)TI

(13.22)

Letting the WACC be equal to the after‐tax minimum attractive rate of return (MARRAT), the cost of capital during a year equals the product of MARRAT and the beginning‐of‐year book value (BOYBV), CoC = MARRAT (BOYBV)

(13.23)

From Equation 9.5, after‐tax cash flow is given by ATCF = (1 − itr)BTCF + itrDWO

Adding and subtracting DWO to Equation 13.24 gives ATCF = (1 − itr)BTCF − DW + itrDWO + DWO

or, after collecting terms, ATCF = (1 − itr)(BTCF − DWO) + DWO

(13.24)

For equity‐based capital investments, BTCF − DWO is taxable income (TI). Therefore, ATCF = (1 − itr)TI + DWO

(13.25)

Adding and subtracting CoC to Equation 13.25 gives ATCF = (1 − itr)TI − CoC + CoC + DWO

or, from Equation 13.22, ATCF = NOPAT − CoC + CoC + DWO

or, from Equation 13.21, ATCF = EVA + MARRAT (BOYBV) + DWO

Therefore, EVA = ATCF − MARRAT (BOYBV) − DWO

The calculation of EVA for an equity‐based capital investment is illustrated in Example 13.14.

(13.26)

EXAMPLE 13.14 Economic Value Added for the SMP Machine Investment A surface‐mount placement (SMP) machine is purchased for $500,000. It qualifies as 5‐year property for MACRS‐GDS depreciation. At the end of the 10‐year planning horizon, it is expected to have a salvage value of $50,000. The use of the SMP machine is expected to reduce manufacturing costs by $108,333.33 each year. With an after‐tax minimum attractive rate of return of 10% and an income‐tax rate of 25%, compute the after‐tax annual worth and the EVA for the investment. Given P = $500,000, itr = 25%, MARRAT = 10%, BTCF (years 1 − 10) = $108,333.33, S = $50,000 Find AWAT, EVA Solution To calculate AWAT, we create a spreadsheet similar to one used in Example 9.4. To calculate EVA, we append four columns to the spreadsheet used to calculate AWAT, as shown in Figure 13.10.

FIGURE 13.10 Annual Worth and Economic Value‐Added Calculations for the SMP Machine Excel® Data File To illustrate the calculations involved, consider the calculation of EVA for year 4. The beginning‐of‐year book value ($144,000) equals the beginning‐of‐year book value the previous year ($240,000) minus the depreciation allowance the previous year ($96,000). Net operating profit after taxes for year 4 ($38,050) equals the product of 1 minus the income‐tax rate (0.75) and taxable income in year 4 ($50,733.33). EVA for year 4 ($23,650) equals the difference in net operating profit after taxes ($38,050) and the product of the after‐tax minimum attractive rate of return (10%) and beginning‐of‐year book value ($144,000). Finally, the overall EVA for the investment ($17,960.83) is obtained by calculating the uniform series equivalent of the EVA values for the 10‐year planning horizon [=PMT(10%,10,−NPV(10%,J2:J11))]. Excel® Video Lesson: The PMT Function Excel® Video Lesson: The NPV Function Exploring the Solution Notice, EVA equals AWAT. Evaluating an equity‐based investment using EVA is equivalent to performing an after‐tax annual worth analysis of the investment. Problems at the end of the chapter will demonstrate EVA analysis can be performed under inflationary conditions, as well as when bonus depreciation and Section 179 expense deductions are available to an investor.

Recall, in Sections 9.3 and 10.3, we calculated after‐tax annual worth when capital investments employed borrowed capital. However, when we calculate EVA, we limit our consideration to equity‐based capital investments, because the calculation of EVA is not straightforward when borrowed capital is used to finance a capital investment. The cost of debt capital is already included in the firm’s after‐tax minimum attractive rate of return. However, because after‐tax annual worth is equivalent to EVA, it can be used to provide insights regarding EVA values when capital investments are financed by using retained earnings versus borrowed capital. As illustrated in the end‐of‐chapter problems, EVA calculations can be performed when inflation exists, when bonus depreciation applies, and when Section 179 expense deduction applies. In the latter case, the expense deduction is treated “like” depreciation in calculating beginning‐of‐year book value.

Concept Check 13.05‐CC001 Economic value added compares the difference between which two of the following? I. Net operating profit after taxes II. Net operating profit before taxes III. Internal rate of return IV. Cost of capital a. I and II b. II and III c. III and IV d. I and IV Correct or Incorrect? Clear

  Check Answer

Notes 1. P. F. Ostwald, and T. S. McLaren, Cost Analysis and Estimating for Engineering and Management, 4th Revised Edition, Prentice Hall, 2003. 2. This example, with slight modification, is taken from E. J. Ferguson and J. E. Shamblin, “Break‐Even Analysis,” The Journal of Industrial Engineering, 18(8), August 1967 with permission of the publisher. 3. Ibid. 4. The FAA Life Cycle Cost Estimating Handbook is available at www.faa.gov. 5. R. D. Stewart, J. D. Johannes, and R. M. Wyskida, Cost Estimator’s Reference Manual, 2nd ed., John Wiley & Sons, 2001. 6. R. S. Kaplan and S. R. Anderson, Time‐Driven Activity Based Costing, Harvard Business School Press, 2007. 7. G. Cokins, Activity‐Based Cost Management, Making It Work, McGraw-Hill, 1996. 8. J. M. Stern, J. S. Shiely, and I. Ross, The EVA Challenge, Wiley, 2001.

Appendix A Time Value of Money Factors Online PDF 0.25% Time Value of Money Factors Discrete Compounding 0.25%

TABLE A-A-1 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series

Gradient Series To Find A To Find F To Find P To Find F To Find A To Find P To Find A To Find P Given G Given P Given F Given A Given F (A|F Given A Given P (A|P Given G (A|G n (F|P i%,n) (P|F i%,n) (F|A i%,n) i%,n) (P|A i%,n) i%,n) (P|G i%,n) i%,n) 1 1.00250 0.99751 1.00000 1.00000 0.99751 1.00250 0.00000 0.00000 2 3 4

1.00501 1.00752 1.01004

0.99502 0.99254 0.99006

2.00250 3.00751 4.01503

0.49938 0.33250 0.24906

1.99252 2.98506 3.97512

0.50188 0.33500 0.25156

0.99502 2.98009 5.95028

0.49938 0.99834 1.49688

5 6 7 8

1.01256 1.01509 1.01763 1.02018

0.98759 0.98513 0.98267 0.98022

5.02506 6.03763 7.05272 8.07035

0.19900 0.16563 0.14179 0.12391

4.96272 5.94785 6.93052 7.91074

0.20150 0.16813 0.14429 0.12641

9.90065 14.82630 20.72235 27.58391

1.99501 2.49272 2.99001 3.48689

9 10 11

1.02273 1.02528 1.02785

0.97778 0.97534 0.97291

9.09053 10.11325 11.13854

0.11000 0.09888 0.08978

8.88852 9.86386 10.83677

0.11250 0.10138 0.09228

35.40614 44.18420 53.91328

3.98335 4.47940 4.97503

12 13 14

1.03042 1.03299 1.03557

0.97048 0.96806 0.96565

12.16638 13.19680 14.22979

0.08219 0.07578 0.07028

11.80725 12.77532 13.74096

0.08469 0.07828 0.07278

64.58858 76.20532 88.75874

5.47025 5.96504 6.45943

15 16 17

1.03816 1.04076 1.04336

0.96324 0.96084 0.95844

15.26537 16.30353 17.34429

0.06551 0.06134 0.05766

14.70420 15.66504 16.62348

0.06801 0.06384 0.06016

102.24409 116.65666 131.99172

6.95339 7.44694 7.94008

18 19 20 21 22

1.04597 1.04858 1.05121 1.05383 1.05647

0.95605 0.95367 0.95129 0.94892 0.94655

18.38765 19.43362 20.48220 21.53341 22.58724

0.05438 0.05146 0.04882 0.04644 0.04427

17.57953 18.53320 19.48449 20.43340 21.37995

0.05688 0.05396 0.05132 0.04894 0.04677

148.24459 165.41059 183.48508 202.46341 222.34096

8.43279 8.92510 9.41698 9.90845 10.39951

23 24 25 26 27 28

1.05911 1.06176 1.06441 1.06707 1.06974 1.07241

0.94419 0.94184 0.93949 0.93714 0.93481 0.93248

23.64371 24.70282 25.76457 26.82899 27.89606 28.96580

0.04229 0.04048 0.03881 0.03727 0.03585 0.03452

22.32414 23.26598 24.20547 25.14261 26.07742 27.00989

0.04479 0.04298 0.04131 0.03977 0.03835 0.03702

243.11313 264.77534 287.32301 310.75160 335.05657 360.23340

10.89014 11.38036 11.87017 12.35956 12.84853 13.33709

29 30 36 40 48 50

1.07510 1.07778 1.09405 1.10503 1.12733 1.13297

0.93015 0.92783 0.91403 0.90495 0.88705 0.88263

30.03821 31.11331 37.62056 42.01320 50.93121 53.18868

0.03329 0.03214 0.02658 0.02380 0.01963 0.01880

27.94004 28.86787 34.38647 38.01986 45.17869 46.94617

0.03579 0.03464 0.02908 0.02630 0.02213 0.02130

386.27760 413.18468 592.49878 728.73988 1040.05520 1125.77667

13.82523 14.31296 17.23058 19.16735 23.02092 23.98016

52

1.13864

0.87824

55.45746

0.01803

48.70484

0.02053

1214.58847

24.93774

Single Sums

n

Uniform Series

Gradient Series To Find A To Find F To Find P To Find F To Find A To Find P To Find A To Find P Given G Given P Given F Given A Given F (A|F Given A Given P (A|P Given G (A|G (F|P i%,n) (P|F i%,n) (F|A i%,n) i%,n) (P|A i%,n) i%,n) (P|G i%,n) i%,n)

55 60 72 75 80 84 90

1.14720 1.16162 1.19695 1.20595 1.22110 1.23335 1.25197

0.87168 0.86087 0.83546 0.82922 0.81894 0.81080 0.79874

58.88194 64.64671 78.77939 82.37922 88.43918 93.34192 100.78845

0.01698 0.01547 0.01269 0.01214 0.01131 0.01071 9.9218E−03

96 100 108 120 132 144

1.27087 1.28362 1.30952 1.34935 1.39040 1.43269

0.78686 0.77904 0.76364 0.74110 0.71922 0.69799

108.34739 113.44996 123.80926 139.74142 156.15817 173.07425

180 240 360 480 600

1.56743 1.82075 2.45684 3.31515 4.47331

0.63799 226.97269 0.54922 328.30200 0.40703 582.73688 0.30165 926.05950 0.22355 1389.32309

51.32644 55.65236 65.81686 68.31075 72.42595 75.68132 80.50382

0.01948 0.01797 0.01519 0.01464 0.01381 0.01321 0.01242

1353.52863 1600.08454 2265.55685 2447.60694 2764.45681 3029.75923 3446.86997

26.37098 28.75142 34.42214 35.83048 38.16942 40.03312 42.81623

9.2296E−03 85.25460 8.8145E−03 88.38248 8.0769E−03 94.54530 7.1561E−03 103.56175 6.4038E−03 112.31206 5.7779E−03 120.80407

0.01173 0.01131 0.01058 9.6561E−03 8.9038E−03 8.2779E−03

3886.28316 4191.24173 4829.01247 5852.11160 6950.01441 8117.41331

45.58444 47.42163 51.07618 56.50843 61.88128 67.19487

4.4058E−03 3.0460E−03 1.7160E−03 1.0798E−03 7.1977E−04

6.9058E−03 5.5460E−03 4.2160E−03 3.5798E−03 3.2198E−03

1.1987E+04 1.9399E+04 3.6264E+04 5.3821E+04 7.0581E+04

82.78122 107.58631 152.89019 192.66991 227.25400

144.80547 180.31091 237.18938 279.34176 310.58071

0.50% Time Value of Money Factors Discrete Compounding 0.50%

TABLE A-A-2 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series

Gradient Series To Find A To Find F To Find P To Find F To Find A To Find P To Find A To Find P Given G Given P Given F Given A Given F (A|F Given A Given P (A|P Given G (A|G (F|P i%,n) (P|F i%,n) (F|A i%,n) i%,n) (P|A i%,n) i%,n) (P|G i%,n) i%,n)

n 1 2

1.00500 1.01003

0.99502 0.99007

1.00000 2.00500

1.00000 0.49875

0.99502 1.98510

1.00500 0.50375

0.00000 0.99007

0.00000 0.49875

3 4

1.01508 1.02015

0.98515 0.98025

3.01502 4.03010

0.33167 0.24813

2.97025 3.95050

0.33667 0.25313

2.96037 5.90111

0.99667 1.49377

5

1.02525

0.97537

5.05025

0.19801

4.92587

0.20301

9.80260

1.99003

6 7

1.03038 1.03553

0.97052 0.96569

6.07550 7.10588

0.16460 0.14073

5.89638 6.86207

0.16960 0.14573

14.65519 20.44933

2.48545 2.98005

8 9

1.04071 1.04591

0.96089 0.95610

8.14141 9.18212

0.12283 0.10891

7.82296 8.77906

0.12783 0.11391

27.17552 34.82436

3.47382 3.96675

10 11

1.05114 1.05640

0.95135 0.94661

10.22803 11.27917

0.09777 0.08866

9.73041 10.67703

0.10277 0.09366

43.38649 52.85264

4.45885 4.95013

12

1.06168

0.94191

12.33556

0.08107

11.61893

0.08607

63.21360

5.44057

13 14

1.06699 1.07232

0.93722 0.93256

13.39724 14.46423

0.07464 0.06914

12.55615 13.48871

0.07964 0.07414

74.46023 86.58346

5.93018 6.41896

15 16

1.07768 1.08307

0.92792 0.92330

15.53655 16.61423

0.06436 0.06019

14.41662 15.33993

0.06936 0.06519

99.57430 113.42380

6.90691 7.39403

17

1.08849

0.91871

17.69730

0.05651

16.25863

0.06151

128.12311

7.88031

18 19

1.09393 1.09940

0.91414 0.90959

18.78579 19.87972

0.05323 0.05030

17.17277 18.08236

0.05823 0.05530

143.66343 160.03602

8.36577 8.85040

20 21

1.10490 1.11042

0.90506 0.90056

20.97912 22.08401

0.04767 0.04528

18.98742 19.88798

0.05267 0.05028

177.23221 195.24341

9.33419 9.81716

22 23

1.11597 1.12155

0.89608 0.89162

23.19443 24.31040

0.04311 0.04113

20.78406 21.67568

0.04811 0.04613

214.06109 233.67676

10.29929 10.78060

24

1.12716

0.88719

25.43196

0.03932

22.56287

0.04432

254.08203

11.26107

25 26

1.13280 1.13846

0.88277 0.87838

26.55912 27.69191

0.03765 0.03611

23.44564 24.32402

0.04265 0.04111

275.26856 297.22805

11.74072 12.21953

27 28

1.14415 1.14987

0.87401 0.86966

28.83037 29.97452

0.03469 0.03336

25.19803 26.06769

0.03969 0.03836

319.95231 343.43317

12.69751 13.17467

29 30

1.15562 1.16140

0.86533 0.86103

31.12439 32.28002

0.03213 0.03098

26.93302 27.79405

0.03713 0.03598

367.66255 392.63241

13.65099 14.12649

36

1.19668

0.83564

39.33610

0.02542

32.87102

0.03042

557.55983

16.96205

40 48

1.22079 1.27049

0.81914 0.78710

44.15885 54.09783

0.02265 0.01849

36.17223 42.58032

0.02765 0.02349

681.33469 959.91881

18.83585 22.54372

50 52

1.28323 1.29609

0.77929 0.77155

56.64516 59.21803

0.01765 0.01689

44.14279 45.68975

0.02265 0.02189

1035.69659 1113.81615

23.46242 24.37781

Single Sums

n

Uniform Series

Gradient Series To Find A To Find F To Find P To Find F To Find A To Find P To Find A To Find P Given G Given P Given F Given A Given F (A|F Given A Given P (A|P Given G (A|G (F|P i%,n) (P|F i%,n) (F|A i%,n) i%,n) (P|A i%,n) i%,n) (P|G i%,n) i%,n)

55 60

1.31563 1.34885

0.76009 0.74137

63.12577 69.77003

0.01584 0.01433

47.98145 51.72556

0.02084 0.01933

1235.26857 1448.64580

25.74471 28.00638

72

1.43204

0.69830

86.40886

0.01157

60.33951

0.01657

2012.34779

33.35041

75 80

1.45363 1.49034

0.68793 0.67099

90.72650 98.06771

0.01102 0.01020

62.41365 65.80231

0.01602 0.01520

2163.75249 2424.64551

34.66794 36.84742

84 90

1.52037 1.56655

0.65773 0.63834

104.07393 113.31094

9.6086E−03 8.8253E−03

68.45304 72.33130

0.01461 0.01383

2640.66405 2976.07688

38.57628 41.14508

96 100

1.61414 1.64667

0.61952 0.60729

122.82854 129.33370

8.1414E−03 7.7319E−03

76.09522 78.54264

0.01314 0.01273

3324.18460 3562.79343

43.68454 45.36126

108

1.71370

0.58353

142.73990

7.0057E−03

83.29342

0.01201

4054.37473

48.67581

120 132

1.81940 1.93161

0.54963 0.51770

163.87935 186.32263

6.1021E−03 5.3670E−03

90.07345 96.45960

0.01110 0.01037

4823.50506 5624.58677

53.55080 58.31029

144 180

2.05075 2.45409

0.48763 0.40748

210.15016 290.81871

4.7585E−03 102.47474 3.4386E−03 118.50351

9.7585E−03 6451.31165 8.4386E−03 9031.33557

62.95514 76.21154

240

3.31020

0.30210

462.04090

2.1643E−03 139.58077

7.1643E−03 1.3416E+04

96.11309

360 480

6.02258 10.95745

0.16604 1004.51504 0.09126 1991.49073

9.9551E−04 166.79161 5.0214E−04 181.74758

5.9955E−03 2.1403E+04 128.32362 5.5021E−03 2.7588E+04 151.79491

600

19.93596

0.05016 3787.19108

2.6405E−04 189.96787

5.2640E−03 3.1974E+04 168.31425

0.75% Time Value of Money Factors Discrete Compounding 0.75%

TABLE A-A-3 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P Given P Given F Given A Given F (A|F Given A Given P (A|P Given G n (F|P i%,n) (P|F i%,n) (F|A i%,n) i%,n) (P|A i%,n) i%,n) (P|G i%,n) 1 1.00750 0.99256 1.00000 1.00000 0.99256 1.00750 0.00000

To Find A Given G (A|G i%,n) 0.00000

2 3

1.01506 1.02267

0.98517 0.97783

2.00750 3.02256

0.49813 0.33085

1.97772 2.95556

0.50563 0.33835

0.98517 2.94083

0.49813 0.99502

4

1.03034

0.97055

4.04523

0.24721

3.92611

0.25471

5.85250

1.49066

5 6

1.03807 1.04585

0.96333 0.95616

5.07556 6.11363

0.19702 0.16357

4.88944 5.84560

0.20452 0.17107

9.70581 14.48660

1.98506 2.47821

7 8

1.05370 1.06160

0.94904 0.94198

7.15948 8.21318

0.13967 0.12176

6.79464 7.73661

0.14717 0.12926

20.18084 26.77467

2.97011 3.46077

9 10

1.06956 1.07758

0.93496 0.92800

9.27478 10.34434

0.10782 0.09667

8.67158 9.59958

0.11532 0.10417

34.25438 42.60641

3.95019 4.43836

11

1.08566

0.92109

11.42192

0.08755

10.52067

0.09505

51.81736

4.92529

12 13

1.09381 1.10201

0.91424 0.90743

12.50759 13.60139

0.07995 0.07352

11.43491 12.34235

0.08745 0.08102

61.87398 72.76316

5.41097 5.89541

14 15

1.11028 1.11860

0.90068 0.89397

14.70340 15.81368

0.06801 0.06324

13.24302 14.13699

0.07551 0.07074

84.47197 96.98758

6.37860 6.86055

16

1.12699

0.88732

16.93228

0.05906

15.02431

0.06656

110.29735

7.34126

17 18

1.13544 1.14396

0.88071 0.87416

18.05927 19.19472

0.05537 0.05210

15.90502 16.77918

0.06287 0.05960

124.38875 139.24940

7.82072 8.29894

19 20

1.15254 1.16118

0.86765 0.86119

20.33868 21.49122

0.04917 0.04653

17.64683 18.50802

0.05667 0.05403

154.86708 171.22969

8.77592 9.25165

21 22

1.16989 1.17867

0.85478 0.84842

22.65240 23.82230

0.04415 0.04198

19.36280 20.21121

0.05165 0.04948

188.32527 206.14200

9.72614 10.19939

23

1.18751

0.84210

25.00096

0.04000

21.05331

0.04750

224.66820

10.67139

24 25

1.19641 1.20539

0.83583 0.82961

26.18847 27.38488

0.03818 0.03652

21.88915 22.71876

0.04568 0.04402

243.89233 263.80295

11.14216 11.61168

26 27

1.21443 1.22354

0.82343 0.81730

28.59027 29.80470

0.03498 0.03355

23.54219 24.35949

0.04248 0.04105

284.38879 305.63869

12.07996 12.54701

28 29

1.23271 1.24196

0.81122 0.80518

31.02823 32.26094

0.03223 0.03100

25.17071 25.97589

0.03973 0.03850

327.54162 350.08668

13.01281 13.47737

30

1.25127

0.79919

33.50290

0.02985

26.77508

0.03735

373.26310

13.94069

36 40

1.30865 1.34835

0.76415 0.74165

41.15272 46.44648

0.02430 0.02153

31.44681 34.44694

0.03180 0.02903

524.99236 637.46933

16.69462 18.50583

48 50

1.43141 1.45296

0.69861 0.68825

57.52071 60.39426

0.01739 0.01656

40.18478 41.56645

0.02489 0.02406

886.84045 953.84863

22.06906 22.94756

52

1.47483

0.67804

63.31107

0.01580

42.92762

0.02330

1022.58522

23.82115

Single Sums

n

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P Given P Given F Given A Given F (A|F Given A Given P (A|P Given G (F|P i%,n) (P|F i%,n) (F|A i%,n) i%,n) (P|A i%,n) i%,n) (P|G i%,n)

To Find A Given G (A|G i%,n)

55

1.50827

0.66301

67.76883

0.01476

44.93161

0.02226

1128.78691

25.12233

60

1.56568

0.63870

75.42414

0.01326

48.17337

0.02076

1313.51888

27.26649

72 75

1.71255 1.75137

0.58392 0.57098

95.00703 100.18331

0.01053 9.9817E−03

55.47685 57.20267

0.01803 0.01748

1791.24629 1917.22249

32.28818 33.51631

80 84

1.81804 1.87320

0.55004 0.53385

109.07253 116.42693

9.1682E−03 8.5891E−03

59.99444 62.15396

0.01667 0.01609

2132.14723 2308.12830

35.53908 37.13566

90 96

1.95909 2.04892

0.51044 0.48806

127.87899 139.85616

7.8199E−03 7.1502E−03

65.27461 68.25844

0.01532 0.01465

2577.99605 2853.93524

39.49462 41.81073

100 108

2.11108 2.24112

0.47369 0.44620

148.14451 165.48322

6.7502E−03 6.0429E−03

70.17462 73.83938

0.01425 0.01354

3040.74530 3419.90409

43.33112 46.31545

120

2.45136

0.40794

193.51428

5.1676E−03

78.94169

0.01267

3998.56214

50.65210

132

2.68131

0.37295

224.17484

4.4608E−03

83.60642

0.01196

4583.57014

54.82318

144 180

2.93284 3.83804

0.34097 0.26055

257.71157 378.40577

3.8803E−03 2.6427E−03

87.87109 98.59341

0.01138 0.01014

5169.58283 6892.60143

58.83144 69.90935

240

6.00915

0.16641

667.88687

1.4973E−03 111.14495

8.9973E−03 9494.11617

85.42103

360

14.73058

0.06789 1830.74348

5.4623E−04 124.28187

8.0462E−03 1.3312E+04 107.11448

480

36.10990

0.02769 4681.32027

2.1361E−04 129.64090

7.7136E−03 1.5513E+04 119.66198

600

88.51826

0.01130 1.1669E+04

8.5696E−05 131.82705

7.5857E−03 1.6673E+04 126.47762

1.00% Time Value of Money Factors Discrete Compounding 1.00%

TABLE A-A-4 Time Value of Money Factors Discrete Compounding Single Sums

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G n (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n) 1 1.01000 0.99010 1.00000 1.00000 0.99010 1.01000 0.00000 0.00000 2

1.02010

0.98030

2.01000

0.49751

1.97040

0.50751

0.98030

0.49751

3

1.03030

0.97059

3.03010

0.33002

2.94099

0.34002

2.92148

0.99337

4

1.04060

0.96098

4.06040

0.24628

3.90197

0.25628

5.80442

1.48756

5 6

1.05101 1.06152

0.95147 0.94205

5.10101 6.15202

0.19604 0.16255

4.85343 5.79548

0.20604 0.17255

9.61028 14.32051

1.98010 2.47098

7

1.07214

0.93272

7.21354

0.13863

6.72819

0.14863

19.91681

2.96020

8

1.08286

0.92348

8.28567

0.12069

7.65168

0.13069

26.38120

3.44777

9

1.09369

0.91434

9.36853

0.10674

8.56602

0.11674

33.69592

3.93367

10 11

1.10462 1.11567

0.90529 0.89632

10.46221 11.56683

0.09558 0.08645

9.47130 10.36763

0.10558 0.09645

41.84350 50.80674

4.41792 4.90052

12

1.12683

0.88745

12.68250

0.07885

11.25508

0.08885

60.56868

5.38145

13

1.13809

0.87866

13.80933

0.07241

12.13374

0.08241

71.11263

5.86073

14

1.14947

0.86996

14.94742

0.06690

13.00370

0.07690

82.42215

6.33836

15 16

1.16097 1.17258

0.86135 0.85282

16.09690 17.25786

0.06212 0.05794

13.86505 14.71787

0.07212 94.48104 0.06794 107.27336

6.81433 7.28865

17

1.18430

0.84438

18.43044

0.05426

15.56225

0.06426 120.78340

7.76131

18

1.19615

0.83602

19.61475

0.05098

16.39827

0.06098 134.99569

8.23231

19 20

1.20811 1.22019

0.82774 0.81954

20.81090 22.01900

0.04805 0.04542

17.22601 18.04555

0.05805 149.89501 0.05542 165.46636

8.70167 9.16937

21

1.23239

0.81143

23.23919

0.04303

18.85698

0.05303 181.69496

9.63542

22

1.24472

0.80340

24.47159

0.04086

19.66038

0.05086 198.56628

10.09982

23

1.25716

0.79544

25.71630

0.03889

20.45582

0.04889 216.06600

10.56257

24 25

1.26973 1.28243

0.78757 0.77977

26.97346 28.24320

0.03707 0.03541

21.24339 22.02316

0.04707 234.18002 0.04541 252.89446

11.02367 11.48312

26

1.29526

0.77205

29.52563

0.03387

22.79520

0.04387 272.19566

11.94092

27

1.30821

0.76440

30.82089

0.03245

23.55961

0.04245 292.07016

12.39707

28

1.32129

0.75684

32.12910

0.03112

24.31644

0.04112 312.50472

12.85158

29 30

1.33450 1.34785

0.74934 0.74192

33.45039 34.78489

0.02990 0.02875

25.06579 25.80771

0.03990 333.48630 0.03875 355.00207

13.30444 13.75566

36

1.43077

0.69892

43.07688

0.02321

30.10751

0.03321 494.62069

16.42848

40

1.48886

0.67165

48.88637

0.02046

32.83469

0.03046 596.85606

18.17761

48

1.61223

0.62026

61.22261

0.01633

37.97396

0.02633 820.14601

21.59759

50 52

1.64463 1.67769

0.60804 0.59606

64.46318 67.76889

0.01551 0.01476

39.19612 40.39419

0.02551 879.41763 0.02476 939.91752

22.43635 23.26863

55

1.72852

0.57853

72.85246

0.01373

42.14719

0.02373 1032.81478

24.50495

Single Sums

n

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60

1.81670

0.55045

81.66967

0.01224

44.95504

0.02224 1192.80614

26.53331

72

2.04710

0.48850

104.70993

9.5502E−03

51.15039

0.01955 1597.86733

31.23861

75 80

2.10913 2.21672

0.47413 0.45112

110.91285 121.67152

9.0161E−03 8.2189E−03

52.58705 54.88821

0.01902 1702.73397 0.01822 1879.87710

32.37934 34.24920

84

2.30672

0.43352

130.67227

7.6527E−03

56.64845

0.01765 2023.31531

35.71704

90

2.44863

0.40839

144.86327

6.9031E−03

59.16088

0.01690 2240.56748

37.87245

96

2.59927

0.38472

159.92729

6.2528E−03

61.52770

0.01625 2459.42979

39.97272

100 108

2.70481 2.92893

0.36971 0.34142

170.48138 192.89258

5.8657E−03 5.1842E−03

63.02888 65.85779

0.01587 2605.77575 0.01518 2898.42028

41.34257 44.01029

120

3.30039

0.30299

230.03869

4.3471E−03

69.70052

0.01435 3334.11485

47.83486

132

3.71896

0.26889

271.89586

3.6779E−03

73.11075

0.01368 3761.69441

51.45200

144

4.19062

0.23863

319.06156

3.1342E−03

76.13716

0.01313 4177.46642

54.86764

180 240

5.99580 10.89255

0.16678 0.09181

499.58020 989.25537

2.0017E−03 1.0109E−03

83.32166 90.81942

0.01200 5330.06592 0.01101 6878.60156

63.96975 75.73933

360

35.94964

0.02782

3494.96413

2.8613E−04

97.21833

0.01029 8720.43230

89.69947

480 118.64773

8.4283E−03 1.1765E+04

8.5000E−05

99.15717

0.01008 9511.15793

95.92002

600 391.58340

2.5537E−03 3.9058E+04

2.5603E−05

99.74463

0.01003 9821.23859

98.46384

1.25% Time Value of Money Factors Discrete Compounding 1.25%

TABLE A-A-5 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

n 1

1.01250

0.98765

1.00000

1.00000

0.98765

1.01250

0.00000

0.00000

2 3

1.02516 1.03797

0.97546 0.96342

2.01250 3.03766

0.49689 0.32920

1.96312 2.92653

0.50939 0.34170

0.97546 2.90230

0.49689 0.99172

4

1.05095

0.95152

4.07563

0.24536

3.87806

0.25786

5.75687

1.48447

5

1.06408

0.93978

5.12657

0.19506

4.81784

0.20756

9.51598

1.97516

6 7

1.07738 1.09085

0.92817 0.91672

6.19065 7.26804

0.16153 0.13759

5.74601 6.66273

0.17403 0.15009

14.15685 19.65715

2.46377 2.95032

8

1.10449

0.90540

8.35889

0.11963

7.56812

0.13213

25.99494

3.43479

9

1.11829

0.89422

9.46337

0.10567

8.46234

0.11817

33.14870

3.91720

10

1.13227

0.88318

10.58167

0.09450

9.34553

0.10700

41.09733

4.39754

11 12

1.14642 1.16075

0.87228 0.86151

11.71394 12.86036

0.08537 0.07776

10.21780 11.07931

0.09787 0.09026

49.82011 59.29670

4.87581 5.35202

13

1.17526

0.85087

14.02112

0.07132

11.93018

0.08382

69.50717

5.82616

14

1.18995

0.84037

15.19638

0.06581

12.77055

0.07831

80.43196

6.29824

15

1.20483

0.82999

16.38633

0.06103

13.60055

0.07353

92.05186

6.76825

16 17

1.21989 1.23514

0.81975 0.80963

17.59116 18.81105

0.05685 0.05316

14.42029 15.22992

0.06935 104.34806 0.06566 117.30207

7.23620 7.70208

18

1.25058

0.79963

20.04619

0.04988

16.02955

0.06238 130.89580

8.16591

19

1.26621

0.78976

21.29677

0.04696

16.81931

0.05946 145.11145

8.62767

20

1.28204

0.78001

22.56298

0.04432

17.59932

0.05682 159.93161

9.08738

21 22

1.29806 1.31429

0.77038 0.76087

23.84502 25.14308

0.04194 0.03977

18.36969 19.13056

0.05444 175.33919 0.05227 191.31742

9.54502 10.00062

23

1.33072

0.75147

26.45737

0.03780

19.88204

0.05030 207.84986

10.45415

24

1.34735

0.74220

27.78808

0.03599

20.62423

0.04849 224.92039

10.90564

25

1.36419

0.73303

29.13544

0.03432

21.35727

0.04682 242.51321

11.35507

26 27

1.38125 1.39851

0.72398 0.71505

30.49963 31.88087

0.03279 0.03137

22.08125 22.79630

0.04529 260.61282 0.04387 279.20402

11.80245 12.24778

28

1.41599

0.70622

33.27938

0.03005

23.50252

0.04255 298.27192

12.69106

29

1.43369

0.69750

34.69538

0.02882

24.20002

0.04132 317.80191

13.13230

30 36

1.45161 1.56394

0.68889 0.63941

36.12907 45.11551

0.02768 0.02217

24.88891 28.84727

0.04018 337.77969 0.03467 466.28302

13.57150 16.16385

40

1.64362

0.60841

51.48956

0.01942

31.32693

0.03192 559.23198

17.85148

48

1.81535

0.55086

65.22839

0.01533

35.93148

0.02783 759.22956

21.12993

50

1.86102

0.53734

68.88179

0.01452

37.01288

0.02702 811.67385

21.92950

52 55

1.90784 1.98028

0.52415 0.50498

72.62710 78.42246

0.01377 0.01275

38.06773 39.60169

0.02627 864.94093 0.02525 946.22770

22.72110 23.89362

n

Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60

2.10718

0.47457

88.57451

0.01129

42.03459

72

2.44592

75

2.53879

80

0.02379 1084.84285

25.80834

0.40884

115.67362

8.6450E−03

0.39389

123.10349

8.1232E−03

47.29247

0.02115 1428.45610

30.20472

48.48897

0.02062 1515.79039

31.26052

2.70148

0.37017

136.11880

7.3465E−03

50.38666

0.01985 1661.86513

32.98225

84 90

2.83911 3.05881

0.35222 0.32692

147.12904 164.70501

6.7968E−03 6.0715E−03

51.82219 53.84606

0.01930 1778.83839 0.01857 1953.83026

34.32581 36.28548

96

3.29551

0.30344

183.64106

5.4454E−03

55.72457

0.01795 2127.52438

38.17929

100

3.46340

0.28873

197.07234

5.0743E−03

56.90134

0.01757 2242.24109

39.40577

108 120

3.82528 4.44021

0.26142 0.22521

226.02255 275.21706

4.4243E−03 3.6335E−03

59.08651 61.98285

0.01692 2468.26361 0.01613 2796.56945

41.77373 45.11844

132

5.15400

0.19402

332.31981

3.0091E−03

64.47807

0.01551 3109.35041

48.22338

144

5.98253

0.16715

398.60208

2.5088E−03

66.62772

0.01501 3404.60974

51.09900

180

9.35633

0.10688

668.50676

1.4959E−03

71.44964

0.01400 4176.90718

58.45945

240 360

19.71549 87.54100

0.05072 0.01142

1497.23948 6923.27961

6.6790E−04 1.4444E−04

75.94228 79.08614

0.01317 5101.52883 0.01264 5997.90267

67.17640 75.84012

480 388.70068

2.5727E−03 3.1016E+04

3.2241E−05

79.79419

0.01253 6284.74422

78.76193

600 1725.91392

5.7940E−04 1.3799E+05

7.2467E−06

79.95365

0.01251 6368.48047

79.65216

1.50% Time Value of Money Factors Discrete Compounding 1.50%

TABLE A-A-6 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

n 1

1.01500

0.98522

1.00000

1.00000

0.98522

1.01500

0.00000

0.00000

2

1.03023

0.97066

2.01500

0.49628

3 4

1.04568 1.06136

0.95632 0.94218

3.04522 4.09090

0.32838 0.24444

1.95588

0.51128

0.97066

0.49628

2.91220 3.85438

0.34338 0.25944

2.88330 5.70985

0.99007 1.48139

5

1.07728

0.92826

5.15227

0.19409

4.78264

0.20909

9.42289

1.97023

6

1.09344

0.91454

6.22955

0.16053

5.69719

0.17553

13.99560

2.45658

7

1.10984

8 9

1.12649 1.14339

0.90103

7.32299

0.13656

6.59821

0.15156

19.40176

2.94046

0.88771 0.87459

8.43284 9.55933

0.11858 0.10461

7.48593 8.36052

0.13358 0.11961

25.61574 32.61248

3.42185 3.90077

10

1.16054

0.86167

10.70272

0.09343

9.22218

0.10843

40.36748

4.37721

11

1.17795

0.84893

11.86326

0.08429

10.07112

0.09929

48.85681

4.85118

12

1.19562

0.83639

13.04121

0.07668

10.90751

0.09168

58.05708

5.32267

13 14

1.21355 1.23176

0.82403 0.81185

14.23683 15.45038

0.07024 0.06472

11.73153 12.54338

0.08524 0.07972

67.94540 78.49944

5.79169 6.25824

15

1.25023

0.79985

16.68214

0.05994

13.34323

0.07494

89.69736

6.72231

16

1.26899

0.78803

17.93237

0.05577

14.13126

0.07077 101.51783

7.18392

17 18

1.28802 1.30734

0.77639 0.76491

19.20136 20.48938

0.05208 0.04881

14.90765 15.67256

0.06708 113.93999 0.06381 126.94349

7.64306 8.09973

19

1.32695

0.75361

21.79672

0.04588

16.42617

0.06088 140.50842

8.55394

20

1.34686

0.74247

23.12367

0.04325

17.16864

0.05825 154.61536

9.00569

21

1.36706

0.73150

24.47052

0.04087

17.90014

0.05587 169.24532

9.45497

22 23

1.38756 1.40838

0.72069 0.71004

25.83758 27.22514

0.03870 0.03673

18.62082 19.33086

0.05370 184.37976 0.05173 200.00058

9.90180 10.34618

24

1.42950

0.69954

28.63352

0.03492

20.03041

0.04992 216.09009

10.78810

25

1.45095

0.68921

30.06302

0.03326

20.71961

0.04826 232.63103

11.22758

26

1.47271

0.67902

31.51397

0.03173

21.39863

0.04673 249.60654

11.66460

27 28

1.49480 1.51722

0.66899 0.65910

32.98668 34.48148

0.03032 0.02900

22.06762 22.72672

0.04532 267.00017 0.04400 284.79585

12.09918 12.53132

29

1.53998

0.64936

35.99870

0.02778

23.37608

0.04278 302.97790

12.96102

30

1.56308

0.63976

37.53868

0.02664

24.01584

0.04164 321.53101

13.38829

36

1.70914

0.58509

47.27597

0.02115

27.66068

0.03615 439.83026

15.90092

40 48

1.81402 2.04348

0.55126 0.48936

54.26789 69.56522

0.01843 0.01437

29.91585 34.04255

0.03343 524.35682 0.02937 703.54615

17.52773 20.66667

50

2.10524

0.47500

73.68283

0.01357

34.99969

0.02857 749.96361

21.42772

52

2.16887

0.46107

77.92489

0.01283

35.92874

0.02783 796.87737

22.17938

55

2.26794

0.44093

84.52960

0.01183

37.27147

0.02683 868.02846

23.28936

Single Sums

n

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60 72

2.44322 2.92116

0.40930 0.34233

96.21465 128.07720

0.01039 7.8078E−03

39.38027 43.84467

0.02539 988.16739 0.02281 1279.79379

25.09296 29.18927

75

3.05459

0.32738

136.97278

7.3007E−03

44.84160

0.02230 1352.56005

30.16306

80

3.29066

0.30389

152.71085

6.5483E−03

46.40732

0.02155 1473.07411

31.74228

84

3.49259

0.28632

166.17264

6.0178E−03

47.57863

0.02102 1568.51404

32.96677

90 96

3.81895 4.17580

0.26185 0.23947

187.92990 211.72023

5.3211E−03 4.7232E−03

49.20985 50.70168

0.02032 1709.54387 0.01972 1847.47253

34.73987 36.43810

100

4.43205

0.22563

228.80304

4.3706E−03

51.62470

0.01937 1937.45061

37.52953

108

4.99267

0.20029

266.17777

3.7569E−03

53.31375

0.01876 2112.13479

39.61708

120

5.96932

0.16752

331.28819

3.0185E−03

55.49845

0.01802 2359.71143

42.51851

132 144

7.13703 8.53316

0.14011 0.11719

409.13539 502.21092

2.4442E−03 1.9912E−03

57.32571 58.85401

0.01744 2588.70855 0.01699 2798.57842

45.15789 47.55119

180

14.58437

0.06857

905.62451

1.1042E−03

62.09556

0.01610 3316.90537

53.41614

240

35.63282

0.02806

2308.85437

4.3312E−04

64.79573

0.01543 3870.69117

59.73682

360 212.70378 480 1269.69754

4.7014E−03 1.4114E+04 7.8759E−04 8.4580E+04

7.0854E−05 1.1823E−05

66.35324 66.61416

0.01507 4310.71648 0.01501 4415.74120

64.96618 66.28833

600 7579.23459

1.3194E−04 5.0522E+05

1.9794E−06

66.65787

0.01500 4438.58047

66.58749

1.75% Time Value of Money Factors Discrete Compounding 1.75%

TABLE A-A-7 Time Value of Money Factors Discrete Compounding Single Sums

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

n 1

1.01750

0.98280

1.00000

1.00000

0.98280

1.01750

0.00000

0.00000

2

1.03531

0.96590

2.01750

0.49566

1.94870

0.51316

0.96590

0.49566

3

1.05342

0.94929

3.05281

0.32757

2.89798

0.34507

2.86447

0.98843

4

1.07186

0.93296

4.10623

0.24353

3.83094

0.26103

5.66334

1.47832

5

1.09062

0.91691

5.17809

0.19312

4.74786

0.21062

9.33099

1.96531

6

1.10970

0.90114

6.26871

0.15952

5.64900

0.17702

13.83671

2.44941

7 8

1.12912 1.14888

0.88564 0.87041

7.37841 8.50753

0.13553 0.11754

6.53464 7.40505

0.15303 0.13504

19.15057 25.24345

2.93062 3.40895

9

1.16899

0.85544

9.65641

0.10356

8.26049

0.12106

32.08698

3.88439

10

1.18944

0.84073

10.82540

0.09238

9.10122

0.10988

39.65354

4.35695

11

1.21026

0.82627

12.01484

0.08323

9.92749

0.10073

47.91623

4.82662

12

1.23144

0.81206

13.22510

0.07561

10.73955

0.09311

56.84886

5.29341

13

1.25299

0.79809

14.45654

0.06917

11.53764

0.08667

66.42596

5.75733

14

1.27492

0.78436

15.70953

0.06366

12.32201

0.08116

76.62270

6.21836

15

1.29723

0.77087

16.98445

0.05888

13.09288

0.07638

87.41495

6.67653

16 17

1.31993 1.34303

0.75762 0.74459

18.28168 19.60161

0.05470 0.05102

13.85050 14.59508

0.07220 98.77919 0.06852 110.69257

7.13182 7.58424

18

1.36653

0.73178

20.94463

0.04774

15.32686

0.06524 123.13283

8.03379

19

1.39045

0.71919

22.31117

0.04482

16.04606

0.06232 136.07832

8.48048

20

1.41478

0.70682

23.70161

0.04219

16.75288

0.05969 149.50799

8.92431

21

1.43954

0.69467

25.11639

0.03981

17.44755

0.05731 163.40134

9.36529

22

1.46473

0.68272

26.55593

0.03766

18.13027

0.05516 177.73847

9.80341

23

1.49036

0.67098

28.02065

0.03569

18.80125

0.05319 192.49999

10.23868

24

1.51644

0.65944

29.51102

0.03389

19.46069

0.05139 207.66706

10.67111

25

1.54298

0.64810

31.02746

0.03223

20.10878

0.04973 223.22138

11.10069

26 27

1.56998 1.59746

0.63695 0.62599

32.57044 34.14042

0.03070 0.02929

20.74573 21.37173

0.04820 239.14512 0.04679 255.42098

11.52744 11.95135

28

1.62541

0.61523

35.73788

0.02798

21.98695

0.04548 272.03215

12.37243

29

1.65386

0.60465

37.36329

0.02676

22.59160

0.04426 288.96226

12.79069

30

1.68280

0.59425

39.01715

0.02563

23.18585

0.04313 306.19544

13.20613

36

1.86741

0.53550

49.56613

0.02018

26.54275

0.03768 415.12498

15.63986

40

2.00160

0.49960

57.23413

0.01747

28.59423

0.03497 492.01087

17.20665

48

2.29960

0.43486

74.26278

0.01347

32.29380

0.03097 652.60539

20.20838

50

2.38079

0.42003

78.90222

0.01267

33.14121

0.03017 693.70101

20.93167

52 55

2.46485 2.59653

0.40570 0.38513

83.70547 91.23016

0.01195 0.01096

33.95972 35.13545

0.02945 735.03220 0.02846 797.33210

21.64424 22.69310

Single Sums

n

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60

2.83182

0.35313

104.67522

9.5534E−03

36.96399

0.02705 901.49545

24.38848

72

3.48721

0.28676

142.12628

7.0360E−03

40.75645

0.02454 1149.11809

28.19476

75

3.67351

0.27222

152.77206

6.5457E−03

41.58748

0.02405 1209.77384

29.08986

80

4.00639

0.24960

171.79382

5.8209E−03

42.87993

0.02332 1309.24819

30.53289

84 90

4.29429 4.76538

0.23287 0.20985

188.24499 215.16462

5.3122E−03 4.6476E−03

43.83614 45.15161

0.02281 1387.15838 0.02215 1500.87981

31.64417 33.24089

96

5.28815

0.18910

245.03739

4.0810E−03

46.33703

0.02158 1610.47158

34.75560

100

5.66816

0.17642

266.75177

3.7488E−03

47.06147

0.02125 1681.08862

35.72112

108

6.51204

0.15356

314.97378

3.1749E−03

48.36790

0.02067 1816.18525

37.54939

120

8.01918

0.12470

401.09620

2.4932E−03

50.01709

0.01999 2003.02686

40.04685

132

9.87514

0.10126

507.15073

1.9718E−03

51.35632

0.01947 2170.82384

42.26985

144

12.16063

0.08223

637.75045

1.5680E−03

52.44385

0.01907 2320.13512

44.24036

180

22.70885

0.04404

1240.50595

8.0612E−04

54.62653

0.01831 2668.57763

48.85131

240 64.30730 360 515.69206

0.01555 3617.56017 1.9391E−03 2.9411E+04

2.7643E−04 3.4001E−05

56.25427 57.03205

0.01778 3001.26781 0.01753 3219.08332

53.35183 56.44341

480 4135.42921

2.4181E−04 2.3625E+05

4.2327E−06

57.12904

0.01750 3257.88395

57.02676

600 3.3163E+04

3.0154E−05 1.8950E+06

5.2772E−07

57.14113

0.01750 3264.17380

57.12476

2.00% Time Value of Money Factors Discrete Compounding 2.00%

TABLE A-A-8 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

n 1

1.02000

0.98039

1.00000

1.00000

0.98039

1.02000

0.00000

0.00000

2 3 4

1.04040

0.96117

2.02000

0.49505

1.94156

1.06121 1.08243

0.94232 0.92385

3.06040 4.12161

0.32675 0.24262

2.88388 3.80773

0.51505

0.96117

0.49505

0.34675 0.26262

2.84581 5.61735

0.98680 1.47525

5

1.10408

0.90573

5.20404

0.19216

4.71346

0.21216

9.24027

1.96040

6

1.12616

0.88797

6.30812

0.15853

5.60143

0.17853

13.68013

2.44226

7

1.14869

0.87056

8

1.17166

0.85349

7.43428

0.13451

6.47199

0.15451

18.90349

2.92082

8.58297

0.11651

7.32548

0.13651

24.87792

3.39608

9

1.19509

0.83676

9.75463

0.10252

8.16224

0.12252

31.57197

3.86805

10

1.21899

0.82035

10.94972

0.09133

8.98259

0.11133

38.95510

4.33674

11

1.24337

0.80426

12.16872

0.08218

9.78685

0.10218

46.99773

4.80213

12

1.26824

0.78849

13.41209

0.07456

10.57534

0.09456

55.67116

5.26424

13 14

1.29361 1.31948

0.77303 0.75788

14.68033 15.97394

0.06812 0.06260

11.34837 12.10625

0.08812 0.08260

64.94755 74.79992

5.72307 6.17862

15

1.34587

0.74301

17.29342

0.05783

12.84926

0.07783

85.20213

6.63090

16

1.37279

0.72845

18.63929

0.05365

13.57771

0.07365

96.12881

7.07990

17

1.40024

0.71416

20.01207

0.04997

14.29187

0.06997 107.55542

7.52564

18

1.42825

0.70016

21.41231

0.04670

14.99203

0.06670 119.45813

7.96811

19

1.45681

0.68643

22.84056

0.04378

15.67846

0.06378 131.81388

8.40732

20

1.48595

0.67297

24.29737

0.04116

16.35143

0.06116 144.60033

8.84328

21

1.51567

0.65978

25.78332

0.03878

17.01121

0.05878 157.79585

9.27599

22

1.54598

0.64684

27.29898

0.03663

17.65805

0.05663 171.37947

9.70546

23 24

1.57690 1.60844

0.63416 0.62172

28.84496 30.42186

0.03467 0.03287

18.29220 18.91393

0.05467 185.33090 0.05287 199.63049

10.13169 10.55468

25

1.64061

0.60953

32.03030

0.03122

19.52346

0.05122 214.25924

10.97445

26

1.67342

0.59758

33.67091

0.02970

20.12104

0.04970 229.19872

11.39100

27

1.70689

0.58586

35.34432

0.02829

20.70690

0.04829 244.43113

11.80433

28

1.74102

0.57437

37.05121

0.02699

21.28127

0.04699 259.93924

12.21446

29

1.77584

0.56311

38.79223

0.02578

21.84438

0.04578 275.70639

12.62138

30

1.81136

0.55207

40.56808

0.02465

22.39646

0.04465 291.71644

13.02512

36

2.03989

0.49022

51.99437

0.01923

25.48884

0.03923 392.04045

15.38087

40 48

2.20804 2.58707

0.45289 0.38654

60.40198 79.35352

0.01656 0.01260

27.35548 30.67312

0.03656 461.99313 0.03260 605.96572

16.88850 19.75559

50

2.69159

0.37153

84.57940

0.01182

31.42361

0.03182 642.36059

20.44198

52

2.80033

0.35710

90.01641

0.01111

32.14495

0.03111 678.78489

21.11638

55

2.97173

0.33650

98.58653

0.01014

33.17479

0.03014 733.35269

22.10572

Single Sums

n

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60

3.28103

0.30478

114.05154

8.7680E−03

34.76089

0.02877 823.69753

23.69610

72 75

4.16114 4.41584

0.24032 0.22646

158.05702 170.79177

6.3268E−03 5.8551E−03

37.98406 38.67711

0.02633 1034.05570 0.02586 1084.63929

27.22341 28.04344

80

4.87544

0.20511

193.77196

5.1607E−03

39.74451

0.02516 1166.78677

29.35718

84

5.27733

0.18949

213.86661

4.6758E−03

40.52552

0.02468 1230.41912

30.36159

90

5.94313

0.16826

247.15666

4.0460E−03

41.58693

0.02405 1322.17008

31.79292

96

6.69293

0.14941

284.64666

3.5131E−03

42.52943

0.02351 1409.29734

33.13699

100

7.24465

0.13803

312.23231

3.2027E−03

43.09835

0.02320 1464.75275

33.98628

108

8.48826

0.11781

374.41288

2.6708E−03

44.10951

0.02267 1569.30251

35.57742

120

10.76516

0.09289

488.25815

2.0481E−03

45.35539

0.02205 1710.41605

37.71142

132 144

13.65283 17.31509

0.07324 0.05775

632.64148 815.75446

1.5807E−03 1.2259E−03

46.33776 47.11235

0.02158 1833.47151 0.02123 1939.79497

39.56755 41.17381

180

35.32083

0.02831

1716.04157

5.8274E−04

48.58440

0.02058 2174.41310

44.75537

240 115.88874

8.6290E−03 5744.43676

1.7408E−04

49.56855

0.02017 2374.87999

47.91102

360 1247.56113

8.0156E−04 6.2328E+04

1.6044E−05

49.95992

0.02002 2483.56794

49.71121

480 1.3430E+04

7.4459E−05 6.7146E+05

1.4893E−06

49.99628

0.02000 2498.02683

49.96426

600 1.4458E+05

6.9167E−06 7.2289E+06

1.3833E−07

49.99965

0.02000 2499.77521

49.99585

3.00% Time Value of Money Factors Discrete Compounding 3.00%

TABLE A-A-9 Time Value of Money Factors Discrete Compounding Single Sums

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G n (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n) 1 1.03000 0.97087 1.00000 1.00000 0.97087 1.03000 0.00000 0.00000 2

1.06090

0.94260

2.03000

0.49261

1.91347

0.52261

0.94260

0.49261

3

1.09273

0.91514

3.09090

0.32353

2.82861

0.35353

2.77288

0.98030

4

1.12551

0.88849

4.18363

0.23903

3.71710

0.26903

5.43834

1.46306

5

1.15927

0.86261

5.30914

0.18835

4.57971

0.21835

8.88878

1.94090

6

1.19405

0.83748

6.46841

0.15460

5.41719

0.18460

13.07620

2.41383

7

1.22987

0.81309

7.66246

0.13051

6.23028

0.16051

17.95475

2.88185

8

1.26677

0.78941

8.89234

0.11246

7.01969

0.14246

23.48061

3.34496

9

1.30477

0.76642

10.15911

0.09843

7.78611

0.12843

29.61194

3.80318

10 11

1.34392 1.38423

0.74409 0.72242

11.46388 12.80780

0.08723 0.07808

8.53020 9.25262

0.11723 0.10808

36.30879 43.53300

4.25650 4.70494

12

1.42576

0.70138

14.19203

0.07046

9.95400

0.10046

51.24818

5.14850

13

1.46853

0.68095

15.61779

0.06403

10.63496

0.09403

59.41960

5.58720

14

1.51259

0.66112

17.08632

0.05853

11.29607

0.08853

68.01413

6.02104

15

1.55797

0.64186

18.59891

0.05377

11.93794

0.08377

77.00020

6.45004

16

1.60471

0.62317

20.15688

0.04961

12.56110

0.07961

86.34770

6.87421

17

1.65285

0.60502

21.76159

0.04595

13.16612

0.07595

96.02796

7.29357

18

1.70243

0.58739

23.41444

0.04271

13.75351

0.07271 106.01367

7.70812

19 20

1.75351 1.80611

0.57029 0.55368

25.11687 26.87037

0.03981 0.03722

14.32380 14.87747

0.06981 116.27882 0.06722 126.79866

8.11788 8.52286

21

1.86029

0.53755

28.67649

0.03487

15.41502

0.06487 137.54964

8.92309

22

1.91610

0.52189

30.53678

0.03275

15.93692

0.06275 148.50939

9.31858

23

1.97359

0.50669

32.45288

0.03081

16.44361

0.06081 159.65661

9.70934

24

2.03279

0.49193

34.42647

0.02905

16.93554

0.05905 170.97108

10.09540

25

2.09378

0.47761

36.45926

0.02743

17.41315

0.05743 182.43362

10.47677

26

2.15659

0.46369

38.55304

0.02594

17.87684

0.05594 194.02598

10.85348

27

2.22129

0.45019

40.70963

0.02456

18.32703

0.05456 205.73090

11.22554

28

2.28793

0.43708

42.93092

0.02329

18.76411

0.05329 217.53197

11.59298

29 30

2.35657 2.42726

0.42435 0.41199

45.21885 47.57542

0.02211 0.02102

19.18845 19.60044

0.05211 229.41367 0.05102 241.36129

11.95582 12.31407

36

2.89828

0.34503

63.27594

0.01580

21.83225

0.04580 313.70284

14.36878

40

3.26204

0.30656

75.40126

0.01326

23.11477

0.04326 361.74994

15.65016

48

4.13225

0.24200

104.40840

9.5778E−03

25.26671

0.03958 455.02547

18.00890

50

4.38391

0.22811

112.79687

8.8655E−03

25.72976

0.03887 477.48033

18.55751

52

4.65089

0.21501

121.69620

8.2172E−03

26.16624

0.03822 499.51915

19.09021

55

5.08215

0.19677

136.07162

7.3491E−03

26.77443

0.03735 531.74111

19.86004

Single Sums

n

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60

5.89160

0.16973

163.05344

6.1330E−03

27.67556

0.03613 583.05261

21.06742

72 75

8.40002

0.11905

246.66724

4.0540E−03

29.36509

0.03405 693.12255

23.60363

9.17893

0.10895

272.63086

3.6680E−03

29.70183

0.03367 717.69785

24.16342

80

10.64089

0.09398

321.36302

3.1117E−03

30.20076

0.03311 756.08652

25.03534

84

11.97642

0.08350

365.88054

2.7331E−03

30.55009

0.03273 784.54337

25.68056

90

14.30047

0.06993

443.34890

2.2556E−03

31.00241

0.03226 823.63021

26.56665

96

17.07551

0.05856

535.85019

1.8662E−03

31.38122

0.03187 858.63770

27.36151

100 108

19.21863 24.34559

0.05203 0.04108

607.28773 778.18627

1.6467E−03 1.2850E−03

31.59891 31.96416

0.03165 879.85405 0.03129 917.60126

27.84445 28.70719

120

34.71099

0.02881

1123.69957

8.8992E−04

32.37302

0.03089 963.86347

29.77366

132

49.48957

0.02021

1616.31893

6.1869E−04

32.65979

0.03062 999.75206

30.61110

144

70.56029

0.01417

2318.67634

4.3128E−04

32.86092

0.03043 1027.33721

31.26319

180 204.50336

4.8899E−03 6783.44532

1.4742E−04

33.17034

0.03015 1076.33852

32.44883

240 1204.85263

8.2998E−04 4.0128E+04

2.4920E−05

33.30567

0.03002 1103.54910

33.13397

360 4.1822E+04

2.3911E−05 1.3940E+06

7.1735E−07

33.33254

0.03000 1110.79761

33.32473

480 1.4517E+06

6.8886E−07 4.8389E+07

2.0666E−08

33.33331

0.03000 1111.09932

33.33300

600 5.0389E+07

1.9846E−08 1.6796E+09

5.9537E−10

33.33333

0.03000 1111.11069

33.33332

4.00% Time Value of Money Factors Discrete Compounding 4.00%

TABLE A-A-10 Time Value of Money Factors Discrete Compounding Single Sums

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

n 1

1.04000

0.96154

1.00000

1.00000

0.96154

1.04000

0.00000

0.00000

2

1.08160

0.92456

2.04000

0.49020

1.88609

0.53020

0.92456

0.49020

3

1.12486

0.88900

3.12160

0.32035

2.77509

0.36035

2.70255

0.97386

4

1.16986

0.85480

4.24646

0.23549

3.62990

0.27549

5.26696

1.45100

5

1.21665

0.82193

5.41632

0.18463

4.45182

0.22463

8.55467

1.92161

6 7

1.26532 1.31593

0.79031 0.75992

6.63298 7.89829

0.15076 0.12661

5.24214 6.00205

0.19076 0.16661

12.50624 17.06575

2.38571 2.84332

8

1.36857

0.73069

9.21423

0.10853

6.73274

0.14853

22.18058

3.29443

9

1.42331

0.70259

10.58280

0.09449

7.43533

0.13449

27.80127

3.73908

10

1.48024

0.67556

12.00611

0.08329

8.11090

0.12329

33.88135

4.17726

11

1.53945

0.64958

13.48635

0.07415

8.76048

0.11415

40.37716

4.60901

12

1.60103

0.62460

15.02581

0.06655

9.38507

0.10655

47.24773

5.03435

13

1.66507

0.60057

16.62684

0.06014

9.98565

0.10014

54.45462

5.45329

14

1.73168

0.57748

18.29191

0.05467

10.56312

0.09467

61.96179

5.86586

15

1.80094

0.55526

20.02359

0.04994

11.11839

0.08994

69.73550

6.27209

16 17

1.87298 1.94790

0.53391 0.51337

21.82453 23.69751

0.04582 0.04220

11.65230 12.16567

0.08582 0.08220

77.74412 85.95809

6.67200 7.06563

18

2.02582

0.49363

25.64541

0.03899

12.65930

0.07899

94.34977

7.45300

19

2.10685

0.47464

27.67123

0.03614

13.13394

0.07614 102.89333

7.83416

20

2.19112

0.45639

29.77808

0.03358

13.59033

0.07358 111.56469

8.20912

21

2.27877

0.43883

31.96920

0.03128

14.02916

0.07128 120.34136

8.57794

22

2.36992

0.42196

34.24797

0.02920

14.45112

0.06920 129.20242

8.94065

23

2.46472

0.40573

36.61789

0.02731

14.85684

0.06731 138.12840

9.29729

24

2.56330

0.39012

39.08260

0.02559

15.24696

0.06559 147.10119

9.64790

25 26

2.66584 2.77247

0.37512 0.36069

41.64591 44.31174

0.02401 0.02257

15.62208 15.98277

0.06401 156.10400 0.06257 165.12123

9.99252 10.33120

27

2.88337

0.34682

47.08421

0.02124

16.32959

0.06124 174.13846

10.66399

28

2.99870

0.33348

49.96758

0.02001

16.66306

0.06001 183.14235

10.99092

29

3.11865

0.32065

52.96629

0.01888

16.98371

0.05888 192.12059

11.31205

30

3.24340

0.30832

56.08494

0.01783

17.29203

0.05783 201.06183

11.62743

36

4.10393

0.24367

77.59831

0.01289

18.90828

0.05289 253.40520

13.40181

40

4.80102

0.20829

95.02552

0.01052

19.79277

0.05052 286.53030

14.47651

48

6.57053

0.15219

139.26321

7.1806E−03

21.19513

0.04718 347.24455

16.38322

50

7.10668

0.14071

152.66708

6.5502E−03

21.48218

0.04655 361.16385

16.81225

52 55

7.68659 8.64637

0.13010 0.11566

167.16472 191.15917

5.9821E−03 5.2312E−03

21.74758 22.10861

0.04598 374.56381 0.04523 393.68897

17.22324 17.80704

n

Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60

10.51963

0.09506

237.99069

4.2018E−03

22.62349

0.04420 422.99665

18.69723

72

16.84226

0.05937

396.05656

2.5249E−03

23.51564

0.04252 481.01697

20.45519

75

18.94525

0.05278

448.63137

2.2290E−03

23.68041

0.04223 493.04083

20.82062

80 84

23.04980 26.96500

0.04338 0.03709

551.24498 649.12512

1.8141E−03 1.5405E−03

23.91539 24.07287

0.04181 511.11614 0.04154 523.94309

21.37185 21.76488

90

34.11933

0.02931

827.98333

1.2078E−03

24.26728

0.04121 540.73692

22.28255

96

43.17184

0.02316

1054.29603

9.4850E−04

24.42092

0.04095 554.93118

22.72360

100

50.50495

0.01980

1237.62370

8.0800E−04

24.50500

0.04081 563.12487

22.98000

108

69.11951

0.01447

1702.98772

5.8720E−04

24.63831

0.04059 576.89491

23.41455

120 110.66256

9.0365E−03 2741.56402

3.6476E−04

24.77409

0.04036 592.24276

23.90573

132 177.17433

5.6442E−03 4404.35813

2.2705E−04

24.85890

0.04023 602.84668

24.25074

144 283.66180

3.5253E−03 7066.54508

1.4151E−04

24.91187

0.04014 610.10550

24.49056

180 1164.12891

8.5901E−04 2.9078E+04

3.4390E−05

24.97852

0.04003 620.59757

24.84525

240 1.2246E+04 360 1.3552E+06

8.1658E−05 3.0613E+05 7.3790E−07 3.3880E+07

3.2666E−06 2.9516E−08

24.99796 24.99998

0.04000 624.45902 0.04000 624.99290

24.98040 24.99973

480 1.4997E+08

6.6680E−09 3.7492E+09

2.6672E−10

25.00000

0.04000 624.99992

25.00000

600 1.6596E+10

6.0255E−11 4.1490E+11

2.4102E−12

25.00000

0.04000 625.00000

25.00000

5.00% Time Value of Money Factors Discrete Compounding 5.00%

TABLE A-A-11 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

n 1

1.05000

0.95238

1.00000

1.00000

0.95238

1.05000

0.00000

0.00000

2

1.10250

0.90703

2.05000

0.48780

1.85941

0.53780

0.90703

0.48780

3 4

1.15763 1.21551

0.86384 0.82270

3.15250 4.31013

0.31721 0.23201

2.72325 3.54595

0.36721 0.28201

2.63470 5.10281

0.96749 1.43905

5

1.27628

0.78353

5.52563

0.18097

4.32948

0.23097

8.23692

1.90252

6

1.34010

0.74622

6.80191

0.14702

5.07569

0.19702

11.96799

2.35790

7

1.40710

0.71068

8.14201

0.12282

5.78637

0.17282

16.23208

2.80523

8

1.47746

0.67684

9.54911

0.10472

6.46321

0.15472

20.96996

3.24451

9

1.55133

0.64461

11.02656

0.09069

7.10782

0.14069

26.12683

3.67579

10

1.62889

0.61391

12.57789

0.07950

7.72173

0.12950

31.65205

4.09909

11

1.71034

0.58468

14.20679

0.07039

8.30641

0.12039

37.49884

4.51444

12 13

1.79586 1.88565

0.55684 0.53032

15.91713 17.71298

0.06283 0.05646

8.86325 9.39357

0.11283 0.10646

43.62405 49.98791

4.92190 5.32150

14

1.97993

0.50507

19.59863

0.05102

9.89864

0.10102

56.55379

5.71329

15

2.07893

0.48102

21.57856

0.04634

10.37966

0.09634

63.28803

6.09731

16

2.18287

0.45811

23.65749

0.04227

10.83777

0.09227

70.15970

6.47363

17

2.29202

0.43630

25.84037

0.03870

11.27407

0.08870

77.14045

6.84229

18

2.40662

0.41552

28.13238

0.03555

11.68959

0.08555

84.20430

7.20336

19

2.52695

0.39573

30.53900

0.03275

12.08532

0.08275

91.32751

7.55690

20

2.65330

0.37689

33.06595

0.03024

12.46221

0.08024

98.48841

7.90297

21

2.78596

0.35894

35.71925

0.02800

12.82115

0.07800 105.66726

8.24164

22 23

2.92526 3.07152

0.34185 0.32557

38.50521 41.43048

0.02597 0.02414

13.16300 13.48857

0.07597 112.84611 0.07414 120.00868

8.57298 8.89706

24

3.22510

0.31007

44.50200

0.02247

13.79864

0.07247 127.14024

9.21397

25

3.38635

0.29530

47.72710

0.02095

14.09394

0.07095 134.22751

9.52377

26

3.55567

0.28124

51.11345

0.01956

14.37519

0.06956 141.25852

9.82655

27

3.73346

0.26785

54.66913

0.01829

14.64303

0.06829 148.22258

10.12240

28

3.92013

0.25509

58.40258

0.01712

14.89813

0.06712 155.11011

10.41138

29

4.11614

0.24295

62.32271

0.01605

15.14107

0.06605 161.91261

10.69360

30

4.32194

0.23138

66.43885

0.01505

15.37245

0.06505 168.62255

10.96914

36 40

5.79182 7.03999

0.17266 0.14205

95.83632 120.79977

0.01043 8.2782E−03

16.54685 17.15909

0.06043 206.62370 0.05828 229.54518

12.48719 13.37747

48

10.40127

0.09614

188.02539

5.3184E−03

18.07716

0.05532 269.24673

14.89431

50

11.46740

0.08720

209.34800

4.7767E−03

18.25593

0.05478 277.91478

15.22326

52

12.64281

0.07910

232.85617

4.2945E−03

18.41807

0.05429 286.10125

15.53372

55

14.63563

0.06833

272.71262

3.6669E−03

18.63347

0.05367 297.51040

15.96645

Single Sums

n

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60 72

18.67919 33.54513

0.05354 0.02981

353.58372 650.90268

2.8282E−03 1.5363E−03

18.92929 19.40379

0.05283 314.34316 0.05154 345.14853

16.60618 17.78769

75

38.83269

0.02575

756.65372

1.3216E−03

19.48497

0.05132 351.07215

18.01759

80

49.56144

0.02018

971.22882

1.0296E−03

19.59646

0.05103 359.64605

18.35260

84

60.24224

0.01660

1184.84483

8.4399E−04

19.66801

0.05084 365.47273

18.58209

90

80.73037

0.01239

1594.60730

6.2711E−04

19.75226

0.05063 372.74879

18.87120

96

108.18641

9.2433E−03 2143.72821

4.6648E−04

19.81513

0.05047 378.55553

19.10436

100 131.50126

7.6045E−03 2610.02516

3.8314E−04

19.84791

0.05038 381.74922

19.23372

108 194.28725

5.1470E−03 3865.74499

2.5868E−04

19.89706

0.05026 386.82363

19.44125

120 348.91199 132 626.59580

2.8661E−03 6958.23971 1.5959E−03 1.2512E+04

1.4371E−04 7.9924E−05

19.94268 19.96808

0.05014 391.97505 0.05008 395.14839

19.65509 19.78900

144 1125.27603

8.8867E−04 2.2486E+04

4.4473E−05

19.98223

0.05004 397.08516

19.87192

180 6517.39184

1.5344E−04 1.3033E+05

7.6730E−06

19.99693

0.05001 399.38626

19.97238

240 1.2174E+05

8.2143E−06 2.4348E+06

4.1072E−07

19.99984

0.05000 399.95729

19.99803

360 4.2476E+07

2.3542E−08 8.4953E+08

1.1771E−09

20.00000

0.05000 399.99982

19.99999

480 1.4821E+10

6.7474E−11 2.9641E+11

3.3737E−12

20.00000

0.05000 400.00000

20.00000

600 5.1711E+12

1.9338E−13 1.0342E+14

9.6692E−15

20.00000

0.05000 400.00000

20.00000

6.00% Time Value of Money Factors Discrete Compounding 6.00%

TABLE A-A-12 Time Value of Money Factors Discrete Compounding Single Sums

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G n (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n) 1 1.06000 0.94340 1.00000 1.00000 0.94340 1.06000 0.00000 0.00000 2

1.12360

0.89000

2.06000

0.48544

1.83339

0.54544

0.89000

0.48544

3

1.19102

0.83962

3.18360

0.31411

2.67301

0.37411

2.56924

0.96118

4

1.26248

0.79209

4.37462

0.22859

3.46511

0.28859

4.94552

1.42723

5

1.33823

0.74726

5.63709

0.17740

4.21236

0.23740

7.93455

1.88363

6

1.41852

0.70496

6.97532

0.14336

4.91732

0.20336

11.45935

2.33040

7

1.50363

0.66506

8.39384

0.11914

5.58238

0.17914

15.44969

2.76758

8

1.59385

0.62741

9.89747

0.10104

6.20979

0.16104

19.84158

3.19521

9 10

1.68948 1.79085

0.59190 0.55839

11.49132 13.18079

0.08702 0.07587

6.80169 7.36009

0.14702 0.13587

24.57677 29.60232

3.61333 4.02201

11

1.89830

0.52679

14.97164

0.06679

7.88687

0.12679

34.87020

4.42129

12

2.01220

0.49697

16.86994

0.05928

8.38384

0.11928

40.33686

4.81126

13

2.13293

0.46884

18.88214

0.05296

8.85268

0.11296

45.96293

5.19198

14

2.26090

0.44230

21.01507

0.04758

9.29498

0.10758

51.71284

5.56352

15

2.39656

0.41727

23.27597

0.04296

9.71225

0.10296

57.55455

5.92598

16

2.54035

0.39365

25.67253

0.03895

10.10590

0.09895

63.45925

6.27943

17

2.69277

0.37136

28.21288

0.03544

10.47726

0.09544

69.40108

6.62397

18

2.85434

0.35034

30.90565

0.03236

10.82760

0.09236

75.35692

6.95970

19 20

3.02560 3.20714

0.33051 0.31180

33.75999 36.78559

0.02962 0.02718

11.15812 11.46992

0.08962 0.08718

81.30615 87.23044

7.28673 7.60515

21

3.39956

0.29416

39.99273

0.02500

11.76408

0.08500

93.11355

7.91508

22

3.60354

0.27751

43.39229

0.02305

12.04158

0.08305

98.94116

8.21662

23

3.81975

0.26180

46.99583

0.02128

12.30338

0.08128 104.70070

8.50991

24

4.04893

0.24698

50.81558

0.01968

12.55036

0.07968 110.38121

8.79506

25

4.29187

0.23300

54.86451

0.01823

12.78336

0.07823 115.97317

9.07220

26

4.54938

0.21981

59.15638

0.01690

13.00317

0.07690 121.46842

9.34145

27

4.82235

0.20737

63.70577

0.01570

13.21053

0.07570 126.85999

9.60294

28 29

5.11169 5.41839

0.19563 0.18456

68.52811 73.63980

0.01459 0.01358

13.40616 13.59072

0.07459 132.14200 0.07358 137.30959

9.85681 10.10319

30

5.74349

0.17411

79.05819

0.01265

13.76483

0.07265 142.35879

10.34221

36

8.14725

0.12274

119.12087

8.3948E−03

14.62099

0.06839 170.03866

11.62977

40

10.28572

0.09722

154.76197

6.4615E−03

15.04630

0.06646 185.95682

12.35898

48

16.39387

0.06100

256.56453

3.8977E−03

15.65003

0.06390 212.03505

13.54854

50

18.42015

0.05429

290.33590

3.4443E−03

15.76186

0.06344 217.45738

13.79643

52

20.69689

0.04832

328.28142

3.0462E−03

15.86139

0.06305 222.48229

14.02666

55

24.65032

0.04057

394.17203

2.5370E−03

15.99054

0.06254 229.32225

14.34112

Single Sums

n

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60

32.98769

0.03031

533.12818

1.8757E−03

16.16143

0.06188 239.04279

14.79095

72

66.37772

0.01507

1089.62859

9.1774E−04

16.41558

0.06092 255.51462

15.56537

75

79.05692

0.01265

1300.94868

7.6867E−04

16.45585

0.06077 258.45274

15.70583

80

105.79599

9.4522E−03 1746.59989

5.7254E−04

16.50913

0.06057 262.54931

15.90328

84

133.56500

7.4870E−03 2209.41674

4.5261E−04

16.54188

0.06045 265.21627

16.03302

90

189.46451

5.2780E−03 3141.07519

3.1836E−04

16.57870

0.06032 268.39461

16.18912

96 268.75903 100 339.30208

3.7208E−03 4462.65050 2.9472E−03 5638.36806

2.2408E−04 1.7736E−04

16.60465 16.61755

0.06022 270.79093 0.06018 272.04706

16.30814 16.37107

108 540.79597

1.8491E−03 8996.59954

1.1115E−04

16.63585

0.06011 273.93570

16.46659

120 1088.18775

9.1896E−04 1.8120E+04

5.5188E−05

16.65135

0.06006 275.68459

16.55629

132 2189.64755

4.5669E−04 3.6477E+04

2.7414E−05

16.65906

0.06003 276.64619

16.60636

144 4406.00107

2.2696E−04 7.3417E+04

1.3621E−05

16.66288

0.06001 277.17002

16.63398

180 3.5897E+04

2.7858E−05 5.9826E+05

1.6715E−06

16.66620

0.06000 277.68647

16.66165

240 1.1842E+06

8.4449E−07 1.9736E+07

5.0669E−08

16.66665

0.06000 277.77417

16.66646

360 1.2886E+09

7.7605E−10 2.1476E+10

4.6563E−11

16.66667

0.06000 277.77777

16.66667

480 1.4022E+12

7.1316E−13 2.3370E+13

4.2789E−14

16.66667

0.06000 277.77778

16.66667

600 1.5259E+15

6.5536E−16 2.5431E+16

3.9322E−17

16.66667

0.06000 277.77778

16.66667

7.00% Time Value of Money Factors Discrete Compounding 7.00%

TABLE A-A-13 Time Value of Money Factors Discrete Compounding Single Sums

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

n 1

1.07000

0.93458

1.00000

1.00000

0.93458

1.07000

0.00000

0.00000

2

1.14490

0.87344

2.07000

0.48309

1.80802

0.55309

0.87344

0.48309

3

1.22504

0.81630

3.21490

0.31105

2.62432

0.38105

2.50603

0.95493

4

1.31080

0.76290

4.43994

0.22523

3.38721

0.29523

4.79472

1.41554

5

1.40255

0.71299

5.75074

0.17389

4.10020

0.24389

7.64666

1.86495

6 7

1.50073 1.60578

0.66634 0.62275

7.15329 8.65402

0.13980 0.11555

4.76654 5.38929

0.20980 0.18555

10.97838 14.71487

2.30322 2.73039

8 9

1.71819 1.83846

0.58201 0.54393

10.25980 11.97799

0.09747 0.08349

5.97130 6.51523

0.16747 0.15349

18.78894 23.14041

3.14654 3.55174

10 11 12

1.96715 2.10485 2.25219

0.50835 0.47509 0.44401

13.81645 15.78360 17.88845

0.07238 0.06336 0.05590

7.02358 7.49867 7.94269

0.14238 0.13336 0.12590

27.71555 32.46648 37.35061

3.94607 4.32963 4.70252

13 14

2.40985 2.57853

0.41496 0.38782

20.14064 22.55049

0.04965 0.04434

8.35765 8.74547

0.11965 0.11434

42.33018 47.37181

5.06484 5.41673

15 16

2.75903 2.95216

0.36245 0.33873

25.12902 27.88805

0.03979 0.03586

9.10791 9.44665

0.10979 0.10586

52.44605 57.52707

5.75829 6.08968

17 18

3.15882 3.37993

0.31657 0.29586

30.84022 33.99903

0.03243 0.02941

9.76322 10.05909

0.10243 0.09941

62.59226 67.62195

6.41102 6.72247

19 20 21

3.61653 3.86968 4.14056

0.27651 0.25842 0.24151

37.37896 40.99549 44.86518

0.02675 0.02439 0.02229

10.33560 10.59401 10.83553

0.09675 0.09439 0.09229

72.59910 77.50906 82.33932

7.02418 7.31631 7.59901

22 23

4.43040 4.74053

0.22571 0.21095

49.00574 53.43614

0.02041 0.01871

11.06124 11.27219

0.09041 0.08871

87.07930 91.72013

7.87247 8.13685

24 25

5.07237 5.42743

0.19715 0.18425

58.17667 63.24904

0.01719 0.01581

11.46933 11.65358

0.08719 96.25450 0.08581 100.67648

8.39234 8.63910

26 27

5.80735 6.21387

0.17220 0.16093

68.67647 74.48382

0.01456 0.01343

11.82578 11.98671

0.08456 104.98137 0.08343 109.16556

8.87733 9.10722

28 29 30

6.64884 7.11426 7.61226

0.15040 0.14056 0.13137

80.69769 87.34653 94.46079

0.01239 0.01145 0.01059

12.13711 12.27767 12.40904

0.08239 113.22642 0.08145 117.16218 0.08059 120.97182

9.32894 9.54270 9.74868

36 40

11.42394 14.97446

0.08754 0.06678

148.91346 199.63511

6.7153E−03 5.0091E−03

13.03521 13.33171

0.07672 141.19902 0.07501 152.29277

10.83213 11.42335

48 50

25.72891 29.45703

0.03887 0.03395

353.27009 406.52893

2.8307E−03 2.4598E−03

13.73047 13.80075

0.07283 169.49812 0.07246 172.90512

12.34467 12.52868

52 55

33.72535 41.31500

0.02965 0.02420

467.50497 575.92859

2.1390E−03 1.7363E−03

13.86212 13.93994

0.07214 176.00368 0.07174 180.12433

12.69673 12.92146

Single Sums

n

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60

57.94643

72 75

0.01726

813.52038

1.2292E−03

14.03918

0.07123 185.76774

13.23209

130.50646 159.87602

7.6625E−03 1850.09222 6.2548E−03 2269.65742

5.4051E−04 4.4060E−04

14.17625 14.19636

0.07054 194.63648 0.07044 196.10351

13.72976 13.81365

80 84

224.23439 293.92554

4.4596E−03 3189.06268 3.4022E−03 4184.65058

3.1357E−04 2.3897E−04

14.22201 14.23711

0.07031 198.07480 0.07024 199.30463

13.92735 13.99895

90 96

441.10298 661.97663

2.2670E−03 6287.18543 1.5106E−03 9442.52329

1.5905E−04 1.0590E−04

14.25333 14.26413

0.07016 200.70420 0.07011 201.70162

14.08122 14.14047

100 867.71633 108 1490.89820 120 3357.78838

1.1525E−03 1.2382E+04 6.7074E−04 2.1284E+04 2.9782E−04 4.7954E+04

8.0765E−05 4.6983E−05 2.0853E−05

14.26925 14.27613 14.28146

0.07008 202.20008 0.07005 202.90990 0.07002 203.51031

14.17034 14.21323 14.24997

132 7562.38275 144 1.7032E+04

1.3223E−04 1.0802E+05 5.8713E−05 2.4330E+05

9.2576E−06 4.1102E−06

14.28383 14.28488

0.07001 203.80529 0.07000 203.94887

14.26826 14.27726

180 1.9457E+05 240 1.1275E+07

5.1395E−06 2.7796E+06 8.8694E−08 1.6107E+08

3.5977E−07 6.2086E−09

14.28564 14.28571

0.07000 204.06737 0.07000 204.08131

14.28479 14.28569

360 3.7858E+10 480 1.2712E+14

2.6414E−11 5.4083E+11 7.8666E−15 1.8160E+15

1.8490E−12 5.5066E−16

14.28571 14.28571

0.07000 204.08163 0.07000 204.08163

14.28571 14.28571

600 4.2684E+17

2.3428E−18 6.0977E+18

1.6400E−19

14.28571

0.07000 204.08163

14.28571

8.00% Time Value of Money Factors Discrete Compounding 8.00%

TABLE A-A-14 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

n 1 2

1.08000 1.16640

0.92593 0.85734

1.00000 2.08000

1.00000 0.48077

0.92593 1.78326

1.08000 0.56077

0.00000 0.85734

0.00000 0.48077

3 4

1.25971 1.36049

0.79383 0.73503

3.24640 4.50611

0.30803 0.22192

2.57710 3.31213

0.38803 0.30192

2.44500 4.65009

0.94874 1.40396

5 6

1.46933 1.58687

0.68058 0.63017

5.86660 7.33593

0.17046 0.13632

3.99271 4.62288

0.25046 0.21632

7.37243 10.52327

1.84647 2.27635

7 8 9

1.71382 1.85093 1.99900

0.58349 0.54027 0.50025

8.92280 10.63663 12.48756

0.11207 0.09401 0.08008

5.20637 5.74664 6.24689

0.19207 0.17401 0.16008

14.02422 17.80610 21.80809

2.69366 3.09852 3.49103

10 11

2.15892 2.33164

0.46319 0.42888

14.48656 16.64549

0.06903 0.06008

6.71008 7.13896

0.14903 0.14008

25.97683 30.26566

3.87131 4.23950

12 13

2.51817 2.71962

0.39711 0.36770

18.97713 21.49530

0.05270 0.04652

7.53608 7.90378

0.13270 0.12652

34.63391 39.04629

4.59575 4.94021

14 15

2.93719 3.17217

0.34046 0.31524

24.21492 27.15211

0.04130 0.03683

8.24424 8.55948

0.12130 0.11683

43.47228 47.88566

5.27305 5.59446

16 17 18

3.42594 3.70002 3.99602

0.29189 0.27027 0.25025

30.32428 33.75023 37.45024

0.03298 0.02963 0.02670

8.85137 9.12164 9.37189

0.11298 0.10963 0.10670

52.26402 56.58832 60.84256

5.90463 6.20375 6.49203

19 20

4.31570 4.66096

0.23171 0.21455

41.44626 45.76196

0.02413 0.02185

9.60360 9.81815

0.10413 0.10185

65.01337 69.08979

6.76969 7.03695

21 22

5.03383 5.43654

0.19866 0.18394

50.42292 55.45676

0.01983 0.01803

10.01680 10.20074

0.09983 0.09803

73.06291 76.92566

7.29403 7.54118

23 24

5.87146 6.34118

0.17032 0.15770

60.89330 66.76476

0.01642 0.01498

10.37106 10.52876

0.09642 0.09498

80.67259 84.29968

7.77863 8.00661

25 26 27

6.84848 7.39635 7.98806

0.14602 0.13520 0.12519

73.10594 79.95442 87.35077

0.01368 0.01251 0.01145

10.67478 10.80998 10.93516

0.09368 0.09251 0.09145

87.80411 91.18415 94.43901

8.22538 8.43518 8.63627

28 29

8.62711 9.31727

0.11591 0.10733

95.33883 103.96594

0.01049 9.6185E−03

11.05108 11.15841

0.09049 97.56868 0.08962 100.57385

8.82888 9.01328

30 36

10.06266 15.96817

0.09938 0.06262

113.28321 187.10215

8.8274E−03 5.3447E−03

11.25778 11.71719

0.08883 103.45579 0.08534 118.28385

9.18971 10.09490

40 48

21.72452 40.21057

0.04603 0.02487

259.05652 490.13216

3.8602E−03 2.0403E−03

11.92461 12.18914

0.08386 126.04220 0.08204 137.44276

10.56992 11.27584

50 52

46.90161 54.70604

0.02132 0.01828

573.77016 671.32551

1.7429E−03 1.4896E−03

12.23348 12.27151

0.08174 139.59279 0.08149 141.51214

11.41071 11.53177

55

68.91386

0.01451

848.92320

1.1780E−03

12.31861

0.08118 144.00645

11.69015

n

Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60 72

101.25706 254.98251

9.8759E−03 1253.21330 3.9218E−03 3174.78140

7.9795E−04 3.1498E−04

12.37655 12.45098

0.08080 147.30001 0.08031 152.10756

11.90154 12.21652

75 80

321.20453 471.95483

3.1133E−03 4002.55662 2.1188E−03 5886.93543

2.4984E−04 1.6987E−04

12.46108 12.47351

0.08025 152.84485 0.08017 153.80008

12.26577 12.33013

84 642.08934 90 1018.91509 96 1616.89019

1.5574E−03 8013.61677 9.8144E−04 1.2724E+04 6.1847E−04 2.0199E+04

1.2479E−04 7.8592E−05 4.9508E−05

12.48053 12.48773 12.49227

0.08012 154.37137 0.08008 154.99254 0.08005 155.41120

12.36897 12.41158 12.44059

100 2199.76126 108 4071.60456

4.5459E−04 2.7485E+04 2.4560E−04 5.0883E+04

3.6384E−05 1.9653E−05

12.49432 12.49693

0.08004 155.61073 0.08002 155.88006

12.45452 12.47347

120 1.0253E+04 132 2.5819E+04

9.7532E−05 1.2815E+05 3.8731E−05 3.2272E+05

7.8034E−06 3.0986E−06

12.49878 12.49952

0.08001 156.08846 0.08000 156.18004

12.48829 12.49489

144 6.5016E+04 180 1.0382E+06

1.5381E−05 8.1269E+05 9.6322E−07 1.2977E+07

1.2305E−06 7.7057E−08

12.49981 12.49999

0.08000 156.21991 0.08000 156.24768

12.49779 12.49983

240 1.0512E+08 360 1.0778E+12 480 1.1051E+16

9.5126E−09 1.3140E+09 9.2779E−13 1.3473E+13 9.0489E−17 1.3814E+17

7.6101E−10 7.4223E−14 7.2391E−18

12.50000 12.50000 12.50000

0.08000 156.24997 0.08000 156.25000 0.08000 156.25000

12.50000 12.50000 12.50000

600 1.1331E+20

8.8257E−21 1.4163E+21

7.0605E−22

12.50000

0.08000 156.25000

12.50000

9.00% Time Value of Money Factors Discrete Compounding 9.00%

TABLE A-A-15 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G n (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n) 1 1.09000 0.91743 1.00000 1.00000 0.91743 1.09000 0.00000 0.00000 2 3

1.18810 1.29503

0.84168 0.77218

2.09000 3.27810

0.47847 0.30505

1.75911 2.53129

0.56847 0.39505

0.84168 2.38605

0.47847 0.94262

4 5 6

1.41158 1.53862 1.67710

0.70843 0.64993 0.59627

4.57313 5.98471 7.52333

0.21867 0.16709 0.13292

3.23972 3.88965 4.48592

0.30867 0.25709 0.22292

4.51132 7.11105 10.09238

1.39250 1.82820 2.24979

7 8

1.82804 1.99256

0.54703 0.50187

9.20043 11.02847

0.10869 0.09067

5.03295 5.53482

0.19869 0.18067

13.37459 16.88765

2.65740 3.05117

9 10

2.17189 2.36736

0.46043 0.42241

13.02104 15.19293

0.07680 0.06582

5.99525 6.41766

0.16680 0.15582

20.57108 24.37277

3.43123 3.79777

11 12

2.58043 2.81266

0.38753 0.35553

17.56029 20.14072

0.05695 0.04965

6.80519 7.16073

0.14695 0.13965

28.24810 32.15898

4.15096 4.49102

13 14 15

3.06580 3.34173 3.64248

0.32618 0.29925 0.27454

22.95338 26.01919 29.36092

0.04357 0.03843 0.03406

7.48690 7.78615 8.06069

0.13357 0.12843 0.12406

36.07313 39.96333 43.80686

4.81816 5.13262 5.43463

16 17

3.97031 4.32763

0.25187 0.23107

33.00340 36.97370

0.03030 0.02705

8.31256 8.54363

0.12030 0.11705

47.58491 51.28208

5.72446 6.00238

18 19

4.71712 5.14166

0.21199 0.19449

41.30134 46.01846

0.02421 0.02173

8.75563 8.95011

0.11421 0.11173

54.88598 58.38679

6.26865 6.52358

20 21

5.60441 6.10881

0.17843 0.16370

51.16012 56.76453

0.01955 0.01762

9.12855 9.29224

0.10955 0.10762

61.77698 65.05094

6.76745 7.00056

22 23 24

6.65860 7.25787 7.91108

0.15018 0.13778 0.12640

62.87334 69.53194 76.78981

0.01590 0.01438 0.01302

9.44243 9.58021 9.70661

0.10590 0.10438 0.10302

68.20475 71.23594 74.14326

7.22322 7.43574 7.63843

25 26

8.62308 9.39916

0.11597 0.10639

84.70090 93.32398

0.01181 0.01072

9.82258 9.92897

0.10181 0.10072

76.92649 79.58630

7.83160 8.01556

27 28

10.24508 11.16714

0.09761 0.08955

102.72313 112.96822

9.7349E−03 8.8520E−03

10.02658 10.11613

0.09973 0.09885

82.12410 84.54191

8.19064 8.35714

29 30

12.17218 13.26768

0.08215 0.07537

124.13536 136.30754

8.0557E−03 7.3364E−03

10.19828 10.27365

0.09806 0.09734

86.84224 89.02800

8.51538 8.66566

36 40 48

22.25123 31.40942 62.58524

0.04494 0.03184 0.01598

236.12472 337.88245 684.28041

4.2350E−03 2.9596E−03 1.4614E−03

10.61176 10.75736 10.93358

0.09424 99.93194 0.09296 105.37619 0.09146 112.96246

9.41709 9.79573 10.33170

50 52

74.35752 88.34417

0.01345 0.01132

815.08356 970.49077

1.2269E−03 1.0304E−03

10.96168 10.98534

0.09123 114.32507 0.09103 115.51926

10.42952 10.51577

55

114.40826

8.7406E−03 1260.09180

7.9359E−04

11.01399

0.09079 117.03621

10.62614

n

Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60

176.03129

5.6808E−03 1944.79213

5.1419E−04

11.04799

0.09051 118.96825

10.76832

72 75 80

495.11702 641.19089 986.55167

2.0197E−03 5490.18906 1.5596E−03 7113.23215 1.0136E−03 1.0951E+04

1.8214E−04 1.4058E−04 9.1319E−05

11.08867 11.09378 11.09985

0.09018 121.59166 0.09014 121.96458 0.09009 122.43064

10.96540 10.99396 11.02994

84 1392.59819 90 2335.52658

7.1808E−04 1.5462E+04 4.2817E−04 2.5939E+04

6.4674E−05 3.8552E−05

11.10313 11.10635

0.09006 122.69793 0.09004 122.97576

11.05075 11.07256

96 3916.91189 100 5529.04079

2.5530E−04 4.3510E+04 1.8086E−04 6.1423E+04

2.2983E−05 1.6281E−05

11.10827 11.10910

0.09002 123.15295 0.09002 123.23350

11.08660 11.09302

108 1.1017E+04 120 3.0987E+04

9.0769E−05 1.2240E+05 3.2272E−05 3.4429E+05

8.1700E−06 2.9045E−06

11.11010 11.11075

0.09001 123.33666 0.09000 123.40978

11.10131 11.10724

132 8.7156E+04 144 2.4514E+05 180 5.4547E+06

1.1474E−05 9.6839E+05 4.0793E−06 2.7238E+06 1.8333E−07 6.0608E+07

1.0326E−06 3.6714E−07 1.6500E−08

11.11098 11.11107 11.11111

0.09000 123.43855 0.09000 123.44976 0.09000 123.45640

11.10960 11.11052 11.11108

240 9.6020E+08 360 2.9754E+13

1.0415E−09 1.0669E+10 3.3609E−14 3.3060E+14

9.3731E−11 3.0248E−15

11.11111 11.11111

0.09000 123.45679 0.09000 123.45679

11.11111 11.11111

480 9.2197E+17 600 2.8569E+22

1.0846E−18 1.0244E+19 3.5003E−23 3.1744E+23

9.7617E−20 3.1502E−24

11.11111 11.11111

0.09000 123.45679 0.09000 123.45679

11.11111 11.11111

10.00% Time Value of Money Factors Discrete Compounding 10.00%

TABLE A-A-16 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G n (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n) 1 1.10000 0.90909 1.00000 1.00000 0.90909 1.10000 0.00000 0.00000 2 3

1.21000 1.33100

0.82645 0.75131

2.10000 3.31000

0.47619 0.30211

1.73554 2.48685

0.57619 0.40211

0.82645 2.32908

0.47619 0.93656

4 5

1.46410 1.61051

0.68301 0.62092

4.64100 6.10510

0.21547 0.16380

3.16987 3.79079

0.31547 0.26380

4.37812 6.86180

1.38117 1.81013

6 7

1.77156 1.94872

0.56447 0.51316

7.71561 9.48717

0.12961 0.10541

4.35526 4.86842

0.22961 0.20541

9.68417 12.76312

2.22356 2.62162

8 9

2.14359 2.35795

0.46651 0.42410

11.43589 13.57948

0.08744 0.07364

5.33493 5.75902

0.18744 0.17364

16.02867 19.42145

3.00448 3.37235

10 11 12

2.59374 2.85312 3.13843

0.38554 0.35049 0.31863

15.93742 18.53117 21.38428

0.06275 0.05396 0.04676

6.14457 6.49506 6.81369

0.16275 0.15396 0.14676

22.89134 26.39628 29.90122

3.72546 4.06405 4.38840

13 14

3.45227 3.79750

0.28966 0.26333

24.52271 27.97498

0.04078 0.03575

7.10336 7.36669

0.14078 0.13575

33.37719 36.80050

4.69879 4.99553

15 16

4.17725 4.59497

0.23939 0.21763

31.77248 35.94973

0.03147 0.02782

7.60608 7.82371

0.13147 0.12782

40.15199 43.41642

5.27893 5.54934

17 18

5.05447 5.55992

0.19784 0.17986

40.54470 45.59917

0.02466 0.02193

8.02155 8.20141

0.12466 0.12193

46.58194 49.63954

5.80710 6.05256

19 20 21

6.11591 6.72750 7.40025

0.16351 0.14864 0.13513

51.15909 57.27500 64.00250

0.01955 0.01746 0.01562

8.36492 8.51356 8.64869

0.11955 0.11746 0.11562

52.58268 55.40691 58.10952

6.28610 6.50808 6.71888

22 23

8.14027 8.95430

0.12285 0.11168

71.40275 79.54302

0.01401 0.01257

8.77154 8.88322

0.11401 0.11257

60.68929 63.14621

6.91889 7.10848

24 25

9.84973 10.83471

0.10153 0.09230

88.49733 98.34706

0.01130 0.01017

8.98474 9.07704

0.11130 0.11017

65.48130 67.69640

7.28805 7.45798

26 27

11.91818 13.10999

0.08391 0.07628

109.18177 121.09994

9.1590E−03 8.2576E−03

9.16095 9.23722

0.10916 0.10826

69.79404 71.77726

7.61865 7.77044

28 29 30

14.42099 15.86309 17.44940

0.06934 0.06304 0.05731

134.20994 148.63093 164.49402

7.4510E−03 6.7281E−03 6.0792E−03

9.30657 9.36961 9.42691

0.10745 0.10673 0.10608

73.64953 75.41463 77.07658

7.91372 8.04886 8.17623

36 40

30.91268 45.25926

0.03235 0.02209

299.12681 442.59256

3.3431E−03 2.2594E−03

9.67651 9.77905

0.10334 0.10226

85.11938 88.95254

8.79650 9.09623

48 50

97.01723 117.39085

0.01031 960.17234 8.5186E−03 1163.90853

1.0415E−03 8.5917E−04

9.89693 9.91481

0.10104 0.10086

94.02168 94.88887

9.50009 9.57041

52 55

142.04293 189.05914

7.0401E−03 1410.42932 5.2894E−03 1880.59142

7.0900E−04 5.3175E−04

9.92960 9.94711

0.10071 0.10053

95.63512 96.56192

9.63132 9.70754

Single Sums

n 60

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n) 304.48164

3.2843E−03 3034.81640

3.2951E−04

9.96716

0.10033

97.70101

9.80229

72 955.59382 75 1271.89537

1.0465E−03 9545.93818 7.8623E−04 1.2709E+04

1.0476E−04 7.8685E−05

9.98954 9.99214

0.10010 0.10008

99.14189 99.33171

9.92458 9.94099

80 2048.40021 84 2999.06275

4.8819E−04 2.0474E+04 3.3344E−04 2.9981E+04

4.8842E−05 3.3355E−05

9.99512 9.99667

0.10005 0.10003

99.56063 99.68657

9.96093 9.97198

90 5313.02261 96 9412.34365

1.8822E−04 5.3120E+04 1.0624E−04 9.4113E+04

1.8825E−05 1.0625E−05

9.99812 9.99894

0.10002 0.10001

99.81178 99.88738

9.98306 9.98980

100 1.3781E+04 108 2.9540E+04 120 9.2709E+04

7.2566E−05 1.3780E+05 3.3852E−05 2.9539E+05 1.0786E−05 9.2708E+05

7.2571E−06 3.3854E−06 1.0787E−06

9.99927 9.99966 9.99989

0.10001 0.10000 0.10000

99.92018 99.96005 99.98598

9.99274 9.99634 9.99871

132 2.9096E+05 144 9.1316E+05

3.4369E−06 2.9096E+06 1.0951E−06 9.1316E+06

3.4369E−07 1.0951E−07

9.99997 9.99999

0.10000 0.10000

99.99512 99.99831

9.99955 9.99984

180 2.8228E+07 240 8.5950E+09

3.5426E−08 2.8228E+08 1.1635E−10 8.5950E+10

3.5426E−09 1.1635E−11

10.00000 10.00000

0.10000 99.99993 0.10000 100.00000

9.99999 10.00000

360 7.9683E+14 480 7.3874E+19

1.2550E−15 7.9683E+15 1.3537E−20 7.3874E+20

1.2550E−16 1.3537E−21

10.00000 10.00000

0.10000 100.00000 0.10000 100.00000

10.00000 10.00000

600 6.8487E+24

1.4601E−25 6.8487E+25

1.4601E−26

10.00000

0.10000 100.00000

10.00000

11.00% Time Value of Money Factors Discrete Compounding 11.00%

TABLE A-A-17 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

n 1 2

1.11000 1.23210

0.90090 0.81162

1.00000 2.11000

1.00000 0.47393

0.90090 1.71252

1.11000 0.58393

0.00000 0.81162

0.00000 0.47393

3 4

1.36763 1.51807

0.73119 0.65873

3.34210 4.70973

0.29921 0.21233

2.44371 3.10245

0.40921 0.32233

2.27401 4.25020

0.93055 1.36995

5 6 7

1.68506 1.87041 2.07616

0.59345 0.53464 0.48166

6.22780 7.91286 9.78327

0.16057 0.12638 0.10222

3.69590 4.23054 4.71220

0.27057 0.23638 0.21222

6.62400 9.29721 12.18716

1.79226 2.19764 2.58630

8 9

2.30454 2.55804

0.43393 0.39092

11.85943 14.16397

0.08432 0.07060

5.14612 5.53705

0.19432 0.18060

15.22464 18.35204

2.95847 3.31441

10 11

2.83942 3.15176

0.35218 0.31728

16.72201 19.56143

0.05980 0.05112

5.88923 6.20652

0.16980 0.16112

21.52170 24.69454

3.65442 3.97881

12 13

3.49845 3.88328

0.28584 0.25751

22.71319 26.21164

0.04403 0.03815

6.49236 6.74987

0.15403 0.14815

27.83878 30.92896

4.28793 4.58216

14 15

4.31044 4.78459

0.23199 0.20900

30.09492 34.40536

0.03323 0.02907

6.98187 7.19087

0.14323 0.13907

33.94489 36.87095

4.86187 5.12747

16 17 18

5.31089 5.89509 6.54355

0.18829 0.16963 0.15282

39.18995 44.50084 50.39594

0.02552 0.02247 0.01984

7.37916 7.54879 7.70162

0.13552 0.13247 0.12984

39.69533 42.40945 45.00743

5.37938 5.61804 5.84389

19 20

7.26334 8.06231

0.13768 0.12403

56.93949 64.20283

0.01756 0.01558

7.83929 7.96333

0.12756 0.12558

47.48563 49.84227

6.05739 6.25898

21 22

8.94917 9.93357

0.11174 0.10067

72.26514 81.21431

0.01384 0.01231

8.07507 8.17574

0.12384 0.12231

52.07712 54.19116

6.44912 6.62829

23 24

11.02627 12.23916

0.09069 0.08170

91.14788 102.17415

0.01097 9.7872E−03

8.26643 8.34814

0.12097 0.11979

56.18640 58.06561

6.79693 6.95552

25 26 27

13.58546 15.07986 16.73865

0.07361 0.06631 0.05974

114.41331 127.99877 143.07864

8.7402E−03 7.8126E−03 6.9892E−03

8.42174 8.48806 8.54780

0.11874 0.11781 0.11699

59.83220 61.49004 63.04334

7.10449 7.24430 7.37539

28 29

18.57990 20.62369

0.05382 0.04849

159.81729 178.39719

6.2571E−03 5.6055E−03

8.60162 8.65011

0.11626 0.11561

64.49652 65.85418

7.49818 7.61310

30 36

22.89230 42.81808

0.04368 0.02335

199.02088 380.16441

5.0246E−03 2.6304E−03

8.69379 8.87859

0.11502 0.11263

67.12098 73.07116

7.72056 8.23004

40 48

65.00087 149.79695

0.01538 581.82607 6.6757E−03 1352.69958

1.7187E−03 7.3926E−04

8.95105 9.03022

0.11172 0.11074

75.77886 79.17988

8.46592 8.76832

50 52 55

184.56483 227.40232 311.00247

5.4182E−03 1668.77115 4.3975E−03 2058.20294 3.2154E−03 2818.20424

5.9924E−04 4.8586E−04 3.5484E−04

9.04165 9.05093 9.06168

0.11060 0.11049 0.11035

79.73405 80.20238 80.77119

8.81853 8.86123 8.91349

n

Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60 524.05724 72 1833.38837

1.9082E−03 4755.06584 5.4544E−04 1.6658E+04

2.1030E−04 6.0031E−05

9.07356 9.08595

0.11021 0.11006

81.44610 82.24254

8.97620 9.05162

75 2507.39877 80 4225.11275 84 6414.01865

3.9882E−04 2.2785E+04 2.3668E−04 3.8401E+04 1.5591E−04 5.8300E+04

4.3888E−05 2.6041E−05 1.7153E−05

9.08728 9.08876 9.08949

0.11004 0.11003 0.11002

82.33975 82.45294 82.51269

9.06099 9.07197 9.07781

90 1.1997E+04 96 2.2439E+04

8.3355E−05 1.0905E+05 4.4565E−05 2.0398E+05

9.1698E−06 4.9024E−06

9.09015 9.09050

0.11001 0.11000

82.56954 82.60205

9.08341 9.08663

100 3.4064E+04 108 7.8502E+04

2.9356E−05 3.0967E+05 1.2738E−05 7.1365E+05

3.2293E−06 1.4013E−06

9.09064 9.09079

0.11000 0.11000

82.61551 82.63107

9.08797 9.08953

120 2.7464E+05 132 9.6080E+05

3.6412E−06 2.4967E+06 1.0408E−06 8.7345E+06

4.0053E−07 1.1449E−07

9.09088 9.09090

0.11000 0.11000

82.64035 82.64329

9.09047 9.09077

144 3.3613E+06 180 1.4392E+08

2.9750E−07 3.0557E+07 6.9481E−09 1.3084E+09

3.2725E−08 7.6429E−10

9.09091 9.09091

0.11000 0.11000

82.64421 82.64462

9.09087 9.09091

240 7.5425E+10 360 2.0714E+16 480 5.6889E+21

1.3258E−11 6.8568E+11 4.8276E−17 1.8831E+17 1.7578E−22 5.1717E+22

1.4584E−12 5.3103E−18 1.9336E−23

9.09091 9.09091 9.09091

0.11000 0.11000 0.11000

82.64463 82.64463 82.64463

9.09091 9.09091 9.09091

600 1.5624E+27

6.4005E−28 1.4203E+28

7.0405E−29

9.09091

0.11000

82.64463

9.09091

12.00% Time Value of Money Factors Discrete Compounding 12.00%

TABLE A-A-18 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G n (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n) 1 1.12000 0.89286 1.00000 1.00000 0.89286 1.12000 0.00000 0.00000 2 3 4

1.25440 1.40493 1.57352

0.79719 0.71178 0.63552

2.12000 3.37440 4.77933

0.47170 0.29635 0.20923

1.69005 2.40183 3.03735

0.59170 0.41635 0.32923

0.79719 2.22075 4.12731

0.47170 0.92461 1.35885

5 6

1.76234 1.97382

0.56743 0.50663

6.35285 8.11519

0.15741 0.12323

3.60478 4.11141

0.27741 0.24323

6.39702 8.93017

1.77459 2.17205

7 8

2.21068 2.47596

0.45235 0.40388

10.08901 12.29969

0.09912 0.08130

4.56376 4.96764

0.21912 0.20130

11.64427 14.47145

2.55147 2.91314

9 10

2.77308 3.10585

0.36061 0.32197

14.77566 17.54874

0.06768 0.05698

5.32825 5.65022

0.18768 0.17698

17.35633 20.25409

3.25742 3.58465

11 12

3.47855 3.89598

0.28748 0.25668

20.65458 24.13313

0.04842 0.04144

5.93770 6.19437

0.16842 0.16144

23.12885 25.95228

3.89525 4.18965

13 14 15

4.36349 4.88711 5.47357

0.22917 0.20462 0.18270

28.02911 32.39260 37.27971

0.03568 0.03087 0.02682

6.42355 6.62817 6.81086

0.15568 0.15087 0.14682

28.70237 31.36242 33.92017

4.46830 4.73169 4.98030

16 17

6.13039 6.86604

0.16312 0.14564

42.75328 48.88367

0.02339 0.02046

6.97399 7.11963

0.14339 0.14046

36.36700 38.69731

5.21466 5.43530

18 19

7.68997 8.61276

0.13004 0.11611

55.74971 63.43968

0.01794 0.01576

7.24967 7.36578

0.13794 0.13576

40.90798 42.99790

5.64274 5.83752

20 21

9.64629 10.80385

0.10367 0.09256

72.05244 81.69874

0.01388 0.01224

7.46944 7.56200

0.13388 0.13224

44.96757 46.81876

6.02020 6.19132

22 23 24

12.10031 13.55235 15.17863

0.08264 0.07379 0.06588

92.50258 104.60289 118.15524

0.01081 9.5600E−03 8.4634E−03

7.64465 7.71843 7.78432

0.13081 0.12956 0.12846

48.55425 50.17759 51.69288

6.35141 6.50101 6.64064

25 26

17.00006 19.04007

0.05882 0.05252

133.33387 150.33393

7.5000E−03 6.6519E−03

7.84314 7.89566

0.12750 0.12665

53.10464 54.41766

6.77084 6.89210

27 28

21.32488 23.88387

0.04689 0.04187

169.37401 190.69889

5.9041E−03 5.2439E−03

7.94255 7.98442

0.12590 0.12524

55.63689 56.76736

7.00491 7.10976

29 30

26.74993 29.95992

0.03738 0.03338

214.58275 241.33268

4.6602E−03 4.1437E−03

8.02181 8.05518

0.12466 0.12414

57.81409 58.78205

7.20712 7.29742

36 40 48

59.13557 93.05097 230.39078

0.01691 484.46312 0.01075 767.09142 4.3405E−03 1911.58980

2.0641E−03 1.3036E−03 5.2312E−04

8.19241 8.24378 8.29716

0.12206 0.12130 0.12052

63.19703 65.11587 67.40684

7.71409 7.89879 8.12408

50 52

289.00219 362.52435

3.4602E−03 2400.01825 2.7584E−03 3012.70289

4.1666E−04 3.3193E−04

8.30450 8.31035

0.12042 0.12033

67.76241 68.05756

8.15972 8.18950

55

509.32061

1.9634E−03 4236.00505

2.3607E−04

8.31697

0.12024

68.40821

8.22513

n

Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60 897.59693 72 3497.01610

1.1141E−03 7471.64111 2.8596E−04 2.9133E+04

1.3384E−04 3.4325E−05

8.32405 8.33095

0.12013 0.12003

68.81003 69.25301

8.26641 8.31274

75 4913.05584 80 8658.48310

2.0354E−04 4.0934E+04 1.1549E−04 7.2146E+04

2.4430E−05 1.3861E−05

8.33164 8.33237

0.12002 0.12001

69.30310 69.35943

8.31806 8.32409

84 1.3624E+04 90 2.6892E+04

7.3398E−05 1.1353E+05 3.7186E−05 2.2409E+05

8.8084E−06 4.4625E−06

8.33272 8.33302

0.12001 0.12000

69.38797 69.41397

8.32717 8.32999

96 5.3080E+04 100 8.3522E+04

1.8840E−05 4.4232E+05 1.1973E−05 6.9601E+05

2.2608E−06 1.4368E−06

8.33318 8.33323

0.12000 0.12000

69.42806 69.43364

8.33152 8.33214

108 2.0680E+05 120 8.0568E+05

4.8356E−06 1.7233E+06 1.2412E−06 6.7140E+06

5.8028E−07 1.4894E−07

8.33329 8.33332

0.12000 0.12000

69.43976 69.44312

8.33281 8.33318

132 3.1389E+06 144 1.2229E+07 180 7.2318E+08

3.1858E−07 2.6158E+07 8.1772E−08 1.0191E+08 1.3828E−09 6.0265E+09

3.8230E−08 9.8126E−09 1.6593E−10

8.33333 8.33333 8.33333

0.12000 0.12000 0.12000

69.44407 69.44434 69.44444

8.33329 8.33332 8.33333

240 6.4912E+11 360 5.2298E+17

1.5405E−12 5.4093E+12 1.9121E−18 4.3582E+18

1.8487E−13 2.2945E−19

8.33333 8.33333

0.12000 0.12000

69.44444 69.44444

8.33333 8.33333

480 4.2136E+23 600 3.3948E+29

2.3733E−24 3.5113E+24 2.9457E−30 2.8290E+30

2.8479E−25 3.5348E−31

8.33333 8.33333

0.12000 0.12000

69.44444 69.44444

8.33333 8.33333

15.00% Time Value of Money Factors Discrete Compounding 15.00%

TABLE A-A-19 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G n (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n) 1 1.15000 0.86957 1.00000 1.00000 0.86957 1.15000 0.00000 0.00000 2 3

1.32250 1.52088

0.75614 0.65752

2.15000 3.47250

0.46512 0.28798

1.62571 2.28323

0.61512 0.43798

0.75614 2.07118

0.46512 0.90713

4 5

1.74901 2.01136

0.57175 0.49718

4.99338 6.74238

0.20027 0.14832

2.85498 3.35216

0.35027 0.29832

3.78644 5.77514

1.32626 1.72281

6 7

2.31306 2.66002

0.43233 0.37594

8.75374 11.06680

0.11424 0.09036

3.78448 4.16042

0.26424 0.24036

7.93678 10.19240

2.09719 2.44985

8 9 10

3.05902 3.51788 4.04556

0.32690 0.28426 0.24718

13.72682 16.78584 20.30372

0.07285 0.05957 0.04925

4.48732 4.77158 5.01877

0.22285 0.20957 0.19925

12.48072 14.75481 16.97948

2.78133 3.09223 3.38320

11 12

4.65239 5.35025

0.21494 0.18691

24.34928 29.00167

0.04107 0.03448

5.23371 5.42062

0.19107 0.18448

19.12891 21.18489

3.65494 3.90820

13 14

6.15279 7.07571

0.16253 0.14133

34.35192 40.50471

0.02911 0.02469

5.58315 5.72448

0.17911 0.17469

23.13522 24.97250

4.14376 4.36241

15 16

8.13706 9.35762

0.12289 0.10686

47.58041 55.71747

0.02102 0.01795

5.84737 5.95423

0.17102 0.16795

26.69302 28.29599

4.56496 4.75225

17 18

10.76126 12.37545

0.09293 0.08081

65.07509 75.83636

0.01537 0.01319

6.04716 6.12797

0.16537 0.16319

29.78280 31.15649

4.92509 5.08431

19 20 21

14.23177 16.36654 18.82152

0.07027 0.06110 0.05313

88.21181 102.44358 118.81012

0.01134 9.7615E−03 8.4168E−03

6.19823 6.25933 6.31246

0.16134 0.15976 0.15842

32.42127 33.58217 34.64479

5.23073 5.36514 5.48832

22 23

21.64475 24.89146

0.04620 0.04017

137.63164 159.27638

7.2658E−03 6.2784E−03

6.35866 6.39884

0.15727 0.15628

35.61500 36.49884

5.60102 5.70398

24 25

28.62518 32.91895

0.03493 0.03038

184.16784 212.79302

5.4298E−03 4.6994E−03

6.43377 6.46415

0.15543 0.15470

37.30232 38.03139

5.79789 5.88343

26 27

37.85680 43.53531

0.02642 0.02297

245.71197 283.56877

4.0698E−03 3.5265E−03

6.49056 6.51353

0.15407 0.15353

38.69177 39.28899

5.96123 6.03190

28 29 30

50.06561 57.57545 66.21177

0.01997 0.01737 0.01510

327.10408 377.16969 434.74515

3.0571E−03 2.6513E−03 2.3002E−03

6.53351 6.55088 6.56598

0.15306 0.15265 0.15230

39.82828 40.31460 40.75259

6.09600 6.15408 6.20663

36 40

153.15185 267.86355

6.5295E−03 1014.34568 3.7332E−03 1779.09031

9.8586E−04 5.6209E−04

6.62314 6.64178

0.15099 0.15056

42.58717 43.28299

6.43006 6.51678

48 819.40071 50 1083.65744

1.2204E−03 5456.00475 9.2280E−04 7217.71628

1.8328E−04 1.3855E−04

6.65853 6.66051

0.15018 0.15014

43.99967 44.09583

6.60802 6.62048

52 1433.13697 55 2179.62218

6.9777E−04 9547.57978 4.5880E−04 1.4524E+04

1.0474E−04 6.8851E−05

6.66201 6.66361

0.15010 0.15007

44.17154 44.25583

6.63036 6.64142

Single Sums

n

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60 4383.99875

2.2810E−04 2.9220E+04

3.4223E−05

6.66515

0.15003

44.34307

6.65298

72 2.3455E+04 75 3.5673E+04

4.2634E−05 1.5636E+05 2.8033E−05 2.3781E+05

6.3954E−06 4.2050E−06

6.66638 6.66648

0.15001 0.15000

44.42209 44.42918

6.66360 6.66456

80 7.1751E+04 84 1.2549E+05

1.3937E−05 4.7833E+05 7.9686E−06 8.3661E+05

2.0906E−06 1.1953E−06

6.66657 6.66661

0.15000 0.15000

44.43639 44.43963

6.66555 6.66600

90 2.9027E+05 96 6.7142E+05 100 1.1743E+06

3.4450E−06 1.9351E+06 1.4894E−06 4.4761E+06 8.5156E−07 7.8287E+06

5.1676E−07 2.2341E−07 1.2773E−07

6.66664 6.66666 6.66666

0.15000 0.15000 0.15000

44.44222 44.44343 44.44384

6.66636 6.66652 6.66658

108 3.5923E+06 120 1.9219E+07

2.7838E−07 2.3948E+07 5.2031E−08 1.2813E+08

4.1757E−08 7.8046E−09

6.66666 6.66667

0.15000 0.15000

44.44423 44.44440

6.66664 6.66666

132 1.0283E+08 144 5.5016E+08

9.7249E−09 6.8553E+08 1.8177E−09 3.6677E+09

1.4587E−09 2.7265E−10

6.66667 6.66667

0.15000 0.15000

44.44444 44.44444

6.66667 6.66667

180 8.4258E+10 240 3.6939E+14

1.1868E−11 5.6172E+11 2.7072E−15 2.4626E+15

1.7802E−12 4.0608E−16

6.66667 6.66667

0.15000 0.15000

44.44444 44.44444

6.66667 6.66667

360 7.0994E+21 480 1.3645E+29

1.4086E−22 4.7329E+22 7.3289E−30 9.0965E+29

2.1129E−23 1.0993E−30

6.66667 6.66667

0.15000 0.15000

44.44444 44.44444

6.66667 6.66667

600 2.6224E+36

3.8133E−37 1.7483E+37

5.7199E−38

6.66667

0.15000

44.44444

6.66667

18.00% Time Value of Money Factors Discrete Compounding 18.00%

TABLE A-A-20 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

n 1 2

1.18000 1.39240

0.84746 0.71818

1.00000 2.18000

1.00000 0.45872

0.84746 1.56564

1.18000 0.63872

0.00000 0.71818

0.00000 0.45872

3 4

1.64303 1.93878

0.60863 0.51579

3.57240 5.21543

0.27992 0.19174

2.17427 2.69006

0.45992 0.37174

1.93545 3.48281

0.89016 1.29470

5 6 7

2.28776 2.69955 3.18547

0.43711 0.37043 0.31393

7.15421 9.44197 12.14152

0.13978 0.10591 0.08236

3.12717 3.49760 3.81153

0.31978 0.28591 0.26236

5.23125 7.08341 8.96696

1.67284 2.02522 2.35259

8 9

3.75886 4.43545

0.26604 0.22546

15.32700 19.08585

0.06524 0.05239

4.07757 4.30302

0.24524 0.23239

10.82922 12.63287

2.65581 2.93581

10 11

5.23384 6.17593

0.19106 0.16192

23.52131 28.75514

0.04251 0.03478

4.49409 4.65601

0.22251 0.21478

14.35245 15.97164

3.19363 3.43033

12 13

7.28759 8.59936

0.13722 0.11629

34.93107 42.21866

0.02863 0.02369

4.79322 4.90951

0.20863 0.20369

17.48106 18.87651

3.64703 3.84489

14 15 16

10.14724 11.97375 14.12902

0.09855 0.08352 0.07078

50.81802 60.96527 72.93901

0.01968 0.01640 0.01371

5.00806 5.09158 5.16235

0.19968 0.19640 0.19371

20.15765 21.32687 22.38852

4.02504 4.18866 4.33688

17 18

16.67225 19.67325

0.05998 0.05083

87.06804 103.74028

0.01149 9.6395E−03

5.22233 5.27316

0.19149 0.18964

23.34820 24.21231

4.47084 4.59161

19 20

23.21444 27.39303

0.04308 0.03651

123.41353 146.62797

8.1028E−03 6.8200E−03

5.31624 5.35275

0.18810 0.18682

24.98769 25.68130

4.70026 4.79778

21 22

32.32378 38.14206

0.03094 0.02622

174.02100 206.34479

5.7464E−03 4.8463E−03

5.38368 5.40990

0.18575 0.18485

26.30004 26.85061

4.88514 4.96324

23 24

45.00763 53.10901

0.02222 0.01883

244.48685 289.49448

4.0902E−03 3.4543E−03

5.43212 5.45095

0.18409 0.18345

27.33942 27.77249

5.03292 5.09498

25 26 27

62.66863 73.94898 87.25980

0.01596 0.01352 0.01146

342.60349 405.27211 479.22109

2.9188E−03 2.4675E−03 2.0867E−03

5.46691 5.48043 5.49189

0.18292 0.18247 0.18209

28.15546 28.49353 28.79149

5.15016 5.19914 5.24255

28 29

102.96656 121.50054

9.7119E−03 566.48089 8.2304E−03 669.44745

1.7653E−03 1.4938E−03

5.50160 5.50983

0.18177 0.18149

29.05371 29.28416

5.28096 5.31489

30 36

143.37064 387.03680

6.9749E−03 790.94799 2.5837E−03 2144.64890

1.2643E−03 4.6628E−04

5.51681 5.54120

0.18126 0.18047

29.48643 30.26771

5.34484 5.46230

40 750.37834 48 2820.56655

1.3327E−03 4163.21303 3.5454E−04 1.5664E+04

2.4020E−04 6.3840E−05

5.54815 5.55359

0.18024 0.18006

30.52692 30.75871

5.50218 5.53853

50 3927.35686 52 5468.45169 55 8984.84112

2.5462E−04 2.1813E+04 1.8287E−04 3.0375E+04 1.1130E−04 4.9910E+04

4.5844E−05 3.2922E−05 2.0036E−05

5.55414 5.55454 5.55494

0.18005 0.18003 0.18002

30.78561 30.80573 30.82675

5.54282 5.54604 5.54943

n

Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60 2.0555E+04 72 1.4980E+05

4.8650E−05 1.1419E+05 6.6757E−06 8.3220E+05

8.7574E−06 1.2016E−06

5.55529 5.55552

0.18001 0.18000

30.84648 30.86132

5.55264 5.55507

75 2.4612E+05 80 5.6307E+05 84 1.0917E+06

4.0630E−06 1.3673E+06 1.7760E−06 3.1281E+06 9.1603E−07 6.0648E+06

7.3135E−07 3.1968E−07 1.6489E−07

5.55553 5.55555 5.55555

0.18000 0.18000 0.18000

30.86238 30.86335 30.86374

5.55525 5.55541 5.55548

90 2.9470E+06 96 7.9556E+06

3.3933E−07 1.6372E+07 1.2570E−07 4.4198E+07

6.1079E−08 2.2626E−08

5.55555 5.55555

0.18000 0.18000

30.86402 30.86413

5.55553 5.55554

100 1.5424E+07 108 5.7977E+07

6.4833E−08 8.5690E+07 1.7248E−08 3.2210E+08

1.1670E−08 3.1047E−09

5.55556 5.55556

0.18000 0.18000

30.86416 30.86419

5.55555 5.55555

120 4.2251E+08 132 3.0791E+09

2.3668E−09 2.3473E+09 3.2477E−10 1.7106E+10

4.2602E−10 5.8458E−11

5.55556 5.55556

0.18000 0.18000

30.86420 30.86420

5.55556 5.55556

144 2.2439E+10 180 8.6848E+12 240 1.7852E+17

4.4565E−11 1.2466E+11 1.1514E−13 4.8249E+13 5.6017E−18 9.9177E+17

8.0216E−12 2.0726E−14 1.0083E−18

5.55556 5.55556 5.55556

0.18000 0.18000 0.18000

30.86420 30.86420 30.86420

5.55556 5.55556 5.55556

360 7.5426E+25 480 3.1869E+34

1.3258E−26 4.1903E+26 3.1379E−35 1.7705E+35

2.3864E−27 5.6482E−36

5.55556 5.55556

0.18000 0.18000

30.86420 30.86420

5.55556 5.55556

600 1.3465E+43

7.4267E−44 7.4805E+43

1.3368E−44

5.55556

0.18000

30.86420

5.55556

20.00% Time Value of Money Factors Discrete Compounding 20.00%

TABLE A-A-21 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G n (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n) 1 1.20000 0.83333 1.00000 1.00000 0.83333 1.20000 0.00000 0.00000 2 3 4

1.44000 1.72800 2.07360

0.69444 0.57870 0.48225

2.20000 3.64000 5.36800

0.45455 0.27473 0.18629

1.52778 2.10648 2.58873

0.65455 0.47473 0.38629

0.69444 1.85185 3.29861

0.45455 0.87912 1.27422

5 6

2.48832 2.98598

0.40188 0.33490

7.44160 9.92992

0.13438 0.10071

2.99061 3.32551

0.33438 0.30071

4.90612 6.58061

1.64051 1.97883

7 8

3.58318 4.29982

0.27908 0.23257

12.91590 16.49908

0.07742 0.06061

3.60459 3.83716

0.27742 0.26061

8.25510 9.88308

2.29016 2.57562

9 10

5.15978 6.19174

0.19381 0.16151

20.79890 25.95868

0.04808 0.03852

4.03097 4.19247

0.24808 0.23852

11.43353 12.88708

2.83642 3.07386

11 12 13

7.43008 8.91610 10.69932

0.13459 0.11216 0.09346

32.15042 39.58050 48.49660

0.03110 0.02526 0.02062

4.32706 4.43922 4.53268

0.23110 0.22526 0.22062

14.23296 15.46668 16.58825

3.28929 3.48410 3.65970

14 15

12.83918 15.40702

0.07789 0.06491

59.19592 72.03511

0.01689 0.01388

4.61057 4.67547

0.21689 0.21388

17.60078 18.50945

3.81749 3.95884

16 17

18.48843 22.18611

0.05409 0.04507

87.44213 105.93056

0.01144 9.4401E−03

4.72956 4.77463

0.21144 0.20944

19.32077 20.04194

4.08511 4.19759

18 19

26.62333 31.94800

0.03756 0.03130

128.11667 154.74000

7.8054E−03 6.4625E−03

4.81219 4.84350

0.20781 0.20646

20.68048 21.24390

4.29752 4.38607

20 21 22

38.33760 46.00512 55.20614

0.02608 0.02174 0.01811

186.68800 225.02560 271.03072

5.3565E−03 4.4439E−03 3.6896E−03

4.86958 4.89132 4.90943

0.20536 0.20444 0.20369

21.73949 22.17423 22.55462

4.46435 4.53339 4.59414

23 24

66.24737 79.49685

0.01509 0.01258

326.23686 392.48424

3.0653E−03 2.5479E−03

4.92453 4.93710

0.20307 0.20255

22.88671 23.17603

4.64750 4.69426

25 26

95.39622 114.47546

0.01048 471.98108 8.7355E−03 567.37730

2.1187E−03 1.7625E−03

4.94759 4.95632

0.20212 0.20176

23.42761 23.64600

4.73516 4.77088

27 28

137.37055 164.84466

7.2796E−03 681.85276 6.0663E−03 819.22331

1.4666E−03 1.2207E−03

4.96360 4.96967

0.20147 0.20122

23.83527 23.99906

4.80201 4.82911

29 30

197.81359 237.37631

5.0553E−03 984.06797 4.2127E−03 1181.88157

1.0162E−03 8.4611E−04

4.97472 4.97894

0.20102 0.20085

24.14061 24.26277

4.85265 4.87308

36 708.80187 40 1469.77157 48 6319.74872

1.4108E−03 3539.00937 6.8038E−04 7343.85784 1.5823E−04 3.1594E+04

2.8256E−04 1.3617E−04 3.1652E−05

4.99295 4.99660 4.99921

0.20028 0.20014 0.20003

24.71078 24.84691 24.95807

4.94914 4.97277 4.99240

50 9100.43815 52 1.3105E+04

1.0988E−04 4.5497E+04 7.6309E−05 6.5518E+04

2.1979E−05 1.5263E−05

4.99945 4.99962

0.20002 0.20002

24.96978 24.97825

4.99451 4.99603

55 2.2645E+04

4.4160E−05 1.1322E+05

8.8324E−06

4.99978

0.20001

24.98675

4.99757

n

Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60 5.6348E+04 72 5.0240E+05

1.7747E−05 2.8173E+05 1.9904E−06 2.5120E+06

3.5495E−06 3.9809E−07

4.99991 4.99999

0.20000 0.20000

24.99423 24.99923

4.99894 4.99986

75 8.6815E+05 80 2.1602E+06

1.1519E−06 4.3407E+06 4.6291E−07 1.0801E+07

2.3038E−07 9.2583E−08

4.99999 5.00000

0.20000 0.20000

24.99954 24.99980

4.99991 4.99996

84 4.4794E+06 90 1.3376E+07

2.2324E−07 2.2397E+07 7.4763E−08 6.6878E+07

4.4648E−08 1.4953E−08

5.00000 5.00000

0.20000 0.20000

24.99990 24.99996

4.99998 4.99999

96 3.9939E+07 100 8.2818E+07

2.5038E−08 1.9970E+08 1.2075E−08 4.1409E+08

5.0076E−09 2.4149E−09

5.00000 5.00000

0.20000 0.20000

24.99999 24.99999

5.00000 5.00000

108 3.5610E+08 120 3.1750E+09 132 2.8309E+10

2.8082E−09 1.7805E+09 3.1496E−10 1.5875E+10 3.5324E−11 1.4154E+11

5.6164E−10 6.2991E−11 7.0649E−12

5.00000 5.00000 5.00000

0.20000 0.20000 0.20000

25.00000 25.00000 25.00000

5.00000 5.00000 5.00000

144 2.5241E+11 180 1.7891E+14

3.9619E−12 1.2620E+12 5.5895E−15 8.9453E+14

7.9237E−13 1.1179E−15

5.00000 5.00000

0.20000 0.20000

25.00000 25.00000

5.00000 5.00000

240 1.0081E+19 360 3.2007E+28

9.9198E−20 5.0404E+19 3.1243E−29 1.6004E+29

1.9840E−20 6.2486E−30

5.00000 5.00000

0.20000 0.20000

25.00000 25.00000

5.00000 5.00000

480 1.0162E+38 600 3.2266E+47

9.8402E−39 5.0812E+38 3.0992E−48 1.6133E+48

1.9680E−39 6.1984E−49

5.00000 5.00000

0.20000 0.20000

25.00000 25.00000

5.00000 5.00000

25.00% Time Value of Money Factors Discrete Compounding 25.00%

TABLE A-A-22 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G n (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n) 1 1.25000 0.80000 1.00000 1.00000 0.80000 1.25000 0.00000 0.00000 2 3

1.56250 1.95313

0.64000 0.51200

2.25000 3.81250

0.44444 0.26230

1.44000 1.95200

0.69444 0.51230

0.64000 1.66400

0.44444 0.85246

4 5

2.44141 3.05176

0.40960 0.32768

5.76563 8.20703

0.17344 0.12185

2.36160 2.68928

0.42344 0.37185

2.89280 4.20352

1.22493 1.56307

6 7

3.81470 4.76837

0.26214 0.20972

11.25879 15.07349

0.08882 0.06634

2.95142 3.16114

0.33882 0.31634

5.51424 6.77253

1.86833 2.14243

8 9 10

5.96046 7.45058 9.31323

0.16777 0.13422 0.10737

19.84186 25.80232 33.25290

0.05040 0.03876 0.03007

3.32891 3.46313 3.57050

0.30040 0.28876 0.28007

7.94694 9.02068 9.98705

2.38725 2.60478 2.79710

11 12

11.64153 14.55192

0.08590 0.06872

42.56613 54.20766

0.02349 0.01845

3.65640 3.72512

0.27349 0.26845

10.84604 11.60195

2.96631 3.11452

13 14

18.18989 22.73737

0.05498 0.04398

68.75958 86.94947

0.01454 0.01150

3.78010 3.82408

0.26454 0.26150

12.26166 12.83341

3.24374 3.35595

15 16

28.42171 35.52714

0.03518 0.02815

109.68684 138.10855

9.1169E−03 7.2407E−03

3.85926 3.88741

0.25912 0.25724

13.32599 13.74820

3.45299 3.53660

17 18 19

44.40892 55.51115 69.38894

0.02252 0.01801 0.01441

173.63568 218.04460 273.55576

5.7592E−03 4.5862E−03 3.6556E−03

3.90993 3.92794 3.94235

0.25576 0.25459 0.25366

14.10849 14.41473 14.67414

3.60838 3.66979 3.72218

20 21

86.73617 108.42022

0.01153 342.94470 9.2234E−03 429.68087

2.9159E−03 2.3273E−03

3.95388 3.96311

0.25292 0.25233

14.89320 15.07766

3.76673 3.80451

22 23

135.52527 169.40659

7.3787E−03 538.10109 5.9030E−03 673.62636

1.8584E−03 1.4845E−03

3.97049 3.97639

0.25186 0.25148

15.23262 15.36248

3.83646 3.86343

24 25

211.75824 264.69780

4.7224E−03 843.03295 3.7779E−03 1054.79118

1.1862E−03 9.4805E−04

3.98111 3.98489

0.25119 0.25095

15.47109 15.56176

3.88613 3.90519

26 27

330.87225 413.59031

3.0223E−03 1319.48898 2.4179E−03 1650.36123

7.5787E−04 6.0593E−04

3.98791 3.99033

0.25076 0.25061

15.63732 15.70019

3.92118 3.93456

28 29 30

516.98788 646.23485 807.79357

1.9343E−03 2063.95153 1.5474E−03 2580.93941 1.2379E−03 3227.17427

4.8451E−04 3.8746E−04 3.0987E−04

3.99226 3.99381 3.99505

0.25048 0.25039 0.25031

15.75241 15.79574 15.83164

3.94574 3.95506 3.96282

36 3081.48791 40 7523.16385

3.2452E−04 1.2322E+04 1.3292E−04 3.0089E+04

8.1156E−05 3.3235E−05

3.99870 3.99947

0.25008 0.25003

15.94808 15.97661

3.98831 3.99468

48 4.4842E+04 50 7.0065E+04

2.2301E−05 1.7936E+05 1.4272E−05 2.8026E+05

5.5753E−06 3.5682E−06

3.99991 3.99994

0.25001 0.25000

15.99536 15.99692

3.99893 3.99929

52 1.0948E+05 55 2.1382E+05

9.1344E−06 4.3790E+05 4.6768E−06 8.5528E+05

2.2836E−06 1.1692E−06

3.99996 3.99998

0.25000 0.25000

15.99795 15.99890

3.99953 3.99974

Single Sums

n

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60 6.5253E+05

1.5325E−06 2.6101E+06

3.8312E−07

3.99999

0.25000

15.99961

3.99991

72 9.4956E+06 75 1.8546E+07

1.0531E−07 3.7982E+07 5.3920E−08 7.4184E+07

2.6328E−08 1.3480E−08

4.00000 4.00000

0.25000 0.25000

15.99997 15.99998

3.99999 4.00000

80 5.6598E+07 84 1.3818E+08

1.7668E−08 2.2639E+08 7.2370E−09 5.5271E+08

4.4171E−09 1.8093E−09

4.00000 4.00000

0.25000 0.25000

15.99999 16.00000

4.00000 4.00000

90 5.2711E+08 96 2.0108E+09 100 4.9091E+09

1.8971E−09 2.1084E+09 4.9732E−10 8.0431E+09 2.0370E−10 1.9636E+10

4.7428E−10 1.2433E−10 5.0926E−11

4.00000 4.00000 4.00000

0.25000 0.25000 0.25000

16.00000 16.00000 16.00000

4.00000 4.00000 4.00000

108 2.9260E+10 120 4.2580E+11

3.4176E−11 1.1704E+11 2.3485E−12 1.7032E+12

8.5439E−12 5.8714E−13

4.00000 4.00000

0.25000 0.25000

16.00000 16.00000

4.00000 4.00000

132 6.1961E+12 144 9.0166E+13

1.6139E−13 2.4785E+13 1.1091E−14 3.6066E+14

4.0348E−14 2.7727E−15

4.00000 4.00000

0.25000 0.25000

16.00000 16.00000

4.00000 4.00000

180 2.7784E+17 240 1.8130E+23

3.5991E−18 1.1114E+18 5.5157E−24 7.2521E+23

8.9978E−19 1.3789E−24

4.00000 4.00000

0.25000 0.25000

16.00000 16.00000

4.00000 4.00000

360 7.7198E+34 480 3.2870E+46 600 1.3996E+58

1.2954E−35 3.0879E+35 3.0422E−47 1.3148E+47 7.1448E−59 5.5984E+58

3.2384E−36 7.6056E−48 1.7862E−59

4.00000 4.00000 4.00000

0.25000 0.25000 0.25000

16.00000 16.00000 16.00000

4.00000 4.00000 4.00000

30.00% Time Value of Money Factors Discrete Compounding 30.00%

TABLE A-A-23 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

n 1 2

1.30000 1.69000

0.76923 0.59172

1.00000 2.30000

1.00000 0.43478

0.76923 1.36095

1.30000 0.73478

0.00000 0.59172

0.00000 0.43478

3 4

2.19700 2.85610

0.45517 0.35013

3.99000 6.18700

0.25063 0.16163

1.81611 2.16624

0.55063 0.46163

1.50205 2.55243

0.82707 1.17828

5 6 7

3.71293 4.82681 6.27485

0.26933 0.20718 0.15937

9.04310 12.75603 17.58284

0.11058 0.07839 0.05687

2.43557 2.64275 2.80211

0.41058 0.37839 0.35687

3.62975 4.66563 5.62183

1.49031 1.76545 2.00628

8 9

8.15731 10.60450

0.12259 0.09430

23.85769 32.01500

0.04192 0.03124

2.92470 3.01900

0.34192 0.33124

6.47995 7.23435

2.21559 2.39627

10 11

13.78585 17.92160

0.07254 0.05580

42.61950 56.40535

0.02346 0.01773

3.09154 3.14734

0.32346 0.31773

7.88719 8.44518

2.55122 2.68328

12 13

23.29809 30.28751

0.04292 0.03302

74.32695 97.62504

0.01345 0.01024

3.19026 3.22328

0.31345 0.31024

8.91732 9.31352

2.79517 2.88946

14 15 16

39.37376 51.18589 66.54166

0.02540 0.01954 0.01503

127.91255 167.28631 218.47220

7.8178E−03 5.9778E−03 4.5772E−03

3.24867 3.26821 3.28324

0.30782 0.30598 0.30458

9.64369 9.91721 10.14263

2.96850 3.03444 3.08921

17 18

86.50416 112.45541

0.01156 285.01386 8.8924E−03 371.51802

3.5086E−03 2.6917E−03

3.29480 3.30369

0.30351 0.30269

10.32759 10.47876

3.13451 3.17183

19 20

146.19203 190.04964

6.8403E−03 483.97343 5.2618E−03 630.16546

2.0662E−03 1.5869E−03

3.31053 3.31579

0.30207 0.30159

10.60189 10.70186

3.20247 3.22754

21 22

247.06453 321.18389

4.0475E−03 820.21510 3.1135E−03 1067.27963

1.2192E−03 9.3696E−04

3.31984 3.32296

0.30122 0.30094

10.78281 10.84819

3.24799 3.26462

23 24 25

417.53905 542.80077 705.64100

2.3950E−03 1388.46351 1.8423E−03 1806.00257 1.4172E−03 2348.80334

7.2022E−04 5.5371E−04 4.2575E−04

3.32535 3.32719 3.32861

0.30072 0.30055 0.30043

10.90088 10.94326 10.97727

3.27812 3.28904 3.29785

26 917.33330 27 1192.53329

1.0901E−03 3054.44434 8.3855E−04 3971.77764

3.2739E−04 2.5178E−04

3.32970 3.33054

0.30033 0.30025

11.00452 11.02632

3.30496 3.31067

28 1550.29328 29 2015.38126

6.4504E−04 5164.31093 4.9618E−04 6714.60421

1.9364E−04 1.4893E−04

3.33118 3.33168

0.30019 0.30015

11.04374 11.05763

3.31526 3.31894

30 2619.99564 36 1.2646E+04

3.8168E−04 8729.98548 7.9075E−05 4.2151E+04

1.1455E−04 2.3724E−05

3.33206 3.33307

0.30011 0.30002

11.06870 11.10074

3.32188 3.33049

40 3.6119E+04 48 2.9463E+05

2.7686E−05 1.2039E+05 3.3941E−06 9.8211E+05

8.3061E−06 1.0182E−06

3.33324 3.33332

0.30001 0.30000

11.10711 11.11053

3.33223 3.33317

50 4.9793E+05 52 8.4150E+05 55 1.8488E+06

2.0083E−06 1.6598E+06 1.1884E−06 2.8050E+06 5.4090E−07 6.1626E+06

6.0250E−07 3.5651E−07 1.6227E−07

3.33333 3.33333 3.33333

0.30000 0.30000 0.30000

11.11075 11.11089 11.11101

3.33323 3.33327 3.33330

Single Sums

n

Uniform Series

Gradient Series

To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60 6.8644E+06 72 1.5993E+08

1.4568E−07 2.2881E+07 6.2529E−09 5.3309E+08

4.3704E−08 1.8759E−09

3.33333 3.33333

0.30000 0.30000

11.11108 11.11111

3.33332 3.33333

75 3.5136E+08 80 1.3046E+09 84 3.7260E+09

2.8461E−09 1.1712E+09 7.6653E−10 4.3486E+09 2.6839E−10 1.2420E+10

8.5383E−10 2.2996E−10 8.0516E−11

3.33333 3.33333 3.33333

0.30000 0.30000 0.30000

11.11111 11.11111 11.11111

3.33333 3.33333 3.33333

90 1.7985E+10 96 8.6808E+10

5.5603E−11 5.9949E+10 1.1520E−11 2.8936E+11

1.6681E−11 3.4559E−12

3.33333 3.33333

0.30000 0.30000

11.11111 11.11111

3.33333 3.33333

100 2.4793E+11 108 2.0225E+12

4.0333E−12 8.2645E+11 4.9444E−13 6.7416E+12

1.2100E−12 1.4833E−13

3.33333 3.33333

0.30000 0.30000

11.11111 11.11111

3.33333 3.33333

120 4.7120E+13 132 1.0978E+15

2.1223E−14 1.5707E+14 9.1091E−16 3.6593E+15

6.3668E−15 2.7327E−16

3.33333 3.33333

0.30000 0.30000

11.11111 11.11111

3.33333 3.33333

144 2.5577E+16 180 3.2345E+20 240 2.2203E+27

3.9098E−17 8.5255E+16 3.0917E−21 1.0782E+21 4.5040E−28 7.4009E+27

1.1729E−17 9.2751E−22 1.3512E−28

3.33333 3.33333 3.33333

0.30000 0.30000 0.30000

11.11111 11.11111 11.11111

3.33333 3.33333 3.33333

360 1.0462E+41 480 4.9296E+54

9.5586E−42 3.4873E+41 2.0286E−55 1.6432E+55

2.8676E−42 6.0857E−56

3.33333 3.33333

0.30000 0.30000

11.11111 11.11111

3.33333 3.33333

600 2.3228E+68

4.3052E−69 7.7427E+68

1.2915E−69

3.33333

0.30000

11.11111

3.33333

40.00% Time Value of Money Factors Discrete Compounding 40.00%

TABLE A-A-24 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G n (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n) 1 1.40000 0.71429 1.00000 1.00000 0.71429 1.40000 0.00000 0.00000 2 3 4

1.96000 2.74400 3.84160

0.51020 0.36443 0.26031

2.40000 4.36000 7.10400

0.41667 0.22936 0.14077

1.22449 1.58892 1.84923

0.81667 0.62936 0.54077

0.51020 1.23907 2.01999

0.41667 0.77982 1.09234

5 6

5.37824 7.52954

0.18593 0.13281

10.94560 16.32384

0.09136 0.06126

2.03516 2.16797

0.49136 0.46126

2.76373 3.42778

1.35799 1.58110

7 8

10.54135 14.75789

0.09486 0.06776

23.85338 34.39473

0.04192 0.02907

2.26284 2.33060

0.44192 0.42907

3.99697 4.47129

1.76635 1.91852

9 10

20.66105 28.92547

0.04840 0.03457

49.15262 69.81366

0.02034 0.01432

2.37900 2.41357

0.42034 0.41432

4.85849 5.16964

2.04224 2.14190

11 12 13

40.49565 56.69391 79.37148

0.02469 0.01764 0.01260

98.73913 139.23478 195.92869

0.01013 7.1821E−03 5.1039E−03

2.43826 2.45590 2.46850

0.41013 0.40718 0.40510

5.41658 5.61060 5.76179

2.22149 2.28454 2.33412

14 15

111.12007 155.56810

8.9993E−03 275.30017 6.4281E−03 386.42024

3.6324E−03 2.5879E−03

2.47750 2.48393

0.40363 0.40259

5.87878 5.96877

2.37287 2.40296

16 17

217.79533 304.91347

4.5915E−03 541.98833 3.2796E−03 759.78367

1.8451E−03 1.3162E−03

2.48852 2.49180

0.40185 0.40132

6.03764 6.09012

2.42620 2.44406

18 19

426.87885 597.63040

2.3426E−03 1064.69714 1.6733E−03 1491.57599

9.3923E−04 6.7043E−04

2.49414 2.49582

0.40094 0.40067

6.12994 6.16006

2.45773 2.46815

20 836.68255 21 1171.35558 22 1639.89781

1.1952E−03 2089.20639 8.5371E−04 2925.88894 6.0979E−04 4097.24452

4.7865E−04 3.4178E−04 2.4407E−04

2.49701 2.49787 2.49848

0.40048 0.40034 0.40024

6.18277 6.19984 6.21265

2.47607 2.48206 2.48658

23 2295.85693 24 3214.19970

4.3557E−04 5737.14232 3.1112E−04 8032.99925

1.7430E−04 1.2449E−04

2.49891 2.49922

0.40017 0.40012

6.22223 6.22939

2.48998 2.49253

25 4499.87958 26 6299.83141

2.2223E−04 1.1247E+04 1.5873E−04 1.5747E+04

8.8911E−05 6.3504E−05

2.49944 2.49960

0.40009 0.40006

6.23472 6.23869

2.49444 2.49587

27 8819.76398 28 1.2348E+04

1.1338E−04 2.2047E+04 8.0987E−05 3.0867E+04

4.5358E−05 3.2397E−05

2.49972 2.49980

0.40005 0.40003

6.24164 6.24382

2.49694 2.49773

29 1.7287E+04 30 2.4201E+04 36 1.8223E+05

5.7848E−05 4.3214E+04 4.1320E−05 6.0501E+04 5.4877E−06 4.5556E+05

2.3140E−05 1.6529E−05 2.1951E−06

2.49986 2.49990 2.49999

0.40002 0.40002 0.40000

6.24544 6.24664 6.24947

2.49832 2.49876 2.49980

40 7.0004E+05 48 1.0331E+07

1.4285E−06 1.7501E+06 9.6795E−08 2.5828E+07

5.7140E−07 3.8718E−08

2.50000 2.50000

0.40000 0.40000

6.24985 6.24999

2.49994 2.50000

50 2.0249E+07 52 3.9688E+07

4.9385E−08 5.0622E+07 2.5197E−08 9.9220E+07

1.9754E−08 1.0079E−08

2.50000 2.50000

0.40000 0.40000

6.24999 6.25000

2.50000 2.50000

55 1.0890E+08

9.1824E−09 2.7226E+08

3.6730E−09

2.50000

0.40000

6.25000

2.50000

n

Single Sums Uniform Series Gradient Series To Find F To Find P To Find F To Find A To Find P To Find A To Find P To Find A Given P Given F (P|F Given A Given F (A|F Given A Given P Given G Given G (F|P i%,n) i%,n) (F|A i%,n) i%,n) (P|A i%,n) (A|P i%,n) (P|G i%,n) (A|G i%,n)

60 5.8571E+08 72 3.3206E+10

1.7073E−09 1.4643E+09 3.0115E−11 8.3015E+10

6.8293E−10 1.2046E−11

2.50000 2.50000

0.40000 0.40000

6.25000 6.25000

2.50000 2.50000

75 9.1118E+10 80 4.9005E+11

1.0975E−11 2.2779E+11 2.0406E−12 1.2251E+12

4.3899E−12 8.1624E−13

2.50000 2.50000

0.40000 0.40000

6.25000 6.25000

2.50000 2.50000

84 1.8826E+12 90 1.4175E+13

5.3118E−13 4.7065E+12 7.0547E−14 3.5438E+13

2.1247E−13 2.8219E−14

2.50000 2.50000

0.40000 0.40000

6.25000 6.25000

2.50000 2.50000

96 1.0673E+14 100 4.1002E+14

9.3693E−15 2.6683E+14 2.4389E−15 1.0250E+15

3.7477E−15 9.7557E−16

2.50000 2.50000

0.40000 0.40000

6.25000 6.25000

2.50000 2.50000

108 6.0510E+15 120 3.4306E+17 132 1.9449E+19

1.6526E−16 1.5128E+16 2.9150E−18 8.5764E+17 5.1416E−20 4.8623E+19

6.6105E−17 1.1660E−18 2.0566E−20

2.50000 2.50000 2.50000

0.40000 0.40000 0.40000

6.25000 6.25000 6.25000

2.50000 2.50000 2.50000

144 1.1026E+21 180 2.0093E+26

9.0691E−22 2.7566E+21 4.9768E−27 5.0233E+26

3.6276E−22 1.9907E−27

2.50000 2.50000

0.40000 0.40000

6.25000 6.25000

2.50000 2.50000

240 1.1769E+35 360 4.0373E+52

8.4971E−36 2.9422E+35 2.4769E−53 1.0093E+53

3.3988E−36 9.9076E−54

2.50000 2.50000

0.40000 0.40000

6.25000 6.25000

2.50000 2.50000

480 1.3850E+70 600 4.7514E+87

7.2201E−71 3.4626E+70 2.1046E−88 1.1878E+88

2.8880E−71 8.4186E−89

2.50000 2.50000

0.40000 0.40000

6.25000 6.25000

2.50000 2.50000

50.00% Time Value of Money Factors Discrete Compounding 50.00%

TABLE A-A-25 Time Value of Money Factors Discrete Compounding Single Sums Uniform Series To Find F Given P (F|P i%,n)

n

To Find P Given F (P|F i%,n)

To Find F Given A (F|A i%,n)

To Find A Given F (A|F i%,n)

Gradient Series

To Find P To Find A To Find P To Find A Given A Given P Given G Given G (P|A (A|P (P|G (A|G i%,n) i%,n) i%,n) i%,n)

1 2 3

1.50000 2.25000 3.37500

0.66667 0.44444 0.29630

1.00000 2.50000 4.75000

1.00000 0.40000 0.21053

0.66667 1.11111 1.40741

1.50000 0.90000 0.71053

0.00000 0.44444 1.03704

0.00000 0.40000 0.73684

4 5

5.06250 7.59375

0.19753 0.13169

8.12500 13.18750

0.12308 0.07583

1.60494 1.73663

0.62308 0.57583

1.62963 2.15638

1.01538 1.24171

6 7

11.39063 17.08594

0.08779 0.05853

20.78125 32.17188

0.04812 0.03108

1.82442 1.88294

0.54812 0.53108

2.59534 2.94650

1.42256 1.56484

8 9

25.62891 38.44336

0.03902 0.02601

49.25781 74.88672

0.02030 0.01335

1.92196 1.94798

0.52030 0.51335

3.21963 3.42773

1.67518 1.75964

10 11 12

57.66504 86.49756 129.74634

0.01734 0.01156 7.7073E−03

113.33008 170.99512 257.49268

8.8238E−03 5.8481E−03 3.8836E−03

1.96532 1.97688 1.98459

0.50882 0.50585 0.50388

3.58380 3.69941 3.78419

1.82352 1.87134 1.90679

13 14

194.61951 291.92926

5.1382E−03 3.4255E−03

387.23901 581.85852

2.5824E−03 1.7186E−03

1.98972 1.99315

0.50258 0.50172

3.84585 3.89038

1.93286 1.95188

15 16

437.89389 656.84084

2.2837E−03 873.78778 1.5224E−03 1311.68167

1.1444E−03 7.6238E−04

1.99543 1.99696

0.50114 0.50076

3.92236 3.94519

1.96567 1.97560

17 985.26125 18 1477.89188

1.0150E−03 1968.52251 6.7664E−04 2953.78376

5.0800E−04 3.3855E−04

1.99797 1.99865

0.50051 0.50034

3.96143 3.97293

1.98273 1.98781

19 2216.83782 20 3325.25673 21 4987.88510

4.5109E−04 4431.67564 3.0073E−04 6648.51346 2.0049E−04 9973.77019

2.2565E−04 1.5041E−04 1.0026E−04

1.99910 1.99940 1.99960

0.50023 0.50015 0.50010

3.98105 3.98677 3.99078

1.99143 1.99398 1.99579

22 7481.82764 23 1.1223E+04

1.3366E−04 1.4962E+04 8.9105E−05 2.2443E+04

6.6838E−05 4.4556E−05

1.99973 1.99982

0.50007 0.50004

3.99358 3.99554

1.99706 1.99795

24 1.6834E+04 25 2.5251E+04

5.9403E−05 3.3666E+04 3.9602E−05 5.0500E+04

2.9703E−05 1.9802E−05

1.99988 1.99992

0.50003 0.50002

3.99691 3.99786

1.99857 1.99901

26 3.7877E+04 27 5.6815E+04

2.6401E−05 7.5752E+04 1.7601E−05 1.1363E+05

1.3201E−05 8.8006E−06

1.99995 1.99996

0.50001 0.50001

3.99852 3.99898

1.99931 1.99952

28 8.5223E+04 29 1.2783E+05 30 1.9175E+05

1.1734E−05 1.7044E+05 7.8226E−06 2.5567E+05 5.2151E−06 3.8350E+05

5.8671E−06 3.9114E−06 2.6076E−06

1.99998 1.99998 1.99999

0.50001 0.50000 0.50000

3.99930 3.99951 3.99967

1.99967 1.99977 1.99984

36 2.1842E+06 40 1.1057E+07

4.5784E−07 4.3683E+06 9.0438E−08 2.2115E+07

2.2892E−07 4.5219E−08

2.00000 2.00000

0.50000 0.50000

3.99997 3.99999

1.99998 2.00000

48 2.8339E+08 50 6.3762E+08

3.5287E−09 5.6677E+08 1.5683E−09 1.2752E+09

1.7644E−09 7.8416E−10

2.00000 2.00000

0.50000 0.50000

4.00000 4.00000

2.00000 2.00000

52 1.4346E+09

6.9703E−10 2.8693E+09

3.4852E−10

2.00000

0.50000

4.00000

2.00000

Single Sums To Find P Given F (P|F i%,n)

Gradient Series

To Find P To Find A To Find P To Find A Given A Given P Given G Given G (P|A (A|P (P|G (A|G i%,n) i%,n) i%,n) i%,n)

To Find F Given A (F|A i%,n)

To Find A Given F (A|F i%,n)

55 4.8419E+09

2.0653E−10 9.6839E+09

1.0326E−10

2.00000

0.50000

4.00000

2.00000

60 3.6768E+10 72 4.7706E+12

2.7197E−11 7.3537E+10 2.0962E−13 9.5411E+12

1.3599E−11 1.0481E−13

2.00000 2.00000

0.50000 0.50000

4.00000 4.00000

2.00000 2.00000

75 1.6101E+13 80 1.2226E+14

6.2109E−14 3.2201E+13 8.1790E−15 2.4453E+14

3.1055E−14 4.0895E−15

2.00000 2.00000

0.50000 0.50000

4.00000 4.00000

2.00000 2.00000

84 6.1896E+14 90 7.0504E+15

1.6156E−15 1.2379E+15 1.4184E−16 1.4101E+16

8.0780E−16 7.0918E−17

2.00000 2.00000

0.50000 0.50000

4.00000 4.00000

2.00000 2.00000

96 8.0308E+16 100 4.0656E+17 108 1.0420E+19

1.2452E−17 1.6062E+17 2.4597E−18 8.1312E+17 9.5972E−20 2.0839E+19

6.2260E−18 1.2298E−18 4.7986E−20

2.00000 2.00000 2.00000

0.50000 0.50000 0.50000

4.00000 4.00000 4.00000

2.00000 2.00000 2.00000

120 1.3519E+21 132 1.7541E+23

7.3969E−22 2.7038E+21 5.7010E−24 3.5081E+23

3.6984E−22 2.8505E−24

2.00000 2.00000

0.50000 0.50000

4.00000 4.00000

2.00000 2.00000

144 2.2758E+25 180 4.9708E+31

4.3940E−26 4.5517E+25 2.0117E−32 9.9416E+31

2.1970E−26 1.0059E−32

2.00000 2.00000

0.50000 0.50000

4.00000 4.00000

2.00000 2.00000

240 1.8277E+42 360 2.4709E+63

5.4714E−43 3.6554E+42 4.0471E−64 4.9418E+63

2.7357E−43 2.0236E−64

2.00000 2.00000

0.50000 0.50000

4.00000 4.00000

2.00000 2.00000

480 3.3404E+84 600 4.5160E+105

2.9936E−85 6.6809E+84 2.2143E−106 9.0320E+105

1.4968E−85 1.1072E−106

2.00000 2.00000

0.50000 0.50000

4.00000 4.00000

2.00000 2.00000

n

To Find F Given P (F|P i%,n)

Uniform Series

4.00% Time Value of Money Factors Geometric Series - Present Worth 4.00%

TABLE A-B-1

j

Time Value of Money Factors Geometric Series - Present Worth 4% 5% 6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

n 1

0.96154

0.96154

0.96154

0.96154

0.96154

2 3

1.92308 2.88462

1.93232 2.91244

1.94157 2.94044

1.97855 3.05424

2.02478 3.20048

4

3.84615

3.90198

3.95853

4.19198

4.50053

5

4.80769

4.90104

4.99619

5.39536

5.93808

6 7

5.76923 6.73077

5.90971 6.92807

6.05381 7.13177

6.66817 8.01441

7.52769 9.28542

8

7.69231

7.95622

8.23046

9.43832

11.22907

9

8.65385

8.99426

9.35028

10.94438

13.37830

10 11

9.61538 10.57692

10.04228 11.10038

10.49163 11.65493

12.53732 14.22217

15.75485 18.38277

12

11.53846

12.16866

12.84060

16.00422

21.28864

13 14

12.50000 13.46154

13.24720 14.33612

14.04907 15.28078

17.88908 19.88268

24.50186 28.05494

15

14.42308

15.43550

16.53618

21.99129

31.98383

16

15.38462

16.54546

17.81573

24.22156

36.32828

17 18

16.34615 17.30769

17.66609 18.79749

19.11988 20.44910

26.58050 29.07553

41.13223 46.44429

19

18.26923

19.93978

21.80389

31.71450

52.31821

20

19.23077

21.09304

23.18474

34.50572

58.81340

21 22

20.19231 21.15385

22.25740 23.43295

24.59214 26.02660

37.45797 40.58055

65.99559 73.93743

23

22.11538

24.61981

27.48865

43.88327

82.71927

24 25

23.07692 24.03846

25.81807 27.02786

28.97882 30.49764

47.37654 51.07134

92.42996 103.16775

26

25.00000

28.24929

32.04567

54.97930

115.04126

27

25.96154

29.48245

33.62347

59.11272

128.17062

28 29

26.92308 27.88462

30.72748 31.98447

35.23162 36.87069

63.48461 68.10872

142.68867 158.74228

30

28.84615

33.25355

38.54128

72.99961

176.49387

36 40

34.61538 38.46154

41.12844 46.63525

49.26152 57.12012

108.87422 140.45012

330.16651 498.11856

48

46.15385

58.30188

74.75307

229.42525

1124.62271

50

48.07692

61.36078

79.59740

258.63957

1377.12975

52 55

50.00000 52.88462

64.47879 69.26914

84.62985 92.54728

291.32200 347.76220

1685.87657 2282.59383

j n

4% 5% 6% 10% 15% To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

60

57.69231

77.56509

106.79116

465.73578

3779.49708

72 75

69.23077 72.11538

99.17238 104.97315

147.05700 158.64571

928.96423 1102.25501

1.2651E+04 1.7109E+04

80

76.92308

115.01897

179.49439

1464.47438

2.8290E+04

84

80.76923

123.40898

197.66360

1837.00923

4.2299E+04

90 96

86.53846 92.30769

136.61179 150.59485

227.64978 261.26658

2578.64714 3617.00726

7.7332E+04 1.4137E+05

100

96.15385

160.37302

285.90974

4530.94462

2.1137E+05

108

103.84615

181.08887

341.20357

7106.25392

4.7246E+05

120 132

115.38462 126.92308

215.29361 253.66061

441.66933 567.93591

1.3946E+04 2.7354E+04

1.5789E+06 5.2762E+06

144

138.46154

296.69635

726.62925

5.3636E+04

1.7632E+07

180 240

173.07692 230.76923

459.85139 894.10062

1491.78806 4784.77465

4.0412E+05 1.1697E+07

6.5799E+08 2.7421E+11

360

346.15385

3034.33576

4.7492E+04

9.7997E+09

4.7624E+16

480

461.53846

9782.36045

4.6745E+05

8.2098E+12

8.2712E+21

600

576.92308

3.1058E+04

4.5971E+06

6.8779E+15

1.4365E+27

5.00% Time Value of Money Factors Geometric Series - Present Worth 5.00%

TABLE A-B-2 Time Value of Money Factors Geometric Series - Present Worth j

4%

5%

6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

n 1 2

0.95238 1.89569

0.95238 1.90476

0.95238 1.91383

0.95238 1.95011

0.95238 1.99546

3

2.83002

2.85714

2.88444

2.99536

3.13789

4

3.75545

3.80952

3.86429

4.09037

4.38912

5 6

4.67206 5.57995

4.76190 5.71429

4.85348 5.85208

5.23753 6.43932

5.75951 7.26042

7

6.47919

6.66667

6.86020

7.69834

8.90426

8 9

7.36986 8.25205

7.61905 8.57143

7.87791 8.90532

9.01731 10.39908

10.70467 12.67654

10

9.12584

9.52381

9.94251

11.84666

14.83622

11

9.99131

10.47619

10.98959

13.36316

17.20157

12 13

10.84854 11.69760

11.42857 12.38095

12.04663 13.11374

14.95189 16.61626

19.79219 22.62955

14

12.53857

13.33333

14.19101

18.35989

25.73712

15

13.37154

14.28571

15.27855

20.18656

29.14066

16 17

14.19657 15.01375

15.23810 16.19048

16.37644 17.48478

22.10020 24.10497

32.86834 36.95104

18

15.82314

17.14286

18.60369

26.20521

41.42257

19 20

16.62482 17.41887

18.09524 19.04762

19.73325 20.87356

28.40546 30.71048

46.31995 51.68376

21

18.20536

20.00000

22.02474

33.12526

57.55840

22

18.98436

20.95238

23.18688

35.65504

63.99254

23 24

19.75593 20.52016

21.90476 22.85714

24.36009 25.54447

38.30528 41.08172

71.03944 78.75749

25

21.27711

23.80952

26.74013

43.99037

87.21058

26

22.02686

24.76190

27.94718

47.03753

96.46873

27 28

22.76946 23.50499

25.71429 26.66667

29.16573 30.39588

50.22980 53.57407

106.60861 117.71419

29

24.23351

27.61905

31.63774

57.07760

129.87745

30 36

24.95510 29.14256

28.57143 34.28571

32.89143 40.66835

60.74796 86.74607

143.19911 254.42803

40

31.80357

38.09524

46.10418

108.57764

370.48860

48

36.82956

45.71429

57.61414

166.54883

777.78913

50 52

38.02707 39.20189

47.61905 49.52381

60.63061 63.70481

184.73840 204.70155

934.98968 1123.55905

55

40.92249

52.38095

68.42678

238.35462

1479.25741

j n

4%

5%

6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

60 72

43.68262 49.79223

57.14286 68.57143

76.60133 97.87584

306.01168 549.73617

2336.99669 6982.21820

75

51.21312

71.42857

103.58345

635.06433

9176.29941

80

53.49248

76.19048

113.46432

806.61043

1.4467E+04

84 90

55.23904 57.73668

80.00000 85.71429

121.71321 134.68804

975.66772 1296.23897

2.0821E+04 3.5946E+04

96

60.09495

91.42857

148.42217

1720.02329

6.2051E+04

100 108

61.59356 64.42406

95.23810 102.85714

158.02193 178.34867

2075.89057 3020.85487

8.9291E+04 1.8488E+05

120

68.28353

114.28571

211.88030

5294.18080

5.5083E+05

132

71.72430

125.71429

249.45136

9267.03237

1.6411E+06

144 180

74.79180 82.13812

137.14286 171.42857

291.54847 450.78476

1.6210E+04 8.6604E+04

4.8891E+06 1.2928E+08

240

89.94066

228.57143

872.69322

1.4120E+06

3.0342E+10

360

96.80953

342.85714

2933.63852

3.7519E+08

1.6714E+15

480 600

98.98810 99.67906

457.14286 571.42857

9361.32092 2.9408E+04

9.9691E+10 2.6489E+13

9.2066E+19 5.0714E+24

6.00% Time Value of Money Factors Geometric Series - Present Worth 6.00%

TABLE A-B-3 Time Value of Money Factors Geometric Series - Present Worth j

4% 5% 6% 10% 15% To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

n 1

0.94340

0.94340

0.94340

0.94340

0.94340

2

1.86899

1.87789

1.88679

1.92239

1.96689

3

2.77712

2.80357

2.83019

2.93833

3.07729

4 5

3.66812 4.54231

3.72052 4.62882

3.77358 4.71698

3.99261 5.08667

4.28196 5.58892

6

5.40000

5.52855

5.66038

6.22201

7.00685

7 8

6.24151 7.06714

6.41979 7.30262

6.60377 7.54717

7.40020 8.62285

8.54517 10.21410

9

7.87720

8.17712

8.49057

9.89164

12.02473

10

8.67197

9.04337

9.43396

11.20831

13.98909

11 12

9.45174 10.21680

9.90146 10.75144

10.37736 11.32075

12.57466 13.99257

16.12024 18.43234

13

10.96743

11.59341

12.26415

15.46399

20.94074

14 15

11.70389 12.42646

12.42743 13.25359

13.20755 14.15094

16.99093 18.57549

23.66213 26.61457

16

13.13539

14.07195

15.09434

20.21985

29.81770

17

13.83095

14.88259

16.03774

21.92626

33.29278

18 19

14.51339 15.18295

15.68559 16.48101

16.98113 17.92453

23.69706 25.53469

37.06293 41.15317

20

15.83987

17.26892

18.86792

27.44166

45.59071

21

16.48440

18.04941

19.81132

29.42059

50.40501

22 23

17.11677 17.73721

18.82252 19.58835

20.75472 21.69811

31.47419 33.60530

55.62808 61.29462

24

18.34594

20.34695

22.64151

35.81682

67.44227

25 26

18.94319 19.52917

21.09839 21.84275

23.58491 24.52830

38.11179 40.49337

74.11190 81.34781

27

20.10409

22.58008

25.47170

42.96482

89.19810

28

20.66816

23.31046

26.41509

45.52953

97.71492

29 30

21.22159 21.76458

24.03394 24.75060

27.35849 28.30189

48.19102 50.95294

106.95487 116.97934

36

24.81401

28.91080

33.96226

69.85616

197.75529

40

26.66171

31.55569

37.73585

85.00510

278.24756

48 50

29.96041 30.70949

36.55392 37.74537

45.28302 47.16981

122.94741 134.32393

544.24595 642.55541

52

31.43057

38.91444

49.05660

146.57525

758.26765

55

32.46196

40.62702

51.88679

166.74105

971.35170

j n

4%

5%

6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

60

34.05522

43.37529

56.60377

205.75398

1465.53354

72

37.31332

49.46326

67.92453

334.90762

3915.15469

75 80

38.01797 39.10649

50.88009 53.15376

70.75472 75.47170

377.20874 459.04485

5002.55747 7524.44829

84

39.90566

54.89669

79.24528

536.34890

1.0428E+04

90

40.99585

57.39024

84.90566

676.05776

1.7012E+04

96 100

41.96830 42.55752

59.74594 61.24360

90.56604 94.33962

850.53743 990.36455

2.7747E+04 3.8444E+04

108

43.60947

64.07384

101.88679

1340.57814

7.3795E+04

120

44.91528

67.93642

113.20755

2104.89599

1.9623E+05

132 144

45.95427 46.78096

71.38371 74.46038

124.52830 135.84906

3297.00464 5156.33978

5.2178E+05 1.3874E+06

180

48.37851

81.84409

169.81132

1.9634E+04

2.6080E+07

240 360

49.48291 49.94742

89.71927 96.70363

226.41509 339.62264

1.8143E+05 1.5459E+07

3.4660E+09 6.1217E+13

480

49.99465

98.94307

452.83019

1.3171E+09

1.0812E+18

600

49.99946

99.66111

566.03774

1.1221E+11

1.9096E+22

10.00% Time Value of Money Factors Geometric Series - Present Worth 10.00%

TABLE A-B-4

j

Time Value of Money Factors Geometric Series - Present Worth 4% 5% 6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

n 1

0.90909

0.90909

0.90909

0.90909

0.90909

2 3

1.76860 2.58122

1.77686 2.60518

1.78512 2.62930

1.81818 2.72727

1.85950 2.85312

4

3.34951

3.39586

3.44278

3.63636

3.89190

5

4.07590

4.15059

4.22668

4.54545

4.97789

6 7

4.76267 5.41198

4.87102 5.55870

4.98207 5.71000

5.45455 6.36364

6.11325 7.30022

8

6.02587

6.21512

6.41145

7.27273

8.54113

9

6.60628

6.84171

7.08740

8.18182

9.83846

10 11

7.15503 7.67385

7.43981 8.01073

7.73877 8.36645

9.09091 10.00000

11.19475 12.61270

12

8.16436

8.55570

8.97130

10.90909

14.09509

13 14

8.62813 9.06659

9.07589 9.57244

9.55417 10.11583

11.81818 12.72727

15.64487 17.26509

15

9.48114

10.04642

10.65708

13.63636

18.95896

16

9.87308

10.49886

11.17864

14.54545

20.72982

17 18

10.24364 10.59398

10.93073 11.34297

11.68123 12.16555

15.45455 16.36364

22.58117 24.51668

19

10.92522

11.73647

12.63226

17.27273

26.54017

20

11.23839

12.11208

13.08199

18.18182

28.65563

21 22

11.53448 11.81442

12.47063 12.81287

13.51538 13.93300

19.09091 20.00000

30.86725 33.17940

23

12.07909

13.13956

14.33543

20.90909

35.59664

24 25

12.32932 12.56590

13.45140 13.74906

14.72324 15.09694

21.81818 22.72727

38.12376 40.76575

26

12.78958

14.03319

15.45705

23.63636

43.52783

27

13.00106

14.30441

15.80407

24.54545

46.41546

28 29

13.20100 13.39003

14.56330 14.81042

16.13846 16.46070

25.45455 26.36364

49.43434 52.59045

30

13.56876

15.04631

16.77122

27.27273

55.89002

36 40

14.45402 14.89870

16.25279 16.88904

18.41108 19.31845

32.72727 36.36364

79.08675 98.36852

48

15.53791

17.85579

20.77553

43.63636

148.91859

50

15.65769

18.04629

21.07717

45.45455

164.62383

52 55

15.76476 15.90444

18.21986 18.45174

21.35728 21.74040

47.27273 50.00000

181.78927 210.57570

j n

4% 5% 6% 10% 15% To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

60

16.09085

18.77305

22.29149

54.54545

267.96474

72 75

16.37292 16.41841

19.29792 19.38937

23.26344 23.44608

65.45455 68.18182

470.90920 540.94029

80

16.47912

19.51610

23.70880

72.72727

680.55528

84

16.51681

19.59826

23.88661

76.36364

816.87970

90 96

16.55964 16.59022

19.69610 19.77012

24.10849 24.28615

81.81818 87.27273

1072.68244 1406.67449

100

16.60558

19.80915

24.38446

90.90909

1684.29792

108 120

16.62767 16.64677

19.86846 19.92473

24.54232 24.70656

98.18182 109.09091

2412.12984 4126.18446

132

16.65652

19.95693

24.81186

120.00000

7048.22688

144

16.66149

19.97535

24.87937

130.90909

1.2030E+04

180 240

16.66598 16.66664

19.99538 19.99972

24.96821 24.99656

163.63636 218.18182

5.9678E+04 8.5952E+05

360

16.66667

20.00000

24.99996

327.27273

1.7819E+08

480

16.66667

20.00000

25.00000

436.36364

3.6941E+10

600

16.66667

20.00000

25.00000

545.45455

7.6581E+12

15.00% Time Value of Money Factors Geometric Series - Present Worth 15.00%

TABLE A-B-5 Time Value of Money Factors Geometric Series - Present Worth j

4%

5%

6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

n 1 2

0.86957 1.65595

0.86957 1.66352

0.86957 1.67108

0.86957 1.70132

0.86957 1.73913

3

2.36712

2.38843

2.40986

2.49692

2.60870

4

3.01027

3.05030

3.09083

3.25792

3.47826

5 6

3.59190 4.11789

3.65462 4.20640

3.71850 4.29706

3.98584 4.68211

4.34783 5.21739

7

4.59357

4.71019

4.83033

5.34810

6.08696

8 9

5.02375 5.41278

5.17017 5.59016

5.32187 5.77494

5.98514 6.59448

6.95652 7.82609

10

5.76460

5.97362

6.19255

7.17733

8.69565

11

6.08277

6.32374

6.57748

7.73484

9.56522

12 13

6.37051 6.63072

6.64342 6.93529

6.93229 7.25933

8.26811 8.77819

10.43478 11.30435

14

6.86604

7.20179

7.56077

9.26609

12.17391

15

7.07885

7.44511

7.83862

9.73278

13.04348

16 17

7.27131 7.44536

7.66728 7.87012

8.09473 8.33080

10.17919 10.60618

13.91304 14.78261

18

7.60276

8.05533

8.54839

11.01460

15.65217

19 20

7.74511 7.87383

8.22443 8.37883

8.74895 8.93381

11.40527 11.77896

16.52174 17.39130

21

7.99025

8.51980

9.10421

12.13639

18.26087

22

8.09553

8.64851

9.26127

12.47829

19.13043

23 24

8.19074 8.27684

8.76603 8.87333

9.40604 9.53948

12.80532 13.11813

20.00000 20.86957

25

8.35471

8.97131

9.66248

13.41734

21.73913

26

8.42513

9.06076

9.77585

13.70355

22.60870

27 28

8.48881 8.54640

9.14243 9.21700

9.88035 9.97667

13.97731 14.23916

23.47826 24.34783

29

8.59849

9.28509

10.06545

14.48963

25.21739

30 36

8.64559 8.84730

9.34725 9.62183

10.14729 10.52003

14.72921 15.96313

26.08696 31.30435

40

8.92797

9.73718

10.68445

16.62072

34.78261

48

9.01801

9.87306

10.88881

17.63200

41.73913

50 52

9.03129 9.04215

9.89418 9.91178

10.92224 10.95065

17.83343 18.01773

43.47826 45.21739

55

9.05485

9.93285

10.98545

18.26521

47.82609

j n

4%

5%

6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

60 72

9.06909 9.08438

9.95739 9.98570

11.02750 11.07967

18.61094 19.18519

52.17391 62.60870

75

9.08608

9.98911

11.08649

19.28691

65.21739

80

9.08799

9.99309

11.09473

19.42902

69.56522

84 90

9.08896 9.08984

9.99520 9.99722

11.09929 11.10386

19.52203 19.63393

73.04348 78.26087

96

9.09032

9.99839

11.10666

19.71963

83.47826

100 108

9.09052 9.09073

9.99888 9.99946

11.10790 11.10944

19.76530 19.83554

86.95652 93.91304

120

9.09086

9.99982

11.11048

19.90353

104.34783

132

9.09089

9.99994

11.11087

19.94341

114.78261

144 180

9.09090 9.09091

9.99998 10.00000

11.11102 11.11111

19.96680 19.99330

125.21739 156.52174

240

9.09091

10.00000

11.11111

19.99953

208.69565

360 480

9.09091 9.09091

10.00000 10.00000

11.11111 11.11111

20.00000 20.00000

313.04348 417.39130

600

9.09091

10.00000

11.11111

20.00000

521.73913

4.00% Time Value of Money Factors Geometric Series - Future Worth 4.00%

TABLE A-C-1 Time Value of Money Factors Geometric Series - Future Worth j

4% 5% 6% 10% 15% To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

n 1

1.00000

1.00000

1.00000

1.00000

1.00000

2

2.08000

2.09000

2.10000

2.14000

2.19000

3

3.24480

3.27610

3.30760

3.43560

3.60010

4 5

4.49946 5.84929

4.56477 5.96287

4.63092 6.07863

4.90402 6.56428

5.26498 7.22458

6

7.29992

7.47766

7.66000

8.43737

9.52492

7 8

8.85723 10.52745

9.11686 10.88864

9.38492 11.26395

10.54642 12.91700

12.21898 15.36776

9

12.31712

12.80164

13.30836

15.57726

19.04150

10

14.23312

14.86503

15.53017

18.55830

23.32103

11 12

16.28269 18.47345

17.08853 19.48241

17.94223 20.55821

21.89438 25.62327

28.29943 34.08380

13

20.81342

22.05756

23.39274

29.78663

40.79740

14 15

23.31103 25.97515

24.82552 27.79847

26.46138 29.78073

34.43036 39.60508

48.58208 57.60107

16

28.81510

30.98933

33.36852

45.36653

68.04218

17

31.84068

34.41178

37.24361

51.77616

80.12149

18 19

35.06221 38.49051

38.08027 42.01010

41.42613 45.93752

58.90168 66.81766

94.08761 110.22657

20

42.13698

46.21746

50.80062

75.60628

128.86740

21

46.01359

50.71945

56.03978

85.35803

150.38864

22 23

50.13290 54.50813

55.53419 60.68082

61.68093 67.75171

96.17260 108.15978

175.22570 203.87947

24

59.15317

66.17958

74.28152

121.44048

236.92611

25 26

64.08260 69.31174

72.05186 78.32029

81.30172 88.84566

136.14783 152.42845

275.02833 318.94842

27

74.85668

85.00877

96.94887

170.44376

369.56315

28

80.73432

92.14258

105.64917

190.37150

427.88099

29 30

86.96240 93.55954

99.74841 107.85449

114.98682 125.00468

212.40736 236.76675

495.06184 572.43977

36

142.05920

168.78836

202.16597

446.81247

1354.98109

40

184.65464

223.89681

274.23487

674.30392

2391.47751

48 50

303.25515 341.66747

383.07414 436.07164

491.16717 565.67355

1507.44509 1838.06949

7389.36531 9786.82507

52

384.32944

495.62196

650.51483

2239.27239

1.2959E+04

55

457.25979

598.92640

800.19773

3006.87959

1.9736E+04

j n

4%

5%

6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

60

606.90158

815.95585

1123.40317

4899.36687

3.9759E+04

72

1166.00278

1670.28717

2476.77264

1.5646E+04

2.1308E+05

75 80

1366.24433 1773.06147

1988.74313 2651.16420

3005.58331 4137.30972

2.0883E+04 3.3756E+04

3.2413E+05 6.5207E+05

84

2177.94269

3327.72366

5329.99997

4.9535E+04

1.1406E+06

90

2952.63462

4661.10317

7767.25889

8.7982E+04

2.6385E+06

96 100

3985.09305 4856.24502

6501.45689 8099.63097

1.1279E+04 1.4440E+04

1.5615E+05 2.2884E+05

6.1034E+06 1.0675E+07

108

7177.79516

1.2517E+04

2.3584E+04

4.9118E+05

3.2656E+07

120

1.2769E+04

2.3825E+04

4.8876E+04

1.5433E+06

1.7472E+08

132 144

2.2488E+04 3.9276E+04

4.4942E+04 8.4161E+04

1.0062E+05 2.0612E+05

4.8464E+06 1.5215E+07

9.3481E+08 5.0015E+09

180

2.0148E+05

5.3533E+05

1.7366E+06

4.7045E+08

7.6598E+11

240 360

2.8260E+06 4.6911E+08

1.0949E+07 4.1121E+09

5.8595E+07 6.4361E+10

1.4325E+11 1.3281E+16

3.3581E+15 6.4540E+22

480

6.9217E+10

1.4671E+12

7.0103E+13

1.2312E+21

1.2404E+30

600

9.5746E+12

5.1545E+14

7.6293E+16

1.1415E+26

2.3840E+37

5.00% Time Value of Money Factors Geometric Series - Future Worth 5.00%

TABLE A-C-2

j

Time Value of Money Factors Geometric Series - Future Worth 4% 5% 6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

n 1

1.00000

1.00000

1.00000

1.00000

1.00000

2 3

2.09000 3.27610

2.10000 3.30750

2.11000 3.33910

2.15000 3.46750

2.20000 3.63250

4

4.56477

4.63050

4.69707

4.97188

5.33500

5

5.96287

6.07753

6.19440

6.68457

7.35076

6 7

7.47766 9.11686

7.65769 9.38067

7.84235 9.65298

8.62931 10.83233

9.72965 12.52919

8

10.88864

11.25680

11.63926

13.32267

15.81567

9

12.80164

13.29710

13.81507

16.13239

19.66548

10 11

14.86503 17.08853

15.51328 17.91784

16.19531 18.79592

19.29696 22.85555

24.16663 29.42052

12

19.48241

20.52407

21.63401

26.85144

35.54394

13 14

22.05756 24.82552

23.34613 26.39909

24.72791 28.09724

31.33244 36.35133

42.67138 50.95774

15

27.79847

29.69897

31.76300

41.96640

60.58133

16

30.98933

33.26285

35.74771

48.24197

71.74746

17 18

34.41178 38.08027

37.10887 41.25633

40.07545 44.77199

55.24904 63.06596

84.69246 99.68834

19

42.01010

45.72577

49.86493

71.77918

117.04821

20 21

46.21746 50.71945

50.53900 55.71925

55.38378 61.36010

81.48404 92.28575

137.13240 160.35555

22

55.53419

61.29118

67.82767

104.30028

187.19485

23

60.68082

67.28100

74.82259

117.65557

218.19934

24 25

66.17958 72.05186

73.71657 80.62750

82.38347 90.55158

132.49265 148.96702

254.00076 295.32598

26

78.32029

88.04523

99.37103

167.25008

343.01123

27

85.00877

96.00316

108.88896

187.53076

398.01859

28 29

92.14258 99.74841

104.53678 113.68375

119.15576 130.22523

210.01729 234.93915

461.45483 534.59318

30

107.85449

123.48407

142.15488

262.54920

618.89830

36 40

168.78836 223.89681

198.57655 268.19005

235.54359 324.57292

502.41729 764.38534

1473.60036 2608.23558

48

383.07414

475.48661

599.26021

1732.31928

8089.99442

50

436.07164

546.06666

695.27545

2118.46906

1.0722E+04

52 55

495.62196 598.92640

626.12003 766.62829

805.40771 1001.46907

2588.00247 3488.47023

1.4205E+04 2.1650E+04

j n

4% 5% 6% 10% 15% To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

60

815.95585

1067.38205

1430.85050

5716.04907

4.3653E+04

72 75

1670.28717 1988.74313

2300.23777 2773.76328

3283.25810 4022.42349

1.8441E+04 2.4661E+04

2.3422E+05 3.5634E+05

80

2651.16420

3776.10980

5623.45524

3.9977E+04

7.1701E+05

84

3327.72366

4819.37931

7332.27629

5.8776E+04

1.2543E+06

90 96

4661.10317 6501.45689

6919.74558 9891.32894

1.0873E+04 1.6057E+04

1.0465E+05 1.8608E+05

2.9019E+06 6.7131E+06

100

8099.63097

1.2524E+04

2.0780E+04

2.7298E+05

1.1742E+07

108 120

1.2517E+04 2.3825E+04

1.9984E+04 3.9876E+04

3.4651E+04 7.3928E+04

5.8691E+05 1.8472E+06

3.5921E+07 1.9219E+08

132

4.4942E+04

7.8772E+04

1.5631E+05

5.8067E+06

1.0283E+09

144

8.4161E+04

1.5432E+05

3.2807E+05

1.8241E+07

5.5016E+09

180 240

5.3533E+05 1.0949E+07

1.1173E+06 2.7826E+07

2.9379E+06 1.0624E+08

5.6443E+08 1.7190E+11

8.4258E+11 3.6939E+15

360

4.1121E+09

1.4563E+10

1.2461E+11

1.5937E+16

7.0994E+22

480

1.4671E+12

6.7751E+12

1.3874E+14

1.4775E+21

1.3645E+30

600

5.1545E+14

2.9549E+15

1.5207E+17

1.3697E+26

2.6224E+37

6.00% Time Value of Money Factors Geometric Series - Future Worth 6.00%

TABLE A-C-3 Time Value of Money Factors Geometric Series - Future Worth j

4%

5%

6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

n 1 2

1.00000 2.10000

1.00000 2.11000

1.00000 2.12000

1.00000 2.16000

1.00000 2.21000

3

3.30760

3.33910

3.37080

3.49960

3.66510

4

4.63092

4.69707

4.76406

5.04058

5.40588

5 6

6.07863 7.66000

6.19440 7.84235

6.31238 8.02935

6.80711 8.82605

7.47924 9.93935

7

9.38492

9.65298

9.92963

11.12717

12.84877

8 9

11.26395 13.30836

11.63926 13.81507

12.02904 14.34463

13.74352 16.71172

16.27972 20.31553

10

15.53017

16.19531

16.89479

20.07237

25.05233

11

17.94223

18.79592

19.69932

23.87045

30.60103

12 13

20.55821 23.39274

21.63401 24.72791

22.77958 26.15855

28.15580 32.98357

37.08948 44.66510

14

26.46138

28.09724

29.86100

38.41486

53.49780

15

29.78073

31.76300

33.91356

44.51725

63.78337

16 17

33.36852 37.24361

35.74771 40.07545

38.34493 43.18598

51.36553 59.04244

75.74744 89.64990

18

41.42613

44.77199

48.46991

67.63945

105.79016

19 20

45.93752 50.80062

49.86493 55.38378

54.23244 60.51199

77.25774 88.00911

124.51302 146.21558

21

56.03978

61.36010

67.34984

100.01716

171.35505

22

61.68093

67.82767

74.79040

113.41844

200.45787

23 24

67.75171 74.28152

74.82259 82.38347

82.88136 91.67399

128.36382 145.01995

234.13009 273.06935

25

81.30172

90.55158

101.22337

163.57088

318.07869

26 27

88.84566 96.94887

99.37103 108.88896

111.58864 122.83334

184.21984 207.19121

370.08236 430.14410

28

105.64917

119.15576

135.02569

232.73267

499.48806

29

114.98682

130.22523

148.23891

261.11763

579.52296

30 36

125.00468 202.16597

142.15488 235.54359

162.55164 276.69912

292.64778 569.13571

671.86979 1611.16222

40

274.23487

324.57292

388.14030

874.33844

2861.97587

48

491.16717

599.26021

742.36400

2015.58405

8922.29822

50 52

565.67355 650.51483

695.27545 805.40771

868.87520 1015.31890

2474.26747 3033.65117

1.1836E+04 1.5694E+04

55

800.19773

1001.46907

1279.02612

4110.22052

2.3944E+04

j n

4%

5%

6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

60 72

1123.40317 2476.77264

1430.85050 3283.25810

1867.22778 4508.67499

6787.34872 2.2230E+04

4.8345E+04 2.5988E+05

75

3005.58331

4022.42349

5593.65006

2.9821E+04

3.9549E+05

80

4137.30972

5623.45524

7984.60328

4.8565E+04

7.9606E+05

84 90

5329.99997 7767.25889

7332.27629 1.0873E+04

1.0584E+04 1.6087E+04

7.1637E+04 1.2809E+05

1.3929E+06 3.2231E+06

96

1.1279E+04

1.6057E+04

2.4340E+04

2.2859E+05

7.4572E+06

100 108

1.4440E+04 2.3584E+04

2.0780E+04 3.4651E+04

3.2010E+04 5.5100E+04

3.3603E+05 7.2498E+05

1.3044E+07 3.9908E+07

120

4.8876E+04

7.3928E+04

1.2319E+05

2.2905E+06

2.1354E+08

132

1.0062E+05

1.5631E+05

2.7267E+05

7.2193E+06

1.1425E+09

144 180

2.0612E+05 1.7366E+06

3.2807E+05 2.9379E+06

5.9855E+05 6.0957E+06

2.2719E+07 7.0481E+08

6.1128E+09 9.3620E+11

240

5.8595E+07

1.0624E+08

2.6811E+08

2.1484E+11

4.1043E+15

360 480

6.4361E+10 7.0103E+13

1.2461E+11 1.3874E+14

4.3763E+11 6.3497E+14

1.9921E+16 1.8468E+21

7.8882E+22 1.5161E+30

600

7.6293E+16

1.5207E+17

8.6370E+17

1.7122E+26

2.9138E+37

10.00% Time Value of Money Factors Geometric Series - Future Worth 10.00%

TABLE A-C-4 Time Value of Money Factors Geometric Series - Future Worth j

4% 5% 6% 10% 15% To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

n 1

1.00000

1.00000

1.00000

1.00000

1.00000

2

2.14000

2.15000

2.16000

2.20000

2.25000

3

3.43560

3.46750

3.49960

3.63000

3.79750

4 5

4.90402 6.56428

4.97188 6.68457

5.04058 6.80711

5.32400 7.32050

5.69812 8.01694

6

8.43737

8.62931

8.82605

9.66306

10.83000

7 8

10.54642 12.91700

10.83233 13.32267

11.12717 13.74352

12.40093 15.58974

14.22606 18.30868

9

15.57726

16.13239

16.71172

19.29230

23.19857

10

18.55830

19.29696

20.07237

23.57948

29.03631

11 12

21.89438 25.62327

22.85555 26.85144

23.87045 28.15580

28.53117 34.23740

35.98549 44.23643

13

29.78663

31.33244

32.98357

40.79957

54.01033

14 15

34.43036 39.60508

36.35133 41.96640

38.41486 44.51725

48.33180 56.96248

65.56415 79.19627

16

45.36653

48.24197

51.36553

66.83597

95.25296

17

51.77616

55.24904

59.04244

78.11454

114.13587

18 19

58.90168 66.81766

63.06596 71.77918

67.63945 77.25774

90.98047 105.63843

136.31073 162.31725

20

75.60628

81.48404

88.00911

122.31818

192.78075

21

85.35803

92.28575

100.01716

141.27750

228.42536

22 23

96.17260 108.15978

104.30028 117.65557

113.41844 128.36382

162.80550 187.22632

270.08942 318.74310

24

121.44048

132.49265

145.01995

214.90326

375.50887

25 26

136.14783 152.42845

148.96702 167.25008

163.57088 184.21984

246.24332 281.70235

441.68493 518.77238

27

170.44376

187.53076

207.19121

321.79077

608.50641

28

190.37150

210.01729

232.73267

367.07984

712.89237

29 30

212.40736 236.76675

234.93915 262.54920

261.11763 292.64778

418.20881 475.89279

834.24722 975.24739

36

446.81247

502.41729

569.13571

1011.68773

2444.78343

40 48

674.30392 1507.44509

764.38534 1732.31928

874.33844 2015.58405

1645.79111 4233.47929

4452.08581 1.4448E+04

50

1838.06949

2118.46906

2474.26747

5335.94786

1.9325E+04

52

2239.27239

2588.00247

3033.65117

6714.75678

2.5822E+04

55

3006.87959

3488.47023

4110.22052

9452.95712

3.9811E+04

j n

4%

5%

6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

60

4899.36687

5716.04907

6787.34872

1.6608E+04

8.1590E+04

72

1.5646E+04

1.8441E+04

2.2230E+04

6.2548E+04

4.5000E+05

75 80

2.0883E+04 3.3756E+04

2.4661E+04 3.9977E+04

2.9821E+04 4.8565E+04

8.6720E+04 1.4897E+05

6.8802E+05 1.3940E+06

84

4.9535E+04

5.8776E+04

7.1637E+04

2.2902E+05

2.4499E+06

90

8.7982E+04

1.0465E+05

1.2809E+05

4.3470E+05

5.6992E+06

96 100

1.5615E+05 2.2884E+05

1.8608E+05 2.7298E+05

2.2859E+05 3.3603E+05

8.2144E+05 1.2528E+06

1.3240E+07 2.3211E+07

108

4.9118E+05

5.8691E+05

7.2498E+05

2.9003E+06

7.1254E+07

120

1.5433E+06

1.8472E+06

2.2905E+06

1.0114E+07

3.8253E+08

132 144

4.8464E+06 1.5215E+07

5.8067E+06 1.8241E+07

7.2193E+06 2.2719E+07

3.4915E+07 1.1954E+08

2.0508E+09 1.0985E+10

180

4.7045E+08

5.6443E+08

7.0481E+08

4.6192E+09

1.6846E+12

240 360

1.4325E+11 1.3281E+16

1.7190E+11 1.5937E+16

2.1484E+11 1.9921E+16

1.8753E+12 2.6078E+17

7.3876E+15 1.4199E+23

480

1.2312E+21

1.4775E+21

1.8468E+21

3.2236E+22

2.7289E+30

600

1.1415E+26

1.3697E+26

1.7122E+26

3.7357E+27

5.2449E+37

15.00% Time Value of Money Factors Geometric Series - Future Worth 15.00%

TABLE A-C-5

j

Time Value of Money Factors Geometric Series - Future Worth 4% 5% 6%

10%

15%

To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

n 1

1.00000

1.00000

1.00000

1.00000

1.00000

2 3

2.19000 3.60010

2.20000 3.63250

2.21000 3.66510

2.25000 3.79750

2.30000 3.96750

4

5.26498

5.33500

5.40588

5.69812

6.08350

5

7.22458

7.35076

7.47924

8.01694

8.74503

6 7

9.52492 12.21898

9.72965 12.52919

9.93935 12.84877

10.83000 14.22606

12.06814 16.19143

8

15.36776

15.81567

16.27972

18.30868

21.28016

9

19.04150

19.66548

20.31553

23.19857

27.53121

10 11

23.32103 28.29943

24.16663 29.42052

25.05233 30.60103

29.03631 35.98549

35.17876 44.50114

12

34.08380

35.54394

37.08948

44.23643

55.82870

13 14

40.79740 48.58208

42.67138 50.95774

44.66510 53.49780

54.01033 65.56415

69.55325 86.13903

15

57.60107

60.58133

63.78337

79.19627

106.13559

16

68.04218

71.74746

75.74744

95.25296

130.19299

17 18

80.12149 94.08761

84.69246 99.68834

89.64990 105.79016

114.13587 136.31073

159.07955 193.70275

19

110.22657

117.04821

124.51302

162.31725

235.13362

20 21

128.86740 150.38864

137.13240 160.35555

146.21558 171.35505

192.78075 228.42536

284.63543 343.69729

22

175.22570

187.19485

200.45787

270.08942

414.07340

23

203.87947

218.19934

234.13009

318.74310

497.82915

24 25

236.92611 275.02833

254.00076 295.32598

273.06935 318.07869

375.50887 441.68493

597.39498 715.62940

26

318.94842

343.01123

370.08236

518.77238

855.89277

27

369.56315

398.01859

430.14410

608.50641

1022.13348

28 29

427.88099 495.06184

461.45483 534.59318

499.48806 579.52296

712.89237 834.24722

1218.98882 1451.90275

30

572.43977

618.89830

671.86979

975.24739

1727.26362

36 40

1354.98109 2391.47751

1473.60036 2608.23558

1611.16222 2861.97587

2444.78343 4452.08581

4794.31884 9316.99291

48

7389.36531

8089.99442

8922.29822

1.4448E+04

3.4201E+04

50

9786.82507

1.0722E+04

1.1836E+04

1.9325E+04

4.7116E+04

52 55

1.2959E+04 1.9736E+04

1.4205E+04 2.1650E+04

1.5694E+04 2.3944E+04

2.5822E+04 3.9811E+04

6.4803E+04 1.0424E+05

j n

4% 5% 6% 10% 15% To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 To Find P Given A1 (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n) (P|A1i%, j%,n)

60

3.9759E+04

4.3653E+04

4.8345E+04

8.1590E+04

2.2873E+05

72 75

2.1308E+05 3.2413E+05

2.3422E+05 3.5634E+05

2.5988E+05 3.9549E+05

4.5000E+05 6.8802E+05

1.4685E+06 2.3265E+06

80

6.5207E+05

7.1701E+05

7.9606E+05

1.3940E+06

4.9914E+06

84

1.1406E+06

1.2543E+06

1.3929E+06

2.4499E+06

9.1664E+06

90 96

2.6385E+06 6.1034E+06

2.9019E+06 6.7131E+06

3.2231E+06 7.4572E+06

5.6992E+06 1.3240E+07

2.2717E+07 5.6049E+07

100

1.0675E+07

1.1742E+07

1.3044E+07

2.3211E+07

1.0211E+08

108 120

3.2656E+07 1.7472E+08

3.5921E+07 1.9219E+08

3.9908E+07 2.1354E+08

7.1254E+07 3.8253E+08

3.3736E+08 2.0055E+09

132

9.3481E+08

1.0283E+09

1.1425E+09

2.0508E+09

1.1803E+10

144

5.0015E+09

5.5016E+09

6.1128E+09

1.0985E+10

6.8890E+10

180 240

7.6598E+11 3.3581E+15

8.4258E+11 3.6939E+15

9.3620E+11 4.1043E+15

1.6846E+12 7.3876E+15

1.3188E+13 7.7089E+16

360

6.4540E+22

7.0994E+22

7.8882E+22

1.4199E+23

2.2224E+24

480

1.2404E+30

1.3645E+30

1.5161E+30

2.7289E+30

5.6952E+31

600

2.3840E+37

2.6224E+37

2.9138E+37

5.2449E+37

1.3682E+39

Appendix B Online PDF

Interest Rate Equations Discrete Compounding When cash flow frequency and/or compounding frequency are not annual, calculations require either the period interest rate approach or the effective annual interest rate approach. When cash flow frequency does not coincide with compounding frequency, the effective rate per cash flow period is required in calculations. Effective Rate per Cash Flow Period

Period Interest Rate

Effective Annual Interest Rate

Period interest rate = r/m

ieff = (1 + r/m)m − 1

i = (1 + r/m)m/k − 1

Where:

Where:

Where:

r = nominal annual interest rate

ieff = effective annual interest rate

i = interest rate per cash flow period

m = number of compounding/interest periods per r = nominal annual interest rate year m = number of compounding/interest periods per year

r = nominal annual interest rate m = number of compounding/interest periods per year k = number of cash flows per year

Continuous Compounding and Continuous Cash Flows When cash flows and compounding are approximated as occurring continuously throughout the year, the following relationships apply. With continuous compounding at nominal annual rate, r, the effective annual rate is ieff = e

r

− 1

¯ ¯ ¯ When A dollars flow uniformly and continuously during a year, the annual equivalent end‐of‐year cash flow, A, is r ¯ ¯ ¯ A (e − 1)

A = r

Useful Excel® Financial Functions Present Worth (Solve for P) 1. To compute the present worth of a single future sum or a uniform series of cash flows or both, use the PV function: =PV(rate,nper,pmt,fv,type). 2. To compute the present worth of multiple cash flows, use the NPV function: =NPV(rate,valuel,value2, . . .) or =NPV(rate,valuel:valueN), noting that the value obtained occurs one time period before value1.

Annual Worth (Solve for A) To compute the uniform series equivalent of a single sum occurring at the present or n periods in the future or both, use the PMT function: =PMT(rate,nper,pv,fv,type).

Future Worth (Solve for F) To compute the future worth equivalent of a uniform series of cash flows or a present worth amount or both, use the FV function: =FV(rate,nper,pmt,pv,type)

Rate of Return (Solve for i) 1. To compute the interest rate that yields a present worth of zero for a given combination of (P & F), (P & A), (A & F) or (P, A & F) for a given value of n, use the RATE function: =RATE(nper,pmt,pv,fv,type,guess) 2. To compute the interest rate that makes the present worth of a series of cash flows equal zero, use the IRR function: =IRR(valuel:valueN,guess) 3. To compute the interest rate that makes the future worth of an investment of capital at time zero equal to the future worth of reinvested returns, when the returns are reinvested at a rate called the minimum attractive rate of return (MARR), use the MIRR function: =MIRR(values,finance_rate,MARR)

Effective Interest Rate (Solve for ieff) To compute the effective interest rate of a nominal rate (r) compounded m times per year, use the EFFECT function: =EFFECT(r,m)

Determining the Number of Payments (Solve for n) To solve for the number of interest periods, given an interest rate, that yields a present worth of zero for a given combination of (P & F), (P & A), (A & F) or (P, A & F), use the NPER function: =NPER(rate,pmt,pv,fv,type) rate: compound interest rate

guess:

nper: number of compound interest periods

finance_rate: interest rate paid on borrowed investment capital (for our purposes, this is generally zero or left blank)

pv:

net present value

fv: net future value pmt: magnitude of uniform series of cash flows type: 0 or omitted denotes end‐of‐period cash flow 1 denotes beginning‐of‐period cash flow Single Cash Flows Present Worth P = F(1 + i)−n of a Future Payment Future Worth F = P(1 + i)n of a Present Payment Series Cash Flows

Formula

Formula

r: m:

best estimate of IRR, if needed

nominal interest rate, usually per year number of compounding periods, usually in a year

Factor Notation P = F(P|F i%,n)

Excel® Notation =PV(i%,n,,–F)

F = P(F|P i%,n)

=FV(i%, n,,−P)

Factor Notation

Excel® Notation

Single Cash Flows Irregular Series Cash Flows

Formula P

=

A1 (1 + i)

−1

+A3 (1 + i)

+ A2 (1 + i)

−3

+An−1 (1 + i) +An (1 + i)

+ …

Factor Notation −2

n

P = ∑ At (P |F  i%,t) t=1

Excel® Notation =NPV(i%,A1,A2,A3, …,An)

−(n−1)

−n

Uniform Series Cash Flows n

(1+i) −1

Present Worth of Uniform Series

P = A[

Uniform Series from Present Value

A = P[

Future Worth of Uniform Series

F = A[

Uniform Series from Future Value

A = F [

i(1+i)

i(1+i)

]

n

n

]

n

(1+i) −1

n

(1+i) −1

]

i

i n

(1+i) −1

]

P = A(P|A i%,n)

=PV(i%,n,−A)

A = P(A|P i%,n)

=PMT(i%,n,−P)

F = A(F|A i%,n)

=FV(i%,n,−A)

A = F(A|F i%,n)

=PMT(i%,n,,−F)

P = G(P|G i%,n)

=NPV(i%,0,G,2G, …,(n−1)G)

A = G(A|G i%,n)

=PMT(i%,n, −NPV(i%,0,G,2G, …,(n−1)G))

Gradient Series Cash Flows Present Worth of a Gradient Series

1−(1+ni)(1+i)

P = G[

i

−n

]

2

n

Uniform Series Equivalent of a Gradient Series

(1+i) −(1+ni)

A = G[

n

i[(1+i) −1]

]

Single Cash Flows

Formula

Factor Notation

n

(1+i) −(1+ni)

Future Worth of a Gradient Series

F = G[

i

]

2

Excel® Notation

F = G(F|G i%,n)

=FV(i%,n,, −NPV(i%,0,G,2G, …,(n−1)G))

P = A1(P|A1 i%,j%,n)

=NPV(i%,A,(1+j)A, (1+j)2A,…, (1+j)n−1A)

A = A1(A|A1 i%,j%,n)

=PMT(i%,n, −NPV(i%,A,(1+j)A, (1+j)2A,…, (1+j)n−1A))

F = A1(F|A1 i%,j%,n)

=FV(i%,n,, −NPV(i%,A,(1+j)A, (1+j)2A,…, (1+j)n−1A))

Geometric Series Cash Flows Present Worth of a Geometric Series Uniform Series Equivalent of a Geometric Series

⎧ ⎪ P = ⎨ ⎩ ⎪

n

1−(1 + j) (1 + i)

A1 [

i  ≠  j

] 

nA1 /(1 + i)

⎧ ⎪ ⎪ A1 [ ⎪ ⎪

i = j

n

1−(1 + j) (1 + i)

−n

i(1+i)

] [

i−j

n

n

(1+i) −1

] 

i  ≠  j

A = ⎨ ⎪ ⎪ ⎪ ⎩ nA1 ( ⎪

i(1 + i)

n−1

n

(1 + i) − 1

n

Future Worth of a Geometric Series

−n

i−j

⎧ F = ⎨ ⎩

(1 + i) − (1 + j)

A1 [

i−j

nA1 (1 + i)

n−1

i = j

)

n

] 

i  ≠  j

i = j

Index A Abbott Laboratories, 245–246 Abbott, Wallace C., 245 Absent nonmonetary criteria, 78n Accelerated depreciation, 160, 163 Accounting, 260, 285–286 activity‐based, 302 balance sheets in, 286–289 cost, 285 cost accounting principles in, 297–302 financial, 285 fundamental accounting equation in, 286 income statements in, 290–292 managerial, 285 ratio analysis in, 292–296 Accounts receivable turnover, 293, 294 Acid test ratio, 293, 294 Acme Brick Company, 194–195 Activity‐based accounting, 302 Activity based costing, 301–302 Actual dollars, 208 Ad valorem taxes, 176

After‐tax analysis, 211–212 before‐tax versus, 79 with bonus depreciation, 196–197 with borrowed capital, 188–193, 213–216 comparison of alternatives with different property classes, 186–187 comparison of manual versus automated solutions, 187 with Excel® for borrowed funds—single alternative, 189 with MACRS depreciation, 183 with nondepreciable expenditures, 184–185 with retained earnings (no borrowing), 180–188 multiple alternatives, 185–187 single alternative, 181–185 with Section 179 expenses deduction, 198–200 After‐tax cash flow, 180 After‐tax present worth, sensitivity of, 226 Alere Inc., 245 Allowances, depreciation, 157 American Institute of Certified Public Accountants, 286 American Tax Cuts and Jobs Act, 177, 198 Amortization, 158. See also Depreciation Analogy, estimating by, 284 Annual interest rate, 46 effective, 46, 48–50 nominal, 46, 48

Annual worth, 78, 114–118 defined, 114 of multiple alternatives, 117–118 of single alternative, 114–116 Assets, 286 current, 288 first cost of an, 262 fixed, 286, 288 Association for the Advancement of Cost Engineering International, 11, 282 Autocorrelation, 232 Average costs, 274 Average values, 276

B Balance sheets, 286–289 Before‐tax analysis, 208–210 Before‐Tax‐and‐Loan Cash Flow, 180 Before‐Tax Cash Flow, 180 Before‐tax versus after‐tax analysis, 79 Beginning‐of‐period cash flows, 18 Benefit‐cost analysis, 87–98 Benefit‐cost calculations for multiple alternatives and unequal lives, 92–98 for single alternative, 88–91 Benefit‐cost ratio method, 78, 88 Benefits minus costs (B ‐C), 88

Best, Daniel, 203 Beta distribution, 231 Bezos, Jeffrey, 155 Binary linear programming (BLP), 249 “Bird in the hand…” philosophy, 3 Bond rate, 69 Bonds, 68–71, 265 debenture, 265 purchase price for, 70 rate of return for, 70–71 selling price for, 69 Bonus depreciation, 196 after‐tax analysis with, 196–197 with Modified Accelerated Cost Recovery System (MACRS), 171–172 Book loss, 184 Book value, 159, 184 Borrowed capital, after‐tax analysis with, 188–193, 213–216 Borrowing money, after‐tax effects of Bottom‐up estimating, 284 Break‐even analysis, 221–225, 270 alternative analysis using, 272–274 Break‐even point, 270–272 Break‐even value, 221 Brown, Donald, 6

BuiltRite Tool and Engineering Company balance sheet of, 289 income statement of, 292 Bureau of Labor Statistics, 205, 207

C CAP EX, 247 Capital borrowed, 213 after‐tax analysis using, 188–193 contributed, 288 cost of, 3, 265 earned, 288 Capital budgeting, 245–257 addressing additional constraints in, 252–253 with divisible investments, 254–257 with indivisible investments, 249–253 solving the two‐constraint, 250–251 Capital investment analysis, 2 Capital investments analysis of, 2 economic justification of, 8–12 Capitalized worth, 78, 106–110 defined, 107 for a single alternative, 107–109 Capital leases, 179

Capital rationing problem, 247 Capital recovery factor, 34 Carry‐back rules, 184 Carry‐forward rules, 184 Carson’s Cultery Company, 293

Cash flow(s), 2 after‐tax, 180 before‐tax, 180 before‐tax‐and‐loan, 180 beginning‐of‐period, 18 compounded, 4n continuous, 18, 52 differences between compounding and frequencies of, 50–51 differences in, among investment alternatives, 6–7 discounted, 4, 16 discrete, 18 end‐of‐period, 18 estimating, 11, 260 geometric series of, 42–45 gradient series of, 38–41 irregular, 27–31 irregular incremental analysis, 142 loan, 188 multiple geometric series of cash flows, 42–45 gradient series of cash flows, 38–41 irregular cash flows, 27–31 uniform series of cash flows, 31–37 negative, 5 opportunity cost approaches to replacement analysis in, 145–147 positive, 5

single, 19–25 Cash flow approach, 147 Cash flow diagrams (CFD), 16–18 arrows on, 18 benefits of, 18 Cash flow profiles comparing equivalence between, 73 comparing worth of, with variable interest rates, 73 Cash flow rate of return, 130 Cash flow series, with variable interest rates, 72 Caterpillar, 203–204 Central Limit Theorem, 233 Challengers, 145–146, 147 Classical budgeting, classical capital budgeting problem, 246–248 Collectively exhaustive, 9 combined rate, 208 Comparative statements, 292 Compounded cash flow, 4n Compounded continuously, 51 Compounding, 18, 19 continuous, 18 differences between cash flow frequencies and, 50–51 discrete, 18 Compounding frequency, 46–54 period interest rate approach in, 46–48 ConocoPhillips, 76–77

Constant dollars, 208 Constant purchasing power dollars, 208 Constant‐value dollars, 208 Constant‐worth dollars, 208 Consumer price index (CPI), 205–207 Continuous cash flow, 18, 52 Continuous compounding, 18 Contributed capital, 288 Corporate income‐tax rates, 177–180 determining taxable income, 177–180

Cost(s) average, 274 direct, 267 equivalent uniform annual, 147 first, of an asset, 262 fixed, 268–269 future, 264 of goods sold, 267 indirect, 267 marginal, 6, 274–275 operating and maintenance, 262 opportunity, 147, 264–265 overhead, 267 past, 263 standard, 302 sunk, 7, 146, 263 total, 269, 275 variable, 268–269 Cost accounting, 285 Cost accounting principles, 297–302 activity based costing, 301–302 standard costs, 302 traditional cost allocation methods, 297–301 Cost‐effectiveness analysis, 87 Cost estimation, 11, 280–285 Costing, activity based, 301–302

Cost of capital, 3, 265 weighed average, 10–11 Cultural development, 87 Cumulative cash flow series, 132 Cumulative depreciation charge, 159 Current ratio, 293, 294

D Data general sources of, 284–285 historical, 282 relationship, 282 Data research, 282 Debenture bonds, 265 Debt to equity ratio, 295 Declining balance depreciation, 160 Defender, 145, 147 Deferred payment loans, 66–69 Deferred payments, interest and equity payments in, 67–68 Depreciable property, 157

Depreciation, 155–173, 176, 267 accelerated, 160, 163, 176 bonus, 171–172, 196–197 declining, 160 defined, 156 double declining balance, 161 in economic analysis, 156–158 language of, 157–158 straight‐line, 159–160, 179–180 Depreciation allowances, 157, 211 Descartes’ rule of signs, Internal rate of return, 131 Detailed estimates, 282 Direct costs, 267 Discounted cash flows, 4 methods of, 7 rate of return in, 130 rules for, 4, 13, 16 Discounted payback period (DPBP), 78, 99–106, 222 Excel® and, 101–102 with the Excel® NPER tool to determine, for a single investment opportunity, 100–101 with the Excel® SOLVER tool to calculate, 102–103 salvage value and, 103 for single alternative, 99–103 as supplemental tool, 101 Discount rate, 3

Discrete cash flows, 18 Discrete compounding, 18 Divisible investments, 254 capital budgeting problem with, 254–257 Documentation of estimate, 283 Dollars constant, 208 then‐current, 208 “Do nothing” alternative, 9 Double declining balance depreciation, 161 switching to straight‐line depreciation with Excel® function, 163–164 Doubling the value of an investment, 23–25

E Earned capital, 288 Earning power, 3 versus inflation, 3 Earnings before interest and taxes (EBIT), 296 Earnings before interest, taxes, depreciation, and amortization (EBITDA), 296 Economic analysis, 2 depreciation in, 156–158 interest rates in, 72 in public‐sector decision making, 2 Economic decision analysis, 2

Economic justification, 2 of capital investments, 8–12 Economic obsolescence, 146 Economic services, 87 Economic Stimulus Act (2008), 171 Economic value added, 305–307 Economic worth before‐tax versus after‐tax analysis of, 79 equal versus unequal lives in, 79 equivalence of methods of, 79 incremental methods of, 78–79 methods comparing, 78 of the minimum attractive rate of return, 226 ranking methods of, 78 single alternative, 79–80 Economies of scale, 274–275 Economies of scope, 275 Effective annual interest rate, 46, 48–51 Efrati, Amir, 219 “Either/or” contingent constraint, 252 Electronic data interchange (EDI), 1 End‐of‐period cash flows, 18

Engineering economic analysis, 260 defined, 2 equivalence in, 58–63 principles of, 5–7, 13–14 1. Money has a time value, 6 2. Make investments that are economically justified, 6 3. Choose the mutually exclusive investment alternative that maximizing economic worth, 6 4. Two investment alternatives are equivalent if they have the same economic worth, 6 5. Marginal revenue must exceed marginal cost, 6 6. Continue to invest as long as each additional increment of investment yields a return that is greater than the investor’s TVOM, 6 7. Consider only differences in cash flows among investment alternatives, 6–7 8. Compare investment alternatives over a common period of time, 7 9. Risks and returns tend to be positively correlated, 7 10. Past costs are irrelevant in engineering economic analyses, unless they impact future costs, 7

Engineering Economics in Practice Abbott Laboratories, 245–246 Amazon, 155–156 Caterpillar, 203–204 ConocoPhillips, 76–77 Intel, 175–176 J.B. Hunt Transport Services, 144–145 Liu, Josh, 113 Motorola Solutions, 128–129 Scheider, Kellie, analysis of net worth, 15–16 Starbucks, 259–260 Uber, 219–220 Walmart, 1–2 Washington, Samuel, taking of loan, 57–58 Engineering economy, 2 Engineering estimating, 284 Entertainment Engineers, Inc., 83 Equal versus unequal lives, 79 Equipment, leasing versus purchasing, 194–195 Equity funds, 265 Equivalence, 58–63, 225 of methods, 78n Equivalent interest rates, determining, 61–63 Equivalent uniform annual cost, 147

Estimates bottom‐up, 284 cash, 11 computing the cost, 283 detailed, 282 determining methodology, 283 developing structure for, 283 documenting and presenting, 283 order‐of‐magnitude, 282 planning, 282 preliminary, 282 researching, collecting, and analyzing data, 282–283 semidetailed, 282 Estimation by analogy, 284 of cash flows, 11 cost, 280–285 engineering, 284 parametric, 283–284 project, 281–283

Excel® in analyzing the after‐tax effects of borrowed funds—single alternative, 189 calculating for future worth calculations, 21 declining balance worksheet functions in, 161 DPBP and, 101–102 in examining the impact of changes in the MARR, 116 Monte Carlo simulation with, 238–242 for present worth calculations, 26 Excel® DB function, 161 Excel® DDB function, 161, 162 Excel® EFFECT function, 48, 71 Excel® FV function, 22, 67, 69 Excel® FVSCHEDULE, 72 Excel® GOAL SEEK function, 24, 60, 102 Excel® IPMT worksheet function, 65 Excel® IRR function, 239 Excel® NPER function, 222 in determining DPBP of a single investment opportunity, 100–101 parameters of, 24 for values of DPBP, 105 Excel® NPV function, 30, 62 Excel® PMT function, 35 Excel® PPMT worksheet function, 65 Excel® PV function, 32–33

Excel® RATE function, 71 Excel® SLN function, 159, 162 Excel® SOLVER function, 24, 25, 60, 63, 102, 246, 252, 255 to calculate DPBP, 102–103 incorporating additional constraints in the capital budgeting problem with the, 252–253 optimizing investment program with, 256 Excel® VDB function, 161, 162. 163, 166 switching from double declining balance to straight‐line depreciation with, 163–164 Excise taxes, 176 Expenditures, nondepreciable, 184–185 External rate of return (ERR), 78, 129, 137–143 multiple alternatives, 141–142 single alternative, 138–140

F Face value, 68 Federal Aviation Administration’s Life Cycle Cost Estimating Handbook, 282 Financial accounting, 285 Financial loan, 179 First cost of an asset, 262 Fixed costs, 268–269 Ford, Henry, 6 Functional obsolescence, 146 Fundamental accounting equation, 286

Future costs, 264 Future dollars, 208 Future value, 4, 21, 119 Future worth, 78, 119–126 calculations of, 19–25 in choosing retirement plan, 124–125 defined, 119 determining of a geometric series, 45 with multiple compounding periods per year, 46–47 of multiple alternatives, 123–125 in portfolio analysis, 125–126 of single alternative, 119–120 Future worth analysis, 114

G General accounting principles, 285–296 General Motors, 6 Geometric series of cash flows, 42–45 determining future worth of, 45 determining present worth of, 43–44 GIGO (garbage‐in, garbage‐out), 226 Goal setting, as reason for future worth analysis, 119 Gradient series of cash flows, 38–41, 59–60 determining present worth of, 40–41 Gross profit, 291

H Half‐year convention, 166 Higher Education Price Index (HEPI), 208 Historical data, 282 Hold, Benjamin, 203 Hurdle rate, 3

I Immediate payment loans, 64–65, 64–66 Income net operating, 296 taxable, 177 Income statement, 290–292 Income taxes, 175–200. See also After‐tax analysis; Before‐tax analysis leasing versus purchasing equipment, 194–195 Income‐tax rates, 179 corporate, determining taxable income, 177–180 Incremental methods, 78–79 Indirect costs, 267 Indivisible investments, 249 capital budgeting problem with, 249–253 Inflated dollars, 208

Inflation, 203–217 defined, 204 versus earning power, 3 measuring, 204–208 with consumer price index (CPI) in measuring, 205–207 with Higher Education Price Index (HEPI), 207 with producer price index (PPI), 207 Inflation‐adjusted interest rate, 208 Inflation‐free dollars, 208 Inflation rate, 208 Inside PV function, 33, 34 Insider’s viewpoint approach, 147 Intangible property, 158 Intel, 175–176 Interest, 64 Interest rate(s), 3, 19 annual, 46 annual nominal, 48 effective annual, 46, 48–51 inflation‐adjusted, 208 period, 46–48 real, 208 variable, 71–72

Internal rate of return, 78, 129, 130–136, 221, 225 Descartes’ rule of signs, 131 multiple alternatives, 134–136 multiple roots, 131–134 Nordstrom’s criterion, 132 single alternatives, 130–131 Inventory turnover, 293, 294 Investments bond, 68–71 comparing, over time, 7, 11–12 divisible, 254 doubling the value of an, 23–25 economically justified, 6 equivalent alternatives in, 6 identifying alternatives, 9 yield return on, 6 Irregular cash flows, 27–31 Irregular incremental analysis cash flows, 142

J J.B. Hunt Transport Services, 144–145

K Kalanick, Travis, 219

L Latin Hypercube simulation, 238

Leases capital, 179, 180 operating, 179, 180 Leasing equipment versus purchasing equipment, 194–195 Life cycle viewpoint, 261–263 Liquidity, 293 Liu, Josh, 113 Lives, equal versus unequal, 79 Loan cash flow (LCF), 188 Loans deferred payment, 66–69 immediate payment, 64–65 principle payment, 64 Logistics, 1 Longest life approach, 10

M Maintenance costs, 262 Managerial accounting, 285 Manufacturing cost structure viewpoint, 265–268 Marginal costs, 6, 274–275 Marginalism, 276 Marginal revenue, 6 Marginal values, 276 Market basket, 205, 207 Market basket rate, 205

Market rate, 208 Market value, 147, 263 MARR. See Minimum attractive rate of return (MARR) Maturity, 68 Microsoft VBA (Visual Basic for Applications) software language, 238 Minimum attractive rate of return (MARR), 3, 10–11, 13 using Excel® to examine the impact of changes in the, 116 Modified Accelerated Cost Recovery System‐Alternative Depreciation System (MACRS‐ADS), 166, 170 Modified Accelerated Cost Recovery System (MACRS) depreciation, 159, 165–166 after‐tax analysis with, 183 with bonus depreciation, 171–172 Modified Accelerated Cost Recovery System‐General Depreciation System (MACRS‐GDS), 166–169 property classes with, 167 Money earning power of, 3 opportunity cost of, 0 time value of, 2–3, 6 Monte Carlo simulation with Excel®, 238–242 Motorola Mobility Holdings, 128 Motorola Solutions, 128–129, 247 Multiparameter sensitivity analysis, 229–230

Multiple alternatives annual worth of, 117–118 benefit‐cost calculations for, with unequal lives, 92–98 capitalized worth for, 109–110 discounted payback period for, 103–106 future worth of, 123–125 internal rate of return, 134–136 present worth of, 81–85 sensitivity analysis for, 227–229 Multiple cash flows geometric series of cash flows, 42–45 gradient series of cash flows, 38–41 irregular cash flows, 27–31 uniform series of cash flows, 31–37 Multiple roots, internal rate of return, 131–134 Multiple use, concept of, 88 Mutually exclusive investments, 8, 9

N Natural resources projects, 87 Negative cash flows, 5 Net operating income, 296 Net present value, 81 Net present worth, 81 Net residual benefit value, 93

Net worth, 286 analysis of, 15 Nominal annual interest rate, 46, 48 Nominal dollars, 208 Nondepreciable expenditures, after‐tax analysis with, 184–185 Nordstrom’s criterion, Internal rate of return, 132 Normal distribution, 231

O Obsolescence, 146 economic, 146 functional, 146 technological, 146 One‐shot investments, 85 present worth of, 85–86 Operating and maintenance costs, 261 Operating costs, 262 Operating earnings, 296 Operating income, 296 Operating income to total assets ratio, 295 Operating leases, 179 Operating profit, 296 Opportunity costs, 147, 264–265 as approach, to replacement analysis, 148–149 of money, 3

Optimum replacement interval, 150–153 behavior of, as parameters change, 151–152 parameters for determining, 150–151 sensitivity of the, 226 Order‐of‐magitude estimates, 282 Outside PV function, 33 Outsider’s viewpoint approach, 147 Overhead costs, 267 Owners’ equity, 286

P Parameters behavior of optimum replacement interval as changes occur, 151–152 for determining optimum replacement interval, 150–151 Parametric estimating, 283–284 Par value, 68 Past costs, 263 Past/future viewpoint, 263–265 Payback period (PBP), 103 “Percentage of prime cost” method, 300 Period interest rate, 46–48 Perpetuity, 107 Personal income tax calculations, 177 Personal property, 158 with MACRS‐GDS, 167

Planning horizon, 79 defining, 9–10 longest life approach, 10 shortest life approach, 10 Portfolio analysis, 125–126 future worth in, 125–126 Positive cash flows, 5 Preferred investments, selecting, 12 Preliminary estimates, 282 Present value, 4, 19, 81 Present worth, 76–112, 78, 81 calculations of, 25–27 comparing alternatives, 77–81 of geometric series, 43–44 of multiple alternatives, 81–85 of one‐shot investments, 85–86 sensitivity of after‐tax, 226 of single alternative, 81–83 Principal payments, 64 Print parameter, 52 Probabilistic statement, 221 Producer price index (PPI), 207 Profit and loss statement, 290 Project estimation, 281–283

Property depreciable, 157 intangible, 158 personal, 158 real, 158 tangible, 158 Protection, 87 Public‐sector decision making, economic analysis in, 2 Purchase price, determining for a bond, 70 Purchasing equipment, versus leasing equipment, 194–195 Pure discount rate, 208

R Random variables, probability distributions for the, 231 Ranking methods, 78 Rate of return, 128–143, 130. See also External rate of return; Internal rate of return determining for bonds, 70–71 economic worth of minimum attractive, 226 external, 129, 137–143 internal, 129, 130–136 specifying the minimum attractive, 10–11 Rate parameter, 51 Ratio analysis, 292–296 Real dollars, 208 Real interest rate, 208 Real property, 158

Real property with MACRS‐GDS, 167 Redeem, 68 Regression analysis, 283–284 Relationship data, 282 Rental agreement, 179 Replacement analysis, 144–153 cash flow and opportunity cost approaches, 147–148 fundamentals of, 145–147 obsolescence and, 146–147 opportunity cost approach to, 148–149 Retail price index, 205 Retained earnings account, 288 Retained earnings, after‐tax analysis using (no borrowing), 180–188 multiple alternatives, 185–187 single alternative, 181–185 Retirement plan, using future worth to choose, 124–125 Return on assets employed, 293 Return on investment (ROI), 130 Return on owner’s equity, 293 Returns, correlation with risks, 7 Revenues, 291 marginal, 6 operating and maintenance, 262 RFID (ratio frequency identification), 1–2 @RISK, 238

Risk analysis, 221, 231–242 analytical solutions, 233–237 defined, 231 simulation solutions, 237–238 Risks, correlation with returns, 7 Rule of 72, 23

S St. Jude Medical, 245 Sales taxes, 176 Salvage value, 147, 160, 183–184, 261, 262, 263 DPBP and, 103 Sam’s Club, 1 Schneider, Kellie, 16 Scrap value, 263 SEAT. See Seven‐step systematic economic analysis technique (SEAT) Section 179 expense deduction, 198 after‐tax analysis with, 198–200 Selling price, determining for a bond, 69 Semidetailed estimates, 282 Sensitivity analysis, 221, 225–230, 231 multiparameter, 229–230 for multiple alternatives, 227–229 for single alternative, 226–227 Sensitivity of after‐tax present worth, 226 Sensitivity of the optimum replacement interval, 226

Seven‐step systematic economic analysis technique (SEAT), 8, 14, 77 1. Identify the investment alternatives, 8, 9 2. Define the planning horizon, 8, 9–10 3. Specify the discount rate, 8, 10–11 4. Estimate the cash flows, 8, 11 5. Compare the alternatives, 8, 11–12, 77–81, 114, 129, 176 6. Perform supplementary analyses, 8, 12 7. Select the preferred investment, 8, 12 Shareholders’ equity, 286 Shorest life approach, 10 Single alternatives, 79–80 annual worth of a, 114–116 benefit‐cost calculations for a, 88–91 capitalized worth for a, 107–109 discounted payback period for a, 99–103 future worth of a, 119–120 internal rate of return, 130–131 present worth of a, 81–83 sensitivity analysis for a, 226–227 using Excel® to analyze after‐tax effects of borrowed funds, 189 Single cash flows, 19–25 future worth calculations in, 19–25 Single investment opportunity, using the Excel® NPER tool to determine DPBP of a, 100–101 Sinking fund factor, 36–37 Standard costs, 302

Starbucks, 259–260, 286 balance sheets of, 287 consolidates statements of earnings, 290 Statement of earnings, 290 Stern Stewart, 304 Straight‐line depreciation, 159–160, 179–180 switching from double declining balance to, with the Excel VDB function, 163–164 Sunk costs, 7, 8, 146, 263 Supplementary analysis, 221 performing, 12

T Tangible property, 158 Taxable income, 177 determining, 177–180 Tax analysis, before‐tax versus after‐tax, 79 ad valorem, 176 Tax Relief Act (2010), 171–172 Technological obsolescence, 146 Terminal value, 119, 262–263 Then‐current dollars, 208 Thompson, Robert, 2 Time now, 81 Times interest earned ratio, 295

Time value of money (TVOM), 6, 2–3, 15–54 cash flow diagrams in, 16–18 compounding frequency in, 46–54 end‐of‐period cash flows in, 18 illustration of, 17 multiple cash flows in geometric series of cash flows in, 42–45 gradient series of cash flows in, 38–41 irregular cash flows in, 27–31 uniform series of cash flows in, 31–37 present worth calculations in, 25–27 single cash flows in, 19–25 Time zero, 81 Today’s dollars, 208 Total costs, 275 Traditional cost allocation methods, 297–301 True rate of return, 130 Trump, Donald, 198 Two‐constraint capital budgeting problem, solving, 250–251

U Uber, 219–220 Umpleby, Jim, 203 Unequal lives, benefit‐cost calculations for multiple alternatives and, 92–98 Unequal versus equal lives, 79 Uniform series equivalency, of a decreasing gradient series, 59–60

Uniform series of cash flows, 18, 31–37 Uniform withdrawals, determining the size of, 35 Unrecovered investment, 159

V Value(s) book, 159 break‐even, 221 doubling, of an investment, 23–25 market, 147 salvage, 147, 160, 183–184, 261, 262 terminal, 262–263 Variable costs, 268–269 Variable interest rates, 71–72 cash flow series with, 72 Variance of present worth, 233

W Walmart Stores, Inc. (WMT), 1–2 Weighted average cost of capital (WACC), 304 “What if” questions, 12 White, Miles D., 245

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