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Fundamentals of Engineering Economic Analysis [Unbound ed.]
 1118633776, 9781118633779

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Fundamentals of

E n g i n ee r i n g E c o n o m i c A n a ly s i s White

Grasman

Ca s e

L a S c o l a Nee d y

P r at t

Formula

Formula

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Uniform Series Cash Flows

Irregular Series Cash Flows

Series Cash Flows

Future Worth of a Present Payment

Present Worth of a Future Payment

Single Cash Flows

 

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Excel Notation

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FUNDAMENTALS OF ENGINEERING ECONOMIC ANALYSIS First Edition

JOHN A. WHITE University of Arkansas

KELLIE S. GRASMAN Missouri University of Science and Technology

KENNETH E. CASE Oklahoma State University

K I M L a S C O L A N E E DY University of Arkansas

D AV I D B . P R AT T Oklahoma State University

VICE PRESIDENT & EXECUTIVE PUBLISHER Don Fowley ACQUISITIONS EDITOR Linda Ratts OPERATIONS MANAGER Melissa Edwards CONTENT EDITOR Wendy Ashenberg DEVELOPMENT EDITOR Laurie McGuire CONTENT MANAGER Kevin Holm SENIOR PRODUCTION EDITOR Jill Spikereit EXECUTIVE MARKETING MANAGER Christopher Ruel PRODUCT DESIGNER Jennifer Welter MEDIA SPECIALIST James Metzger DESIGN DIRECTOR Harry Nolan DESIGNER Wendy Lai PHOTO EDITOR Sheena Goldstein ASSISTANT EDITOR Jennifer Lartz EDITORIAL ASSISTANT Hope Ellis COVER DESIGN Wendy Lai COVER PHOTO © Tomislav Zivkovic/iStockphoto; © Dominik Pabis/iStockphoto; © David Jones/iStockphoto; © Orlando Rosu/iStockphoto MARKETING ASSISTANT Marissa Carroll CONTENT ASSISTANT Helen Seachrist This book was set in 11/13 Times LT Std by Aptara®, Inc. and printed and bound by Quad Graphics Versailles. The cover was printed by Quad Graphics Versailles. This book is printed on acid free paper. ∞ Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship. Copyright © 2014 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201)748-6011, fax (201)748-6008, website http://www.wiley.com/go/permissions. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at www.wiley.com/go/returnlabel. Outside of the United States, please contact your local representative. Library of Congress Cataloging in Publication Data White, John A., 1939Fundamentals of engineering economic analysis/John A. White, Universisty of Arkansas, Kellie 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. pages cm Includes index. cover ISBN 978-1-118-41470-5 (hardback) 1. Engineering economy. I. Title. TA177.4.W47 2013 658.15—dc23 2013023515 The inside back cover will contain printing identification and country of origin if omitted from this page. In addition, if the ISBN on the back cover differs from the ISBN on this page, the one on the back cover is correct. Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

CONTENTS CHAPTER 1 An Overview of Engineering Economic Analysis

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3

Learning Objectives 4 Introduction 4 1-1 Time Value of Money 4 1-2 Engineering Economy Principles 7 1-3 Economic Justification of Capital Investments 10 1.3.1 SEAT Step 1 of 7: Identify the Investment Alternatives 11 1.3.2 SEAT Step 2 of 7: Defining the Planning Horizon 11 1.3.3 SEAT Step 3 of 7: Specifying the Minimum Attractive Rate of Return 13 1.3.4 SEAT Step 4 of 7: Estimate the Cash Flows 14 1.3.5 SEAT Step 5 of 7: Comparing Alternatives 14 1.3.6 SEAT Step 6 of 7: Performing Supplementary Analyses 14 1.3.7 SEAT Step 7 of 7: Select the Preferred Investment 15 Summary 15 Key Terms 16 FE-Like Problems 17 Problems 18

CHAPTER 2 Time Value of Money Calculations

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23

Learning Objectives 23 Introduction 23 2-1 Cash Flow Diagrams 24 2-2 Single Cash Flows 26 2.2.1 Future Worth Calculations (F |P) 26 2.2.2 Present Worth Calculations (P |F ) 33 2-3 Multiple Cash Flows 35 2.3.1 Irregular Cash Flows 35 2.3.2 Uniform Series of Cash Flows 39 2.3.3 Gradient Series of Cash Flows 46 2.3.4 Geometric Series of Cash Flows 51 iii

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Contents

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Compounding Frequency 55 2.4.1 Period Interest Rate Approach 56 2.4.2 Effective Annual Interest Rate 58 2.4.3 When Compounding and Cash Flow Frequencies Differ Summary 62 Key Terms 66 FE-Like Problems 67 Problems 69

CHAPTER 3 Equivalence, Loans, and Bonds

03 04

61

93

Learning Objectives 93 Introduction 93 3-1 Equivalence 94 3-2 Interest Payments and Principal Payments 100 3.2.1 Immediate Payment Loans 101 3.2.2 Deferred Payment Loans 104 3-3 Bond Investments 106 3-4 Variable Interest Rates 110 3-5 Annual Percentage Rate 111 Summary 113 Key Terms 114 FE-Like Problems 114 Problems 116

CHAPTER 4 Present Worth 125

Learning Objectives 125 Introduction 126 4-1 Comparing Alternatives 126 4.1.1 Methods of Comparing Economic Worth 127 4.1.2 Ranking and Incremental Methods of Economic Worth 127 4.1.3 Equivalence of Methods 128 4.1.4 Before-Tax versus After-Tax Analysis 128 4.1.5 Equal versus Unequal Lives 129 4.1.6 A Single Alternative 129 4-2 Present Worth Calculations 130 4.2.1 Present Worth of a Single Alternative 130 4.2.2 Present Worth of Multiple Alternatives 132 4.2.3 Present Worth of One-Shot Investments 135

Contents

4-3

Benefit-Cost Analysis 137 4.3.1 Benefit-Cost Calculations for a Single Alternative 137 4.3.2 Benefit-Cost Calculations for Multiple Alternatives 139 4-4 Discounted Payback Period 144 4.4.1 Discounted Payback Period for a Single Alternative 144 4.4.2 Discounted Payback Period for Multiple Alternatives 149 4-5 Capitalized Worth 152 4.5.1 Capitalized Worth for a Single Alternative 153 4.5.2 Capitalized Worth for Multiple Alternatives 156 Summary 157 Key Terms 159 FE-Like Problems 159 Problems 162

CHAPTER 5 Annual Worth and Future Worth

05 06

179

Learning Objectives 179 Introduction 179 5-1 Annual Worth 180 5.1.1 Annual Worth of a Single Alternative 180 5.1.2 Annual Worth of Multiple Alternatives 183 5-2 Future Worth 185 5.2.1 Future Worth of a Single Alternative 185 5.2.2 Future Worth of a Multiple Alternatives 189 5.2.3 Portfolio Analysis 192 Summary 194 Key Terms 194 FE-Like Problems 195 Problems 196

CHAPTER 6 Rate of Return 213

Learning Objectives 213 Introduction 214 6-1 Internal Rate of Return Calculations 215 6.1.1 Single Alternative 215 6.1.2 Multiple Roots 217 6.1.3 Multiple Alternatives 220

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Contents

6-2

External Rate of Return Calculations 224 6.2.1 Single Alternative 224 6.2.2 Multiple Alternatives 228 Summary 231 Key Terms 232 FE-Like Problems 232 Problems 235

CHAPTER 7 Replacement Analysis

07 08

257

Learning Objectives 258 Introduction 258 7-1 Fundamentals of Replacement Analysis 258 7-2 Cash Flow and Opportunity Cost Approaches 261 7-3 Optimum Replacement Interval 264 Summary 268 Key Terms 269 FE-Like Problems 269 Problems 271

CHAPTER 8 Depreciation 285

Learning Objectives 286 Introduction 286 8-1 The Role of Depreciation in Economic Analysis 287 8-2 Language of Depreciation 288 8-3 Straight-Line and Declining Balance Depreciation Methods 289 8.3.1 Straight-Line Depreciation (SLN) 289 8.3.2 Declining Balance and Double Declining Balance Depreciation (DB and DDB) 291 8.3.3 Switching from DDB to SLN with the Excel® VDB Function 294 8-4 Modified Accelerated Cost Recovery System (MACRS) 296 8.4.1 MACRS-GDS 297 8.4.2 MACRS-ADS 301 Summary 301 Key Terms 303 FE-Like Problems 303 Problems 305

Contents

CHAPTER 9 Income Taxes 317

09 10 11

Learning Objectives 317 Introduction 317 9-1 Corporate Income Tax Rates 319 9-2 After-Tax Analysis Using Retained Earnings (No Borrowing) 322 9.2.1 Single Alternative 323 9.2.2 Multiple Alternatives 328 9-3 After-Tax Analysis Using Borrowed Capital 330 9-4 Leasing versus Purchasing Equipment 336 Summary 338 Key Terms 340 FE-Like Problems 340 Problems 342

CHAPTER 10 Inflation 359

Learning Objectives 359 Introduction 360 10-1 The Meaning and Measure of Inflation 360 10-2 Before-Tax Analysis 364 10-3 After-Tax Analysis 367 10-4 After-Tax Analysis with Borrowed Capital 369 Summary 373 Key Terms 375 FE-Like Problems 375 Problems 377

C H A P T E R 11 Break-Even, Sensitivity, and Risk Analysis Learning Objectives 391 Introduction 392 11-1 Break-Even Analysis 393 11-2 Sensitivity Analysis 398 11-3 Risk Analysis 405 11.3.1 Analytical Solutions 407 11.3.2 Simulation Solutions 412 Summary 418 Key Terms 419 FE-Like Problems 419 Problems 421

391

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CHAPTER 12 Capital Budgeting

12

437

Learning Objectives 438 Introduction 438 12-1 The Classical Capital Budgeting Problem 438 12-2 Capital Budgeting Problem with Indivisible Investments 440 12-3 Capital Budgeting Problem with Divisible Investments 446 12-4 Practical Considerations in Capital Budgeting 449 Summary 450 Key Terms 451 FE-Like Problems 451 Problems 453

APPENDIX A a Single Sums, Uniform Series, and Gradient Series Interest Factors b Geometric Series Present Worth Interest Factors 496 c Geometric Series Future Worth Interest Factors 501

APPENDIX B Obtaining and Estimating Cash Flows

507

B-1 Introduction 506 B-2 Cost Terminology 508 B.2.1 Life Cycle Viewpoint 508 B.2.2 Past/Future Viewpoint 510 B.2.3 Manufacturing Cost Structure Viewpoint 514 B.2.4 Fixed and Variable Viewpoint 517 B.2.5 Average and Marginal Viewpoint 527 B-3 Cost Estimation 530 B.3.1 Project Estimation 532 B.3.2 Estimating Methodologies 535 B.3.3 General Sources of Data 536 B-4 General Accounting Principles 537 B.4.1 Balance Sheet 538 B.4.2 Income Statement 542 B.4.3 Ratio Analysis 545 B-5 Cost Accounting Principles 550 B.5.1 Traditional Cost Allocation Methods 551 B.5.2 Activity Based Costing 556 B.5.3 Standard Costs 557 B.5.4 Economic Value Added 558 Summary 561 FE-Like Problems 562 Problems 563

ANSWERS TO SELECTED EVEN NUMBERED PROBLEMS INDEX

599

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471

PREFACE INTRODUCTION We are pleased to present Fundamentals of Engineering Economic Analysis, a new text to serve the undergraduate engineering economics course. Three overarching goals informed the development of this book and its corresponding WileyPLUS course: •





Streamlined: This text was carefully optimized to serve a 1semester, 1–3 credit-hour course without sacrificing rigor or essential content. As engineering programs are increasingly squeezed to cover more content without extending the 4-year curriculum, Engineering Economics courses are being reduced from two semesters to one semester and/or from 3 credit hours down to 2 or even 1 credit hour. Fundamentals of Engineering Economic Analysis is built to meet the needs of the course as it is today, tailored down to a content set that can be covered within the allotted timeframe. Accessible: Students who enter this course rarely have any background in accounting or finance. This text has been designed to support a broad spectrum of engineering students with the best-integrated pedagogy, in both print and electronic formats. It also provides real-world vignettes to motivate and deepen students’ engagement with the course topics. Unique digital solution: The WileyPLUS course that accompanies this text provides a wealth of homework problems that are automatically graded to provide instant feedback to students, along with secure solutions. The course also provides students with an abundance of learning support features that are available 2437, such as videos that show an instructor walking through the solution of a typical problem.

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. But the coverage has been carefully streamlined to serve the always time-challenged one-semester course. In addition, through the efforts of new co-authors Kellie Grasman and Kim LaScola Needy, Fundamentals of Engineering Economic Analysis provides thoughtful and wellintegrated pedagogical support to help a wide range of students better grasp—and retain—this critical material. ix

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Preface

We believe this synthesis of proven content and consistent pedagogy is ideally suited to help students master a course that will inform the rest of their lives as working engineers.

APPROACH OF THE TEXT Fundamentals of Engineering Economic Analysis 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 realworld 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. Finally, each chapter ends with a set of FE-Like Problems to help students who want to prepare for the FE Exam, as well as an abundant variety of homework problems at various levels of difficulty. 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 Evaluation and 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

Preface

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 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. 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 Topical coverage is focused on core and essential topics for those who will perform engineering economics analysis, including: • •

Principles and methodology for engineering economic analyses considering the time value of money. Time value of money including all commonly used cash flow profiles.

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Preface

• •

• • • • • • •

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. 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.

IMPORTANT STUDENT SUPPORT FEATURES Pedagogical features in both the print text and WileyPLUS are designed to support and reinforce student learning. In the print text:



• •

• •



Learning objectives set student expectations at the beginning of the chapter, are restated at the appropriate point in the body of the chapter, and are reviewed in the chapter Summary. Key terms are defined in margins and listed and cross-referenced in the chapter summary for easy review. Real-world vignettes and discussion questions begin each chapter, placing the content in a practical perspective and opening it up for further exploration. Summary sections at ends of chapters highlight key concepts, equations, learning objectives, and key terms in an easily accessible format. 816 end-of-chapter problems are organized by text section to facilitate assignment and self-study. Answers to even-numbered problems are provided at the end of the book. 138 FE-like problems at ends of chapters give students an opportunity to practice for the FE exam, as well as check their understanding of chapter contents.

Preface

In WileyPLUS:

• •





Video Lessons are brief lectures explaining concepts from the text, so that students have multiple means of better understanding core material. Video Solutions to selected problems give students an online “office hours” experience for problems they may struggle with, and model a solution strategy for other homework problems. Algorithmic problems are extensions of print problems, but with different quantitative information. This gives instructors additional exercises to assign and students more opportunity to apply their understanding. GO Tutorial problems provide students with a step-by-step guide on how to approach a problem, so they can review it (at instructor’s discretion) and then go back and try the problem again on their own.

On the companion Web site:



Useful Excel spreadsheet utilities that assist in performing specific engineering economy analyses.

INSTRUCTOR SUPPORT The following instructor support materials are available on the Web site for this text: http://www.wiley.com/WileyCDA/WileyTitle/productCd-1118414705. html •





PowerPoint lecture slides 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. 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 password-protected, and accessible only to instructors adopting this text. Please visit the instructor section of the website at http://bcs.wiley. com/he-bcs/Books?action=index&itemId=1118414705&bcsId=8241 to register for a password. Useful Excel spreadsheet utilities that assist in performing specific engineering economy analyses.

WILEYPLUS WileyPLUS combines the complete, dynamic online text with all of the teaching and learning resources you need, in one easy-to-use system. WileyPLUS offers today’s engineering students the iterative and visual learning materials

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they need to help them grasp difficult concepts—and apply what they’ve learned to solve problems. A robust variety of examples and exercises enables students to work problems, see their results, and obtain instant feedback including hints and reading references linked directly to the online text. For instructors, WileyPLUS: • •

Makes the course more manageable—automatic grading and real-time reporting saves instructors time and energy Allows instructors to focus on maximizing their impact on students

WileyPLUS provides an abundance of resources, both for students and instructors: • •



A complete online version of the textbook A wealth of practice problems with automatic grading and immediate feedback for students ° Secure solutions make homework meaningful—problems are algorithmic so students receive unique values in each problem ° Integration with Blackboard Course Management Systems enables the ability to have all course materials in one place 2437 support for students to ensure positive learning outcomes ° Video Solutions demonstrate how to solve a problem, both by hand and with Excel ° Mini-lecture videos by co-author Kellie Grasman explain concepts from the text, so that students may have multiple means of better understanding the topic

Contact your local Wiley representative or visit www.wileyplus.com for more information about using WileyPLUS in your course.

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

Preface

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 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 Alberto Marquez, Lamar University Gary Maul, Ohio State University R. Eugene McGinnis, Christian Brothers University Mamunur Rashid, Georgia Southern 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 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. J. B. Hunt Transport Services, Inc: Jerry W. Walton, Executive Vice President and Chief Financial Officer, Donald G. Cope, Sr. Vice President, Controller, and Chief Accounting Officer, and David G. Mee, Executive Vice President and Chief Financial Officer. Motorola Solutions: Dr. Thomas F. Davis, Corporate Vice President and Chief Economist (retired). Schneider National, Inc.: Dr. Christopher B. Lofgren, President and Chief Executive Officer. Wal-Mart Stores: Michael T. Duke, President and Chief Executive Officer.

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Real-world vignettes show how practicing engineers use the concepts

E N G I N E E R I N G E C O N O M I C S I N P R AC T I C E : K E L L I E S C H N E I D E R A N A LY Z E S H E R N E T WO RT H Immediately after receiving her engineering degree, Kellie Schneider began employment with a multinational company. Her initial salary was $60,000 per year. Kellie decided she would invest 10 percent of her gross salary each month. Based on discussions with the company’s personnel and with a person in human resources, she believed her salary would increase annually at a rate ranging from 3 percent to 15 percent, depending on her performance. After analyzing various investment opportunities, she anticipated she would be able to earn between 5 percent and 10 percent 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 percent to 5 percent 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 engage students in a thought-provoking discussion

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? c02TimeValueOfMoneyCalculations.indd Page 23

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TIME VALUE OF MONEY CALCULATIONS

Learning objectives help students measure their progress

LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

1. Construct a cash flow diagram (CFD) depicting the cash inflows and outflows for an investment alternative. (Section 2.1) 2. Perform time value of money calculations for single cash flows with annual compounding. (Section 2.2) 3. Distinguish between uniform, irregular, gradient and geometric series of cash flows. (Section 2.3) 4. Perform time value of money calculations for multiple cash flows including: • Irregular series of cash flows with annual compounding. (Section 2.3.1) • Uniform series of cash flows with annual compounding. (Section 2.3.2) • Gradient series of cash flows with annual compounding. (Section 2.3.3) • Geometric series of cash flows with annual compounding. (Section 2.3.4)

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A Walkthrough of the Text

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CASH FLOW DIAGRAMS

LEA R N IN G O B JEC TIV E: Construct a cash flow diagram (CFD) depicting the

Video Lesson: Cash Flow Diagrams

Key terms and definitions are highlighted

Cash Flow Diagram (CFD) A diagram depicting the magnitude and timing of cash flowing in and out of the investment alternative.

cash inflows and outflows for an investment alternative.

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.

Worked Examples

Many examples are linked to others in context. That is, the/208/WB00898/XXXXXXXXXXXX/ch02/text_s same scenario is re-visited in multiple Examples to implement progressively complex concepts and techniques. This approach also facilitates comparison of results under different situations, such as no tax versus tax and various loan plans.

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Who Wants to Be a Millionaire?

EXAMPLE

Examples follow a consistent problemsolving methodology

Suppose Crystal Wilson wants to accumulate $1,000,000 by the time she retires in 40 years. If she earns 10 percent on her investments, how much must she invest each year in order to realize her goal? KEY DATA

SOLUTION

Shows the handcalculation

Given: F 5 $1,000,000; i 5 10%; n 5 40 Find: A Applying Equation 2.27, A 5 $1,000,0001A Z F,10%,402 5 $1,000,00010.00225942 5 $2,259.40/year Using the Excel® PMT worksheet function gives A 5PMT110%,40,,210000002 5 $2,259.41/year

Shows how to solve with Excel

xviii A Walkthrough of the Text

Chapter Summaries SUMMARY

KEY CONCEPTS

Brief summary of key concepts, linked to Learning Objectives

1. Learning Objective: 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.

KEY TERMS

Reviews Key Terms c02TimeValueOfMoneyCalculations.indd Page 67 5/3/13 9:28 PM f-481

Capital Recovery Factor, p. 42 Cash Flow Diagram (CFD), p. 24 Compounding, p. 26

Interest Rate, p. 26 Nominal Annual Interest Rate, p. 55 Period Interest Rate, p. 56

Effective Annual Interest Rate, p. 58 Future Value, p. 28

Present Value, p. 27 Sinking Fund Factor, p. 45

Geometric Series, p. 51 Gradient Series, p. 46

Uniform Series, p. 26

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FE-LIKE PROBLEMS 1.

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 c. 32 years b. 19 years d. Cannot be determined without knowing the amount invested.

FE-Like Problems for exam preparation

c02TimeValueOfMoneyCalculations.indd 2. Let F be the accumulated sum, P the principal invested, i the annual Page com-69 5/3/13 9:28 PM f-481

pound interest rate, and n the number of years. Which of the following correctly relates these quantities? a. F 5 P (1 1 in) c. F 5 P (1 1 n)i b. F 5 P (1 1 i)n d. F 5 P (1 1 ni)n 2 1

PROBLEMS Introduction 1.

Abundant end-of-chapter problems at various difficulty levels and keyed to chapter sections

You are offered $200 now plus $100 a year from now for your used computer. Since 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?

2. 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 con-

sider fair? b. It is suggested that a single fair amount be paid one year from now. What

do you consider fair? 3. What words comprise the abbreviation “DCF”? Tell/describe/define what it means in 10 words or less. 4. State the four DCF rules. Section 2.1

Cash-Flow Diagrams

5. 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.

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ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : WAL M A RT In fiscal year (FY) 2011, Walmart Stores, Inc. (WMT) employed more than 2.1 million people worldwide, including 1.4 million in the United States. Its annual sales increased 3.47 percent from the previous year to a record $419 billion, and income increased 6.3 percent to a record $15.4 billion. As of January 31, 2011, Walmart had a total of 8,907 retail outlets worldwide, including 3,804 Supercenters, discount stores, and Neighborhood Markets as well as 609 Sam’s Clubs in the United States and 4,557 retail outlets in stores in 14 countries outside the United States. Walmart imports billions of dollars of products from 70 countries. Its imports are greater than those of most countries in the world. In a 15-year period (from 1990 to 2005) the number of Walmart stores in the United States increased by more than 2,000; in the next 6-year period (from 2005 to 2011), the number of stores increased by 3,800. 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 logistics operations employ more than 85,000 associates. Its distribution network includes approximately 150 distribution centers and 50 transportation offices. A major user of transportation services from external sources, Walmart has the fourth-largest private fleet of trucks in the United States, including 7,200 tractors, 53,000 trailers, and 8,000 drivers. Highly regarded for its rapid response to the September 11, 2001 attacks on the World Trade Center, to Hurricane Katrina in 2005, and to the 2010 oil spill in the Gulf of Mexico, Walmart has 9 disaster distribution centers, strategically located across the United States and stocked 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 2

AN OVERVIEW OF ENGINEERING ECONOMIC ANALYSIS

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 gives priority attention to reducing life-cycle costs, resulting in a 19.2 percent return on investment (ROI) in FY 2011.

ENGINEER ING EC ONOM I C S I N P R AC T I C E : ROB ERT THOM PS ON’S FI N AN C I AL P L AN N I N G 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? 3

4

Chapter 1

An Overview of Engineering Economic Analysis

LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

1. Apply the four discounted cash flow (DCF) rules to simple time value of money (TVOM) situations (Section 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).

3. Identify the seven steps of the systematic economic analysis technique (SEAT) used to perform engineering economic analyses (Section 1.3).

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

LEARN I N G O B JEC T I V E : Apply the four discounted cash flow (DCF) rules to

Video Lesson: Time Value of Money

simple time value of money (TVOM) situations.

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

1-1 Time Value of Money

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.

The Time Value of Money Illustrated

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.

EXAMPLE

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? 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 percent.

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

5

SOLUTION

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Chapter 1

An Overview of Engineering Economic Analysis

Discounted Cash Flow (DCF) The movement of money forward or backward in time.

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. 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. 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. 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: Money has a time value. 2. Quantities of money cannot be added or subtracted unless they occur at the same point in time. 1.

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

1-2

Engineering Economy Principles

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.

Applying the Four DCF Rules

EXAMPLE

Recall the previous example, where the student’s TVOM was 20 percent. 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? 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 percent 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 percent 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 2$1,200 1 $1,100, or 2$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 percent instead of 20 percent. 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.

1-2

ENGINEERING ECONOMY PRINCIPLES

LEARNING O BJECTI VE: 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.

SOLUTION

7

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Chapter 1

An Overview of Engineering Economic Analysis

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

2.

3.

4.

5.

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.) 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. 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 discussed in Chapters 4, 5, and 6.) 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 well-behaved cash flow profiles, the equivalence holds only for the TVOM that equates their economic worths. (Chapters 2 and 6 examine this principle more closely.) 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

1-2

6.

7.

8.

9.

10.

Engineering Economy Principles

used in comparing marginal revenues and marginal costs if they occur at different points in time. 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 said about this principle in Chapters 6, 9, 10, and 12.) 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.) 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.) 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. 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,

9

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An Overview of Engineering Economic Analysis

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.

1-3

ECONOMIC JUSTIFICATION OF CAPITAL INVESTMENTS

LEARN I N G O B JEC T I V E : Identify the seven steps of the systematic economic

Video Lesson: Systematic Economic Analysis Technique (SEAT)

analysis technique (SEAT) used to perform engineering economic analyses.

In performing engineering economic analyses, it is helpful to follow a consistent methodology. The following seven-step systematic economic analysis technique (SEAT) is recommended: 1. 2. 3. 4. 5. 6. 7.

Identify the investment alternatives. Define the planning horizon. Specify the discount rate. Estimate the cash flows. Compare the alternatives. Perform supplementary analyses. Select the preferred investment.

1-3

Economic Justification of Capital Investments

11

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 to 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. 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

Planning Horizon The period of time or width of the “window” over which the economic performance of each investment alternative will be viewed.

12

Chapter 1

An Overview of Engineering Economic Analysis

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. 2. 3. 4. 5.

Set the planning horizon equal to the shortest life among the alternatives. Set the planning horizon equal to the longest life among the alternatives. Set the planning horizon equal to the least common multiple of the lives of the various alternatives. Use a standard length horizon equal to the period of time that best fits the organization’s need, such as 10 years. 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

Economic Justification of Capital Investments

13

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! 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. 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 5 (E/V ) ie 1 (D/ V) id (1 2 itr)

(1.1)

where E 5 a firm’s total equity, expressed in dollars D 5 a firm’s total debt and leases, expressed in dollars V 5 E 1 D, a firm’s total invested capital ie 5 cost of equity or expected rate of return on equity, expressed in percent id 5 cost of debt or expected rate of return on borrowing, expressed in percent itr 5 corporate tax rate

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.

14

Chapter 1

An Overview of Engineering Economic Analysis

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 2 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

Summary 15

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.

KEY CONCEPTS 1. Learning Objective: 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.

SUMMARY

16

Chapter 1

An Overview of Engineering Economic Analysis

2. 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. (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. 3. Learning Objective: Identify the seven steps of the systematic economic analysis technique (SEAT) used to perform engineering economic analyses. (Section 1.3)

The seven-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.

KEY TERMS Discounted Cash Flow (DCF), p. 6 Time Value of Money (TVOM), p. 5 Minimum Attractive Rate of Return, p. 13 Weighted Average Cost of Capital (WACC), p. 13 Planning Horizon, p. 11

Summary

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1.

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 c. Economy of scale b. Marginal return d. Pareto principle

2.

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 c. Independent b. Collectively exhaustive d. Mutually exclusive

3.

Risks and returns are generally __________ correlated. a. inversely c. not b. negatively d. positively

4.

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

5.

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

6.

If a student’s time value of money rate is 30 percent, then the student would be indifferent between $100 today and how much in 1 year? a. $30 c. $103 b. $100 d. $130

7.

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.

17

18

Chapter 1

An Overview of Engineering Economic Analysis

PROBLEMS Section 1.1 Time Value of Money 1.

Wylie has been offered the choice of receiving $5,000 today or an agreedupon 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?

2. RT is about to loan his granddaughter Cynthia $20,000 for 1 year. RT’s

TVOM, based upon his current investment earnings, is 8 percent. Cynthia’s TVOM, based upon earnings on investments, is 12 percent. 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? 3.

RT is about to loan his granddaughter Cynthia $20,000 for 1 year. RT’s TVOM, based upon his current investment earnings, is 12 percent, and he has no desire to loan money for a lower rate. Cynthia is currently earning 8 percent 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?

4. If your TVOM is 15 percent and your friend’s is 20 percent, 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 their 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 5.

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 1 2 3

4 5

Investment A small FCC-licensed commercial radio station New antenna and hardline from transmitter New amplifier to boost from 5,000 watts to 30,000 watts New lightningprotection equipment New control console

Cost in Today’s $

Revenues in Today’s $

$180,000

$270,000

$30,000

$50,000

$40,000

$30,000

$30,000

$60,000

$15,000

$5,000

Summary 19

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 well illustrated by this problem? c. Of the systematic economic analysis technique’s seven steps, which one(s)

is well illustrated by this problem? 6. 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 situa-

tion 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 well illustrated by this problem? e. Of the systematic economic analysis technique’s seven steps, which one(s) is well illustrated by this problem? 7. 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 well illustrated by this problem? c. Of the systematic economic analysis technique’s seven steps, which one(s) is well illustrated by this problem? 8. 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.

20

Chapter 1

An Overview of Engineering Economic Analysis

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

is well illustrated by this problem? 9.

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 well illustrated by this problem? c. Of the systematic economic analysis technique’s seven steps, which one(s) is well illustrated by this problem?

10. 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. 11.

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 well illustrated by this problem? c. Of the systematic economic analysis technique’s seven steps, which one(s)

is well illustrated by this problem? 12. 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: You Invest, $

Chance of Success

A

$100,000

95%

$110,789.47

5%

$95,000

B

$100,000

60%

$150,000.00

40%

$50,000

C

$100,000

20%

$510,000.01

80%

$10,000

Alternative

a. b. c. d.

Returned to You Chance of Returned to You If Success, $ Failure If Failure, $

Which would you select? Why would you make this selection? Of the ten principles, which one(s) is well illustrated by this problem? Of the systematic economic analysis technique’s seven steps, which one(s) is well illustrated by this problem?

Summary 21

13. 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 well illustrated by this problem? c. Of the systematic economic analysis technique’s seven steps, which one(s)

is well illustrated by this problem? 14.

Reconsider the county commissioner’s evaluation of three projects in Problem 13. 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 well illustrated by this problem? c. Of the systematic economic analysis technique’s seven steps, which one(s)

is well illustrated by this problem?

ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : K ELLIE S C HN E I DE R AN ALY Z E S HER NE T WO RT H Immediately after receiving her engineering degree, Kellie Schneider began employment with a multinational company. Her initial salary was $60,000 per year. Kellie decided she would invest 10 percent of her gross salary each month. Based on discussions with the company’s personnel and with a person in human resources, she believed her salary would increase annually at a rate ranging from 3 percent to 15 percent, depending on her performance. After analyzing various investment opportunities, she anticipated she would be able to earn between 5 percent and 10 percent 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 percent to 5 percent 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? 22

TIME VALUE OF MONEY CALCULATIONS

LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

1. Construct a cash flow diagram (CFD) depicting the cash inflows and outflows for an investment alternative. (Section 2.1)

2. Perform time value of money calculations for single cash flows with annual compounding. (Section 2.2)

3. Distinguish between uniform, irregular, gradient and geometric series of cash flows. (Section 2.3)

4. Perform time value of money calculations for multiple cash flows including: • Irregular series of cash flows with annual compounding. (Section 2.3.1) • Uniform series of cash flows with annual compounding. (Section 2.3.2) • Gradient series of cash flows with annual compounding. (Section 2.3.3) • Geometric series of cash flows with annual compounding. (Section 2.3.4) 5. Perform time value of money calculations for multiple compounding periods per year using the period interest rate and the effective annual interest rate. (Section 2.4)

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. 23

24

Chapter 2

Time Value of Money Calculations

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

LEARN I N G O B JEC T I V E : Construct a cash flow diagram (CFD) depicting the

Video Lesson: Cash Flow Diagrams

Cash Flow Diagram (CFD) A diagram depicting the magnitude and timing of cash flowing in and out of the investment alternative.

cash inflows and outflows for an investment alternative.

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. $5,000

$5,000

$5,000

(+) 0 (–)

1

2

3

4

Time $2,000 $3,000

$4,000 FIGURE 2 . 1 A Cash Flow Diagram (CFD)

5

2-1

Cash Flow Diagrams

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 investor’s perspective: Downward arrows denote expenditures, and upward arrows denote receipts.

A Simple Illustration of the Time Value of Money

EXAMPLE

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. $3,000

(+)

0

1

(–)

$3,000

2 3 End of Year

4

$3,000

5

6

7

Alternative A $6,000 $3,000

(+)

0 (–)

1

2 3 End of Year

$3,000

4

5

$3,000

6

7

Alternative B $6,000 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. Since we prefer to receive the $3,000 sooner, Alternative A is preferred to Alternative B.

25

26

Chapter 2

Time Value of Money Calculations

Compounding The practice of calculating interest on both the original principal and any interest accumulated to date.

Uniform Series A series characterized by cash flows of equal (or uniform) value for each period.

In this chapter, we emphasize end-of-period cash flows and end-ofperiod compounding. 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 the financial institution. Beginning-of-period cash flows can be handled easily by noting that the end of period t is the beginning of period t 1 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-year cash flows are assumed unless otherwise noted. 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.

2-2 Video Lesson: Transformations—Single Cash Flow

LEARN I N G O B JEC T I V E : Perform time value of money calculations for single cash flows with annual compounding.

Our analysis of the time value of money begins with the simplest scenario: a single cash flow. 2.2.1

Interest Rate The rate of change in the accumulated value of money.

SINGLE CASH FLOWS

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 5 a iFt21 t51

(2.1)

2-2

where t increments the years from 1 to n, F0 5 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 Fn 5 Fn21 11 1 i2

Single Cash Flows

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

(2.2)

Compounding Interest for 5 Years

EXAMPLE

Suppose you loan $10,000 for 1 year to an individual who agrees to pay you interest at a compound rate of 10 percent/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

Given: Loan amount 5 $10,000; interest rate 5 10%/year; length of loan 5 5 years Find: Value of the loan amount after 5 years

KEY DATA

As shown in Table 2.1, the $10,000 owed, compounded over a 5-year period at 10 percent annual compound interest, totals $16,105.10.

SOLUTION

TABLE 2.1

Tabular Solution to Example 2.2

Unpaid Balance at the Beginning of the Year

Annual Interest

$10,000.00

$1,000.00

2

$11,000.00

3

$12,100.00

4 5

Year 1

Payment

Unpaid Balance at the End of the Year

$0.00

$11,000.00

$1,100.00

$0.00

$12,100.00

$1,210.00

$0.00

$13,310.00

$13,310.00

$1,331.00

$0.00

$14,641.00

$14,641.00

$1,464.10

$16,105.10

$0.00

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 5 P11 1 i2 n

(2.3)

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 1 i)n is tabulated in Appendix A for

27

28

Chapter 2

Time Value of Money Calculations

TABLE 2.2

Derivation of Equation 2.3 (A)

(B)

(C) 5 (A) 1 (B)

End of Period

Amount Owed

Interest for Next Period

Amount Owed for Next Period*

0

P

Pi

1

P 11 1 i 2

P 11 1 i 2i

2

P 11 1 i 2 2

3

P 11 1 i 2 3

P 11 1 i 2 2i

A

A

A

n21 n

P 11 1 i 2

n21

P 11 1 i 2 n

P 11 1 i 2 1 P 11 1 i 2i 5 P 11 1 i 2 2

P 11 1 i 2 2 1 P 11 1 i 2 2i 5 P 11 1 i 2 3

P 11 1 i 2 3i

P 11 1 i 2

P 1 Pi 5 P 11 1 i 2

P 11 1 i 2 3 1 P 11 1 i 2 3i 5 P 11 1 i 2 4 A

n21

i

P 11 1 i 2

n21

1 P 11 1 i 2 n21i 5 P 11 1 i 2 n

*Notice, the value in column (C) for the end of period (n 2 1) provides the value in column (A) for the end of period n.

various values of i and n. The quantity (1 1 i)n is referred to as the single sum, future worth factor. It is denoted (F Z P i%,n) and reads ‘‘the F, given P factor at i% for n periods.’’ The above discussion is summarized as follows. Let

P 5 the equivalent value of an amount of money at a present point in time, or present worth. F 5 the equivalent value of an amount of money at a future point in time, or future worth. i 5 the interest rate per interest period. n 5 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 5 P1F Z P i%,n2

Future Value The value of money at some future point in time.

(2.4)

A CFD depicting the relationship between F and P is given in Figure 2.3. Remember, F occurs n periods after P. In addition to solving numerically for the value of (1 1 i)n and using tabulated values in Appendix A of the single sum, future worth factor, (F Z 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), A comma may be placed between the factor identifier and the interest rate. Thus, (F Z P, i%,n) and (F Z P i%,n) are equivalent representations of (1 1 i)n. 2 Microsoft’s Excel® software is used throughout the text. Where a computer is used to generate a solution, software functions are shown in blue boldface type in the text. 1

2-2

Single Cash Flows

29

F

0

1

2 End of Period

n–1

n

F occurs n periods after P

P

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

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). 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: 5FV(i,n,,2P). 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, since no equal cash flow per period applies, and type is not included, since endof-period cash flows apply. Finally, notice that a negative value is entered for P, since 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. Recall Example 2.2, in which $10,000 was loaned for 5 years at 10 percent 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

The Excel® FV Financial Functions Used with a Single Sum of Money

F I GU R E 2 . 4

CFD of the Time Relationship Between P and F

F I G U RE 2 . 3

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or as a percentage, and the value of P can be entered as either a positive or a negative amount, since 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.

Repaying a 5-Year Loan with a Single Payment

EXAMPLE

Dia St. John borrows $1,000 at 12 percent compounded annually. The loan is to be paid back after 5 years. How much should she repay? KEY DATA

Given: Loan amount 5 $1,000; interest rate 5 12% compounded annually; length of loan 5 5 years Find: Value of the loan amount after 5 years

SOLUTION

Using the compound interest tables in Appendix A for 12 percent and 5 periods, the value of the single sum, future worth factor 1F Z P 12%,52 is shown to be 1.76234. Thus, F 5 P1F Z P 12%,52 5 $1,00011.762342 5 $1,762.34  

Using the Excel® FV worksheet function, F 5FV112%, 5,,210002 5 1762.34

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. 2.

3. 4. 5. 6.

Obtain an approximation by using the Rule of 72. Consult the tables in Appendix A for the stated interest rate and find the value of n that makes the (F Z P i%,n) factor equal 2, then interpolate as necessary. Solve mathematically for the value of n that makes (1 1 i)n equal 2. Use the Excel® NPER worksheet function. Use the Excel® GOAL SEEK tool. Use the Excel® SOLVER tool.

Example 2.4 explores each of these methods.

2-2

Doubling Your Money

Single Cash Flows

EXAMPLE

How long does it take for an investment to double in value if it earns (a) 2 percent, (b) 4 percent, or (c) 12 percent annual compound interest? 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 5 36 yrs b. n < 72/4 5 18 yrs c. n < 72/12 5 6 yrs The second approach is to consult Appendix A and, for each value of i, determine the value of n for which 1F Z P i%,n2 5 2. a. For i 5 2%, n is between 30 and 36; interpolating gives

n < 30 1 612.0000 2 1.811362/ 12.03989 2 1.811362 5 34.953 yrs. b. n < 17 1 12.0000 2 1.947902/ 12.02582 2 1.947902 5 17.669 yrs; c. n < 6 1 12.0000 2 1.973822/ 12.21068 2 1.973822 5 6.111 yrs. With the third approach, we solve mathematically for n such that (1 1 i)n 5 2 gives n 5 log 2/log11 1 i2. Therefore, the correct values of n (to 3 decimal places) are a. n 5 log 2/log 1.02 5 35.003 yrs; b. n 5 log 2/log 1.04 5 17.673 yrs; and c. n 5 log 2/log 1.12 5 6.116 yrs. 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 5 NPER12%,,21,22 5 35.003 yrs b. n 5 NPER14%,, 21,22 5 17.673 yrs and c. n 5 NPER112%,, 21,22 5 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  

SOLUTION

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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 5 2. The results obtained by GOAL SEEK are shown in Figure 2.5. Namely, a. n 5 34.999 yrs b. n 5 17.672 yrs c.

n 5 6.116 yrs

FIGURE 2 . 5

The Excel® Goal Seek Tools Used to Solve Example 2.4

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: s Max s Min d Value of: 2 By Changing Cells: Bx 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 5 2, where x denotes the row number in the spreadsheet. The results obtained by SOLVER are shown in Figure 2.6. Namely, a. n 5 35.003 yrs b. n 5 17.673 yrs c.

n 5 6.116 yrs

2-2

FIGURE 2.6

Single Cash Flows

The Excel® Solver Tool Used to Solve Example 2.4

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)

Since 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 5 P11 1 i2 n

(2.5)

Dividing both sides of Equation 2.5 by (1 1 i)n, we have the relation, P 5 F11 1 i2 2n

(2.6)

EXPLORING THE SOLUTION

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or P 5 F1P ƒ F i%,n2  

(2.7)

where (1 1 i)–n and (P Z F i%,n) are referred to as the single sum, present worth factor. In addition to solving numerically for the value of (1 1 i)–n and using tabulated values of the single sum, present worth factor, (P Z 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). To solve for P when given i, n, and F, the following can be entered in any cell in an Excel® spreadsheet: 5PV(i,n,,2F). Notice, as with the FV function, there are no spaces between the equal sign and the closing parenthesis; also, as before, since 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.

Saving Money

EXAMPLE

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 percent compounded annually. How much must be deposited today? KEY DATA

SOLUTION

Given: F 5 $10,000; i 5 5% compounded annually; n 5 4 years Find: P Using the compound interest tables in Appendix A for 5 percent and 4 periods, the value of the single sum, present worth factor, 1P Z F 5%,42, is shown to be 0.82270. Thus, P 5 F1P Z F 5%,42  

 

 

5 $10,00010.822702

5 $8,227.00 As shown in Figure 2.7, using the Excel® PV worksheet function, P 5PV(5%,4,,210000) 5 $8227.02

2-3 Multiple Cash Flows

35

The Excel® PV Financial Function Used with a Single Sum of Money

FIGURE 2.7

2-3

MULTIPLE CASH FLOWS

LEARNING OBJECTIVES: Distinguish between uniform, irregular, gradient and geometric series of cash flows. Perform time value of money calculations for multiple cash flows with annual compounding.

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. 2.3.1 Irregular Cash Flows LEARNING O BJECTI VE: Perform time value of money calculations for irreg-

ular cash flows with annual compounding.

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. If we let At denote the magnitude of a cash flow (receipt or disbursement) at the end of time period t, then P 5 A1 11 1 i2 21 1 A2 11 1 i2 22 1 A3 11 1 i2 23 1 p 1 An21 11 1 i2 21n212 1 An 11 1 i2 2n

(2.8)

or, using summation notation, n

P 5 a At 11 1 i2 2t t51

(2.9)

Video Lesson: Transformations—Multiple Cash Flows

36

Chapter 2

Time Value of Money Calculations

or, equivalently, n

P 5 a At 1P Z F i%,t2

(2.10)

 

t51

Computing the Present Worth of a Series of Cash Flows

EXAMPLE

Video Example

KEY DATA

SOLUTION

Consider the series of cash flows depicted by the CFD given in Figure 2.8. Using an interest rate of 6 percent per interest period, what is the present worth equivalent of cash flows? Given: Cash flow series in Figure 2.8; i 5 6% per interest period. Find: P The present worth equivalent is given by P 5 $3001P Z F 6%,12 2 $3001P Z F 6%,32 1 $2001P Z F 6%,42 1 $4001P Z F 6%,62 1 $2001P Z F 6%,82 5 $30010.943402 2 $30010.839622 1 $20010.792092 1 $40010.704962 1 $20010.627412 5 $597.02  

 

 

 

 

$400 $300

(+)

$200 0

1

2

3

4 5 End of Period

$200 6

7

8

(–) $300 FIGURE 2.8

CFD of Multiple Cash Flows

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 5 A1 11 1 i2 n21 1 A2 11 1 i2 n22 1 A3 11 1 i2 n23 1 p 1 An22 11 1 i2 2 1 An21 11 1 i2 1 An

(2.11)

2-3 Multiple Cash Flows

or, using summation notation, n

F 5 a At 11 1 i2 n2t

(2.12)

t51

or, equivalently, n

F 5 a At 1F Z P i%,n 2 t2

(2.13)

 

t51

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, since we know the value of future worth is given by F 5 P11 1 i2 n

(2.14)

substituting Equation 2.9 into Equation 2.14 yields n

F 5 11 1 i2 n a At 11 1 i2 2t t51

Hence, n

F 5 a At 1F Z P i%,n 2 t2

(2.15)

 

t51

Determining the Future Worth of a Series of Cash Flows

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 percent per interest period. Given: Cash flow series in Figure 2.8; i 5 6% per interest period. Find: F F 5 $3001F Z P 6%,72 2 $3001F Z P 6%,52 1 $2001F Z P 6%,42 1 $4001F Z P 6%,22 1 $200 5 $30011.503632 2 $30011.338232 1 $20011.262482 1 $40011.123602 1 $200 5 $951.56  

 

 

 

Video Example

KEY DATA

SOLUTION

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Alternately, since we know the present worth is equal to $597.02, F 5 $597.021F Z P 6%,82 5 $597.0211.593852 5 $951.56  

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.

Using the Excel NPV Function

EXAMPLE

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, since 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

FIGURE 2 . 9

Example 2.6

The Excel® NPV and FV Financial Functions Used to Solve

2-3

Multiple Cash Flows

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.

2.3.2

Uniform Series of Cash Flows

LEARNING O BJECTI VE: Perform time value of money calculations for a

uniform series of cash flows with annual compounding.

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 n

P 5 a A11 1 i2 2t

(2.16)

t51

where A is the magnitude of an individual cash flow in the series. Letting X 5 (1 1 i)21 and bringing A outside the summation yields n

P 5 A a Xt t51 n

5 AX a X t21 t51

Letting h 5 t 2 1 gives the geometric series n21

P 5 AX a Xh

(2.17)

h50

Since 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 n21 h aX 5

h50

1 2 Xn 12X

(2.18)

Replacing X with (1 1 i)21 yields the following relationship between P and A 11 1 i2 n 2 1 (2.19) P5A i11 1 i2 n more commonly expressed as P 5 A1P Z A i%,n2

(2.20)

where (P Z 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.

Video Lesson: Transformations— Uniform Series

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Computing the Present Worth of a Uniform Series of Cash Flows

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 percent compounded annually, how much should he deposit? KEY DATA

SOLUTION

Given: A 5 $2,000; i 5 5%; n 5 5 Find: P 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 5 $2,0001P Z A 5%,52 5 $2,00014.329482 5 $8,658.96  

0

A

A

1

2 End of Period

A

A

n–1

n

P occurs one period before the first A

P

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

FIGURE 2.10

CFD of the Relationship Between P and A in a Loan Transaction

Thus, if $8,658.96 is deposited in a fund paying 5 percent compounded annually, then five equal annual withdrawals of $2,000 can be made. After the fifth withdrawal, the fund will be depleted. 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: 5PV1i,n,2A2. 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 5PV15%,5,220002 5 8658.95

2-3 Multiple Cash Flows

Computing the Present Worth of a Delayed Uniform Series of Cash Flows

EXAMPLE

In Example 2.9, suppose the first withdrawal does not occur until 3 years after the deposit. How much should be deposited? Given: A 5 $2000; i 5 5%; n 5 5 Find: P at time 0

KEY DATA

As depicted in Figure 2.11, the value of P to be determined occurs at t 5 0, whereas a straightforward application of the (P Z A 5%,5) factor will yield a single sum equivalent at t 5 2, one period before the first A. Hence, to determine the present worth, the value obtained after using the (P Z A 5%,5) factor must be moved backward in time 2 years. The latter operation is performed using the (P Z F 5%,2) factor. Therefore,

SOLUTION

P 5 $2,0001P Z A 5%,52 1P Z F 5%,22 5 $2,00014.329482 10.907032 5 $7,853.94  

(+)

0

$2,000 1

(–)

2

 

$2,000

3 4 End of Year

$2,000

$2,000

$2,000

5

6

7

i = 5%

FIGURE 2.11

CFD for Example 2.10

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 5PV15%,2,,2PV15%,5,220002 2 5 $7853.93 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.)

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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. Since 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.

The reciprocal relationship between P and A can be expressed as i11 1 i2 n A5P 11 1 i2 n 2 1

(2.21)

or as A 5 P1A Z P i%,n2  

(2.22)

Capital Recovery Factor The amount of annual savings or recovery funds required to justify a capital investment over some period of time, n.

The expression (A Z P i%,n) is called the capital recovery factor, since it provides the annual savings or recovery of funds (A) required to justify the capital investment (P). The (A Z P i%,n) factor is used frequently in both personal financing and in comparing economic investment alternatives.

EXAMPLE

What Size Uniform Withdrawals Can Occur?

Video Example

KEY DATA

SOLUTION

Suppose Rachel Townsley deposits $10,000 into an account that pays 8 percent 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? Given: P 5 $10,000; i 5 8% compounded annually; n 5 10 Find: A Since we know that A and P are related by A 5 P1A Z P i%,n2  

then A 5 $10,0001A Z P 8%,102 5 $10,00010.149032 5 $1490.30  

2-3 Multiple Cash Flows

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. Using the Excel® PMT worksheet function, A 5PMT18%,10,2100002 5 1490.29

Determining the Size of Delayed Uniform Withdrawals

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?

(+)

0

1

(–)

2

A

A

A

A

A

A

A

A

A

A

3

4

5 6 7 End of Year

8

9

10

11

12

i = 8%

$10,000 FIGURE 2.12

CFD for the Deferred Payment Example

Given: P 5 $10,000; i 5 8% compounded annually; n 5 10 Find: Amount in the fund at t 5 2 (V2); then find A

KEY DATA

The amount in the fund at t 5 2 equals

SOLUTION

V2 5 $10,0001F Z P 8%,22 5 $10,00011.166402 5 $11,664  

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Therefore, the size of the equal annual withdrawal will be A 5 $11,664.001A Z P 8%,102 5 $11,664.0010.149032 5 $1,738.29  

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 5PMT18%,10,2FV18%,2,,2100002 2 5 $1738.28

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 percent annual compound interest. Instead of using the embedded approach, separate Excel® calculations could be performed, as shown below: V2 5FV18%,2,,2100002 5 $11,664.00 A 5PMT18%,10,2116642 5 $1738.28

The future worth of a uniform series is obtained by recalling that F 5 P11 1 i2 n

(2.23)

Substituting Equation 2.19 into Equation 2.23 for P and reducing yields 11 1 i2 n 2 1 F5A (2.24) i or, equivalently, F 5 A1F Z A i%,n2  

(2.25)

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

EXAMPLE

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 percent interest compounded annually?

2-3

Given: A 5 $1000; i 5 6% compounded annually; n 5 30 Find: F

Multiple Cash Flows

KEY DATA

F 5 $1,0001F Z A 6%,302 5 $1,000179.058192 5 $79,058.19

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 5FV16%,30,10002 5 $78,058.19

The reciprocal relationship between A and F is easily obtained from Equation 2.24. Specifically, we find that A5F

i 11 1 i2 n 2 1

(2.26)

or, equivalently, A 5 F1A Z F i%,n2

(2.27)

 

The expression (A Z F i%,n) is referred to as the sinking fund factor, since 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.

F occurs at the same time as the last A

F

F = A(F|A i%,n) A = F(A|F i%,n) End of Period 0

1

2

n–1

A

A

A

n A

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

FIGURE 2.13

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

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Who Wants to Be a Millionaire?

EXAMPLE

Suppose Crystal Wilson wants to accumulate $1,000,000 by the time she retires in 40 years. If she earns 10 percent on her investments, how much must she invest each year in order to realize her goal? KEY DATA

SOLUTION

Given: F 5 $1,000,000; i 5 10%; n 5 40 Find: A Applying Equation 2.27, A 5 $1,000,0001A Z F,10%,402 5 $1,000,00010.00225942 5 $2,259.40/year Using the Excel® PMT worksheet function gives A 5PMT110%,40,,210000002 5 $2,259.41/year

2.3.3 Gradient Series of Cash Flows LEARNING OBJECTIVE: Perform time value of money calculations for a

Video Lesson: Transformations—Gradient and Geometric Series Gradient Series A series characterized by cash flows that increase by a constant amount (G) each period.

gradient series of cash flows with annual compounding.

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 5 1t 2 12G  t 5 1, p , n

(2.28)

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.

2-3

Composite series A1 + (n – 2)G

A1

0

1

A1 + G

2

A1 + (n – 1)G

A1 + 2G

...

3 End of Period

n–1

n

P = P1 + P2

=

0

A1

A1

1

2

Uniform series A1

...

3 End of Period

A1 n–1

n

+

P1

Gradient series (n – 2)G G 0

A1

1

2

(n – 1)G

2G

...

3 End of Period

n–1

n

P2

CFD of a Combination of Uniform and Gradient Series in an Investment Scenario

FIGURE 2.14

The present worth equivalent of a gradient series is obtained by recalling n

P 5 a At 11 1 i2 2t t51

(2.29)

Multiple Cash Flows

47

48

Chapter 2

Time Value of Money Calculations

Substituting Equation 2.28 into Equation 2.29 gives n

P 5 G a 1t 2 12 11 1 i2 2t

(2.30)

t51

which reduces to P 5 Gc

1 2 11 1 ni2 11 1 i2 2n d i2

(2.31)

or, equivalently, P 5 G1P Z G i%,n2  

(2.32)

where (P Z G i%,n) is the gradient series, present worth factor and is tabulated in Appendix A. The uniform series equivalent to the gradient series is obtained by multiplying the value of the gradient series, present worth factor by the value of the (A Z P i%,n) factor: A 5 Gc

n 1 2 1A Z F i%,n2 d i i  

or, equivalently, A 5 G1A Z G i%,n2  

(2.33)

where the factor (A Z 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, multiply the value of the (A Z G i%,n) factor by the value of the (F Z A i%,n) factor to obtain the (F Z G i%,n) factor: 11 1 i2 n 2 11 1 ni2 d F 5 Gc i2

(2.34)

The (F Z 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.

2-3 Multiple Cash Flows

Determining the Present Worth of a Gradient Series (Sitting on Top of a Uniform Series) 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 percent compounded annually, determine the present worth equivalent for the maintenance costs. 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. Given: The maintenance costs—a composite series of cash flows as shown in Figure 2.15 Find: P of the composite series

Composite series 0

1

2

3

4

5

$3,000 $4,000 $5,000

$6,000 $7,000

= Uniform series 0

1

2

3

4

5

$3,000

$3,000

$3,000

$3,000

$3,000

+ Gradient series 0

1

2 $1,000

3

4

5

$2,000 $3,000 $4,000

FIGURE 2.15

CFD for Example 2.15

EXAMPLE

Video Example

KEY DATA

49

50

Chapter 2

Time Value of Money Calculations

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 5 $1,0001A Z G 8%,52 5 $1,00011.846472 5 $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 5 $1,846.47 1 $3,000, or $4,846.47. Therefore, converting the uniform series to its present worth equivalent, P 5 $4,846.471P Z A 8%,52 5 $4,846.4713.992712 5 $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 5 $4,846.471F Z A 8%,52 5 $4,846.4715.866602 5 $28,432.30  

Similarly, using the present worth equivalent, the future worth equivalent is F 5 $19,350.551F Z P 8%,52 5 $19,350.5511.469332 5 $28,432.34  

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: 5NPV18%,3000,4000,5000,6000,70002 or 51000*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 5PMT18%,5,219350.562 5 $4846.47 F 5FV18%,5,,219350.562 5 $28,432.32

2-3

2.3.4

Multiple Cash Flows

51

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 flow’s size from one period to the next, the size of the tth cash flow can be given by At 5 At21 11 1 j2  t 5 2, p , n

or, more conveniently,

At 5 A1 11 1 j2 t21  t 5 1, p , n

(2.35)

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 percent a year, then the resulting series representation of labor costs will be geometric. A1 (1 + j)(n – 1) A1 (1 + j)n – 2

A1

0

1

A1 (1 + j)

A1 (1 + j)2

...

2

3 End of Period

n–1

n

P FIGURE 2.16

CFD for an Increasing Geometric Series

The present worth equivalent of the cash flow series is obtained by substituting Equation 2.35 into Equation 2.9 to obtain n

P 5 A1 11 1 j2 21 a 11 1 j2 t 11 1 i2 2t

(2.36)

t51

This can be written as: 1 2 11 1 j2 n 11 1 i2 2n d i2j P5 • n A1/ 11 1 i2 A1 c

  i ? j   i 5 j

(2.37)

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

52

Chapter 2

Time Value of Money Calculations

or P 5 A1 1P Z A1 i%, j%,n2  

(2.38)

where (P Z 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.

Determining the Present Worth of a Geometric Series

EXAMPLE

A company is considering purchasing a new machine tool. In addition to the initial purchase and installation costs, management is concerned about the machine’s maintenance costs, which are expected to be $1,000 at the end of the first year of the machine’s life and increase 8 percent/year thereafter. The machine tool’s expected life is 15 years. Company management would like to know the present worth equivalent for expected costs. If the firm’s time value of money is 10 percent/year compounded annually, what is the present worth equivalent? KEY DATA

Given: Machine maintenance costs – a geometric series with A1 5 $1,000, j 5 8%, i 5 10%, n 5 15 Find: P – the present worth equivalent of all maintenance costs

SOLUTION

P 5 $1,0001P Z A1 10%,8%,152 5 $1,000112.030402 5 $12,030.40 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: 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, 5NPV ≥ ¥ 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 51000*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)

2-3 Multiple Cash Flows

FIGURE 2.17

Excel® Solution to Example 2.16

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.)

A uniform series equivalent to the geometric series is obtained by multiplying the value of the geometric series, present worth factor by the value of the (A Z P i%,n) factor resulting in 1 2 11 1 j2 n 11 1 i2 2n i11 1 i2 n i?j dc d i2j 11 1 i2 n 2 1 ¥      (2.39) A5 ≥ i11 1 i2 n21 i 5 j nA1a b 11 1 i2 n 2 1 A1 c

or A 5 A1(A Z A1 i%,j%,n)

(2.40)

53

54

Chapter 2

Time Value of Money Calculations

where (A Z A1 i%,j%,n) is the geometric-to-uniform series conversion factor. This factor is not tabulated in Appendix A. The future worth equivalent of the geometric series is obtained by multiplying the value of the geometric series, present worth factor by the (F Z P i%,n) factor to obtain 11 1 i2 n 2 11 1 j2 n d A1 c i2j F5 µ nA1 11 1 i2 n21

i?j i5j

(2.41)

or F 5 A1 1F Z A1 i%,j%,n2 where (F Z A1 i%,j%,n) is the geometric series, future worth factor and is tabulated in Appendix A.

Determining the Future Worth of a Geometric Series

EXAMPLE

Video Example

KEY DATA

SOLUTION

Mattie Bookhout receives an annual bonus and deposits it in a savings account that pays 8 percent compounded annually. The size of her bonus increases by 10 percent each year; her initial deposit is $500. Determine how much will be in the fund immediately after her tenth deposit. Given: A1 5 $500, i 5 8%, j 5 10%, and n 5 10 Find: F  

The value of F is given by F 5 $5001F ZA1 8%,10%,102 5 $500121.740872 5 $10,870.44 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.

2-4 Compounding Frequency

FIGURE 2.18

2-4

55

Excel® Solution to Example 2.17

COMPOUNDING FREQUENCY

LEARNING O BJECTI VE: Perform time value of money calculations for mul-

tiple compounding periods per year using the period interest rate and the effective annual interest rate.

Thus far, when referring to an interest rate, we said that 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 percent 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 percent 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 percent per month, an alternate way of expressing the interest rate is 18 percent per annum compounded monthly, or 18 percent per year per month. When expressed in this form, 18 percent is known as the nominal annual interest rate. We designate the nominal rate3 as r. 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.

Nominal Annual Interest Rate The annual interest rate without adjustments for compounding.

56

Chapter 2

Time Value of Money Calculations

Period Interest Rate The nominal annual interest rate divided by the number of interest periods per year.

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 interest rate approach. 2.4.1 Period Interest Rate Approach

To utilize the period interest rate approach, we must define a new term— the period interest rate: Period interest rate 5

Nominal annual 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.

Determining Future Worth with Multiple Compounding Periods per Year

EXAMPLE

Two thousand dollars is invested in an account. What is the account balance after 3 years when the interest rate is: a. 12 percent per year compounded monthly? b. 12 percent per year compounded semiannually? c. KEY DATA

SOLUTION

12 percent per year compounded quarterly?

Given: P 5 $2,000; nominal interest rate 5 12%; duration of investment 5 3 years Find: For each scenario: period interest rate; number of interest periods; F a. Nominal annual interest rate 5 12%/year

Number of interest periods/year 5 12 months/year Period interest rate 5  

12%/year 5 1%/month 12 months/year  

2-4 Compounding Frequency

Number of interest periods 5 3 years(12 months/year) 5 36 months F 5 $2,0001F Z P 1%,362 5 $2,00011.430772 5 $2,861.54  

b. Nominal annual interest rate 5 12%/year

Number of interest periods/year 5 2 semiannual periods/year Period interest rate 5  

12%/year 5 6%/semiannual period 2 semiannual periods/year  

Number of interest periods 5 3 years (2 semiannual periods/year) 5 6 semiannual periods F 5 $2,0001F Z P 6%,62 5 $2,00011.418522 5 $2,837.04  

c.

Nominal annual interest rate 5 12%/year Number of interest periods/year 5 4 quarters/year Period interest rate 5  

12%/year 5 3%/quarter 4 quarters/year  

Number of interest periods 5 3 years (4 quarters/year) 5 12 quarters F 5 $2,0001F Z P 3%,122 5 $2,00011.425762 5 $2,851.54  

Determining Car Payments Rebecca Carlson purchases a car for $25,000 and finances her purchase by borrowing the money at 8 percent 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?

EXAMPLE

Video Example

57

58

Chapter 2

Time Value of Money Calculations

KEY DATA

SOLUTION

Given: P 5 $25,000; nominal annual interest rate 5 8%/year; duration of loan 5 5 years; number of interest periods/year 5 12 months/year Find: Period interest rate; number of interest periods; A Period interest rate 5  

8%/year 5 0.66667%/month 12 months/year  

Number of interest periods 5 5 years(12 months/year) 5 60 months A 5 $25,0001A Z P 0.66667%,602 5 $25,000 £  

0.006666711.00666672 60 § 11.00666672 60 2 1

5 $506.91/month Using the Excel® PMT worksheet function, A 5PMT10.08/12,60,2250002 5 $506.91

2.4.2 Effective Annual Interest Rate Video Lesson: Effective Annual Interest Rate Effective Annual Interest Rate The annual interest rate that is equivalent to the period 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. For example, if the interest rate is 12 percent per year compounded quarterly, then the nominal annual interest rate is 12 percent, and there are four interest periods per year. Thus, the period interest rate is 3 percent per quarter. Hence, $1 invested for 1 year at 3 percent per quarter has a future worth of F 5 $1(F Z P 3%,4) 5 $1(1.12551) 5 $1.12551 To obtain the same value in 1 year requires an annual compound interest rate of 12.551 percent. This value is called the effective annual interest rate and is given by (1.03)4 2 1 5 0.12551, or 12.551 percent. 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).

2-4 Compounding Frequency

The general equation for the effective annual interest rate, ieff, is ieff 5 11 1 r/m2 m 2 1 5 EFFECT1r,m2

(2.43)

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

Calculating the Effective Annual Interest Rate

EXAMPLE

Calculate the effective annual interest rate for each of the following cases: (a) 12 percent per year compounded quarterly; (b) 12 percent per year compounded monthly; (c) and 12 percent per year compounded every minute. a. Twelve percent per year compounded quarterly: r 5 12% ; m 5 4

From Equation 2.43, ieff 5 (1 1 0.12/4)4 2 1 5 (1.03)4 2 1 5 0.12551 5 12.551% Using the Excel® EFFECT worksheet function gives the same result: ieff 5EFFECT112%,42 5 12.551% b. Twelve percent per year compounded monthly: r 5 12%; m 5 12

From Equation 2.43, ieff 5 (1 1 0.12/12)12 2 1 5 (1.01)12 2 1 5 0.12683 5 12.683% Using the Excel® EFFECT worksheet function gives the same result: ieff 5EFFECT112%,122 5 12.683% c.

Twelve percent per year compounded every minute: r 5 12%; m 5 525,600 From Equation 2.43, to eight decimal places, ieff 5 11 1 0.12/525,6002 525,600 2 1 5 0.12749684 5 12.749684%

Using the Excel® EFFECT worksheet function gives ieff 5EFFECT112%,5256002 5 12.749684%

SOLUTION

59

60

Chapter 2

Time Value of Money Calculations

Making Monthly House Payments

EXAMPLE

Greg Wilhelm 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 percent compounded monthly. The closing fee on the loan is $2,000. a. What is Greg’s monthly payment on the loan? b. What is Greg’s monthly payment if he chooses to finance the closing

fee along with the loan? c. What is the effective annual interest rate on the loan? KEY DATA

Given: Loan amount 5 $100,000; closing cost 5 $2,000; nominal annual r 5 6% (compounded monthly); n 5 360 Find: A, ieff

SOLUTION

a. In order to calculate Greg’s monthly payment we first need to deter-

mine the period interest rate: period interest rate 5

6%/year 5 0.5%/month 12 months/year

Greg’s monthly payment on the 30-year loan is calculated as follows: A 5 $100,0001A Z P 0.5%,3602 5 $100,00010.00599552 5 $599.55  

or, using the Excel® PMT worksheet function, A 5PMT10.06/12,360,21000002 5 $599.55 b. If Greg chooses to finance the closing costs also, then his monthly pay-

ment will be A 5 $102,0001A Z P 0.5%,3602 5 $102,00010.00599552 5 $611.54  

or A 5PMT10.06/12,360,21020002 5 $611.54

2-4 Compounding Frequency

c.

If Greg made 360 payments of $611.54/month for $102,000, then the effective annual interest rate on the loan would be ieff 5EFFECT(12*RATE(360,611.54,2102000),12) 5 6.1678%

2.4.3 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? 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 5 11 1 r/m2 m/k 2 1

Video Lesson: Effective Rate per Payment Period

(2.44)

Equation 2.44 results from setting the effective annual interest rate for the stated compounding frequency of money equal to that for the cash flow frequency: 11 1 i2 k 2 1 5 11 1 r/m2 m 2 1

(2.45)

and solving for i.

When Cash Flow Frequency Does Not Match Compounding Frequency

EXAMPLE

What size monthly payments should occur when $10,000 is borrowed at 8 percent per year compounded quarterly and the loan is repaid with 36 equal monthly payments? Given: P 5 $10,000; r 5 8% (compounded quarterly); n 5 36 Find: A

KEY DATA

61

62

Chapter 2

Time Value of Money Calculations

SOLUTION

From Equation 2.44, r 5 0.08, k 5 12, and m 5 4. Therefore, i 5 11 1 0.08/42 4/12 2 1 5 0.006623 or 0.6623%/month  

Knowing the effective monthly interest rate, the monthly payment can be determined: A 5 $10,0001A Z P 0.6623%,362 5 $10,0003 10.0066232 11.0066232 36 y11.0066232 36 2 1 4 5 $313.12  

Using the Excel® PMT worksheet function, A 5PMT11.02^11/3221,36,2100002 5 $313.12.

SUMMARY

KEY CONCEPTS 1. Learning Objective: 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.

$5,000

$5,000

$5,000

(+) 0 (–)

1

2

3

4

Time $2,000 $3,000

$4,000 FIGURE 2 . 19

An Example Cash Flow Diagram

5

Summary 63

2. Learning Objective: 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

Formula

P 5 F 11 1 i 2

Factor Notation 2n

F 5 P 11 1 i 2 n

Cash Flow Diagram

P 5 F 1P Z F i%,n2

F

 

F 5 P 1F Z P i%,n2  

0

1

2 n–1 End of Period

n

P

3. Learning Objective: Distinguish between uniform, irregular, gradient, and geometric series of cash flows. (Section 2.3)

Series of cash flows may exhibit a pattern or be irregular. Recognizing patterns when they occur can facilitate analysis of a particular investment alternative. Three common patterns of cash flows are uniform, gradient, and geometric: ■





Uniform cash flows involve sums of equal (or uniform) value for each period. An example is the uniform payments of a monthly car loan or home mortgage payment. Gradient series cash flows increase by a constant value in each period. For example, maintenance costs for a piece of equipment may increase by a constant value each year. Geometric series cash flows increase or decrease by a constant percentage each period. For example, labor costs may increase by a constant percentage each year.

The use of a CFD is helpful in illustrating the conversion of these positive and negative cash flows into a single value. 4. Learning Objective: Perform time value of money calculations for multiple cash flows with annual compounding. (Section 2.3)

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 for irregular, gradient, and geometric series are summarized in the table that follows.

64

A5Pc F 5 Ac A5Fc

• Uniform Series from Present Value

• Future Worth of Uniform Series

• Uniform Series from Future Value

P 5 Gc A 5 Gc

F 5 Gc

• Uniform Series Equivalent of a Gradient Series

• Future Worth of a Gradient Series

d

d

d

i2

11 1 i 2 n 2 11 1 ni 2

 

n 1 2 1A Z F i%,n2 d i i

i2

d

1 2 11 1 ni 211 1 i 2 2n

i d 11 1 i 2 n 2 1

i

11 1 i 2 n 2 1

11 1 i 2 n 2 1

i 11 1 i 2 n

i 11 1 i 2

n

11 1 i 2 2 1

• Present Worth of a Gradient Series

Gradient Series Cash Flows

P 5 Ac

• Present Worth of Uniform Series

n

d

P 5 A1 11 1 i 2 21 1 A 2 11 1 i 2 22 1 A 3 11 1 i 2 23 1 p 1 An21 11 1 i 2 21n212 1 An 11 1 i 2 2n

Irregular Series Cash Flows

Uniform Series Cash Flows

Formula

Series Cash Flows  

 

F 5 G 1F Z G i%,n2

 

A 5 G 1A Z G i%,n2

 

P 5 G 1P Z G i%,n2

 

A 5 F1A Z F i%,n2

 

F 5 A1F Z A i%,n2

 

A 5 P1A Z P i%,n2

P 5 A1P Z A i%,n2

t51

P 5 a At 1P Z F i%,t 2

n

Factor Notation

P

0

0

P

0

P

0

1

A

1

1

A

1

A1

3

2

G

A

n–1

A

...

A

n –1

3 End of Period

2G

End of Period 2

... End of Period

2 End of Period

A

A2

2

A3

n A

F

n

A

n–1

An – 1

n–1

(n – 2)G

Cash Flow Diagram

n

(n – 1)G

An

n

• Future Worth of a Geometric Series

• Uniform Series Equivalent of a Geometric Series

• Present Worth of a Geometric Series

Geometric Series Cash Flows

nA1 y11 1 i 2

i2j

1 2 11 1 j 2 n 11 1 i 2 2n d i5j

iZj

F5 µ i2j

11 1 i 2 n 2 11 1 j 2 n

nA1 11 1 i 2 n21

A1 c

A1 c

d i5j

i?j

i 11 1 i 2 n 1 2 11 1 j 2 n 11 1 i 2 2n 2 d c d i2j 11 1 i 2 n 2 1 A5 µ i 11 1 i 2 n21 nA1 a b 11 1 i 2 n 2 1

P5 µ

A1 c

i5j

iZj

F 5 A1 1F Z A1 i%,j%,n2

A 5 A1(A Z A1 i%,j%,n)

 

P 5 A1 1P Z A1 i%,j%,n2

P

0

1

A1

2

... 3 End of Period

A1 (1 + j)

A1 (1 + j)2

n–1

A1 (1 +

n

A1 (1 + j)(n – 1) j)n – 2

65

66

Chapter 2

Time Value of Money Calculations

5. Learning Objective: Perform time value of money calculations for multiple compounding periods per year by utilizing the period interest rate and the effective annual interest rate. (Section 2.4)

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: Period interest rate 5

b.

Nominal annual interest rate Number of interest periods per year

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 5 11 1 rym2 m 2 1 5 EFFECT1r,m2

(2.43)

When cash flow and compounding frequencies differ, the effective interest rate per cash-flow period i is obtained by the equation i 5 11 1 rym2 m/k 2 1

(2.44)

where r 5 the nominal annual interest rate for money, m 5 the number of compounding periods in a year, and k 5 the number of cash flows in a year.

KEY TERMS Capital Recovery Factor, p. 42 Cash Flow Diagram (CFD), p. 24 Compounding, p. 26 Effective Annual Interest Rate, p. 58 Future Value, p. 28 Geometric Series, p. 51 Gradient Series, p. 46

Interest Rate, p. 26 Nominal Annual Interest Rate, p. 55 Period Interest Rate, p. 56 Present Value, p. 27 Sinking Fund Factor, p. 45 Uniform Series, p. 26

Summary 67

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1.

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 c. 32 years b. 19 years d. Cannot be determined without knowing the amount invested.

2.

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 5 P (1 1 in) c. F 5 P (1 1 n)i n b. F 5 P (1 1 i) d. F 5 P (1 1 ni)n 2 1

3.

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? 200 X

0 a. $246 b. $165

1

X

2

3

c. $200 d. $146

4.

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 c. $12,043 b. $9,359 d. $10,043

5.

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 c. $9,542 b. $10,258 d. $943

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6.

If you invest $5,000 three years from now, how much will be in the account 15 years from now if i 5 10% compounded annually? a. $8,053 c. $20,886 b. $15,692 d. $27,800

7.

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 c. infinite series b. geometric series d. uniform series

8.

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 c. $4,843,380 b. $2,281,830 d. $5,281,830

9.

What is the effective annual interest rate if the nominal annual interest rate is 24% per year compounded monthly? a. 2.00% c. 26.82% b. 24.00% d. 27.12%

10.

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

11.

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 c. $3,940 b. $3,840 d. $4,170

12.

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

13.

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 c. 25.0 years b. 15.5 years d. 18.0 years

Summary 69

PROBLEMS Introduction 1.

You are offered $200 now plus $100 a year from now for your used computer. Since 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?

2. 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? 3. What words comprise the abbreviation “DCF”? Tell/describe/define what it means in 10 words or less. 4. State the four DCF rules. Section 2.1 Cash-Flow Diagrams 5. Pooi Phan needs $2,000 to pay off her bills. She borrows this amount from a

6.

7.

8.

9.

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. 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. 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 5 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. 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. GO Tutorial 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

70

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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. 10. 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. 11. 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 5 0 (now), purchase blender for $62,000. At t 5 0, install at cost of $8,000. At t 5 1, savings generated by blender is $10,000. At t 5 1, maintenance costs of $800. At t 5 2, savings generated by blender are $12,000. At t 5 2, maintenance costs of $1,200. At t 5 3, savings generated by blender is $18,000. At t 5 3, maintenance costs of $1,600. At t 5 4, 5, 6, 7, 8, 9, 10, savings generated are $24,000 and maintenance is $4,000. At t 5 10, the blender is sold for $8,000. At t 5 10, blender removal costs are $1,600. 12.

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 2.2.1

Single Cash Flows—Future Worth Calculations

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? 14. Video Solution How much will a $25,000 investment today be worth in 10 years if it earns 7% annual compound interest? 15. 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% d. 15% b. 7% e. 20% c. 10% 13.

Summary 71

16. 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? 17.

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?

18.

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?

19.

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

20. At what interest rate will money: a. Double itself in 10 years? b. Triple itself in 10 years? c. Quadruple itself in 10 years? 21.

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?

22. If you invest $1,500 today and withdraw $2,500 in 3 years, what interest rate

was earned? 23.

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.

24. 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? 25.

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?

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Section 2.2.2

Single Cash Flows—Present Worth Calculations

26. Develop a single spreadsheet that allows you to calculate F Z P and P Z F fac-

tors. 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 Z 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. 27.

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?

28. How much money would have to be deposited today to accumulate: a. b. c. d.

$10,000 after 6 years if the investment earns 5%/year compounded annually? $6,500 after 4 years if the investment earns 8%/year compounded annually? $3,400 after 12 years if the investment earns 6%/year compounded annually? $13,500 after 5 years if the investment earns 10%/year compounded annually?

29.

GO Tutorial What deposit today is required for it to be worth $150,000 in 25 years if the deposit earns 5% annual compound interest?

30.

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®.

31.

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?

32. What is the present value of the following future receipts? a. b. c. d. e.

$19,000 5 years from now at 9% compounded annually $8,300 12 years from now at 15% compounded annually $6,200 53 years from now at 12% compounded annually $13,000 18 years from now at 19.2% compounded annually $5,000 10 years from now at 8% compounded annually

33. Calculate using the interest formula the factor (PZF11.5%,37). Compare that

to the result obtained using Excel®’s PV function. 34. 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? 35. How much money today is equivalent to $10,000 in 12 years, with interest

at 10% compounded annually?

Summary 73

36. 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 Z F formula directly in Excel® or your calculator. c. Use an appropriate Excel® function. 37. If a fund pays 12 percent compounded annually, what single deposit now will

accumulate $12,000 at the end of the 10th year? If the fund pays 6 percent compounded annually, what single deposit now is required in order to accumulate $6,000 at the end of the tenth year? Section 2.3.1

Irregular Series Cash Flows

38. 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

$0

4

1

2$500

5

$500

2

$200

6

3

$400

7

2$200 $100

2$300

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. 39.

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%. Video Solution An investment has the following cash flow series where

40.

interest is 8%: End of Year

Cash Flow

End of Year

Cash Flow

0 1 2 3

$300 $300 $600

$0 $800 $700

2$500

5 6 7 8

4

2$300

$600

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. 41.

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?

74

Chapter 2

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42. Renaldo borrows $8,000 from his aunt today to help pay for college ex-

penses. 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

$8,000

5

2$X

1

$0

6

2($X 1 100)

2

$0

7

2($X 1 400)

3

$0

8

2$2X

4

$0

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. 43.

The cash flow profile for an investment is given below and the interest rate is 8% compounded annually. End of Year 0 1 2 3

Net Cash Flow $0 $500 2$200 $600

End of Year

Net Cash Flow

4 5

2$100

$300

6

$200

Find the present worth of this series using the actual cash flows. Find the future worth of this cash flow series using the actual cash flows. Find the future worth using the present worth. Find the worth of the series at EOY 4 using the individual cash flows. Find the present worth using the worth at EOY 4. 44. 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. 45. 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: a. b. c. d. e.

End of Year

Cash Flow

End of Year

Cash Flow

0

$20,000

5

2$2X

1

$0

6

2$3X

2

$0

7

2$4X

3

$0

8

2$5X

4

2$X

Summary 75

a. b. c. d.

Draw the cash flow diagram from Ken’s perspective. Find the value of X such that the loan is fully repaid with the last payment. What is the dollar amount of each of the 5 payments. 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?

46. Maria deposits $1,200, $500, and $2,000 at t 5 1, 2, and 3, respectively. If

the fund pays 8 percent compounded per period, what sum will be accumulated in the fund at (a) t 5 3, and (b) t 5 6? 47. Juan deposits $1,000 in a savings account that pays 8 percent 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 2.3.2

Uniform Series Cash Flows

48. A debt of $1,000 is incurred at t 5 0. What is the amount of four equal

payments at t 5 1, 2, 3, and 4 that will repay the debt if money is worth 10 percent compounded per period? 49. Develop a single spreadsheet that allows you to calculate any of the P Z A, A Z P

F Z A, or A Z 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 Z 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. 50. 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? 51.

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?

52. Five deposits of $500 each are made a t 5 1, 2, 3, 4, and 5 into a fund paying

6 percent compounded per period. How much will be accumulated in the fund at (a) t 5 5, and (b) t 5 10? 53.

GO Tutorial 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?

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Chapter 2

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54. 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? 55.

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%.

56. 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? 57. What equal annual deposits must be made at t 5 2, 3, 4, 5, and 6 in order

to accumulate $25,000 at t 5 8 if money is worth 10 percent compounded annually? 58. 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? 59.

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?

60. If annual deposits of $1,000 are made into a fund paying 12 percent interest

compounded annually, how much money will be in the fund immediately after the 5th deposit? 61. 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? 62. 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? 63.

You deposit $1,000 in a fund at the end of each year for a 10-year period. The fund pays 5% compounded annually. How much money is available to withdraw immediately after your last deposit?

64. 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.)

Summary 77

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? 65.

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?

66. 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? 67.

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?

68. 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? 69. 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? 70. 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 com-

pounded 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%? 71.

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?

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Chapter 2

Time Value of Money Calculations

b. $12,000 is invested at the end of each of 10 years at 10%/year com-

pounded 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? 72.

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?

73.

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?

74. Determine the equivalent annual cash flow of this series at 10% interest: End of Year

75.

Cash Flow

End of Year

Cash Flow

0

2$2,500

5

$0

1

$3,000

6

2 3 4

$4,500 $0

7 8

2$1,000 $7,000 $3,000

2$5,000

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?

76. 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? 77.

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.

Summary 79

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? 78. 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-of-year 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 5 $1,000, n 5 5, and i 5 10% for each approach.

Section 2.3.3 79.

Gradient Series Cash Flows

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%.

80. Develop a single spreadsheet that allows you to calculate the P Z G, F Z G, and

A Z 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 Z G factor, your control panel needs to include i and n. Test the P Z G and A Z G factors against the tables in Appendix A to ensure they work right. Keep this handy. It will be useful to you. 81. John borrows $10,000 at 18 percent 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? 82. Solve problem 81 for the case in which each successive payment is $700 less

than the previous payment. 83.

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?

84.

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?

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85.

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?

86. Consider the following cash flow profile: EOY 0 1 2 3 4

Cash Flow

EOY

2$75,000 $3,000 $6,000 $9,000 $12,000

Cash Flow

5

$15,000

6 7 8

$18,000 $21,000 $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% 87.

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.

88. 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. 89.

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 5 10 b. What uniform annual series of deposits (n 5 10) would result in the same accumulated balance at the end of year 10.

90. Consider the following cash flow profile: EOY

Cash Flow

EOY

0

2$45,000 $12,000 $11,000 $10,000 $9,000

5

$8,000

6 7 8

$7,000 $6,000 $5,000

1 2 3 4

Cash Flow

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?

Summary 81

91. In Problem 90, what uniform annual series over [4,7] will be equivalent to

the cash flow profile if money is worth 6% compounded annually? 92. In Problem 90, 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? Consider a loan of $10,000 and the following pattern of cash flows.

93.

End of Year 0

Cash Flow 2$10,000 $3,000 $4,000

1 2

End of Year

Cash Flow

3

$5,000

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 2$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? 94. Consider the following cash flow profile:

EOY

Cash Flow

EOY

0

2$50,000 $13,000 $12,000 $11,000 $10,000

5

1 2 3 4

6 7 8

Cash Flow

$9,000 $8,000 $7,000 $6,000

What is the present worth equivalent for the cash flow series with an interest rate of 12%? 95. In Problem 94, using an interest rate of 10%, what uniform series over the

closed interval [1,8] is equivalent to the cash flow profile shown? 96. In Problem 94, using an interest rate of 8%, what single sum of money oc-

curring at the end-of-year 8 is equivalent to the cash flow profile shown? 97. In Problem 94, 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 5 1 and a value of 8X at EOY 5 8? 98. 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 5 25? b. What uniform annual series of deposits (n 5 25) would result in the same accumulated balance at t 5 25?

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99. 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 1 X INR at the end of the 2nd year, 2 million 1 2X INR at the end of the 3rd year, 2 million 1 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? 100. Miller Machining needs to purchase a piece of machinery to be able to com-

pete 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? 101.

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 5 27,000 1 [1,850 1 200(A Z G 8%,6)](P Z A 8%,6)(P Z F 8%,4)

102. 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 1 19,500 at time 40). Interest is 8% compounded annually. What is X? 103. Your friend claims that the following series of payments is absolutely worth-

less since they add up to $0: End of Year

Cash Flow

0

$100 $80 $60 $40 $20 $0

1 2 3 4 5

End of Year 6 7 8 9 10

Cash Flow

2$20 2$40 2$60 2$80 2$100

The time value of money is 18%. Determine the present value of these cash flows.

Summary 83

104. 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 5 800 1 950(P Z A i%,4) 2 450(P Z G i%,4) 2 600(P Z A i%,3)(P Z F i%,4) 105. 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 percent annual compound interest rate. Each payment is to be $3,000 more than the previous payment. Determine the size of the last payment. 106. Solve Problem 105 for the case in which each successive payment is to be

$3,000 less than the previous payment. 107. 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. 108. 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 Z G i%,n) factor. Develop an (F Z G i%,n) factor. a. Express the (F Z G i%,n) factor in terms of the existing factors listed in the tables of Appendix A. b. Express the (F Z 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 5 and i 5 10% for each approach. 109. 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 percent compounded annually over the 15-year period, how much was withdrawn at each withdrawal point? 110. An easy payment plan offered by a local electronics store for your new au-

dio 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 2.3.4

Geometric Series Cash Flows

111. Solve Problem 81 for the case in which each successive payment is to be

10% greater than the previous payment. 112. Solve problem 81 for the case in which each successive payment is to be

10% less than the previous payment.

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113. Solve Problem 105 for the case in which each successive payment is to be

12% greater than the previous payment. 114. Solve problem 105 for the case in which each successive payment is to be

20% less than the previous payment. 115. Consider the following cash flow profile: EOY

Cash Flow

EOY

0

2$45,000 $12,000 $11,000 $10,000 $9,000

5

$8,000

6 7 8

$7,000 $6,000 $5,000

1 2 3 4

Cash Flow

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%. 116.

GO Tutorial 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?

117. In Problem 116, how much will be in the fund immediately after the 30th de-

posit if the fund pays 6% compounded annually and each successive deposit is 6% greater than the preceding deposit? 118. Develop a single spreadsheet that allows you to calculate the (P Z A1 i%, j%,n),

(F Z A1, i%, j%,n), and (A Z A1 i%, j%,n) factors. Note that the (A Z 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 Z A1 i%,j%,n) factor, your control panel needs to include i, j, and n. Test the first two factors using i 5 7%, j 5 5% and n 5 10 against the tables in Appendix A to ensure they work right. Then, using A1 5 $1,000, i 5 7%, and j 5 5%, calculate A using the third factor and check it against (P Z A1 i%,j%,n)*(A Z P i%,n). Keep this handy. It will be useful to you.

119.

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%.

120. 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.

Summary 85

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? 121.

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?

122. 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 5 12% and the net revenue is assumed to be an endof-year cash flow, what is the present value of the cash flow series over the 6 years? 123. On your child’s first birthday, you open an account to fund her college edu-

cation. 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? 124. A cash flow series is increasing geometrically at the rate of 6% per year. The

initial cash flow at t 5 1 is $1,000. The increasing payments end at t 5 20. The interest rate in effect is 15% compounded annually. Find the present amount at t 5 0 that is equivalent to this cash flow series. 125. 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? 126. 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. 127. 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? 128. 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?

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129.

GO Tutorial 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?

130. 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. 131. In a new, highly automated factory, labor costs are expected to decrease at an

annual compound rate of 5 percent; material costs are expected to increase at an annual compound rate of 6 percent; and energy costs are expected to increase at an annual compound rate of 3 percent. 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 percent compounded annually, what uniform annual costs over a 5-year period would be equivalent to the cumulative labor, material, and energy costs? 132.

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?

133.

In Problem 132, 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?

134.

In Problem 132, what uniform annual investment is required to achieve the same account balance?

135. Carlson Photography receives royalties based on the use of their photo-

graphs 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 2.4.1

Period Interest Rate

136. 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?

Summary 87

137.

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?

138. 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? 139.

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)?

140. 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? 141.

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)?

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142. Yavuz wishes to make a single deposit P at time t 5 0 into a fund paying

15 percent compounded quarterly such that $1,000 payments are received at t 5 1, 2, 3, and 4 (periods are 3-month intervals), and a single payment of $7,500 is received at t 5 12. What single deposit is required? 143.

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?

144.

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?

145.

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?

146. Lynn borrows $5,000 at 15 percent 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 percent greater than the previous payment. What is the size of her 6th payment? 147. Solve problem 146 for the case in which the size of each payment is $60

greater than the previous payment. 148. 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 percent 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? 149.

GO Tutorial 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 2.4.2

Effective Annual Interest Rate

150. 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. 151.

What is the effective annual interest rate for 10% compounded (a) semiannually, (b) every four months, (c) quarterly, (d) every other month, (e) monthly?

Summary 89

152.

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?

153.

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?

154. How much money must be invested in an account that pays 6% per year in-

terest 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? 155.

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 5 $21,080 (A Z P 1.25,60) 5 $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?

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156. 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? 157. Barbara makes four consecutive annual deposits of $2,000 in a savings ac-

count that pays interest at a rate of 10 percent compounded semiannually. How much money will be in the account 2 years after the last deposit? 158.

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. b. c. d.

How much money does Mr. Shark hand you? How much weekly interest are you really paying? What is the nominal annual interest rate? What is the effective annual interest rate?

Section 2.4.3 159.

Differing Frequencies of Compounding and Cash Flows

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?

160. 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? 161.

GO Tutorial 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?

162.

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?

3 ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E — S AM U E L WAS HI N G TO N TAK E S 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 percent for a 5-year loan period or 5.5 percent for a 10-year loan period. Over the past 10 years, Samuel earned an average of 5.25 percent 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. 92

EQUIVALENCE, LOANS, AND BONDS

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?

LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

1. Compare the equivalence between two or more cash flow profiles (Section 3.1).

2. Analyze immediate payment and deferred payment loans, including payment amount, remaining balance, and interest and principal per payment (Section 3.2).

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

4. Calculate the worth of a cash flow profile with variable interest rates (Section 3.4).

5. Explain the Annual Percentage Rate (APR) commonly calculated for a home mortgage (Section 3.5).

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 93

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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

LEARN I N G O B JEC T I V E : Compare the equivalence between two or more

cash flow profiles. Equivalence The state of being equal in value, such as with two cash flow profiles.

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 percent compounded annually? 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 PW112 5 $1,322.501P Z F 15%,22 5 $1,322.5011.152 22 5 $1,322.5010.756142 5 $1,000 5PV115%,2,,21322.52 5 $1,000 PW122 5 $1,000 Because the present worths are equal, the two cash flow profiles are equivalent. Notice, we did not obtain the value of the (P Z 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 5 2 and t 5 6.

3-1 Equivalence

A Uniform Series Equivalency of a Decreasing Gradient Series

EXAMPLE

Using an 8 percent discount rate, what uniform series over five periods, [1, 5], is equivalent to the cash flow profile given in Figure 3.1?

Video Example

i = 8%

$500

$400 $300 $200 $100 0

1

2

3

4

5

6

7

The Decreasing Gradient CFD for Example 3.1

FIGURE 3.1

The cash flow profile in Figure 3.1 consists of the difference in a uniform series of $500 and a gradient series, with G 5 $100. A uniform series equivalent of the cash flow profile can be obtained over the interval [2, 6] as follows:

SOLUTION

A32, 64 5 $500 2 $1001A Z G 8%,52 5 $500 2 $10011.846472 5 $315.35

$315.35

$315.35

$315.35

$400

$315.35

i = 8%

$500

$315.35

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

2

3

4

5

6

$300 $200 $100 0

1

FIGURE 3.2

2

3

4

5

6

7

=

0

1

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 A31, 54 5 $315.351P Z F 8%,12 5 $315.3510.925932 5 $291.99

7

95

$300

$291.99

$400

$291.99

i = 8%

$500

$291.99

Equivalence, Loans, and Bonds

$291.99

Chapter 3

$291.99

96

1

2

3

4

5

$200 $100 0

1

FIGURE 3.3

2

3

4

5

6

=

7

0

6

7

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 percent interest rate.)

Determining an Equivalent Gradient Step

EXAMPLE

Consider Figure 3.4. Determine the value of X that makes the two cash flow profiles equivalent using a TVOM of 15 percent. i = 15% 0

1

2

3

4

=

$200

0

1

2

3

4

$200 $300

$300

$200 + X $400

$200 + 2X $200 + 3X

FIGURE 3 . 4

SOLUTION

CFDs for Example 3.2

Equating the future worths of the two cash flow profiles at t 5 4 gives $2001F Z A 15%,42 1 $1001F Z A 15%,32 1 $100 5 3$200 1 X1A Z G 15%,42 4 1F Z A 15%,42

3-1 Equivalence

Eliminating the common term of $2001F Z A 15%,42 yields

$10013.472502 1 $100 5 X11.326262 14.993382

Solving for X gives a value of $67.53. 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.

FIGURE 3.5

Using the Excel® SOLVER Tool to Solve Example 3.2

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 lefthand 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.

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Determining an Equivalent Interest Rate

EXAMPLE

For what interest (discount) rate are the two cash flow profiles shown in Figure 3.6 equivalent?

i=?

$3,500 $3,000 $2,500 $2,000 $1,500

$1,500 $1,500 $1,500 $1,500 $1,500 0

1

2

3

4

5

=

0

1

2

3

4

5

$4,000

$7,000 FIGURE 3 . 6

SOLUTION

CFDs for Example 3.3

Converting each cash flow profile to a uniform series over the interval [1, 5] gives 2$4,0001A Z P i%,52 1 1,500 5 27,0001A Z P i%,52 1 1,500 1 $5001A Z G i%,52 or $3,0001A Z P i%,52 5 $5001A Z G i%,52 which reduces to

1A Z G i%,52 5 61A Z P i%,52

Searching through the interest tables at n 5 5, the value of the (A Z G i%,5) factor is six times the value of the (A Z P i%,5) factor for an interest rate between 12 percent and 15 percent. Specifically, with a 12 percent interest rate, 1A Z G 12%,52 2 61A Z P 12%,52 5 1.77459 2 610.277412 5 0.11013

and, using a 15 percent interest rate,

1A Z G 15%,52 2 61A Z P 15%,52 5 1.72281 2 610.298322 5 20.06711

3-1 Equivalence

Interpolating for i gives i 5 0.12 1 10.15 2 0.122 10.110132y10.11013 1 0.067112 or i 5 0.13864 Therefore, using a discount rate of approximately 13.864 percent will establish an equivalence relationship between the cash flow profiles given in Figure 3.6. 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.

FIGURE 3.7

Using the Excel® SOLVER Tool to Solve Example 3.3

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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 percent. After applying SOLVER, as shown in Figure 3.8, a value of 13.8677 percent 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

3-2

Solution to Example 3.3

INTEREST PAYMENTS AND PRINCIPAL PAYMENTS

LEARN I N G O B JEC T I V E : Analyze immediate payment and deferred payment

Video Lesson: Loans—Interest and Principal 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.

loans, including payment amount, remaining balance, and interest and principal per payment.

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. 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

3-2

Interest Payments and Principal Payments

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 percent/ 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 1 n 2 1, where t $ 1. The payments to be made are denoted by At, At11, . . . , At1n21 . 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 2 k payments remaining to be paid. Likewise, if you decide to pay off the loan immediately after making the k 1 1st payment, the amount owed is equal to the present worth of the n 2 k 2 1 payments remaining to be paid. The difference in the two present worth values is the amount of principal in the k 1 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 5 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 5 1), then the magnitude of each payment, $A, is given by $A 5 $P(A Z 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 5 A1P Z A i%,n 2 t2 where A, n, and i are defined as noted above.

(3.1)

Immediate Payment Loans When the loan payment begins one interest period after receipt of the principal.

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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. (3.2) Payofft 5 A 1 Ut 5 A 1 A1P Z A i%,n 2 t2 5 A31 1 1P Z A i%,n 2 t2 4

(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 (3.4) It 5 iUt21 Substituting the results of Equation 3.1 into Equation 3.4, the following relationship can be derived. (3.5) It 5 A 31 2 1P Z F i%,n 2 t 1 12 4 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 or

Pt 5 A 2 It

(3.6)

Pt 5 A1P Z F i%,n 2 t 1 12

(3.7)

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. EXAMPLE

Purchasing a Car 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 percent 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?

3-2

Interest Payments and Principal Payments

Given: P 5 $20,000; i 5 6%; compounded monthly; n 5 48 Find: A, U24, Payoff36, I12, P12 a.

A 5 P1A Z P i%,n2 Period interest rate 5 6 percent/12 months 5 0.5 percent/month A 5 $20,0001A Z P 0.5%,482 5 $20,00010.023492 5 $469.80 5 PMT10.5%,48,2200002 5 $469.70

b.

Ut 5 A1P Z A i%,n 2 t2   n 5 48  t 5 24   n 2 t 5 48 2 245 24

U24 5 A1P Z A 0.5%,242 5 $469.80122.562872 5 $10,600.04 5 PV10.5%,24,2PMT10.5%,48,2200002 2 5 $10,597.79 c.

d.

Payofft 5 A 1 A1P Z A i%,n 2 t2 n 5 48   t 5 36  n 2 t 5 48 2 36 5 12 Payoff36 5 A 1 A1P Z A 0.5%,122 5 $469.80 1 $469.80111.618932 5 $5,928.37 5 PV10.5%,12,2PMT10.5%,48,2200002 2 1 PMT10.5%,48,2200002 5 $5,927.12

It 5 iUt21 n 5 48   t 5 12  t 2 1 5 12 2 1 5 11 U11 5 A1P Z A 0.5%,372 5 $469.80133.702502 5 $15,833.44 I12 5 0.0051$15,833.442 5 $79.17 5 IPMT10.5%,12,48,2200002 5 $79.15

e.

Pt 5 A 2 It t 5 12 P12 5 A 2 I12 5 $469.80 2 $79.17 5 $390.63 5 PPMT10.5%,12,48,2200002 5 $390.55

KEY DATA

SOLUTION

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3.2.2 Deferred Payment Loans

Deferred Payment Loans When the loan payment begins more than one interest period after receipt of the principal.

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. 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 IR UB UIB

5 loan principal 5 interest rate on the loan 5 unpaid balance at the beginning of the interest period 5 unpaid accumulated interest immediately before making a payment UIA 5 unpaid accumulated interest immediately after making a payment AO 5 amount owed just before making a payment Int 5 interest earned during the period 5 UB 3 IR 5 size of the deferred payment Ad IPmt 5 amount of the payment that is an interest payment PPmt 5 amount of the payment that is a principal payment 5 Ad 2 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): IPmt 5 min1UIB; Ad 2

(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

Interest and Equity in Deferred Payments The owner of a small business borrows $10,000 at 15 percent 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?

3-2

Interest Payments and Principal Payments

Given: P 5 $10,000; IR 5 15%; n 5 5 Find: Deferred payment amount (Ad); principal and interest amounts for each payment

KEY DATA

The size of the new, deferred payments will be

SOLUTION

105

Ad 5 $10,0001F Z P 15%,32 1A Z P 15%,52 5 $10,00011.520882 10.298322 5 $4,537.09 5 PMT115%,5,2FV115%,3,,2100002 5 $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 Z 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. 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. TABLE 3.1

Year

Interest and Equity Payments in a Deferred Payment Loan

Unpaid Balance Before Payment (UB)

Interest During Year (Int)

Unpaid Interest Before Payment (UIB)

Amount Owed (AO)

Loan Payment (Ad )

Interest Payment (IPmt)

Principal Payment (PPmt)

Unpaid Interest After Payment (UIA)

Unpaid Balance After Payment (UBA)

1

$10,000.00

$1,500.00

$1,500.00

$11,500.00

$0.00

$0.00

$0.00

$1,500.00

$11,500.00

2

$11,500.00

$1,725.00

$3,225.00

$13,225.00

$0.00

$0.00

$0.00

$3,225.00

$13,225.00

3

$13,225.00

$1,983.75

$5,208.75

$15,208.75

$0.00

$0.00

$0.00

$5,208.75

$15,208.75

4

$15,208.75

$2,281.31

$7,490.06

$17,490.06

$4,537.01

$4,537.01

$0.00

$2,953.06

$12,953.06

5

$12,953.06

$1,942.96

$4,896.01

$14,896.01

$4,537.01

$4,537.01

$0.00

$359.01

$10,359.01

6

$10,359.01

$1,553.85

$1,912.86

$11,912.86

$4,537.01

$1,912.86

$2,624.15

$0.00

$7,375.85

7

$7,375.85

$1,106.38

$1,106.38

$8,482.23

$4,537.01

$1,106.38

$3,430.63

$0.00

$3,945.22

8

$3,945.22

$591.78

$591.78

$4,537.01

$4,537.01

$591.78

$3,945.22

$0.00

UBt 5 UBAt–1

Intt 5 IRt 3 UBt

UIBt 5 Intt 1 UIAt21

AOt 5 UBt 1 Intt

Adt

IPmt 5 min(UIBt; Adt)

PPmtt 5 Adt 2 IPmtt

UIAt 5 UIBt 2 IPmtt

$0.00 UBAt 5 AOt 2 Adt

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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.

3-3 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.

BOND INVESTMENTS

LEARN I N G O B JEC T I V E : Analyze investments in bonds and determine the purchase price, selling price, and return on such investments.

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 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. 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 percent per year payable quarterly, 8 percent per year payable semiannually, 9 percent 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. We now employ the following notation: P 5 the purchase price of a bond F 5 the sales price (or redemption value) of a bond

3-3

Bond Investments

V 5 the par or face value of a bond r 5 the bond rate (coupon rate) per interest period i 5 the yield rate (return on investment or rate of return) per interest period n 5 the number of interest (coupon) payments received by the bondholder A 5 Vr 5 the interest or coupon payment received per interest period The general expression relating these terms is P 5 Vr1P Z A i%,n2 1 F1P Z F i%,n2

(3.9)

Three types of bond problems are considered: 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. 1.

Each of these cases is illustrated in the following examples.

Determining the Selling Price for a Bond On January 1, 2011, Austin plans to pay $1,050 for a $1,000, 12 percent semiannual bond. He will keep the bond for 3 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 percent compounded semiannually?

EXAMPLE

Video Example

Given: P 5 $1,050; V 5 $1,000; n 5 6; r 5 6%; i 5 5% Find: F

KEY DATA

The present worth for the bond investment is given by

SOLUTION

P 5 Vr1P Z A 5%,62 1 F1P Z F 5%,62 or $1,050 5 1$1,0002 10.062 15.075692 1 F10.746222 Solving for F yields a value of $998.98. As long as the selling price at the end of 3 years is at least $998.98, the bond will yield a return of at least 10 percent compounded semiannually.

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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 5FV15%,6,60,210502 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.)

Determining the Purchase Price for a Bond

EXAMPLE

Emma plans to purchase a $1,000, 12 percent semiannual bond, hold it for 3 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 percent compounded semiannually on her investment? KEY DATA

SOLUTION

Given: F 5 $1,000; V 5 $1,000; n 5 6; r 5 6%; i 5 7% Find: P The present worth for the bond investment is given by P 5 Vr1P Z A 7%,62 1 F1P Z F 7%,62 or P 5 1$1,0002 10.062 14.766542 1 $1,00010.666342 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 percent compounded semiannually. The Excel® PV worksheet function matches perfectly with this bond calculation. Recall the PV syntax is PV(rate,nper,pmt,fv,type). Therefore, entering 5PV17%,6,260,210002 in any cell in an Excel® spreadsheet will give the answer sought: $952.33.

3-3

Determining the Rate of Return for a Bond Investment

Bond Investments

EXAMPLE

Charlotte purchased a $1,000, 12 percent quarterly bond for $1,020, kept it for 3 years, received twelve coupon payments, and sold it for $950. What was her quarterly yield on her bond investment? What was her effective annual rate of return? Given: P 5 $1,020; V 5 $1,000; F 5 $950; r 5 3% Find: i, ieff

KEY DATA

Setting the present worth of her investment equal to 0 gives

SOLUTION

2$1,020 1 1$1,0002 10.032 1P Z A i%,122 1 $9501P Z F i%,122 5 $0 Solving by trial and error, we first let i 5 2 percent/quarter (8 percent compounded quarterly): 2$1,020 1 $30110.575342 1 $95010.788492 . $0 2 $46.32570 . $0 Next, letting i 5 2½ percent/quarter (10 percent compounded quarterly): 2$1,020 1 $30110.257782 1 $95010.743562 , $0 2 $5.88520 , $0 Linear interpolation gives 0.02 1 0.0051$46.325702/ 1$46.32570 1 5.885202 5 0.02444 or 2.444 percent/quarter (9.776 percent compounded quarterly). The effective annual return is ieff 5 3 11 1 0.024442 4 2 14100% 5 10.140% 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 5 RATE112,30,21020,9502 5 2.442% Embedding the RATE function in the EFFECT function allows us to calculate the effective annual return on the bond transaction: ieff 5 EFFECT14*RATE112,30,21020,9502,42 5 10.132%

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3-4

VARIABLE INTEREST RATES

LEARN I N G O B JEC T I V E : Calculate the worth of a cash flow profile with vari-

able 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 5 P11 1 i1 2 11 1 i2 2 . . . 11 1 in21 2 11 1 in 2 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 5 A1 11 1 i1 2 21 1 A2 11 1 i1 2 21 11 1 i2 2 21 1 . . . 1 An11 1 i1 2 21 11 1 i2 2 21 . . . 11 1 in 2 21 (3.14) The future worth of a series of cash flows with variable interest rates can be given as F 5 An 1 An21 11 1 in 2 1 An22 11 1 in21 2 11 1 in 2 1 . . . 1 A1 11 1 i2 2 11 1 i3 2 . . . 11 1 in21 2 11 1 in 2

EXAMPLE

(3.15)

A Cash Flow Series with Variable Interest Rates 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.

$300 $200 (+)

10% 0

$200

10% 1

8% 2

8% 3

12% 4

5

(–) $200 FIGURE 3 . 9

CFD for Variable Interest Rate Example 3.9

3-5 Annual Percentage Rate

Computing the present worth gives SOLUTION P 5 $2001P Z F 10%,12 2 $2001P Z F 10%,12 1P Z F 10%,12 1 $3001P Z F 8%,12 1P Z F 10%,12 1P Z F 10%,12 1 $2001P Z F 12%,12 1P Z F 8%,12 1P Z F 8%,12 1P Z F 10%,12 1P Z F 10%,12 5 $2001P Z F 10%,12 2 $2001P Z F 10%,22 1 $3001P Z F 8%,12 1P Z F 10%,22 1 $2001P Z F 12%,12 1P Z F 8%,22 1P Z F 10%,22 5 $372.63  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The future worth is given by F 5 $200 1 $3001F Z P 8%,12 1F Z P 12%,12 2 $2001F Z P 8%,22 1F Z P 12%,12 1 $2001F Z P 10%,12 1F Z P 8%,22 1F Z P 12%,12 5 $200 1 $30011.080002 11.120002 2 $20011.166402 11.120002 1 $20011.100002 11.166402 11.120002 5 $589.01  

 

 

The uniform series equivalent is obtained as follows: P 5 A1P Z F 10%,12 1 A1P Z F 10%,22 1 A1P Z F 8%,12 1P Z F 10%,22 1 A1P Z F 8%,22 1P Z F 10%,22 1 A1P Z F 12%,12 1P Z F 8%,22 1P Z F 10%,22 $372.63 5 A3 10.909092 1 10.826452 1 10.925932 10.826452 1 10.857342 10.826452 1 10.892862 10.857342 10.826452 4 5 3.841958A A 5 $96.99  

 

 

 

 

 

 

 

 

Thus, $96.99/time period for five time periods is equivalent to the original cash flow series. 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 5 FVSCHEDULE1200,50.1,0.08,0.08,0.1262 2FVSCHEDULE1200,50.08,0.08,0.1262 1 FVSCHEDULE1300,50.08,0.1262 1 200 5 $589.01

3-5

ANNUAL PERCENTAGE RATE

LEARNING O BJECTI VE: Explain the Annual Percentage Rate (APR) commonly calculated for a home mortgage.

In Chapter 2, we defined the effective annual interest rate as the annual compounding rate that was equivalent to the stated interest rate. Specifically, we

111

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Annual Percentage Rate (APR) A term used by lenders to disclose the annual interest rate to borrowers. This term is not calculated in a consistent way from lender to lender, and may reflect a nominal annual interest rate, an effective annual interest rate, or something else.

showed that 12 percent per year compounded quarterly had an effective annual interest rate of 12.551 percent. Unfortunately, in loan transactions lenders are not required to provide the effective annual interest rate to borrowers. Instead, they are required to provide the annual percentage rate for the financial transaction. Denoted APR, the annual percentage rate specified might be a nominal annual interest rate, an effective annual interest rate, or something quite different from either the nominal or effective annual rate. An APR is commonly calculated for home mortgages. Many variations exist in how APR values are calculated. The differences in results may well be due to differences in various assumptions, including the following: ■









when the first payment is to be made (immediately, in 30 days, or in 45 days) whether interest is prepaid (at the beginning of the month) or postpaid (at the end of the month) whether points are applied to only the loan principal or to the sum of the loan principal and ‘‘other’’ closing costs the number of days considered in a year (365 versus 360—the product of 30 days per month and 12 months per year) the number of decimal points included in the calculations

Two basic approaches are used to compute the APR for a home mortgage, although there are many variations on each. The first approach, which we call additive, assumes that the buyer finances the closing costs; hence, the payments are based on the sum of the closing costs and the amount needed to purchase the house. Because the net cash flow for the buyer at the time of closing is the purchase price for the house, the APR is the nominal annual rate that equates the monthly payment to the lender with the amount needed to purchase the house. The second approach, which we call subtractive, assumes that the buyer pays all closing costs at the time of closing and makes mortgage payments based only on the price of the house purchased. Because the net cash flow for the buyer at the time of closing is the difference in what is obtained to purchase the house and the closing costs paid to the lender, the APR is the nominal annual rate that equates monthly payments made to the lender with the buyer’s net cash flow at the time of closing. The additive approach is more commonly used than the subtractive approach. Among the APRadd calculators available is eFunda’s (www. efunda.com) for fixed rate loans. A Web-based APR calculator for adjustable rate mortgages is available at http://www.dinkytown.net.

Summary 113

KEY CONCEPTS 1. Learning Objective: Compare the equivalence between 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. 2. Learning Objective: 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. Learning Objective: 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 5 Vr1P Z A i%,n2 1 F1P Z F i%,n2 where P 5 the purchase price of a bond F 5 the sales price (or redemption value) of a bond V 5 the par or face value of a bond r 5 the bond rate (coupon rate) per interest period i 5 the yield rate (return on investment or rate of return) per interest period n 5 the number of interest (coupon) payments received by the bondholder A 5 Vr 5 the interest or coupon payment received per interest period

SUMMARY

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4. Learning Objective: 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. 5. Learning Objective: Explain the Annual Percentage Rate (APR) commonly calculated for a home mortgage. (Section 3.5)

There are many variations in calculating the APR including the additive and subtractive approach discussed in this chapter. In addition, the APR calculation is dependent on various assumptions including when the first payment is to be made, whether interest is prepaid or postpaid, how points are applied, the number of days in a year, and the precision of the calculations. Caution should be taken when considering an APR given these variations. A Web-based additive calculator can be used to illustrate calculating the APR under various assumptions.

KEY TERMS Annual Percentage Rate (APR), p. 112 Bond, p. 106 Bond Rate, p. 106 Deferred Payment Loans, p. 104 Equivalence, p. 94

Immediate Payment Loans, p. 101 Maturity, p. 106 Par Value, p. 106 Principal Payment, p. 100 Redeem, p. 106

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1.

What single sum of money at t 5 4 is equivalent to receiving $5,000 at t 5 1, $6,000 at t 5 2, $7,000 at t 5 3, and $8,000 at t 5 4 if money is compounded at a rate of 8% per time period? a. $28,857 c. $30,892 b. $26,000 d. $33,363

Summary 115

2.

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 1 0.1) 5 $3,300 c. [$5,000(F Z P 10%,4) 1 $4,000(F Z P 10%,3) 1 $3,000(F Z P 10%,2) 1 $2,000(F Z P 10%,1) 1 1,000]/5 d. $5,000 2 $1,000(A Z G 10%,5)

3.

You borrow $5,000 at 10% per year and will pay off the loan in 3 equal annual payments starting one 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.

4.

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 5 $6,661.08; principal 5 $0.00 b. Interest 5 $2,281.31; principal 5 $4,379.77 c. Interest 5 $1,500.00; principal 5 $5,161.08 d. Interest 5 $0.00; principal 5 $6,661.08

5.

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 2$9,600, $300, $300, $300, and $9,900, and the bond will earn more than 10% b. Semiannual cash flows will be 2$9,600, $300, $300, $300, and $9,900, and the bond will earn less than 10% c. Semiannual cash flows will be 2$9,600, $300, $300, $300, and $10,300, and the bond will earn more than 10% d. Semiannual cash flows will be 2$9,600, $300, $300, $300, and $10,300, and the bond will earn less than 10%

6.

Consider a cash flow and interest profile as shown:

Cash Flow at End of Year Interest Rate During Year

Year 0

Year 1

Year 2

Year 3

2$1,000

$3,000

$2,000

$1,000

NA

6%

8%

10%

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

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7.

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 c. $174,000 b. $171,000 d. $177,000

8.

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 c. $120.23 b. $682.73 d. The answer cannot be determined without more information.

9.

$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 c. $28,804.50 b. $28,614.29 d. $29,552.62

PROBLEMS Section 3.1

Equivalence

1. If $7,000 is borrowed and repaid with four quarterly payments of $600 dur-

ing 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? 2. Quarterly deposits of $1,000 are made at t 5 1, 2, 3, 4, 5, 6, and 7. Then,

withdrawals of size A are made at t 5 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? 3. What uniform annual series of cash flows over a 12-year period is equivalent

to an investment of $5,000 at t 5 0, followed by receipts of $600 per year for 11 years and a final receipt of $1,600 at t 5 12 if the investor’s time value of money is 6% per time period? 4. 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 two 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?

Summary 117

b. What is the effective annual interest rate at which these two payment

options are equivalent? 5.

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 5 6.5%. b. If your TVOM is 8%, would you be indifferent between these two series of payments? If not, which do you prefer? c. If your TVOM is 5%, would you be indifferent between these two series of payments? If not, which do you prefer?

6. 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 a their TVOM of 14%. Determine the values of Q and R.

End of Year 0 1 2 3 4 5

Option 1

Option 2

Option 3

$0 1,800 1,800 1,800 1,800 1,800

$0 Q 2Q 3Q 4Q 5Q

$0 R (1.1)R (1.1)2R (1.1)3R (1.1)4R

Consider the following three cash flow series:

7.

End of Year 0 1 2 3 4 5

Cash Flow Series A

Cash Flow Series B

Cash Flow Series C

2$1,000 X 1.5X 2.0X 2.5X 3.0X

2$2,500 3,000 2,500 2,000 1,500 1,000

Y Y Y 2Y 2Y 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.

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8. Consider the following three cash flow series: End of Year 0 1 2 3 4 5

Cash Flow Series A

Cash Flow Series B

Cash Flow Series C

$3.0X 2.5X 2.0X 1.5X 1.0X

$1,000 1,500 2,000 2,500 3,000

21,000

22,500

2Y 2Y 2Y Y Y Y

Determine the values of X and Y so that you are indifferent between all three cash flows if your TVOM is 11% per year compounded yearly. 9.

Zetterberg Builders is given two options for making payments on a brush hog. Find the value of X such that they would be indifferent between the two cash flow profiles if their TVOM is 12% per year compounded yearly. End of Year 0 1 2 3 4 5

Section 3.2

Series 1

Series 2

$150 $200 $250 $300 $0 $0

$0 $0 $35X $25X $15X $5X

Interest Payments and Principal Payments

10.

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?

11.

CTL (Concrete Testing Lab) borrowed $80,000 for new equipment at 8% per year, compounded quarterly. It is to be paid back over three 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?

12. 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

Summary 119

rate was 6.0%, 2 percent 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 them pay off the loan immediately for the amount owed at the end of year 2, less $X. What is the range of $X that would be beneficial to both the bank and Med Diagnostics? 13.

GO Tutorial 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. b. c. d. e.

What nominal interest rate is Aerotron paying on the loan? What effective interest rate are they paying? How much of the 14th month’s payment is interest? How much is principal? How much of the 18th month’s payment is interest? How much is principal? How much of the 22nd month’s payment is interest? How much is principal?

14. BioElectroMechanical Systems (BEMS) is a startup company with high po-

tential and little available cash. They obtain $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 they pay? 15. $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 one year after receiving the $100,000, how

much of the third payment will be an interest payment? b. If the first payment is made four years after receiving the $100,000, how

much of the first payment will be an interest payment? c. If the first payment is made four years after receiving the $100,000, how

much of the last payment will be an interest payment? 16. $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? 17. $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?

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Section 3.3

Bond Investments

18. 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? 19.

You have just purchased ten 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?

20. 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? 21.

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.

22. 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. 23. 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. 24. 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

Summary 121

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? 25.

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?

26. One hundred $1,000 bonds having bond coupon rates of 8% per year pay-

able 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. 27. 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? 28. 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. 29. Five $1,000 bonds having a bond rate of 8% per year payable quarterly are pur-

chased 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. 30. Twenty-five $1,000 bonds having a bond rate of 8% per year payable quar-

terly 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 31. 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. 32. 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. 33. Based on the interest rates and cash flows shown in the cash flow diagram,

determine the value of $X.

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$25,000 $X 0

1 4%

2 4%

3 4%

4 3%

5 3%

6 3%

7 5%

8 5%

9 5%

$50,000 34. Based on the interest rates and cash flows shown in the cash flow diagram,

determine the value of $X. $10,000 $X 0

1 4%

2 4%

3 3%

4 5%

$150,000 35. Based on the interest rates and cash flows shown in the cash flow diagram,

determine the value of $X. $X $10,000 0

1 3%

2 3%

3 3%

4 5%

5 5%

6 5%

7 5%

8 4%

9 4%

10 4%

$150,000 36. 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? 37. In Problem 36, 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)? 38. 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

Summary 123

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? 39.

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? 40.

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? 41.

GO Tutorial 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)?

Consider the following cash flows and interest rates:

42.

End of Year 0 1 2 3

Interest Rate During Period 10% 8% 12%

Cash Flow at End of Period $0 $2,000 2$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.

ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : C ONO C O P HI L L I P S In 2011 ConocoPhillips was one of the largest integrated energy companies and refiners in the United States and abroad, based on market capitalization and oil and natural gas proved reserves and production.The company is known globally for its technological expertise in exploration and production, reservoir management and exploitation, 3-D seismic technology, high-grade petroleum coke upgrading, and sulfur removal. Headquartered in Houston, Texas, ConocoPhillips operates in more than 30 countries. As of December 31, 2011, the company had 29,800 employees worldwide and assets of $153.2 billion. Its capital expenditures and investments in 2011 totaled $13.3 billion; total revenues were $251.2 billion. Its debtto-capital ratio at the end of 2011 was 26 percent, compared to 25 percent at the end of 2010. The company’s core business activities include petroleum exploration and production; petroleum refining, marketing, supply, and transportation; and natural gas gathering, processing, and marketing. Also, the company invests in emerging businesses—power generation, carbon-to-liquids, technology solutions, and emerging technologies such as renewable fuels and alternative energy sources—that provide current and potential future growth opportunities. In 2010, the company initiated a three-year strategic plan to reposition itself by implementing a spinoff of its downstream businesses into a new company, Phillips 66. In this move, ConocoPhillips will become an independent exploration and production company. These changes are expected to be complete in 2012. In its Form 10-K for 2011, 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 and end-of-year costs (adjusted only for existing contractual changes), appropriate statutory tax rates and a prescribed 10 percent discount factor. Twelve-month average 124

PRESENT WORTH

prices are calculated on the unweighted arithmetic average of the firstday-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 become 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 their business might ConocoPhillips rely on present worth analysis?

LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

1. List eight DCF methods for comparing economic alternatives (Section 4.1.1). 125

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2. Differentiate between ranking methods and incremental methods of determining economic worth. (Section 4.1.2)

3. 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. Perform an economic analysis of public investments utilizing the benefitcost analysis for a given interest rate. (Section 4.3)

5. 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)

6. Calculate the capitalized worth (CW) of an investment for a given interest rate when the planning horizon is infinitely long. (Section 4.5)

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. 2. 3. 4. 5. 6. 7.

4-1

Identify the investment alternatives. Define the planning horizon. Specify the discount rate. Estimate the cash flows. Compare the alternatives. Perform supplementary analyses. Select the preferred investment.

COMPARING ALTERNATIVES

LEARN I N G O B JEC T I V E : List eight DCF methods for comparing economic

Video Lesson: Project Analysis Methods

alternatives.

This section summarizes the eight DCF methods that are covered in detail in Chapters 4–6. It also touches on several important considerations when

4-1

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. 2.

3.

4. 5.

6. 7.

8.

The present worth (PW) method converts all cash flows to a single sum equivalent at time zero using i 5 MARR. The benefit-cost ratio (B/C) method determines the ratio of the present worth of benefits (savings or positive-valued cash flows) to the negative of the present worth of the investment(s) (or negative-valued cash flows) using i 5 MARR. The discounted payback period (DPBP) method determines how long it takes for the cumulative present worth to be positive using i 5 MARR. The capitalized worth (CW) method determines the present worth (using i 5 MARR) when the planning horizon is infinitely long. The annual worth (AW) method converts all cash flows to an equivalent uniform annual series of cash flows over the planning horizon using i 5 MARR. The future worth (FW) method converts all cash flows to a single sum equivalent at the end of the planning horizon using i 5 MARR. The internal rate of return (IRR) method determines the interest rate that yields a future worth (or present worth or annual worth) of zero. 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

LEARNING O BJECTI VE: Differentiate between ranking methods and incre-

mental methods of determining economic worth.

The eight DCF methods can be divided into two groups: ranking and incremental methods.

Comparing Alternatives

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Ranking Methods

Present worth, future worth, annual worth, and capitalized worth are ranking methods; as such, the alternative having the greatest PW, FW, 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 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, and CW. However, there seems to be little reason to employ the more cumbersome incremental solution, since it is much simpler to compute the value of the PW, FW, AW, or CW and recommend the one having the greatest value over the planning horizon. 4.1.3

Equivalence of Methods

Six of the eight analysis methods are equivalent: PW, FW, AW, IRR, ERR, 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 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.

4-1

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 aftertax (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, and AW, 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 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.

Comparing Alternatives

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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.

4-2 Video Lesson: Present Worth Present Worth (PW) The value of all cash flows converted to a single sum equivalent at time zero using i 5 MARR.

PRESENT WORTH CALCULATIONS

LEARN I N G O B JEC T I V E : Calculate a present worth (PW) converting all cash flows to a single sum equivalent at time zero for a given interest rate.

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.” 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

PW1i%2 5 a At 11 1 i2 2t

(4.1)

i50

Present Worth of a Single Investment Opportunity

EXAMPLE

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 percent after-tax MARR, should the investment be made? KEY DATA

Given: The cash flows outlined in Figure 4.1; MARR 5 10%; planning horizon 5 10 years Find: PW of the investment

4-2

Present Worth Calculations

$50,000 (+) $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500

0

1

2

3

4

5

6

7

8

9

10

(–)

$500,000 FIGURE 4.1

CFD for Example 4.1

From our work in Chapter 2, we compute the present worth for the investment:

SOLUTION

PW 5 2$500,000 1 $92,5001P Z A 10%,102 1 $50,0001P Z F 10%,102 5 2$500,000 1 $92,50016.144572 1 $50,00010.385542 5 $87,649.73 or, using Excel®, PW 5 PV110%,10,292500,2500002 2 500000 5 87,649.62 Since PW . $0, the investment is recommended. 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 2$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 ninth year that the cumulative present worth becomes positive. Therefore, if something unforeseen occurred, causing the company to abandon the investment before the ninth year, would the investment still be

EXPLORING THE SOLUTION

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FIGURE 4 . 2

Spreadsheet of Cumulative Present Worth Over the Planning Horizon

profitable? The answer to the question depends on the SMP machine’s salvage value at the time it is abandoned. 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 5 a Ajt 11 1 MARR2 2t ;j

(4.2)

t50

Present Worth of Two Alternatives

EXAMPLE

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

4-2

Present Worth Calculations

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 percent is used. Which alternative design, if either, should the theme park select? Given: The cash flows outlined in Figure 4.3; MARR 5 10%; planning horizon 5 10 years Find: PW of each alternative

KEY DATA

$55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 (+) 0

1

2

3

4

5

6

7

8

9

10

(–)

Alternative A

$300,000

$80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 (+) 0

1

2

3

4

5

(–)

Alternative B

$450,000 FIGURE 4.3

CFDs for Example 4.2

6

7

8

9

10

133

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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 5 2$300,000 1 $55,0001P Z A 10%,102 5 2$300,000 1 $55,00016.144572

5 $37,951.35 . $0.00 1therefore, A is better than doing nothing2 or, using Excel®, PWA 5 PV110%,10,2550002 2 30000

5 37,951.19 . $0.00 1therefore, A is better than doing nothing2

Alternative B PWB 5 2$450,000 1 $80,0001P Z A 10%,102 5 2$450,000 1 $80,00016.144572

5 $41,565.60 . $37,951.35 1therefore, B is better than A2 or, using Excel®, PWB 5 PV110%,10,2800002 2 450000

5 41,565.37 + 37,951.19 1therefore, B is better than A2

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 percent, but, thereafter, Alternative A has the greatest present worth. The manager also noticed that for MARR values greater than 13 percent, neither alternative has a positive present worth.

4-2 Present Worth Calculations

135

$400,000 $350,000 $300,000

Present Worth

$250,000 $200,000 $150,000 $100,000 $50,000 $0 0%

2%

4%

6%

8%

10%

12%

14%

16%

18%

20%

–$50,000 –$100,000 –$150,000 MARR Alternative A FIGURE 4.4

Alternative B

Plot of Present Worths for Example 4.2

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.

Selecting from Among Multiple One-Shot Investments 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 percent, reflects the opportunity to reinvest recovered capital. Which alternative is preferred?

One-Shot Investment An investment that is available only once.

EXAMPLE

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$6000 $5000

$4500

$4000

$3500 $3500 $3500

$3000 $2000 $1000

(+) 0

1

2

3

4

0

1

2

3

4

5

6

(–) Alternative 2

Alternative 1 $4000

MARR = 15% $5000

FIGURE 4.5

KEY DATA

CFDs for Example 4.3

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 5 15%; planning horizon 5 6 years Find: PW of each alternative

SOLUTION

Using present worth analysis, the following results are obtained: PW1 115%2 5 2$4,000 1 $3,5001P Z A 15%,42 1 $1,0001P Z F 15%,42 5 2$4,000 1 $3,50012.854982 1 $1,00010.571752 5 $6,564.18 5 NPV115%,3500,3500,3500,45002 2 4000 5 6564.18 PW2 115%2 5 2$5,000 1 $1,0001P Z A 15%,62 1 $1,0001P Z G 15%,62 5 2$5,000 1 $1,00013.784482 1 $1,00017.936782 5 $6,721.26 5 NPV115%,1000,2000,3000,4000,5000,60002 2 5000 5 6721.26 Thus, we would recommend Alternative 2.

4-3

4-3

Benefit-Cost Analysis

BENEFIT-COST ANALYSIS

LEARNING O BJECTI VE: Perform an economic analysis of public invest-

ments utilizing the benefit-cost analysis for a given interest rate.

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. Four classes cover the spectrum of projects that government enters: 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 assistance for some, provides transportation and power generation for others, and also relates to natural resource benefits. Many 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 multiple-use concept is common, as in wildland management projects, where economic (timber), wildlife preservation (deer, squirrel), and recreation projects (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 of specific individuals or groups. Also, there are often multiple government agencies that have an interest in a project. For example, there is often federal support of state road projects. Finally, some public-sector projects are not easily evaluated due to difficulty in estimating benefits. Also, it may be many years before their benefits are realized. 4.3.1

Benefit-Cost Calculations for a Single Alternative

Benefit-cost analysis is recommended for a formal economic analysis of government programs or projects. Its two most common forms are the

Video Lesson: Benefit-Cost Analysis

137

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Present Worth

Benefit-Cost Ratio (B/C) The ratio of the present worth of benefits (savings or positive-valued cash flows) to the negative of the present worth of the investment(s) (or negative-valued cash flows) using i 5 MARR.

benefit-cost ratio (B/C) or, equivalently, a measure of benefits minus costs (B 2 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, using an appropriate discount rate for the time value of money. The mathematics of benefit-cost analysis is quite simple, although its application and quantification of benefits can be challenging. If Bjt 5 net public benefits associated with alternative j during year t, t 5 1;2; . . . ; n Cjt 5 net government costs associated with alternative j during year t, t 5 0;1;2; . . . ; n, and i 5 appropriate interest or discount rate, 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

B/Cj 1i2 5

a Bjt 11 1 i2

2t

a Cjt 11 1 i2

2t

t51 n

(4.3)

t50

Although Equation 4.3 is a ratio of present worths of benefits to costs, it could just as well be a ratio of annual worths of benefits to costs. The benefits minus costs criterion (B 2 C) is expressed as n

1B 2 C2 j 1i2 5 a 1Bjt 2 Cjt 2 11 1 i2 2t

(4.4)

t50

which is similar to 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.

Benefit-Cost Analysis for a Single Alternative

EXAMPLE

Video Example

Consider a 10-year public investment program 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 percent. Is this program desirable?

4-3 Benefit-Cost Analysis

FIGURE 4.6

Costs and Benefits for Public Investment Program, i 5 7%

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 $360,115. The program is desirable when considered alone. It is also common to calculate the ratio of benefits to costs, resulting in B/C 5 $1,424,102/1,063,987 5 1.33. When the B/C ratio is greater than 1.00, the program is desirable when considered alone.

4.3.2 Benefit-Cost Calculations for Multiple Alternatives

When two or more project alternatives are being compared using a B/C ratio, the analysis should be done on an incremental basis—that is, let the alternative with the lower present worth of costs be Alternative 1 and let the other be Alternative 2. Then, the incremental benefits of the second alternative over the first, DB2–1(i), are divided by the incremental costs of the second over the first, DC2–1(i)—that is, n

2t a 1B2t 2 B1t 2 11 1 i2 DB221 1i2 t51 DB/C221 1i2 5 5 n DC221 1i2 2t a 1C2t 2 C1t 2 11 1 i2

(4.5)

t50

Note that if the first alternative is to do nothing, the incremental B/C ratio is also the straight B/C ratio for the second alternative. As long as DB/C2–1(i) exceeds 1.0, Alternative 2 is preferable to Alternative 1; otherwise, Alternative 1 is preferred to Alternative 2. The winner of these is then compared on an incremental basis with another alternative. These pairwise comparisons continue until all alternatives have been exhausted and only the best project remains.

SOLUTION

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With the B 2 C criterion, an incremental basis may be used following the same rules as for the B/C ratio, preferring Alternative 2 to Alternative 1 as long as the following condition holds: n

D 1B 2 C2 221 1i2 5 DB221 1i2 2 DC221 1i2 5 a 3B2t 1i2 2 B1t 1i2 4 11 1 i2 2t n

t51

2 a 3C2t 1i2 2 C1t 1i2 4 11 1 i2 2t $ 0

(4.6)

t50

The incremental basis of Equation 4.6 is not required when using the B 2 C criterion, as long as 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

Benefit-Cost Analysis of Three Routes The state of Washington must decide between 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

4-3

Benefit-Cost Analysis

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. 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) 1 (250)(1.10) 1 (80)(0.30) 1 (3,320)(0.60)] (26)(365) 5 $24,066,640/year Route B: [(350)(0.70) 1 (250)(1.10) 1 (80)(0.30) 1 (3320)(0.60)] (22)(365) 5 $20,364,080/yr Route C: [(350)(0.70) 1 (250)(1.10) 1 (80)(0.30) 1 (3320)(0.60)] (20.5)(365) 5 $18,975,620/yr Annual public time costs: Route A: [(350)(1/45)(25) 1 (250)(1/35)(25) 1 (80)(1/45)(10) 1 (3,320)(0.25)(1/45)(25) 1 (3,320)(0.75)(1/45)(10)] (26)(365) 5 $13,335,710/year Route B: [(350)(1/50)(25) 1 (250)(1/40)(25) 1 (80)(1/50)(10) 1 (3,320)(0.25)(1/50))(25) 1 (3,320)(0.75)(1/50)(10)] (22)(365) 5 $10,119,808/year Route C: [(350)(1/50)(25) 1 (250)(1/40)(25) 1 (80)(1/50)(10) 1 (3,320)(0.25)(1/50))(25) 1 (3,320)(0.75)(1/50)(10)] (20.5)(365) 5 $9,429,821/year Annual accident costs per vehicle: Route A: (105)(18,000) 5 $1,890,000/year Route B: (75)(18,000) 5 $1,350,000/year Route C: (50)(18,000) 5 $900,000/year All relevant government and public costs are summarized in Figure 4.7.

SOLUTION

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FIGURE 4 . 7

Summary of Data for Three Routes of Example 4.5

Since 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 real discount rate of 8 percent are given as follows: DBB2A 18%2 5 Public costsA 18%2 2 Public costsB 18%2 $39,292,350 2 $31,833,888 5 $7,458,462

DCB2A 18%2 5 Government costsB 18%2 2 Government costsA 18%2 $8,775,501 2 $8,009,533 5 $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 DB/CB2A 18%2 5

DBB2A 18%2 $7,458,462 5 5 9.74 DCB2A 18%2 $765,968

4-3 Benefit-Cost Analysis

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 DB/C ratio may now be calculated to determine whether or not Route C is preferable to Route B: DBC2B 18%2 5 Public costsB 18%2 2 Public costsC 18%2 $31,833,888 2 $29,305,441 5 $2,528,447

DCC2B 18%2 5 Government costsC 18%2 2 Government costsB 18%2 $10,162,134 2 $8,775,501 5 $1,386,633 DBC2B 18%2 5

DBC2B 18%2 $2,528,447 5 5 1.82 DCB2A 18%2 $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 4.8 illustrates the benefit-cost calculations using Excel®.

FIGURE 4.8

Benefit-Cost Calculations for Example 4.5

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4-4

DISCOUNTED PAYBACK PERIOD

LEARN I N G O B JEC T I V E : Calculate the discounted payback period (DPBP) for

Video Lesson: Payback Period and Discounted Payback Period

Discounted Payback Period (DPBP) The amount of time required for the cumulative present worth to become positive using i 5 MARR.

a given interest rate to determine how long it takes for the cumulative present worth (PW) to be positive.

The time required for an investment to be fully recovered, including the time value of money, is the amount of time required for the cumulative present worth to become positive using i 5 MARR; it is called the discounted payback period (DPBP) for an investment. Note that the DPBP may be a fractional or whole-year amount. 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.

Computing the DPBP of a Single Investment Opportunity

EXAMPLE

Video Example

KEY DATA

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. Given: Net cash flows as shown in Table 4.1; MARR 5 10%, planning horizon 5 7 years Find: DPBP Net Cash Flows for Example 4.6 TABLE 4.1

EOY

Net Cash Flow

0

–$400,000

1

150,000

2

125,000

3

100,000

4

75,000

5

50,000

6

25,000

7

0

4-4 Discounted Payback Period

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 continuous cash flows requires interpolation to calculate the DPBP of 5 1 (2,927.02/14,111.85) 5 5.21 years. If we assume an end-of-year cash flow convention, the DBPB is 6 years.

SOLUTION

Present Worth and Cumulative Present Worth of Example 4.6

TABLE 4.2

Cash Flows EOY

Cumulative PW of CF

Net CF

(P/F 10%, n)

PW of CF

0

2$400,000

1.00000

2400,000

2400,000

1

150,000

0.90909

136,363.64

2263,636.36

2

125,000

0.82645

103,305.79

2160,330.58

3

100,000

0.75131

75,131.48

285,199.10

4

75,000

0.68301

51,226.01

233,973.09

5

50,000

0.62092

31,046.07

22,927.02

6

25,000

0.56447

14,111.85

11,184.83

7

0

0.51316

0

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

Using the Excel® NPER Tool to Determine DPBP of a Single Investment Opportunity

EXAMPLE

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. Given: The cash flows shown in Figure 4.1; MARR 5 10%; planning horizon 5 10 years Find: DPBP using the Excel NPER function

KEY DATA

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.

SOLUTION

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First, the engineer used the Excel® NPER worksheet function, assuming a negligible salvage value: yrs to recover the investment 5 NPER 110%, 92500,25000002 5 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 percent; then, he 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 4.9, the

FIGURE 4.9

Plot of Cumulative Present Worths for Example 4.7

4-4 Discounted Payback Period

engineer calculated the present worth, assuming the SMP machine was kept for 1, 2, . . . , 10 years. Based on the salvage value decreasing at a geometric rate of 20.6 percent each year, as shown in Figure 4.9, 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.

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. As illustrated in the 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 5 NPER(MARR,A,2P) 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.

Using the Excel® SOLVER Tool to Calculate the DPBP

EXAMPLE

Use the Excel® SOLVER tool to determine the DPBP for the SMP investment. Figure 4.10 shows a spreadsheet with the relevant input parameters.

KEY DATA

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Excel® SOLVER Set Up to Calculate the DPBP in Example 4.8 with Gradient Decrease in Salvage Value

FIGURE 4.10

SOLUTION

When salvage value decreases at an annual rate of 20.6 percent, 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 4.11, SOLVER yields a value of 6.95 years when salvage value decreases geometrically over time. 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 4.11, SOLVER yielded a value of 2.17 years when salvage value decreases by $45,000 per year.

4-4 Discounted Payback Period

FIGURE 4.11 Excel® SOLVER Solutions for the DPBP in Example 4.8 with Geometric and Gradient Decreases in 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. 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.

Using DPBP When Choosing Among Three Alternatives

EXAMPLE

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? Given: The cash flows shown in Figure 4.12; MARR 5 10% Find: DPBP for each alternative

KEY DATA

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$55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000

(+) 0

1

2

3

4

5

6

7

8

9

10

(–) Alternative A

$300,000

$80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000

(+) 0

1

2

3

4

5

6

7

8

9

10

(–)

Alternative B

$450,000 $45,000

(+)

0

1

$40,000

2

$35,000

3

$30,000

4

$25,000

$20,000

5

$15,000

6

7

$10,000 8

$5,000 9

$0 10

(–)

Alternative C

$150,000 FIGURE 4.12

SOLUTION

CFDs for the Alternatives in Example 4.9

The present worth values for the three alternatives (A, B, and C), for DPBP years are given by PWA 5 2$300,000 1 $55,0001P Z A 10%, DPBPA 2 PWB 5 2$450,000 1 $80,0001P Z A 10%, DPBPB 2 PWC 5 2$150,000 1 $45,0001PZ A 10%, DPBPC 2 2 $5,0001P ZG 10%, DPBPC 2

4-4 Discounted Payback Period

The Excel® NPER function can be used to solve for the values of DPBPA and DPBPB that equates the present worth to 0: DPBPA 5 NPER110%,255000,3000002 5 8.273 years DPBPB 5 NPER110%,280000, 4500002 5 8.674 years The Excel® SOLVER tool is used to obtain the value of DPBPC , as shown in Figure 4.13.

DPBPC 5 6.273 years Thus, alternative C has the smallest DPBP.

FIGURE 4.13

Excel® SOLVER Solution for DPBPC in Example 4.9

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 5 2$150,000 1 $45,0001P Z A 10%,102 2 $5,0001P Z G 10%,102 5 2$150,000 1 $45,00016.144572 2 $5,000122.891342 5 $12,048.95 or, using Excel®, PWC 5 1000*NPV110%,45,40,35,30,25,20,15,10,5,02 2 150000 5 12,048.81 whereas, using Excel®, PWA 5 37,951.19 and PWB 5 41,565.37. Figure 4.14 illustrates how the three present worths change over the duration of the planning horizon.

EXPLORING THE SOLUTION

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$50,000 $0 –$50,000

0

1

2

3

4

5

6

7

8

9

10

Present Worth

–$100,000 –$150,000 –$200,000 –$250,000 –$300,000 –$350,000 –$400,000 –$450,000 Investment Year Alternative A FIGURE 4. 1 4

Alternative B

Alternative C

Discounted Payback Period Analysis for Example 4.9

As Example 4.9 illustrates, the DPBP can lead to wrong conclusions regarding the investment alternative that maximizes economic worth.

4-5 Video Lesson: Capitalized Worth

Capitalized Worth (CW) The value of all cash flows converted to a single sum equivalent at time zero using i 5 MARR when the planning horizon is infinitely long.

CAPITALIZED WORTH

LEARN I NG O B JEC T I V E : Calculate the capitalized worth (CW) of an investment for a given interest rate when the planning horizon is infinitely long.

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. The capitalized worth method is applicable only if there is reason to believe a series of cash flows will extend indefinitely into the future.

4-5 Capitalized Worth

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 CW 5 A1P Z A i%,q2 5

A i

(4.7)

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.

Capitalized Cost of Building Maintenance

EXAMPLE

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 percent MARR, what is the capitalized cost for refurbishing the capitol dome? Given: Cash flow profile in Figure 4.15 Find: Capitalized Cost (CC)

KEY DATA

0

10

20

30

n=∞

$750,000

$750,000

$750,000

$750,000

$750,000

(+)

(–)

FIGURE 4.15

CFD for Example 4.10

CC 5 $750,0001A Z P 5%,102/0.05 5 $750,00010.129502/0.05 5 $1,942,500

SOLUTION

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Alternatively, CC 5 $750,000 1 $750,0001A Z F 5%,102/0.05 5 $750,000 1 $750,00010.079502/0.05 5 $1,942,500 The alternative approach yielded the same result because of the following relationship between the A Z P and A Z F factors: (A Z P i%,n) 5 (A Z F i%,n) 1 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 percent annual compound interest, then $750,000 can be paid out every 10 years, forever, to cover the cost of cleaning the capitol dome.

Capitalized Cost for Highway Construction

EXAMPLE

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 percent. KEY DATA

Given: The relevant data is organized in Table 4.3 TABLE 4.3

Data for Highway Construction Costs

Paving

Initial Cost

Duration

Replacement Cost $150/ft

$150/ft

5 yrs

$9,000/mi

1 yr

Ditches (install)

$7.75/ft

15 yrs

(cleaning/maintenance)

$3.75/ft

1 yr

4 3 $8,000/mi 4 3 $1,250/mi

10 yrs 1 yr

Maintenance

Culverts (install 1 replace) (cleaning)

$7.75/ft 4 3 $10,000/mi 4 3 $1,250/mi

4-5

Capitalized Worth

Find: CC(highway/mile) 5 CC(paving/mile) 1 CC(maintenance/mile) 1 CC(ditches/mile) 1 CC(culverts/mile) 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. CC1paving/mile2 5 5,280 ft/mi3 $150/ft1A Z P 5%,52 1 $7.75/ft1A Z P 5%,152 4 /10.052 5 $3,737,409

5 1PMT115%,5,25280*1502 1 PMT15%,15,25280*7.7522/0.05

5 $3,737,487 2. CC1paving/mile2 5 5,280 ft/mi5$150/ft 3 1 1 1A Z F 5%,52 4 1 $7.75/ft 3 1 1 1A Z F 5%,152 4 6/10.052 5 $3,737,409 5 5280*1150 1 PMT15%,5,,21502/0.05 1 7.75 1 PMT15%,15, ,27.75/0.05 5 $3,737,487 To complete the example, we add the capitalized cost for maintaining the highway and ditches, plus the installation and maintenance of the culverts: CC1maintenance/mile2 5 $9,000/0.05 5 $180,000 CC1ditches/mile2 5 3 122 15,280 ft/mi2 1$3.75/ft2 4/ 10.052 5 $792,000 CC1culverts/mile2 5 142 3$8,000 1 $1,250/0.05 1 $10,0001A Z F 5%,102/0.05 4 5 $195,600 5 4* 18000 1 1250/0.05 1 PMT15%,10,, 2100002/0.052 5 $195,604

SOLUTION

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Therefore, CC1highway/mile2 5 $3,737,487 1 $180,000 1 $792,000 1 $195,604 5 $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 4.10, since 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.

Capitalized Cost for Water Delivery

EXAMPLE

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.

Summary 157

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 5 percent MARR and an infinitely long planning horizon, which alternative has the lowest capitalized cost? Given:

KEY DATA 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 Pipeline CC 5 $125,000,000 1 3$10,000,0001A Z F 5%,152 1 $5,000,0004/0.05 5 $234,268,000.00 5 125000000 1 1PMT15%,15,2100000002 1 50000002/0.05 5 $234,268,457.52

SOLUTION

Canal CC 5 $200,000,000 1 3$5,000,0001A Z F 5%,102 1 $1,000,0004/0.05 5 $227,950,000.00 5 200000000 1 1PMT15%,10, ,250000002 1 10000002/0.05 5 $227,950,457.50 The canal has the smaller capitalized cost and would be recommended.

KEY CONCEPTS 1. Learning Objective: List eight DCF methods for comparing economic alternatives. (Section 4.1.1)

The eight DCF methods for comparing economic alternatives are: 1. Present worth (PW) 2. Benefit-cost ratio (B/C) 3. Discounted payback period (DPBP)

SUMMARY

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Chapter 4

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4. 5. 6. 7. 8.

Capitalized worth (CW) Annual worth (AW) Future worth (FW) Internal rate of return (IRR) External rate of return (ERR)

2. Learning Objective: Differentiate between ranking methods and incremental methods of determining economic worth. (Section 4.1.2)

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 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). 3. Learning Objective: 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 5 a Ajt 11 1 MARR2 2t ;j

(4.2)

t50

4. Learning Objective: 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

B/C j 1i2 5

2t a Bjt 11 1 i2

t51 n

a Cjt 11 1 i2

t50

(4.3) 2t

Summary 159

5. 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. (Section 4.4)

The DPBP shows the 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. 6. Learning Objective: 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 5 A1P Z A i%,q2 5 A/i (4.7)

KEY TERMS Benefit-cost Ratio (B/C), p. 138 Capitalized Worth (CW), p. 152 Discounted Payback Period (DPBP), p. 144

One-Shot Investment, p. 135 Present Worth (PW), p. 130

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1.

When using present worth to evaluate the attractiveness of a single investment alternative, what value is the calculated PW compared to? a. 0.0 c. 1.0 b. MARR d. WACC

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2.

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 percent, which of the following most closely represents the present worth of the well? a. $1,770,000 c. $1,253,000 b. $1,508,000 d. $1,082,000

3.

The present worth of a multiyear investment with all positive cash flows (incomes) other than the initial investment is PW . 0 at MARR 5 i%. If MARR changes to (i 1 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

4.

Consider the following cash flow diagram. Which of the expressions is not valid for the present worth? $250 $250 $250 $200 $150 $100 +0 – a. b. c. d.

1

2

3

4

5

6

10%/yr

P 5 1001P Z A 10%,62 1 501P Z G 10%,32 1 1501P Z A 10%,32 1P Z F 10%,32 P 5 1001P Z A 10%,32 1 501P Z G 10%,32 1 2501P Z A 10%,32 1P Z F 10%,32 P 5 2501P Z A 10%,62 2 501P Z G 10%,32 P 5 1001P Z A 10%,42 1 501P Z G 10%,42 1 2501P Z A 10%,22 1P Z F 10%,42  

 

 

 

 

 

 

 

 

 

 

 

 

 

5.

Ivan, an industrial engineering student, is working on a homework problem for Engineering Economy. He needs to calculate the PW at 12 percent of a cash flow series with $1,000 at t 5 3, $1,500 at t 5 4, and $2,000 at t 5 5. If Ivan uses the equation P 5 1,000(P Z A 12%,3) 1 500(P Z G 12%,3), where is the P now located? a. t 5 4 c. t 5 1 b. t 5 2 d. t 5 0

6.

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 percent, what one-time fee should the owner charge for the perpetual care service? a. $96 c. $1,500 b. $120 d. $12,000

7.

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

Summary 161

8 percent. What is the present equivalent cost of the investment if the planning horizon is 30 years? a. $335,000 c. $360,000 b. $344,500 d. $395,500 8.

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 percent 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 percent. a. $412 c. $1,044 b. $812 d. $1,834

9.

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 percent. Compute the maximum expenditure justifiable for the purifier. a. $24 b. $33

c. $300 d. $1,875

10.

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 percent on the special account? a. $1,000 c. $18,000 b. $17,000 d. $37,000

11.

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 1)%, what will be the relationship between PW(A) and PW(B)? a. PW(A) . PW(B) c. PW(A) , PW(B) b. PW(A) 5 PW(B) d. Cannot be determined without the cash flow profiles

12.

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 percent, what is the benefit-cost ratio of the shelving system? a. 1.51 c. 1.73 b. 2.24 d. 1.56

13.

When using the benefit-cost ratio measure of worth, what benchmark is the calculated ratio compared to in determining if an individual investment is attractive? a. 0.0 c. 1.0 b. MARR d. IRR

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14.

When using a benefit cost ratio 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

15.

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 percent? Purchase Cost: $7,000 Annual Savings: $1,500 Life: 15 years a. 3.21 c. 1.76 b. 1.83 d. 1.34

PROBLEMS Section 4.1 1.

Comparing Alternatives

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)

(b)

(2)

(c)

Discounted payback period Capitalized worth

(d)

External rate of return

(4)

(e)

Future worth

(5)

(f)

Internal rate of return

(6)

(g)

Present worth

(7)

(3)

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

2. GeoWorld Systems uses a subset of the following questions during the inter-

view 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 percent. a. The present worth of the project is $1,367. b. The internal rate of return of the project is 12.9 percent. c. The annual worth of the project is −$632.

Summary 163

d. e. f. g. h. i. j. k. l. m. n. o. p. q.

The benefit cost ratio of the project is 1.08. The future worth of the project is $3.75. The external rate of return of the project is 15.3 percent. The present worth of the project is −$47. The internal rate of return of the project is 14.7 percent. The annual worth of the project is $6,775. The benefit cost ratio of the project is 0.97. The future worth of the project is −$13,470. The external rate of return of the project is 3.7 percent. For the cases (if any) in which “do nothing” was preferred, what assumption is being made about the return generated by the “uninvested” funds? Is it possible that the values stated in (a) and (b) were correctly calculated on the same project? If not, why? Is it possible that the values stated in (c) and (d) were correctly calculated on the same project? If not, why? What do you know must be true about the present worth for the project in (j)? What do you know must be true about the internal rate of return for the project in (j)?

3. 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 (b) Benefit-cost ratio (c) External rate of return (d) Future worth (e) Internal rate of return (f) Present worth

Section 4.2.1 4.

(1) Measure of worth is greater than 0 (2) Measure of worth is greater than 1 (3) Measure of worth is greater than MARR

Present Worth of a Single Alternative

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 percent/year. End of Year 0 1 2 3 4 5

Receipts

Disbursements

$0 $0 $2,000 $4,000 $3,000 $1,600

$5,000 $200 $300 $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?

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5.

Bailey, Inc., is considering buying a new gang punch that would allow them 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 percent/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?

6. Repeat Problem 5 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. 7. Repeat Problem 5 assuming that the cost of capital has increased due to weak

market conditions and MARR is now 6 percent/year. 8.

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 percent/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.

9. Repeat parts (a) through (c) of Problem 8 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. 10. 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 percent/year. Using a present worth approach, determine the maximum amount Fabco should be willing to pay for the valves. 11. 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 percent/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? 12. Quilts R Us (QRU) is considering an investment in a new patterning attach-

ment with the cash flow profile shown in the table below. QRU’s MARR is 13.5 percent/year.

Summary 165

EOY 0

Cash Flow 2$1,400 $0 $500 $500 $500 $500 $0 $500

1 2 3 4 5 6 7

EOY

Cash Flow

8

$600

9 10 11 12 13 14 15

$700 $800 $900 −$1,000 −$2,000 −$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? 13.

GO Tutorial Galvanized Products is considering purchasing a new computer system for their 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 percent 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 percent/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?

14. 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 percent/year. EOY 0

Cash Flow (K$)

EOY

Cash Flow (K$)

2$2

6

1

2$10

7

2$10

2

2$12

8

2$12

$200

3

2$14

9

2$14

4

2$16

10

2$100

5

2$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?

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15.

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 percent/year. EOY

Cash Flow (M$)

EOY

Cash Flow (M$)

0

2$12

4

$5

1

2$1 $5 $2

5

$5

6 7

$2 $5

2 3

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? 16. Home Innovation is evaluating a new product design. The estimated receipts

and disbursements associated with the new product are shown below. MARR is 10 percent/year. End of Year

Receipts

Disbursements

0 1 2 3 4

$0 $600 $600 $700 $700

$1,000 $300 $300 $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? 17.

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 percent compounded annually. Aerotron Electronics’ MARR is 10 percent 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?

18.

Video Solution 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

Summary 167

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 percent/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 19.

Present Worth of Multiple Alternatives

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 percent 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? 20.

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 percent of the loss and will cost $0.40 per foot. A 2-inch thick insulated covering will eliminate 92 percent 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 percent/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.

21.

GO Tutorial 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 percent/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?

22. Quantum Logistics, Inc., a wholesale distributor, is considering the construc-

tion 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 percent/year.

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Present Worth

City

Initial Cost

Net Annual Income

Lagrange Auburn Anniston

$1,260,000 $1,000,000 $1,620,000

$480,000 $410,000 $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? 23.

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 percent/year. a. What is the present worth of each project? b. Which project should be recommended?

24. DelRay Foods must purchase a new gumdrop machine. Two machines are avail-

able. 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 percent/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? 25.

Xanadu Mining is considering three mutually exclusive alternatives, as shown in the table below. MARR is 10 percent/year. EOY

A001

B002

C003

0

2$210 $80 $90 $100 $110

2$110 $60 $60 $60 $70

2$160 $80 $80 $80 $80

1 2 3 4

a. What is the present worth of each alternative? b. Which alternative should be recommended? 26. Two storage structures, given code names Y and Z, are being considered for a

military base located in Sontaga. The military uses a 5 percent/year expected

Summary 169

rate of return and a 24-year life for decisions of this type. The relevant characteristics for each structure are shown below.

First Cost Estimated Life Estimated Salvage Value Annual Maintenance Cost

Structure Y

Structure Z

$4,500 12 years None $1,000

$10,000 24 years $1,800 $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? 27. 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 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 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 percent/year MARR, should it accept the man’s offer? Use a present worth analysis. 28. 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 percent/year. a. What is the present worth of each polisher? b. Which polisher should be recommended? 29.

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 percent/year.

30. 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.

170 Chapter 4

Present Worth

Useful Life First Cost Salvage Value Annual Revenue Annual Expenses

T1

T2

T3

T4

10 years $600,000 $70,000 $400,000 $130,000

10 years $800,000 $130,000 $600,000 $270,000

10 years $470,000 $65,000 $260,000 $70,000

10 years $540,000 $200,000 $320,000 $120,000

a. Determine the preferred telescope if MARR is 25 percent/year. b. Determine the preferred telescope if MARR is 42 percent/year. 31.

GO Tutorial 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 percent, 60 percent, 70 percent, and 80 percent for years 1 through 4, respectively, and 90 percent for the fifth through fifteenth years, MARR of 12 percent/year, 365 operating days/year, and ignoring the cost of land, should the motel be built? Base your decision on a present worth analysis.

32. 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 percent/ 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 percent/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 33.

Present Worth of One-Shot Investments

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 percent/year. Based on a present worth analysis, which alternative is preferred? EOY 0 1 2 3 4 5 6 7

Alternative W

Alternative X

2$100,000 $20,000 $20,000 $50,000 $80,000

2$150,000 $40,000 $45,000 $50,000 $55,000

$110,000

$60,000 $65,000 $70,000

Summary 171

34.

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 percent/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 Expected Annual Return

35.

Option 1

Option 2

$100,000 12 years $16,500

$75,000 9 years $14,300

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 percent/year. Based on a present worth analysis, which machine is preferred?

Initial Investment Estimated Life End of Life Salvage Annual Income Annual Expense

Section 4.3.1

E Series

M Series

$40,000 7 $10,000 $19,400 $10,000

$60,000 5 $0 $26,000 $6,000

Benefit-Cost Calculations for a Single Alternative

36.

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 percent. 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?

37.

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 were removed is $6. Each stone can be removed at a cost of $1. The interest rate is 2 percent. a. Compute the benefit-cost ratio for the individual if he alone removed the 100 stones.

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b. Compute the benefit-cost ratio for the individual if the task was under-

taken 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? 38. 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 percent, the planning horizon is 10 years, and the building renovation is expected to have a salvage value of 30 percent, 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 calcu-

late B 2 C. b. What is the smallest benefit per person that will make this project desirable? 39.

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 percent 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 percent of initial cost. Interest is 7 percent. An estimated 1.5 million visits per year are made to the zoo, and the cost remains free yearround. How much additional benefit per visit, on average, must the visitors perceive to justify the renovation?

40. The Boundary Waters Canoe Area Wilderness (BWCA) located in north-

eastern 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

Summary 173

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 2 C? Section 4.3.2 41.

Benefit-Cost Calculations for Multiple Alternatives

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

A

B

C

Benefits Disbenefits Costs Savings

$200,000 $37,000 $150,000 $15,000

$300,000 $69,000 $234,000 $31,000

$400,000 $102,000 $312,000 $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 2 C for each alternative. 42. 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 percent interest rate, a 15-year study period, and a salvage value of 50 percent of first cost, determine which road should be constructed. 43.

GO Tutorial 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.

Benefits Disbenefits Costs Savings

Option 1

Option 2

Option 3

$400,000 $78,000 $235,000 $25,000

$550,000 $125,000 $390,000 $65,000

$575,000 $180,000 $480,000 $90,000

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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 2 C for each option. d. At what value of Option 2 costs are you indifferent between Option 1 and

Option 2? 44.

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 percent of initial cost, respectively. Using a MARR of 7 percent and a 20-year study period, which should be constructed? a. Use an incremental B/C analysis. b. Use a B 2 C analysis. c. Which route is preferred if 0 percent interest is used?

45.

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 5 7 percent. 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.

46. 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 two-minute cycle with 60 seconds of green-yellow and 60 seconds of red for each direction. Approximately 10 percent of the 12,000 vehicles using the intersection each day are held up for an extra 2 full minutes and average 3 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

Summary 175

extra waiting and start-stops can be reduced by 90 percent 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 percent 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. Section 4.4.1

Discounted Payback Period for a Single Alternative

47.

Reconsider Problem 5 (repeated here). Bailey, Inc., is considering buying a new gang punch that would allow them 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 percent/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?

48.

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 percent/year. End of Year 0 1 2 3 4 5

Receipts

Disbursements

$0 $600 $600 $700 $700 $700

$1,000 $300 $300 $300 $300 $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 49.

Discounted Payback Period for Multiple Alternatives

Reconsider Problem 19 (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 percent 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 21?

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50.

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 percent/ year. Model 127B

Model 334A

$50,000 10 $10,000 $19,400 $10,000

$80,000 5 $0 $26,000 $6,000

Initial Investment Estimated Life End of Life Salvage Annual Income Annual Expense

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

DPBP? 51.

Reconsider Problem 22 (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 percent/year. City

Initial Cost

Net Annual Income

Lagrange Auburn Anniston

$1,260,000 $1,000,000 $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 recom-

mendation in Problem 22? Section 4.5

Capitalized Worth

52. 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 percent/year TVOM, what is the capitalized cost for the water treatment plant? 53.

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 percent/year interest rate, determine the capitalized cost for each design and recommend which should be chosen.

Summary 177

54.

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 percent/year, determine the capitalized cost.

55.

GO Tutorial 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 percent/year/month on its investments and there is exactly one winning ticket.

56. 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 given out each year under each of the following conditions. The state university earns 4 percent 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. 57. Reconsider the data from Problem 56. 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. 58. A prospective venture has the following cash flow profile over a 5-year

horizon. What is the capitalized worth at 6 percent/year assuming the pattern repeats indefinitely? End of Year 0 1 2 3 4 5

59.

Receipts

Disbursements

$100 $300 $500 $400 $350 $250

$1,100 $50 $250 $150 $100 $0

You decide to open an individual retirement account (IRA) at your local bank that pays 8 percent/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?

5 ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : J OS H L I U 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 interests in pursuing an MBA. Due to his interests in 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?

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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?

LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

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)

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)

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.

179

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Annual Worth and Future Worth

Systematic Economic Analysis Technique 1. 2. 3. 4. 5. 6. 7.

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 5 MARR.

Identify the investment alternatives Define the planning horizon Specify the discount rate Estimate the cash flows Compare the alternatives Perform supplementary analyses 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 portfolio 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.

5-1

ANNUAL WORTH

LEARN I N G O B JEC T I V E : Calculate an annual worth (AW) converting all cash

Video Lesson: Annual Worth

flows to an equivalent uniform annual series of cash flows over the planning horizon for a given interest rate.

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: n

AW 5 c a At 11 1 MARR2 n2t d 1A Z F MARR%,n2 t50

(5.1)

5-1 Annual Worth

Of course, the annual worth can also be calculated using Equation 4.1 to compute the present worth and multiplying the result by (A Z 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.

Annual Worth of a Single Alternative

EXAMPLE

We continue to use the acquisition of the 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%. Given: The cash flows outlined in Figure 5.1; MARR 5 10%; planning horizon 5 10 years Find: AW of the investment. Is this investment recommended?

KEY DATA

$50,000 (+) $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500

0

1

2

3

4

5

6

7

8

9

10

(–)

$500,000 FIGURE 5.1

CFD for Example 5.1

The annual worth for the investment will be AW 5 2$500,0001A Z P 10%,102 1 $92,500 1 $50,0001A Z F 10%,102 5 2$500,00010.162752 1 $92,500 1 $50,00010.062752 5 $14,262.50 or, using Excel®, AW 5 PMT110%,10,500000,2500002 1 92500 5 14,264.57

SOLUTION

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Since 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

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 percent to 20 percent. The plot of annual worth for the 10-year planning horizon is provided in Figure 5.2.

FIGURE 5.2

Analyzing the Effect on AW of Changes in MARR for Example 5.2

5-1 Annual Worth

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 percent. (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.)

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 n

Maximize AWj 5 c a At j 11 1 MARR2 n2t d 1A Z F MARR%,n2 (5.2) ;j t50

Using Annual Worth to Choose Between Two Alternatives 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 percent MARR and using an annual worth measure, which design, if either, should be chosen? Given: The cash flows outlined in Figure 5.3; MARR 5 10%; planning horizon 5 10 years Find: AW of each investment alternative. Which investment is recommended?

EXAMPLE

Video Example

KEY DATA

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Annual Worth and Future Worth

$55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 (+) 0

1

2

3

4

5

6

7

8

9

10

(–)

Alternative A

$300,000

$80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 (+) 0

1

2

3

4

5

6

7

8

9

(–)

Alternative B

$450,000 FIGURE 5.3

CFDs for Example 5.3

AWA 5 2$300,0001A Z P 10%,102 1 $55,000 5 2$300,00010.162752 1 $55,000 5 $6,175.00 5 PMT110%,10,3000002 1 55000 5 6176.38

SOLUTION

Likewise, AWB 5 2$450,0001A Z P 10%,102 1 $80,000 5 2$450,00010.162752 1 $80,000 5 $6,762.50 5 PMT110%,10,4500002 1 80000 5 6764.57 Since AWB . AWA . AWDN, Design B is recommended.

10

5-2

5-2

Future Worth

185

FUTURE WORTH

LEARNING O BJECTI VE: 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.’’ 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. We also learn how to determine the point in time when an investment begins to ‘‘make money.’’ 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.

Future Worth (FW) The value of all cash flows converted to a single sum equivalent at the end of the planning horizon using i 5 MARR.

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 5 a At 11 1 MARR2 n2t

(5.3)

t50

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.

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Future Worth of a Single Investment

EXAMPLE

Previously, we considered the acquisition of a new surface mount placement (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 percent after-tax MARR and future worth analysis, should the investment be made? KEY DATA

SOLUTION

Given: The cash flows outlined in Figure 5.1; MARR 5 10%; planning horizon 5 10 years Find: FW of the investment. Is this investment recommended? The future worth for the investment will be FW 5 2$500,0001F Z P 10%,102 1 $92,5001F Z A 10%,102 1 $50,000 5 2$500,00012.593742 1 $92,500115.937422 1 $50,000 5 $227,341.40 or, using Excel®, FW 5 FV110%,10,292500,5000002 1 50000 5 227,340.55 Since 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.

Using Future Worth to Achieve a Financial Goal

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 35-year period and if the annual return on the engineer’s investments is approximately 6.5 percent, what gradient step is required to achieve the goal if the first of the 36 investments equals $5,000? KEY DATA

Given: F 5 $5 million; i 5 6.5%; A1 5 $5,000; n 5 36 Find: G

5-2

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 3$5,000,0001A Z F 6.5%,362 2 $5,000 4/ 1A Z G 6.5%,362 where

Future Worth

187

SOLUTION

1A Z F 6.5%,362 5 0.065/ 3 11.0652 36 2 1 4 5 0.0075133

and 1A Z G 6.5%,362 5 511.0652 36 2 31 1 3610.06524 6/50.065 311.0652 36 2 14 6 5 11.22339 Therefore, G 5 3$5,000,00010.00751332 2 $5,000 4/11.22339 5 $2,901.66 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

Set Up to Use the Excel® SOLVER Tool to Determine the Gradient Step Needed to Achieve a Financial Goal

F I G U RE 5 . 4

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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 5 $3,000, the final balance is $5,146,882.19.) 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 5 $2,901.67, then a $5 million balance will occur in the investment account after 36 years.

FIGURE 5.5

EXPLORING THE SOLUTION

Excel® SOLVER Solution to Example 5.5

What if the investments do not earn 6.5 percent? 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 percent, the engineer anticipates the size of the investment portfolio will be between $4.6 million and $6.4 million after the thirty-sixth deposit.

5-2

FIGURE 5.6

Future Worth

Impact of Annual Returns on an Investment Portfolio

5.2.2 Future Worth of a 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 5 a Ajt 11 1 MARR2 n2t ;j

(5.4)

t50

Using Future Worth to Choose Between Two Alternatives 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

EXAMPLE

Video Example

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Annual Worth and Future Worth

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 percent 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 5 10%; planning horizon 5 10 years Find: FW of each investment alternative. Which investment is recommended?

SOLUTION

FWA 5 2$300,0001F Z P 10%,102 1 $55,0001F Z A 10%,102 5 2$300,00012.593742 1 $55,000115.937422 5 $98,436.10 or, using Excel®, FWA 5 FV110%,10,255000,3000002 5 $98,435.62 Likewise, FWB 5 2$450,0001F Z P 10%,102 1 $80,0001F Z A 10%,102 5 2$450,00012.593742 1 $80,000115.937422 5 $107,810.60 or, using Excel®, FWB 5 FV110%,10,280000,450000 5 $107,809.86 Since FWB . FWA . $0, Design Alternative B is recommended.

EXAMPLE

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,

5-2

Future Worth

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 percent 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 percent 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 percent. 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 percent of their annual deposit. The company still matches up to 4 percent 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 percent 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 percent, which retirement plan should she choose? Given: The employee and plan details summarized below. First Plan

KEY DATA

Second Plan

4% of annual salary invested 1 4% company match

4% of annual salary invested 1 4% company match

Current annual salary 5 $55,000

Current annual salary 5 $55,000

Annual salary increase 5 5%

Annual salary increase 5 5%

Historical portfolio annual growth 5 6%

Historical portfolio annual growth 5 2% to 12%

No management fees

Annual management fee 5 1.5%

Find: FW of each plan. Under the first plan, her investment portfolio will have the following value after 40 years: FW1 5 210.042 1$55,0002 1F Z A16%,5%,402 5 $4,4003 11.062 40 2 11.052 40 4/ 10.06 2 0.052 5 $1,428,120.90

SOLUTION

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Under the second plan, in the pessimistic case (earns 2 percent/year), her investment portfolio will have the following value after 40 years: FW2p 5 210.042 1$55,0002 10.9852 1F Z A12%,5%,402 5 $4,334 3 11.022 40 2 11.052 40 4/ 10.02 2 0.052 5 $698,055.57

Under the second plan, in the optimistic case (earns 12 percent/year), her investment portfolio will have the following value after 40 years: FW2o 5 210.042 1$55,0002 10.9852 1F Z A112%,5%,402 5 $4,334 3 11.122 40 2 11.052 40 4/ 10.12 2 0.052 5 $5,325,308.50

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 impact on investment returns of the global economic collapse of 2008, 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.

EXAMPLE

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 percent 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.

5-2

Future Worth

Given: The cash flows outlined in Figure 5.3; MARR 5 10%; planning horizon 5 10 years Find: FW of the investment portfolio for the do-nothing alternative, Alternative A, and Alternative B (including the investment and residual capital)

KEY DATA

The future worth for the do-nothing alternative (DN) would be

SOLUTION

FWDN 5 $450,0001F Z P 10%,102 5 $450,00012.593742 5 $1,167,183.00 or, using Excel®, FWDN 5 FV110%,10,,2450,0002 5 $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 5 $80,0001F Z A 10%,102 5 $80,000115.937422 5 $1,274,993.60 or, using Excel®, FWB 5 FV110%,10,2800002 5 $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 5 $55,0001F Z A 10%,102 5 $55,000115.937422 5 $876,558.10 or, using Excel®, FWA 5 FV110%,10,2550002 5 $876,558.35 In addition, since $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 percent. Therefore, the $150,000 will grow at a rate of 10 percent compounded annually to achieve a value of FWRC 5 $150,0001F Z P 10%,102 5 $150,00012.593742 5 $389,061 or, using Excel®, FWRC 5 FV110%,10,,21500002 5 $389,061.37 Hence, if Design Alternative A is chosen, the investment portfolio will total $876,558.10 1 $389,061.00, or $1,265,619.10. Or, using Excel®, the investment portfolio will total $876,558.35 1 $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

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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 2 $1,265,619.72, or $9374.25 greater than if Design A is purchased.

SUMMARY

KEY CONCEPTS 1. 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. (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: n

Maximize AWj 5 c a At j 11 1 MARR2 n2t d 1A Z F MARR%,n2 (5.2) ;j t50

2. 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. (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 5 a Ajt 11 1 MARR2 n2t ;j t50

KEY TERMS Annual Worth (AW), p. 180

Future Worth (FW), p. 185

(5.4)

Summary 195

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1.

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 percent. What is the annual equivalent cost of the investment if the planning horizon is 30 years? a. $29,760 c. $31,980 b. $30,600 d. $35,130

2.

When using annual worth to evaluate the attractiveness of a single alternative, what value is the calculated AW compared to? a. PW c. 0.0 b. FW d. MARR

3.

The annual worth of an alternative is 0. Which of the following is (are) also true? (1) PW 5 0 (2) FW 5 0 a. (1) only c. Both (1) and (2) b. (1) only d. Neither (1) nor (2)

4.

The overhead costs in a highly automated factory are expected to increase at an annual compound rate of 10 percent 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 percent/year. a. $263,250 c. $200,000 b. $231,520 d. $187,020

5.

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 percent/year compounded annually. a. $11,000 c. $13,423 b. $4,223 d. $15,223

6.

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

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year into a savings account that earns 8 percent annual interest per year. How much must each annual deposit be? a. $244 c. $280 b. $259 d. $357 7.

On the day your daughter is born, you deposit $1,000 in a college savings account that earns 8 percent compounded annually. On each of her birthdays thereafter, up to and including her eighteenth birthday, you deposit an additional $1,000. How much money is in the college account the day after her eighteenth birthday? a. $37,450 c. $41,450 b. $38,950 d. $46,800

8.

A deposit of $800 is planned for the end of each year into an account paying 8 percent/year compounded annually. The deposits were not made for the tenth and eleventh 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 c. $59,537 b. $55,397 d. $53,597

9.

If you invest $3,000 three years from now, how much will be in the account 15 years from now if i 5 8 percent compounded annually? a. $3,500 c. $9,415 b. $7,555 d. $9,516

10.

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 percent/year. What is the future equivalent cost of the investment if the planning horizon is 30 years? a. $3,371,000 c. $3,623,000 b. $3,467,000 d. $3,980,000

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 5.1.1 Annual Worth of a Single Alternative 1.

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 percent/year.

Summary 197

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? 2. DuraTech Manufacturing is evaluating a process improvement project. The

estimated receipts and disbursements associated with the project are shown below. MARR is 6 percent/year. End of Year 0 1 2 3 4 5

Receipts

Disbursements

$0 $0 $2,000 $4,000 $3,000 $1,600

$5,000 $200 $300 $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? 3.

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 percent/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?

4. 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 percent/year. Using an annual worth approach, determine the maximum amount Fabco should be willing to pay for the valves. 5.

Galvanized Products is considering the purchase of a new computer system for their enterprise data management system. The vendor has quoted a purchase price of $100,000. Galvanized Products is planning to borrow onefourth of the purchase price from a bank at 15 percent 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 percent/year to evaluate investments.

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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? 6. Quilts R Us (QRU) is considering an investment in a new patterning attach-

ment with the cash flow profile shown in the table below. QRU’s MARR is 13.5 percent/year. EOY

Cash Flow

EOY 8

$600

1 2 3 4

2$1,400 $0 $500 $500 $500

9 10 11 12

$700 $800 $900 2$1,000

5

$500

13

2$2,000

6

$0

14

7

$500

15

2$3,000 $1,400

0

Cash Flow

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? 7.

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 percent/year. EOY

Cash Flow (M$)

EOY

Cash Flow (M$)

0

2$12

4

$5

1

2$1 $5 $2

5

$5

6 7

$2 $5

2 3

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? 8. 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 percent/year. EOY

Cash Flow (K$)

EOY

Cash Flow (K$)

0

2$2

6

$200

1

2$10

7

2$10 2$12

2

2$12

8

3

2$14

9

2$14

4

2$16

10

2$100

5

2$18

Summary 199

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? 9.

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 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 percent compounded annually. Aerotron Electronics’ MARR is 10 percent 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? 10. Home Innovation is evaluating a new product design. The estimated receipts

and disbursements associated with the new product are shown below. MARR is 10 percent/year.

End of Year 0 1 2 3 4 5

Receipts

Disbursements

$0 $600 $600 $700 $700 $700

$1,000 $300 $300 $300 $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? 11.

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 percent/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?

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12. 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 percent/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 13.

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 percent per 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 ERP system should Manuel purchase? 14.

GO Tutorial 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 percent of the loss and will cost $0.40 per foot. A 2-inch thick insulated covering will eliminate 92 percent 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 percent/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.

15.

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 percent/year. a. What is the annual worth of each polisher? b. Which polisher should be recommended?

16.

GO Tutorial Quantum Logistics, Inc., a wholesale distributor, is considering the construction of a new warehouse to serve the southeastern geographic

Summary 201

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 percent/year. City

Initial Cost

Net Annual Income

Lagrange Auburn Anniston

$1,260,000 $1,000,000 $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? 17.

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 percent/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?

18.

GO Tutorial 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 percent/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?

19.

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 percent/year. a. What is the annual worth of each polisher? b. Which polisher should be recommended?

20. 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

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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 percent/year. a. What is the annual worth of each project? b. Which project should be recommended? 21.

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 percent/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.

22. 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 percent/year. Based on an annual worth analysis, determine if the packaging equipment should be purchased or leased. 23.

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 percent/year. Based on an annual worth analysis, which alternative (if any) should be implemented? End of Year

Light Bar

Sliding Spots

Reflected Beam

0

2$6,000 $2,000 $2,000 $2,000 $2,000 $2,000 $2,000

2$14,000 $3,500 $3,500 $3,500 $3,500 $3,500 $3,500

2$20,000 $0 $2,300 $4,600 $6,900 $9,200 $11,500

1 2 3 4 5 6

24. 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 percent/year/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 percent/year for

Summary 203

economic decision making. Based on an annual worth analysis, is the purchase of the new sprinkler system economically attractive? Section 5.2.1 25.

Future Worth of a Single Alternative

An investment has the following cash flow profile. MARR is 12 percent/ 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

2$30,000 $6,000 $13,500 $X $13,500

1 2 3 4

26. A 22-year-old engineering graduate wants to accumulate $2,000,000 to be avail-

able 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 percent per year. Since 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 percent 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? 27. Reconsider the situation described in Problem 26. Assume that rather than an-

nual 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 percent 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 percent annual raise each year throughout her career (starting 1 year from today) and adjusts her monthly deposits by the same 4 percent each year, how much will be in the account immediately after the last deposit? 28. 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 percent per year.

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Birthday 10 11 12 13 14

Deposit $1,000 $1,000 $2,000 $1,500

Birthday

Deposit

15 16 17 18

$3,000 $2,500 $2,000 $2,000

2$1,500

a. How much is in the account immediately after the deposit on your child’s

eighteenth birthday? b. How much would you have needed to deposit on each birthday to accu-

mulate the same total if you had started on your child’s tenth 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 first birthday and made equal annual deposits with no withdrawals? 29.

You decide to open an individual retirement account (IRA) at your local bank that pays 8 percent/year/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?

30. You decide to open an individual retirement account (IRA) at your local

stockbroker that pays 10 percent/year/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 forty-two $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? 31.

On your child’s first 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 percent larger than the previous. The account pays interest at 5 percent/year compounded annually. How much money is in the account immediately after the deposit on his eighteenth birthday?

32.

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 sixty-fifth birthday and that deposits start on the birthday shown and continue annually in the same amount on each birthday up to and including your sixtyfifth birthday. Assume that interest is compounded annually at 10 percent/year.

Summary 205

Birthday of First Deposit

Amount of Required Annual Deposit

25 30 35 40 45 50 55 60 65

33.

$1,000,000

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 sixty-fifth birthday and that deposits start on your twenty-sixth birthday and continue annually in the same amount on each birthday up to and including your sixty-fifth birthday. Interest Rate Earned

Amount of Required Annual Deposit

4 percent/year 5 percent/year 6 percent/year 7 percent/year 8 percent/year 9 percent/year

34.

Video Solution Bailey, Inc., is considering buying a new gang punch that would allow them 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 percent/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?

35.

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 percent/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?

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36. DuraTech Manufacturing is evaluating a process improvement project. The

estimated receipts and disbursements associated with the project are shown below. MARR is 6 percent/year. End of Year 0 1 2 3 4 5

Receipts

Disbursements

$0 $0 $2,000 $4,000 $3,000 $1,600

$5,000 $200 $300 $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? 37.

Galvanized Products is considering the purchase of a new computer system for their enterprise data management system. The vendor has quoted a purchase price of $100,000. Galvanized Products is planning to borrow onefourth of the purchase price from a bank at 15 percent 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 percent/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?

38. Quilts RUs (QRU) is considering an investment in a new patterning attach-

ment with the cash flow profile shown in the table below. QRU’s MARR is 13.5 percent/year. EOY 0 1 2 3 4 5 6 7

Cash Flow 2$1,400 $0 $500 $500 $500 $500 $0 $500

EOY

Cash Flow

8

$600

9 10 11 12 13 14 15

$700 $800 $900 −$1,000 −$2,000 −$3,000 $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?

Summary 207

39.

Video Solution 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 percent/year. EOY

Cash Flow (M$)

EOY

Cash Flow (M$)

0

2$12

4

$5

1

2$1 $5 $2

5

$5

6 7

$2 $5

2 3

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? 40. 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 percent/year. EOY

Cash Flow (K$)

EOY

Cash Flow (K$)

0

2$2

6

$200

1

2$10

7

2$10 2$12

2

2$12

8

3

2$14

9

2$14

4

2$16

10

2$100

5

2$18

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 41.

Future Worth of a Multiple Alternatives

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 percent/year. End of Year 1 2 3 4

Investment 1

Investment 2

Investment 3

$8,000 $9,000 $10,000 $11,000

$11,000 $10,000 $9,000 $8,000

$9,500 $9,500 $9,500 $9,500

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Annual Worth and Future Worth

42. The engineering team at Manuel’s Manufacturing, Inc., is planning to pur-

chase 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 percent per 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 ERP system should Manuel purchase? 43.

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 percent of the loss and will cost $0.40 per foot. A 2-inch thick insulated covering will eliminate 92 percent 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 percent/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.

44. Nadine Chelesvig has patented her invention. She is offering a potential man-

ufacturer 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 percent/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? 45.

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 percent/year. City

Initial Cost

Net Annual Income

Lagrange Auburn Anniston

$1,260,000 $1,000,000 $1,620,000

$480,000 $410,000 $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?

Summary 209

46.

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 percent/year. a. What is the future worth of each polisher? b. Which polisher should be recommended?

47.

Xanadu Mining is considering three mutually exclusive alternatives, as shown in the table below. MARR is 10 percent/year. EOY 0 1 2 3 4

A001

B002

C003

2$210 $80 $90 $100 $110

2$110 $60 $60 $60 $70

2$160 $80 $80 $80 $80

a. What is the future worth of each alternative? b. Which alternative should be recommended? 48. 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 percent per year compounded monthly. Use a future worth analysis to make a recommendation to Yani. 49.

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 percent/year.

Initial Investment Estimated Life Estimated Salvage Value Annual Operating Cost Annual Maintenance Cost

Drill Press T

Drill Press M

$20,000 10 years $5,000 $12,000 $2,000

$30,000 10 years $7,000 $6,000 $4,000

a. What is the future worth of each drill press? b. Which drill press should be recommended?

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Annual Worth and Future Worth

50.

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 zero 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 percent/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.

51.

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 percent/year, determine the preferred gate. End of Year

Gate 1

Gate 2

Gate 3

0

2$15,000

2$19,000

2$24,000

1

2$6,500

2$5,600

2$4,000

2

2$6,500

2$5,600

2$4,000

3

2$6,500

2$5,600

2$4,000

4

2$6,500

2$5,600

2$4,000

5

2$6,500 1 $0

2$5,600 1 $2,000

2$4,000 1 $5,000

52. 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 percent/ 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 percent/year for economic decision making. Based on a future worth analysis, is the purchase of the new sprinkler system economically attractive? 53.

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 percent/ year. Based on a future worth analysis, which alternative (if any) should be implemented?

Summary 211

End of Year

Light Bar

Sliding Spots

Reflected Beam

0

2$6,000 $2,000 $2,000 $2,000 $2,000 $2,000 $2,000

2$14,000 $3,500 $3,500 $3,500 $3,500 $3,500 $3,500

2$20,000 $0 $2,300 $4,600 $6,900 $9,200 $11,500

1 2 3 4 5 6

54. 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 percent/year. Based on a future worth analysis, determine which design should be chosen. Section 5.2.3 55.

Portfolio Analysis

An investor has $100,000 to invest in a business venture, or she can earn 10 percent/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 percent/year. Based on a future worth analysis, what should be done with the $100,000? End of Year 0 1 2 3 4

BV01

BV02

BV03

2$35,000 $0 $0 $0 $50,000

2$80,000 $10,000 $10,000 $10,000 $90,000

2$60,000 $0 $40,000 $0 $40,000

56. Reconsider data from Problem 53 (Orpheum Productions lighting enhance-

ment). 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.

6 ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : MOTORO L A S O L U T I O N S Motorola was founded in Illinois in 1928. From car radios, to televisions, to two-way 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., and subsequently were sold to Google in 2012. Following the separation of Motorola Mobility, the original company was renamed Motorola Solutions. In its 2011 Annual Report, Motorola Solutions Chairman & CEO Greg Brown stated, “Our strategy is focused on providing mission-critical communication solutions to government and enterprise customers.” The key words are mission-critical communication solutions. First responders depend on Motorola Solutions products in “moments that matter”—from military maneuvers, to emergency response, to rescue missions. 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 the delivery very likely was communicated using Motorola Solutions products. In addition, there are numerous behind-the-scenes applications of Motorola’s products, including barcode readers in distribution centers. Motorola Solutions, Inc. had annual revenues in 2011 of $8.2 billion. As of December 31, 2011, it and its subsidiaries had approximately 23,000 employees in 65 countries. During 2011, it had more than 100,000 customers in 100 countries. Approximately 7,000 engineers and scientists were involved in performing research and development; R&D expenditures totaled about $1 billion. 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 return on a particular capital investment is 212

RATE OF RETURN

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 an 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 their 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 their 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 their 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)?

LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

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) 213

214

Chapter 6

Rate of Return

2. Apply Descartes’ rule of signs and Norstrom’s criterion to test for multiple roots when using the IRR method. (Section 6.1.2)

3. 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)

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. 2. 3. 4. 5. 6. 7.

Identify the investment alternatives Define the planning horizon Specify the discount rate Estimate the cash flows Compare the alternatives Perform supplementary analyses 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,

6-1 Internal Rate of Return Calculations

215

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 O BJECTI VE: 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*j , satisfies the following equality:

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.

n

0 5 a Ajt 11 1 i*j 2 n2t

(6.1)

t50

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 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.

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

EXAMPLE

216

Chapter 6

Rate of Return

horizon, it was estimated that the SMP machine would be worth $50,000. Using a 10 percent MARR and internal rate of return analysis, should the investment be made? Given: The cash flows shown in Figure 6.1; MARR 5 10%; planning horizon 5 10 years Find: The IRR of the investment. Is this investment recommended?

KEY DATA

$50,000 (+) $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500

0

1

2

3

4

5

6

7

8

9

10

(–)

$500,000 FIGURE 6.1

SOLUTION

CFD for Example 6.1

Setting the future worth for the investment equal to 0 gives 0 5 2$500,0001F Z P i*%,102 1 $92,5001F Z A i*%,102 1 $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 5 1$500,000 2 $50,0002 1A Z P i*,102 1 $50,000i* 2 $92,500

For i 5 12%, $450,0001A Z P 12%,102 1 $50,00010.122 2 $92,500 5 2$6,850 For i 5 15%, $450,0001A Z P 15%,102 1 $50,00010.152 2 $92,500 5 $4,685 Interpolating for i* gives 13.78%.

6-1 Internal Rate of Return Calculations

Alternately, one can use the Excel® RATE worksheet function to solve for i* in this example: i* 5 RATE110,92500,2500000,500002 5 13.8% Because i* . 10 percent, the investment is recommended.

Example 6.1 illustrates one of the attractions of the IRR method. We can say “Invest $500,000 to obtain a 13.8 percent return on your investment.” This is often more informative or appealing than a dollar figure such as present or future worth. 6.1.2 Multiple Roots LEARNING O BJECTI VE: Apply Descartes’ rule of signs and Norstrom’s criterion to test for multiple roots when using the IRR method.

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.

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 percent interest rate. FW1 120%2 5 2$4,00011.22 3 1 $16,40011.22 2 2 $22,32011.22 1 $10,080 5 0 3 2 FW2 140%2 5 2$4,00011.42 1 $16,40011.42 2 $22,32011.42 1 $10,080 5 0 3 2 FW3 150%2 5 2$4,00011.52 1 $16,40011.52 2 $22,32011.52 1 $10,080 5 0

EXAMPLE

217

Chapter 6

Rate of Return

TABLE 6.1

Cash Flow Profile

EOY

CF

0

−$4,000

1

$16,400

2

−$22,320

3

$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%. $60 $40 Future Worth

218

$20 $0 10%

20%

30%

40%

50%

60%

–$20 –$40 MARR FIGURE 6 . 2

Plot of Future Worth for Example 6.2

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 positive-valued 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.

6-1 Internal Rate of Return Calculations

Rate of Return with Multiple Sign Changes and a Single Root

EXAMPLE

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 sixth 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? 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.

KEY DATA

From the plot of future worth shown in Figure 6.3, it is evident that a single root exists, at i* 5 19.12%, and using the Excel® SOLVER tool, FW is maximized when MARR equals 8.5469%. For the investment IRR 5 19.12%.

SOLUTION

FIGURE 6.3

IRR for a Natural Gas Investment

219

220

Chapter 6

Rate of Return

6.1.3 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 ij*, satisfies the following equality: n

0 5 a Ajt 11 1 i*j 2 n2t t50

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.

IRR Analysis with Mutually Exclusive Alternatives

EXAMPLE

Video Example

KEY DATA

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 aftertax 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 percent MARR and an IRR comparison, which design (if either) should be chosen? Given: The cash flows outlined in Figure 6.4; MARR 5 10%; planning horizon 5 10 years Find: Recommended investment using incremental IRR analysis.

6-1

Internal Rate of Return Calculations

$55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 (+) 0

1

2

3

4

5

6

7

8

9

10

(–)

Alternative A

$300,000

$80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 (+) 0

1

2

3

4

5

6

7

8

9

10

(–)

Alternative B

$450,000 FIGURE 6.4

CFDs for Example 6.4

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. iA* 5 RATE110,255000,3000002 5 12.87% . MARR 5 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 in initial investment in Design A): * 5 RATE110,225000,1500002 iB2A 5 10.56% . MARR 5 10%

SOLUTION

221

222

Chapter 6

Rate of Return

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: iB* 5 RATE110,280000,4500002 5 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 well-behaved 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.

IRR Analysis with Regular Cash Flow Series for Alternative Investments, but Irregular Incremental Analysis Cash Flows

EXAMPLE

Two mutually exclusive investment alternatives are being considered. The MARR is 12 percent. 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.

6-1

TABLE 6.2 EOY

Data for Example 6.5 CF(1)

CF(2)

CF(2-1)

0

2$100,000.00

2$104,000.00

2$4,000.00

1

$33,600.00

$50,000.00

$16,400.00

2

$72,320.00

$50,000.00

2$22,320.00

3

$39,920.00

$50,000.00

$10,080.00

IRR 5

Internal Rate of Return Calculations

20.8122%

20.0000%

IRR 5

40.0000%

IRR 5

50.0000%

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 5 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, 2$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. $60

Future Worth

$40 $20 $0 10%

20%

30%

40%

50%

60%

–$20 –$40 MARR FIGURE 6.5 Plot of Future Worth of Incremental Investment CF(2-1) for Example 6.5

SOLUTION

223

224

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Rate of Return

6-2

EXTERNAL RATE OF RETURN CALCULATIONS

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.

LEARN I N G O B JEC T I V E

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.

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

n

t50

t50

n2t 5 a Ct 11 1 i¿2 n2t a Rt 11 1 r2

(6.2)

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 i9 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 5 MARR, then IRR 5 ERR 5 MARR. 6.2.1

Single Alternative

We begin by considering ERR for a single-alternative, the now-familiar SMP machine.

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.

6-2 External Rate of Return Calculations

Computing the External Rate of Return for a Single Alternative

EXAMPLE

The investment of $500,000 in a surface mount placement 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? Given: The cash flows outlined in Figure 6.1; MARR 5 10%; planning horizon 5 10 years Find: The ERR of the investment. Is this investment recommended?

KEY DATA

The only negative-valued cash flow is the initial investment. Therefore, Equation 6.2 becomes

SOLUTION

$500,0001F Z P i¿,102 5 $92,5001F Z A 10%,102 1 $50,000 $500,00011 1 i¿2 10 5 $92,5001F Z A 10%,102 1 $50,000 11 1 i¿ 2 10 5 3 $92,500115.937422 1 $50,000 4 /$500,000 5 3.048423 i¿ 5 11.79117% or i¿ 5 RATE110,,2500000,FV110%,10,29250021500002 5 11.79117% Since i9 . 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 5MIRR(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 figure for the ERR: 11.79118%. Recall, for this example, IRR 5 13.8%. 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. Define two new cash flow series: CF(1) and CF(2). Using the Excel® IF function, separate the original cash flow roots into two series: {Rt, t 5 0, . . . ,10} and {Ct, t 5 0, . . . ,10}. 2. Next, develop a new cash flow series {CFt, t 5 0, . . . ,10}, which, except for t 5 10, is the negative of {Ct}. 1.

225

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Rate of Return

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 negative-valued 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 percent. 5. Recalling the definition of the external rate of return (the interest rate that makes the absolute value of the future worth of the negativevalued 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 5 11.79118%.

FIGURE 6 . 6

EXAMPLE

ERR Solution to Example 6.6

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 5 0, 1, 2, and 3: 2$4,000, $16,400, 2$22,320, and $10,080, respectively. Using a MARR of 12 percent, find the ERR.

6-2 External Rate of Return Calculations

From Equation 6.2, $4,0001F Z P i¿,32 1 $22,3201F Z P i¿,12 5 $16,4001F Z P 12%,22 1 $10,080 To determine the value of i9 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. 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

SOLUTION

227

228

Chapter 6

Rate of Return

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%

4.2999%

24%

23.9837%

44%

44.0049%

6%

6.2338%

26%

25.9805%

46%

46.0052%

8%

8.1775%

28%

27.9799%

48%

48.0037%

10%

10.1302%

30%

29.9812%

50%

50.0000%

12%

12.0911%

32%

31.9840%

52%

51.9939%

14%

14.0592%

34%

33.9877%

54%

53.9850%

16%

16.0399%

36%

35.9919%

56%

55.9732%

18%

18.0144%

38%

37.9962%

58%

57.9581%

6.2.2 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 (ij9), there is a common reinvestment rate (r). Mathematically, for alternative j, the following equality must hold: n

n

t50

t50

n2t 5 a Cjt 11 1 ij¿2 n2t a Rjt 11 1 r2

(6.3)

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

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 percent MARR, which alternative is preferred?

6-2

External Rate of Return Calculations

Given: The cash flows outlined in Figure 6.4; MARR 5 10%; planning horizon 5 10 years Find: ERR for each alternative and the incremental investment.

KEY DATA

The following steps are taken: 1. Compute the ERR for the smaller initial investment alternative, Design A:

SOLUTION

$300,00011 1 i¿A 2 10 5 $55,0001F Z A 10%,102 11 1 i¿A 2 10 5 $55,000115.937422/$300,000 i¿A 5 11.31814% . MARR 5 10% or i¿A 5 RATE110,,2300000,FV110%,10,2550002 2 i¿A 5 11.31814% . MARR 5 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: $150,00011 1 i¿B 2A 2 10 5 $25,0001F Z A 10%,102 11 1 i¿B 2A 2 10 5 $25,000115.93742/$150,000 i¿B 2A 5 10.26219% . MARR 5 10% or i¿B 2A 5 RATE110,,2150000,FV110%,10,2250002 2 5 10.26219% . MARR 5 10% The incremental investment to accept Design B is justified economically. 3. Design Alternative B will have an overall external rate of return of $450,00011 1 i¿B 2 10 5 $80,0001F Z A 10%,102 11 1 i¿B 2 10 5 $80,000115.93742/$450,000 i¿B 5 10.97611% or i¿B 5 RATE110,,2450000,FV110%,10,2800002 2 5 10.97611%

229

230

Chapter 6

Rate of Return

ERR Analysis with Regular Cash Flow Series for Alternative Investments, but Irregular Incremental Analysis Cash Flows

EXAMPLE

Recall Example 6.5 where two mutually exclusive alternative investments are considered. The MARR is 12 percent. 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 three years. It also has a regular cash flow, with a single change of sign. Using ERR, which alternative is preferred?

FIGURE 6 . 8

SOLUTION

Data and ERR Analysis for Example 6.9

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 5 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, 2$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%. Since ERR221 5 12.0911% . MARR 5 12%, Alternative 2 is selected.

Summary 231

SUMMARY

KEY CONCEPTS 1. 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. (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 ij*, satisfies the following equality: n

0 5 a Ajt 11 1 ij* 2 n2t

(6.1)

t50

In words, the internal rate of return is the interest rate that makes the future worth of an investment equal 0. 2. Learning Objective: Apply Descartes’ rule of signs and Norstrom’s criterion to test for multiple roots when using the IRR method. (Section 6.1.2)

Descartes’ rule of signs and Norstrom’s criterion are common methods used to test for multiple solutions with the IRR method. Descartes’ rule of signs states that there will be at most as many positive rates of return as there are sign changes in the cash flow profile. Norstrom’s criterion is applied to determine if there is at most one real positive IRR. 3. 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. (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

n

t50

t50

n2t 5 a Ct 11 1 i¿2 n2t a Rt 11 1 r2

(6.2)

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Chapter 6

Rate of Return

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 i9 is the external rate of return.

KEY TERMS External Rate of Return (ERR), p. 224

Internal Rate of Return (IRR), p. 215

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1.

Consider the following cash flow diagram. What is the value of X if the internal rate of return is 15%? X

0

1

X

2

$400 a. $246 b. $255 2.

c. $281 d. $290

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? PW 5 0 AW 5 0 a. I only c. Neither I nor II b. II only d. Both I and II

Summary 233

3.

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 c. a unique value for ERR d. a positive PW at MARR

4.

What is the internal rate of return of the following cash flow diagram?

0

1

$30

$31

2

3

$15

$30

a. 20.0% b. 18.2%

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% c. 12% to 14% b. 10% to 12% d. Greater than 14% The next two questions are based on the following “present worth versus interest rate” graph for a well-behaved investment. A Present Worth

5.

c. 17.5% d. 15.0%

E

C

B Interest Rate

D

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Chapter 6

Rate of Return

6.

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

7.

The IRR of this investment is located at which point? a. A c. D b. C d. E

8.

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

9.

Using an incremental internal rate of return (IRR) analysis the decision to replace the “current best” by the “challenger” is based on what decision rule? a. the internal rate of return of the increment is greater than the external rate of return b. the internal rate of return of the increment is greater than the internal rate of return of the previous increment c. the internal rate of return of the increment is greater than zero d. the internal rate of return of the increment is greater than MARR

10.

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

11.

When conducting an incremental analysis, what step must always be taken immediately prior to beginning the pairwise comparisons? a. order the alternatives from highest to lowest initial investment b. order the alternatives from lowest to highest present worth c. order the alternatives from lowest to highest internal rate of return d. order the alternatives from lowest to highest initial investment

12.

Consider the IRR and ERR measures of worth. If we define a “root” to mean a value for the measure that results in PW 5 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

Summary 235

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 13.

Consider the calculation of an external rate of return (ERR). The positive cash flows in the cash flow profile are moved forward to t 5 n using what value of i in the (FZP,i,n–t) factors? a. 0 b. the unknown value of ERR (i9) 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. Introduction 1. 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 (b) External Rate of Return (c) Future Worth (d) Internal Rate of Return (e) Present Worth

(1) Ranking Approach (2) Incremental Approach

2. Match the measures of worth in the first column with the appropriate unit of

measure that results from the analysis. Measure of Worth

Resulting Units of Measure

(a) Annual Worth (b) External Rate of Return (c) Future Worth (d) Internal Rate of Return (e) Present Worth

Sections 6.1.1 and 6.1.2

(1) dollars (2) percentage

IRR Calculations—Single Alternative

3. Draw a cash flow diagram of any investment that has both of the following

properties: (1) the investment has a four-year life, (2) the investment has a 10%/yr internal rate of return

236

Chapter 6

Rate of Return

4. 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

2$30,000 $6,000 $13,500 $X $13,500

1 2 3 4

a. MARR is 12%/yr b. MARR is 15%/yr c. MARR is 24%/yr 5.

d. MARR is 8%/yr e. MARR is 0%/yr

Nu Things, Inc. is considering an investment in a business venture with the following anticipated cash flow results: EOY

Cash Flow

EOY

Cash Flow

0

2$70,000 20,000 19,000 18,000 17,000 16,000 15,000

7

14,000

14

7,000

8 9 10 11 12 13

13,000 12,000 11,000 10,000 9,000 8,000

15 16 17 18 19 20

6,000 5,000 4,000 3,000 2,000 1,000

1 2 3 4 5 6

EOY

Cash Flow

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. 6. 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? 7.

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.

Summary 237

8. 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%/yr. EOY

Cash Flow (M$)

0

2$12

1

2$1 $5 $2 $5 $5 $2 $5

2 3 4 5 6 7

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? 9. 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 0 1 2 3 4 5

Receipts

Disbursements

$0 $0 $2,000 $4,000 $3,000 $1,600

$5,000 $200 $300 $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? 10. 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?

238

Chapter 6

Rate of Return

11. 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 0 1 2 3 4 5

Receipts

Disbursements

$0 $600 $600 $700 $700 $700

$1,000 $300 $300 $300 $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? 12. 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 0 1 2 3 4 5 6

NCF 2$70,000 $30,000 $30,000 $30,000 $30,000 $30,000 $30,000 1 $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? 13.

Bailey, Inc. is considering buying a new gang punch that would allow them 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?

14. 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

Summary 239

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? GO Tutorial 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? 16. Brock Associates invested $80,000 in a business venture with the following results: 15.

EOY 0 1 2 3 4

CF 2$80,000 10,000 16,000 22,000 28,000

EOY

CF

5

$28,000

6 7 8

22,000 16,000 10,000

MARR is 12%. Determine the internal rate of return and whether or not this is a desirable venture. 17. Shrewd Endeavors, Inc. invested $70,000 in a business venture with the

following cash flow results: EOY

CF

0

2$70,000 20,000 19,000 18,000 17,000 16,000 15,000 14,000 13,000 12,000 11,000

1 2 3 4 5 6 7 8 9 10

EOY

CF

11

10,000

12 13 14 15 16 17 18 19 20

9,000 8,000 7,000 6,000 5,000 4,000 3,000 2,000 1,000

MARR is 10%. Determine the internal rate of return and whether or not this is a desirable venture.

240

Chapter 6

Rate of Return

18. An investment of $20,000 for a new condenser is being considered. Esti-

mated 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. 19. 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 5 10%. 20.

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. b. c. d. e. f. g. h. i.

21.

the present worth of the project is greater than zero the present worth of the project is equal to zero the present worth of the project is less than zero the future worth of the project is greater than zero the future worth of the project is equal to zero the future worth of the project is less than zero the annual worth of the project is greater than zero the annual worth of the project is equal to zero the annual worth of the project is less than zero

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 fifth through fifteenth 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.

22. 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,

Summary 241

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. 23.

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 of 12%/yr/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?

24. Galvanized Products is considering the purchase of a new computer system

for their enterprise data management system. The vendor has quoted a purchase price of $100,000. Galvanized Products is planning to borrow onefourth 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 five 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 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? 25. Aerotron Electronics is considering the purchase of a water filtration sys-

tem 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 two 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?

242

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Rate of Return

26. 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 four equal annual payments (first payment at t 5 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? Consider the following cash flow profile and assume MARR is 10%/yr.

27.

EOY 0

2$100 $25 $25 $60

1 2 3

a. b. c. d.

NCF

EOY

NCF

4

2$30 $60 $25

5 6

What does Descartes’ rule of signs tell us about the IRR(s) of this project? What does Norstrom’s criterion tell us about the IRR(s) of this project? Determine the IRR(s) for this project. Is this project economically attractive? Video Solution Consider the following cash flow profile and assume

28.

MARR is 10%/yr. EOY 0

2$100 $25 $25 $25

1 2 3

a. b. c. d.

NCF

EOY

NCF

4

$25

5 6

$25 $25

What does Descartes’ rule of signs tell us about the IRR(s) of this project? What does Norstrom’s criterion tell us about the IRR(s) of this project? Determine the IRR(s) for this project. Is this project economically attractive? Consider the following cash flow profile and assume MARR is 10%/yr.

29.

EOY 0 1 2 3

a. b. c. d.

NCF 2$100 $15 $15 $15

EOY

NCF

4

$15

5 6

$15 $15

What does Descartes’ rule of signs tell us about the IRR(s) of this project? What does Norstrom’s criterion tell us about the IRR(s) of this project? Determine the IRR(s) for this project. Is this project economically attractive?

Summary 243

30. Consider the following cash flow profile and assume MARR is 10%/yr. EOY 0

NCF

EOY

NCF

2$100 $25

4

$250

5

2$200

2

$200

6

2$100

3

2$100

1

a. b. c. d.

What does Descartes’ rule of signs tell us about the IRR(s) of this project? What does Norstrom’s criterion tell us about the IRR(s) of this project? Determine the IRR(s) for this project. Is this project economically attractive? Consider the following cash flow profile and assume MARR is 10%/yr.

31.

EOY 0 1 2

EOY

NCF

4

2$950 $700

5 6

2$750 $900

3

a. b. c. d.

NCF 2$100 $800

2$800

What does Descartes’ rule of signs tell us about the IRR(s) of this project? What does Norstrom’s criterion tell us about the IRR(s) of this project? Determine the IRR(s) for this project. Is this project economically attractive?

32. Consider the following cash flow profile and assume MARR is 10%/yr. EOY 0 1 2 3

NCF

EOY

2$101 $411 2$558 $253

NCF

4

$2

5 6

$8 2$14

a. Determine the IRR(s) for this project. b. Is this project economically attractive? 33. 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

2$1,400 $0 $500 $500 $500

8

$600

1 2 3 4

9 10 11 12

$700 $800 $900 2$1,000

5

$500

13

2$2,000

6

$0

14

7

$500

15

2$3,000 $1,400

0

Cash Flow

244

Chapter 6

Rate of Return

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

34. Quantum Logistics, Inc., a wholesale distributor, is considering the construc-

tion 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 Auburn Anniston

$1,260,000 $1,000,000 $1,620,000

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

35. Two mutually exclusive proposals, each with a life of 5 years, are under con-

sideration. MARR is 12%. Each proposal has the following cash flow profile: EOY 0 1 2 3 4 5

NCF(A)

NCF(B)

2$30,000 $9,300 $9,300 $9,300 $9,300 $9,300

2$42,000 $12,625 $12,625 $12,625 $12,625 $12,625

Determine which alternative the decision maker should select using the internal rate of return method. 36. Chingos and Daughters Construction is considering three investment propos-

als: 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 5 15%.

Initial investment Planning horizon Salvage values Annual receipts Annual disbursements

NCF(A)

NCF(B)

NCF(C)

$600,000 10 years $70,000 $400,000 $130,000

$800,000 10 years $130,000 $600,000 $270,000

$470,000 10 years $65,000 $260,000 $70,000

Determine which alternative should be selected using the internal rate of return method.

Summary 245

37. 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 null A B C D

Initial Investment

IRR

$0 $600,000 $800,000 $470,000 $540,000

0.0% 44.0% 40.0% 39.2% 36.0%

Incremental IRR(s) Increment

Incremental IRR

B-A B-D B-C A-D A-C D-C

28.3% 48.8% 41.4% 116.5% 61.0% 18.4%

For each of the values of MARR below indicate which project is preferred based on an incremental IRR analysis. a. MARR 5 50% b. MARR 5 41% c. MARR 5 25% 38. 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.

Project A B C D E F G

Incremental Rate of Return of “Row”—“Column”

Initial Investment

Null

A

B

C

D

E

F

$10,000 12,000 13,000 15,000 16,000 18,000 23,000

10% 9 8 7 6 5 7

7% 2 9 5 8 3

0.1% 5 1 2 8

9% 6 5 7

3% 5 4

5% 3

2%

246

Chapter 6

Rate of Return

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 5 12% d. MARR 5 3.5% b. MARR 5 9.5% e. MARR 5 1.5% c. MARR 5 8% 39.

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.

Useful Life First Cost Salvage Value Annual Revenue Annual Expenses

T1

T2

T3

T4

10 years $600,000 $70,000 $400,000 $130,000

10 years $800,000 $130,000 $600,000 $270,000

10 years $470,000 $65,000 $260,000 $70,000

10 years $540,000 $200,000 $320,000 $120,000

a. Determine the preferred telescope if MARR is 25%/yr. b. Determine the preferred telescope if MARR is 42%/yr. 40. 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

2$6,000 $2,000 $2,000 $2,000 $2,000 $2,000 $2,000

2$14,000 $3,500 $3,500 $3,500 $3,500 $3,500 $3,500

2$20,000 $0 $2,300 $4,600 $6,900 $9,200 $11,500

1 2 3 4 5 6

41.

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.

Summary 247

42. On-Site Testing Service has received four investment proposals for consid-

eration. 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 0 1 through 8

X1

X2

Y1

Y2

2$10,000 $1,600

2$15,000 $2,600

2$6,000 $2,500

2$9,000 $3,500

a. List all the alternatives to be considered. b. Determine which (if any) proposals On-Site Testing should accept. 43. Arnold Engineering has available two mutually exclusive investment propos-

als, 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

2$40,000 8,000 8,000 8,000 8,000 8,000 8,000 8,000 8,000 8,000 8,000

2$30,000 9,000 8,500 8,000 7,500 7,000 6,500 6,000 5,500 5,000 4,500

1 2 3 4 5 6 7 8 9 10

Determine which alternative the decision maker should select using an internal rate of return approach. 44. 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%.

Initial investment Planning horizon Annual receipts Annual disbursements Salvage value

NCF(A)

NCF(B)

NCF(C)

NCF(D)

$400,000 10 years $205,000 $110,000 $50,000

$400,000 10 years $215,000 $125,000 $50,000

$600,000 10 years $260,000 $120,000 $100,000

$300,000 10 years $230,000 $150,000 $50,000

Determine which alternative the decision maker should select. Use the internal rate of return method.

248

Chapter 6

Rate of Return

45.

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

Gate 1

Gate 2

Gate 3

0

2$15,000

2$19,000

2$24,000

1

2$6,500

2$5,600

2$4,000

2

2$6,500

2$5,600

2$4,000

3

2$6,500

2$5,600

2$4,000

4

2$6,500

2$5,600

2$4,000

5

2$6,500 1 $0

2$5,600 1 $2,000

2$4,000 1 $5,000

46. Five projects form the mutually exclusive, collectively exhaustive set under

consideration. The cash flow profiles for the five projects are given in the table below.

Life Initial Investment Salvage Value Annual Revenues Annual Expenses

null

A

B

C

D

10 years 0 0 0 0

10 years $600,000 $70,000 $400,000 $130,000

10 years $800,000 $130,000 $600,000 $270,000

10 years $470,000 $65,000 $260,000 $70,000

10 years $540,000 $200,000 $320,000 $120,000

Information on each project 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” Project

Null

C

D

A

C D A B

39% 36% 44% 40%

18% XXXXX 41%

117% 49%

28%

a. Specify the incremental cash flow profile that must be analyzed to deter-

mine 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 project is preferred? d. Based on the data in the table, if MARR is 40%, specify whether the present

worth of each project would be positive, negative, or zero when evaluated at MARR?

Summary 249

47. DelRay Foods must purchase a new gumdrop machine. Two machines are avail-

able. 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 should be recommended? 48.

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?

49.

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?

50. 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

2$210 $80 $90 $100 $110

2$110 $60 $60 $60 $70

2$160 $80 $80 $80 $80

1 2 3 4

51.

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

250

Chapter 6

Rate of Return

a life expectancy of 10 years. Determine which insulation (if any) should be purchased using an internal rate of return analysis. 52. 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. 53.

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 five years at an expected cost of $150,000 per year to 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 five 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?

54. 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 6.2.1

ERR Calculations—Single Alternative

55. Consider the following cash flow profile and assume MARR is 10%/yr. EOY 0 1 2 3

NCF 2$100 $800 2$750 $900

EOY 4 5 6

NCF 2$950 $700 2$800

Summary 251

a. Determine the ERR for this project. b. Is this project economically attractive?

Consider the following cash flow profile and assume MARR is 10%/yr.

56.

EOY 0

NCF 2$100 $25 $25 $25

1 2 3

EOY

NCF

4

$25

5 6

$25 $25

a. Determine the ERR for this project. b. Is this project economically attractive? 57. Consider the following cash flow profile and assume MARR is 10%/yr. EOY 0

NCF 2$100 $25 $25 $60

1 2 3

EOY

NCF

4

2$30 $60 $25

5 6

a. Determine the ERR for this project. b. Is this project economically attractive? 58. Consider the following cash flow profile and assume MARR is 10%/yr. EOY 0

NCF 2$101 $411

1 2

2$558 $253

3

EOY

NCF

4

$2

5 6

2$14

$8

a. Determine the ERR for this project. b. Is this project economically attractive?

Consider the following cash flow profile and assume MARR is 10%/yr.

59.

EOY 0 1 2 3

NCF 2$100 $15 $15 $15

EOY

NCF

4

$15

5 6

$15 $15

a. Determine the ERR for this project. b. Is this project economically attractive?

252

Chapter 6

Rate of Return

60. Consider the following cash flow profile and assume MARR is 10%/yr. EOY 0

NCF

EOY

NCF

2$100 $25

4

$250

5

2$200

2

$200

6

2$100

3

2$100

1

a. Determine the ERR for this project. b. Is this project economically attractive? 61.

Nu Things, Inc. is considering an investment in a business venture with the following anticipated cash flow results: EOY 0 1 2 3 4 5 6

Cash Flow

EOY

Cash Flow

2$70,000 20,000 19,000 18,000 17,000 16,000 15,000

7

14,000

14

EOY

Cash Flow 7,000

8 9 10 11 12 13

13,000 12,000 11,000 10,000 9,000 8,000

15 16 17 18 19 20

6,000 5,000 4,000 3,000 2,000 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. 62. Solve problem 16, except use the external rate of return approach. 63. Solve problem 17, except use the external rate of return approach. 64. Solve problem 18, except use the external rate of return approach. 65. Solve problem 19, except use the external rate of return approach. 66. Quilts R Us (QRU) is considering an investment in a new patterning attach-

ment with the cash flow profile shown in the table below. QRU’s MARR is 13.5%/yr. EOY 0 1 2 3 4 5 6 7

Cash Flow 2$1,400 $0 $500 $500 $500 $500 $0 $500

EOY

Cash Flow

8

$600

9 10 11 12 13 14 15

$700 $800 $900 −$1,000 −$2,000 −$3,000 $1,400

Summary 253

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? 67.

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 two 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 Fit buy the water filtration system? 68. 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 1 2 3 4

$0 $600 $600 $700 $700

$1,000 $300 $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? 69.

GO Tutorial Galvanized Products is considering the purchase of a new computer system for their 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 five years and has a salvage value of $5,000 at that time. Over the five year period, Galvanized Products expects to pay a technician $25,000 per year to maintain the system but will

254

Chapter 6

Rate of Return

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

70. 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

71.

Gate 1

Gate 2

Gate 3

0

2$15,000

2$19,000

2$24,000

1

2$6,500

2$5,600

2$4,000

2

2$6,500

2$5,600

2$4,000

3

2$6,500

2$5,600

2$4,000

4

2$6,500

2$5,600

2$4,000

5

2$6,500 1 $0

2$5,600 1 $2,000

2$4,000 1 $5,000

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 five years at an expected cost of $150,000 per year to 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 five 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?

Summary 255

72. 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? 73. Solve problem 35, except use the external rate of return approach. 74. Solve problem 36, except use the external rate of return approach. 75. Solve problem 43, except use the external rate of return approach. 76. Solve problem 44, except use the external rate of return approach.

ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : J. B. HU N T T R AN S P O RT S E RVI C E S J. B. Hunt Transport Services Inc. (JBHT) is one of the largest surface transportation companies in North America. A full truckload transportation firm that delivers goods all over the United States, its approximately 10,500 drivers make short- and long-haul deliveries. In 2011, JBHT’s consolidated revenue was $4.5 billion, spread across its four business segments: intermodal (JBI), dedicated contract services (DCS), truck (JBT), and integrated capacity solutions (ICS). Intermodal is a partnership between JBHT and most major North American rail carriers to deliver freight throughout the continental United States, Canada, and Mexico; full containers are delivered by JBI between customers and railroads. In 2011, JBI generated 59 percent of the company’s total revenue. Dedicated contract services involve partnerships with major corporations such as Home Depot, Wal-Mart, Procter & Gamble, and Coca-Cola; DCS provides logistics services for its partners, often driving trucks and trailers showing the customers’ logos rather than JBHT’s logo; in 2011, DCS accounted for 22 percent of total revenue. The truck segment, consisting of truckload dry-van shipping, is the most visible, with its JBHT trucks regularly traveling U.S. interstates and highways; in 2011, JBT represented 11 percent of total revenue. The newest segment, integrated capacity solutions, provides transportation solutions to customers by utilizing third-party carriers, as well as JBHT-owned equipment. In a sense, ICS serves as a freight broker; it purchases transportation services on behalf of its customers and bills its customers for services provided. ICS services include flatbed, refrigerated, less-than-truckload (LTL), and a variety of dry-van and intermodal solutions. In 2011, ICS generated 8 percent of total revenue. With more than 9,100 company-owned tractors and 75,000 trailers and containers used to ship nearly 3.4 million loads in 2011, JBHT is constantly faced with decisions regarding the acquisition and replacement of tractors

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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 $100,000. In addition, JBHT annually purchases a large quantity of trailers and vans at an average cost of approximately $25,000. Also, JBHT annually purchases several hundred intermodal containers for approximately $15,000 each, plus the chassis to haul the containers (each chassis costs approximately $13,000). Tractors tend to be replaced after driving approximately 600,000 miles; during the life of a tractor, it incurs annual repair and maintenance expenses in excess of $7,000. Trade-in values on used tractors vary depending on use, but it is typical for JBHT to receive approximately $25,000 for a used tractor and $10,000 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?

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LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

1. Explain the terms defender, challenger, obsolescence, and sunk costs as they pertain to replacement analysis. (Section 7.1)

2. Perform a replacement analysis using a cash flow approach. (Section 7.2) 3. Perform a replacement analysis using an opportunity cost approach. (Section 7.2)

4. Determine the optimum replacement interval in cases where an asset will be used for many years. (Section 7.3)

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

FUNDAMENTALS OF REPLACEMENT ANALYSIS

LEARN I N G O B JEC T I V E : Explain the terms challenger, defender, obsoles-

cence, and sunk costs as they pertain to 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

7-1

Fundamentals of Replacement Analysis

replacement analysis when referring to 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: 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 thirdparty suppliers; and d. increased demand that cannot be met with the current equipment— either supplementary or replacement equipment is required to meet demand. 1.

Obsolescence is a frequently cited reason for replacing an asset. Various types of obsolescence can occur, including the following: 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. 1.

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

259

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.

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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.)

Systematic Economic Analysis Technique 1. 2. 3. 4. 5. 6. 7.

Identify the investment alternatives Define the planning horizon Specify the discount rate Estimate the cash flows Compare the alternatives Perform supplementary analyses 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 2. 3.

4.

5.

6.

to invest in new technology. The current equipment still works and produces a product of acceptable quality—an “if it isn’t broken, don’t replace it” attitude prevails. 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!”) 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. 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). 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.

7-2

Cash Flow and Opportunity Cost Approaches

261

7. The psychological impact of sunk costs—it is very hard for some peo-

ple 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 since 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.

7-2

CASH FLOW AND OPPORTUNITY COST APPROACHES

LEARNING O BJECTI VES: Perform a replacement analysis using a cash flow

approach. Perform a replacement analysis using an opportunity cost approach.

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.

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Equivalent Uniform Annual Cost The EUAC is the equivalent uniform annual cost to keep or operate an asset.

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.

EXAMPLE

Cash Flow Approach to Replacement Analysis

Video Example

KEY DATA

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%. Given: MARR 5 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

2$50,000.00

2$105,000.00

2$5,000.00

1–4

2$35,000.00

2$15,000.00

2$38,000.00

5

2$35,000.00

$25,000.00

2$18,000.00

Find: EUAC for each alternative. SOLUTION

The EUACs for each alternative are as follows: a) EUAC(Keep) 5 $35,000 1 $50,000(A Z P 10%,5) 5 $35,000 1 $50,000 (0.26380) 5 $48,190.00

5 PMT110%,5,2500002 1 35000 5 48,189.87 b) EUAC(Replace) 5 $15,000 1 $105,000(A Z P 10%,5) – $40,000 (A Z F 10%,5) 5 $15,000 1 $105,000(0.26380) – $40,000(0.16380) 5 $36,147.00 5 PMT110%,5,2105000,400002 1 15000 5 $36,146.84

7-2 Cash Flow and Opportunity Cost Approaches

c) EUAC(Lease) 5 $18,000 1 $20,000(F Z P 10%,1) – $15,000(A Z P 10%,5) 5 $18,000 1 $20,000(1.10) – $15,000(0.26380) 5 $36,043.00 5 PMT110%,5,150002 1 18000 2 FV110%,1,,200002 5 $36,043.04 Notice, the $100,000 purchase price for the mixer is a sunk cost. It does not affect the cash flows over the planning horizon.

Opportunity Cost Approach to Replacement Analysis

EXAMPLE

The replacement problem in Example 7.1 can also be addressed by the opportunity cost approach. Given: MARR 5 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

2$65,000.00

2$120,000.00

2$20,000.00

1–4

2$35,000.00

2$15,000.00

2$38,000.00

5

2$35,000.00

$25,000.00

2$18,000.00

KEY DATA

Find: EUAC for each alternative. The EUACs for each alternative are as follows: a) EUAC(Keep) 5 $35,000 1 $65,000(A Z P 10%,5) 5 $35,000 1 $65,000 (0.26380) 5 $52,147.00

5 PMT110%,5,2650002 1 35000 5 $52,146.84 b) EUAC(Replace) 5 $15,000 1 $120,000(A Z P 10%,5) – $40,000 (A Z F 10%,5) 5 $15,000 1 $120,000(0.26380) – $40,000(0.16380) 5 $40,104.00 5 PMT110%,5,2120000,400002 1 15000 5 $40,103.80 c) EUAC(Lease) 5 $18,000 1 $20,000(F Z P 10%,1) 5 $18,000 1 $20,000 (1.10) 5 $40,000.00 5 18000 2 FV(10%,1,,20000) 5 $40,000.00

SOLUTION

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EXPLORING THE SOLUTION

Notice, the same conclusion is reached with the two approaches: leasing has the lowest equivalent uniform annual cost. Furthermore, since the EUAC values obtained using the two approaches differ by $15,000(A Z P10%,5) 5 $3,957.00 or 5 PMT(10%,5,215000) 5 $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) 5 $52,147.00 – $40,104.00 5 $12,043.00 (opportunity cost approach) 5 $48,190.00 – $36,147.00 5 $12,043.00 (cash flow approach) EUAC(K–L) 5 $52,147.00 – $40,000.00 5 $12,147.00 (opportunity cost approach) 5 $48,190.00 – $36,043.00 5 $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.

7-3

OPTIMUM REPLACEMENT INTERVAL

LEARN I N G O B JEC T I V E : Determine the optimum replacement interval in

cases where an asset will be used for many years.

Optimum Replacement Interval (ORI) The interval at which an asset should be replaced to minimize cost (or maximize worth).

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. As equipment ages, operating and maintenance (O&M) costs tend to increase. At the same time, the cost of ownership tends to decrease with

7-3

Optimum Replacement Interval

Dollars/year

Equivalent uniform annual cost

Annual operating and maintenance cost Capital recovery cost Life, years

EUAC Components Used to Determine the Optimum Replacement Interval

FIGURE 7.1

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. The cost of ownership is often called the capital recovery cost and is computed as follows: CR 5 P1A Z P i%,n2 2 F1A Z F i%,n2

(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%. 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.

Computing the Optimum Replacement Interval 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*)?

Video Lesson: Capital Recovery Cost Capital Recovery Cost represents the cost of ownership, a value that typically decreases with increased usage.

EXAMPLE

Video Example

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KEY DATA

SOLUTION

Given: Initial investment (P) 5 $500,000, Salvage value (F ) 5 $500,000(0.75n), O&M 5 $125,000 in year one, increasing 15% per year Find: ORI and EUAC* For the example problem, EUAC 5 [$500,000 1 $125,000(P Z A1 10%,15%,n)](A Z P 10%,n) – $500,000(0.75n)(A Z F 10%,n). As shown below, ORI 5 5 years and EUAC* 5 $276,611. n

(A|P 10%,n)

(P |A1 10%,15%,n)

(A|F 10%,n)

EUAC

4

0.31547

3.89190

0.21547

$277,119

5

0.26380

4.97789

0.16380

$276,611

6

0.22961

6.11325

0.12691

$278,729

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. 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. 1.

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.

7-3

Optimum Replacement Interval

Optimum Replacement Timing with a Given Planning Horizon

EXAMPLE

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. Since 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}? Given: Initial investment (P) 5 $500,000, Salvage value (F ) 5 $500,000(0.75n), O&M 5 $125,000 in year one, increasing 15% per year, Planning horizon 5 9 years Find: The replacement sequence that minimizes PW of costs given a 9-year planning horizon.

KEY DATA

Shown below is the EUAC for values of n ranging from 1 to 9.

SOLUTION

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} 5 EUAC(x)(P Z A 10%,x) 1 EUARC(y)(P Z A 10%,y)(P Z 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} {7,2}

$1,608,014.16 $1,634,857.37

{8,1} {9,0}

$1,671,520.06 $1,713,885.95

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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).

SUMMARY

KEY CONCEPTS 1. Learning Objective: Explain the terms defender, challenger, obsolescence, and sunk costs as they pertain to replacement analysis. (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. 2. Learning Objective: Perform a replacement analysis utilizing a cash flow approach. (Section 7.2)

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. 3. Learning Objective: Perform a replacement analysis utilizing an opportunity cost approach. (Section 7.2)

In the opportunity cost approach, also called the outsider’s point 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. 4. Learning Objective: Determine the optimum replacement interval in cases where an asset will be used for many years. (Section 7.3)

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

Summary 269

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.

KEY TERMS Capital Recovery Cost, p. 265 Challenger, p. 259 Defender, p. 259 Economic Obsolescence, p. 259 Equivalent Uniform Annual Cost (EUAC), p. 262 Functional Obsolescence, p. 259

Opportunity Cost, p. 261 Optimum Replacement Interval (ORI), p. 264 Replacement Analysis, p. 259 Salvage Value, p. 261 Sunk Costs, p. 259 Technological Obsolescence, p. 259

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1.

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 c. $390,000 b. $370,000 d. $450,000

2.

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 c. $390,000 b. $370,000 d. $450,000

3.

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?

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n

EUAC(n)

1 2 3 4

1582.00 1550.00 1575.00 1580.00

a. 1 year b. 2 years 4.

A radiology clinic is considering buying a new $700,000 x-ray machine, which will have no salvage value after installation since 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 ten 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. b. c. d.

5.

c. 3 years d. 4 years

One year, since it will have no salvage after installation Five years, since the maintenance costs are constant Ten years, because maintenance costs don’t increase Cannot be determined from the information given Which of the following is not an approach to replacement analysis? c. outsider viewpoint d. supply chain approach

a. cash flow approach b. insider viewpoint 6.

The most common approach to determining the planning horizon for replacement analysis is which of the following? a. Shortest life c. Longest life b. Median Life d. Least Common Multiple

7.

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 a before tax analysis and an outsider’s point of view, what is the first cost of keeping the old lathe? a. $29,000 c. $20,000 b. $45,000 d. $16,000

8.

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 a before tax analysis and an insider’s point of view, what is the first cost of the new lathe? a. $29,000 c. $25,000 b. $45,000 d. $16,000

9.

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

Summary 271

10.

Increasing the magnitude of the initial investment tends to ____________ the optimum replacement interval. a. Decrease c. Reverse b. Increase d. Not affect

PROBLEMS Section 7.1 Fundamentals of Replacement Analysis 1. 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 tell for each challenger the main reason you would consider it. 2. Give two examples each of (i) functional obsolescence, (ii) technological

obsolescence, and (iii) economic obsolescence for items that you or your family own. 3. 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. Section 7.2 Cash Flow and Opportunity Cost Approaches 4. 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. Analyze this using an EUAC measure and a MARR of 20% to perform a before-tax analysis to see if the automatic painting machine should be replaced if the old automatic painting machine is taken in 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). 5.

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

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have a salvage value of $45,000. Its annual O&M cost will be only $40,000 due to newer technology. Analyze this using 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). 6. Online Educators (OE), a not-for-profit firm exempt from taxes, is consider-

ing replacement of some electronic equipment associated with its distance learning (DL) facility. They spent $100,000 on the equipment 3 years ago and have depreciated it to a current book value of $40,000. Its end-of-year O&M costs are $9000 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-of-year 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. 7.

A currently-owned shredder for use in a refuse-powered electrical generating plant has a present net realizable value of $200,000 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 1500 hours and 250 days per year. Using a 4 year planning horizon, a before-tax analysis, 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. 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).

Summary 273

8. 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). 9.

Allen Construction purchased a crane 6 years ago for $130,000. They need a crane of this capacity for the next five 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 $8000 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).

10. A division of Raytheon owns a 5-year old turret lathe that has a non-tax

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.

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11.

GO Tutorial 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, 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 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.

12. 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 their reputation for highly functional, high quality, cost effective designs and construction. They are building a new wing at their 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. They 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/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.

Summary 275

c. Clearly show the cash flow profile for each alternative using an opportu-

nity cost approach (outsider’s viewpoint approach). d. Using a PW analysis and an opportunity cost approach, decide which is

the more favorable alternative. 13.

Video Solution Clear Water Company has a down-hole 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 trade-in allowance for the existing auger is $17,000. The “treated auger” will have a $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 their equipment and services to perform the drilling that would normally be done with the existing auger. The competitor requires a beginning-ofyear 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 and a cash flow approach, decide which is the more favor-

able alternative. c. Clearly show the cash flow profile for each alternative using an opportu-

nity cost approach (outsider’s viewpoint approach). d. Using an EUAC comparison and an opportunity cost approach, decide

which is the more favorable alternative. 14. 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.

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15. 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. 16. 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 $4000. 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% before-tax 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. 17. 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 a $8000 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. 18. 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 year ago for a cost of $20,000. The present market value is determined to be $8,000. Estimated remaining life

Summary 277

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 average annual expenses expected are: Fuel Labor (including maintenance) Payroll taxes Taxes and insurance on furnace Other

$35,000 $40,000 10% of direct labor costs 1% of purchase price $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 Labor (operating) Payroll taxes Taxes and insurance on furnace Other

$29,000 $30,000 10% of direct labor costs 1% of purchase price $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. 19. 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 cash flow approach. b. Use the opportunity cost approach

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20. A machine was purchased 5 years ago at $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 $5000. The new machine has a list price of $15,000, an estimated life of 10 years, annual operating plus maintenance costs of $7,500, annual revenues of $13,000, and salvage values at the end of the jth year according to Sj 5 $15,000 2 $1,500j,   for j 5 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. 21. A building supplies distributor purchased a gasoline-powered fork-lift 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 operating expenses for year j have been Cj 5 $2,000 1 $4001 j 2 12

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 operating expenses are expected to be $1,200. If a MARR of 10% is assumed and a 4-year planning horizon is adopted, based on a cash flow approach should the replacement be made now? 22. 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. 23. A firm is contemplating replacing a computer they purchased 3 years ago

for $400,000. It will have a salvage value of $20,000 in 4 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.

Summary 279

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. 24. National Chemicals has an automatic chemical mixture 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 a cash flow approach, what do you recommend? 25. The Ajax Specialty Items Corporation has received a 5-year contract to pro-

duce 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 five 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. 26. 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

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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. 27. A highway construction firm purchased a particular earth-moving 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. 28. 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, should the equipment be replaced? Section 7.3 29.

Optimum Replacement Interval

You plan to purchase a car for $28,000. Its market value will decrease by 20% per year. You have determined that the IRS-allowed mileage reimbursement rate for business travel is about right for fuel and maintenance at $0.485 per mile in the first year. You anticipate that it will go up 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%.

30. Griffin Dewatering purchases a wellpoint pump connected to a skid-mounted

diesel engine for $14,000. Its market value for salvage purposes decrease by 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 by $600 each year thereafter. What is the optimum replacement interval if MARR 5 15%? 31. Polaris Industries wishes to purchase a multiple-use in-plant “road test”

simulator that can be used for ATVs, motorcycles, and snow mobiles. 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

Summary 281

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 by 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 by 25% each subsequent year. MARR is 15%. What is the optimum replacement interval? 32. A granary purchases a conveyor used in the manufacture of grain for trans-

porting, 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 $3,000. Operation and maintenance is expected to cost $14,000 in the first year, increasing by $1,000 per year thereafter. The granary uses a MARR of 15%. What is the optimum replacement interval for the conveyor? 33.

GO Tutorial 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 drop by $40,000 per year thereafter to a minimum of $30,000. Operating costs are $20,000 the first year, increasing by 8% per year. Maintenance costs are only $8,000 the first year but will increase by 35% each year thereafter. Milliken’s MARR is 20%. Determine the optimum replacement interval for the dyer.

34. 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 Operating cost Salvage value

1

2

3

4

5

6

$1,500 8,000

$1,800 6,000

$2,100 5,000

$2,400 5,000

$2,700 3,000

$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 Operating cost Salvage value

1

2

3

4

5

6

$ 1,000 13,000

$ 1,100 11,000

$ 1,200 10,000

$1,300 9,000

$1,400 8,000

$1,500 7,000

If interest is at 0%, when should the new machine be purchased? 35. 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 if constant worth dollar estimates are used. The appropriate discount rate is 15%. Operating and

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maintenance costs for a unit of equipment in its tth year of service, denoted by Ct, are as follows: t

Ct

t

1 2 3 4 5

$6,000 7,500 9,150 10,950 12,900

6 7 8 9 10

Ct $15,000 17,250 19,650 22,200 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 5 10 years.) b. Determine the optimum replacement interval assuming a finite planning horizon of 15 years, with Ct11 5 Ct 1 $1,500 1 $150(t 2 1) for t 5 10, 11, . . . 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? 36. 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 5 $4,00011.102 t 

t 5 1, 2, . . . , 12

Fn 5 $44,00010.502 n   n 5 0, 1, 2, . . . , 12

Determine the optimum replacement interval assuming a MARR of (a) 0%, (b) 10%. (Maximum life 5 12 years.) 37. $100,000 is invested in equipment having a negligible salvage value regard-

less 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? 38. $100,000 is invested in equipment having a negligible salvage value regard-

less 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? 39. $250,000 is invested in equipment having a negligible salvage value regard-

less 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? 40. $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?

Summary 283

41. $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? 42. $500,000 is invested in equipment having a negligible salvage value regard-

less 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? 43. $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? 44. $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? 45. $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? 46. 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. 47. 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. 48. 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. 49. True or False: Consider an optimum replacement interval problem of the type

considered in this chapter. If O&M costs are a geometric series, then decreasing the magnitude of the O&M cost in the first year will tend to decrease the optimum replacement interval. 50. 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.

8 ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : S C HNE I DE R N AT I O N AL Based on 2011 revenues, Schneider National is a $3.4 billion privately held company in the transportation and logistics sector. Headquartered in Green Bay, Wisconsin, its customers include many of the global companies on Fortune magazine’s list of top 100 companies. Schneider operates within North America and China, it employs 17,400 associates, and it is structured into three principal business units—truckload, intermodal, and logistics. The largest business unit, truckload, includes 9,860 company drivers and 1,775 owner-operators (independent drivers with their own tractors), operating 11,800 tractors and 31,500 trailers. Intermodal, a business that combines the geographic reach and delivery of trucking with the efficient long-haul transportation of railroads, is the second largest business unit: It employs 1,350 company drivers, 140 owner-operators, and 13,500 containers to service customers in the intermodal market. Schneider Logistics has 2,000 associates worldwide and provides a variety of value-added services to customers, including Transportation Management (Brokerage), Port Logistics (Transloading and Distribution, Inland Logistics Management, and China Solutions), Supply Chain Management (Transportation Network Design), Supplier Management, Integrated Delivery Services (LTL and Cross-Dock), and Schneider Consulting. Each of Schneider’s businesses has a different level of asset intensity. The entire portfolio, however, results in $1 billion of invested capital showing on its balance sheet. As a result, it is critical that each increment of investment produce positive contributions to after-tax income, ensuring value creation for shareholders commensurate with its level of risk. Engineering economic analysis principles and tools are used by Schneider in allocating capital and human resources decisions. Three recent examples of economic analyses performed include the design of a trailer fleet, the use of trailer-tracking devices, and a strategic acquisition. ■

284

Economic analyses showed that increasing the purchase price of duraplate 53-foot Wabash trailers by adding a 22-inch “high rail” aluminum

DEPRECIATION

strip at the bottom of the trailer’s sides reduced a significant amount of damage from lift-truck forks and mishandling by trailer-pool hostlers. The analysis included reductions in maintenance expense, plus savings by reducing the number of trailers due to less repair time required. ■

The incorporation of economic analysis at the project’s beginning significantly influenced the ultimate design and functionality of a trailer-tracking device installed on all of Schneider’s trailers. The initial results evidenced that knowing where trailers were at all times added significant value for Schneider and its customers.



In pursuing a strategic acquisition, Schneider required that the investment produce a series of cash flows that meet or exceed its weighted average cost of capital, adjusted for risk. Because of the inevitability of future downside considerations, as well as the cyclical nature of the transportation market, the economic justifications were required to include well-demonstrated and documented scenario and risk-management plans showing how the returns on the target acquisitions would be impacted.

Because Schneider National is privately owned, it is not required to satisfy United States Securities and Exchange Commission rules that apply to publicly traded corporations. As a result, less pressure exists to meet or exceed estimates of quarterly earnings developed by financial analysts. Financial decisions at Schneider are based on after-tax returns on investment. In contrast to Schneider, U.S. publicly traded corporations generally maintain two sets of books, or accounting records—and it is perfectly legal. Consequently, two different depreciation schedules are maintained: one for financial reporting and the other for computing income taxes to be paid. We consider both in this chapter. In the following chapter, we address the tax consequences of various depreciation allowances. When a firm’s capital assets are distributed around the world, 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.

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DISCUSSION QUESTIONS: 1. What impact does depreciation have on capital investment decision making at Schneider?

2. What will be the impact of these capital investment decisions on aftertax income for Schneider?

3. Although maintaining two sets of books is legal, what concerns, if any, do you have about this practice?

4. In the previous chapter on replacement analysis, we introduced J. B. Hunt Transport Services Inc. (JBHT), another transportation and logistics company. What differences do you see between JBHT and Schneider with respect to depreciation of their significant volume of assets? Recall that JBHT is a public company based in North America.

LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

1. Describe the impact that depreciation has on design decisions from an income-tax perspective. (Section 8.1)

2. Define a depreciable property and distinguish whether this property is tangible or intangible, and if tangible, whether the property is personal or real. (Section 8.2)

3. Calculate depreciation using the Straight-Line (SLN), Declining Balance (DB), and Double Declining Balance (DDB) methods. (Section 8.3)

4. Determine at what point it is optimal to switch from a declining balance rate to a straight-line rate. (Section 8.3.3)

5. Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS). (Section 8.4).

INTRODUCTION Depreciation The concept that takes into account the fact that most property decreases in value with use and time.

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. 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

8-1 The Role of Depreciation in Economic Analysis

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

THE ROLE OF DEPRECIATION IN ECONOMIC ANALYSIS

LEARNING O BJECTI VE: Describe the impact that depreciation has on design

decisions from an income-tax perspective.

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 in 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 out 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.

Video Lesson: Depreciation

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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.

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. 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 after-tax 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.

8-2

LANGUAGE OF DEPRECIATION

LEARN I N G O B JE C T I V E : Define a depreciable property and distinguish whether this property is tangible or intangible, and if tangible, whether the property is personal or real.

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.

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 speaking, 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: 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. 1.

8-3

Straight-Line and Declining Balance Depreciation Methods

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. 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.’’ (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.)

8-3

STRAIGHT-LINE AND DECLINING BALANCE DEPRECIATION METHODS

LEARNING O BJECTI VE: 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 discussed 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 discussed later in this chapter. 8.3.1

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

289

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 straightline 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.

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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 5 1P 2 F2/n

5 SLN1P,F,n2

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.

(8.1)

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. 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 5 P 2 td

5 P 2 t 1P 2 F2/n

(8.2)

Hence, book value is a straight line. Further, since 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: 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. 1.

Straight-Line Depreciation Applied to the SMP Machine

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 5 $500,000; Salvage Value F 5 $50,000; Useful Life n 5 10 yrs Find: d5, B5

8-3

Straight-Line and Declining Balance Depreciation Methods

The following values occur for t 5 5, with SLN:

SOLUTION

d5 5 1$500,000 2 $50,0002/10 5 $45,000 5 SLN1500000,50000,102 5 45,000 B5 5 $500,000 2 51$45,0002 5 $275,000

8.3.2

291

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. 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, aftertax 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 5 pBt21 5 pP11 2 p2 t21

(8.3)

and Bt 5 P11 2 p2 t

(8.4)

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 discuss the incometax 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 percent, 150 percent, and 200 percent (the maximum amount allowed by the IRS) of the straight-line rate, which

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.

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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 5 P(1 2 p)n Solving for p gives

p 5 1 2 1F Z P2 1/n 5 2RATE1n,,2P,F2

Double Declining Balance (DDB) A type of accelerated depreciation method that is twice, or 200 percent, the straightline rate.

(8.5)

Notice the minus sign between the equal sign and the Excel® RATE worksheet function. Also, when p 5 2/n, which is twice, or 200 percent, the straight-line rate, the DB plan is called double declining balance (DDB) depreciation. 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)

where cost 5 the cost basis of the asset salvage 5 the salvage value of the asset life 5 the cost recovery period of the asset period 5 the time period for which the depreciation charge is desired factor 5 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 5 the starting period for which depreciation is calculated; must use the same units as life end_period 5 the ending period for which depreciation is calculated; must use the same units as life no_switch 5 a logical value specifying whether to switch to straight-line depreciation when the straight-line depreciation is greater than the declining balance calculation The Excel® DDB function uses a depreciation rate equal to 2/n. When using the Excel® VDB function, switching from declining balance to

8-3 Straight-Line and Declining Balance Depreciation Methods

straight-line depreciation is an option; if the declining balance depreciation rate is not specified, it is assumed to equal 2/n. We address the concept of switching depreciation methods in the next section. Since 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.

Declining Balance Depreciation Applied to the SMP Machine

EXAMPLE

For the SMP machine in Example 8.1, what are the depreciation charge and book value after year 5 using declining balance depreciation? Given: Cost Basis P 5 $500,000; Salvage Value F 5 $50,000; Useful Life n 5 10 years Find: d5, B5

KEY DATA

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,

SOLUTION

p 5 1 2 1$50,000/$500,00020.1 5 20.5672% 5 2RATE110,,500000,500002 5 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 5 2/10, from Equations 8.3 and 8.4, d5 5 0.201$500,0002 10.802 4 5 $40,960.00 B5 5 $500,00010.802 5 5 $163,840.00

Using the Excel® DDB worksheet function gives d5 5 DDB1500000,50000,10,52 5 $40,960.00

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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 5, d5 5 VDB1500000,50000,10,4,5,22 5 $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 5 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 t

Depreciation Allowances and Book Values for the SMP Investment SLN dt

0

SLN Bt

DDB dt

$500,000.00

1

$45,000.00

2 3

DDB Bt

VDB dt

$500,000.00

VDB Bt $500,000.00

$455,000.00

$100,000.00

$400,000.00

$100,000.00

$400,000.00

$45,000.00

$410,000.00

$80,000.00

$320,000.00

$80,000.00

$320,000.00

$45,000.00

$365,000.00

$64,000.00

$256,000.00

$64,000.00

$256,000.00

4

$45,000.00

$320,000.00

$51,200.00

$204,800.00

$51,200.00

$204,800.00

5

$45,000.00

$275,000.00

$40,960.00

$163,840.00

$40,960.00

$163,840.00

6

$45,000.00

$230,000.00

$32,768.00

$131,072.00

$32,768.00

$131,072.00

7

$45,000.00

$185,000.00

$26,214.40

$104,857.60

$26,214.40

$104,857.60

8

$45,000.00

$140,000.00

$20,971.52

$83,886.08

$20,971.52

$83,886.08

9

$45,000.00

$95,000.00

$16,777.22

$67,108.86

$16,943.04

$66,943.04

10

$45,000.00

$50,000.00

$13,421.77

$53,687.09

$16,943.04

$50,000.00

8.3.3

Switching from DDB to SLN with the Excel® VDB Function

LEARN I N G O B JEC T I V E : 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

8-3 Straight-Line and Declining Balance Depreciation Methods

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 straight-line depreciation occurs at the first year for which Bt21 2 F . pBt21 n 2 1t 2 12

(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.

Switching from DDB to SLN with the SMP Machine

EXAMPLE

For the SMP machine in previous examples, what is the optimal point for the depreciation method to switch from DDB to SLN? Given: P 5 $500,000, F 5 $50,000, n 5 10, and p 5 0.2 for DDB. Find: The t for which

KEY DATA

Bt21 2 F . pBt21 n 2 1t 2 12

When t 5 1, the left-hand side (LHS) of Equation 8.6 is ($500,000 2 $50,000)/10 5 $45,000; the right-hand side (RHS) of Equation 8.6 is 0.20 ($500,000) 5 $100,000. Because LHS , RHS, try t 5 2. When t 5 2, the LHS of Equation 8.6 is ($400,000 2 $50,000)/9 5 $38,888.89; the RHS of Equation 8.6 is 0.20 ($400,000) 5 $80,000. Because LHS , RHS, try t 5 3.

SOLUTION

295

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This iterative approach continues until the LHS of equation 8.6 exceeds the RHS as it does in year t 5 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 5 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

8-4

Depreciation and Book Value Using DDB Switching to SLN

MODIFIED ACCELERATED COST RECOVERY SYSTEM (MACRS)

LEARN I N G O B JEC T I V E : Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS).

Modified Accelerated Cost Recovery System (MACRS) The only depreciation method approved by the U.S. IRS for computing income-tax liability.

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.

8-4

Modified Accelerated Cost Recovery System (MACRS) 297

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. 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 that it is used for half of the last year of service. 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. 8.4.1

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. 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 5 1, F 5 0, n 5 5, start_period 5 0.0, end_period 5 0.5, and factor 5 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 20-year 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.

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 straightline depreciation, with a longer recovery than GDS. 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.

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Depreciation

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, general-purpose 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

Real Property

27.5-Year 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

39-Year Property

Nonresidential real property: depreciable property with a class life of 27.5 years or more and that is not residential rental property

MACRS-GDS percentages for 3-, 5-, 7-, and 10-year property are 200% DBSLH and 15- and 20-year property are 150% DBSLH TABLE 8.3

EOY

3-Year Property

5-Year Property

7-Year Property

10-Year Property

15-Year Property

20-Year Property

1 2

33.33 44.45

20.00 32.00

14.29 24.49

10.00 18.00

5.00 9.50

3.750 7.219

3 4

14.81 7.41

19.20 11.52

17.49 12.49

14.40 11.52

8.55 7.70

6.677 6.177

11.52 5.76

8.93 8.92

9.22 7.37

6.93 6.23

5.713 5.285

8.93 4.46

6.55 6.55

5.90 5.90

4.888 4.522

9 10

6.56 6.55

5.91 5.90

4.462 4.461

11 12

3.28

5.91 5.90

4.462 4.461

13 14

5.91 5.90

4.462 4.461

15 16

5.91 2.95

4.462 4.461

5 6 7 8

17 18

4.462 4.461

19 20

4.462 4.461

21

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

1.970% 1.667%

8

9

1.364% 1.061%

10

11

12

1

3.485% 3.182% 2.879% 2.576% 2.273%

2–9

3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636%

0.758% 0.455% 0.152%

10–26*

3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637%

11–27** 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 28

1.970% 2.273% 2.258% 2.879% 3.182% 3.485% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636%

29

0.152% 0.455% 0.758%

1.061% 1.364% 1.667%

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

1.391% 1.177%

8

9

11

2–39

2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564%

40

0.107% 0.321% 0.535% 0.749% 0.963%

1.605% 1.819%

0.535% 0.321%

12

2.461% 2.247% 2.033% 1.819% 1.605%

1.177% 1.391%

0.963% 0.749%

10

1

0.107%

2.033% 2.247% 2.461%

*Even-numbered year. **Odd-numbered year.

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For a given property class, the sum of the depreciation rates equals 100 percent. 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 dt 5 pt P

(8.7)

and t

t

j51

j51

Bt 5 P 2 a dj 5 P a1 2 a pj b

(8.8)

Applying MACRS with the SMP Machine Investment

EXAMPLE

Video Example

What are the depreciation allowances and book values for the SMP machine using MACRS depreciation and property classes?

KEY DATA

Use Tables 8.2 and 8.3 to determine the property class and MACRS percentages for the SMP machine. Given: P 5 $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 d2 d3 d4 d5 d6

5 0.201$500,0002 5 $100,000  B1 5 $500,000 2 $100,000 5 $400,000 5 0.321$500,0002 5 $160,000  B2 5 $400,000 2 $160,000 5 $240,000 5 0.1921$500,0002 5 $96,000  B3 5 $240,000 2 $96,000 5 $144,000 5 0.11521$500,0002 5 $57,600 B4 5 $144,000 2 $57,600 5 $86,400 5 0.11521$500,0002 5 $57,600 B5 5 $86,400 2 $57,600 5 $28,800 5 0.05761$500,0002 5 $28,800 B6 5 $28,800 2 $28,800 5 $0

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

Summary 301

consequences of the design decisions that engineers make. Because depreciation affects income taxes, engineers must understand how depreciation allowances are determined. 8.4.2 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 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.

KEY CONCEPTS 1. Learning Objective: Describe the impact that depreciation has on design decisions from an income-tax perspective. (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. 2. Learning Objective: Define a depreciable property and distinguish whether this property is tangible or intangible, and if it is tangible, whether the property is personal or real. (Section 8.2)

The Internal Revenue Service requires 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.

SUMMARY

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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. 3. Learning Objective: Calculate depreciation using the Straight-Line (SLN), Declining Balance (DB), and Double Declining Balance (DDB) methods. (Section 8.3)

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 the 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

Depreciation Value

Book Value

Straight Line

dt 5 (P 2 F )/n

Bt 5 P – td 5 P 2 t (P 2 F)/n

Declining Balance

dt 5 pBt21 5 pP(1 2 p)t21

Bt 5 P (1 2 p)t

4. Learning Objective: Determine at what point it is optimal to switch from a declining balance rate to a straight-line rate. (Section 8.3.3)

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 specified using the declining balance rate. The equation that defines this point is: Bt21 2 F . pBt21 n 2 1t 2 12

(8.6)

5. Learning Objective: Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS). (Section 8.4)

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.

Summary 303

KEY TERMS Amortization, p. 289 Book Value, p. 290 Cost Basis, p. 289 Declining Balance (DB), p. 291 Depreciable Property, p. 288 Depreciation, p. 286 Depreciation Allowances, p. 288 Double Declining Balance (DDB), p. 292

Modified Accelerated Cost Recovery System (MACRS), p. 296 Modified Accelerated Cost Recovery System–Alternative Depreciation System (MACRS-ADS), p. 297 Modified Accelerated Cost Recovery System–General Depreciation System (MACRS-GDS), p. 297 Personal Property, p. 289 Real Property, p. 289

Half-Year Convention, p. 297 Intangible Property, p. 289

Straight-Line Depreciation, p. 289 Tangible Property, p. 289

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1. a. b. c. d.

Which of the following is not a requirement for an asset to be depreciable? It must have a life longer than 1 year It must have a basis (initial purchase plus installation cost) greater than $1,000 It must be held with the intent to produce income It must wear out or get used up

2.

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 c. $17,500.00 b. $20,000.00 d. $25,000.00

3.

Which of the following is not true about depreciation? Depreciation is not a cash flow To be depreciable, an asset must have a life longer than one year A 5-year property will generate a regular MACRS-GDS depreciation deductions in six fiscal years. For MACRS-GDS an estimate of the salvage values is required

a. b. c. d.

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4.

The depreciation deduction for year 11 of an asset with a 20-year 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 c. $72,682 b. $67,682 d. $80,000

5.

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 c. $72,682 b. $67,682 d. $80,000

6.

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

7.

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 c. $72,576 b. $78,687 d. $48,510

8.

The depreciation allowance for a $100,000 MACRS-GDS asset was $8,550 after its third year. What was the depreciation allowance after its second year? a. $8,550 c. $18,000 b. $9,500 d. Cannot be determined with the information given

9.

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 c. $25,007.50 b. $20,000.00 d. $28,580.00

10.

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 c. $3,916 b. $1,728 d. $4,272

Summary 305

11.

Which of the following is (are) required to calculate MACRS-GDS depreciation deductions? I. Property Class III. First Cost II. Salvage Value IV. Annual Maintenance Costs a. I and III only c. I, II, and III b. II and III only d. I, II, III, and IV

PROBLEMS Section 8.1 The Role of Depreciation in Economic Analysis 1. 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? 2. 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 since they are not cash flows? 3. 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 seven 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? 4. 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.

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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 8.2 5.

Language of Depreciation

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.

6. 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. 7.

For each of the following assets, state whether the asset is tangible/ intangible property, personal/real property, and depreciable/nondepreciable property. a. A tractor (part of tractor-trailer rig - 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.

8. 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. Section 8.3 9.

Straight-Line and Declining Balance Depreciation Methods

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 using a rate that ensures the book value equals the salvage value.

Summary 307

c. Use double declining balance depreciation. d. Use double declining balance switching to straight line depreciation. 10. 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 they 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 using 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. 11.

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 using 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.

12.

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 using 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.

13.

GO Tutorial 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 using 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.

14. 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

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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 three 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. 15. 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. 16. 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 8.4

Modified Accelerated Cost Recovery System (MACRS)

17. Which of the following statements are FALSE? a. MACRS is the only depreciation method approved by the IRS for comput-

ing 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 straight line depreciation. d. MACRS-ADS is based on straight line depreciation. 18. 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 1).

Summary 309

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. 19. A mold for manufacturing medical supplies (MACRS-GDS 5-year 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. 20. 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. 21. 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 MACRSGDS allowances. 22. 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. 23. Repeat the previous problem (problem 22) if the material-handling equipment

is removed just after the tax year, again using MACRS-GDS allowances. 24. 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. 25. 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. 26. 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. 27. 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

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depreciation deduction during each of the 6 tax years involved using MACRSGDS allowances. 28. 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 MACRSGDS allowances. 29. 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 percent 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. 30. 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 percent 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. 31. 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 percent 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. 32. 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 percent 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. 33.

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.

Summary 311

34. Bell’s Amusements purchased an expensive ride for their theme and amuse-

ment 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 MACRS-GDS). 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. 35.

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.

36. 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 8 year life. c. Determine the unrecovered investment at the end of each of the 8 years. 37.

GO Tutorial 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.

38. 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 MACRS-GDS depreciation deduction during each of the tax years involved assuming: a. What is the MACRS-GDS property class? b. The equipment is removed one day before the end of the tax year. c. The equipment is removed one day after the end of the tax year.

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39.

GO Tutorial 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?

40. 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. 41. 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. 42. 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 salvage value of nothing 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. 43. 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 10 year life. c. Determine the unrecovered investment at the end of each of the 10 years. 44.

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

Summary 313

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 5th year. c. Assume the line equipment is sold during the 3rd year of use. 45.

GO Tutorial 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.

46. 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. 47. 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. 48. 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. 49.

Suppose the IRS has decided to institute a new MACRS-GDS 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.

50. 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 MACRSGDS percentages for each year. 51.

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.

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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? 52. Now that you are making the big bucks, your spouse has decided to venture

into the rental property business. He purchases a rental house and after making some improvements it has a basis of $85,000. He 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. 53.

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.

54. 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. 55.

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.

56. A building with business offices, a reception area, and numerous small diag-

nosis 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.

Summary 315

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? 57. Ultra-clean special handling devices used in the filling process for the manu-

facture 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 salvage value of nothing 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.

ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : I N T E L With revenue of $54 billion in 2011, Intel is the world’s largest semiconductor chip maker. The chances are great that your computer includes one or more of Intel’s products: chips, printed circuit boards, and other semiconductor components. More specifically, Intel’s products include microprocessors, chipsets, motherboards, flash memory, wired and wireless connectivity products, communications infrastructure components (including network processors), application and cellular baseband processors, and products for networked storage. Intel’s products appear in computers, as well as in servers and networking and communications products. Intel has long been a leader in silicon process technology and manufacturing. As of December 31, 2011, Intel employed 100,100 people worldwide, with approximately 55 percent of those employees located in the United States. A substantial number of Intel’s employees are engineers or computer scientists. Intel invests heavily in R&D and is committed to investing in world-class technology development, especially in the design and production of integrated circuits. In 2011, R&D expenditures totaled $8.4 billion. 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 one of its most significant strategic advantages. Intel’s plans for 2011 included building and expanding their 22 nm process technology manufacturing capacity.

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 “lowertech” company? 316

INCOME TAXES

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?

LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

1. Calculate corporate income taxes, including the effective, incremental, and marginal tax rates. (Section 9.1)

2. Analyze the after-tax implications for investment alternatives using retained earnings. (Section 9.2)

3. Analyze the after-tax implications for investment alternatives using borrowed capital. (Section 9.3)

4. Illustrate the different tax consequences for the common example of a lease vs. purchase alternative. (Section 9.4)

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 317

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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. 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. Aftertax 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 after-tax analyses are emphasized here. Furthermore, due to the diversity of state laws, only federal income taxes are considered.

Systematic Economic Analysis Technique 1. 2. 3. 4. 5. 6. 7.

Identify the investment alternatives Define the planning horizon Specify the discount rate Estimate the cash flows Compare the alternatives Perform supplementary analyses Select the preferred investment

9-1

9-1

Corporate Income Tax Rates

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CORPORATE INCOME TAX RATES

LEARNING O BJECTI VE: Calculate corporate income taxes, including the

effective, incremental, and marginal tax rates.

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. Ordinary federal income-tax rates imposed on corporations are given in Table 9.1. The tax rates shown are current for 2012; in fact, they have not changed since 1993. As a matter of interest, the highest corporate tax bracket has ranged from 1 percent in 1915 to 12 percent in 1918, to 13.5 percent in 1927, to 19 percent in 1939, to 24 percent in 1940, to 40 percent in 1945, to 52.8 percent in 1969, to 46 percent from 1979 to 1986, to 34 percent from 1988 to 1992, and, most recently, to 35 percent from 1993 to 2012. Because income tax laws do change and may differ from those described in the text, consult the U.S. Internal Revenue Service website (http://www. irs.gov) for up-to-date information on U.S. tax laws. TABLE 9.1

Corporate Income Tax Rates for Tax Years Beginning January 1, 1993 and Beyond

Taxable Income (TI), in $

Tax Rate (t)

Income Tax (T)

0 , TI # 50,000

0.15

0.15(TI)

50,000 , TI # 75,000

0.25

7,500 1 0.25(TI 2 50,000)

75,000 , TI # 100,000

0.34

13,750 1 0.34(TI 2 75,000)

100,000 , TI # 335,000

0.39(0.34 1 0.05)

22,250 1 0.39(TI 2 100,000)

335,000 , TI # 10,000,000

0.34

113,900 1 0.34(TI 2 335,000)

10,000,000 , TI # 15,000,000

0.35

3,400,000 1 0.35(TI 2 10,000,000)

15,000,000 , TI # 18,333,333

0.38(0.35 1 0.03)

5,150,000 1 0.38(TI 2 15,000,000)

18,333,333 , TI

0.35

0.35(TI)

Computing Income Taxes for Corporations A small company is currently forecasting a taxable income of $50,000 for the year. The company’s owner is considering an investment that will increase taxable income by $45,000. If the investment is pursued and the anticipated return occurs, what will be the magnitude of the increase in income taxes caused by the new investment? What will it be if the company is currently forecasting a taxable income of $400,000 for the year?

EXAMPLE

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KEY DATA

Given: Base taxable income 1 5 $50,000; Base taxable income 2 5 $400,000; Increase in annual income with investment 5 $45,000 Find: Increase in income taxes for each scenario

SOLUTION

From Table 9.1, with a base taxable income of $50,000, the federal income tax will be 0.15($50,000), or $7,500. The income tax for a taxable income of $95,000 will be $13,750 1 0.34($95,000 2 $75,000) 5 $13,750 1 0.34($20,000) 5 $20,550. The magnitude increase in income tax caused by the new investment is therefore $20,550 2 $7,500 5 $13,050. Again, from Table 9.1, with a base taxable income of $400,000, the federal income tax will be $113,900 1 0.34($400,000 2 $335,000) 5 $113,900 1 0.34($65,000) 5 $136,000, or 34 percent of $400,000. In this case, every dollar of the additional $45,000 in taxable income will be taxed at 34 percent, so the increase in taxable income will be 0.34($45,000) 5 $15,300, for a total tax of $151,300.

Effective Tax Rate The income tax divided by the taxable income. Also referred to as the average tax rate. Incremental Tax Rate The incremental income tax divided by the incremental investment.

From Example 9.1, we can define three tax rates for the corporation: 1.

The effective tax rate, or average tax rate, which equals the income tax divided by the taxable income

2. The incremental tax rate, which is the average rate charged to the

incremental investment

Marginal Tax Rate The tax rate that will apply to the last dollar included in taxable income.

3. The marginal tax rate, which is the tax rate that will apply to the last

EXAMPLE

Computing Effective, Incremental, and Marginal Income Tax Rates

dollar included in taxable income

In Example 9.1, what are the effective, incremental, and marginal tax rates when the base taxable income is $50,000? KEY DATA

Given: Base taxable income 5 $50,000; Income tax on base taxable income 5 $7,500; Increase in taxable income with investment 5 $45,000; Taxable income with investment 5 $95,000; Income tax on taxable income with investment 5 $20,550; Increase on taxable income with investment 5 $13,050 Find: Effective, incremental, and marginal tax rates

9-1

When the base taxable income was $50,000, the corporation had an effective tax rate of 15 percent: $7,500/$50,000. With the addition of $45,000 in taxable income, the new effective tax rate became $20,550/$95,000 5 21.63 percent. The $45,000 increase in taxable income caused taxes to increase from $7,500 to $20,550. Therefore, $13,050/$45,000, or 29 percent, is the incremental tax rate. The marginal tax rate was 34 percent, because taxable income of $95,000 is in the 34 percent tax bracket. Therefore, the last $1 added to taxable income contributed 34¢ to the company’s income taxes.

Conservatively, when performing economic justifications, an investment alternative’s after-tax economic worth should be computed using the marginal tax rate. 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.

Gross income (income less tax exemptions)

Taxable income

Depreciation allowance FIGURE 9.1

Deductions (other than interest on borrowed money and depreciation)

Interest on borrowed money

Pictorial Representation of Taxable Income

Corporate Income Tax Rates

SOLUTION

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9-2

AFTER-TAX ANALYSIS USING RETAINED EARNINGS (NO BORROWING)

LEARN I N G O B JEC T I V E : 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 11 2 effective income tax rate2

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).

The base elements needed to calculate after-tax cash flow (ATCF) for an alternative are summarized in Figure 9.2, 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. The following notation will be used for after-tax analysis of an investment alternative’s economic worth, without borrowed capital:

BTCF 5 before-tax cash flow DWO 5 depreciation write-off or allowance TI 5 taxable income itr 5 income tax rate T 5 income tax ATCF 5 after-tax cash flow

Gross income (income less tax exemptions)

After-tax cash flow

Income taxes FIGURE 9 . 2

Deductions (other than interest on borrowed money and depreciation allowance)

Interest on borrowed money

Pictorial Representation of After-Tax Cash Flow (ATCF)

9-2 After-Tax Analysis Using Retained Earnings (No Borrowing)

323

The following equation holds for investments in depreciable property: ATCF 5 BTCF 2 T

(9.1)

T 5 itr1TI2

(9.2)

TI 5 BTCF 2 DWO

(9.3)

ATCF 5 BTCF11 2 itr2 1 itr1DWO2

(9.4)

where where Therefore, Because state income taxes are generally deductible expenses when computing federal income taxes, the overall income tax rate (itr) should equal itr 5 str 1 ftr11 2 str2

(9.5)

where str 5 state tax rate, and ftr 5 federal tax rate. In the text, a value of 40 percent will be used for the combined income tax rate in most examples and end-of-chapter problems. The 40 percent tax rate is obtained using a 7 percent state income-tax rate and a 35 percent federal income tax rate. From Equation 9.5, 0.07 1 0.3511 2 0.072 5 0.3955   or   39.55%. Depending on the particular state (and country) in which the investment is made, a different marginal 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. State Tax Rate (str) The state income-tax rate. Federal Tax Rate (ftr) The federal income-tax rate.

9.2.1 Single Alternative

In Examples 9.3 through 9.5 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.

After-Tax Analysis with SLN Depreciation

EXAMPLE

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 40 percent income-tax rate and will perform an after-tax analysis using a 10 percent after-tax MARR. Given: P 5 $500,000, itr 5 40%, MARRAT 5 10%, ATCF (years 1–10) 5 $92,500, S 5 $50,000 Find: PWAT, AWAT, FWAT, IRRAT, ERRAT

KEY DATA

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SOLUTION

TABLE 9.2

In the previous chapters, the $92,500 net revenue was given to be an aftertax 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 $124,166.67 results in an ATCF of $92,500 when straight-line depreciation is used with a $50,000 salvage value.

Before-Tax and After-Tax Analysis of the SMP Investment with SLN Depreciation

EOY

BTCF

0 1

2$500,000.00 $124,166.67

DWO $45,000.00

2 3

$124,166.67 $124,166.67

$45,000.00 $45,000.00

4 5

$124,166.67 $124,166.67

6 7 8 9

TI

T

ATCF

$79,166.67

$31,666.67

2$500,000.00 $92,500.00

$79,166.67 $79,166.67

$31,666.67 $31,666.67

$92,500.00 $92,500.00

$45,000.00 $45,000.00

$79,166.67 $79,166.67

$31,666.67 $31,666.67

$92,500.00 $92,500.00

$124,166.67 $124,166.67

$45,000.00 $45,000.00

$79,166.67 $79,166.67

$31,666.67 $31,666.67

$92,500.00 $92,500.00

$124,166.67 $124,166.67

$45,000.00 $45,000.00

$79,166.67 $79,166.67

$31,666.67 $31,666.67

$92,500.00 $92,500.00

10 MARRBT 5

$174,166.67 16.667%

$45,000.00

$79,166.67

$31,666.67 MARRAT 5

$142,500.00 10%

PWBT 5 FWBT 5

$96,229.49 $449,547.99

PWAT 5 FWAT 5

$87,649.62 $227,340.55

AWBT 5 IRRBT 5

$20,406.41 21.64%

AWAT 5 IRRAT 5

$14,264.57 13.80%

ERRBT 5

18.74%

ERRAT 5

11.79%

The computation of ATCF is as shown in Table 9.2. In the tenth 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 nine 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

9-2 After-Tax Analysis Using Retained Earnings (No Borrowing)

325

the after-tax minimum attractive rate of return (MARRAT) by 1 minus the income tax rate. Hence, with a 40 percent tax rate and a 10 percent MARRAT, a MARRBT of 16.667 percent would be used.

After-Tax Analysis with MACRS Depreciation

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 40% tax rate and a 10 percent MARRAT? Given: P 5 $500,000, itr 5 40%, MARRAT 5 10%, ATCF (years 1–10) 5 $92,500, S 5 $50,000 Find: PWAT, AWAT, FWAT, IRRAT, ERRAT

KEY DATA

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.)

SOLUTION

TABLE 9.3

After-Tax Analysis of the SMP Investment with MACRS Depreciation

EOY

BTCF

0 1

2$500,000.00 $124,166.67

DWO

TI

T

$100,000.00

$24,166.67

$9,666.67

2$500,000.00 $114,500.00

ATCF

2 3

$124,166.67 $124,166.67

$160,000.00 $96,000.00

2$35,833.33 $28,166.67

−$14,333.33 $11,266.67

$138,500.00 $112,900.00

4 5

$124,166.67 $124,166.67

$57,600.00 $57,600.00

$66,566.67 $66,566.67

$26,626.67 $26,626.67

$97,540.00 $97,540.00

6 7

$124,166.67 $124,166.67

$28,800.00 $0.00

$95,366.67 $124,166.67

$38,146.67 $49,666.67

$86,020.00 $74,500.00

8 9

$124,166.67 $124,166.67

$0.00 $0.00

$124,166.67 $124,166.67

$49,666.67 $49,666.67

$74,500.00 $74,500.00

10 MARRBT 5

$174,166.67 16.667%

$0.00

$174,166.67

$69,666.67

$104,500.00

PWBT 5 FWBT 5

$96,229.49 $449,547.99

PWAT 5 FWAT 5

$123,988.64 $321,594.61

AWBT 5 IRRBT 5

$20,406.41 21.64%

AWAT 5 IRRAT 5

$20,178.58 16.12%

ERRBT 5

18.74%

ERRAT 5

12.46%

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Examples 9.3 and 9.4 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 9.3 using SLN depreciation, there are no tax implications. If the salvage value exceeds the book value, as it does in Example 9.4 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 (40%). 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 5 BTCFn 3including Fn 4 2 DWOn 2 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. 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 sixth tax year and it is 7-year property, then the MACRS depreciation allowance percentages are 14.29 percent, 24.49 percent, 17.49 percent, 12.49 percent, 8.93 percent, and 4.46 percent; normally, the depreciation percentage for the sixth year would be 8.92 percent. Not all expenditures are used to acquire depreciable property. Consider, for example, an expenditure on a consulting study to identify costsaving 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.

9-2 After-Tax Analysis Using Retained Earnings (No Borrowing)

After-Tax Analysis with Non-Depreciable Expenditures

EXAMPLE

In Example 9.4, $500,000 was invested in depreciable property to obtain before-tax annual revenues of $124,166.67 each of the following 9 years and $174,166.67 the tenth year. Now suppose the expenditure is on a consulting study that identifies cost-saving opportunities that will produce the same BTCF. With a 40 percent income-tax rate and an after-tax MARR of 10 percent, what will be the after-tax measures of economic worth for the investment? Given: P 5 $500,000, itr 5 40%, MARRAT 5 10%, BTCF (years 1–9) 5 $124,166.67, BTCF (year 10) 5 $174,166.67 Find: PWAT, AWAT, FWAT, IRRAT, ERRAT

KEY DATA

Table 9.4 provides the results of the after-tax analysis.

SOLUTION

After-Tax Analysis of a $500,000 Investment in a Consulting Study TABLE 9.4

EOY

BTCF

TI

T

ATCF

0

2$500,000.00

2$500,000.00

2$200,000.00

2$300,000.00

1

$124,166.67

$124,166.67

$49,666.67

$74,500.00

2

$124,166.67

$124,166.67

$49,666.67

$74,500.00

3

$124,166.67

$124,166.67

$49,666.67

$74,500.00

4

$124,166.67

$124,166.67

$49,666.67

$74,500.00

5

$124,166.67

$124,166.67

$49,666.67

$74,500.00

6

$124,166.67

$124,166.67

$49,666.67

$74,500.00

7

$124,166.67

$124,166.67

$49,666.67

$74,500.00

8

$124,166.67

$124,166.67

$49,666.67

$74,500.00

9

$124,166.67

$124,166.67

$49,666.67

$74,500.00

10

$174,166.67

$174,166.67

$69,666.67

$104,500.00

PWAT 5

$169,336.56

FWAT 5

$439,215.43

AWAT 5

$27,558.75

IRRAT 5

21.64%

ERRAT 5

15.03%

Now, Examples 9.4 and 9.5 show some interesting observations. Note that the values at the bottom of Table 9.4 when the investment is expensed

327

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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.

After-Tax Comparison of Alternatives with Different Property Classes

EXAMPLE

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 $70,000 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 savings of $102,000 per year over the 10-year horizon. An after-tax MARR of 10 percent applies, and with a 40 percent income tax rate this is equivalent to a before-tax MARR of 16.67 percent. Which alternative is preferred? KEY DATA

Given: Alternative A: P 5 $300,000; 3-year property; itr 5 40%; MARRAT 5 10%; BTCF (years 1–10) 5 $70,000 Alternative B: P 5 $450,000; 7-year property; itr 5 40%; MARRAT 5 10%; BTCF (years 1–10) 5 $102,000 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 $57,940.19 versus $55,908.46 for Alternative B. On a before-tax basis, however, Alternative B is preferred to Alternative A.

9-2 After-Tax Analysis Using Retained Earnings (No Borrowing)

TABLE 9.5

329

After-Tax Comparison of Investment Alternatives with Different Property Classes

EOY

BTCF(A)

0

2$300,000.00

1

$70,000.00

2 3

DWO(A)

Tl(A)

T(A)

ATCF(A)

$99,990.00

2$29,990.00

2$11,996.00

$81,996.00

$70,000.00

$133,350.00

2$63,350.00

2$25,340.00

$95,340.00

$70,000.00

$44,430.00

$25,570.00

$10,228.00

$59,772.00

4

$70,000.00

$22,230.00

$47,770.00

$19,108.00

$50,892.00

5

$70,000.00

$0.00

$70,000.00

$28,000.00

$42,000.00

6

$70,000.00

$0.00

$70,000.00

$28,000.00

$42,000.00

7

$70,000.00

$0.00

$70,000.00

$28,000.00

$42,000.00 $42,000.00

2$300,000.00

8

$70,000.00

$0.00

$70,000.00

$28,000.00

9

$70,000.00

$0.00

$70,000.00

$28,000.00

$42,000.00

10

$70,000.00

$0.00

$70,000.00

$28,000.00

$42,000.00

BTPW(A) 5

$30,095.51

ATPW(A) 5

$57,940.19

EOY

BTCF(B)

0

2$450,000.00

1

$102,000.00

$64,305.00

$37,695.00

$15,078.00

$86,922.00

2

$102,000.00

$110,205.00

2$8,205.00

2$3,282.00

$105,282.00

3

$102,000.00

$78,705.00

$23,295.00

$9,318.00

$92,682.00

4

$102,000.00

$56,205.00

$45,795.00

$18,318.00

$83,682.00

5

$102,000.00

$40,185.00

$61,815.00

$24,726.00

$77,274.00

6

$102,000.00

$40,140.00

$61,860.00

$24,744.00

$77,256.00

7

$102,000.00

$40,185.00

$61,815.00

$24,726.00

$77,274.00

8

$102,000.00

$20,070.00

$81,930.00

$32,772.00

$69,228.00

9

$102,000.00

$0.00

$102,000.00

$40,800.00

$61,200.00

10

$102,000.00

$0.00

$102,000.00

$40,800.00

$61,200.00

ATPW(B) 5

$55,908.46

BTPW(B) 5

DWO(B)

Tl(B)

T(B)

ATCF(B) 2$450,000.00

$30,996.31

After-Tax Comparison of Manual versus Automated Solutions 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

EXAMPLE

Video Example

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$125,000 and will have annual operating costs of $500. The robot qualifies as 3-year property. If a tax rate of 40 percent and a MARR of 10 percent 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.

TABLE 9.6 EOY

KEY DATA

Given: Manual Alternative: A 5 $50,000/year; itr 5 40%; MARR 5 10%; N 5 5 Robotic Alternative: P 5 $125,000; A 5 $500; 3-year property; itr 5 40%; MARR 5 10%; N 5 5; SV 5 $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(0.60), or $30,000. Robotic palletizing: As shown in Table 9.6, the equivalent uniform annual cost of robotic palletizing equals $19,840.63. Therefore, the robot is justified economically.

After-Tax Comparison of Manual versus Robotic Palletizing BTCF

DWO

TI

T

ATCF

0

−$125,000.00

1

−$500.00

$41,662.50

−$42,162.50

−$16,865.00

$16,365.00

2

−$500.00

$55,562.50

−$56,062.50

−$22,425.00

$21,925.00

3

−$500.00

$18,512.50

−$19,012.50

−$7,605.00

$7,105.00

4

−$500.00

$9,262.50

−$9,762.50

−$3,905.00

$3,405.00

5

$24,500.00

$0.00

$24,500.00

$9,800.00

$14,700.00

EUACBT 5

$29,379.75

EUACAT 5

$19,840.63

9-3

−$125,000.00

AFTER-TAX ANALYSIS USING BORROWED CAPITAL

LEARN I N G O B JEC T I V E : 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

9-3 After-Tax Analysis Using Borrowed Capital

331

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 5 principal payment IPMT 5 interest payment LCF 5 loan cash flow 5 PPMT 1 IPMT The following formulas apply when borrowed funds are used: ATCF 5 BT&LCF 2 LCF 2 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 5 itr 1TI 2

but, now, TI 5 BT&LCF 2 IPMT 2 DWO

(9.8)

Taxable income equals before-tax-and-loan cash flow, less the interest paid, less the depreciation allowance. Therefore, ATCF 5 BT&LCF11 2 itr2 2 LCF 1 itr 1DWO 1 IPMT2

(9.9)

Finally, in the year of the depreciable property’s disposal, TIn 5 BT&LCFn 1including Fn 2 2 IPMTn 2 DWOn 2 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 5 i 11 2 itr2

(9.11)

MARRBT is therefore given by MARRAT /(1 – itr). Hence, if the incometax rate is 40 percent, and MARRAT is 10 percent, then MARRBT is 0.10/0.60, or 16.667 percent. When the interest rate, for example i 5 12 percent, 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 9.8.

Loan Cash Flow (LCF) The cash flow reflecting the sum of the principal payment and the interest payment.

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Using Excel to Analyze After-Tax Effects of Borrowed Funds— Single Alternative

EXAMPLE

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-ofperiod payments; and (d) Plan 4—make a single payment of principal and interest at the end of the loan period. The loan interest rate 5 12%. KEY DATA

Given: Loan amount 5 $300,000; P 5 $500,000; itr 5 40%; MARRAT 5 10%; BT&LCF (years 1–9) 5 $124,166.67; BT&LCF (year 10) 5 $174,166.67 Find: PWAT, AWAT, FWAT, IRRAT, ERRAT for Plans 1, 2, 3, and 4

SOLUTION

a) Figure 9.3 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.

F IGURE 9.3

Using Plan 1

After-Tax Analysis of the SMP Investment with $300,000 of Borrowed Capital Repaid

9-3 After-Tax Analysis Using Borrowed Capital

333

As shown, the after-tax present worth (PWAT), based on a 12 percent interest rate on the $300,000 loan and a 10 percent MARRAT, is $175,603.01. When no borrowed funds were used, PWAT was $123,988.64 (see Example 9.4). Borrowing the $300,000 at 12 percent compound annual interest increased PWAT, because interest is deducted from taxable income. 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 –18.62 percent and at 41.54 percent. 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 9.3, ERRAT equals 17.16 percent. 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 $156,374.28. 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. After-Tax Analysis of the SMP Investment with $300,000 of Borrowed Capital Repaid Using Plan 2 TABLE 9.7

MARRAT 5 10% Income Tax Rate 5 40% EOY

BT&LCF

PPMT

0

2$500,000.00

2$300,000.00

1

$124,166.67

2 3 4

Interest Rate 5 12% IPMT

DWO

TI

Tax

ATCF

$30,000.00

$36,000.00

$100,000.00

−$11,833.33

−$4,733.33

$62,900.00

$124,166.67

$30,000.00

$32,400.00

$160,000.00

−$68,233.33

−$27,293.33

$89,060.00

$124,166.67

$30,000.00

$28,800.00

$96,000.00

−$633.33

−$253.33

$65,620.00

$124,166.67

$30,000.00

$25,200.00

$57,600.00

$41,366.67

$16,546.67

$52,420.00 $54,580.00

2$200,000.00

5

$124,166.67

$30,000.00

$21,600.00

$57,600.00

$44,966.67

$17,986.67

6

$124,166.67

$30,000.00

$18,000.00

$28,800.00

$77,366.67

$30,946.67

$45,220.00

7

$124,166.67

$30,000.00

$14,400.00

$0.00

$109,766.67

$43,906.67

$35,860.00

8

$124,166.67

$30,000.00

$10,800.00

$0.00

$113,366.67

$45,346.67

$38,020.00

9

$124,166.67

$30,000.00

$7,200.00

$0.00

$116,966.67

$46,786.67

$40,180.00

10

$174,166.67

$30,000.00

$3,600.00

$0.00

$170,566.67

$68,226.67

$72,340.00

PWAT 5

$156,374.28

FWAT 5

$405,594.61

AWAT 5

$25,449.19

IRRAT 5

28.51%

ERRAT 5

16.54%

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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 $160,734.89. 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. After-Tax Analysis of the SMP Investment with $300,000 of Borrowed Capital Repaid Using Plan 3 TABLE 9.8

MARRAT 5 10% Income Tax Rate 5 40% EOY

BT&LCF

PPMT

0

−$500,000.00

−$300,000.00

1

$124,166.67

2

Interest Rate 5 12% IPMT

DWO

TI

Tax

ATCF

$17,095.25

$36,000.00

$100,000.00

−$11,833.33

−$4,733.33

$75,804.75

$124,166.67

$19,146.68

$33,948.57

$160,000.00

−$69,781.90

−$27,912.76

$98,984.18

3

$124,166.67

$21,444.28

$31,650.97

$96,000.00

−$3,484.30

−$1,393.72

$72,465.14

4

$124,166.67

$24,017.59

$29,077.65

$57,600.00

$37,489.02

$14,995.61

$56,075.81

5

$124,166.67

$26,899.71

$26,195.54

$57,600.00

$40,371.13

$16,148.45

$54,922.97

6

$124,166.67

$30,127.67

$22,967.58

$28,800.00

$72,399.09

$28,959.64

$42,111.78

7

$124,166.67

$33,742.99

$19,352.26

$0.00

$104,814.41

$41,925.76

$29,145.66

8

$124,166.67

$37,792.15

$15,303.10

$0.00

$108,863.57

$43,545.43

$27,525.99

9

$124,166.67

$42,327.21

$10,768.04

$0.00

$113,398.63

$45,359.45

$25,711.97

10

$174,166.67

$47,406.47

$5,688.78

$0.00

$168,477.89

$67,391.16

$53,680.26

−$200,000.00

PWAT 5

$160,734.89

FWAT 5

$416,904.92

AWAT 5

$26,158.86

IRRAT 5

31.65%

ERRAT 5

16.68%

d) Figure 9.4 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 $162,184.44, Plan 4 ranks second to Plan 1 in maximizing PWAT. 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 9.4, the two roots occur at –3.65 percent and at 53.49 percent. 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 16.73 percent. Notice, PWAT is maximized when MARRAT equals

9-3 After-Tax Analysis Using Borrowed Capital

F IGURE 9.4

After-Tax Analysis of SMP Investment with $300,000 of Borrowed Capital Repaid Using Plan 4

9.328 percent. (We used the Excel® SOLVER tool to obtain the MARRAT value that maximized PWAT.) Given the results obtained for the four payment plans, it is anticipated that after-tax PW will increase as the amount borrowed increases. Figure 9.5 presents the results for Plan 1 if the investment is entirely paid for using borrowed funds. As expected, the PWAT of $210,012.58 is greater than the PWAT of $175,603.01 that occurred when only $300,000 was borrowed. When 100 percent of investment capital is obtained by borrowing, no initial investment occurs at the beginning of the ATCF series. As shown in Figure 9.5, 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 5 14.332 percent.)

EXPLORING THE SOLUTION

335

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FIGURE 9.5

Income Taxes

After-Tax Analysis of $500,000 Investment with 100% Borrowed Capital Repaid Using Plan 1

Example 9.8 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. Examples 9.6, 9.7, and 9.8 present ample guidance for after-tax analysis of multiple alternatives using borrowed capital.

9-4

LEASING VERSUS PURCHASING EQUIPMENT

LEARN I N G O B JEC T I V E : 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.

9-4 Leasing Versus Purchasing Equipment

After-Tax Comparison of Leasing versus Purchasing

337

EXAMPLE

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 percent and an income-tax rate of 40 percent, should the company lease the lift trucks? Given Buy: P 5 $18,000; A 5 $3,750 per truck; SV 5 $3,000 per truck at end of Year 5; itr 5 40%; MARRAT 5 12% Lease: A 5 $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 3-year property?

KEY DATA

As shown in Table 9.9, an after-tax analysis indicates the lift trucks should be leased. The present worth cost of purchasing is $96,486.74, whereas the

SOLUTION

TABLE 9.9 EOY

After-Tax Comparison of Purchasing versus Leasing Lift Trucks BTCF(P)

DWO(P)

Tl(P)

T(P)

ATCF(P)

0

2$90,000.00

1

2$18,750.00

$29,997.00

2$48,747.00

2$19,498.80

$748.80

2

2$18,750.00

$40,005.00

2$58,755.00

2$23,502.00

$4,752.00

3

2$18,750.00

$13,329.00

2$32,079.00

2$12,831.60

2$5,918.40

4

2$18,750.00

$6,669.00

2$25,419.00

2$10,167.60

2$8,582.40

5

2$3,750.00

$0.00

2$3,750.00

2$1,500.00

2$2,250.00

PWBT(P) 5 EOY

2$90,000.00

PWAT(P) 5

2$140,045.81 BTCF(L)

DWO(L)

Tl(L)

2$96,486.74 T(L)

ATCF(L)

0

2$29,500.00

2$29,500.00

2$11,800.00

2$17,700.00

1

2$38,500.00

$0.00

2$38,500.00

2$15,400.00

2$23,100.00

2

2$38,500.00

$0.00

2$38,500.00

2$15,400.00

2$23,100.00

3

2$38,500.00

$0.00

2$38,500.00

2$15,400.00

2$23,100.00

4

2$38,500.00

$0.00

2$38,500.00

2$15,400.00

2$23,100.00

5

$9,000.00

$0.00

2$9,000.00

2$3,600.00

2$5,400.00

PWBT(L) 5

2$132,783.18

PWAT(L) 5

2$90,926.87

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present worth cost of leasing is $90,926.87. The $5,559.87 difference in present worth costs is probably great enough for the company to lease the lift trucks. Interestingly, as shown in Table 9.8, if the decision had been made using a before-tax MARR of 0.12/(1 2 0.4), or 20 percent, the before-tax present worth cost of purchasing is $140,045.81, and the present worth cost of leasing is $132,783.18. Again, the $7,262.63 difference in the present worths is probably great enough for the company to decide to lease the lift trucks.

SUMMARY

KEY CONCEPTS 1. Learning Objective: Calculate corporate income taxes, including the effective, incremental, and marginal tax rates. (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. 2. Learning Objective: 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 5 BTCF 11 2 itr2 1 itr 1DWO2 where BTCF 5 before-tax cash flow DWO 5 depreciation write-off or allowance itr 5 income tax rate ATCF 5 after-tax cash flow

(9.4)

Summary 339

The overall income tax rate can be expressed as itr 5 str 1 ftr 11 2 str2

(9.5)

where str 5 state income-tax rate, and ftr 5 federal income-tax rate. 3. Learning Objective: 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 5 BT&LCF11 2 itr2 2 LCF 1 itr 1DWO 1 IPMT2

(9.9)

where BT&LCF 5 before-tax-and-loan cash flow LCF 5 loan cash flow IPMT 5 interest payment In the year of the depreciable property’s disposal, TIn 5 BT&LCFn 1including Fn 2 2 IPMTn 2 DWOn 2 Bn

(9.10)

where TIn 5 taxable income in the year of property’s disposal Bn 5 book value at time of disposal Fn 5 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 5 i 11 2 itr2

(9.11)

4. Learning Objective: 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.

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KEY TERMS After-Tax Cash Flow (ATCF), p. 322 Before-Tax-and-Loan Cash Flow (BT&LCF), p. 322 Before-Tax Cash Flow (BTCF), p. 322 Effective Tax Rate, p. 320 Federal Tax Rate (ftr), p. 323

Income Tax Rate (itr), p. 323 Incremental Tax Rate, p. 320 Loan Cash Flow (LCF), p. 331 Marginal Tax Rate, p. 320 State Tax Rate (str), p. 323

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1.

The marginal tax rate on a corporate income of $87,000 is closest to which of the following? (Table 9.1 or similar is required for this question.) a. 15.0% c. 25.0% b. 20.5% d. 34.0%

2.

The average tax rate on a corporate income of $87,000 is closest to which of the following? (Table 9.1 or similar is required for this question.) a. 15.0% c. 25.0% b. 20.5% d. 34.0%

3.

The correctly calculated taxes due on a corporate taxable income of $13,000,000 are closest to which of the following? (Table 9.1 or similar is required for this question.) a. $3,400,000 c. $4,450,000 b. $4,420,000 d. $4,550,000

4.

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

5.

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?

Summary 341

a. b. c. d. 6.

MARR MARR * (1 1 tax rate) MARR/(1 2 tax rate) MARR * (1 2 tax rate)

Consider the following data extracted from an after-tax cash flow calculation. Before-Tax Cash Flow 5 22,500 Loan Principal Payment 5 7,434 Loan Interest Payment 5 892 MACRS Depreciation Deduction 5 7,405 Taxes Due 5 5,397 Which of the following is closest to the after-tax cash flow? a. $1,372 c. $8,806 b. $8,777 d. $16,211

7.

Consider the following data extracted from an after-tax cash flow calculation. Before Tax Cash Flow 5 $22,500 Loan Principal Payment 5 $5,926 Loan Interest Payment 5 $2,400 MACRS Depreciation Deduction 5 $16,665 Which of the following is closest to the Taxable Income? a. 2$2,491 c. $3,435 b. 2$91 d. $14,174

8.

Consider the following data for 2012 from an after-tax cash flow analysis. What is the loan interest payment for 2012? Before Tax Cash Flow 5 $20,000 Loan Principal Payment 5 $4,018 Depreciation Deduction 5 $8,920 Taxable Income 5 $8,018 Taxes Due 5 $2,726 After Tax Cash Flow 5 $10,194 a. $1,274 c. $7,062 b. $3,062 d. $11,080

9.

Consider the following data for 2011 from an after tax cash flow analysis. What is the taxable income for 2011? Before Tax Cash Flow 5 $23,000 Loan Principal Payment 5 $3,203 Loan Interest Payment 5 $3,877 Depreciation Deduction 5 $12,490 Taxes Due 5 $2,255 After Tax Cash Flow 5 $21,530 a. $40,000 c. $6,633 b. $35,540 d. $28,460

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10.

Consider the following data for 2010 from an after tax cash flow analysis. What is the after tax cash flow for 2010? Before Tax Cash Flow 5 $23,000 Loan Principal Payment 5 $3,203 Loan Interest Payment 5 $3,877 Depreciation Deduction 5 $12,490 Taxable Income 5 $6,633 Taxes Due 5 $2,255 a. $20,744 c. $3,430 b. $13,665 d. $1,175

PROBLEMS Introduction 1. What are the various taxes paid in the USA by corporations? Which of those

has the most significant impact upon economic analyses? 2. 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 3. 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) 4. Explain the primary effect of (a) the Economic Recovery Act of 1981,

(b) the Tax Reform Act of 1986, and (c) the Omnibus Reconciliation Act of 1993. 5. Explain why it is generally preferred to do economic analyses on an after-tax

basis, rather than on a before-tax basis.

Summary 343

Section 9.1 Corporate Income Tax Rates 6. What is the federal income tax for each of the following corporate taxable

incomes? $25,000 $70,000 $95,000 $200,000

a. b. c. d.

e. f. g. h.

$1,000,000 $12,000,000 $17,000,000 $25,000,000

7.

Calculate the corporate income tax for each of the following corporate taxable incomes. For each, determine the effective (average) tax rate and also the marginal tax rate. a. $15,000 d. $400,000 b. $88,000 e. $16,700,000 c. $180,000

8.

Video Solution Calculate the corporate income tax for each of the following corporate taxable incomes. For each, determine the effective (average) tax rate and also the marginal tax rate. a. $12,000 d. $1,000,000 b. $65,000 e. $19,300,000 c. $220,000

9.

Determine the smallest taxable income on which: a. The very last dollar is taxed at 25% or more. b. Every dollar is taxed at 25% or more. c. Every dollar is taxed at 34% or more.

10. Determine the smallest taxable income on which: a. The very last dollar is taxed at 35% or more. b. Every dollar is taxed at 34% or more. c. Every dollar is taxed at 35% or more. 11.

TenTec in Sevierville, TN makes commercial and amateur radio equipment including receivers, transceivers, antenna tuners, linear amplifiers, etc. Their taxable income last year was $720,000. They have established a new line of electronic equipment that is on sale this year, and expected to add another $140,000 to taxable income. a. Determine the effective (average) tax rate on all of last year’s taxable income. b. Determine the effective (average) tax rate on all of this year’s taxable income. c. Determine the incremental tax rate. d. Determine the marginal tax rate this year.

12. An engineer named Don maintains a small incorporated civil engineering

consulting practice. Last year, Don’s firm’s taxable income was $41,000. During that year, he spent a great deal of non-billable time developing computerized tools for analyzing changes in water runoff due to new highway,

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building, and parking lot construction. Given his expertise and his ability to sign off as a licensed professional engineer (he took the FE exam in college), this year his company has been given the opportunity to take on two projects that will increase taxable income by $175,000. a. Determine the effective (average) tax rate on all of last year’s taxable income. b. Determine the effective (average) tax rate on all of this year’s taxable income. c. Determine the incremental tax rate. d. Determine the marginal tax rate this year. 13.

Noise Sniffers Inc (NSI) is a contractor to public utilities providing electrical service to homes and businesses. Most small- and mid-sized municipal utilities do not have the expertise or the equipment to go out and “sniff ” noise caused by foreign objects on power lines, poor grounding on poles, cracked or carbonized lightning arrestors, etc. Success means reduction in energy loss to the utility and reduction in static to those using radios nearby. Early on, ham radio operator K5KC suggested NSI personnel get training from the American Radio Relay League and purchase the right detection equipment. NSI now travels throughout the state to help utilities in need. Taxable income last year was only about $30,000 for this sporadic part-time work. This year, NSI is adding a larger adjoining state as a customer and will enjoy an increase of about $40,000 in taxable income. a. Determine the effective (average) tax rate on all of last year’s taxable

income. b. Determine the effective (average) tax rate on all of this year’s taxable

income. c. Determine the incremental tax rate. d. Determine the marginal tax rate this year. 14. Matrix Service Company is an industrial service contractor with a strong

reputation in refining, power, petrochemicals, gas/LNG, and related areas. Their base taxable income is $19.2 million on sales of $490 million. They have landed two large above-ground storage tank projects and expect taxable income to increase to $25.7 million on sales of $540 million next year. a. Determine the effective (average) tax rate on all of last year’s taxable income. b. Determine the effective (average) tax rate on all of this year’s taxable income. c. Determine the incremental tax rate. d. Determine the marginal tax rate for next year. Section 9.2.1 After-Tax Analysis Using Retained Earnings—Single Alternative (No Borrowing) 15.

West Mountain Radio, an electronics firm specializing in equipment interfaces, had been in a marginal tax bracket of 39% and then this year had an

Summary 345

increase in taxable income of $150,000. Their increase in taxes was exactly $54,750. For each of the following, determine the smallest amount that meets the conditions given above. a. What was their taxable income last year? b. What was their total income tax this year? 16. Pneumatics, a startup firm of four persons, had an increase in taxable income

of $20,000 this year. Their increase in taxes was exactly $6350. a. What was their taxable income last year? b. What was their total income tax this year? 17.

Acme Universal is a micro-cap that had a $3,000,000 drop in taxable income this year. Their drop in taxes was $1,030,000. For each of the following, determine the largest amount that meets the conditions given above. a. What was their taxable income last year? b. What was their total income tax this year?

18. Westwood Specialties had an increase in taxable income of $5,000,000 this

year. Their increase in taxes was exactly $1,720,000. a. What was their taxable income last year? b. What was their total income tax this year? 19. 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 taxes are 40 percent. 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). 20. 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 taxes are 40 percent. 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). 21. 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 40 percent. Find the after-tax cash flow for each year using MACRS-GDS allowances.

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22. Raytheon wishes to use an automated environmental chamber in the manu-

facture 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 taxes are 40%. 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). 23.

GO Tutorial 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 40% 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). 24. 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 40% 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 23(b) where MACRS-GDS is used. Explain the differences. 25. 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 40% and the after-tax MARR is 9%. There is some concern that the need for this equipment will last only 10 years and 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).

Summary 347

26. Bell’s Amusements purchased an expensive ride for their theme and amuse-

ment 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 taxes are 40% 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). 27.

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 taxes are 40% 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).

28. Henredon can spend $190,000 now for a design portfolio with a different fur-

niture 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 marginal tax rate is 40% 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 27(b) where MACRS-GDS is used. Explain the differences. 29.

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 marginal tax rate is 40% and MARR is 10% on after-tax cash flows. Suppose that, to AgriGrow’s

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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). 30. 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 marginal tax rate is 40% 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 29(b) where MACRS-GDS is used. Explain the differences. 31. Specialized production equipment is purchased for $125,000. The equip-

ment 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 40 percent 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 percent. EOY 0 1 2 3 4 5

BTCF –$125,000 50,000 60,000 70,000 80,000 120,000

32. 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 40 percent income tax rate, determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR using an 8 percent MARRAT. EOY 0 1 2 3 4 5

BTCF –$800,000 100,000 200,000 300,000 400,000 700,000

Summary 349

33. An investment of $800,000 is made in equipment that qualifies as 3-year equip-

ment 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 40 percent tax rate applies and the after-tax MARR is 8 percent. Determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR. EOY 0 1 2 3 4 5

BTCF 2$800,000 100,000 200,000 300,000 400,000 700,000

34. In Problem 33, suppose the equipment still qualifies as MACRS-GDS 3-year

property, but is sold for $300,000 after 3 years of use. A 40 percent tax rate applies and the after-tax MARR is 8 percent. 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) 35.

GO Tutorial 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 marginal tax rate is 40% 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?

36. 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 marginal tax rate is 40% 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.

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37.

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 $2000 at that time. The annual cost of use and maintenance will be $9500. Alternatively, a more automated instrument (same property class) available from the manufacturer costs $29,000, with use and maintenance costs of only $7500 and salvage value after 6 years of $3000. The marginal tax rate is 40% and MARR is an after-tax 12%. Determine which alternative is less costly, based upon comparison of after-tax annual worth.

38. Two investments involving a granary qualify for different property classes.

Investment A costs $70,000 with $3000 salvage value after 16 years and is depreciated as MACRS-GDS in the 10-year property class. Investment B costs $110,000 with a $4000 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 marginal tax rate is 40%, 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? 39.

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 $3000 salvage value after 16 years. The other can be purchased and installed for $110,000 with $4000 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 marginal tax rate of 40%, 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?

40. 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 40 percent, for what value of $X will the firm be indifferent between the two investment alternatives? The after-tax MARR is 15 percent. (Hint: Excel® SOLVER would be an excellent way to determine the answer.) 41. 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 percent, and a 40 percent tax

Summary 351

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 42. What is the difference or distinction being made when we speak of before-tax

cash flows and before-tax and loan cash flows. Be precise in your answer. 43.

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 marginal tax rate

of 40%. c. What is the after-tax cash flow for the first year? 44. 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 marginal tax rate of 40%. c. Determine the after tax cash flow for the year. 45.

GO Tutorial 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 40% marginal tax rate. c. Determine the after-tax cash flow for the first year.

46. 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

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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 marginal tax rate of 40%? c. What is the after-tax cash flow for the first year? Note: Problems 47–55 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.

47. An investment of $250,000 is made in equipment that qualifies as MACRS-

GDS 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 percent of the investment capital at an annual compound interest rate of 18 percent and the loan is repaid over a 5-year period. A 40 percent tax rate and an MARRAT of 7 percent apply. EOY 0 1 2 3 4 5

BTCF –$250,000 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. Select 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. 48. 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 percent. 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 40 percent. An MARRAT of 10% applies.

Summary 353

EOY 0

BTCF 2$800,000 100,000 200,000 300,000 400,000 700,000

1 2 3 4 5

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. 49. Specialized production equipment is purchased for $125,000. The equip-

ment qualifies as 5-year equipment for MACRS-GDS depreciation. Suppose 40 percent of the investment capital is borrowed at an annual compound rate of 18 percent 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 40 percent 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 percent. EOY 0 1 2 3 4 5

BTCF –$125,000 50,000 60,000 70,000 80,000 120,000

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. 50. 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 marginal tax rate is 40% and their MARR is 9% after taxes. Determine the PW, FW, AW, IRR, and ERR for the investment if: a. b. c. d.

The loan is paid back using Plan 1. The loan is paid back using Plan 2. The loan is paid back using Plan 3. The loan is paid back using Plan 4.

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Income Taxes

51.

GO Tutorial 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 marginal tax rate is 40%, 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. b. c. d.

The loan is paid back using Plan 1. The loan is paid back using Plan 2. The loan is paid back using Plan 3. The loan is paid back using Plan 4.

52. A Boeing contractor responsible for producing a portion of the landing gear

for huge airliners experienced a storm-related power glitch during the multiaxis 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 marginal tax rate is 40%, 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. b. c. d. 53.

The loan is paid back using Plan 1. The loan is paid back using Plan 2. The loan is paid back using Plan 3. The loan is paid back using Plan 4.

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 taxes are 40%. 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,d. The loan is paid back using Plan 1. b,e. The loan is paid back using Plan 2. c,f. The loan is paid back using Plan 3.

Summary 355

54. 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 40% and the after-tax MARR is 9%. There is some concern that the need for this equipment will last only 10 years and 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,d. The loan is paid back using Plan 1. b,e. The loan is paid back using Plan 2. c,f. The loan is paid back using Plan 3. 55.

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 12.5% 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%, 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.

56. 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. LOAN INT 1 RCVD 2 PAID

LOAN PRINC 1 RCVD 2 PAID

BTLCF 1 RCVD 2 PAID

LOAN 1 RCVD 2 PAID

0

2$90,000.00

$30,000.00

1

$35,000.00

2$4,500.00 2$8,639.31

2

$35,000.00

2$3,204.10

3

$40,000.00

EOY

a. b. c. d. e.

DEPR

BK VALUE

TI

TAX 1 RCVD 2 PAID

$90,000.00 $12,861.00

$7,870.50

ATCF 1 RCVD 2 PAID

PWATCF

2$60,000.00

2$60,000.00

$77,139.00 $17,639.00

2$7,408.38

$14,452.31

$12,903.85

$9,754.90

2$4,097.06

$17,763.63

$14,161.06

$47,227.50

What is the property class of the asset? What is the value of MARR? What is the salvage value received at the end of year 3? What is the loan interest rate? Determine the value of the entries in the BOLD BORDERED cells.

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57.

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 (Note! The shaded cells would normally have something in them but have been erased for this exercise.):

BTLCF 1 IN; 2 OUT

LOAN PRINC 1 IN; 2 OUT

LOAN PRINC REMAIN

0

2$180,000.00

$70,000.00

$70,000.00

1

$40,000.00

2$9,785.71

$60,214.29

2$4,900.00 $36,000.00

2

$40,000.00

$49,743.59

2$4,215.00

3

$40,000.00

EOY

4

LOAN INT 1 IN; 2 OUT

DEPR

BK VALUE

TI

$180,000.00

2$2,697.80 $10,368.00

a. b. c. d. e.

ATCF 1 IN; 2 OUT

PWATCF

2$110,000.00 2$110,000.00

$144,000.00

$25,638.29

$22,891.33

$7,853.40 $1,957.95

2$38,539.93

TAX 1 IN; 2 OUT

2$704.62

$24,609.43

$41,472.00 $15,462.20

Fill in all of the blanks having BOLD BORDERS. What is the value of MARR? What is the loan interest rate? What is the MACRS property class? What is the tax rate?

Section 9.3 After-Tax Analysis Using Borrowed Capital—Multiple Alternatives

Note: Problems 58–63 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. 58. Recall problem 36. Suppose both models will require a loan of $40,000 at an

interest rate of 17 percent, with payment using Plan 1 over a period of 4 years. Rework the problem. 59. Recall problem 37. Suppose both alternatives will require a loan of 50 percent of

the investment cost, an annual interest rate of 18 percent, with payment using Plan 3 over the 6-year planning horizon. Rework the problem. 60. Recall problem 38. Suppose both investments will require a loan of $14,000

at an interest rate of 10 percent, with payment using Plan 4 over a period of 3 years. Rework the problem.

Summary 357

61. Recall problem 39. Suppose both conveyors will require a loan of $14,000

at an interest rate of 10 percent, with payment using Plan 2 over a period of 3 years. Rework the problem. 62. Recall problem 40. Suppose both alternatives will involve a loan of 40 percent

of the investment cost, with payment over 10 years using Plan 4 and a sweet Federal government loan rate of 2 percent. Rework the problem. 63. Recall problem 41. Suppose both options will involve a loan of 40 percent 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 percent. Rework the problem. Section 9.4 Purchasing versus Leasing Equipment 64. 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 $4000 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 end-ofyear, with an immediate credit through salvage of their present equipment for $45,000. Marginal taxes are 40% 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. 65.

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. Marginal taxes are 40% 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.

ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : C ATER P I L L AR In 1890, Benjamin Holt 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 the 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 chairman and CEO, Douglas R. Oberhelman, stated in the 2011 Annual Report, “Business has changed. The industries we serve have changed. Competition is tough, customer requirements are demanding, and the industries we serve are booming, and they are complicated.” In 2011, the company had sales and revenues totaling $60.1 billion, an increase of 41 percent from 2010. Caterpillar is a global company with seventy percent of sales and revenue in 2011 occurring outside the United States. In 2011, Caterpillar’s worldwide employment was 125,099 of which 53,236 were within the United States and 71,863 were located outside the United States. Caterpillar’s profits were $7.15 billion, compared with $3.96 billion in 2010, and capital expenditures totaled $2.6 billion, up from $1.58 billion in 2010. R&D expenditures in 2011 were $2.3 billion, representing 3.8 percent of sales. In its 2011 Annual Report, the company acknowledges that its business is highly sensitive to global economic conditions and 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, regions that have experienced tremendous inflation over the past decade. 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.

358

INFLATION

DISCUSSION QUESTIONS: 1. In 2011, Caterpillar generated 70 percent of its sales and revenue 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?

LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

1. Explain the effects of inflation on purchasing power and how the inflation rate is estimated. (Section 10.1)

2. Conduct a before-tax analysis with inflation by correctly incorporating inflationary effects in both the cash flows and the discount rate. (Section 10.2)

3. Conduct an after-tax analysis with inflation by correctly adjusting for situations in which depreciation allowances are not permitted to increase with inflation. (Section 10.3)

4. Conduct an after-tax analysis with inflation using borrowed capital. (Section 10.4) 359

360

Chapter 10

Inflation

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.

10-1

Select the preferred investment

THE MEANING AND MEASURE OF INFLATION

LEARN I N G O B JEC T I V E : Explain the effects of inflation on purchasing power and how the rate is estimated.

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.

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 can have such a significant impact on an investment’s economic worth that it should be considered when comparing investment

10-1 The Meaning and Measure of Inflation 361

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 percent 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. 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. The CPI is a market basket rate, in that it is based on 80,000 prices that are recorded in 87 urban areas. The so-called 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.

Consumer Price Index (CPI) A value developed 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.

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Table 10.1 shows CPI values from 1913 through 2012. 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 2007 through 2012, annual inflation rates have been 2.8%, 3.8%, 20.4%, 1.6%, 3.2%, and 2.1% respectively. TABLE 10.1 Year

CPI

1913

9.9

1914

10.0

1915

10.1

1916

CPI Data 1913–2012 % Change

Year

CPI

Year

CPI

1946

19.5

% Change 8.3%

1979

72.6

% Change 11.3%

1.0%

1947

22.3

14.4%

1980

82.4

13.5%

1.0%

1948

24.1

8.1%

1981

90.9

10.3%

10.9

7.9%

1949

23.8

21.2%

1982

96.5

6.2%

1917

12.8

17.4%

1950

24.1

1.3%

1983

99.6

3.2%

1918

15.1

18.0%

1951

26.0

7.9%

1984 103.9

4.3%

1919

17.3

14.6%

1952

26.5

1.9%

1985 107.6

3.6%

1920

20.0

15.6%

1953

26.7

0.8%

1986 109.6

1.9%

1921

17.9

210.5%

1954

26.9

0.7%

1987 113.6

3.6%

1922

16.8

26.1%

1955

26.8

20.4%

1988 118.3

4.1%

1923

17.1

1.8%

1956

27.2

1.5%

1989 124.0

4.8%

1924

17.1

0.0%

1957

28.1

3.3%

1990 130.7

5.4%

1925

17.5

2.3%

1958

28.9

2.8%

1991 136.2

4.2%

1926

17.7

1.1%

1959

29.1

0.7%

1992 140.3

3.0%

1927

17.4

21.7%

1960

29.6

1.7%

1993 144.5

3.0%

1928

17.1

21.7%

1961

29.9

1.0%

1994 148.2

2.6%

1929

17.1

0.0%

1962

30.2

1.0%

1995 152.4

2.8%

1930

16.7

22.3%

1963

30.6

1.3%

1996 156.9

3.0%

1931

15.2

29.0%

1964

31.0

1.3%

1997 160.5

2.3%

1932

13.7

29.9%

1965

31.5

1.6%

1998 163.0

1.6%

1933

13.0

25.1%

1966

32.4

2.9%

1999 166.6

2.2%

1934

13.4

3.1%

1967

33.4

3.1%

2000 172.2

3.4%

1935

13.7

2.2%

1968

34.8

4.2%

2001 177.1

2.8%

1936

13.9

1.5%

1969

36.7

5.5%

2002 179.9

1.6%

1937

14.4

3.6%

1970

38.8

5.7%

2003 184.0

2.3%

1938

14.1

22.1%

1971

40.5

4.4%

2004 188.9

2.7%

1939

13.9

21.4%

1972

41.8

3.2%

2005 195.3

3.4%

1940

14.0

0.7%

1973

44.4

6.2%

2006 201.6

3.2%

1941

14.7

5.0%

1974

49.3

11.0%

2007 207.3

2.8%

1942

16.3

10.9%

1975

53.8

9.1%

2008 215.3

3.8%

1943

17.3

6.1%

1976

56.9

5.8%

2009 214.5

20.4%

1944

17.6

1.7%

1977

60.6

6.5%

2010 218.1

1.6%

1945

18.0

2.3%

1978

65.2

7.6%

2011 224.9

3.2%

2012 229.6

2.1%

10-1 The Meaning and Measure of Inflation 363

Percent Annual Change

20.0% 15.0% 10.0% 5.0% 0% –5.0%

–10.0% 2012

2005

1998

1991

1984

1977

1970

1963

1956

1949

1942

1935

1928

1921

1914

–15.0% Year FIGURE 10.1

CPI Annual Percent Changes 1913–2012

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

364

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Inflation

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.

production areas. Price increases in one segment of the economy are generally felt in all segments of the economy. 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. 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

10-2

BEFORE-TAX ANALYSIS

LEARN I N G O B JEC T I V E : Conduct a before-tax analysis with inflation by correctly incorporating inflationary effects in both the cash flows and the discount rate.

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.

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. 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).

10-2 Before-Tax Analysis

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. 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: 11 1 ic 2 5 11 1 ir 2 11 1 f 2

(10.1)

i c 5 ir 1 f 1 ir 1 f 2

(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 5 MARRr 1 f 1 MARRr 1 f 2

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.

(10.3)

When given ic and f, the value of ir can be obtained from Equation 10.2. Specifically, ir 5 1ic 2 f 2/ 11 1 f 2

(10.4)

MARRr 5 1MARRc 2 f 2/ 11 1 f 2

(10.5)

Likewise,

Calculating the Real Interest Rate

EXAMPLE

If inflation averages 4 percent per year and your return on an investment, based on then-current dollars, is 10 percent, what is your real return on investment? Given: f 5 4%, ic 5 10% Find: ir

365

KEY DATA

366

Chapter 10

Inflation

SOLUTION

From Equation 10.4,

ir 5 1ic 2 f 2/ 11 1 f 2 5 10.10 2 0.042/ 11 1 0.042 5 0.057692 or 5.7692%

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: or Likewise, or

$Tk 5 $Ck 11 1 f 2 k

(10.6)

$Tk 5 $Ck 1F Z P f%,k2

(10.7)

$Ck 5 $Tk 11 1 f 2 2k

(10.8)

$Ck 5 $Tk 1P Z F f %,k2

(10.9)

Before-Tax Impact of Inflation on the Present Worth of Raw Material Cost

EXAMPLE

Video Example

KEY DATA

SOLUTION

A small business spent $125,000 for raw material during 2012. It anticipates the cost of raw material will increase, due to inflation, at a rate of 3 percent per year. It also anticipates that the amount of raw material required will increase at a rate of 8 percent per year due to increased demand for the firm’s products. If the business has a 10 percent real required return on its investments, what is the present worth of the cost of raw material over the 5-year period (2013–2017)? Given: Cash flows shown in Table 10.2, f 5 3%, MARRr 5 10% Find: PW of raw material cost Table 10.2 shows the yearly cash flows for this example in constant and thencurrent dollars. The constant-dollar cash flows equal $125,000(1.08)k 2 2012 and the then-current cash flows equal $125,000[1.08(1.03)] k 2 2012. 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% 1 3% 1 10%(3%) or 13.3% MARRc.

10-3 After-Tax Analysis 367

Alternatively, the present worths can be calculated by hand using calculated or tabulated values, as shown in Table 10.2. TABLE 10.2

k

Cash Flows and Present Worth for Example 10.2

$Ck

(P |F, 10%,n)

PWk in Constant $

$Tk

(P |F, 13.3%,n)

PWk in ThenCurrent $

2013

$135,000.00 0.90909

$122,727.27 $139,050.00 0.88261

$122,727.27

2014

$145,800.00 0.82645 $120,495.87 $154,679.22 0.77900

$120,495.87

2015

$157,464.00 0.75131

$118,305.03 $172,065.16 0.68756

$118,305.03

2016

$170,061.12

0.68301

$116,154.03 $191,405.29 0.60685

$116,154.03

2017

$183,666.01 0.62092

$114.042.14 $212,919.24 0.53561

$114.042.14

PWtotal 5 $591,724.35

PWtotal 5 $591,724.35

10-3

AFTER-TAX ANALYSIS

LEARNING O BJECTI VE: Conduct an after-tax analysis with inflation by cor-

rectly adjusting for situations in which depreciation allowances are not permitted to increase 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.

Using Excel to Analyze the After-Tax Impact of Inflation on the SMP Investment Recall the $500,000 investment in a surface-mount placement machine that produces net annual savings, before taxes, of $124,166.67 for 10 years, plus a $50,000 salvage value at the end of the 10-year period.

EXAMPLE

368

Chapter 10

Inflation

The estimates were in constant dollars. The SMP machine qualified as 5-year property for MACRS-GDS depreciation. A 40 percent income tax rate and a MARRr of 10 percent were used to perform an after-tax analysis. To illustrate the effect of inflation in after-tax analyses, suppose annual inflation equals 4 percent. What are the PW and IRR values of the investment in then-current dollars? KEY DATA

Given: Cash flows and depreciation as outlined in Figure 10.2, MARRr 5 10%, f 5 4%, income tax 5 40% 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 SOLUTION

As shown in Figure 10.2, then-current, before-tax cash flows ($TBTCF) are obtained by increasing constant dollar, before-tax cash flows ($CBTCF) at an annual compound rate of 4 percent; depreciation allowances are in then-current dollars. The resulting then-current, after-tax present worth is $109,201.18, compared to $123,988.64 if inflation is negligible. Likewise, when inflation is negligible, the after-tax internal rate of return is 16.12 percent. With 4 percent inflation, the combined after-tax internal rate of return is 19.99 percent; therefore, the real aftertax internal rate of return is IRRr 5 10.1999 2 0.042/ 11.042 5 0.1538, or 15.38%.

10-4 After-Tax Analysis with Borrowed Capital

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 $109,201.18, which is identical to the then-current result. The real after-tax internal rate of return is 15.38 percent.

FIGURE 10.3

After-Tax, Constant Dollar Analysis of the SMP Investment with

4% Inflation

10-4

AFTER-TAX ANALYSIS WITH BORROWED CAPITAL

LEARNING O BJECTI VE: 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

EXPLORING THE SOLUTION

369

370

Chapter 10

Inflation

dollar approach, however, then the principal payments, interest payments, and depreciation allowances must be converted to constantdollar cash flows. Equation 10.8 or 10.9 can be used to perform the conversion.

Using Excel to Analyze the Impact of Inflation on the SMP Investment When Money Is Borrowed

EXAMPLE

Suppose $300,000 is borrowed at a 12 percent 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 9.8, but we now include inflation of 4 percent/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. Which of the payment plans should be used if the MARRr equals 10 percent? (As before, a 40 percent income tax rate is used.) KEY DATA

Given: i 5 12%, f 5 4%, MARRr 5 10%, tax rate 5 40%, $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 $220,132.29; Plan 2 yields an after-tax present worth of $182,165.68; Plan 3 yields an after-tax present worth of $191,157.40; Plan 4 yields an after-tax present worth of $232,335.64.

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 9.8, the ERR is calculated using the Excel® IRR worksheet function with the modified ATCF column.

10-4 After-Tax Analysis with Borrowed Capital

371

FIGURE 10.4 After-Tax Analysis of the SMP Investment with $300,000 Borrowed, Repaid with Plan 1, and 4% Inflation

After-Tax Analysis of the SMP Investment; $300,000 Borrowed and Repaid with Plan 2; 4% Inflation

TABLE 10.3

MARRr 5 10% Income Tax Rate 5 40% EOY 0 1 2 3 4 5 6 7 8 9 10

$TBTCF

$TPPMT

2$500,000.00 $129,133.34 $134,298.67 $139,670.62 $145,257.44 $151,067.74 $157,110.45 $163,394.87 $169,930.66 $176,727.89 $257,809.22

2$300,000.00 $30,000.00 $30,000.00 $30,000.00 $30,000.00 $30,000.00 $30,000.00 $30,000.00 $30,000.00 $30,000.00 $30,000.00

Interest Rate 5 12% Inflation Rate 5 4% $TIPMT $36,000.00 $32,400.00 $28,800.00 $25,200.00 $21,600.00 $18,000.00 $14,400.00 $10,800.00 $7,200.00 $3,600.00

$TDWO

$TTI

$TTax

$TATCF

$100,000.00 $160,000.00 $96,000.00 $57,600.00 $57,600.00 $28,800.00 $0.00 $0.00 $0.00 $0.00

2$6,866.66 2$58,101.33 $14,870.62 $62,457.44 $71,867.74 $110,310.45 $148,994.87 $159,130.66 $169,527.89 $254,209.22

2$2,746.67 2$23,240.53 $5,948.25 $24,982.98 $28,747.10 $44,124.18 $59,597.95 $63,652.26 $67,811.16 $101,683.69 PW$T 5 FW$T 5 AW$T 5 IRRc 5 ERRc 5 IRRr 5 ERRr 5

2$200,000.00 $65,880.00 $95,139.20 $74,922.37 $65,074.47 $70,720.64 $64,986.27 $59,396.92 $65,478.40 $71,716.73 $122,525.53 $182,165.68 $699,401.91 $35,470.47 35.34% 22.05% 30.13% 17.36%

372

Chapter 10

Inflation

After-Tax Analysis of the SMP Investment; $300,000 Borrowed and Repaid with Plan 3; 4% Inflation

TABLE 10.4

MARRr 5 10% Income Tax Rate 5 40% EOY 0 1 2 3 4 5 6 7 8 9 10

Interest Rate 5 12% Inflation Rate 5 4%

$TBTCF

$TPPMT

2$500,000.00 $129,133.34 $134,298.67 $139,670.62 $145,257.44 $151,067.74 $157,110.45 $163,394.87 $169,930.66 $176,727.89 $257,809.22

2$300,000.00 $17,095.25 $19,146.68 $21,444.28 $24,017.59 $26,899.71 $30,127.67 $33,742.99 $37,792.15 $42,327.21 $47,406.47

TABLE 10.5

$TIPMT $36,000.00 $33,948.57 $31,650.97 $29,077.65 $26,195.54 $22,967.58 $19,352.26 $15,303.10 $10,768.04 $5,688.78

$TDWO

$TTI

$100,000.00 $160,000.00 $96,000.00 $57,600.00 $57,600.00 $28,800.00 $0.00 $0.00 $0.00 $0.00

2$6,866.66 2$59,649.90 $12,019.65 $58,579.79 $67,272.20 $105,342.87 $144,042.61 $154,627.56 $165,959.85 $252,120.44

$TTax

$TATCF

2$2,746.67 2$23,859.96 $4,807.86 $23,431.91 $26,908.88 $42,137.15 $57,617.04 $61,851.02 $66,383.94 $100,848.18 PW$T 5 FW$T 5 AW$T 5 IRRc 5 ERRc 5 IRRr 5 ERRr 5

2$200,000.00 $78,784.75 $105,063.38 $81,767.51 $68,730.28 $71,063.61 $61,878.05 $52,682.57 $54,984.39 $57,248.70 $103,865.79 $191,157.40 $733,924.46 $37,221.30 38.58% 22.34% 33.25% 17.63%

After-Tax Analysis of the SMP Investment; $300,000 Borrowed and Repaid with Plan 4; 4% Inflation

MARRr 5 10% Income Tax Rate 5 40% EOY

$TBTCF

$TPPMT

0 1 2 3 4 5 6 7 8 9 10

2$500,000.00 $129,133.34 $134,298.67 $139,670.62 $145,257.44 $151,067.74 $157,110.45 $163,394.87 $169,930.66 $176,727.89 $257,809.22

2$300,000.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $300,000.00

Interest Rate 5 12% Inflation Rate 5 4% $TIPMT $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $631,754.46

$TDWO

$TTI

$100,000.00 $29,133.34 $160,000.00 2$25,701.33 $96,000.00 $43,670.62 $57,600.00 $87,657.44 $57,600.00 $93,467.74 $28,800.00 $128,310.45 $0.00 $163,394.87 $0.00 $169,930.66 $0.00 $176,727.89 $0.00 2$373,945.24

$TTax

$TATCF

2$200,000.00 $11,653.33 $117,480.00 2$10,280.53 $144,579.20 $17,468.25 $122,202.37 $35,062.98 $110,194.47 $37,387.10 $113,680.64 $51,324.18 $105,786.27 $65,357.95 $98,036.92 $67,972.26 $101,958.40 $70,691.16 $106,036.73 2$149,578.10 2$524,367.15 $232,335.64 PW$T 5 $892,023.06 FW$T 5 $45,239.33 AW$T 5 58.76% IRRc 5 23.57% ERRc 5 52.66% IRRr 5 18.82% ERRr 5

$TATCF 2$200,000.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $1,659,897.55

Summary 373

In Chapter 9, PWAT was $175,603.01 for Plan 1; $156,374.28 for Plan 2; $160,734.89 for Plan 3; and $162,184.44 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. Notice the impact of inflation on the preferred plan. With no inflation, Plan 1 is preferred; with 4 percent inflation, Plan 4 is preferred.

EXPLORING THE SOLUTION

Comparison of PWAT for Four Payment Plans, with and Without Inflation TABLE 10.6

Plan

PWAT (0% Inflation)

PWAT (4% Inflation)

Difference

1

$175,603.01

$220,132.29

$44,529.28

2

$156,374.28

$182,165,68

$25,791.40

3

$160,734.89

$191,157.40

$30,422.51

4

$162,184.44

$232,335.64

$70,151.20

KEY CONCEPTS 1. Learning Objective: Explain the effects of inflation on purchasing power and how the rate is estimated. (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.

SUMMARY

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2. Learning Objective: Conduct a before-tax analysis with inflation by correctly incorporating inflationary effects in both the cash flows and the discount rate. (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 5 ir 1 f 1 ir 1 f 2

(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 5 MARRr 1 f 1 MARRr 1 f 2

(10.3)

Two key equations relating constant-dollars ($Ck ) and then-current dollars ($Tk ) at the end of year k are:

and

$Tk 5 $Ck 11 1 f 2 k

(10.6)

$Ck 5 $Tk 11 1 f 2 2k

(10.8)

3. Learning Objective: Conduct an after-tax analysis with inflation by correctly adjusting for the situation that in the U.S. depreciation allowances are not permitted to increase 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. 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

Summary 375

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.

KEY TERMS Combined Rate, p. 365 Constant Dollars, p. 364 Consumer Price Index (CPI), p. 361 Higher Education Price Index (HEPI), p. 364

Inflation, p. 360 Inflation Rate, p. 365 Producer Price Index (PPI), p. 364 Real Interest Rate, p. 365 Then-Current Dollars, p. 364

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1.

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 c. j 1 d b. d d. j 1 d 1 dj

2.

Mike’s Veneer Shop owns a vacuum press that requires annual maintenance. Mike has a contract to cover the maintenance expenses for the next five 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

3.

Mike’s Veneer Shop owns a vacuum press that requires annual maintenance. Mike has a contract to cover the maintenance expenses for the next

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five 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 4.

An economist has predicted that there will be a 7% per year inflation of prices during the next ten years. If this prediction proves to be correct, an item that presently sells for $10 would sell for what price in ten years? a. $5.08 c. $17.00 b. $10.70 d. $19.67

5.

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% c. 7.0% b. 17.7% d. 10.7%

6.

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 constant-worth dollars? a. 10.0% c. 7.0% b. 17.7% d. 10.7%

7.

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

8.

Ten years ago Jennifer bought an investment property for $100,000. Over the ten 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

9.

As reported by the Bureau of Labor Statistics, the CPI for 2005 was 585.0 (using a Base Year of 1967 5 100). The CPI for 2006 was 603.9. Based on this data, what was the inflation rate for 2006? a. 3.23% c. 6.04% b. 5.85% d. 18.9%

Summary 377

PROBLEMS Section 10.1 The Meaning and Measure of Inflation 1. What is a good working definition of “inflation” in 10 words or less? 2. Give four examples of goods or services that have exhibited inflation in recent

years. 3. What is the relationship between “inflation” and “deflation”? Give an example

of deflation experienced in your everyday life. 4. Give two examples of goods or services that you have seen inflate dramati-

cally and also deflate dramatically over the past few years. 5. What is meant by a “market basket rate?” 6. What are the differences between the Consumer Price Index and the Producer

Price Index? 7. 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? 8. 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. 9.

The CPI-U (U.S. city average, all items) has the following annual averages: Year

Index

2002 2003 2004 2005 2006

179.9 184.0 188.9 195.3 201.6

a. For each year from 2003 to 2006 determine the annual inflation rate in

percent to two decimal places. b. Since inflation, like interest, is compounded from period to period (e.g.,

year to year), estimate the overall annual inflation rate per year from 2002 to 2006. Suggestion! Do not simply average the rates of part (a). 10. 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 %

2001 2002 2003 2004 2005

1.6 2.4 1.9 3.3 3.4

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a. Assuming the index value in year 2000 was 100.0, determine the index for

each year from 2001 to 2005 to one place after the decimal. b. Since inflation, like interest, is compounded from period to period (e.g.,

year to year), estimate the overall annual inflation rate per year from 2002 to 2006. Suggestion! Do not simply average the rates given above. 11.

The Korean manufacturing output per hour index is given as follows for years 2001–2005 as follows: Year

Index

2001 2002 2003 2004 2005

214.8 235.8 252.2 281.2 305.1

a. For each year from 2002 to 2005 determine the rate of increase in Korean

manufacturing output per hour to two decimal places. b. Since 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 2002 to 2005. Suggestion! Do not simply average the rates of part (a). 12. The Korean hourly compensation rates of increase are presented as follows: Year

Rate %

2001 2002 2003 2004 2005

–14.84 13.75 10.09 18.87 18.79

a. Assuming the index value in year 2000 was 165.9, determine the index for

each year from 2001 to 2005 to one place after the decimal. b. Since 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 2001 to 2005. Suggestion! Do not simply average the rates given above. Section 10.2 13.

Before-Tax Analysis

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?

14. 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? 15.

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.

Summary 379

16. 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? 17.

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 percent, determine the combined rate of interest you must earn.

18. 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%. 19.

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 five years using: a. Then-current dollars. b. Constant-worth dollars.

20. You invested $10,000 on January 1, 2004, at 7% interest compounded annu-

ally. You have not touched the investment since that date. You are planning to take your money and close out the investment on January 1, 2014. 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? 21.

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.

22. 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 three 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? 23.

Video Solution 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, 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.

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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. 24. 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. 25.

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 23%?

26. A 24-year old December 2012 graduate has decided he wants the equivalent of

$2,500,000 in January 1, 2013 buying power to be available exactly 40 years later on January 1, 2053. He plans to make his first investment of $Z on January 1, 2014 and every year thereafter, with the last payment of $Z on January 1, 2053. 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? 27.

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.

28. 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%/year over the award period and the winner’s TVOM (real interest rate) is 3.5%/year.

Summary 381

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? 29. A family wishes 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. The plan to make investments on the child’s birthday (USA-style—first birthday is one year after date of birth), every year from age 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 they put up? 30. Labor costs over a 4-year period have been forecast in then-current 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. 31. 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 then-current and constant dollars. 32. 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? 33. Mellin Transformers, Inc. uses a required return of 15% in all economic jus-

tifications. Inflation is anticipated to be 5% over the foreseeable future. What real discount rate are they implicitly using? 34. 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 35. 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, $15,000, $2,000, and $1,000 in then-current dollars, during those years, determine the present worth equivalent for labor cost.

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36. 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 percent 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 percent, and a real return of 3.6 percent is desired. What is the present worth of this project using a. a then-current analysis? b. a constant-worth analysis? 37. 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. 38. $90,000 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 equales 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. 39. A rental car agency rents cars for an average price of $45/day. The num-

ber of cars owned in 2010 totals 150. The number of days rented per car in 2010 equals 200. Hence, the rental revenue in 2010 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 2015 if the agency’s real MARR is 10%? (Allow for a fractional valued number of cars.) 40. In 2005, Motorola introduced a new smart phone on the market at a retail

price of $250; the cost of producing the phone was $50/unit. During 2005, 5 million smart phones were sold. The number of smart phones sold increased at an annual rate of 10% through 2009; in 2010, Apple introduced a new product that caused the demand for the Motorola smart phone to equal one-half the number sold in 2009. Over the 5-year period (2006, 2007, 2008, 2009, and 2010), the retail price for the Motorola smart phone (measured

Summary 383

in then-current dollars) decreased at an annual rate of 20% and the cost of producing the smart phone (measured in constant or 2005 dollars) decreased at an annual rate of 25%. Over the 5-year period, annual inflation was 2%. Motorola has a real required return on investment of 10%. What was the cumulative present worth of the profits for the Motorola smart phone for the years 2005 through 2010? 41. 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%. 42. 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. 43. True or False: if the real MARR is 10% and inflation is 4%, then an invest-

ment that yields a combined external rate of return of 13.5% is not justified economically. Section 10.3 After-Tax Analysis 44. An economic analysis is being performed in real (not actual) dollars. The com-

pany’s combined MARR is 10 percent, and the inflation rate is 4 percent. The asset has a first cost of $10,000. It will be depreciated as MACRS 3-year property using rates of 33.33 percent, 44.45 percent, 14.81 percent, and 7.41 percent. What depreciation amount will be shown in year 3 of the analysis? 45. 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 40 percent, the real MARR is 18 percent, and inflation is 4 percent. 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. 46. Reconsider Problem 45 exactly as written. a. b. c. d. e. f. g. h.

Determine the real after-tax cash flows for each year. Determine the PW of the after-tax cash flows. Determine the AW of the after-tax cash flows. Determine the FW of the after-tax cash flows. Determine the real IRR of the after-tax cash flows. Determine the real ERR of the after-tax cash flows. Determine the combined IRR of the after-tax cash flows. Determine the combined ERR of the after-tax cash flows.

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47.

Video Solution 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 real dollars. Depreciation follows MACRS 7-year property, taxes are 40 percent, the real after-tax MARR is 10 percent, and inflation is 3.9 percent. 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.

48. Reconsider Problem 47 exactly as written. a. b. c. d. e. f. g. h. 49.

Determine the real after-tax cash flows for each year. Determine the PW of the after-tax cash flows. Determine the AW of the after-tax cash flows. Determine the FW of the after-tax cash flows. Determine the real IRR of the after-tax cash flows. Determine the real ERR of the after-tax cash flows. Determine the combined IRR of the after-tax cash flows. Determine the combined ERR of the after-tax cash flows.

GO Tutorial 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 5-year property, taxes are 40 percent, the real after-tax MARR is 10 percent, and inflation is 4.2 percent. 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.

50. Reconsider Problem 49 exactly as written. a. b. c. d. e. f.

Determine the real after-tax cash flows for each year. Determine the PW of the after-tax cash flows. Determine the AW of the after-tax cash flows. Determine the FW of the after-tax cash flows. Determine the real IRR of the after-tax cash flows. Determine the real ERR of the after-tax cash flows.

Summary 385

g. Determine the combined IRR of the after-tax cash flows. h. Determine the combined ERR of the after-tax cash flows. 51. Henredon purchases a high-precision programmable router for shaping fur-

niture 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 40 percent, the actual after-tax MARR is 14.62 percent, and inflation is 4.2 percent. 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. 52. Reconsider Problem 51 exactly as written. a. b. c. d. e. f. g. h.

Determine the actual after-tax cash flows for each year. Determine the PW of the after-tax cash flows. Determine the AW of the after-tax cash flows. Determine the FW of the after-tax cash flows. Determine the combined IRR of the after-tax cash flows. Determine the combined ERR of the after-tax cash flows. Determine the real IRR of the after-tax cash flows. Determine the real ERR of the after-tax cash flows.

53. 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 $100,000 salvage value at the end of the 4-year planning horizon. A 40% 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. 54. An investment of $600,000 is made in equipment that qualifies as 3-year equip-

ment 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 40% 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. 55. 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

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salvage value occurs at the end of the 5-year planning horizon. A 40% 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. 56. 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 40% 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. 57. 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. A 40% 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. 58. 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 then-current salvage value occurs at the end of the 5-year planning horizon. A 40% 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. Section 10.4 After-Tax Analysis with Borrowed Capital 59. 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 percent. Inflation is running 3.8 percent. Her combined MARR is 9 percent. 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 then-current 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)?

Summary 387

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)? 60. For the situation stated in Problem 59, let the interest on borrowed money

go from 5 percent to 15 percent in 1 percent increments. For each borrowing rate, which payment plan is preferred? Do your answers match what is predicted in the text? 61. 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 percent, and the loan is available to them at a rate of 10 percent. Their combined MARR is 17 percent. 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)? 62. For the situation stated in Problem 61, let the interest on borrowed money

go from 7 percent to 20 percent in 1 percent increments. For each borrowing rate, which payment plan is preferred? Do your answers match what is predicted in the text? 63. 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 percent 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 dollars. Depreciation follows MACRS 5-year property, taxes are 40 percent, the real MARR is 18 percent, and inflation is 4 percent. 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. 64. Reconsider Problem 63 exactly as written. a. Determine the real after-tax cash flows for each year. b. Determine the PW of the after-tax cash flows.

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c. d. e. f. g. h. 65.

Determine the AW of the after-tax cash flows. Determine the FW of the after-tax cash flows. Determine the real IRR of the after-tax cash flows. Determine the real ERR of the after-tax cash flows. Determine the combined IRR of the after-tax cash flows. Determine the combined ERR of the after-tax cash flows.

Video Solution 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. Henredon will borrow $100,000 at 13 percent over 6 years, paying only interest each year and paying all the principal in the sixth year. 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 40 percent, the real after-tax MARR is 10 percent, and inflation is 3.9 percent. a. b. c. d. e. f. g. h.

Determine the actual after-tax cash flows for each year. Determine the PW of the after-tax cash flows. Determine the AW of the after-tax cash flows. Determine the FW of the after-tax cash flows. Determine the combined IRR of the after-tax cash flows. Determine the combined ERR of the after-tax cash flows. Determine the real IRR of the after-tax cash flows. Determine the real ERR of the after-tax cash flows.

66. Reconsider Problem 65 exactly as written. a. b. c. d. e. f. g. h. 67.

Determine the real after-tax cash flows for each year. Determine the PW of the after-tax cash flows. Determine the AW of the after-tax cash flows. Determine the FW of the after-tax cash flows. Determine the real IRR of the after-tax cash flows. Determine the real ERR of the after-tax cash flows. Determine the combined IRR of the after-tax cash flows. Determine the combined ERR of the after-tax cash flows.

GO Tutorial 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 percent 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 40 percent, the real after-tax MARR is 10 percent, and inflation is 4.2 percent. 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 discussed in the chapter: a. Plan 1 d. Plan 4 b. Plan 2 e. Which is the preferred plan c. Plan 3 for Raytheon?

Summary 389

68. 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 percent 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 40 percent, the real MARR is 18 percent, and inflation is 4 percent. 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 discussed in the chapter: a. Plan 1 d. Plan 4 b. Plan 2 e. Which is the preferred plan c. Plan 3 for Electronic Games? 69. 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 40% tax rate applies. Inflation is 4%/yr. The real ATMARR is 8%. Among the four payment plans discussed 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. 70. 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. $200,000 of the investment capital is obtained by borrowing money at an annual compound interest rate of 18% and the loan is repaid over a 5-year period. A 40% tax rate and 3% inflation rate apply. The ATMARR, is 7%. Among the four payment plans discussed in the chapter, use the one that maximizes the after-tax present worth and determine the after-tax present worth for the investment. 71. 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. $400,000 of the investment capital is obtained by borrowing at an annual compound interest rate of 10% and repaying the loan over a 5-year period. A 40% tax rate and 3% inflation rate apply. The real ATMARR is 10%. Among the four payment plans discussed 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.

ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : B P 2010 was a difficult year for BP, one of the world’s largest oil and gas companies. The Gulf of Mexico oil spill in April claimed eleven lives, injured multiple people, affected the incomes of thousands, and damaged the environment immeasurably. Carl-Henric Svanberg, BP’s Chairman, wrote in the 2010 Annual Report, “In the days after the accident in the Gulf of Mexico the company faced a complex and fast-changing crisis. With oil escaping into the ocean, uncertainty grew around our ability to seal the well and restore the areas affected. This was an intense period, with the situation worsening almost daily. Our meeting with President Obama on 16 June 2010 provided reassurance to the U.S. government that BP would do the right thing in the Gulf, and this marked a turning point. Through diligence and invention, our teams stopped the flow of oil in July and completed relief-well operations in September.” Subsequent investigations concluded that what happened in the Gulf of Mexico was a complex accident involving multiple causes and multiple parties. In financial terms, BP paid out over $26 billion in 2010 and 2011 in pre-tax dollars to cover oil spill response costs, meet claims and litigation expenses, support research, promote tourism, and help restore the environment. 2011 was a year of recovery, consolidation, and change for BP. In 2011, the board outlined three priorities for the company related to putting safety first, regaining trust and creating value. The company developed a 10-point strategic plan to support these priorities. The first point, as outlined in their 2011 Annual Report, states, “We are determined that BP will deliver world-class performance in safety, risk management and operational discipline. We will be a company that systematically applies our global standards as a single team.” As successful as BP had been until April of 2010, it cannot accurately predict what the future holds for it and its product portfolio. The uncertainty is due mainly to a host of exogenous factors, such as political instability in the Middle East, weather, earthquakes, the overall economy, and environmental 390

BREAK-EVEN, SENSITIVITY, AND RISK ANALYSIS

concerns, among others. BP is not alone in this regard. No business or government is immune to the impact of unknown and unknowable factors on economic outcomes of capital investments. In fact, the Form 10-K filings for publicly traded companies include extensive lists of risks that can significantly affect the profitability of the companies.

DISCUSSION QUESTIONS: 1. Is the risk experienced by BP limited to financial risk? If not, what other risks are present?

2. Is the exposure to risk at BP limited to their shareholders and employees? If not, who else is at risk?

3. What are some specific steps that BP can take to limit their risks? 4. How might sensitivity analysis be used to mitigate the risk faced by BP? 5. Given the complex investment decisions made by a company such as BP, 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?

LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

1. Calculate the break-even value using a break-even analysis approach. (Section 11.1)

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) 391

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3. Perform risk analysis using analytical solutions to derive exact values or estimates for the expected value and standard deviation of the measure of economic worth. (Section 11.3.1)

4. Perform risk analysis using simulation solutions to derive exact values or estimates for the expected value and standard deviation of the measure of economic worth. (Section 11.3.2)

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 Sandy, which devastated parts of the Caribbean and northeastern and midAtlantic portions of the United States in late October 2012; 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 sixth step in the systematic economic analysis technique: perform supplementary analyses.

Systematic Economic Analysis Technique 1. 2. 3. 4. 5. 6. 7.

Identify the investment alternatives Define the planning horizon Specify the discount rate Estimate the cash flows Compare the alternatives Perform supplementary analyses Select the preferred investment

11-1

Break-Even Analysis

393

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 O BJECTI VE: Calculate the break-even value using a break-even

analysis approach.

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 breakeven 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. 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

Video Lesson: Break-Even Analysis 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.

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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.) 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

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

$17.50/unit

Equipment cost

$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 percent

11-1

Break-Even Analysis

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? Given: Product cost parameters as defined in the chart above

KEY DATA

Find: Break-even value of units sold annually (X units/year) 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:

SOLUTION

AW115%2 5 2$250,0001A Z P 15%,52 2 $50,000 2 $150,000 2 0.11$252X 2 $0.50X 1 $17.50X 5 2$274,578.89 1 $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.

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.

EXPLORING THE SOLUTION

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$600,000

$500,000

Profit region $400,000 Annual cost $300,000 Loss region $200,000

Annual revenue $100,000 Break-Even point 18,937 units $0 0

FIGURE 11.1

3,000

6,000

9,000

12,000 15,000 18,000 Annual Sales (units)

21,000 24,000 27,000

30,000

Break-Even Analysis for the Annual Sales of a New Product

Break-Even Analysis with Excel for the SMP Investment

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 percent 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 follow-

ing annual worth relationship must hold: $500,000 1A Z P 10%,102 5 X 1 $50,000 1A Z F 10%,102

11-1

or

X 5 $500,000 1A Z P 10%,102 2 $50,000 1A Z F 10%,102

Using the Excel® PMT function, X 5 PMT110%,10,2500000,500002 5 $78,235.43 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 $500,0001A Z P 10%,n2 5 $92,500 1 1$500,000 2 $45,000n2 1A Z F 10%,n2. 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.

Using the Excel® SOLVER Tool to Determine the Break-Even Value for the Planning Horizon

FIGURE 11.2

Break-Even Analysis

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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 5 RATE110,,2500000,500002 5 220.5672% Therefore,

$500,0001A Z P 10%,n2 5 $92,500 1 $500,00010.794328n 2 1A Z F 10%,n2

As shown in Figure 11.2, using the Excel® SOLVER tool yields a value of 6.95 years for the break-even 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 5 NPER192500,2500000,500002 5 7.57668 years Likewise, if the salvage value is negligible, regardless of investment duration, the break-even value of n is given by n 5 NPER192500,25000002 5 8.15972 years

11-2 SENSITIVITY ANALYSIS LEARN I N G O B JEC T I V E : Perform a sensitivity analysis to examine the impact

Video Lesson: Sensitivity Analysis 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.

on the measure of economic worth and the robustness of the economic decision when values of one or more parameters vary over specified ranges.

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. 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;

11-2









Sensitivity Analysis

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.

Sensitivity Analysis for a Single Alternative 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 percent 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 650 percent for each parameter, what is the impact on AW? If it is assumed that all estimates are correct except that for annual savings, the alternative’s annual worth can be given as AW110%2 5 2$500,0001A Z P 10%,102 1 $50,0001A Z F 10%,102 1 $92,50011 1 X2 where X denotes the percent error (decimal equivalent) in estimating the value for annual savings. Plotting annual worth as a function of the

EXAMPLE

Video Example

SOLUTION

399

400 Chapter 11

Break-Even, Sensitivity, and Risk Analysis

$70,000

$50,000

Initial investment

$40,000

Annual Worth

$60,000 Annual savings

$30,000 $20,000 $10,000

–50%

–40%

Planning horizon

–30%

–20%

–10%

$0 0%

MARR 10%

20%

Salvage value

30%

40%

50%

–$10,000 –$20,000 –$30,000 –$40,000

FIGURE 11.3

Sensitivity Analysis for Example 11.3

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. (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

11-2 Sensitivity Analysis

savings, and the planning horizon. It is insensitive to changes in salvage value and is moderately sensitive to changes in the MARR. 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, 2$177,340.55 for the salvage value, 7.5767 years for the investment’s duration, and 13.8 percent 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 break-even value for the investment duration, and the internal rate of return (IRR) is the breakeven value for the MARR.)

Sensitivity Analysis for Multiple Alternatives

EXPLORING THE SOLUTION

EXAMPLE

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 percent 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. Given: The cash flows outlined in Figure 11.4; MARR 5 10%; planning horizon 5 10 years Find: Sensitivity of design choice to estimation errors in annual revenue

KEY DATA

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:

SOLUTION

PWA 5 2$300,000 1 $55,00011 1 x2 1P Z A 10%,102 and PWB 5 2$450,000 1 $80,00011 1 y2 1P Z A 10%,102

401

402

Chapter 11

Break-Even, Sensitivity, and Risk Analysis

$55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000

(+) 0

1

2

3

4

5

6

7

8

9

10

(–)

$300,000

Alternative A

$80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000

(+) 0

1

3

2

4

5

6

7

8

9

10

(–)

$450,000

Alternative B

FIGURE 11.4

CFDs for Example 11.4

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 2$300,000 1 $55,000(1 1 x)(P Z A 10%,10) . 2$450,000 1 $80,000(1 1 y)(P Z A 10%,10) or $3,614.18 2 $337,951.20x 1 $491,565.00y , $0 Solving for y, y , 20.007352385 1 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

11-2

Sensitivity Analysis

Estimation errors for Design B

40%

Choose Design B

–50%

–40%

–30%

–20%

–10%

30%

20%

10%

0% 0%

10% 20% 30% 40% Estimation errors for Design A

–10%

–20%

–30%

Choose Design A

–40% FIGURE 11.5

Sensitivity Analysis for Example 11.4

preferred for error combinations above the indifference diagonal. If the estimates are correct for Design B, an increase of 1.07 percent 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 percent 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.

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.

50%

403

404

Chapter 11

Break-Even, Sensitivity, and Risk Analysis

Multiparameter Sensitivity Analysis

EXAMPLE

For Example 11.4, 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? 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.

TABLE 11.1 Optimistic, Pessimistic, and Most Likely Estimates for Four Scream Machine Parameters Parameter

Pessimistic

Most Likely

Initial Investment (A)

Optimistic $285,000

$310,000

$300,000

Initial Investment (B)

$400,000

$510,000

$450,000

Annual Revenue (A)

$65,000

$40,000

$55,000

Annual Revenue (B)

$85,000

$70,000

$80,000

PWA

$114,396.86

2$64,217.32

$37,951.19

PWB

$122,288.20

2$79,880.30

$41,565.37

It seems unlikely that pessimistic estimates will occur for all four parameters and for both designs. Likewise, it is unlikely that the optimistic estimates will occur for all four parameters for both designs. And, unfortunately, neither is it likely that the most likely estimates will be realized for each of the eight parameters (four for each design). Instead, combinations of pessimistic, optimistic, and most likely values generally will occur. There are 81 possible combinations of the four parameters and three estimates (34 5 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. Some might conclude that Design A is best because more than 50 percent 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

11-3 Risk Analysis

405

parameters and conduct a risk analysis. This process is demonstrated in Example 11.7 of the next section.

FIGURE 11.6

Considering 81 Possible Combinations of 3 Estimates and 4 Parameters

11-3 RISK ANALYSIS 1 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. 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 1

A basic understanding of probability theory and Monte Carlo simulation is assumed for this section.

Risk Analysis The process of incorporating explicitly random variation in one or more parameters.

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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. 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

0.135% 2.145%

13.59%

34.13%

34.13%

68.26% 95.45% 99.73% FIGURE 11. 7

Normal Distribution

13.59%

2.145%

11-3

FIGURE 11.8

Risk Analysis

407

Sample Beta Distributions

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. 11.3.1 Analytical Solutions LEARNING O BJECTI VE: Perform risk analysis using analytical solutions to

derive exact values or estimates for the expected value and standard deviation of the measure of economic worth.

To illustrate the use of an analytical approach to develop the probability distribution for present worth, consider the following present worth formulation: n

PW 5 a At 11 1 i2 2t

(11.1)

t50

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 n

E1PW2 5 a E 3At 11 1 i2 2t 4 t50

(11.2)

Autocorrelated Correlated with itself over time.

408

Chapter 11

Break-Even, Sensitivity, and Risk Analysis

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

E1PW2 5 a E1At 2 11 1 i2 2t

(11.3)

t50

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

V1PW2 5 a V1At 2 11 1 i2 22t

(11.4)

t50

When the annual cash flows are statistically independent, based on the Central Limit Theorem, we can expect the distribution of present worth to approximate a normal distribution. The Excel® NORMSDIST function can be used to approximate the probability of present worth being greater than 0. The syntax for the NORMSDIST 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 5NORMSDIST((z-m)/s) 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 5NORMSDIST((m-z)/s) in any cell. Hence, for z 5 0, Pr(PW . 0) 5NORMSDIST(E(PW)/SD(PW))

(11.5)

where E(PW) and SD(PW) denote the expected value and standard deviation of present worth, respectively.

EXAMPLE

Risk Analysis for the SMP Investment 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;

11-3

Risk Analysis

409

and 10 percent 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)

A2p(A)

SV

p(SV)

SVp(SV)

SV2p(SV)

$75,000

0.070

$5,250

393,750,000

$40,000

0.10

$4,000

160,000,000

$80,000

0.095

$7,600

608,000,000

$45,000

0.20

$9,000

405,000,000

$85,000

0.131

$11,135

946,475,000

$50,000

0.40

$20,000

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

0.183

$17,385

1,651,575,000

$60,000

0.10

$6,000

360,000,000

$100,000

0.181

$18,100

1,809,999,998

Sum

1.00

$50,000

2,530,000,000

$105,000

0.162

$17,010

1,786,050,000

Sum

1.000

$92,500

8,637,649,998

E(A) 5

$92,500

E (SV ) 5

$50,000

V (A) 5

81,400,001.7

V (SV ) 5

30,000,000

SD(A) 5

$9,022.19

S D (SV ) 5

$5,477.23

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 V1A2 5 E1A2 2 2 3E1A2 2 4 5 192,5002 2 2 8,637,649,998 5 81,400,001.7 and the variance for salvage value is V1SA2 5 E1SV2 2 2 3E1SV2 4 2 5 150,0002 2 2 2,530,000,000 5 30,000,000 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 tenth year equals the sum of the expected annual savings and the expected salvage value; likewise, the variance of the cash flow in the tenth year equals the sum of the annual savings variance and the salvage value variance.

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Break-Even, Sensitivity, and Risk Analysis

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) 5 V (A) 5 SD(A) 5 EOY(t)

$92,500 81,400,001.7 $9,022.19 E (CF)

(1.10)⫺t

E (CF) (1.10)⫺t

E(SV) 5 V (SV) 5 SD(SV) 5

$50,000 30,000,000 $5,477.23

V (CF)

(1.10)⫺2t

V (CF) (1.10)⫺2t

0

2$500,000

1.0000

2$500,000.00

0.0

1.0000

0.00

1

$92,500

0.9091

$84,090.91

81,400,001.7

0.8264

67,272,728.68

2

$92,500

0.8264

$76,446.28

81,400,001.7

0.6830

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

$92,500

0.6209

$57,435.22

81,400,001.7

0.3855

31,383,224.41

6

$92,500

0.5645

$52,213.84

81,400,001.7

0.3186

25,936,549.10

7

$92,500

0.5132

$47,467.13

81,400,001.7

0.2633

21,435,164.55

8

$92,500

0.4665

$43,151.93

81,400,001.7

0.2176

17,715,012.02

9

$92,500

0.4241

$39,229.03

81,400,001.7

0.1799

14,640,505.80

10

$142,500

0.3855

$54,939.92

111,400,001.7

0.1486

16,558,900.41

E(PW) 5

$87,649.62

V (PW) 5 SD(PW) 5 Pr(PW + 0)* 5

334,461,261.82 $18,288.28 0.999999178

*Central Limit Theorem approximation

EXAMPLE

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 percent 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?

11-3

Given: MARR 5 10%; planning horizon 5 10 years

Risk Analysis

411

KEY DATA

Design A: Initial cost 5 $300,000; Annual Revenue is normally distributed with mean 5 $55,000 & standard deviation 5 $5,000 Design B: Initial cost 5 $450,000; Annual Revenue is normally distributed with mean 5 $80,000 & standard deviation 5 $7,500 Find: Pr[PW(A) . 0], Pr[PW(B) . 0], Pr[PW(B) . PW(A)] 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 2 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 2 A) . 0].

TABLE 11.4

EOY(t)

SOLUTION

Risk Analysis for the Selection of the Scream Machine Design (1.10)⫺2t

V (At) (1.10)⫺2t

V (At)

V (Bt) (1.10)⫺2t

V (Bt)

V (Bt 2 At)

V (Bt 2 At) (1.10)⫺2t

0

1.0000

0

0.0000

0

0.0000

1

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

0.5645

25,000,000

14111848.2513

56,250,000

31751658.5655

81,250,000

45863506.8169

4

0.4665

25,000,000

11662684.5052

56,250,000

26241040.1368

81,250,000

37903724.6420

5

0.3855

25,000,000

9638582.2357

56,250,000

21686810.0304

81,250,000

31325392.2661

6

0.3186

25,000,000

7965770.4428

56,250,000

17922983.4962

81,250,000

25888753.9390

7

0.2633

25,000,000

6583281.3577

56,250,000

14812383.0547

81,250,000

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

0.1486

25,000,000

3716090.7006

56,250,000

8361204.0764

81,250,000

10

V [PW(A)] 5 SD[PW(A)] 5 Pr[PW(A) + 0]14 5

101351949.0447 $10,067.37 0.999918 V [PW(B)] 5 SD[PW(B)] 5 Pr[PW(B) + 0]* 5

*Central Limit Theorem approximations

V [PW(B 2 A)] 5 SD[PW(B 2 A)] 5 Pr[PW(B 2 A) + 0]* 5 228041885.3507 $15,101.06 0.997043

12077294.7770 329393834.3954 $18,149.21 0.578922

412

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Break-Even, Sensitivity, and Risk Analysis

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 percent chance that Design B is best and a 42 percent chance that Design A is best. The probabilities were calculated using the Excel® NORMSDIST function as follows: Pr3PW1A2 . 0 4 5 NORMSDIST137951.19/10067.372 5 0.999918 Pr3PW1B2 . 04 5 NORMSDIST141565.37/15101.062 5 0.997043 Pr3PW1B 2 A2 . 0 4 5 NORMSDIST13614.18/18149.212 5 0.578922

11.3.2

Simulation Solutions2

LEARN I N G O B JEC T I V E : Perform risk analysis using simulation solutions to derive exact values or estimates for the expected value and standard deviation of the measure of economic worth.

Some of the major reasons for using simulation in risk analysis include the following: 1. 2. 3. 4. 5.

2

Except for the simplest problems, analytical solutions are difficult to obtain. Simulation is useful in selling a system modification to management. Simulation can be used as a verification of analytical solutions. Simulation is very versatile. Less background in mathematical analysis and probability theory is generally required.

We assume the reader is familiar with Monte Carlo simulation. 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.

11-3 Risk Analysis

Some of the major disadvantages of simulation are the following: 1. 2. 3. 4. 5.

Simulation can be quite time-consuming to formulate. Simulations introduce a source of randomness not present in analytical solutions (sampling error). Monte Carlo simulations do not reproduce the input distribution exactly (especially the tails of the distribution).3 Validation of the simulation model is easily overlooked. 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 4.5 simulation software in performing engineering economic analyses.

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.

3

Latin hypercube simulation, included in Pallisade Corporation’s @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.

EXAMPLE

413

Chapter 11

Break-Even, Sensitivity, and Risk Analysis

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

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. This should not be surprising, because the expected number of instances of present worth being less than 0 is (0.00000082)(100,000), or 0.082. The histogram obtained from the Monte Carlo simulation is provided in Figure 11.9.4 Distribution for PW(SMP)/B13 2.500 Values in 10^ -5

414

Mean=87677.55

2.000 1.500 1.000 0.500 0.000

0

40 5%

80

120

160

Values in Thousands 90% 5% 56.9239 117.3692

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.)

FIGURE 11. 9

4

Many of the figures and tables in this section were generated with the help of @RISK 4.5, a software product of Palisade Corporation, Ithaca, NY: www.palisade.com, or call 800-432RISK (7475).

11-3 Risk Analysis

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) 5 Pr(PW , 0), we already know how many times IRR is less than 10 percent during the 100,000 simulation trials—zero!) The Monte Carlo simulation yielded an estimate of 13.80126 percent 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 percent. 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. Distribution for IRR(SMP)/B14 50 45 40 35 30 25 20 15 10 5 0 0.1

Mean=0.1380126

0.1175 5%

0.135

0.1525 5%

90% .1247

0.17

.1508

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.)

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

EXAMPLE

415

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Chapter 11

Break-Even, Sensitivity, and Risk Analysis

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 2 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. Recall, solving analytically for the mean and variance for PW(A), PW(B), and PW(B 2 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. 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.

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.)

FIGURE 11.11

11-3

Risk Analysis

Values in 10^ –5

Distribution for PW(A)/B13 4.000 3.500 3.000 2.500 2.000 1.500 1.000 0.500 0.000 –10

Mean=37963.06

15

40

Values in Thousands 5% 90% 21.3354

65

90 5%

54.4994

PW Histogram from 100,000 Simulated Investments in Design A

F I G U RE 11. 1 2 a

Distribution for PW(B)/C13 3.000

Mean=41564.11

Values in 10^ –5

2.500 2.000 1.500 1.000 0.500 0.000 –40

0

40

80

Values in Thousands 5% 90% 16.9294

66.3692

120 5%

PW Histogram from 100,000 Simulated Investments in Design B

F I G U RE 11. 1 2 b

Distribution for PW(B – A)/D13 2.500

Mean=3601.05

Values in 10^ –5

2.000 1.500

0.500 0.000 –80

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.)

F I G U RE 11. 1 2 c

1.000

–60

–40

–20

0

Values in Thousands 90% 5% –26.1902

20

40

33.4125

60 5%

80

100

417

418

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Break-Even, Sensitivity, and Risk Analysis

The Monte Carlo simulation results produced slightly smaller estimates for the probabilities of a positive-valued 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.

SUMMARY

KEY CONCEPTS 1. Learning Objective: 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, break-even analysis is often used to determine the level of annual savings required to justify a particular capital investment. 2. 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. (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. 3. Learning Objective: Perform risk analysis using analytical solutions to derive exact values or estimates for the expected value and standard deviation of the measure of economic worth. (Section 11.3.1)

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

Summary 419

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. 4. Learning Objective: Perform risk analysis using simulation solutions to derive exact values or estimates for the expected value and standard deviation of the measure of economic worth. (Section 11.3.2)

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. KEY TERMS Autocorrelated, p. 407 Break-Even Analysis, p. 393 Break-Even Value, p. 393

Risk Analysis, p. 405 Sensitivity Analysis, p. 398

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1.

If the total cost for producing widgets can be represented by TC 5 8,000 1 0.75*X, where X is the number of widgets produced and total revenue can be represented by TR 5 4.00*x, what is the break-even value for number of widgets produced? a. 1,684 c. 2,462 b. 2,000 d. 3,763

Chapter 11

Break-Even, Sensitivity, and Risk Analysis

The next five questions refer to the following sensitivity graph:

PW

420

–30

–20

–10

3500 3000 2500 2000 1500 1000 500 0 –500 0 –1000 –1500

10

20

30

Percent Change Annual Revenue

Initial Investment

Salvage Value

a. b. c. d.

The analysis is most sensitive to changes in which component? Annual Revenue Initial Investment Salvage Value Cannot be determined from the information given

a. b. c. d.

The analysis is least sensitive to changes in which component? Annual Revenue Initial Investment Salvage Value Cannot be determined from the information given

2.

3.

4.

What is the numeric value of the present worth of the original project (i.e., no changes)? a. 210 b. 20 c. 1,000 d. Cannot be determined from the information given

5.

What percentage change in initial investment would cause the project to become unattractive? a. b. c. d.

6.

210 120 11,000 Cannot be determined from the information given

If a line for “annual expenses” was to be 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)

Summary 421

7.

Which of the following is not a method typically used for supplementary analysis of engineering economy problems? a. break-even analysis c. risk analysis b. depreciation analysis d. sensitivity analysis

8.

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 five years? Value of Crop Damage

Probability

$0 $100,000 $200,000 $300,000

a. $57,000 b. $167,000

60% 25% 13% 2%

c. $228,000 d. $285,000

9.

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 c. 4,667 b. 3,500 d. 7,000

10.

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 c. 1.00 b. 0.50 d. 1.50

PROBLEMS Introduction 1. 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

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2. Match the terms in the first column with an appropriate definition from the

second column. Terms

Definitions

(a) break-even analysis

(b) sensitivity analysis

(c) risk analysis

Section 11.1 3.

(1) Determining how the worth of a investment changes with changes in one or more parameters (2) Determining probabilistic statements about the worth of an investment based on probabilistic values assigned to one or more parameters (3) Determining the indifference value of a particular parameter

Break-Even Analysis

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.

4. Reconsider Problem 3. 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. 5.

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 5 years $0 $6.00

$10,000 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 alter-

native is preferred and what will be the total annual cost? 6. A manufacturer offers an inventor the choice of two contracts for the exclu-

sive 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.

Summary 423

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? 7.

A pipeline contractor can purchase a needed truck for $40,000. Its estimated life is six 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.

8.

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?

9.

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?

10. The Cooper Company is considering investing in a recuperator. The recu-

perator 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.

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.

12. A small utility company is considering the purchase of a power driven post

hole digger. The equipment would 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

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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%. 13.

GO Tutorial 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.

14. 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? 15.

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 units sold per day for each of the following values of MARR. a. MARR is 5% b. MARR is 10% c. MARR is 15%

16. 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 break-even sales volume? 17.

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

Summary 425

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 CT 1 CT 2 CT 3

Fixed Cost per Order

Variable Cost per Unit

$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? 18. Two 100-horsepower motors are under consideration by the Mighty Machin-

ery Company. Motor Q costs $5000 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 5 1 horsepower.) 19. 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 breakeven value for annual sales if the selling price per unit is (a) $0.40, (b) $0.30, and (c) $0.20. 20. Owners of a nationwide motel chain are considering locating a new mo-

tel 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 15-year 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.) 21. 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

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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.) 22. 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 2 3 4 5 6

$230,000 185,000 145,000 110,000 80,000 55,000

$300,000 245,000 200,000 160,000 125,000 95,000

23. 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%. 24. 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 5 1 horsepower.)

Summary 427

25. A machine can be purchased at t 5 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 5 10%, what must the average annual revenues be in order to be indifferent between (a) purchasing the machine, or (b) doing nothing? 26. Two condensers are being considered by the Ajax Company. A copper con-

denser 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 27.

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 tradeoffs 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.

28. 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 seven 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: 280%, 260%, 240%, 220%, 120%, 140%. c. Determine the percentage change in net annual savings that causes a reversal in the decision regarding the attractiveness of the project. 29. Plot a sensitivity graph for annual worth versus initial cost, annual revenue,

and salvage value for the data in the table on the next page. Vary only one

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parameter at a time, each within the range of 220% to 120%. MARR is 3%/yr. Project life is 4 years. Initial Cost Annual Revenue Salvage Value

$120,000 $25,000 $35,000

Based on your graph, which parameter shows the most sensitivity? the least? 30.

Video Solution A new project will cost $80,000 initially and will last for seven 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 1/2 50%.

31. In problem 20 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.

Building cost Furnishings cost Annual operating and maintenance costs

Pessimistic

Most Likely

Optimistic

$7,500,000

$5,000,000

$4,000,000

3,000,000

1,875,000

1,000,000

200,000

125,000

75,000

35/day

55/day

75/day

Average rate

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. 32. 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 220% to 120%. MARR is 20%/yr. Project life is 10 years. Initial Cost Annual Revenue Salvage Value

$800,000 $330,000 $130,000

Based on your graph, which parameter shows the most sensitivity? the least? 33. 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.

Summary 429

34.

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 Annual Revenue

$350,000

$175,000

$150,000

$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. Based on a beta approximation of expected value estimates, is the new

motel economically attractive? 35.

GO Tutorial Initial estimates of the parameters for an investment are given

below. Parameter Initial Investment Net Annual Receipts Project Life Salvage Value MARR

Initial Estimate

Sensitivity

$15,000

none

$2,500

230%, 0%, 130%

10 years

220%, 0%, 120%

$500

250%, 0%, 150%

15% per year

none

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. 36. Reconsider the data in Problem 35. 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 37. 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 not autocorrelated.

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38. 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 2 3 4 5 6 7

0.10 0.15 0.20 0.25 0.15 0.10 0.05

Analytically determine the probability of the investment being profitable. 39. In problem 38 suppose the MARR is not known with certainty and the

following probability distribution is anticipated to hold: i

p(N )

0.10 0.12 0.15

0.20 0.60 0.20

Use analytical methods to determine the probability of the investment being profitable. 40. In problem 38 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 $4,500 $4,000

0.20 0.60 0.20

Use analytical methods to determine the probability of the investment’s being profitable. 41. In problem 38 suppose the minimum attractive rate of return is distributed as

given in problem 39 and suppose the annual receipts are distributed as given in problem 40. Analytically determine the probability that the investment will be profitable. 42. In problem 41 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. 43. Suppose n 5 4, i 5 0%, a $11,000 investment is made, and the receipt in year

j, j 5 1, . . . , 4, is statistically independent and distributed as in problem 40. Determine the probability distribution for present worth.

Summary 431

44. 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(Cj )

V(Cj )

0

625 3 104

1

2$22,500 4,000

2

5,000

25 3 104

3

6,000

36 3 104

4

7,000

49 3 104

5

8,000

64 3 104

6

9,000

81 3 104

16 3 104

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. 45. Solve problem 44 using a discount rate of (a) 0%, (b) 15%. 46. 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 end-ofyear 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? 47. In problem 46, suppose the MARR were 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? 48. 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 8 10

$5,000 3,000 1,000

0.2 0.4 0.4

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49.

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) 4 5 6 7 8

Probability 0.10 0.25 0.45 0.15 0.05

For the following questions, determine 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. 50. 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 0 1 2 3 4 5

Expected Net Cash Flow

Standard Deviation of Net Cash Flow

2$19,700,000 $1,200,000 $3,600,000 $6,000,000 $9,600,000 $11,000,000

$120,000 $240,000 $650,000 $750,000 $1,080,000

$0

For the following questions, determine 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 positive present worth. 51. 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.

Summary 433

For the following questions, determine 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 positive present worth. 52. One of two mutually exclusive alternatives must be selected for implementa-

tion. 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. A A A B

Parameter Initial Cost Annual Maintenance Salvage Value End-of-Year Lease Payment

Mean

Std. Dev.

Distribution

$13,000 $5,000 $2,000 $7,500

None $500 $800 $750

None Normal Normal Normal

For the following questions, determine 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 positive present worth. 53. An investment of $5,000 is expected to generate the probabilistic returns

shown in the table below over its three year life. Assume the annual cash flows are independent and that the distribution of present worth is normal. EOY 1 2 3

Annual Return $2,500 with probability 0.4 or $1,800 with probability 0.6 $3,000 with probability 0.5 or $2,000 with probability 0.5 $3,500 with probability 0.7 or $2,500 with probability 0.3

For the following questions, determine 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. 54. A project under consideration has a 10 year projected life. The initial invest-

ment 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%.

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For the following questions, determine 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. A proposed project has the following cash flow estimates.

55.

End of Year 0 1 2 3 4 5

Mean Net Cash Flow

Standard Deviation of Cash Flow

2$32,000 $4,000 $8,000 $12,000 $12,000 $12,000

$1,000 $2,000 $3,000 $5,000 $6,000 $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, determine 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. 56. A proposed project has the following cash flow estimates.

End of Year 0 1 2

Mean Net Cash Flow

Standard Deviation of Cash Flow

2$800,000 $1,000,000 $1,000,000

$250,000 $450,000 $600,000

Assuming 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, determine 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.

Summary 435

57. A $5,000 process improvement project is expected to increase annual ex-

penses 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, determine 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.

ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : AB B OTT L AB O R ATO R I E S Founded by Dr. Wallace C. Abbott more than a century ago, incorporated in 1900, and headquartered in Abbott Park, Illinois, Abbott Laboratories is a global, broad-based health care products company with four reportable revenue segments: pharmaceutical, nutritional, diagnostic, and vascular products. Its chairman and CEO, Miles D. White, a mechanical engineering graduate from Stanford, noted, “There’s one fundamental fact that everyone at Abbott Laboratories understands: the purpose of our company is to improve lives.” 2011 marked an extraordinary year for the company, in which it announced plans to separate Abbott into two leading healthcare companies by the end of 2012. The new company is a research-based pharmaceutical company named AbbVie Inc. that will focus on branded drugs. According to a report by CNBC on November 29, 2012, “this split is meant to free Abbott from the risk and obligations of developing pharmaceutical drugs, leaving it with a more predictable business built around nutritional formula, generic drugs and heart stents.” In preparation for this split, according to its 2011 annual report, the company has recently expanded its presence in emerging markets and is working to rebuild its pharmaceutical pipeline. These strategic actions also include significant growth globally, and in 2011 the company reported that its sales outside the United States exceeded domestic sales. As of December 31, 2011, Abbott employed approximately 91,922 people around the world. Abbott owns or leases 19 plants in the United States and Puerto Rico, and has manufacturing facilities in 15 other countries including Argentina, Brazil, Canada, England, France, Germany, Ireland, Italy, Japan, Mexico, Pakistan, Singapore, Spain, Sweden, and the Netherlands. Abbott’s sales in 2011 totaled $38.9 billion, up from $35.2 billion in 2010. Capital expenditures increased from $1.02 billion in 2010 to $1.5 billion in 436

CAPITAL BUDGETING

2011 in support of upgrading and expanding manufacturing, R&D, and investments in information technology and administrative support facilities in all four business segments. As is true of health care innovators, Abbott invests heavily in R&D: $4.1 billion in 2011, up from $3.7 billion in 2010, the majority of which was concentrated in pharmaceutical products. 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 announced spin-off of the pharmaceutical portion of Abbott’s business is meant to reduce risk for investors. What other benefits might be realized?

2. Abbott invests heavily in capital. How might this be affected with the spin-off and what changes do you predict?

3. In addition to a sharp increase in capital expenditures from 2010 to 2011, we notice a similar sharp increase in R&D. 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? 437

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LEARNING OBJECTIVES When you have finished studying this chapter, you should be able to:

1. Explain the importance of capital budgeting. (Section 12.1) 2. Solve the capital budgeting problem with independent, indivisible investments as a binary linear programming problem. (Section 12.2)

3. Solve the capital budgeting problem with independent, divisible investments. (Section 12.3)

4. State some of the practical considerations in capital budgeting. (Section 12.4)

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. 2. 3. 4. 5. 6. 7.

Identify the investment alternatives Define the planning horizon Specify the discount rate Estimate the cash flows Compare the alternatives Perform supplementary analyses Select the preferred investment

12-1 THE CLASSICAL CAPITAL BUDGETING PROBLEM LEARN I N G O B JEC T I V E : Explain the importance of capital budgeting.

Our focus in this chapter is on quantitative approaches to optimize financial returns when capital is limited. Choosing from among a set of investments

12-1 The Classical Capital Budgeting Problem

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. 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. 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 debt-to-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

439

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.

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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. In this chapter, we show 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. 1.

12-2 CAPITAL BUDGETING PROBLEM WITH INDIVISIBLE INVESTMENTS LEARN I N G O B JEC T I V E : Solve the capital budgeting problem with independent, indivisible investments as a binary linear programming problem.

Indivisible Investment An investment that must be pursued entirely or not at all.

Investments are called indivisible when they must be pursued entirely or not at all. Hence, a binary decision variable (xj 5 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

12-2 Capital Budgeting Problem with Indivisible Investments

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: Maximize subject to

PW1x1 1 PW2x2 1 . . . 1 PWn21xn21 1 PWn xn

(12.1)

c1x1 1 c2x2 1 . . . 1 cn21xn21 1 cn xn ⱕ C

(12.2)

xj 5 10, 12   j 5 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 LorieSavage 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. 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.

Solving a Capital Budgeting Problem with the Excel® SOLVER Tool 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 5-year planning horizon. Two investments (5 and 6) terminate in fewer

EXAMPLE

Video Example

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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. SOLUTION

FIGURE 12.1

In Figure 12.1, for row k, k 5 2, . . . , 7, the entry in cell Hk is obtained using the Excel® worksheet function, SUMPRODUCT(Bk:Gk,B9:G9); for row k, k 5 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 column j, j 5 B, . . . , G, the entries in cells j10 and j11 equal the product of the entries in cells j 2 and j9 and cells j 8 and j 9, respectively. 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%.

SOLVER set up for Example 12.1

12-2 Capital Budgeting Problem with Indivisible Investments

FIGURE 12.2

SOLVER parameters for Example 12.1

FIGURE 12.3

SOLVER solution to Example 12.1

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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 1 xh # 1. The constraint ensures that both xg and xh cannot equal 1; 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. 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 1 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.

Incorporating Additional Constraints in the Capital Budgeting Problem with the Excel® SOLVER Tool

EXAMPLE

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

12-2 Capital Budgeting Problem with Indivisible Investments

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

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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 1/2 the original value such as 1/2 10%, 1/2 20%, etc., so that the decision maker can gain insight into the robustness of the solution.

12-3 CAPITAL BUDGETING PROBLEM WITH DIVISIBLE INVESTMENTS LEARN I N G O B JEC T I V E : Solve the capital budgeting problem with independent, divisible investments.

Divisible Investment An investment that may be pursed in a fractional portion.

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. Certain oil and gas exploration investments are often divisible, because it is not unusual for several individuals to go 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 5 1) versus not investing at all (xj 5 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 subject to

PW1 p1 1 PW2 p2 1 . . . 1 PWn21 pn21 1 PWn pn

(12.4)

c1 p1 1 c2 p2 1 . . . 1 cn21 pn21 1 cn pn ⱕ C

(12.5)

0 ⱕ pj ⱕ 1   j 5 1, . . . , n

(12.6)

Optimizing the Investment Portfolio with the Excel® SOLVER Tool When Investments Are Divisible

EXAMPLE

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

12-3 Capital Budgeting Problem with Divisible Investments

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 ,5 1 and B9:G9 .5 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

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When partial funding of investments is allowed, salvage value equals the initial investment, and annual returns are a uniform annual series, 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.

Optimizing the Investment Portfolio with the Excel® SOLVER Tool When Investments Are Divisible and Have a Special Structure

EXAMPLE

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 five 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 five 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 5 p3 5 p4 5 100%, p5 5 64.29% (45/70), and p2 5 0%. The IRR for the portfolio is $3,750 1 $9,250 1 $11,250 1 $14,250 (45/70) 5 $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 Initial Investment

1

2

3

4

5 $70,000

$15,000

$25,000

$40,000

$50,000

Annual Return

$3,750

$5,000

$9,250

$11,250

$14,250

Salvage Value

$15,000

$25,000

$40,000

$50,000

$70,000

Internal Rate of Return

25.00%

20.00%

23.13%

22.50%

20.36%

12-4

Practical Considerations in Capital Budgeting

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.

12-4 PRACTICAL CONSIDERATIONS IN CAPITAL BUDGETING LEARNING O BJECTI VE: State some of the practical considerations in capital

budgeting.

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. In closing, 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

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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.

SUMMARY

KEY CONCEPTS 1. Learning Objective: 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. 2. Learning Objective: 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. 3. Learning Objective: 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

Summary 451

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. 4. Learning Objective: State some of the practical considerations in capital budgeting. (Section 12.4)

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.

KEY TERMS CAP EX, p. 439 Capital Budgeting Problem, p. 439

Divisible Investment, p. 446 Indivisible Investment, p. 440

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1.

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

2.

Consider a capital budgeting formulation where the binary variables x1 and x2 are used to represent the acceptance (xi 5 1) or rejection (xi 5 0) of each alternative. A mutual exclusivity constraint between the two alternatives can be represented by which of the following? a. x1 1 x2 ,5 1 c. x1 1 x2 .5 1 b. x2 ,5 x1 d. x1 ,5 x2

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3.

Consider a capital budgeting formulation where the binary variables x1 and x2 are used to represent the acceptance (xi 5 1) or rejection (xi 5 0) of each alternative. The requirement that x2 is contingent upon x1 can be represented by which of the following? a. x1 1 x2 ,5 1 c. x1 1 x2 .5 1 b. x2 ,5 x1 d. x1 ,5 x2

4.

Consider a capital budgeting formulation where the binary variables x1 and x2 are used to represent the acceptance (xi 5 1) or rejection (xi 5 0) of each alternative. The requirement that the null alternative is not feasible can be represented by which of the following? a. x1 1 x2 ,5 1 c. x1 1 x2 .5 1 b. x2 ,5 x1 d. x1 ,5 x2

5.

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 c. Annual worth b. Future worth d. Internal rate of return

6.

If six investment proposals are under consideration, how many investment combinations must be evaluated if a complete enumeration approach is being used? a. 6 c. 62 5 36 b. 2*6 5 12 d. 26 5 64

7.

Which of the following is not an approach which can be used to perform a capital budgeting economic analysis? a. Box-Jenkins algorithm b. Excel® SOLVER c. Exhaustive enumeration d. Lori-Savage formulation

8.

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 c. IRR b. Initial investment d. PW

9.

Consider the following binary linear programming formulation of a capital budgeting problem. Max 1,200 x1 1 600 x2 1 950 x3 1 1,650 x4 s.t. 15,000 x1 1 20,000 x2 1 25,000 x3 1 30,000 x4 ,5 70,000 x1 1 x2 ,5 1 x4 ,5 x3 x1, x2, x3, x4 5 (0, 1) The first cost of project x3 is a. $70,000 c. $950 b. $25,000 d. $x4

Summary 453

10.

Consider the following binary linear programming formulation of a capital budgeting problem. Max 1,200 x1 1 600 x2 1 950 x3 1 1,650 x4 s.t. 15,000 x1 1 20,000 x2 1 25,000 x3 1 30,000 x4 ,5 70,000 x1 1 x2 ,5 1 x4 ,5 x3 x1, x2, x3, x4 5 (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

11.

Consider the following binary linear programming formulation of a capital budgeting problem. Max 1,200 x1 1 600 x2 1 950 x3 1 1,650 x4 s.t. 15,000 x1 1 20,000 x2 1 25,000 x3 1 30,000 x4 ,5 70,000 x1 1 x2 ,5 1 x4 ,5 x3 x1, x2, x3, x4 5 (0, 1) The capital budget limit is a. $90,000 c. $30,000 b. $70,000 d. $4,400

PROBLEMS Section 12.2

Capital Budgeting Problem with Indivisible Investments

1. 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. 2. Aerotron Radio Inc. has $250,000 available and its engineering staff has pro-

posed 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 1 2 3 4 5

Initial Investment

Annual Return

$75,000 $65,000 $50,000 $80,000 $100,000

$10,800 $12,000 $7,500 $13,750 $15,750

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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) plus 20%, and (3) minus 20%. 3.

Polaris Industries has $1,250,000 available for additional innovations on the Victory Vision motorcycle. These include the five 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 1 2 3 4 5

Initial Investment

Annual Return

$350,000 $300,000 $250,000 $500,000 $400,000

$90,000 $85,000 $75,000 $130,000 $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) plus 20%, and (3) minus 20%.

Summary 455

4. 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 1 2 3 4 5

Initial Investment

Annual Return

$25,000 $40,000 $85,000 $100,000 $65,000

$7,500 $12,000 $20,000 $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) plus 20%, and (3) minus 20%. 5.

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 1 2 3 4 5

Initial Investment

Annual Return

$150,000 $130,000 $100,000 $160,000 $200,000

$24,000 $22,000 $15,000 $25,000 $30,000

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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) plus 20%, and (3) minus 20%. 6. Yaesu America wishes to enhance their already fine line of electronic equip-

ment 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 1 2 3 4 5

Initial Investment

Annual Return

$400,000 $300,000 $200,000 $600,000 $500,000

$50,000 $36,000 $25,000 $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) plus 20%, and (3) minus 20%.

Summary 457

7.

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 1 2 3 4 5

Initial Investment

Annual Return

$45,000 $60,000 $85,000 $100,000 $75,000

$4,000 $7,000 $9,000 $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) plus 20%, and (3) minus 20%. 8. 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 2 3 4 5

Initial Investment

Annual Receipts

Annual Disbursements

Salvage Value

$300,000 $400,000 $450,000 $500,000 $600,000

$205,000 $230,000 $245,000 $260,000 $290,000

$125,000 $130,000 $140,000 $135,000 $150,000

$50,000 $50,000 $60,000 $75,000 $75,000

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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) plus 20%, and (3) minus 20%. 9. A division of Conoco-Phillips is involved in their periodic capital budgeting

activity and the engineering and operations staffs have identified ten 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 1 2 3 4 5 6 7 8 9 10

Initial Investment

Annual Return

Salvage Value

$150,000 $200,000 $225,000 $275,000 $350,000 $400,000 $475,000 $500,000 $550,000 $600,000

$35,000 $38,000 $45,000 $60,000 $75,000 $95,000 $110,000 $85,000 $120,000 $125,000

$25,000 $50,000 $22,500 $27,500 $55,000 $75,000 $50,000 $100,000 $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 five 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?

Summary 459

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) plus 20%, and (3) minus 20%. 10. A lending firm is considering 6 independent and indivisible investment alter-

natives 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 since the initial investment and the salvage value are the same, and the annual returns are constant each year.): Alternative 1 2 3 4 5 6

Initial Investment

Annual Return

$25,000 $35,000 $30,000 $40,000 $60,000 $50,000

$2,600 $3,750 $3,050 $4,775 $6,750 $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) plus 20%, and (3) minus 20%. 11.

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.

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EOY 0 1 2 3 4 5 6 7 8 9 10

CF(1)

CF(2)

CF(3)

CF(4)

2$50 2$100 $50 $50 $50 $50 $50 $50 $50 $50 $75

2$125 2$75 $70 $70 $70 $70 $70 $70 $70 $70 $100

2$200 $50 $50 $50 $50 $75 $75 $75 $75 $75 $75

2$250 $75 $75 $75 $75 $75 $85 $85 $85 $85 $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%, and (3) minus 20%. e. determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) plus 20%, and (3) minus 20%. 12. 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

$75,000 $60,000 $80,000 $55,000 $90,000

$8,750 $7,250 $9,750 $6,500 $10,750

A B C D E

13. 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

$8,000 $15,000 $10,000 $20,000 $19,000 $12,000

$3,200 $4,750 $3,070 $5,950 $5,150 $4,250

$1,000 $1,750 $1,100 $2,000 $2,100 $1,200

A B C D E F

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.

Summary 461

14. Consider the six indivisible investment alternatives shown below. The planning

horizon is 8 years. The MARR is 15%. $60,000 is available for investment. Initial Investment

Annual Return

Salvage Value

M N O P Q

$8,000 $15,000 $10,000 $20,000 $19,000

$3,200 $4,750 $3,070 $5,950 $5,150

$1,000 $1,750 $1,100 $2,000 $2,100

R

$12,000

$4,250

$1,200

Alternative

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. 15. The City of Clyde, Ohio is using the binary linear programming (BLP) for-

mulation of a capital budgeting problem shown below. Assume $175,000 is available for investment and all investments are indivisible; MARR 5 12%; n 5 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.

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Specify the contents of the target cell. Specify the contents for By Changing Cells. Specify all constraints to be added. 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. a. b. c. d.

16. 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.

Summary 463

Specify the contents of the target cell. Specify the contents for By Changing Cells. Specify all constraints to be added. 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 3 investments must be pursued. Show how you would incorporate that constraint in SOLVER. a. b. c. d.

Section 12.3 17.

Capital Budgeting Problem with Divisible Investments

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.”

18. 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.” 19. Aerotron Radio Inc. has $250,000 available and its engineering staff has pro-

posed 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 1 2 3 4 5

Initial Investment

Annual Return

$75,000 $65,000 $50,000 $80,000 $100,000

$10,800 $12,000 $7,500 $13,750 $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.

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c. Determine the optimum portfolio and its PW, specifying which invest-

ments 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 1, 2, and 3 are divisible and investments 4 and 5 are indivisible. Use Excel® and SOLVER. 20. 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 1 2 3 4 5

Initial Investment

Annual Return

$350,000 $300,000 $250,000 $500,000 $400,000

$90,000 $85,000 $75,000 $130,000 $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) plus 20%, and (3) minus 20%. 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. 21.

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 1 2 3 4 5

Initial Investment

Annual Return

$25,000 $40,000 $85,000 $100,000 $65,000

$7,500 $12,000 $20,000 $22,000 $17,000

Summary 465

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 1, 2, and 5 are divisible and investments 3 and 4 are indivisible. Use Excel® and SOLVER. 22. 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 1 2 3 4 5

Initial Investment

Annual Return

$150,000 $130,000 $100,000 $160,000 $200,000

$24,000 $22,000 $15,000 $25,000 $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) plus 20%, and (3) minus 20%. 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. 23.

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

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is projected to receive an annual return as noted below. They have available $1,250,000 to invest and their MARR is 10%. Investment 1 2 3 4 5

Initial Investment

Annual Return

$400,000 $300,000 $200,000 $600,000 $500,000

$50,000 $36,000 $25,000 $69,000 $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) 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 divisible and investments 1, 3, and 5 are indivisible. Use Excel® and SOLVER. 24. 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 1 2 3 4 5

Initial Investment

Annual Return

$45,000 $60,000 $85,000 $100,000 $75,000

$4,000 $7,000 $9,000 $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.

Summary 467

d. Determine the optimum investment portfolio and its PW when invest-

ments 2 and 4 are indivisible and investments 1, 3, and 5 are divisible. Use Excel® and SOLVER. 25. A laboratory within Bayer is considering the five divisible investment propos-

als 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 2 3 4 5

Initial Investment

Annual Receipts

Annual Disbursements

Salvage Value

$300,000 $400,000 $450,000 $500,000 $600,000

$205,000 $230,000 $245,000 $260,000 $290,000

$125,000 $130,000 $140,000 $135,000 $150,000

$50,000 $50,000 $60,000 $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) plus 20%, and (3) minus 20%. 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. 26. A division of Conoco-Phillips is involved in their periodic capital budgeting

activity and the engineering and operations staffs have identified ten 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 1 2 3 4 5 6 7 8 9 10

Initial Investment

Annual Return

Salvage Value

$150,000 $200,000 $225,000 $275,000 $350,000 $400,000 $475,000 $500,000 $550,000 $600,000

$35,000 $38,000 $45,000 $60,000 $75,000 $95,000 $110,000 $85,000 $120,000 $125,000

$25,000 $50,000 $22,500 $27,500 $55,000 $75,000 $50,000 $100,000 $75,000 $75,000

468

Chapter 12

Capital Budgeting

a. Determine the optimum portfolio, including which investments are fully

b.

c.

d.

e.

f.

27.

or partially (if partial, give percentage) selected. You may use Excel®; do not use SOLVER. 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. 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. 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. 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. 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 five investments must be made, albeit partially; investments 1 through 5 are indivisible and investments 6 through 10 are divisible. Use Excel® and SOLVER.

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 since the initial investment and the salvage value are the same, and the annual returns are constant each year.): Initial Investment

Annual Return

1 2 3 4 5

Investment

$25,000 $35,000 $30,000 $40,000 $60,000

$2,600 $3,750 $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.

Summary 469

c. Determine the optimum portfolio and its PW, specifying which invest-

ments 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 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. 28. 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 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

2$50

2$125

1

2$100 $50 $50 $50 $50 $50 $50 $50 $50 $75

2$75 $70 $70 $70 $70 $70 $70 $70 $70 $100

2$200 $50

2$250 $75

$50 $50 $50 $75 $75 $75 $75 $75 $75

$75 $75 $75 $75 $85 $85 $85 $85 $100

2 3 4 5 6 7 8 9 10

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) 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, 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.

470

Chapter 12

Capital Budgeting

29. Consider the six 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 B C D E

$8,000 $15,000 $10,000 $20,000 $19,000

$3,200 $4,750 $3,070 $5,950 $5,150

$1,000 $1,750 $1,100 $2,000 $2,100

F

$12,000

$4,250

$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. 30. Consider the six divisible investment alternatives shown below. The planning

horizon is 8 years. The MARR is 15%. $60,000 is available for investment. Initial Investment

Annual Return

Salvage Value

M N O P Q

Alternative

$8,000 $15,000 $10,000 $20,000 $19,000

$3,200 $4,750 $3,070 $5,950 $5,150

$1,000 $1,750 $1,100 $2,000 $2,100

R

$12,000

$4,250

$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.

APPENDIX A Single Sums

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.00250 1.00501 1.00752 1.01004 1.01256 1.01509 1.01763 1.02018 1.02273 1.02528 1.02785 1.03042 1.03299 1.03557 1.03816 1.04076 1.04336 1.04597 1.04858 1.05121 1.05383 1.05647 1.05911 1.06176 1.06441 1.06707 1.06974 1.07241 1.07510 1.07778 1.09405 1.10503 1.12733 1.13297 1.13864 1.14720 1.16162 1.19695 1.20595 1.22110 1.23335 1.25197 1.27087 1.28362 1.30952 1.34935 1.39040 1.43269 1.56743 1.82075 2.45684 3.31515 4.47331

0.99751 0.99502 0.99254 0.99006 0.98759 0.98513 0.98267 0.98022 0.97778 0.97534 0.97291 0.97048 0.96806 0.96565 0.96324 0.96084 0.95844 0.95605 0.95367 0.95129 0.94892 0.94655 0.94419 0.94184 0.93949 0.93714 0.93481 0.93248 0.93015 0.92783 0.91403 0.90495 0.88705 0.88263 0.87824 0.87168 0.86087 0.83546 0.82922 0.81894 0.81080 0.79874 0.78686 0.77904 0.76364 0.74110 0.71922 0.69799 0.63799 0.54922 0.40703 0.30165 0.22355

1.00000 2.00250 3.00751 4.01503 5.02506 6.03763 7.05272 8.07035 9.09053 10.11325 11.13854 12.16638 13.19680 14.22979 15.26537 16.30353 17.34429 18.38765 19.43362 20.48220 21.53341 22.58724 23.64371 24.70282 25.76457 26.82899 27.89606 28.96580 30.03821 31.11331 37.62056 42.01320 50.93121 53.18868 55.45746 58.88194 64.64671 78.77939 82.37922 88.43918 93.34192 100.78845 108.34739 113.44996 123.80926 139.74142 156.15817 173.07425 226.97269 328.30200 582.73688 926.05950 1389.32309

1.00000 0.49938 0.33250 0.24906 0.19900 0.16563 0.14179 0.12391 0.11000 0.09888 0.08978 0.08219 0.07578 0.07028 0.06551 0.06134 0.05766 0.05438 0.05146 0.04882 0.04644 0.04427 0.04229 0.04048 0.03881 0.03727 0.03585 0.03452 0.03329 0.03214 0.02658 0.02380 0.01963 0.01880 0.01803 0.01698 0.01547 0.01269 0.01214 0.01131 0.01071 9.9218E203 9.2296E203 8.8145E203 8.0769E203 7.1561E203 6.4038E203 5.7779E203 4.4058E203 3.0460E203 1.7160E203 1.0798E203 7.1977E204

0.99751 1.99252 2.98506 3.97512 4.96272 5.94785 6.93052 7.91074 8.88852 9.86386 10.83677 11.80725 12.77532 13.74096 14.70420 15.66504 16.62348 17.57953 18.53320 19.48449 20.43340 21.37995 22.32414 23.26598 24.20547 25.14261 26.07742 27.00989 27.94004 28.86787 34.38647 38.01986 45.17869 46.94617 48.70484 51.32644 55.65236 65.81686 68.31075 72.42595 75.68132 80.50382 85.25460 88.38248 94.54530 103.56175 112.31206 120.80407 144.80547 180.31091 237.18938 279.34176 310.58071

1.00250 0.50188 0.33500 0.25156 0.20150 0.16813 0.14429 0.12641 0.11250 0.10138 0.09228 0.08469 0.07828 0.07278 0.06801 0.06384 0.06016 0.05688 0.05396 0.05132 0.04894 0.04677 0.04479 0.04298 0.04131 0.03977 0.03835 0.03702 0.03579 0.03464 0.02908 0.02630 0.02213 0.02130 0.02053 0.01948 0.01797 0.01519 0.01464 0.01381 0.01321 0.01242 0.01173 0.01131 0.01058 9.6561E203 8.9038E203 8.2779E203 6.9058E203 5.5460E203 4.2160E203 3.5798E203 3.2198E203

0.00000 0.99502 2.98009 5.95028 9.90065 14.82630 20.72235 27.58391 35.40614 44.18420 53.91328 64.58858 76.20532 88.75874 102.24409 116.65666 131.99172 148.24459 165.41059 183.48508 202.46341 222.34096 243.11313 264.77534 287.32301 310.75160 335.05657 360.23340 386.27760 413.18468 592.49878 728.73988 1040.05520 1125.77667 1214.58847 1353.52863 1600.08454 2265.55685 2447.60694 2764.45681 3029.75923 3446.86997 3886.28316 4191.24173 4829.01247 5852.11160 6950.01441 8117.41331 1.1987E104 1.9399E104 3.6264E104 5.3821E104 7.0581E104

0.00000 0.49938 0.99834 1.49688 1.99501 2.49272 2.99001 3.48689 3.98335 4.47940 4.97503 5.47025 5.96504 6.45943 6.95339 7.44694 7.94008 8.43279 8.92510 9.41698 9.90845 10.39951 10.89014 11.38036 11.87017 12.35956 12.84853 13.33709 13.82523 14.31296 17.23058 19.16735 23.02092 23.98016 24.93774 26.37098 28.75142 34.42214 35.83048 38.16942 40.03312 42.81623 45.58444 47.42163 51.07618 56.50843 61.88128 67.19487 82.78122 107.58631 152.89019 192.66991 227.25400

471

Time Value of Money Factors Discrete Compounding 0.25%

n

Uniform Series

0.25%

TABLE A-a-1

0.50%

Time Value of Money Factors Discrete Compounding 0.50%

472 Appendix A

TABLE A-a-2 Single Sums

Uniform Series

Gradient Series

n

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

1.00500 1.01003 1.01508 1.02015 1.02525 1.03038 1.03553 1.04071 1.04591 1.05114 1.05640 1.06168 1.06699 1.07232 1.07768 1.08307 1.08849 1.09393 1.09940 1.10490 1.11042 1.11597 1.12155 1.12716 1.13280 1.13846 1.14415 1.14987 1.15562 1.16140 1.19668 1.22079 1.27049 1.28323 1.29609 1.31563 1.34885 1.43204 1.45363 1.49034 1.52037 1.56655 1.61414 1.64667 1.71370 1.81940 1.93161 2.05075 2.45409 3.31020 6.02258 10.95745 19.93596

0.99502 0.99007 0.98515 0.98025 0.97537 0.97052 0.96569 0.96089 0.95610 0.95135 0.94661 0.94191 0.93722 0.93256 0.92792 0.92330 0.91871 0.91414 0.90959 0.90506 0.90056 0.89608 0.89162 0.88719 0.88277 0.87838 0.87401 0.86966 0.86533 0.86103 0.83564 0.81914 0.78710 0.77929 0.77155 0.76009 0.74137 0.69830 0.68793 0.67099 0.65773 0.63834 0.61952 0.60729 0.58353 0.54963 0.51770 0.48763 0.40748 0.30210 0.16604 0.09126 0.05016

1.00000 2.00500 3.01502 4.03010 5.05025 6.07550 7.10588 8.14141 9.18212 10.22803 11.27917 12.33556 13.39724 14.46423 15.53655 16.61423 17.69730 18.78579 19.87972 20.97912 22.08401 23.19443 24.31040 25.43196 26.55912 27.69191 28.83037 29.97452 31.12439 32.28002 39.33610 44.15885 54.09783 56.64516 59.21803 63.12577 69.77003 86.40886 90.72650 98.06771 104.07393 113.31094 122.82854 129.33370 142.73990 163.87935 186.32263 210.15016 290.81871 462.04090 1004.51504 1991.49073 3787.19108

1.00000 0.49875 0.33167 0.24813 0.19801 0.16460 0.14073 0.12283 0.10891 0.09777 0.08866 0.08107 0.07464 0.06914 0.06436 0.06019 0.05651 0.05323 0.05030 0.04767 0.04528 0.04311 0.04113 0.03932 0.03765 0.03611 0.03469 0.03336 0.03213 0.03098 0.02542 0.02265 0.01849 0.01765 0.01689 0.01584 0.01433 0.01157 0.01102 0.01020 9.6086E203 8.8253E203 8.1414E203 7.7319E203 7.0057E203 6.1021E203 5.3670E203 4.7585E203 3.4386E203 2.1643E203 9.9551E204 5.0214E204 2.6405E204

0.99502 1.98510 2.97025 3.95050 4.92587 5.89638 6.86207 7.82296 8.77906 9.73041 10.67703 11.61893 12.55615 13.48871 14.41662 15.33993 16.25863 17.17277 18.08236 18.98742 19.88798 20.78406 21.67568 22.56287 23.44564 24.32402 25.19803 26.06769 26.93302 27.79405 32.87102 36.17223 42.58032 44.14279 45.68975 47.98145 51.72556 60.33951 62.41365 65.80231 68.45304 72.33130 76.09522 78.54264 83.29342 90.07345 96.45960 102.47474 118.50351 139.58077 166.79161 181.74758 189.96787

1.00500 0.50375 0.33667 0.25313 0.20301 0.16960 0.14573 0.12783 0.11391 0.10277 0.09366 0.08607 0.07964 0.07414 0.06936 0.06519 0.06151 0.05823 0.05530 0.05267 0.05028 0.04811 0.04613 0.04432 0.04265 0.04111 0.03969 0.03836 0.03713 0.03598 0.03042 0.02765 0.02349 0.02265 0.02189 0.02084 0.01933 0.01657 0.01602 0.01520 0.01461 0.01383 0.01314 0.01273 0.01201 0.01110 0.01037 9.7585E203 8.4386E203 7.1643E203 5.9955E203 5.5021E203 5.2640E203

0.00000 0.99007 2.96037 5.90111 9.80260 14.65519 20.44933 27.17552 34.82436 43.38649 52.85264 63.21360 74.46023 86.58346 99.57430 113.42380 128.12311 143.66343 160.03602 177.23221 195.24341 214.06109 233.67676 254.08203 275.26856 297.22805 319.95231 343.43317 367.66255 392.63241 557.55983 681.33469 959.91881 1035.69659 1113.81615 1235.26857 1448.64580 2012.34779 2163.75249 2424.64551 2640.66405 2976.07688 3324.18460 3562.79343 4054.37473 4823.50506 5624.58677 6451.31165 9031.33557 1.3416E104 2.1403E104 2.7588E104 3.1974E104

0.00000 0.49875 0.99667 1.49377 1.99003 2.48545 2.98005 3.47382 3.96675 4.45885 4.95013 5.44057 5.93018 6.41896 6.90691 7.39403 7.88031 8.36577 8.85040 9.33419 9.81716 10.29929 10.78060 11.26107 11.74072 12.21953 12.69751 13.17467 13.65099 14.12649 16.96205 18.83585 22.54372 23.46242 24.37781 25.74471 28.00638 33.35041 34.66794 36.84742 38.57628 41.14508 43.68454 45.36126 48.67581 53.55080 58.31029 62.95514 76.21154 96.11309 128.32362 151.79491 168.31425

Appendix A 473

Single Sums To Find P Given F (P|F i %,n)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

1.00750 1.01506 1.02267 1.03034 1.03807 1.04585 1.05370 1.06160 1.06956 1.07758 1.08566 1.09381 1.10201 1.11028 1.11860 1.12699 1.13544 1.14396 1.15254 1.16118 1.16989 1.17867 1.18751 1.19641 1.20539 1.21443 1.22354 1.23271 1.24196 1.25127 1.30865 1.34835 1.43141 1.45296 1.47483 1.50827 1.56568 1.71255 1.75137 1.81804 1.87320 1.95909 2.04892 2.11108 2.24112 2.45136 2.68131 2.93284 3.83804 6.00915 14.73058 36.10990 88.51826

0.99256 0.98517 0.97783 0.97055 0.96333 0.95616 0.94904 0.94198 0.93496 0.92800 0.92109 0.91424 0.90743 0.90068 0.89397 0.88732 0.88071 0.87416 0.86765 0.86119 0.85478 0.84842 0.84210 0.83583 0.82961 0.82343 0.81730 0.81122 0.80518 0.79919 0.76415 0.74165 0.69861 0.68825 0.67804 0.66301 0.63870 0.58392 0.57098 0.55004 0.53385 0.51044 0.48806 0.47369 0.44620 0.40794 0.37295 0.34097 0.26055 0.16641 0.06789 0.02769 0.01130

To Find F Given A (F|A i %,n) 1.00000 2.00750 3.02256 4.04523 5.07556 6.11363 7.15948 8.21318 9.27478 10.34434 11.42192 12.50759 13.60139 14.70340 15.81368 16.93228 18.05927 19.19472 20.33868 21.49122 22.65240 23.82230 25.00096 26.18847 27.38488 28.59027 29.80470 31.02823 32.26094 33.50290 41.15272 46.44648 57.52071 60.39426 63.31107 67.76883 75.42414 95.00703 100.18331 109.07253 116.42693 127.87899 139.85616 148.14451 165.48322 193.51428 224.17484 257.71157 378.40577 667.88687 1830.74348 4681.32027 1.1669E104

Gradient Series

To Find A Given F (A|F i%,n)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.00000 0.49813 0.33085 0.24721 0.19702 0.16357 0.13967 0.12176 0.10782 0.09667 0.08755 0.07995 0.07352 0.06801 0.06324 0.05906 0.05537 0.05210 0.04917 0.04653 0.04415 0.04198 0.04000 0.03818 0.03652 0.03498 0.03355 0.03223 0.03100 0.02985 0.02430 0.02153 0.01739 0.01656 0.01580 0.01476 0.01326 0.01053 9.9817E203 9.1682E203 8.5891E203 7.8199E203 7.1502E203 6.7502E203 6.0429E203 5.1676E203 4.4608E203 3.8803E203 2.6427E203 1.4973E203 5.4623E204 2.1361E204 8.5696E205

0.99256 1.97772 2.95556 3.92611 4.88944 5.84560 6.79464 7.73661 8.67158 9.59958 10.52067 11.43491 12.34235 13.24302 14.13699 15.02431 15.90502 16.77918 17.64683 18.50802 19.36280 20.21121 21.05331 21.88915 22.71876 23.54219 24.35949 25.17071 25.97589 26.77508 31.44681 34.44694 40.18478 41.56645 42.92762 44.93161 48.17337 55.47685 57.20267 59.99444 62.15396 65.27461 68.25844 70.17462 73.83938 78.94169 83.60642 87.87109 98.59341 111.14495 124.28187 129.64090 131.82705

1.00750 0.50563 0.33835 0.25471 0.20452 0.17107 0.14717 0.12926 0.11532 0.10417 0.09505 0.08745 0.08102 0.07551 0.07074 0.06656 0.06287 0.05960 0.05667 0.05403 0.05165 0.04948 0.04750 0.04568 0.04402 0.04248 0.04105 0.03973 0.03850 0.03735 0.03180 0.02903 0.02489 0.02406 0.02330 0.02226 0.02076 0.01803 0.01748 0.01667 0.01609 0.01532 0.01465 0.01425 0.01354 0.01267 0.01196 0.01138 0.01014 8.9973E203 8.0462E203 7.7136E203 7.5857E203

0.00000 0.98517 2.94083 5.85250 9.70581 14.48660 20.18084 26.77467 34.25438 42.60641 51.81736 61.87398 72.76316 84.47197 96.98758 110.29735 124.38875 139.24940 154.86708 171.22969 188.32527 206.14200 224.66820 243.89233 263.80295 284.38879 305.63869 327.54162 350.08668 373.26310 524.99236 637.46933 886.84045 953.84863 1022.58522 1128.78691 1313.51888 1791.24629 1917.22249 2132.14723 2308.12830 2577.99605 2853.93524 3040.74530 3419.90409 3998.56214 4583.57014 5169.58283 6892.60143 9494.11617 1.3312E104 1.5513E104 1.6673E104

0.00000 0.49813 0.99502 1.49066 1.98506 2.47821 2.97011 3.46077 3.95019 4.43836 4.92529 5.41097 5.89541 6.37860 6.86055 7.34126 7.82072 8.29894 8.77592 9.25165 9.72614 10.19939 10.67139 11.14216 11.61168 12.07996 12.54701 13.01281 13.47737 13.94069 16.69462 18.50583 22.06906 22.94756 23.82115 25.12233 27.26649 32.28818 33.51631 35.53908 37.13566 39.49462 41.81073 43.33112 46.31545 50.65210 54.82318 58.83144 69.90935 85.42103 107.11448 119.66198 126.47762

Time Value of Money Factors Discrete Compounding 0.75%

n

To Find F Given P (F|P i %,n)

Uniform Series

0.75%

TABLE A-a-3

1.00%

Time Value of Money Factors Discrete Compounding 1.00%

474 Appendix A

TABLE A-a-4 Single Sums

Uniform Series

Gradient Series

n

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i %,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

1.01000 1.02010 1.03030 1.04060 1.05101 1.06152 1.07214 1.08286 1.09369 1.10462 1.11567 1.12683 1.13809 1.14947 1.16097 1.17258 1.18430 1.19615 1.20811 1.22019 1.23239 1.24472 1.25716 1.26973 1.28243 1.29526 1.30821 1.32129 1.33450 1.34785 1.43077 1.48886 1.61223 1.64463 1.67769 1.72852 1.81670 2.04710 2.10913 2.21672 2.30672 2.44863 2.59927 2.70481 2.92893 3.30039 3.71896 4.19062 5.99580 10.89255 35.94964 118.64773 391.58340

0.99010 0.98030 0.97059 0.96098 0.95147 0.94205 0.93272 0.92348 0.91434 0.90529 0.89632 0.88745 0.87866 0.86996 0.86135 0.85282 0.84438 0.83602 0.82774 0.81954 0.81143 0.80340 0.79544 0.78757 0.77977 0.77205 0.76440 0.75684 0.74934 0.74192 0.69892 0.67165 0.62026 0.60804 0.59606 0.57853 0.55045 0.48850 0.47413 0.45112 0.43352 0.40839 0.38472 0.36971 0.34142 0.30299 0.26889 0.23863 0.16678 0.09181 0.02782 8.4283E203 2.5537E203

1.00000 2.01000 3.03010 4.06040 5.10101 6.15202 7.21354 8.28567 9.36853 10.46221 11.56683 12.68250 13.80933 14.94742 16.09690 17.25786 18.43044 19.61475 20.81090 22.01900 23.23919 24.47159 25.71630 26.97346 28.24320 29.52563 30.82089 32.12910 33.45039 34.78489 43.07688 48.88637 61.22261 64.46318 67.76889 72.85246 81.66967 104.70993 110.91285 121.67152 130.67227 144.86327 159.92729 170.48138 192.89258 230.03869 271.89586 319.06156 499.58020 989.25537 3494.96413 1.1765E104 3.9058E104

1.00000 0.49751 0.33002 0.24628 0.19604 0.16255 0.13863 0.12069 0.10674 0.09558 0.08645 0.07885 0.07241 0.06690 0.06212 0.05794 0.05426 0.05098 0.04805 0.04542 0.04303 0.04086 0.03889 0.03707 0.03541 0.03387 0.03245 0.03112 0.02990 0.02875 0.02321 0.02046 0.01633 0.01551 0.01476 0.01373 0.01224 9.5502E203 9.0161E203 8.2189E203 7.6527E203 6.9031E203 6.2528E203 5.8657E203 5.1842E203 4.3471E203 3.6779E203 3.1342E203 2.0017E203 1.0109E203 2.8613E204 8.5000E205 2.5603E205

0.99010 1.97040 2.94099 3.90197 4.85343 5.79548 6.72819 7.65168 8.56602 9.47130 10.36763 11.25508 12.13374 13.00370 13.86505 14.71787 15.56225 16.39827 17.22601 18.04555 18.85698 19.66038 20.45582 21.24339 22.02316 22.79520 23.55961 24.31644 25.06579 25.80771 30.10751 32.83469 37.97396 39.19612 40.39419 42.14719 44.95504 51.15039 52.58705 54.88821 56.64845 59.16088 61.52770 63.02888 65.85779 69.70052 73.11075 76.13716 83.32166 90.81942 97.21833 99.15717 99.74463

1.01000 0.50751 0.34002 0.25628 0.20604 0.17255 0.14863 0.13069 0.11674 0.10558 0.09645 0.08885 0.08241 0.07690 0.07212 0.06794 0.06426 0.06098 0.05805 0.05542 0.05303 0.05086 0.04889 0.04707 0.04541 0.04387 0.04245 0.04112 0.03990 0.03875 0.03321 0.03046 0.02633 0.02551 0.02476 0.02373 0.02224 0.01955 0.01902 0.01822 0.01765 0.01690 0.01625 0.01587 0.01518 0.01435 0.01368 0.01313 0.01200 0.01101 0.01029 0.01008 0.01003

0.00000 0.98030 2.92148 5.80442 9.61028 14.32051 19.91681 26.38120 33.69592 41.84350 50.80674 60.56868 71.11263 82.42215 94.48104 107.27336 120.78340 134.99569 149.89501 165.46636 181.69496 198.56628 216.06600 234.18002 252.89446 272.19566 292.07016 312.50472 333.48630 355.00207 494.62069 596.85606 820.14601 879.41763 939.91752 1032.81478 1192.80614 1597.86733 1702.73397 1879.87710 2023.31531 2240.56748 2459.42979 2605.77575 2898.42028 3334.11485 3761.69441 4177.46642 5330.06592 6878.60156 8720.43230 9511.15793 9821.23859

0.00000 0.49751 0.99337 1.48756 1.98010 2.47098 2.96020 3.44777 3.93367 4.41792 4.90052 5.38145 5.86073 6.33836 6.81433 7.28865 7.76131 8.23231 8.70167 9.16937 9.63542 10.09982 10.56257 11.02367 11.48312 11.94092 12.39707 12.85158 13.30444 13.75566 16.42848 18.17761 21.59759 22.43635 23.26863 24.50495 26.53331 31.23861 32.37934 34.24920 35.71704 37.87245 39.97272 41.34257 44.01029 47.83486 51.45200 54.86764 63.96975 75.73933 89.69947 95.92002 98.46384

Appendix A 475

Single Sums

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

1.01250 1.02516 1.03797 1.05095 1.06408 1.07738 1.09085 1.10449 1.11829 1.13227 1.14642 1.16075 1.17526 1.18995 1.20483 1.21989 1.23514 1.25058 1.26621 1.28204 1.29806 1.31429 1.33072 1.34735 1.36419 1.38125 1.39851 1.41599 1.43369 1.45161 1.56394 1.64362 1.81535 1.86102 1.90784 1.98028 2.10718 2.44592 2.53879 2.70148 2.83911 3.05881 3.29551 3.46340 3.82528 4.44021 5.15400 5.98253 9.35633 19.71549 87.54100 388.70068 1725.91392

Gradient Series

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)

To Find P Given A (P|A i %,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

0.98765 0.97546 0.96342 0.95152 0.93978 0.92817 0.91672 0.90540 0.89422 0.88318 0.87228 0.86151 0.85087 0.84037 0.82999 0.81975 0.80963 0.79963 0.78976 0.78001 0.77038 0.76087 0.75147 0.74220 0.73303 0.72398 0.71505 0.70622 0.69750 0.68889 0.63941 0.60841 0.55086 0.53734 0.52415 0.50498 0.47457 0.40884 0.39389 0.37017 0.35222 0.32692 0.30344 0.28873 0.26142 0.22521 0.19402 0.16715 0.10688 0.05072 0.01142 2.5727E203 5.7940E204

1.00000 2.01250 3.03766 4.07563 5.12657 6.19065 7.26804 8.35889 9.46337 10.58167 11.71394 12.86036 14.02112 15.19638 16.38633 17.59116 18.81105 20.04619 21.29677 22.56298 23.84502 25.14308 26.45737 27.78808 29.13544 30.49963 31.88087 33.27938 34.69538 36.12907 45.11551 51.48956 65.22839 68.88179 72.62710 78.42246 88.57451 115.67362 123.10349 136.11880 147.12904 164.70501 183.64106 197.07234 226.02255 275.21706 332.31981 398.60208 668.50676 1497.23948 6923.27961 3.1016E104 1.3799E105

1.00000 0.49689 0.32920 0.24536 0.19506 0.16153 0.13759 0.11963 0.10567 0.09450 0.08537 0.07776 0.07132 0.06581 0.06103 0.05685 0.05316 0.04988 0.04696 0.04432 0.04194 0.03977 0.03780 0.03599 0.03432 0.03279 0.03137 0.03005 0.02882 0.02768 0.02217 0.01942 0.01533 0.01452 0.01377 0.01275 0.01129 8.6450E203 8.1232E203 7.3465E203 6.7968E203 6.0715E203 5.4454E203 5.0743E203 4.4243E203 3.6335E203 3.0091E203 2.5088E203 1.4959E203 6.6790E204 1.4444E204 3.2241E205 7.2467E206

0.98765 1.96312 2.92653 3.87806 4.81784 5.74601 6.66273 7.56812 8.46234 9.34553 10.21780 11.07931 11.93018 12.77055 13.60055 14.42029 15.22992 16.02955 16.81931 17.59932 18.36969 19.13056 19.88204 20.62423 21.35727 22.08125 22.79630 23.50252 24.20002 24.88891 28.84727 31.32693 35.93148 37.01288 38.06773 39.60169 42.03459 47.29247 48.48897 50.38666 51.82219 53.84606 55.72457 56.90134 59.08651 61.98285 64.47807 66.62772 71.44964 75.94228 79.08614 79.79419 79.95365

1.01250 0.50939 0.34170 0.25786 0.20756 0.17403 0.15009 0.13213 0.11817 0.10700 0.09787 0.09026 0.08382 0.07831 0.07353 0.06935 0.06566 0.06238 0.05946 0.05682 0.05444 0.05227 0.05030 0.04849 0.04682 0.04529 0.04387 0.04255 0.04132 0.04018 0.03467 0.03192 0.02783 0.02702 0.02627 0.02525 0.02379 0.02115 0.02062 0.01985 0.01930 0.01857 0.01795 0.01757 0.01692 0.01613 0.01551 0.01501 0.01400 0.01317 0.01264 0.01253 0.01251

0.00000 0.97546 2.90230 5.75687 9.51598 14.15685 19.65715 25.99494 33.14870 41.09733 49.82011 59.29670 69.50717 80.43196 92.05186 104.34806 117.30207 130.89580 145.11145 159.93161 175.33919 191.31742 207.84986 224.92039 242.51321 260.61282 279.20402 298.27192 317.80191 337.77969 466.28302 559.23198 759.22956 811.67385 864.94093 946.22770 1084.84285 1428.45610 1515.79039 1661.86513 1778.83839 1953.83026 2127.52438 2242.24109 2468.26361 2796.56945 3109.35041 3404.60974 4176.90718 5101.52883 5997.90267 6284.74422 6368.48047

0.00000 0.49689 0.99172 1.48447 1.97516 2.46377 2.95032 3.43479 3.91720 4.39754 4.87581 5.35202 5.82616 6.29824 6.76825 7.23620 7.70208 8.16591 8.62767 9.08738 9.54502 10.00062 10.45415 10.90564 11.35507 11.80245 12.24778 12.69106 13.13230 13.57150 16.16385 17.85148 21.12993 21.92950 22.72110 23.89362 25.80834 30.20472 31.26052 32.98225 34.32581 36.28548 38.17929 39.40577 41.77373 45.11844 48.22338 51.09900 58.45945 67.17640 75.84012 78.76193 79.65216

Time Value of Money Factors Discrete Compounding 1.25%

n

To Find F Given P (F|P i %,n)

Uniform Series

1.25%

TABLE A-a-5

1.50%

Time Value of Money Factors Discrete Compounding 1.50%

476 Appendix A

TABLE A-a-6 Single Sums

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Uniform Series

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.01500 1.03023 1.04568 1.06136 1.07728 1.09344 1.10984 1.12649 1.14339 1.16054 1.17795 1.19562 1.21355 1.23176 1.25023 1.26899 1.28802 1.30734 1.32695 1.34686 1.36706 1.38756 1.40838 1.42950 1.45095 1.47271 1.49480 1.51722 1.53998 1.56308 1.70914 1.81402 2.04348 2.10524 2.16887 2.26794 2.44322 2.92116 3.05459 3.29066 3.49259 3.81895 4.17580 4.43205 4.99267 5.96932 7.13703 8.53316 14.58437 35.63282 212.70378 1269.69754 7579.23459

0.98522 0.97066 0.95632 0.94218 0.92826 0.91454 0.90103 0.88771 0.87459 0.86167 0.84893 0.83639 0.82403 0.81185 0.79985 0.78803 0.77639 0.76491 0.75361 0.74247 0.73150 0.72069 0.71004 0.69954 0.68921 0.67902 0.66899 0.65910 0.64936 0.63976 0.58509 0.55126 0.48936 0.47500 0.46107 0.44093 0.40930 0.34233 0.32738 0.30389 0.28632 0.26185 0.23947 0.22563 0.20029 0.16752 0.14011 0.11719 0.06857 0.02806 4.7014E203 7.8759E204 1.3194E204

1.00000 2.01500 3.04522 4.09090 5.15227 6.22955 7.32299 8.43284 9.55933 10.70272 11.86326 13.04121 14.23683 15.45038 16.68214 17.93237 19.20136 20.48938 21.79672 23.12367 24.47052 25.83758 27.22514 28.63352 30.06302 31.51397 32.98668 34.48148 35.99870 37.53868 47.27597 54.26789 69.56522 73.68283 77.92489 84.52960 96.21465 128.07720 136.97278 152.71085 166.17264 187.92990 211.72023 228.80304 266.17777 331.28819 409.13539 502.21092 905.62451 2308.85437 1.4114E104 8.4580E104 5.0522E105

1.00000 0.49628 0.32838 0.24444 0.19409 0.16053 0.13656 0.11858 0.10461 0.09343 0.08429 0.07668 0.07024 0.06472 0.05994 0.05577 0.05208 0.04881 0.04588 0.04325 0.04087 0.03870 0.03673 0.03492 0.03326 0.03173 0.03032 0.02900 0.02778 0.02664 0.02115 0.01843 0.01437 0.01357 0.01283 0.01183 0.01039 7.8078E203 7.3007E203 6.5483E203 6.0178E203 5.3211E203 4.7232E203 4.3706E203 3.7569E203 3.0185E203 2.4442E203 1.9912E203 1.1042E203 4.3312E204 7.0854E205 1.1823E205 1.9794E206

0.98522 1.95588 2.91220 3.85438 4.78264 5.69719 6.59821 7.48593 8.36052 9.22218 10.07112 10.90751 11.73153 12.54338 13.34323 14.13126 14.90765 15.67256 16.42617 17.16864 17.90014 18.62082 19.33086 20.03041 20.71961 21.39863 22.06762 22.72672 23.37608 24.01584 27.66068 29.91585 34.04255 34.99969 35.92874 37.27147 39.38027 43.84467 44.84160 46.40732 47.57863 49.20985 50.70168 51.62470 53.31375 55.49845 57.32571 58.85401 62.09556 64.79573 66.35324 66.61416 66.65787

1.01500 0.51128 0.34338 0.25944 0.20909 0.17553 0.15156 0.13358 0.11961 0.10843 0.09929 0.09168 0.08524 0.07972 0.07494 0.07077 0.06708 0.06381 0.06088 0.05825 0.05587 0.05370 0.05173 0.04992 0.04826 0.04673 0.04532 0.04400 0.04278 0.04164 0.03615 0.03343 0.02937 0.02857 0.02783 0.02683 0.02539 0.02281 0.02230 0.02155 0.02102 0.02032 0.01972 0.01937 0.01876 0.01802 0.01744 0.01699 0.01610 0.01543 0.01507 0.01501 0.01500

0.00000 0.97066 2.88330 5.70985 9.42289 13.99560 19.40176 25.61574 32.61248 40.36748 48.85681 58.05708 67.94540 78.49944 89.69736 101.51783 113.93999 126.94349 140.50842 154.61536 169.24532 184.37976 200.00058 216.09009 232.63103 249.60654 267.00017 284.79585 302.97790 321.53101 439.83026 524.35682 703.54615 749.96361 796.87737 868.02846 988.16739 1279.79379 1352.56005 1473.07411 1568.51404 1709.54387 1847.47253 1937.45061 2112.13479 2359.71143 2588.70855 2798.57842 3316.90537 3870.69117 4310.71648 4415.74120 4438.58047

0.00000 0.49628 0.99007 1.48139 1.97023 2.45658 2.94046 3.42185 3.90077 4.37721 4.85118 5.32267 5.79169 6.25824 6.72231 7.18392 7.64306 8.09973 8.55394 9.00569 9.45497 9.90180 10.34618 10.78810 11.22758 11.66460 12.09918 12.53132 12.96102 13.38829 15.90092 17.52773 20.66667 21.42772 22.17938 23.28936 25.09296 29.18927 30.16306 31.74228 32.96677 34.73987 36.43810 37.52953 39.61708 42.51851 45.15789 47.55119 53.41614 59.73682 64.96618 66.28833 66.58749

Appendix A 477

Single Sums

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.01750 1.03531 1.05342 1.07186 1.09062 1.10970 1.12912 1.14888 1.16899 1.18944 1.21026 1.23144 1.25299 1.27492 1.29723 1.31993 1.34303 1.36653 1.39045 1.41478 1.43954 1.46473 1.49036 1.51644 1.54298 1.56998 1.59746 1.62541 1.65386 1.68280 1.86741 2.00160 2.29960 2.38079 2.46485 2.59653 2.83182 3.48721 3.67351 4.00639 4.29429 4.76538 5.28815 5.66816 6.51204 8.01918 9.87514 12.16063 22.70885 64.30730 515.69206 4135.42921 3.3163E104

0.98280 0.96590 0.94929 0.93296 0.91691 0.90114 0.88564 0.87041 0.85544 0.84073 0.82627 0.81206 0.79809 0.78436 0.77087 0.75762 0.74459 0.73178 0.71919 0.70682 0.69467 0.68272 0.67098 0.65944 0.64810 0.63695 0.62599 0.61523 0.60465 0.59425 0.53550 0.49960 0.43486 0.42003 0.40570 0.38513 0.35313 0.28676 0.27222 0.24960 0.23287 0.20985 0.18910 0.17642 0.15356 0.12470 0.10126 0.08223 0.04404 0.01555 1.9391E203 2.4181E204 3.0154E205

1.00000 2.01750 3.05281 4.10623 5.17809 6.26871 7.37841 8.50753 9.65641 10.82540 12.01484 13.22510 14.45654 15.70953 16.98445 18.28168 19.60161 20.94463 22.31117 23.70161 25.11639 26.55593 28.02065 29.51102 31.02746 32.57044 34.14042 35.73788 37.36329 39.01715 49.56613 57.23413 74.26278 78.90222 83.70547 91.23016 104.67522 142.12628 152.77206 171.79382 188.24499 215.16462 245.03739 266.75177 314.97378 401.09620 507.15073 637.75045 1240.50595 3617.56017 2.9411E104 2.3625E105 1.8950E106

1.00000 0.49566 0.32757 0.24353 0.19312 0.15952 0.13553 0.11754 0.10356 0.09238 0.08323 0.07561 0.06917 0.06366 0.05888 0.05470 0.05102 0.04774 0.04482 0.04219 0.03981 0.03766 0.03569 0.03389 0.03223 0.03070 0.02929 0.02798 0.02676 0.02563 0.02018 0.01747 0.01347 0.01267 0.01195 0.01096 9.5534E203 7.0360E203 6.5457E203 5.8209E203 5.3122E203 4.6476E203 4.0810E203 3.7488E203 3.1749E203 2.4932E203 1.9718E203 1.5680E203 8.0612E204 2.7643E204 3.4001E205 4.2327E206 5.2772E207

0.98280 1.94870 2.89798 3.83094 4.74786 5.64900 6.53464 7.40505 8.26049 9.10122 9.92749 10.73955 11.53764 12.32201 13.09288 13.85050 14.59508 15.32686 16.04606 16.75288 17.44755 18.13027 18.80125 19.46069 20.10878 20.74573 21.37173 21.98695 22.59160 23.18585 26.54275 28.59423 32.29380 33.14121 33.95972 35.13545 36.96399 40.75645 41.58748 42.87993 43.83614 45.15161 46.33703 47.06147 48.36790 50.01709 51.35632 52.44385 54.62653 56.25427 57.03205 57.12904 57.14113

1.01750 0.51316 0.34507 0.26103 0.21062 0.17702 0.15303 0.13504 0.12106 0.10988 0.10073 0.09311 0.08667 0.08116 0.07638 0.07220 0.06852 0.06524 0.06232 0.05969 0.05731 0.05516 0.05319 0.05139 0.04973 0.04820 0.04679 0.04548 0.04426 0.04313 0.03768 0.03497 0.03097 0.03017 0.02945 0.02846 0.02705 0.02454 0.02405 0.02332 0.02281 0.02215 0.02158 0.02125 0.02067 0.01999 0.01947 0.01907 0.01831 0.01778 0.01753 0.01750 0.01750

0.00000 0.96590 2.86447 5.66334 9.33099 13.83671 19.15057 25.24345 32.08698 39.65354 47.91623 56.84886 66.42596 76.62270 87.41495 98.77919 110.69257 123.13283 136.07832 149.50799 163.40134 177.73847 192.49999 207.66706 223.22138 239.14512 255.42098 272.03215 288.96226 306.19544 415.12498 492.01087 652.60539 693.70101 735.03220 797.33210 901.49545 1149.11809 1209.77384 1309.24819 1387.15838 1500.87981 1610.47158 1681.08862 1816.18525 2003.02686 2170.82384 2320.13512 2668.57763 3001.26781 3219.08332 3257.88395 3264.17380

0.00000 0.49566 0.98843 1.47832 1.96531 2.44941 2.93062 3.40895 3.88439 4.35695 4.82662 5.29341 5.75733 6.21836 6.67653 7.13182 7.58424 8.03379 8.48048 8.92431 9.36529 9.80341 10.23868 10.67111 11.10069 11.52744 11.95135 12.37243 12.79069 13.20613 15.63986 17.20665 20.20838 20.93167 21.64424 22.69310 24.38848 28.19476 29.08986 30.53289 31.64417 33.24089 34.75560 35.72112 37.54939 40.04685 42.26985 44.24036 48.85131 53.35183 56.44341 57.02676 57.12476

Time Value of Money Factors Discrete Compounding 1.75%

n

Uniform Series

1.75%

TABLE A-a-7

2.00%

Time Value of Money Factors Discrete Compounding 2.00%

478 Appendix A

TABLE A-a-8 Single Sums

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Uniform Series

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.02000 1.04040 1.06121 1.08243 1.10408 1.12616 1.14869 1.17166 1.19509 1.21899 1.24337 1.26824 1.29361 1.31948 1.34587 1.37279 1.40024 1.42825 1.45681 1.48595 1.51567 1.54598 1.57690 1.60844 1.64061 1.67342 1.70689 1.74102 1.77584 1.81136 2.03989 2.20804 2.58707 2.69159 2.80033 2.97173 3.28103 4.16114 4.41584 4.87544 5.27733 5.94313 6.69293 7.24465 8.48826 10.76516 13.65283 17.31509 35.32083 115.88874 1247.56113 1.3430E104 1.4458E105

0.98039 0.96117 0.94232 0.92385 0.90573 0.88797 0.87056 0.85349 0.83676 0.82035 0.80426 0.78849 0.77303 0.75788 0.74301 0.72845 0.71416 0.70016 0.68643 0.67297 0.65978 0.64684 0.63416 0.62172 0.60953 0.59758 0.58586 0.57437 0.56311 0.55207 0.49022 0.45289 0.38654 0.37153 0.35710 0.33650 0.30478 0.24032 0.22646 0.20511 0.18949 0.16826 0.14941 0.13803 0.11781 0.09289 0.07324 0.05775 0.02831 8.6290E203 8.0156E204 7.4459E205 6.9167E206

1.00000 2.02000 3.06040 4.12161 5.20404 6.30812 7.43428 8.58297 9.75463 10.94972 12.16872 13.41209 14.68033 15.97394 17.29342 18.63929 20.01207 21.41231 22.84056 24.29737 25.78332 27.29898 28.84496 30.42186 32.03030 33.67091 35.34432 37.05121 38.79223 40.56808 51.99437 60.40198 79.35352 84.57940 90.01641 98.58653 114.05154 158.05702 170.79177 193.77196 213.86661 247.15666 284.64666 312.23231 374.41288 488.25815 632.64148 815.75446 1716.04157 5744.43676 6.2328E104 6.7146E105 7.2289E106

1.00000 0.49505 0.32675 0.24262 0.19216 0.15853 0.13451 0.11651 0.10252 0.09133 0.08218 0.07456 0.06812 0.06260 0.05783 0.05365 0.04997 0.04670 0.04378 0.04116 0.03878 0.03663 0.03467 0.03287 0.03122 0.02970 0.02829 0.02699 0.02578 0.02465 0.01923 0.01656 0.01260 0.01182 0.01111 0.01014 8.7680E203 6.3268E203 5.8551E203 5.1607E203 4.6758E203 4.0460E203 3.5131E203 3.2027E203 2.6708E203 2.0481E203 1.5807E203 1.2259E203 5.8274E204 1.7408E204 1.6044E205 1.4893E206 1.3833E207

0.98039 1.94156 2.88388 3.80773 4.71346 5.60143 6.47199 7.32548 8.16224 8.98259 9.78685 10.57534 11.34837 12.10625 12.84926 13.57771 14.29187 14.99203 15.67846 16.35143 17.01121 17.65805 18.29220 18.91393 19.52346 20.12104 20.70690 21.28127 21.84438 22.39646 25.48884 27.35548 30.67312 31.42361 32.14495 33.17479 34.76089 37.98406 38.67711 39.74451 40.52552 41.58693 42.52943 43.09835 44.10951 45.35539 46.33776 47.11235 48.58440 49.56855 49.95992 49.99628 49.99965

1.02000 0.51505 0.34675 0.26262 0.21216 0.17853 0.15451 0.13651 0.12252 0.11133 0.10218 0.09456 0.08812 0.08260 0.07783 0.07365 0.06997 0.06670 0.06378 0.06116 0.05878 0.05663 0.05467 0.05287 0.05122 0.04970 0.04829 0.04699 0.04578 0.04465 0.03923 0.03656 0.03260 0.03182 0.03111 0.03014 0.02877 0.02633 0.02586 0.02516 0.02468 0.02405 0.02351 0.02320 0.02267 0.02205 0.02158 0.02123 0.02058 0.02017 0.02002 0.02000 0.02000

0.00000 0.96117 2.84581 5.61735 9.24027 13.68013 18.90349 24.87792 31.57197 38.95510 46.99773 55.67116 64.94755 74.79992 85.20213 96.12881 107.55542 119.45813 131.81388 144.60033 157.79585 171.37947 185.33090 199.63049 214.25924 229.19872 244.43113 259.93924 275.70639 291.71644 392.04045 461.99313 605.96572 642.36059 678.78489 733.35269 823.69753 1034.05570 1084.63929 1166.78677 1230.41912 1322.17008 1409.29734 1464.75275 1569.30251 1710.41605 1833.47151 1939.79497 2174.41310 2374.87999 2483.56794 2498.02683 2499.77521

0.00000 0.49505 0.98680 1.47525 1.96040 2.44226 2.92082 3.39608 3.86805 4.33674 4.80213 5.26424 5.72307 6.17862 6.63090 7.07990 7.52564 7.96811 8.40732 8.84328 9.27599 9.70546 10.13169 10.55468 10.97445 11.39100 11.80433 12.21446 12.62138 13.02512 15.38087 16.88850 19.75559 20.44198 21.11638 22.10572 23.69610 27.22341 28.04344 29.35718 30.36159 31.79292 33.13699 33.98628 35.57742 37.71142 39.56755 41.17381 44.75537 47.91102 49.71121 49.96426 49.99585

Appendix A 479

Single Sums

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.03000 1.06090 1.09273 1.12551 1.15927 1.19405 1.22987 1.26677 1.30477 1.34392 1.38423 1.42576 1.46853 1.51259 1.55797 1.60471 1.65285 1.70243 1.75351 1.80611 1.86029 1.91610 1.97359 2.03279 2.09378 2.15659 2.22129 2.28793 2.35657 2.42726 2.89828 3.26204 4.13225 4.38391 4.65089 5.08215 5.89160 8.40002 9.17893 10.64089 11.97642 14.30047 17.07551 19.21863 24.34559 34.71099 49.48957 70.56029 204.50336 1204.85263 4.1822E104 1.4517E106 5.0389E107

0.97087 0.94260 0.91514 0.88849 0.86261 0.83748 0.81309 0.78941 0.76642 0.74409 0.72242 0.70138 0.68095 0.66112 0.64186 0.62317 0.60502 0.58739 0.57029 0.55368 0.53755 0.52189 0.50669 0.49193 0.47761 0.46369 0.45019 0.43708 0.42435 0.41199 0.34503 0.30656 0.24200 0.22811 0.21501 0.19677 0.16973 0.11905 0.10895 0.09398 0.08350 0.06993 0.05856 0.05203 0.04108 0.02881 0.02021 0.01417 4.8899E203 8.2998E204 2.3911E205 6.8886E207 1.9846E208

1.00000 2.03000 3.09090 4.18363 5.30914 6.46841 7.66246 8.89234 10.15911 11.46388 12.80780 14.19203 15.61779 17.08632 18.59891 20.15688 21.76159 23.41444 25.11687 26.87037 28.67649 30.53678 32.45288 34.42647 36.45926 38.55304 40.70963 42.93092 45.21885 47.57542 63.27594 75.40126 104.40840 112.79687 121.69620 136.07162 163.05344 246.66724 272.63086 321.36302 365.88054 443.34890 535.85019 607.28773 778.18627 1123.69957 1616.31893 2318.67634 6783.44532 4.0128E104 1.3940E106 4.8389E107 1.6796E109

1.00000 0.49261 0.32353 0.23903 0.18835 0.15460 0.13051 0.11246 0.09843 0.08723 0.07808 0.07046 0.06403 0.05853 0.05377 0.04961 0.04595 0.04271 0.03981 0.03722 0.03487 0.03275 0.03081 0.02905 0.02743 0.02594 0.02456 0.02329 0.02211 0.02102 0.01580 0.01326 9.5778E203 8.8655E203 8.2172E203 7.3491E203 6.1330E203 4.0540E203 3.6680E203 3.1117E203 2.7331E203 2.2556E203 1.8662E203 1.6467E203 1.2850E203 8.8992E204 6.1869E204 4.3128E204 1.4742E204 2.4920E205 7.1735E207 2.0666E208 5.9537E210

0.97087 1.91347 2.82861 3.71710 4.57971 5.41719 6.23028 7.01969 7.78611 8.53020 9.25262 9.95400 10.63496 11.29607 11.93794 12.56110 13.16612 13.75351 14.32380 14.87747 15.41502 15.93692 16.44361 16.93554 17.41315 17.87684 18.32703 18.76411 19.18845 19.60044 21.83225 23.11477 25.26671 25.72976 26.16624 26.77443 27.67556 29.36509 29.70183 30.20076 30.55009 31.00241 31.38122 31.59891 31.96416 32.37302 32.65979 32.86092 33.17034 33.30567 33.33254 33.33331 33.33333

1.03000 0.52261 0.35353 0.26903 0.21835 0.18460 0.16051 0.14246 0.12843 0.11723 0.10808 0.10046 0.09403 0.08853 0.08377 0.07961 0.07595 0.07271 0.06981 0.06722 0.06487 0.06275 0.06081 0.05905 0.05743 0.05594 0.05456 0.05329 0.05211 0.05102 0.04580 0.04326 0.03958 0.03887 0.03822 0.03735 0.03613 0.03405 0.03367 0.03311 0.03273 0.03226 0.03187 0.03165 0.03129 0.03089 0.03062 0.03043 0.03015 0.03002 0.03000 0.03000 0.03000

0.00000 0.94260 2.77288 5.43834 8.88878 13.07620 17.95475 23.48061 29.61194 36.30879 43.53300 51.24818 59.41960 68.01413 77.00020 86.34770 96.02796 106.01367 116.27882 126.79866 137.54964 148.50939 159.65661 170.97108 182.43362 194.02598 205.73090 217.53197 229.41367 241.36129 313.70284 361.74994 455.02547 477.48033 499.51915 531.74111 583.05261 693.12255 717.69785 756.08652 784.54337 823.63021 858.63770 879.85405 917.60126 963.86347 999.75206 1027.33721 1076.33852 1103.54910 1110.79761 1111.09932 1111.11069

0.00000 0.49261 0.98030 1.46306 1.94090 2.41383 2.88185 3.34496 3.80318 4.25650 4.70494 5.14850 5.58720 6.02104 6.45004 6.87421 7.29357 7.70812 8.11788 8.52286 8.92309 9.31858 9.70934 10.09540 10.47677 10.85348 11.22554 11.59298 11.95582 12.31407 14.36878 15.65016 18.00890 18.55751 19.09021 19.86004 21.06742 23.60363 24.16342 25.03534 25.68056 26.56665 27.36151 27.84445 28.70719 29.77366 30.61110 31.26319 32.44883 33.13397 33.32473 33.33300 33.33332

Time Value of Money Factors Discrete Compounding 3.00%

n

Uniform Series

3.00%

TABLE A-a-9

4.00%

Time Value of Money Factors Discrete Compounding 4.00%

480 Appendix A

TABLE A-a-10 Single Sums

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Uniform Series

To Find F Given P (F|P i %,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)

1.04000 1.08160 1.12486 1.16986 1.21665 1.26532 1.31593 1.36857 1.42331 1.48024 1.53945 1.60103 1.66507 1.73168 1.80094 1.87298 1.94790 2.02582 2.10685 2.19112 2.27877 2.36992 2.46472 2.56330 2.66584 2.77247 2.88337 2.99870 3.11865 3.24340 4.10393 4.80102 6.57053 7.10668 7.68659 8.64637 10.51963 16.84226 18.94525 23.04980 26.96500 34.11933 43.17184 50.50495 69.11951 110.66256 177.17433 283.66180 1164.12891 1.2246E104 1.3552E106 1.4997E108 1.6596E110

0.96154 0.92456 0.88900 0.85480 0.82193 0.79031 0.75992 0.73069 0.70259 0.67556 0.64958 0.62460 0.60057 0.57748 0.55526 0.53391 0.51337 0.49363 0.47464 0.45639 0.43883 0.42196 0.40573 0.39012 0.37512 0.36069 0.34682 0.33348 0.32065 0.30832 0.24367 0.20829 0.15219 0.14071 0.13010 0.11566 0.09506 0.05937 0.05278 0.04338 0.03709 0.02931 0.02316 0.01980 0.01447 9.0365E203 5.6442E203 3.5253E203 8.5901E204 8.1658E205 7.3790E207 6.6680E209 6.0255E211

1.00000 2.04000 3.12160 4.24646 5.41632 6.63298 7.89829 9.21423 10.58280 12.00611 13.48635 15.02581 16.62684 18.29191 20.02359 21.82453 23.69751 25.64541 27.67123 29.77808 31.96920 34.24797 36.61789 39.08260 41.64591 44.31174 47.08421 49.96758 52.96629 56.08494 77.59831 95.02552 139.26321 152.66708 167.16472 191.15917 237.99069 396.05656 448.63137 551.24498 649.12512 827.98333 1054.29603 1237.62370 1702.98772 2741.56402 4404.35813 7066.54508 2.9078E104 3.0613E105 3.3880E107 3.7492E109 4.1490E111

1.00000 0.49020 0.32035 0.23549 0.18463 0.15076 0.12661 0.10853 0.09449 0.08329 0.07415 0.06655 0.06014 0.05467 0.04994 0.04582 0.04220 0.03899 0.03614 0.03358 0.03128 0.02920 0.02731 0.02559 0.02401 0.02257 0.02124 0.02001 0.01888 0.01783 0.01289 0.01052 7.1806E203 6.5502E203 5.9821E203 5.2312E203 4.2018E203 2.5249E203 2.2290E203 1.8141E203 1.5405E203 1.2078E203 9.4850E204 8.0800E204 5.8720E204 3.6476E204 2.2705E204 1.4151E204 3.4390E205 3.2666E206 2.9516E208 2.6672E210 2.4102E212

To Find P Given A (P|A i%,n) 0.96154 1.88609 2.77509 3.62990 4.45182 5.24214 6.00205 6.73274 7.43533 8.11090 8.76048 9.38507 9.98565 10.56312 11.11839 11.65230 12.16567 12.65930 13.13394 13.59033 14.02916 14.45112 14.85684 15.24696 15.62208 15.98277 16.32959 16.66306 16.98371 17.29203 18.90828 19.79277 21.19513 21.48218 21.74758 22.10861 22.62349 23.51564 23.68041 23.91539 24.07287 24.26728 24.42092 24.50500 24.63831 24.77409 24.85890 24.91187 24.97852 24.99796 24.99998 25.00000 25.00000

Gradient Series To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.04000 0.53020 0.36035 0.27549 0.22463 0.19076 0.16661 0.14853 0.13449 0.12329 0.11415 0.10655 0.10014 0.09467 0.08994 0.08582 0.08220 0.07899 0.07614 0.07358 0.07128 0.06920 0.06731 0.06559 0.06401 0.06257 0.06124 0.06001 0.05888 0.05783 0.05289 0.05052 0.04718 0.04655 0.04598 0.04523 0.04420 0.04252 0.04223 0.04181 0.04154 0.04121 0.04095 0.04081 0.04059 0.04036 0.04023 0.04014 0.04003 0.04000 0.04000 0.04000 0.04000

0.00000 0.92456 2.70255 5.26696 8.55467 12.50624 17.06575 22.18058 27.80127 33.88135 40.37716 47.24773 54.45462 61.96179 69.73550 77.74412 85.95809 94.34977 102.89333 111.56469 120.34136 129.20242 138.12840 147.10119 156.10400 165.12123 174.13846 183.14235 192.12059 201.06183 253.40520 286.53030 347.24455 361.16385 374.56381 393.68897 422.99665 481.01697 493.04083 511.11614 523.94309 540.73692 554.93118 563.12487 576.89491 592.24276 602.84668 610.10550 620.59757 624.45902 624.99290 624.99992 625.00000

0.00000 0.49020 0.97386 1.45100 1.92161 2.38571 2.84332 3.29443 3.73908 4.17726 4.60901 5.03435 5.45329 5.86586 6.27209 6.67200 7.06563 7.45300 7.83416 8.20912 8.57794 8.94065 9.29729 9.64790 9.99252 10.33120 10.66399 10.99092 11.31205 11.62743 13.40181 14.47651 16.38322 16.81225 17.22324 17.80704 18.69723 20.45519 20.82062 21.37185 21.76488 22.28255 22.72360 22.98000 23.41455 23.90573 24.25074 24.49056 24.84525 24.98040 24.99973 25.00000 25.00000

Appendix A 481

Single Sums

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

To Find F Given P (F|P i %,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)

1.05000 1.10250 1.15763 1.21551 1.27628 1.34010 1.40710 1.47746 1.55133 1.62889 1.71034 1.79586 1.88565 1.97993 2.07893 2.18287 2.29202 2.40662 2.52695 2.65330 2.78596 2.92526 3.07152 3.22510 3.38635 3.55567 3.73346 3.92013 4.11614 4.32194 5.79182 7.03999 10.40127 11.46740 12.64281 14.63563 18.67919 33.54513 38.83269 49.56144 60.24224 80.73037 108.18641 131.50126 194.28725 348.91199 626.59580 1125.27603 6517.39184 1.2174E105 4.2476E107 1.4821E110 5.1711E112

0.95238 0.90703 0.86384 0.82270 0.78353 0.74622 0.71068 0.67684 0.64461 0.61391 0.58468 0.55684 0.53032 0.50507 0.48102 0.45811 0.43630 0.41552 0.39573 0.37689 0.35894 0.34185 0.32557 0.31007 0.29530 0.28124 0.26785 0.25509 0.24295 0.23138 0.17266 0.14205 0.09614 0.08720 0.07910 0.06833 0.05354 0.02981 0.02575 0.02018 0.01660 0.01239 9.2433E203 7.6045E203 5.1470E203 2.8661E203 1.5959E203 8.8867E204 1.5344E204 8.2143E206 2.3542E208 6.7474E211 1.9338E213

1.00000 2.05000 3.15250 4.31013 5.52563 6.80191 8.14201 9.54911 11.02656 12.57789 14.20679 15.91713 17.71298 19.59863 21.57856 23.65749 25.84037 28.13238 30.53900 33.06595 35.71925 38.50521 41.43048 44.50200 47.72710 51.11345 54.66913 58.40258 62.32271 66.43885 95.83632 120.79977 188.02539 209.34800 232.85617 272.71262 353.58372 650.90268 756.65372 971.22882 1184.84483 1594.60730 2143.72821 2610.02516 3865.74499 6958.23971 1.2512E104 2.2486E104 1.3033E105 2.4348E106 8.4953E108 2.9641E111 1.0342E114

1.00000 0.48780 0.31721 0.23201 0.18097 0.14702 0.12282 0.10472 0.09069 0.07950 0.07039 0.06283 0.05646 0.05102 0.04634 0.04227 0.03870 0.03555 0.03275 0.03024 0.02800 0.02597 0.02414 0.02247 0.02095 0.01956 0.01829 0.01712 0.01605 0.01505 0.01043 8.2782E203 5.3184E203 4.7767E203 4.2945E203 3.6669E203 2.8282E203 1.5363E203 1.3216E203 1.0296E203 8.4399E204 6.2711E204 4.6648E204 3.8314E204 2.5868E204 1.4371E204 7.9924E205 4.4473E205 7.6730E206 4.1072E207 1.1771E209 3.3737E212 9.6692E215

To Find P Given A (P|A i%,n) 0.95238 1.85941 2.72325 3.54595 4.32948 5.07569 5.78637 6.46321 7.10782 7.72173 8.30641 8.86325 9.39357 9.89864 10.37966 10.83777 11.27407 11.68959 12.08532 12.46221 12.82115 13.16300 13.48857 13.79864 14.09394 14.37519 14.64303 14.89813 15.14107 15.37245 16.54685 17.15909 18.07716 18.25593 18.41807 18.63347 18.92929 19.40379 19.48497 19.59646 19.66801 19.75226 19.81513 19.84791 19.89706 19.94268 19.96808 19.98223 19.99693 19.99984 20.00000 20.00000 20.00000

Gradient Series To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.05000 0.53780 0.36721 0.28201 0.23097 0.19702 0.17282 0.15472 0.14069 0.12950 0.12039 0.11283 0.10646 0.10102 0.09634 0.09227 0.08870 0.08555 0.08275 0.08024 0.07800 0.07597 0.07414 0.07247 0.07095 0.06956 0.06829 0.06712 0.06605 0.06505 0.06043 0.05828 0.05532 0.05478 0.05429 0.05367 0.05283 0.05154 0.05132 0.05103 0.05084 0.05063 0.05047 0.05038 0.05026 0.05014 0.05008 0.05004 0.05001 0.05000 0.05000 0.05000 0.05000

0.00000 0.90703 2.63470 5.10281 8.23692 11.96799 16.23208 20.96996 26.12683 31.65205 37.49884 43.62405 49.98791 56.55379 63.28803 70.15970 77.14045 84.20430 91.32751 98.48841 105.66726 112.84611 120.00868 127.14024 134.22751 141.25852 148.22258 155.11011 161.91261 168.62255 206.62370 229.54518 269.24673 277.91478 286.10125 297.51040 314.34316 345.14853 351.07215 359.64605 365.47273 372.74879 378.55553 381.74922 386.82363 391.97505 395.14839 397.08516 399.38626 399.95729 399.99982 400.00000 400.00000

0.00000 0.48780 0.96749 1.43905 1.90252 2.35790 2.80523 3.24451 3.67579 4.09909 4.51444 4.92190 5.32150 5.71329 6.09731 6.47363 6.84229 7.20336 7.55690 7.90297 8.24164 8.57298 8.89706 9.21397 9.52377 9.82655 10.12240 10.41138 10.69360 10.96914 12.48719 13.37747 14.89431 15.22326 15.53372 15.96645 16.60618 17.78769 18.01759 18.35260 18.58209 18.87120 19.10436 19.23372 19.44125 19.65509 19.78900 19.87192 19.97238 19.99803 19.99999 20.00000 20.00000

Time Value of Money Factors Discrete Compounding 5.00%

n

Uniform Series

5.00%

TABLE A-a-11

6.00%

Time Value of Money Factors Discrete Compounding 6.00%

482 Appendix A

TABLE A-a-12 Single Sums

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Uniform Series

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.06000 1.12360 1.19102 1.26248 1.33823 1.41852 1.50363 1.59385 1.68948 1.79085 1.89830 2.01220 2.13293 2.26090 2.39656 2.54035 2.69277 2.85434 3.02560 3.20714 3.39956 3.60354 3.81975 4.04893 4.29187 4.54938 4.82235 5.11169 5.41839 5.74349 8.14725 10.28572 16.39387 18.42015 20.69689 24.65032 32.98769 66.37772 79.05692 105.79599 133.56500 189.46451 268.75903 339.30208 540.79597 1088.18775 2189.64755 4406.00107 3.5897E104 1.1842E106 1.2886E109 1.4022E112 1.5259E115

0.94340 0.89000 0.83962 0.79209 0.74726 0.70496 0.66506 0.62741 0.59190 0.55839 0.52679 0.49697 0.46884 0.44230 0.41727 0.39365 0.37136 0.35034 0.33051 0.31180 0.29416 0.27751 0.26180 0.24698 0.23300 0.21981 0.20737 0.19563 0.18456 0.17411 0.12274 0.09722 0.06100 0.05429 0.04832 0.04057 0.03031 0.01507 0.01265 9.4522E203 7.4870E203 5.2780E203 3.7208E203 2.9472E203 1.8491E203 9.1896E204 4.5669E204 2.2696E204 2.7858E205 8.4449E207 7.7605E210 7.1316E213 6.5536E216

1.00000 2.06000 3.18360 4.37462 5.63709 6.97532 8.39384 9.89747 11.49132 13.18079 14.97164 16.86994 18.88214 21.01507 23.27597 25.67253 28.21288 30.90565 33.75999 36.78559 39.99273 43.39229 46.99583 50.81558 54.86451 59.15638 63.70577 68.52811 73.63980 79.05819 119.12087 154.76197 256.56453 290.33590 328.28142 394.17203 533.12818 1089.62859 1300.94868 1746.59989 2209.41674 3141.07519 4462.65050 5638.36806 8996.59954 1.8120E104 3.6477E104 7.3417E104 5.9826E105 1.9736E107 2.1476E110 2.3370E113 2.5431E116

1.00000 0.48544 0.31411 0.22859 0.17740 0.14336 0.11914 0.10104 0.08702 0.07587 0.06679 0.05928 0.05296 0.04758 0.04296 0.03895 0.03544 0.03236 0.02962 0.02718 0.02500 0.02305 0.02128 0.01968 0.01823 0.01690 0.01570 0.01459 0.01358 0.01265 8.3948E203 6.4615E203 3.8977E203 3.4443E203 3.0462E203 2.5370E203 1.8757E203 9.1774E204 7.6867E204 5.7254E204 4.5261E204 3.1836E204 2.2408E204 1.7736E204 1.1115E204 5.5188E205 2.7414E205 1.3621E205 1.6715E206 5.0669E208 4.6563E211 4.2789E214 3.9322E217

0.94340 1.83339 2.67301 3.46511 4.21236 4.91732 5.58238 6.20979 6.80169 7.36009 7.88687 8.38384 8.85268 9.29498 9.71225 10.10590 10.47726 10.82760 11.15812 11.46992 11.76408 12.04158 12.30338 12.55036 12.78336 13.00317 13.21053 13.40616 13.59072 13.76483 14.62099 15.04630 15.65003 15.76186 15.86139 15.99054 16.16143 16.41558 16.45585 16.50913 16.54188 16.57870 16.60465 16.61755 16.63585 16.65135 16.65906 16.66288 16.66620 16.66665 16.66667 16.66667 16.66667

1.06000 0.54544 0.37411 0.28859 0.23740 0.20336 0.17914 0.16104 0.14702 0.13587 0.12679 0.11928 0.11296 0.10758 0.10296 0.09895 0.09544 0.09236 0.08962 0.08718 0.08500 0.08305 0.08128 0.07968 0.07823 0.07690 0.07570 0.07459 0.07358 0.07265 0.06839 0.06646 0.06390 0.06344 0.06305 0.06254 0.06188 0.06092 0.06077 0.06057 0.06045 0.06032 0.06022 0.06018 0.06011 0.06006 0.06003 0.06001 0.06000 0.06000 0.06000 0.06000 0.06000

0.00000 0.89000 2.56924 4.94552 7.93455 11.45935 15.44969 19.84158 24.57677 29.60232 34.87020 40.33686 45.96293 51.71284 57.55455 63.45925 69.40108 75.35692 81.30615 87.23044 93.11355 98.94116 104.70070 110.38121 115.97317 121.46842 126.85999 132.14200 137.30959 142.35879 170.03866 185.95682 212.03505 217.45738 222.48229 229.32225 239.04279 255.51462 258.45274 262.54931 265.21627 268.39461 270.79093 272.04706 273.93570 275.68459 276.64619 277.17002 277.68647 277.77417 277.77777 277.77778 277.77778

0.00000 0.48544 0.96118 1.42723 1.88363 2.33040 2.76758 3.19521 3.61333 4.02201 4.42129 4.81126 5.19198 5.56352 5.92598 6.27943 6.62397 6.95970 7.28673 7.60515 7.91508 8.21662 8.50991 8.79506 9.07220 9.34145 9.60294 9.85681 10.10319 10.34221 11.62977 12.35898 13.54854 13.79643 14.02666 14.34112 14.79095 15.56537 15.70583 15.90328 16.03302 16.18912 16.30814 16.37107 16.46659 16.55629 16.60636 16.63398 16.66165 16.66646 16.66667 16.66667 16.66667

Appendix A 483

Single Sums

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.07000 1.14490 1.22504 1.31080 1.40255 1.50073 1.60578 1.71819 1.83846 1.96715 2.10485 2.25219 2.40985 2.57853 2.75903 2.95216 3.15882 3.37993 3.61653 3.86968 4.14056 4.43040 4.74053 5.07237 5.42743 5.80735 6.21387 6.64884 7.11426 7.61226 11.42394 14.97446 25.72891 29.45703 33.72535 41.31500 57.94643 130.50646 159.87602 224.23439 293.92554 441.10298 661.97663 867.71633 1490.89820 3357.78838 7562.38275 1.7032E104 1.9457E105 1.1275E107 3.7858E110 1.2712E114 4.2684E117

0.93458 0.87344 0.81630 0.76290 0.71299 0.66634 0.62275 0.58201 0.54393 0.50835 0.47509 0.44401 0.41496 0.38782 0.36245 0.33873 0.31657 0.29586 0.27651 0.25842 0.24151 0.22571 0.21095 0.19715 0.18425 0.17220 0.16093 0.15040 0.14056 0.13137 0.08754 0.06678 0.03887 0.03395 0.02965 0.02420 0.01726 7.6625E203 6.2548E203 4.4596E203 3.4022E203 2.2670E203 1.5106E203 1.1525E203 6.7074E204 2.9782E204 1.3223E204 5.8713E205 5.1395E206 8.8694E208 2.6414E211 7.8666E215 2.3428E218

1.00000 2.07000 3.21490 4.43994 5.75074 7.15329 8.65402 10.25980 11.97799 13.81645 15.78360 17.88845 20.14064 22.55049 25.12902 27.88805 30.84022 33.99903 37.37896 40.99549 44.86518 49.00574 53.43614 58.17667 63.24904 68.67647 74.48382 80.69769 87.34653 94.46079 148.91346 199.63511 353.27009 406.52893 467.50497 575.92859 813.52038 1850.09222 2269.65742 3189.06268 4184.65058 6287.18543 9442.52329 1.2382E104 2.1284E104 4.7954E104 1.0802E105 2.4330E105 2.7796E106 1.6107E108 5.4083E111 1.8160E115 6.0977E118

1.00000 0.48309 0.31105 0.22523 0.17389 0.13980 0.11555 0.09747 0.08349 0.07238 0.06336 0.05590 0.04965 0.04434 0.03979 0.03586 0.03243 0.02941 0.02675 0.02439 0.02229 0.02041 0.01871 0.01719 0.01581 0.01456 0.01343 0.01239 0.01145 0.01059 6.7153E203 5.0091E203 2.8307E203 2.4598E203 2.1390E203 1.7363E203 1.2292E203 5.4051E204 4.4060E204 3.1357E204 2.3897E204 1.5905E204 1.0590E204 8.0765E205 4.6983E205 2.0853E205 9.2576E206 4.1102E206 3.5977E207 6.2086E209 1.8490E212 5.5066E216 1.6400E219

0.93458 1.80802 2.62432 3.38721 4.10020 4.76654 5.38929 5.97130 6.51523 7.02358 7.49867 7.94269 8.35765 8.74547 9.10791 9.44665 9.76322 10.05909 10.33560 10.59401 10.83553 11.06124 11.27219 11.46933 11.65358 11.82578 11.98671 12.13711 12.27767 12.40904 13.03521 13.33171 13.73047 13.80075 13.86212 13.93994 14.03918 14.17625 14.19636 14.22201 14.23711 14.25333 14.26413 14.26925 14.27613 14.28146 14.28383 14.28488 14.28564 14.28571 14.28571 14.28571 14.28571

1.07000 0.55309 0.38105 0.29523 0.24389 0.20980 0.18555 0.16747 0.15349 0.14238 0.13336 0.12590 0.11965 0.11434 0.10979 0.10586 0.10243 0.09941 0.09675 0.09439 0.09229 0.09041 0.08871 0.08719 0.08581 0.08456 0.08343 0.08239 0.08145 0.08059 0.07672 0.07501 0.07283 0.07246 0.07214 0.07174 0.07123 0.07054 0.07044 0.07031 0.07024 0.07016 0.07011 0.07008 0.07005 0.07002 0.07001 0.07000 0.07000 0.07000 0.07000 0.07000 0.07000

0.00000 0.87344 2.50603 4.79472 7.64666 10.97838 14.71487 18.78894 23.14041 27.71555 32.46648 37.35061 42.33018 47.37181 52.44605 57.52707 62.59226 67.62195 72.59910 77.50906 82.33932 87.07930 91.72013 96.25450 100.67648 104.98137 109.16556 113.22642 117.16218 120.97182 141.19902 152.29277 169.49812 172.90512 176.00368 180.12433 185.76774 194.63648 196.10351 198.07480 199.30463 200.70420 201.70162 202.20008 202.90990 203.51031 203.80529 203.94887 204.06737 204.08131 204.08163 204.08163 204.08163

0.00000 0.48309 0.95493 1.41554 1.86495 2.30322 2.73039 3.14654 3.55174 3.94607 4.32963 4.70252 5.06484 5.41673 5.75829 6.08968 6.41102 6.72247 7.02418 7.31631 7.59901 7.87247 8.13685 8.39234 8.63910 8.87733 9.10722 9.32894 9.54270 9.74868 10.83213 11.42335 12.34467 12.52868 12.69673 12.92146 13.23209 13.72976 13.81365 13.92735 13.99895 14.08122 14.14047 14.17034 14.21323 14.24997 14.26826 14.27726 14.28479 14.28569 14.28571 14.28571 14.28571

Time Value of Money Factors Discrete Compounding 7.00%

n

Uniform Series

7.00%

TABLE A-a-13

8.00%

Time Value of Money Factors Discrete Compounding 8.00%

484 Appendix A

TABLE A-a-14 Single Sums

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Uniform Series

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.08000 1.16640 1.25971 1.36049 1.46933 1.58687 1.71382 1.85093 1.99900 2.15892 2.33164 2.51817 2.71962 2.93719 3.17217 3.42594 3.70002 3.99602 4.31570 4.66096 5.03383 5.43654 5.87146 6.34118 6.84848 7.39635 7.98806 8.62711 9.31727 10.06266 15.96817 21.72452 40.21057 46.90161 54.70604 68.91386 101.25706 254.98251 321.20453 471.95483 642.08934 1018.91509 1616.89019 2199.76126 4071.60456 1.0253E104 2.5819E104 6.5016E104 1.0382E106 1.0512E108 1.0778E112 1.1051E116 1.1331E120

0.92593 0.85734 0.79383 0.73503 0.68058 0.63017 0.58349 0.54027 0.50025 0.46319 0.42888 0.39711 0.36770 0.34046 0.31524 0.29189 0.27027 0.25025 0.23171 0.21455 0.19866 0.18394 0.17032 0.15770 0.14602 0.13520 0.12519 0.11591 0.10733 0.09938 0.06262 0.04603 0.02487 0.02132 0.01828 0.01451 9.8759E203 3.9218E203 3.1133E203 2.1188E203 1.5574E203 9.8144E204 6.1847E204 4.5459E204 2.4560E204 9.7532E205 3.8731E205 1.5381E205 9.6322E207 9.5126E209 9.2779E213 9.0489E217 8.8257E221

1.00000 2.08000 3.24640 4.50611 5.86660 7.33593 8.92280 10.63663 12.48756 14.48656 16.64549 18.97713 21.49530 24.21492 27.15211 30.32428 33.75023 37.45024 41.44626 45.76196 50.42292 55.45676 60.89330 66.76476 73.10594 79.95442 87.35077 95.33883 103.96594 113.28321 187.10215 259.05652 490.13216 573.77016 671.32551 848.92320 1253.21330 3174.78140 4002.55662 5886.93543 8013.61677 1.2724E104 2.0199E104 2.7485E104 5.0883E104 1.2815E105 3.2272E105 8.1269E105 1.2977E107 1.3140E109 1.3473E113 1.3814E117 1.4163E121

1.00000 0.48077 0.30803 0.22192 0.17046 0.13632 0.11207 0.09401 0.08008 0.06903 0.06008 0.05270 0.04652 0.04130 0.03683 0.03298 0.02963 0.02670 0.02413 0.02185 0.01983 0.01803 0.01642 0.01498 0.01368 0.01251 0.01145 0.01049 9.6185E203 8.8274E203 5.3447E203 3.8602E203 2.0403E203 1.7429E203 1.4896E203 1.1780E203 7.9795E204 3.1498E204 2.4984E204 1.6987E204 1.2479E204 7.8592E205 4.9508E205 3.6384E205 1.9653E205 7.8034E206 3.0986E206 1.2305E206 7.7057E208 7.6101E210 7.4223E214 7.2391E218 7.0605E222

0.92593 1.78326 2.57710 3.31213 3.99271 4.62288 5.20637 5.74664 6.24689 6.71008 7.13896 7.53608 7.90378 8.24424 8.55948 8.85137 9.12164 9.37189 9.60360 9.81815 10.01680 10.20074 10.37106 10.52876 10.67478 10.80998 10.93516 11.05108 11.15841 11.25778 11.71719 11.92461 12.18914 12.23348 12.27151 12.31861 12.37655 12.45098 12.46108 12.47351 12.48053 12.48773 12.49227 12.49432 12.49693 12.49878 12.49952 12.49981 12.49999 12.50000 12.50000 12.50000 12.50000

1.08000 0.56077 0.38803 0.30192 0.25046 0.21632 0.19207 0.17401 0.16008 0.14903 0.14008 0.13270 0.12652 0.12130 0.11683 0.11298 0.10963 0.10670 0.10413 0.10185 0.09983 0.09803 0.09642 0.09498 0.09368 0.09251 0.09145 0.09049 0.08962 0.08883 0.08534 0.08386 0.08204 0.08174 0.08149 0.08118 0.08080 0.08031 0.08025 0.08017 0.08012 0.08008 0.08005 0.08004 0.08002 0.08001 0.08000 0.08000 0.08000 0.08000 0.08000 0.08000 0.08000

0.00000 0.85734 2.44500 4.65009 7.37243 10.52327 14.02422 17.80610 21.80809 25.97683 30.26566 34.63391 39.04629 43.47228 47.88566 52.26402 56.58832 60.84256 65.01337 69.08979 73.06291 76.92566 80.67259 84.29968 87.80411 91.18415 94.43901 97.56868 100.57385 103.45579 118.28385 126.04220 137.44276 139.59279 141.51214 144.00645 147.30001 152.10756 152.84485 153.80008 154.37137 154.99254 155.41120 155.61073 155.88006 156.08846 156.18004 156.21991 156.24768 156.24997 156.25000 156.25000 156.25000

0.00000 0.48077 0.94874 1.40396 1.84647 2.27635 2.69366 3.09852 3.49103 3.87131 4.23950 4.59575 4.94021 5.27305 5.59446 5.90463 6.20375 6.49203 6.76969 7.03695 7.29403 7.54118 7.77863 8.00661 8.22538 8.43518 8.63627 8.82888 9.01328 9.18971 10.09490 10.56992 11.27584 11.41071 11.53177 11.69015 11.90154 12.21652 12.26577 12.33013 12.36897 12.41158 12.44059 12.45452 12.47347 12.48829 12.49489 12.49779 12.49983 12.50000 12.50000 12.50000 12.50000

Appendix A 485

Single Sums

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.09000 1.18810 1.29503 1.41158 1.53862 1.67710 1.82804 1.99256 2.17189 2.36736 2.58043 2.81266 3.06580 3.34173 3.64248 3.97031 4.32763 4.71712 5.14166 5.60441 6.10881 6.65860 7.25787 7.91108 8.62308 9.39916 10.24508 11.16714 12.17218 13.26768 22.25123 31.40942 62.58524 74.35752 88.34417 114.40826 176.03129 495.11702 641.19089 986.55167 1392.59819 2335.52658 3916.91189 5529.04079 1.1017E104 3.0987E104 8.7156E104 2.4514E105 5.4547E106 9.6020E108 2.9754E113 9.2197E117 2.8569E122

0.91743 0.84168 0.77218 0.70843 0.64993 0.59627 0.54703 0.50187 0.46043 0.42241 0.38753 0.35553 0.32618 0.29925 0.27454 0.25187 0.23107 0.21199 0.19449 0.17843 0.16370 0.15018 0.13778 0.12640 0.11597 0.10639 0.09761 0.08955 0.08215 0.07537 0.04494 0.03184 0.01598 0.01345 0.01132 8.7406E203 5.6808E203 2.0197E203 1.5596E203 1.0136E203 7.1808E204 4.2817E204 2.5530E204 1.8086E204 9.0769E205 3.2272E205 1.1474E205 4.0793E206 1.8333E207 1.0415E209 3.3609E214 1.0846E218 3.5003E223

1.00000 2.09000 3.27810 4.57313 5.98471 7.52333 9.20043 11.02847 13.02104 15.19293 17.56029 20.14072 22.95338 26.01919 29.36092 33.00340 36.97370 41.30134 46.01846 51.16012 56.76453 62.87334 69.53194 76.78981 84.70090 93.32398 102.72313 112.96822 124.13536 136.30754 236.12472 337.88245 684.28041 815.08356 970.49077 1260.09180 1944.79213 5490.18906 7113.23215 1.0951E104 1.5462E104 2.5939E104 4.3510E104 6.1423E104 1.2240E105 3.4429E105 9.6839E105 2.7238E106 6.0608E107 1.0669E110 3.3060E114 1.0244E119 3.1744E123

1.00000 0.47847 0.30505 0.21867 0.16709 0.13292 0.10869 0.09067 0.07680 0.06582 0.05695 0.04965 0.04357 0.03843 0.03406 0.03030 0.02705 0.02421 0.02173 0.01955 0.01762 0.01590 0.01438 0.01302 0.01181 0.01072 9.7349E203 8.8520E203 8.0557E203 7.3364E203 4.2350E203 2.9596E203 1.4614E203 1.2269E203 1.0304E203 7.9359E204 5.1419E204 1.8214E204 1.4058E204 9.1319E205 6.4674E205 3.8552E205 2.2983E205 1.6281E205 8.1700E206 2.9045E206 1.0326E206 3.6714E207 1.6500E208 9.3731E211 3.0248E215 9.7617E220 3.1502E224

0.91743 1.75911 2.53129 3.23972 3.88965 4.48592 5.03295 5.53482 5.99525 6.41766 6.80519 7.16073 7.48690 7.78615 8.06069 8.31256 8.54363 8.75563 8.95011 9.12855 9.29224 9.44243 9.58021 9.70661 9.82258 9.92897 10.02658 10.11613 10.19828 10.27365 10.61176 10.75736 10.93358 10.96168 10.98534 11.01399 11.04799 11.08867 11.09378 11.09985 11.10313 11.10635 11.10827 11.10910 11.11010 11.11075 11.11098 11.11107 11.11111 11.11111 11.11111 11.11111 11.11111

1.09000 0.56847 0.39505 0.30867 0.25709 0.22292 0.19869 0.18067 0.16680 0.15582 0.14695 0.13965 0.13357 0.12843 0.12406 0.12030 0.11705 0.11421 0.11173 0.10955 0.10762 0.10590 0.10438 0.10302 0.10181 0.10072 0.09973 0.09885 0.09806 0.09734 0.09424 0.09296 0.09146 0.09123 0.09103 0.09079 0.09051 0.09018 0.09014 0.09009 0.09006 0.09004 0.09002 0.09002 0.09001 0.09000 0.09000 0.09000 0.09000 0.09000 0.09000 0.09000 0.09000

0.00000 0.84168 2.38605 4.51132 7.11105 10.09238 13.37459 16.88765 20.57108 24.37277 28.24810 32.15898 36.07313 39.96333 43.80686 47.58491 51.28208 54.88598 58.38679 61.77698 65.05094 68.20475 71.23594 74.14326 76.92649 79.58630 82.12410 84.54191 86.84224 89.02800 99.93194 105.37619 112.96246 114.32507 115.51926 117.03621 118.96825 121.59166 121.96458 122.43064 122.69793 122.97576 123.15295 123.23350 123.33666 123.40978 123.43855 123.44976 123.45640 123.45679 123.45679 123.45679 123.45679

0.00000 0.47847 0.94262 1.39250 1.82820 2.24979 2.65740 3.05117 3.43123 3.79777 4.15096 4.49102 4.81816 5.13262 5.43463 5.72446 6.00238 6.26865 6.52358 6.76745 7.00056 7.22322 7.43574 7.63843 7.83160 8.01556 8.19064 8.35714 8.51538 8.66566 9.41709 9.79573 10.33170 10.42952 10.51577 10.62614 10.76832 10.96540 10.99396 11.02994 11.05075 11.07256 11.08660 11.09302 11.10131 11.10724 11.10960 11.11052 11.11108 11.11111 11.11111 11.11111 11.11111

Time Value of Money Factors Discrete Compounding 9.00%

n

Uniform Series

9.00%

TABLE A-a-15

10.00%

Time Value of Money Factors Discrete Compounding 10.00%

486 Appendix A

TABLE A-a-16 Single Sums

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Uniform Series

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

To Find P Given G (P|G i%,n)

To Find A Given G (A|G i%,n)

1.10000 1.21000 1.33100 1.46410 1.61051 1.77156 1.94872 2.14359 2.35795 2.59374 2.85312 3.13843 3.45227 3.79750 4.17725 4.59497 5.05447 5.55992 6.11591 6.72750 7.40025 8.14027 8.95430 9.84973 10.83471 11.91818 13.10999 14.42099 15.86309 17.44940 30.91268 45.25926 97.01723 117.39085 142.04293 189.05914 304.48164 955.59382 1271.89537 2048.40021 2999.06275 5313.02261 9412.34365 1.3781E104 2.9540E104 9.2709E104 2.9096E105 9.1316E105 2.8228E107 8.5950E109 7.9683E114 7.3874E119 6.8487E124

0.90909 0.82645 0.75131 0.68301 0.62092 0.56447 0.51316 0.46651 0.42410 0.38554 0.35049 0.31863 0.28966 0.26333 0.23939 0.21763 0.19784 0.17986 0.16351 0.14864 0.13513 0.12285 0.11168 0.10153 0.09230 0.08391 0.07628 0.06934 0.06304 0.05731 0.03235 0.02209 0.01031 8.5186E203 7.0401E203 5.2894E203 3.2843E203 1.0465E203 7.8623E204 4.8819E204 3.3344E204 1.8822E204 1.0624E204 7.2566E205 3.3852E205 1.0786E205 3.4369E206 1.0951E206 3.5426E208 1.1635E210 1.2550E215 1.3537E220 1.4601E225

1.00000 2.10000 3.31000 4.64100 6.10510 7.71561 9.48717 11.43589 13.57948 15.93742 18.53117 21.38428 24.52271 27.97498 31.77248 35.94973 40.54470 45.59917 51.15909 57.27500 64.00250 71.40275 79.54302 88.49733 98.34706 109.18177 121.09994 134.20994 148.63093 164.49402 299.12681 442.59256 960.17234 1163.90853 1410.42932 1880.59142 3034.81640 9545.93818 1.2709E104 2.0474E104 2.9981E104 5.3120E104 9.4113E104 1.3780E105 2.9539E105 9.2708E105 2.9096E106 9.1316E106 2.8228E108 8.5950E110 7.9683E115 7.3874E120 6.8487E125

1.00000 0.47619 0.30211 0.21547 0.16380 0.12961 0.10541 0.08744 0.07364 0.06275 0.05396 0.04676 0.04078 0.03575 0.03147 0.02782 0.02466 0.02193 0.01955 0.01746 0.01562 0.01401 0.01257 0.01130 0.01017 9.1590E203 8.2576E203 7.4510E203 6.7281E203 6.0792E203 3.3431E203 2.2594E203 1.0415E203 8.5917E204 7.0900E204 5.3175E204 3.2951E204 1.0476E204 7.8685E205 4.8842E205 3.3355E205 1.8825E205 1.0625E205 7.2571E206 3.3854E206 1.0787E206 3.4369E207 1.0951E207 3.5426E209 1.1635E211 1.2550E216 1.3537E221 1.4601E226

0.90909 1.73554 2.48685 3.16987 3.79079 4.35526 4.86842 5.33493 5.75902 6.14457 6.49506 6.81369 7.10336 7.36669 7.60608 7.82371 8.02155 8.20141 8.36492 8.51356 8.64869 8.77154 8.88322 8.98474 9.07704 9.16095 9.23722 9.30657 9.36961 9.42691 9.67651 9.77905 9.89693 9.91481 9.92960 9.94711 9.96716 9.98954 9.99214 9.99512 9.99667 9.99812 9.99894 9.99927 9.99966 9.99989 9.99997 9.99999 10.00000 10.00000 10.00000 10.00000 10.00000

1.10000 0.57619 0.40211 0.31547 0.26380 0.22961 0.20541 0.18744 0.17364 0.16275 0.15396 0.14676 0.14078 0.13575 0.13147 0.12782 0.12466 0.12193 0.11955 0.11746 0.11562 0.11401 0.11257 0.11130 0.11017 0.10916 0.10826 0.10745 0.10673 0.10608 0.10334 0.10226 0.10104 0.10086 0.10071 0.10053 0.10033 0.10010 0.10008 0.10005 0.10003 0.10002 0.10001 0.10001 0.10000 0.10000 0.10000 0.10000 0.10000 0.10000 0.10000 0.10000 0.10000

0.00000 0.82645 2.32908 4.37812 6.86180 9.68417 12.76312 16.02867 19.42145 22.89134 26.39628 29.90122 33.37719 36.80050 40.15199 43.41642 46.58194 49.63954 52.58268 55.40691 58.10952 60.68929 63.14621 65.48130 67.69640 69.79404 71.77726 73.64953 75.41463 77.07658 85.11938 88.95254 94.02168 94.88887 95.63512 96.56192 97.70101 99.14189 99.33171 99.56063 99.68657 99.81178 99.88738 99.92018 99.96005 99.98598 99.99512 99.99831 99.99993 100.00000 100.00000 100.00000 100.00000

0.00000 0.47619 0.93656 1.38117 1.81013 2.22356 2.62162 3.00448 3.37235 3.72546 4.06405 4.38840 4.69879 4.99553 5.27893 5.54934 5.80710 6.05256 6.28610 6.50808 6.71888 6.91889 7.10848 7.28805 7.45798 7.61865 7.77044 7.91372 8.04886 8.17623 8.79650 9.09623 9.50009 9.57041 9.63132 9.70754 9.80229 9.92458 9.94099 9.96093 9.97198 9.98306 9.98980 9.99274 9.99634 9.99871 9.99955 9.99984 9.99999 10.00000 10.00000 10.00000 10.00000

Appendix A 487

Single Sums

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

1.11000 1.23210 1.36763 1.51807 1.68506 1.87041 2.07616 2.30454 2.55804 2.83942 3.15176 3.49845 3.88328 4.31044 4.78459 5.31089 5.89509 6.54355 7.26334 8.06231 8.94917 9.93357 11.02627 12.23916 13.58546 15.07986 16.73865 18.57990 20.62369 22.89230 42.81808 65.00087 149.79695 184.56483 227.40232 311.00247 524.05724 1833.38837 2507.39877 4225.11275 6414.01865 1.1997E104 2.2439E104 3.4064E104 7.8502E104 2.7464E105 9.6080E105 3.3613E106 1.4392E108 7.5425E110 2.0714E116 5.6889E121 1.5624E127

0.90090 0.81162 0.73119 0.65873 0.59345 0.53464 0.48166 0.43393 0.39092 0.35218 0.31728 0.28584 0.25751 0.23199 0.20900 0.18829 0.16963 0.15282 0.13768 0.12403 0.11174 0.10067 0.09069 0.08170 0.07361 0.06631 0.05974 0.05382 0.04849 0.04368 0.02335 0.01538 6.6757E203 5.4182E203 4.3975E203 3.2154E203 1.9082E203 5.4544E204 3.9882E204 2.3668E204 1.5591E204 8.3355E205 4.4565E205 2.9356E205 1.2738E205 3.6412E206 1.0408E206 2.9750E207 6.9481E209 1.3258E211 4.8276E217 1.7578E222 6.4005E228

1.00000 2.11000 3.34210 4.70973 6.22780 7.91286 9.78327 11.85943 14.16397 16.72201 19.56143 22.71319 26.21164 30.09492 34.40536 39.18995 44.50084 50.39594 56.93949 64.20283 72.26514 81.21431 91.14788 102.17415 114.41331 127.99877 143.07864 159.81729 178.39719 199.02088 380.16441 581.82607 1352.69958 1668.77115 2058.20294 2818.20424 4755.06584 1.6658E104 2.2785E104 3.8401E104 5.8300E104 1.0905E105 2.0398E105 3.0967E105 7.1365E105 2.4967E106 8.7345E106 3.0557E107 1.3084E109 6.8568E111 1.8831E117 5.1717E122 1.4203E128

1.00000 0.47393 0.29921 0.21233 0.16057 0.12638 0.10222 0.08432 0.07060 0.05980 0.05112 0.04403 0.03815 0.03323 0.02907 0.02552 0.02247 0.01984 0.01756 0.01558 0.01384 0.01231 0.01097 9.7872E203 8.7402E203 7.8126E203 6.9892E203 6.2571E203 5.6055E203 5.0246E203 2.6304E203 1.7187E203 7.3926E204 5.9924E204 4.8586E204 3.5484E204 2.1030E204 6.0031E205 4.3888E205 2.6041E205 1.7153E205 9.1698E206 4.9024E206 3.2293E206 1.4013E206 4.0053E207 1.1449E207 3.2725E208 7.6429E210 1.4584E212 5.3103E218 1.9336E223 7.0405E229

0.90090 1.71252 2.44371 3.10245 3.69590 4.23054 4.71220 5.14612 5.53705 5.88923 6.20652 6.49236 6.74987 6.98187 7.19087 7.37916 7.54879 7.70162 7.83929 7.96333 8.07507 8.17574 8.26643 8.34814 8.42174 8.48806 8.54780 8.60162 8.65011 8.69379 8.87859 8.95105 9.03022 9.04165 9.05093 9.06168 9.07356 9.08595 9.08728 9.08876 9.08949 9.09015 9.09050 9.09064 9.09079 9.09088 9.09090 9.09091 9.09091 9.09091 9.09091 9.09091 9.09091

1.11000 0.58393 0.40921 0.32233 0.27057 0.23638 0.21222 0.19432 0.18060 0.16980 0.16112 0.15403 0.14815 0.14323 0.13907 0.13552 0.13247 0.12984 0.12756 0.12558 0.12384 0.12231 0.12097 0.11979 0.11874 0.11781 0.11699 0.11626 0.11561 0.11502 0.11263 0.11172 0.11074 0.11060 0.11049 0.11035 0.11021 0.11006 0.11004 0.11003 0.11002 0.11001 0.11000 0.11000 0.11000 0.11000 0.11000 0.11000 0.11000 0.11000 0.11000 0.11000 0.11000

To Find P Given G (P|G i%,n) 0.00000 0.81162 2.27401 4.25020 6.62400 9.29721 12.18716 15.22464 18.35204 21.52170 24.69454 27.83878 30.92896 33.94489 36.87095 39.69533 42.40945 45.00743 47.48563 49.84227 52.07712 54.19116 56.18640 58.06561 59.83220 61.49004 63.04334 64.49652 65.85418 67.12098 73.07116 75.77886 79.17988 79.73405 80.20238 80.77119 81.44610 82.24254 82.33975 82.45294 82.51269 82.56954 82.60205 82.61551 82.63107 82.64035 82.64329 82.64421 82.64462 82.64463 82.64463 82.64463 82.64463

To Find A Given G (A|G i%,n) 0.00000 0.47393 0.93055 1.36995 1.79226 2.19764 2.58630 2.95847 3.31441 3.65442 3.97881 4.28793 4.58216 4.86187 5.12747 5.37938 5.61804 5.84389 6.05739 6.25898 6.44912 6.62829 6.79693 6.95552 7.10449 7.24430 7.37539 7.49818 7.61310 7.72056 8.23004 8.46592 8.76832 8.81853 8.86123 8.91349 8.97620 9.05162 9.06099 9.07197 9.07781 9.08341 9.08663 9.08797 9.08953 9.09047 9.09077 9.09087 9.09091 9.09091 9.09091 9.09091 9.09091

Time Value of Money Factors Discrete Compounding 11.00%

n

Uniform Series

11.00%

TABLE A-a-17

12.00%

Time Value of Money Factors Discrete Compounding 12.00%

488 Appendix A

TABLE A-a-18 Single Sums

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Uniform Series

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i %,n)

To Find A Given P (A|P i%,n)

1.12000 1.25440 1.40493 1.57352 1.76234 1.97382 2.21068 2.47596 2.77308 3.10585 3.47855 3.89598 4.36349 4.88711 5.47357 6.13039 6.86604 7.68997 8.61276 9.64629 10.80385 12.10031 13.55235 15.17863 17.00006 19.04007 21.32488 23.88387 26.74993 29.95992 59.13557 93.05097 230.39078 289.00219 362.52435 509.32061 897.59693 3497.01610 4913.05584 8658.48310 1.3624E104 2.6892E104 5.3080E104 8.3522E104 2.0680E105 8.0568E105 3.1389E106 1.2229E107 7.2318E108 6.4912E111 5.2298E117 4.2136E123 3.3948E129

0.89286 0.79719 0.71178 0.63552 0.56743 0.50663 0.45235 0.40388 0.36061 0.32197 0.28748 0.25668 0.22917 0.20462 0.18270 0.16312 0.14564 0.13004 0.11611 0.10367 0.09256 0.08264 0.07379 0.06588 0.05882 0.05252 0.04689 0.04187 0.03738 0.03338 0.01691 0.01075 4.3405E203 3.4602E203 2.7584E203 1.9634E203 1.1141E203 2.8596E204 2.0354E204 1.1549E204 7.3398E205 3.7186E205 1.8840E205 1.1973E205 4.8356E206 1.2412E206 3.1858E207 8.1772E208 1.3828E209 1.5405E212 1.9121E218 2.3733E224 2.9457E230

1.00000 2.12000 3.37440 4.77933 6.35285 8.11519 10.08901 12.29969 14.77566 17.54874 20.65458 24.13313 28.02911 32.39260 37.27971 42.75328 48.88367 55.74971 63.43968 72.05244 81.69874 92.50258 104.60289 118.15524 133.33387 150.33393 169.37401 190.69889 214.58275 241.33268 484.46312 767.09142 1911.58980 2400.01825 3012.70289 4236.00505 7471.64111 2.9133E104 4.0934E104 7.2146E104 1.1353E105 2.2409E105 4.4232E105 6.9601E105 1.7233E106 6.7140E106 2.6158E107 1.0191E108 6.0265E109 5.4093E112 4.3582E118 3.5113E124 2.8290E130

1.00000 0.47170 0.29635 0.20923 0.15741 0.12323 0.09912 0.08130 0.06768 0.05698 0.04842 0.04144 0.03568 0.03087 0.02682 0.02339 0.02046 0.01794 0.01576 0.01388 0.01224 0.01081 9.5600E203 8.4634E203 7.5000E203 6.6519E203 5.9041E203 5.2439E203 4.6602E203 4.1437E203 2.0641E203 1.3036E203 5.2312E204 4.1666E204 3.3193E204 2.3607E204 1.3384E204 3.4325E205 2.4430E205 1.3861E205 8.8084E206 4.4625E206 2.2608E206 1.4368E206 5.8028E207 1.4894E207 3.8230E208 9.8126E209 1.6593E210 1.8487E213 2.2945E219 2.8479E225 3.5348E231

0.89286 1.69005 2.40183 3.03735 3.60478 4.11141 4.56376 4.96764 5.32825 5.65022 5.93770 6.19437 6.42355 6.62817 6.81086 6.97399 7.11963 7.24967 7.36578 7.46944 7.56200 7.64465 7.71843 7.78432 7.84314 7.89566 7.94255 7.98442 8.02181 8.05518 8.19241 8.24378 8.29716 8.30450 8.31035 8.31697 8.32405 8.33095 8.33164 8.33237 8.33272 8.33302 8.33318 8.33323 8.33329 8.33332 8.33333 8.33333 8.33333 8.33333 8.33333 8.33333 8.33333

1.12000 0.59170 0.41635 0.32923 0.27741 0.24323 0.21912 0.20130 0.18768 0.17698 0.16842 0.16144 0.15568 0.15087 0.14682 0.14339 0.14046 0.13794 0.13576 0.13388 0.13224 0.13081 0.12956 0.12846 0.12750 0.12665 0.12590 0.12524 0.12466 0.12414 0.12206 0.12130 0.12052 0.12042 0.12033 0.12024 0.12013 0.12003 0.12002 0.12001 0.12001 0.12000 0.12000 0.12000 0.12000 0.12000 0.12000 0.12000 0.12000 0.12000 0.12000 0.12000 0.12000

To Find P Given G (P|G i%,n) 0.00000 0.79719 2.22075 4.12731 6.39702 8.93017 11.64427 14.47145 17.35633 20.25409 23.12885 25.95228 28.70237 31.36242 33.92017 36.36700 38.69731 40.90798 42.99790 44.96757 46.81876 48.55425 50.17759 51.69288 53.10464 54.41766 55.63689 56.76736 57.81409 58.78205 63.19703 65.11587 67.40684 67.76241 68.05756 68.40821 68.81003 69.25301 69.30310 69.35943 69.38797 69.41397 69.42806 69.43364 69.43976 69.44312 69.44407 69.44434 69.44444 69.44444 69.44444 69.44444 69.44444

To Find A Given G (A|G i%,n) 0.00000 0.47170 0.92461 1.35885 1.77459 2.17205 2.55147 2.91314 3.25742 3.58465 3.89525 4.18965 4.46830 4.73169 4.98030 5.21466 5.43530 5.64274 5.83752 6.02020 6.19132 6.35141 6.50101 6.64064 6.77084 6.89210 7.00491 7.10976 7.20712 7.29742 7.71409 7.89879 8.12408 8.15972 8.18950 8.22513 8.26641 8.31274 8.31806 8.32409 8.32717 8.32999 8.33152 8.33214 8.33281 8.33318 8.33329 8.33332 8.33333 8.33333 8.33333 8.33333 8.33333

Appendix A 489

Single Sums

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

1.15000 1.32250 1.52088 1.74901 2.01136 2.31306 2.66002 3.05902 3.51788 4.04556 4.65239 5.35025 6.15279 7.07571 8.13706 9.35762 10.76126 12.37545 14.23177 16.36654 18.82152 21.64475 24.89146 28.62518 32.91895 37.85680 43.53531 50.06561 57.57545 66.21177 153.15185 267.86355 819.40071 1083.65744 1433.13697 2179.62218 4383.99875 2.3455E104 3.5673E104 7.1751E104 1.2549E105 2.9027E105 6.7142E105 1.1743E106 3.5923E106 1.9219E107 1.0283E108 5.5016E108 8.4258E110 3.6939E114 7.0994E121 1.3645E129 2.6224E136

0.86957 0.75614 0.65752 0.57175 0.49718 0.43233 0.37594 0.32690 0.28426 0.24718 0.21494 0.18691 0.16253 0.14133 0.12289 0.10686 0.09293 0.08081 0.07027 0.06110 0.05313 0.04620 0.04017 0.03493 0.03038 0.02642 0.02297 0.01997 0.01737 0.01510 6.5295E203 3.7332E203 1.2204E203 9.2280E204 6.9777E204 4.5880E204 2.2810E204 4.2634E205 2.8033E205 1.3937E205 7.9686E206 3.4450E206 1.4894E206 8.5156E207 2.7838E207 5.2031E208 9.7249E209 1.8177E209 1.1868E211 2.7072E215 1.4086E222 7.3289E230 3.8133E237

1.00000 2.15000 3.47250 4.99338 6.74238 8.75374 11.06680 13.72682 16.78584 20.30372 24.34928 29.00167 34.35192 40.50471 47.58041 55.71747 65.07509 75.83636 88.21181 102.44358 118.81012 137.63164 159.27638 184.16784 212.79302 245.71197 283.56877 327.10408 377.16969 434.74515 1014.34568 1779.09031 5456.00475 7217.71628 9547.57978 1.4524E104 2.9220E104 1.5636E105 2.3781E105 4.7833E105 8.3661E105 1.9351E106 4.4761E106 7.8287E106 2.3948E107 1.2813E108 6.8553E108 3.6677E109 5.6172E111 2.4626E115 4.7329E122 9.0965E129 1.7483E137

1.00000 0.46512 0.28798 0.20027 0.14832 0.11424 0.09036 0.07285 0.05957 0.04925 0.04107 0.03448 0.02911 0.02469 0.02102 0.01795 0.01537 0.01319 0.01134 9.7615E203 8.4168E203 7.2658E203 6.2784E203 5.4298E203 4.6994E203 4.0698E203 3.5265E203 3.0571E203 2.6513E203 2.3002E203 9.8586E204 5.6209E204 1.8328E204 1.3855E204 1.0474E204 6.8851E205 3.4223E205 6.3954E206 4.2050E206 2.0906E206 1.1953E206 5.1676E207 2.2341E207 1.2773E207 4.1757E208 7.8046E209 1.4587E209 2.7265E210 1.7802E212 4.0608E216 2.1129E223 1.0993E230 5.7199E238

0.86957 1.62571 2.28323 2.85498 3.35216 3.78448 4.16042 4.48732 4.77158 5.01877 5.23371 5.42062 5.58315 5.72448 5.84737 5.95423 6.04716 6.12797 6.19823 6.25933 6.31246 6.35866 6.39884 6.43377 6.46415 6.49056 6.51353 6.53351 6.55088 6.56598 6.62314 6.64178 6.65853 6.66051 6.66201 6.66361 6.66515 6.66638 6.66648 6.66657 6.66661 6.66664 6.66666 6.66666 6.66666 6.66667 6.66667 6.66667 6.66667 6.66667 6.66667 6.66667 6.66667

1.15000 0.61512 0.43798 0.35027 0.29832 0.26424 0.24036 0.22285 0.20957 0.19925 0.19107 0.18448 0.17911 0.17469 0.17102 0.16795 0.16537 0.16319 0.16134 0.15976 0.15842 0.15727 0.15628 0.15543 0.15470 0.15407 0.15353 0.15306 0.15265 0.15230 0.15099 0.15056 0.15018 0.15014 0.15010 0.15007 0.15003 0.15001 0.15000 0.15000 0.15000 0.15000 0.15000 0.15000 0.15000 0.15000 0.15000 0.15000 0.15000 0.15000 0.15000 0.15000 0.15000

To Find P Given G (P|G i%,n) 0.00000 0.75614 2.07118 3.78644 5.77514 7.93678 10.19240 12.48072 14.75481 16.97948 19.12891 21.18489 23.13522 24.97250 26.69302 28.29599 29.78280 31.15649 32.42127 33.58217 34.64479 35.61500 36.49884 37.30232 38.03139 38.69177 39.28899 39.82828 40.31460 40.75259 42.58717 43.28299 43.99967 44.09583 44.17154 44.25583 44.34307 44.42209 44.42918 44.43639 44.43963 44.44222 44.44343 44.44384 44.44423 44.44440 44.44444 44.44444 44.44444 44.44444 44.44444 44.44444 44.44444

To Find A Given G (A|G i%,n) 0.00000 0.46512 0.90713 1.32626 1.72281 2.09719 2.44985 2.78133 3.09223 3.38320 3.65494 3.90820 4.14376 4.36241 4.56496 4.75225 4.92509 5.08431 5.23073 5.36514 5.48832 5.60102 5.70398 5.79789 5.88343 5.96123 6.03190 6.09600 6.15408 6.20663 6.43006 6.51678 6.60802 6.62048 6.63036 6.64142 6.65298 6.66360 6.66456 6.66555 6.66600 6.66636 6.66652 6.66658 6.66664 6.66666 6.66667 6.66667 6.66667 6.66667 6.66667 6.66667 6.66667

Time Value of Money Factors Discrete Compounding 15.00%

n

Uniform Series

15.00%

TABLE A-a-19

18.00%

Time Value of Money Factors Discrete Compounding 18.00%

490 Appendix A

TABLE A-a-20 Single Sums

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Uniform Series

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

1.18000 1.39240 1.64303 1.93878 2.28776 2.69955 3.18547 3.75886 4.43545 5.23384 6.17593 7.28759 8.59936 10.14724 11.97375 14.12902 16.67225 19.67325 23.21444 27.39303 32.32378 38.14206 45.00763 53.10901 62.66863 73.94898 87.25980 102.96656 121.50054 143.37064 387.03680 750.37834 2820.56655 3927.35686 5468.45169 8984.84112 2.0555E104 1.4980E105 2.4612E105 5.6307E105 1.0917E106 2.9470E106 7.9556E106 1.5424E107 5.7977E107 4.2251E108 3.0791E109 2.2439E110 8.6848E112 1.7852E117 7.5426E125 3.1869E134 1.3465E143

0.84746 0.71818 0.60863 0.51579 0.43711 0.37043 0.31393 0.26604 0.22546 0.19106 0.16192 0.13722 0.11629 0.09855 0.08352 0.07078 0.05998 0.05083 0.04308 0.03651 0.03094 0.02622 0.02222 0.01883 0.01596 0.01352 0.01146 9.7119E203 8.2304E203 6.9749E203 2.5837E203 1.3327E203 3.5454E204 2.5462E204 1.8287E204 1.1130E204 4.8650E205 6.6757E206 4.0630E206 1.7760E206 9.1603E207 3.3933E207 1.2570E207 6.4833E208 1.7248E208 2.3668E209 3.2477E210 4.4565E211 1.1514E213 5.6017E218 1.3258E226 3.1379E235 7.4267E244

1.00000 2.18000 3.57240 5.21543 7.15421 9.44197 12.14152 15.32700 19.08585 23.52131 28.75514 34.93107 42.21866 50.81802 60.96527 72.93901 87.06804 103.74028 123.41353 146.62797 174.02100 206.34479 244.48685 289.49448 342.60349 405.27211 479.22109 566.48089 669.44745 790.94799 2144.64890 4163.21303 1.5664E104 2.1813E104 3.0375E104 4.9910E104 1.1419E105 8.3220E105 1.3673E106 3.1281E106 6.0648E106 1.6372E107 4.4198E107 8.5690E107 3.2210E108 2.3473E109 1.7106E110 1.2466E111 4.8249E113 9.9177E117 4.1903E126 1.7705E135 7.4805E143

1.00000 0.45872 0.27992 0.19174 0.13978 0.10591 0.08236 0.06524 0.05239 0.04251 0.03478 0.02863 0.02369 0.01968 0.01640 0.01371 0.01149 9.6395E203 8.1028E203 6.8200E203 5.7464E203 4.8463E203 4.0902E203 3.4543E203 2.9188E203 2.4675E203 2.0867E203 1.7653E203 1.4938E203 1.2643E203 4.6628E204 2.4020E204 6.3840E205 4.5844E205 3.2922E205 2.0036E205 8.7574E206 1.2016E206 7.3135E207 3.1968E207 1.6489E207 6.1079E208 2.2626E208 1.1670E208 3.1047E209 4.2602E210 5.8458E211 8.0216E212 2.0726E214 1.0083E218 2.3864E227 5.6482E236 1.3368E244

0.84746 1.56564 2.17427 2.69006 3.12717 3.49760 3.81153 4.07757 4.30302 4.49409 4.65601 4.79322 4.90951 5.00806 5.09158 5.16235 5.22233 5.27316 5.31624 5.35275 5.38368 5.40990 5.43212 5.45095 5.46691 5.48043 5.49189 5.50160 5.50983 5.51681 5.54120 5.54815 5.55359 5.55414 5.55454 5.55494 5.55529 5.55552 5.55553 5.55555 5.55555 5.55555 5.55555 5.55556 5.55556 5.55556 5.55556 5.55556 5.55556 5.55556 5.55556 5.55556 5.55556

1.18000 0.63872 0.45992 0.37174 0.31978 0.28591 0.26236 0.24524 0.23239 0.22251 0.21478 0.20863 0.20369 0.19968 0.19640 0.19371 0.19149 0.18964 0.18810 0.18682 0.18575 0.18485 0.18409 0.18345 0.18292 0.18247 0.18209 0.18177 0.18149 0.18126 0.18047 0.18024 0.18006 0.18005 0.18003 0.18002 0.18001 0.18000 0.18000 0.18000 0.18000 0.18000 0.18000 0.18000 0.18000 0.18000 0.18000 0.18000 0.18000 0.18000 0.18000 0.18000 0.18000

To Find P Given G (P|G i%,n) 0.00000 0.71818 1.93545 3.48281 5.23125 7.08341 8.96696 10.82922 12.63287 14.35245 15.97164 17.48106 18.87651 20.15765 21.32687 22.38852 23.34820 24.21231 24.98769 25.68130 26.30004 26.85061 27.33942 27.77249 28.15546 28.49353 28.79149 29.05371 29.28416 29.48643 30.26771 30.52692 30.75871 30.78561 30.80573 30.82675 30.84648 30.86132 30.86238 30.86335 30.86374 30.86402 30.86413 30.86416 30.86419 30.86420 30.86420 30.86420 30.86420 30.86420 30.86420 30.86420 30.86420

To Find A Given G (A|G i%,n) 0.00000 0.45872 0.89016 1.29470 1.67284 2.02522 2.35259 2.65581 2.93581 3.19363 3.43033 3.64703 3.84489 4.02504 4.18866 4.33688 4.47084 4.59161 4.70026 4.79778 4.88514 4.96324 5.03292 5.09498 5.15016 5.19914 5.24255 5.28096 5.31489 5.34484 5.46230 5.50218 5.53853 5.54282 5.54604 5.54943 5.55264 5.55507 5.55525 5.55541 5.55548 5.55553 5.55554 5.55555 5.55555 5.55556 5.55556 5.55556 5.55556 5.55556 5.55556 5.55556 5.55556

Appendix A 491

Single Sums

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

1.20000 1.44000 1.72800 2.07360 2.48832 2.98598 3.58318 4.29982 5.15978 6.19174 7.43008 8.91610 10.69932 12.83918 15.40702 18.48843 22.18611 26.62333 31.94800 38.33760 46.00512 55.20614 66.24737 79.49685 95.39622 114.47546 137.37055 164.84466 197.81359 237.37631 708.80187 1469.77157 6319.74872 9100.43815 1.3105E104 2.2645E104 5.6348E104 5.0240E105 8.6815E105 2.1602E106 4.4794E106 1.3376E107 3.9939E107 8.2818E107 3.5610E108 3.1750E109 2.8309E110 2.5241E111 1.7891E114 1.0081E119 3.2007E128 1.0162E138 3.2266E147

0.83333 0.69444 0.57870 0.48225 0.40188 0.33490 0.27908 0.23257 0.19381 0.16151 0.13459 0.11216 0.09346 0.07789 0.06491 0.05409 0.04507 0.03756 0.03130 0.02608 0.02174 0.01811 0.01509 0.01258 0.01048 8.7355E203 7.2796E203 6.0663E203 5.0553E203 4.2127E203 1.4108E203 6.8038E204 1.5823E204 1.0988E204 7.6309E205 4.4160E205 1.7747E205 1.9904E206 1.1519E206 4.6291E207 2.2324E207 7.4763E208 2.5038E208 1.2075E208 2.8082E209 3.1496E210 3.5324E211 3.9619E212 5.5895E215 9.9198E220 3.1243E229 9.8402E239 3.0992E248

1.00000 2.20000 3.64000 5.36800 7.44160 9.92992 12.91590 16.49908 20.79890 25.95868 32.15042 39.58050 48.49660 59.19592 72.03511 87.44213 105.93056 128.11667 154.74000 186.68800 225.02560 271.03072 326.23686 392.48424 471.98108 567.37730 681.85276 819.22331 984.06797 1181.88157 3539.00937 7343.85784 3.1594E104 4.5497E104 6.5518E104 1.1322E105 2.8173E105 2.5120E106 4.3407E106 1.0801E107 2.2397E107 6.6878E107 1.9970E108 4.1409E108 1.7805E109 1.5875E110 1.4154E111 1.2620E112 8.9453E114 5.0404E119 1.6004E129 5.0812E138 1.6133E148

1.00000 0.45455 0.27473 0.18629 0.13438 0.10071 0.07742 0.06061 0.04808 0.03852 0.03110 0.02526 0.02062 0.01689 0.01388 0.01144 9.4401E203 7.8054E203 6.4625E203 5.3565E203 4.4439E203 3.6896E203 3.0653E203 2.5479E203 2.1187E203 1.7625E203 1.4666E203 1.2207E203 1.0162E203 8.4611E204 2.8256E204 1.3617E204 3.1652E205 2.1979E205 1.5263E205 8.8324E206 3.5495E206 3.9809E207 2.3038E207 9.2583E208 4.4648E208 1.4953E208 5.0076E209 2.4149E209 5.6164E210 6.2991E211 7.0649E212 7.9237E213 1.1179E215 1.9840E220 6.2486E230 1.9680E239 6.1984E249

0.83333 1.52778 2.10648 2.58873 2.99061 3.32551 3.60459 3.83716 4.03097 4.19247 4.32706 4.43922 4.53268 4.61057 4.67547 4.72956 4.77463 4.81219 4.84350 4.86958 4.89132 4.90943 4.92453 4.93710 4.94759 4.95632 4.96360 4.96967 4.97472 4.97894 4.99295 4.99660 4.99921 4.99945 4.99962 4.99978 4.99991 4.99999 4.99999 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000

1.20000 0.65455 0.47473 0.38629 0.33438 0.30071 0.27742 0.26061 0.24808 0.23852 0.23110 0.22526 0.22062 0.21689 0.21388 0.21144 0.20944 0.20781 0.20646 0.20536 0.20444 0.20369 0.20307 0.20255 0.20212 0.20176 0.20147 0.20122 0.20102 0.20085 0.20028 0.20014 0.20003 0.20002 0.20002 0.20001 0.20000 0.20000 0.20000 0.20000 0.20000 0.20000 0.20000 0.20000 0.20000 0.20000 0.20000 0.20000 0.20000 0.20000 0.20000 0.20000 0.20000

To Find P Given G (P|G i%,n) 0.00000 0.69444 1.85185 3.29861 4.90612 6.58061 8.25510 9.88308 11.43353 12.88708 14.23296 15.46668 16.58825 17.60078 18.50945 19.32077 20.04194 20.68048 21.24390 21.73949 22.17423 22.55462 22.88671 23.17603 23.42761 23.64600 23.83527 23.99906 24.14061 24.26277 24.71078 24.84691 24.95807 24.96978 24.97825 24.98675 24.99423 24.99923 24.99954 24.99980 24.99990 24.99996 24.99999 24.99999 25.00000 25.00000 25.00000 25.00000 25.00000 25.00000 25.00000 25.00000 25.00000

To Find A Given G (A|G i%,n) 0.00000 0.45455 0.87912 1.27422 1.64051 1.97883 2.29016 2.57562 2.83642 3.07386 3.28929 3.48410 3.65970 3.81749 3.95884 4.08511 4.19759 4.29752 4.38607 4.46435 4.53339 4.59414 4.64750 4.69426 4.73516 4.77088 4.80201 4.82911 4.85265 4.87308 4.94914 4.97277 4.99240 4.99451 4.99603 4.99757 4.99894 4.99986 4.99991 4.99996 4.99998 4.99999 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000

Time Value of Money Factors Discrete Compounding 20.00%

n

Uniform Series

20.00%

TABLE A-a-21

25.00%

Time Value of Money Factors Discrete Compounding 25.00%

492 Appendix A

TABLE A-a-22 Single Sums

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Uniform Series

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

1.25000 1.56250 1.95313 2.44141 3.05176 3.81470 4.76837 5.96046 7.45058 9.31323 11.64153 14.55192 18.18989 22.73737 28.42171 35.52714 44.40892 55.51115 69.38894 86.73617 108.42022 135.52527 169.40659 211.75824 264.69780 330.87225 413.59031 516.98788 646.23485 807.79357 3081.48791 7523.16385 4.4842E104 7.0065E104 1.0948E105 2.1382E105 6.5253E105 9.4956E106 1.8546E107 5.6598E107 1.3818E108 5.2711E108 2.0108E109 4.9091E109 2.9260E110 4.2580E111 6.1961E112 9.0166E113 2.7784E117 1.8130E123 7.7198E134 3.2870E146 1.3996E158

0.80000 0.64000 0.51200 0.40960 0.32768 0.26214 0.20972 0.16777 0.13422 0.10737 0.08590 0.06872 0.05498 0.04398 0.03518 0.02815 0.02252 0.01801 0.01441 0.01153 9.2234E203 7.3787E203 5.9030E203 4.7224E203 3.7779E203 3.0223E203 2.4179E203 1.9343E203 1.5474E203 1.2379E203 3.2452E204 1.3292E204 2.2301E205 1.4272E205 9.1344E206 4.6768E206 1.5325E206 1.0531E207 5.3920E208 1.7668E208 7.2370E209 1.8971E209 4.9732E210 2.0370E210 3.4176E211 2.3485E212 1.6139E213 1.1091E214 3.5991E218 5.5157E224 1.2954E235 3.0422E247 7.1448E259

1.00000 2.25000 3.81250 5.76563 8.20703 11.25879 15.07349 19.84186 25.80232 33.25290 42.56613 54.20766 68.75958 86.94947 109.68684 138.10855 173.63568 218.04460 273.55576 342.94470 429.68087 538.10109 673.62636 843.03295 1054.79118 1319.48898 1650.36123 2063.95153 2580.93941 3227.17427 1.2322E104 3.0089E104 1.7936E105 2.8026E105 4.3790E105 8.5528E105 2.6101E106 3.7982E107 7.4184E107 2.2639E108 5.5271E108 2.1084E109 8.0431E109 1.9636E110 1.1704E111 1.7032E112 2.4785E113 3.6066E114 1.1114E118 7.2521E123 3.0879E135 1.3148E147 5.5984E158

1.00000 0.44444 0.26230 0.17344 0.12185 0.08882 0.06634 0.05040 0.03876 0.03007 0.02349 0.01845 0.01454 0.01150 9.1169E203 7.2407E203 5.7592E203 4.5862E203 3.6556E203 2.9159E203 2.3273E203 1.8584E203 1.4845E203 1.1862E203 9.4805E204 7.5787E204 6.0593E204 4.8451E204 3.8746E204 3.0987E204 8.1156E205 3.3235E205 5.5753E206 3.5682E206 2.2836E206 1.1692E206 3.8312E207 2.6328E208 1.3480E208 4.4171E209 1.8093E209 4.7428E210 1.2433E210 5.0926E211 8.5439E212 5.8714E213 4.0348E214 2.7727E215 8.9978E219 1.3789E224 3.2384E236 7.6056E248 1.7862E259

0.80000 1.44000 1.95200 2.36160 2.68928 2.95142 3.16114 3.32891 3.46313 3.57050 3.65640 3.72512 3.78010 3.82408 3.85926 3.88741 3.90993 3.92794 3.94235 3.95388 3.96311 3.97049 3.97639 3.98111 3.98489 3.98791 3.99033 3.99226 3.99381 3.99505 3.99870 3.99947 3.99991 3.99994 3.99996 3.99998 3.99999 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000

1.25000 0.69444 0.51230 0.42344 0.37185 0.33882 0.31634 0.30040 0.28876 0.28007 0.27349 0.26845 0.26454 0.26150 0.25912 0.25724 0.25576 0.25459 0.25366 0.25292 0.25233 0.25186 0.25148 0.25119 0.25095 0.25076 0.25061 0.25048 0.25039 0.25031 0.25008 0.25003 0.25001 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000

To Find P Given G (P|G i%,n) 0.00000 0.64000 1.66400 2.89280 4.20352 5.51424 6.77253 7.94694 9.02068 9.98705 10.84604 11.60195 12.26166 12.83341 13.32599 13.74820 14.10849 14.41473 14.67414 14.89320 15.07766 15.23262 15.36248 15.47109 15.56176 15.63732 15.70019 15.75241 15.79574 15.83164 15.94808 15.97661 15.99536 15.99692 15.99795 15.99890 15.99961 15.99997 15.99998 15.99999 16.00000 16.00000 16.00000 16.00000 16.00000 16.00000 16.00000 16.00000 16.00000 16.00000 16.00000 16.00000 16.00000

To Find A Given G (A|G i%,n) 0.00000 0.44444 0.85246 1.22493 1.56307 1.86833 2.14243 2.38725 2.60478 2.79710 2.96631 3.11452 3.24374 3.35595 3.45299 3.53660 3.60838 3.66979 3.72218 3.76673 3.80451 3.83646 3.86343 3.88613 3.90519 3.92118 3.93456 3.94574 3.95506 3.96282 3.98831 3.99468 3.99893 3.99929 3.99953 3.99974 3.99991 3.99999 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000

Appendix A 493

Single Sums

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

1.30000 1.69000 2.19700 2.85610 3.71293 4.82681 6.27485 8.15731 10.60450 13.78585 17.92160 23.29809 30.28751 39.37376 51.18589 66.54166 86.50416 112.45541 146.19203 190.04964 247.06453 321.18389 417.53905 542.80077 705.64100 917.33330 1192.53329 1550.29328 2015.38126 2619.99564 1.2646E104 3.6119E104 2.9463E105 4.9793E105 8.4150E105 1.8488E106 6.8644E106 1.5993E108 3.5136E108 1.3046E109 3.7260E109 1.7985E110 8.6808E110 2.4793E111 2.0225E112 4.7120E113 1.0978E115 2.5577E116 3.2345E120 2.2203E127 1.0462E141 4.9296E154 2.3228E168

0.76923 0.59172 0.45517 0.35013 0.26933 0.20718 0.15937 0.12259 0.09430 0.07254 0.05580 0.04292 0.03302 0.02540 0.01954 0.01503 0.01156 8.8924E203 6.8403E203 5.2618E203 4.0475E203 3.1135E203 2.3950E203 1.8423E203 1.4172E203 1.0901E203 8.3855E204 6.4504E204 4.9618E204 3.8168E204 7.9075E205 2.7686E205 3.3941E206 2.0083E206 1.1884E206 5.4090E207 1.4568E207 6.2529E209 2.8461E209 7.6653E210 2.6839E210 5.5603E211 1.1520E211 4.0333E212 4.9444E213 2.1223E214 9.1091E216 3.9098E217 3.0917E221 4.5040E228 9.5586E242 2.0286E255 4.3052E269

1.00000 2.30000 3.99000 6.18700 9.04310 12.75603 17.58284 23.85769 32.01500 42.61950 56.40535 74.32695 97.62504 127.91255 167.28631 218.47220 285.01386 371.51802 483.97343 630.16546 820.21510 1067.27963 1388.46351 1806.00257 2348.80334 3054.44434 3971.77764 5164.31093 6714.60421 8729.98548 4.2151E104 1.2039E105 9.8211E105 1.6598E106 2.8050E106 6.1626E106 2.2881E107 5.3309E108 1.1712E109 4.3486E109 1.2420E110 5.9949E110 2.8936E111 8.2645E111 6.7416E112 1.5707E114 3.6593E115 8.5255E116 1.0782E121 7.4009E127 3.4873E141 1.6432E155 7.7427E168

1.00000 0.43478 0.25063 0.16163 0.11058 0.07839 0.05687 0.04192 0.03124 0.02346 0.01773 0.01345 0.01024 7.8178E203 5.9778E203 4.5772E203 3.5086E203 2.6917E203 2.0662E203 1.5869E203 1.2192E203 9.3696E204 7.2022E204 5.5371E204 4.2575E204 3.2739E204 2.5178E204 1.9364E204 1.4893E204 1.1455E204 2.3724E205 8.3061E206 1.0182E206 6.0250E207 3.5651E207 1.6227E207 4.3704E208 1.8759E209 8.5383E210 2.2996E210 8.0516E211 1.6681E211 3.4559E212 1.2100E212 1.4833E213 6.3668E215 2.7327E216 1.1729E217 9.2751E222 1.3512E228 2.8676E242 6.0857E256 1.2915E269

0.76923 1.36095 1.81611 2.16624 2.43557 2.64275 2.80211 2.92470 3.01900 3.09154 3.14734 3.19026 3.22328 3.24867 3.26821 3.28324 3.29480 3.30369 3.31053 3.31579 3.31984 3.32296 3.32535 3.32719 3.32861 3.32970 3.33054 3.33118 3.33168 3.33206 3.33307 3.33324 3.33332 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333

1.30000 0.73478 0.55063 0.46163 0.41058 0.37839 0.35687 0.34192 0.33124 0.32346 0.31773 0.31345 0.31024 0.30782 0.30598 0.30458 0.30351 0.30269 0.30207 0.30159 0.30122 0.30094 0.30072 0.30055 0.30043 0.30033 0.30025 0.30019 0.30015 0.30011 0.30002 0.30001 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000 0.30000

To Find P Given G (P|G i%,n) 0.00000 0.59172 1.50205 2.55243 3.62975 4.66563 5.62183 6.47995 7.23435 7.88719 8.44518 8.91732 9.31352 9.64369 9.91721 10.14263 10.32759 10.47876 10.60189 10.70186 10.78281 10.84819 10.90088 10.94326 10.97727 11.00452 11.02632 11.04374 11.05763 11.06870 11.10074 11.10711 11.11053 11.11075 11.11089 11.11101 11.11108 11.11111 11.11111 11.11111 11.11111 11.11111 11.11111 11.11111 11.11111 11.11111 11.11111 11.11111 11.11111 11.11111 11.11111 11.11111 11.11111

To Find A Given G (A|G i%,n) 0.00000 0.43478 0.82707 1.17828 1.49031 1.76545 2.00628 2.21559 2.39627 2.55122 2.68328 2.79517 2.88946 2.96850 3.03444 3.08921 3.13451 3.17183 3.20247 3.22754 3.24799 3.26462 3.27812 3.28904 3.29785 3.30496 3.31067 3.31526 3.31894 3.32188 3.33049 3.33223 3.33317 3.33323 3.33327 3.33330 3.33332 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333 3.33333

Time Value of Money Factors Discrete Compounding 30.00%

n

Uniform Series

30.00%

TABLE A-a-23

40.00%

Time Value of Money Factors Discrete Compounding 40.00%

494 Appendix A

TABLE A-a-24 Single Sums

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Uniform Series

Gradient Series

To Find F Given P (F|P i %,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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

1.40000 1.96000 2.74400 3.84160 5.37824 7.52954 10.54135 14.75789 20.66105 28.92547 40.49565 56.69391 79.37148 111.12007 155.56810 217.79533 304.91347 426.87885 597.63040 836.68255 1171.35558 1639.89781 2295.85693 3214.19970 4499.87958 6299.83141 8819.76398 1.2348E104 1.7287E104 2.4201E104 1.8223E105 7.0004E105 1.0331E107 2.0249E107 3.9688E107 1.0890E108 5.8571E108 3.3206E110 9.1118E110 4.9005E111 1.8826E112 1.4175E113 1.0673E114 4.1002E114 6.0510E115 3.4306E117 1.9449E119 1.1026E121 2.0093E126 1.1769E135 4.0373E152 1.3850E170 4.7514E187

0.71429 0.51020 0.36443 0.26031 0.18593 0.13281 0.09486 0.06776 0.04840 0.03457 0.02469 0.01764 0.01260 8.9993E203 6.4281E203 4.5915E203 3.2796E203 2.3426E203 1.6733E203 1.1952E203 8.5371E204 6.0979E204 4.3557E204 3.1112E204 2.2223E204 1.5873E204 1.1338E204 8.0987E205 5.7848E205 4.1320E205 5.4877E206 1.4285E206 9.6795E208 4.9385E208 2.5197E208 9.1824E209 1.7073E209 3.0115E211 1.0975E211 2.0406E212 5.3118E213 7.0547E214 9.3693E215 2.4389E215 1.6526E216 2.9150E218 5.1416E220 9.0691E222 4.9768E227 8.4971E236 2.4769E253 7.2201E271 2.1046E288

1.00000 2.40000 4.36000 7.10400 10.94560 16.32384 23.85338 34.39473 49.15262 69.81366 98.73913 139.23478 195.92869 275.30017 386.42024 541.98833 759.78367 1064.69714 1491.57599 2089.20639 2925.88894 4097.24452 5737.14232 8032.99925 1.1247E104 1.5747E104 2.2047E104 3.0867E104 4.3214E104 6.0501E104 4.5556E105 1.7501E106 2.5828E107 5.0622E107 9.9220E107 2.7226E108 1.4643E109 8.3015E110 2.2779E111 1.2251E112 4.7065E112 3.5438E113 2.6683E114 1.0250E115 1.5128E116 8.5764E117 4.8623E119 2.7566E121 5.0233E126 2.9422E135 1.0093E153 3.4626E170 1.1878E188

1.00000 0.41667 0.22936 0.14077 0.09136 0.06126 0.04192 0.02907 0.02034 0.01432 0.01013 7.1821E203 5.1039E203 3.6324E203 2.5879E203 1.8451E203 1.3162E203 9.3923E204 6.7043E204 4.7865E204 3.4178E204 2.4407E204 1.7430E204 1.2449E204 8.8911E205 6.3504E205 4.5358E205 3.2397E205 2.3140E205 1.6529E205 2.1951E206 5.7140E207 3.8718E208 1.9754E208 1.0079E208 3.6730E209 6.8293E210 1.2046E211 4.3899E212 8.1624E213 2.1247E213 2.8219E214 3.7477E215 9.7557E216 6.6105E217 1.1660E218 2.0566E220 3.6276E222 1.9907E227 3.3988E236 9.9076E254 2.8880E271 8.4186E289

0.71429 1.22449 1.58892 1.84923 2.03516 2.16797 2.26284 2.33060 2.37900 2.41357 2.43826 2.45590 2.46850 2.47750 2.48393 2.48852 2.49180 2.49414 2.49582 2.49701 2.49787 2.49848 2.49891 2.49922 2.49944 2.49960 2.49972 2.49980 2.49986 2.49990 2.49999 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000

1.40000 0.81667 0.62936 0.54077 0.49136 0.46126 0.44192 0.42907 0.42034 0.41432 0.41013 0.40718 0.40510 0.40363 0.40259 0.40185 0.40132 0.40094 0.40067 0.40048 0.40034 0.40024 0.40017 0.40012 0.40009 0.40006 0.40005 0.40003 0.40002 0.40002 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000 0.40000

To Find P Given G (P|G i%,n) 0.00000 0.51020 1.23907 2.01999 2.76373 3.42778 3.99697 4.47129 4.85849 5.16964 5.41658 5.61060 5.76179 5.87878 5.96877 6.03764 6.09012 6.12994 6.16006 6.18277 6.19984 6.21265 6.22223 6.22939 6.23472 6.23869 6.24164 6.24382 6.24544 6.24664 6.24947 6.24985 6.24999 6.24999 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000 6.25000

To Find A Given G (A|G i%,n) 0.00000 0.41667 0.77982 1.09234 1.35799 1.58110 1.76635 1.91852 2.04224 2.14190 2.22149 2.28454 2.33412 2.37287 2.40296 2.42620 2.44406 2.45773 2.46815 2.47607 2.48206 2.48658 2.48998 2.49253 2.49444 2.49587 2.49694 2.49773 2.49832 2.49876 2.49980 2.49994 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000 2.50000

Appendix A 495

Single Sums

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

1.50000 2.25000 3.37500 5.06250 7.59375 11.39063 17.08594 25.62891 38.44336 57.66504 86.49756 129.74634 194.61951 291.92926 437.89389 656.84084 985.26125 1477.89188 2216.83782 3325.25673 4987.88510 7481.82764 1.1223E104 1.6834E104 2.5251E104 3.7877E104 5.6815E104 8.5223E104 1.2783E105 1.9175E105 2.1842E106 1.1057E107 2.8339E108 6.3762E108 1.4346E109 4.8419E109 3.6768E110 4.7706E112 1.6101E113 1.2226E114 6.1896E114 7.0504E115 8.0308E116 4.0656E117 1.0420E119 1.3519E121 1.7541E123 2.2758E125 4.9708E131 1.8277E142 2.4709E163 3.3404E184 4.5160E1105

Gradient Series

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)

To Find P Given A (P|A i%,n)

To Find A Given P (A|P i%,n)

0.66667 0.44444 0.29630 0.19753 0.13169 0.08779 0.05853 0.03902 0.02601 0.01734 0.01156 7.7073E203 5.1382E203 3.4255E203 2.2837E203 1.5224E203 1.0150E203 6.7664E204 4.5109E204 3.0073E204 2.0049E204 1.3366E204 8.9105E205 5.9403E205 3.9602E205 2.6401E205 1.7601E205 1.1734E205 7.8226E206 5.2151E206 4.5784E207 9.0438E208 3.5287E209 1.5683E209 6.9703E210 2.0653E210 2.7197E211 2.0962E213 6.2109E214 8.1790E215 1.6156E215 1.4184E216 1.2452E217 2.4597E218 9.5972E220 7.3969E222 5.7010E224 4.3940E226 2.0117E232 5.4714E243 4.0471E264 2.9936E285 2.2143E2106

1.00000 2.50000 4.75000 8.12500 13.18750 20.78125 32.17188 49.25781 74.88672 113.33008 170.99512 257.49268 387.23901 581.85852 873.78778 1311.68167 1968.52251 2953.78376 4431.67564 6648.51346 9973.77019 1.4962E104 2.2443E104 3.3666E104 5.0500E104 7.5752E104 1.1363E105 1.7044E105 2.5567E105 3.8350E105 4.3683E106 2.2115E107 5.6677E108 1.2752E109 2.8693E109 9.6839E109 7.3537E110 9.5411E112 3.2201E113 2.4453E114 1.2379E115 1.4101E116 1.6062E117 8.1312E117 2.0839E119 2.7038E121 3.5081E123 4.5517E125 9.9416E131 3.6554E142 4.9418E163 6.6809E184 9.0320E1105

1.00000 0.40000 0.21053 0.12308 0.07583 0.04812 0.03108 0.02030 0.01335 8.8238E203 5.8481E203 3.8836E203 2.5824E203 1.7186E203 1.1444E203 7.6238E204 5.0800E204 3.3855E204 2.2565E204 1.5041E204 1.0026E204 6.6838E205 4.4556E205 2.9703E205 1.9802E205 1.3201E205 8.8006E206 5.8671E206 3.9114E206 2.6076E206 2.2892E207 4.5219E208 1.7644E209 7.8416E210 3.4852E210 1.0326E210 1.3599E211 1.0481E213 3.1055E214 4.0895E215 8.0780E216 7.0918E217 6.2260E218 1.2298E218 4.7986E220 3.6984E222 2.8505E224 2.1970E226 1.0059E232 2.7357E243 2.0236E264 1.4968E285 1.1072E2106

0.66667 1.11111 1.40741 1.60494 1.73663 1.82442 1.88294 1.92196 1.94798 1.96532 1.97688 1.98459 1.98972 1.99315 1.99543 1.99696 1.99797 1.99865 1.99910 1.99940 1.99960 1.99973 1.99982 1.99988 1.99992 1.99995 1.99996 1.99998 1.99998 1.99999 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000

1.50000 0.90000 0.71053 0.62308 0.57583 0.54812 0.53108 0.52030 0.51335 0.50882 0.50585 0.50388 0.50258 0.50172 0.50114 0.50076 0.50051 0.50034 0.50023 0.50015 0.50010 0.50007 0.50004 0.50003 0.50002 0.50001 0.50001 0.50001 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000 0.50000

To Find P Given G (P|G i%,n) 0.00000 0.44444 1.03704 1.62963 2.15638 2.59534 2.94650 3.21963 3.42773 3.58380 3.69941 3.78419 3.84585 3.89038 3.92236 3.94519 3.96143 3.97293 3.98105 3.98677 3.99078 3.99358 3.99554 3.99691 3.99786 3.99852 3.99898 3.99930 3.99951 3.99967 3.99997 3.99999 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000 4.00000

To Find A Given G (A|G i%,n) 0.00000 0.40000 0.73684 1.01538 1.24171 1.42256 1.56484 1.67518 1.75964 1.82352 1.87134 1.90679 1.93286 1.95188 1.96567 1.97560 1.98273 1.98781 1.99143 1.99398 1.99579 1.99706 1.99795 1.99857 1.99901 1.99931 1.99952 1.99967 1.99977 1.99984 1.99998 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000

Time Value of Money Factors Discrete Compounding 50.00%

n

To Find F Given P (F|P i %,n)

Uniform Series

50.00%

TABLE A-a-25

4.00% Time Value of Money Factors Geometric Series - Present Worth 4.00%

496 Appendix A

TABLE A-b-1 j

4%

5%

6%

10%

15%

n

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

0.96154 1.92308 2.88462 3.84615 4.80769 5.76923 6.73077 7.69231 8.65385 9.61538 10.57692 11.53846 12.50000 13.46154 14.42308 15.38462 16.34615 17.30769 18.26923 19.23077 20.19231 21.15385 22.11538 23.07692 24.03846 25.00000 25.96154 26.92308 27.88462 28.84615 34.61538 38.46154 46.15385 48.07692 50.00000 52.88462 57.69231 69.23077 72.11538 76.92308 80.76923 86.53846 92.30769 96.15385 103.84615 115.38462 126.92308 138.46154 173.07692 230.76923 346.15385 461.53846 576.92308

0.96154 1.93232 2.91244 3.90198 4.90104 5.90971 6.92807 7.95622 8.99426 10.04228 11.10038 12.16866 13.24720 14.33612 15.43550 16.54546 17.66609 18.79749 19.93978 21.09304 22.25740 23.43295 24.61981 25.81807 27.02786 28.24929 29.48245 30.72748 31.98447 33.25355 41.12844 46.63525 58.30188 61.36078 64.47879 69.26914 77.56509 99.17238 104.97315 115.01897 123.40898 136.61179 150.59485 160.37302 181.08887 215.29361 253.66061 296.69635 459.85139 894.10062 3034.33576 9782.36045 3.1058E104

0.96154 1.94157 2.94044 3.95853 4.99619 6.05381 7.13177 8.23046 9.35028 10.49163 11.65493 12.84060 14.04907 15.28078 16.53618 17.81573 19.11988 20.44910 21.80389 23.18474 24.59214 26.02660 27.48865 28.97882 30.49764 32.04567 33.62347 35.23162 36.87069 38.54128 49.26152 57.12012 74.75307 79.59740 84.62985 92.54728 106.79116 147.05700 158.64571 179.49439 197.66360 227.64978 261.26658 285.90974 341.20357 441.66933 567.93591 726.62925 1491.78806 4784.77465 4.7492E104 4.6745E105 4.5971E106

0.96154 1.97855 3.05424 4.19198 5.39536 6.66817 8.01441 9.43832 10.94438 12.53732 14.22217 16.00422 17.88908 19.88268 21.99129 24.22156 26.58050 29.07553 31.71450 34.50572 37.45797 40.58055 43.88327 47.37654 51.07134 54.97930 59.11272 63.48461 68.10872 72.99961 108.87422 140.45012 229.42525 258.63957 291.32200 347.76220 465.73578 928.96423 1102.25501 1464.47438 1837.00923 2578.64714 3617.00726 4530.94462 7106.25392 1.3946E104 2.7354E104 5.3636E104 4.0412E105 1.1697E107 9.7997E109 8.2098E112 6.8779E115

0.96154 2.02478 3.20048 4.50053 5.93808 7.52769 9.28542 11.22907 13.37830 15.75485 18.38277 21.28864 24.50186 28.05494 31.98383 36.32828 41.13223 46.44429 52.31821 58.81340 65.99559 73.93743 82.71927 92.42996 103.16775 115.04126 128.17062 142.68867 158.74228 176.49387 330.16651 498.11856 1124.62271 1377.12975 1685.87657 2282.59383 3779.49708 1.2651E104 1.7109E104 2.8290E104 4.2299E104 7.7332E104 1.4137E105 2.1137E105 4.7246E105 1.5789E106 5.2762E106 1.7632E107 6.5799E108 2.7421E111 4.7624E116 8.2712E121 1.4365E127

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Appendix A 497

4%

5%

6%

10%

15%

n

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

0.95238 1.89569 2.83002 3.75545 4.67206 5.57995 6.47919 7.36986 8.25205 9.12584 9.99131 10.84854 11.69760 12.53857 13.37154 14.19657 15.01375 15.82314 16.62482 17.41887 18.20536 18.98436 19.75593 20.52016 21.27711 22.02686 22.76946 23.50499 24.23351 24.95510 29.14256 31.80357 36.82956 38.02707 39.20189 40.92249 43.68262 49.79223 51.21312 53.49248 55.23904 57.73668 60.09495 61.59356 64.42406 68.28353 71.72430 74.79180 82.13812 89.94066 96.80953 98.98810 99.67906

0.95238 1.90476 2.85714 3.80952 4.76190 5.71429 6.66667 7.61905 8.57143 9.52381 10.47619 11.42857 12.38095 13.33333 14.28571 15.23810 16.19048 17.14286 18.09524 19.04762 20.00000 20.95238 21.90476 22.85714 23.80952 24.76190 25.71429 26.66667 27.61905 28.57143 34.28571 38.09524 45.71429 47.61905 49.52381 52.38095 57.14286 68.57143 71.42857 76.19048 80.00000 85.71429 91.42857 95.23810 102.85714 114.28571 125.71429 137.14286 171.42857 228.57143 342.85714 457.14286 571.42857

0.95238 1.91383 2.88444 3.86429 4.85348 5.85208 6.86020 7.87791 8.90532 9.94251 10.98959 12.04663 13.11374 14.19101 15.27855 16.37644 17.48478 18.60369 19.73325 20.87356 22.02474 23.18688 24.36009 25.54447 26.74013 27.94718 29.16573 30.39588 31.63774 32.89143 40.66835 46.10418 57.61414 60.63061 63.70481 68.42678 76.60133 97.87584 103.58345 113.46432 121.71321 134.68804 148.42217 158.02193 178.34867 211.88030 249.45136 291.54847 450.78476 872.69322 2933.63852 9361.32092 2.9408E104

0.95238 1.95011 2.99536 4.09037 5.23753 6.43932 7.69834 9.01731 10.39908 11.84666 13.36316 14.95189 16.61626 18.35989 20.18656 22.10020 24.10497 26.20521 28.40546 30.71048 33.12526 35.65504 38.30528 41.08172 43.99037 47.03753 50.22980 53.57407 57.07760 60.74796 86.74607 108.57764 166.54883 184.73840 204.70155 238.35462 306.01168 549.73617 635.06433 806.61043 975.66772 1296.23897 1720.02329 2075.89057 3020.85487 5294.18080 9267.03237 1.6210E104 8.6604E104 1.4120E106 3.7519E108 9.9691E110 2.6489E113

0.95238 1.99546 3.13789 4.38912 5.75951 7.26042 8.90426 10.70467 12.67654 14.83622 17.20157 19.79219 22.62955 25.73712 29.14066 32.86834 36.95104 41.42257 46.31995 51.68376 57.55840 63.99254 71.03944 78.75749 87.21058 96.46873 106.60861 117.71419 129.87745 143.19911 254.42803 370.48860 777.78913 934.98968 1123.55905 1479.25741 2336.99669 6982.21820 9176.29941 1.4467E104 2.0821E104 3.5946E104 6.2051E104 8.9291E104 1.8488E105 5.5083E105 1.6411E106 4.8891E106 1.2928E108 3.0342E110 1.6714E115 9.2066E119 5.0714E124

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

Time Value of Money Factors Geometric Series - Present Worth 5.00%

j

5.00%

TABLE A-b-2

6.00% Time Value of Money Factors Geometric Series - Present Worth 6.00%

498 Appendix A

TABLE A-b-3 j

4%

5%

6%

10%

15%

n

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

0.94340 1.88679 2.83019 3.77358 4.71698 5.66038 6.60377 7.54717 8.49057 9.43396 10.37736 11.32075 12.26415 13.20755 14.15094 15.09434 16.03774 16.98113 17.92453 18.86792 19.81132 20.75472 21.69811 22.64151 23.58491 24.52830 25.47170 26.41509 27.35849 28.30189 33.96226 37.73585 45.28302 47.16981 49.05660 51.88679 56.60377 67.92453 70.75472 75.47170 79.24528 84.90566 90.56604 94.33962 101.88679 113.20755 124.52830 135.84906 169.81132 226.41509 339.62264 452.83019 566.03774

0.94340 1.92239 2.93833 3.99261 5.08667 6.22201 7.40020 8.62285 9.89164 11.20831 12.57466 13.99257 15.46399 16.99093 18.57549 20.21985 21.92626 23.69706 25.53469 27.44166 29.42059 31.47419 33.60530 35.81682 38.11179 40.49337 42.96482 45.52953 48.19102 50.95294 69.85616 85.00510 122.94741 134.32393 146.57525 166.74105 205.75398 334.90762 377.20874 459.04485 536.34890 676.05776 850.53743 990.36455 1340.57814 2104.89599 3297.00464 5156.33978 1.9634E104 1.8143E105 1.5459E107 1.3171E109 1.1221E111

0.94340 1.96689 3.07729 4.28196 5.58892 7.00685 8.54517 10.21410 12.02473 13.98909 16.12024 18.43234 20.94074 23.66213 26.61457 29.81770 33.29278 37.06293 41.15317 45.59071 50.40501 55.62808 61.29462 67.44227 74.11190 81.34781 89.19810 97.71492 106.95487 116.97934 197.75529 278.24756 544.24595 642.55541 758.26765 971.35170 1465.53354 3915.15469 5002.55747 7524.44829 1.0428E104 1.7012E104 2.7747E104 3.8444E104 7.3795E104 1.9623E105 5.2178E105 1.3874E106 2.6080E107 3.4660E109 6.1217E113 1.0812E118 1.9096E122

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

0.94340 1.86899 2.77712 3.66812 4.54231 5.40000 6.24151 7.06714 7.87720 8.67197 9.45174 10.21680 10.96743 11.70389 12.42646 13.13539 13.83095 14.51339 15.18295 15.83987 16.48440 17.11677 17.73721 18.34594 18.94319 19.52917 20.10409 20.66816 21.22159 21.76458 24.81401 26.66171 29.96041 30.70949 31.43057 32.46196 34.05522 37.31332 38.01797 39.10649 39.90566 40.99585 41.96830 42.55752 43.60947 44.91528 45.95427 46.78096 48.37851 49.48291 49.94742 49.99465 49.99946

0.94340 1.87789 2.80357 3.72052 4.62882 5.52855 6.41979 7.30262 8.17712 9.04337 9.90146 10.75144 11.59341 12.42743 13.25359 14.07195 14.88259 15.68559 16.48101 17.26892 18.04941 18.82252 19.58835 20.34695 21.09839 21.84275 22.58008 23.31046 24.03394 24.75060 28.91080 31.55569 36.55392 37.74537 38.91444 40.62702 43.37529 49.46326 50.88009 53.15376 54.89669 57.39024 59.74594 61.24360 64.07384 67.93642 71.38371 74.46038 81.84409 89.71927 96.70363 98.94307 99.66111

Appendix A 499

4%

5%

6%

10%

15%

n

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

0.90909 1.81818 2.72727 3.63636 4.54545 5.45455 6.36364 7.27273 8.18182 9.09091 10.00000 10.90909 11.81818 12.72727 13.63636 14.54545 15.45455 16.36364 17.27273 18.18182 19.09091 20.00000 20.90909 21.81818 22.72727 23.63636 24.54545 25.45455 26.36364 27.27273 32.72727 36.36364 43.63636 45.45455 47.27273 50.00000 54.54545 65.45455 68.18182 72.72727 76.36364 81.81818 87.27273 90.90909 98.18182 109.09091 120.00000 130.90909 163.63636 218.18182 327.27273 436.36364 545.45455

0.90909 1.85950 2.85312 3.89190 4.97789 6.11325 7.30022 8.54113 9.83846 11.19475 12.61270 14.09509 15.64487 17.26509 18.95896 20.72982 22.58117 24.51668 26.54017 28.65563 30.86725 33.17940 35.59664 38.12376 40.76575 43.52783 46.41546 49.43434 52.59045 55.89002 79.08675 98.36852 148.91859 164.62383 181.78927 210.57570 267.96474 470.90920 540.94029 680.55528 816.87970 1072.68244 1406.67449 1684.29792 2412.12984 4126.18446 7048.22688 1.2030E104 5.9678E104 8.5952E105 1.7819E108 3.6941E110 7.6581E112

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

0.90909 1.76860 2.58122 3.34951 4.07590 4.76267 5.41198 6.02587 6.60628 7.15503 7.67385 8.16436 8.62813 9.06659 9.48114 9.87308 10.24364 10.59398 10.92522 11.23839 11.53448 11.81442 12.07909 12.32932 12.56590 12.78958 13.00106 13.20100 13.39003 13.56876 14.45402 14.89870 15.53791 15.65769 15.76476 15.90444 16.09085 16.37292 16.41841 16.47912 16.51681 16.55964 16.59022 16.60558 16.62767 16.64677 16.65652 16.66149 16.66598 16.66664 16.66667 16.66667 16.66667

0.90909 1.77686 2.60518 3.39586 4.15059 4.87102 5.55870 6.21512 6.84171 7.43981 8.01073 8.55570 9.07589 9.57244 10.04642 10.49886 10.93073 11.34297 11.73647 12.11208 12.47063 12.81287 13.13956 13.45140 13.74906 14.03319 14.30441 14.56330 14.81042 15.04631 16.25279 16.88904 17.85579 18.04629 18.21986 18.45174 18.77305 19.29792 19.38937 19.51610 19.59826 19.69610 19.77012 19.80915 19.86846 19.92473 19.95693 19.97535 19.99538 19.99972 20.00000 20.00000 20.00000

0.90909 1.78512 2.62930 3.44278 4.22668 4.98207 5.71000 6.41145 7.08740 7.73877 8.36645 8.97130 9.55417 10.11583 10.65708 11.17864 11.68123 12.16555 12.63226 13.08199 13.51538 13.93300 14.33543 14.72324 15.09694 15.45705 15.80407 16.13846 16.46070 16.77122 18.41108 19.31845 20.77553 21.07717 21.35728 21.74040 22.29149 23.26344 23.44608 23.70880 23.88661 24.10849 24.28615 24.38446 24.54232 24.70656 24.81186 24.87937 24.96821 24.99656 24.99996 25.00000 25.00000

Time Value of Money Factors Geometric Series - Present Worth 10.00%

j

10.00%

TABLE A-b-4

15.00%

Time Value of Money Factors Geometric Series - Present Worth 15.00%

500 Appendix A

TABLE A-b-5 j

4%

5%

6%

10%

15%

n

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

To Find P Given A1 (P | A1 i %, j %,n)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

0.86957 1.65595 2.36712 3.01027 3.59190 4.11789 4.59357 5.02375 5.41278 5.76460 6.08277 6.37051 6.63072 6.86604 7.07885 7.27131 7.44536 7.60276 7.74511 7.87383 7.99025 8.09553 8.19074 8.27684 8.35471 8.42513 8.48881 8.54640 8.59849 8.64559 8.84730 8.92797 9.01801 9.03129 9.04215 9.05485 9.06909 9.08438 9.08608 9.08799 9.08896 9.08984 9.09032 9.09052 9.09073 9.09086 9.09089 9.09090 9.09091 9.09091 9.09091 9.09091 9.09091

0.86957 1.66352 2.38843 3.05030 3.65462 4.20640 4.71019 5.17017 5.59016 5.97362 6.32374 6.64342 6.93529 7.20179 7.44511 7.66728 7.87012 8.05533 8.22443 8.37883 8.51980 8.64851 8.76603 8.87333 8.97131 9.06076 9.14243 9.21700 9.28509 9.34725 9.62183 9.73718 9.87306 9.89418 9.91178 9.93285 9.95739 9.98570 9.98911 9.99309 9.99520 9.99722 9.99839 9.99888 9.99946 9.99982 9.99994 9.99998 10.00000 10.00000 10.00000 10.00000 10.00000

0.86957 1.67108 2.40986 3.09083 3.71850 4.29706 4.83033 5.32187 5.77494 6.19255 6.57748 6.93229 7.25933 7.56077 7.83862 8.09473 8.33080 8.54839 8.74895 8.93381 9.10421 9.26127 9.40604 9.53948 9.66248 9.77585 9.88035 9.97667 10.06545 10.14729 10.52003 10.68445 10.88881 10.92224 10.95065 10.98545 11.02750 11.07967 11.08649 11.09473 11.09929 11.10386 11.10666 11.10790 11.10944 11.11048 11.11087 11.11102 11.11111 11.11111 11.11111 11.11111 11.11111

0.86957 1.70132 2.49692 3.25792 3.98584 4.68211 5.34810 5.98514 6.59448 7.17733 7.73484 8.26811 8.77819 9.26609 9.73278 10.17919 10.60618 11.01460 11.40527 11.77896 12.13639 12.47829 12.80532 13.11813 13.41734 13.70355 13.97731 14.23916 14.48963 14.72921 15.96313 16.62072 17.63200 17.83343 18.01773 18.26521 18.61094 19.18519 19.28691 19.42902 19.52203 19.63393 19.71963 19.76530 19.83554 19.90353 19.94341 19.96680 19.99330 19.99953 20.00000 20.00000 20.00000

0.86957 1.73913 2.60870 3.47826 4.34783 5.21739 6.08696 6.95652 7.82609 8.69565 9.56522 10.43478 11.30435 12.17391 13.04348 13.91304 14.78261 15.65217 16.52174 17.39130 18.26087 19.13043 20.00000 20.86957 21.73913 22.60870 23.47826 24.34783 25.21739 26.08696 31.30435 34.78261 41.73913 43.47826 45.21739 47.82609 52.17391 62.60870 65.21739 69.56522 73.04348 78.26087 83.47826 86.95652 93.91304 104.34783 114.78261 125.21739 156.52174 208.69565 313.04348 417.39130 521.73913

Appendix A

501

4%

5%

6%

10%

15%

n

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

1.00000 2.09000 3.27610 4.56477 5.96287 7.47766 9.11686 10.88864 12.80164 14.86503 17.08853 19.48241 22.05756 24.82552 27.79847 30.98933 34.41178 38.08027 42.01010 46.21746 50.71945 55.53419 60.68082 66.17958 72.05186 78.32029 85.00877 92.14258 99.74841 107.85449 168.78836 223.89681 383.07414 436.07164 495.62196 598.92640 815.95585 1670.28717 1988.74313 2651.16420 3327.72366 4661.10317 6501.45689 8099.63097 1.2517E104 2.3825E104 4.4942E104 8.4161E104 5.3533E105 1.0949E107 4.1121E109 1.4671E112 5.1545E114

1.00000 2.10000 3.30760 4.63092 6.07863 7.66000 9.38492 11.26395 13.30836 15.53017 17.94223 20.55821 23.39274 26.46138 29.78073 33.36852 37.24361 41.42613 45.93752 50.80062 56.03978 61.68093 67.75171 74.28152 81.30172 88.84566 96.94887 105.64917 114.98682 125.00468 202.16597 274.23487 491.16717 565.67355 650.51483 800.19773 1123.40317 2476.77264 3005.58331 4137.30972 5329.99997 7767.25889 1.1279E104 1.4440E104 2.3584E104 4.8876E104 1.0062E105 2.0612E105 1.7366E106 5.8595E107 6.4361E110 7.0103E113 7.6293E116

1.00000 2.14000 3.43560 4.90402 6.56428 8.43737 10.54642 12.91700 15.57726 18.55830 21.89438 25.62327 29.78663 34.43036 39.60508 45.36653 51.77616 58.90168 66.81766 75.60628 85.35803 96.17260 108.15978 121.44048 136.14783 152.42845 170.44376 190.37150 212.40736 236.76675 446.81247 674.30392 1507.44509 1838.06949 2239.27239 3006.87959 4899.36687 1.5646E104 2.0883E104 3.3756E104 4.9535E104 8.7982E104 1.5615E105 2.2884E105 4.9118E105 1.5433E106 4.8464E106 1.5215E107 4.7045E108 1.4325E111 1.3281E116 1.2312E121 1.1415E126

1.00000 2.19000 3.60010 5.26498 7.22458 9.52492 12.21898 15.36776 19.04150 23.32103 28.29943 34.08380 40.79740 48.58208 57.60107 68.04218 80.12149 94.08761 110.22657 128.86740 150.38864 175.22570 203.87947 236.92611 275.02833 318.94842 369.56315 427.88099 495.06184 572.43977 1354.98109 2391.47751 7389.36531 9786.82507 1.2959E104 1.9736E104 3.9759E104 2.1308E105 3.2413E105 6.5207E105 1.1406E106 2.6385E106 6.1034E106 1.0675E107 3.2656E107 1.7472E108 9.3481E108 5.0015E109 7.6598E111 3.3581E115 6.4540E122 1.2404E130 2.3840E137

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

1.00000 2.08000 3.24480 4.49946 5.84929 7.29992 8.85723 10.52745 12.31712 14.23312 16.28269 18.47345 20.81342 23.31103 25.97515 28.81510 31.84068 35.06221 38.49051 42.13698 46.01359 50.13290 54.50813 59.15317 64.08260 69.31174 74.85668 80.73432 86.96240 93.55954 142.05920 184.65464 303.25515 341.66747 384.32944 457.25979 606.90158 1166.00278 1366.24433 1773.06147 2177.94269 2952.63462 3985.09305 4856.24502 7177.79516 1.2769E104 2.2488E104 3.9276E104 2.0148E105 2.8260E106 4.6911E108 6.9217E110 9.5746E112

Time Value of Money Factors Geometric Series - Future Worth 4.00%

j

4.00%

TABLE A-c-1

5.00% Time Value of Money Factors Geometric Series - Future Worth 5.00%

502 Appendix A

TABLE A-c-2 j

4%

5%

6%

10%

15%

n

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

1.00000 2.10000 3.30750 4.63050 6.07753 7.65769 9.38067 11.25680 13.29710 15.51328 17.91784 20.52407 23.34613 26.39909 29.69897 33.26285 37.10887 41.25633 45.72577 50.53900 55.71925 61.29118 67.28100 73.71657 80.62750 88.04523 96.00316 104.53678 113.68375 123.48407 198.57655 268.19005 475.48661 546.06666 626.12003 766.62829 1067.38205 2300.23777 2773.76328 3776.10980 4819.37931 6919.74558 9891.32894 1.2524E104 1.9984E104 3.9876E104 7.8772E104 1.5432E105 1.1173E106 2.7826E107 1.4563E110 6.7751E112 2.9549E115

1.00000 2.11000 3.33910 4.69707 6.19440 7.84235 9.65298 11.63926 13.81507 16.19531 18.79592 21.63401 24.72791 28.09724 31.76300 35.74771 40.07545 44.77199 49.86493 55.38378 61.36010 67.82767 74.82259 82.38347 90.55158 99.37103 108.88896 119.15576 130.22523 142.15488 235.54359 324.57292 599.26021 695.27545 805.40771 1001.46907 1430.85050 3283.25810 4022.42349 5623.45524 7332.27629 1.0873E104 1.6057E104 2.0780E104 3.4651E104 7.3928E104 1.5631E105 3.2807E105 2.9379E106 1.0624E108 1.2461E111 1.3874E114 1.5207E117

1.00000 2.15000 3.46750 4.97188 6.68457 8.62931 10.83233 13.32267 16.13239 19.29696 22.85555 26.85144 31.33244 36.35133 41.96640 48.24197 55.24904 63.06596 71.77918 81.48404 92.28575 104.30028 117.65557 132.49265 148.96702 167.25008 187.53076 210.01729 234.93915 262.54920 502.41729 764.38534 1732.31928 2118.46906 2588.00247 3488.47023 5716.04907 1.8441E104 2.4661E104 3.9977E104 5.8776E104 1.0465E105 1.8608E105 2.7298E105 5.8691E105 1.8472E106 5.8067E106 1.8241E107 5.6443E108 1.7190E111 1.5937E116 1.4775E121 1.3697E126

1.00000 2.20000 3.63250 5.33500 7.35076 9.72965 12.52919 15.81567 19.66548 24.16663 29.42052 35.54394 42.67138 50.95774 60.58133 71.74746 84.69246 99.68834 117.04821 137.13240 160.35555 187.19485 218.19934 254.00076 295.32598 343.01123 398.01859 461.45483 534.59318 618.89830 1473.60036 2608.23558 8089.99442 1.0722E104 1.4205E104 2.1650E104 4.3653E104 2.3422E105 3.5634E105 7.1701E105 1.2543E106 2.9019E106 6.7131E106 1.1742E107 3.5921E107 1.9219E108 1.0283E109 5.5016E109 8.4258E111 3.6939E115 7.0994E122 1.3645E130 2.6224E137

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

1.00000 2.09000 3.27610 4.56477 5.96287 7.47766 9.11686 10.88864 12.80164 14.86503 17.08853 19.48241 22.05756 24.82552 27.79847 30.98933 34.41178 38.08027 42.01010 46.21746 50.71945 55.53419 60.68082 66.17958 72.05186 78.32029 85.00877 92.14258 99.74841 107.85449 168.78836 223.89681 383.07414 436.07164 495.62196 598.92640 815.95585 1670.28717 1988.74313 2651.16420 3327.72366 4661.10317 6501.45689 8099.63097 1.2517E104 2.3825E104 4.4942E104 8.4161E104 5.3533E105 1.0949E107 4.1121E109 1.4671E112 5.1545E114

Appendix A 503

4%

5%

6%

10%

15%

n

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

1.00000 2.11000 3.33910 4.69707 6.19440 7.84235 9.65298 11.63926 13.81507 16.19531 18.79592 21.63401 24.72791 28.09724 31.76300 35.74771 40.07545 44.77199 49.86493 55.38378 61.36010 67.82767 74.82259 82.38347 90.55158 99.37103 108.88896 119.15576 130.22523 142.15488 235.54359 324.57292 599.26021 695.27545 805.40771 1001.46907 1430.85050 3283.25810 4022.42349 5623.45524 7332.27629 1.0873E104 1.6057E104 2.0780E104 3.4651E104 7.3928E104 1.5631E105 3.2807E105 2.9379E106 1.0624E108 1.2461E111 1.3874E114 1.5207E117

1.00000 2.12000 3.37080 4.76406 6.31238 8.02935 9.92963 12.02904 14.34463 16.89479 19.69932 22.77958 26.15855 29.86100 33.91356 38.34493 43.18598 48.46991 54.23244 60.51199 67.34984 74.79040 82.88136 91.67399 101.22337 111.58864 122.83334 135.02569 148.23891 162.55164 276.69912 388.14030 742.36400 868.87520 1015.31890 1279.02612 1867.22778 4508.67499 5593.65006 7984.60328 1.0584E104 1.6087E104 2.4340E104 3.2010E104 5.5100E104 1.2319E105 2.7267E105 5.9855E105 6.0957E106 2.6811E108 4.3763E111 6.3497E114 8.6370E117

1.00000 2.16000 3.49960 5.04058 6.80711 8.82605 11.12717 13.74352 16.71172 20.07237 23.87045 28.15580 32.98357 38.41486 44.51725 51.36553 59.04244 67.63945 77.25774 88.00911 100.01716 113.41844 128.36382 145.01995 163.57088 184.21984 207.19121 232.73267 261.11763 292.64778 569.13571 874.33844 2015.58405 2474.26747 3033.65117 4110.22052 6787.34872 2.2230E104 2.9821E104 4.8565E104 7.1637E104 1.2809E105 2.2859E105 3.3603E105 7.2498E105 2.2905E106 7.2193E106 2.2719E107 7.0481E108 2.1484E111 1.9921E116 1.8468E121 1.7122E126

1.00000 2.21000 3.66510 5.40588 7.47924 9.93935 12.84877 16.27972 20.31553 25.05233 30.60103 37.08948 44.66510 53.49780 63.78337 75.74744 89.64990 105.79016 124.51302 146.21558 171.35505 200.45787 234.13009 273.06935 318.07869 370.08236 430.14410 499.48806 579.52296 671.86979 1611.16222 2861.97587 8922.29822 1.1836E104 1.5694E104 2.3944E104 4.8345E104 2.5988E105 3.9549E105 7.9606E105 1.3929E106 3.2231E106 7.4572E106 1.3044E107 3.9908E107 2.1354E108 1.1425E109 6.1128E109 9.3620E111 4.1043E115 7.8882E122 1.5161E130 2.9138E137

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

1.00000 2.10000 3.30760 4.63092 6.07863 7.66000 9.38492 11.26395 13.30836 15.53017 17.94223 20.55821 23.39274 26.46138 29.78073 33.36852 37.24361 41.42613 45.93752 50.80062 56.03978 61.68093 67.75171 74.28152 81.30172 88.84566 96.94887 105.64917 114.98682 125.00468 202.16597 274.23487 491.16717 565.67355 650.51483 800.19773 1123.40317 2476.77264 3005.58331 4137.30972 5329.99997 7767.25889 1.1279E104 1.4440E104 2.3584E104 4.8876E104 1.0062E105 2.0612E105 1.7366E106 5.8595E107 6.4361E110 7.0103E113 7.6293E116

Time Value of Money Factors Geometric Series - Future Worth 6.00%

j

6.00%

TABLE A-c-3

10.00% Time Value of Money Factors Geometric Series - Future Worth 10.00%

504 Appendix A

TABLE A-c-4 j

4%

5%

6%

10%

15%

n

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

1.00000 2.15000 3.46750 4.97188 6.68457 8.62931 10.83233 13.32267 16.13239 19.29696 22.85555 26.85144 31.33244 36.35133 41.96640 48.24197 55.24904 63.06596 71.77918 81.48404 92.28575 104.30028 117.65557 132.49265 148.96702 167.25008 187.53076 210.01729 234.93915 262.54920 502.41729 764.38534 1732.31928 2118.46906 2588.00247 3488.47023 5716.04907 1.8441E104 2.4661E104 3.9977E104 5.8776E104 1.0465E105 1.8608E105 2.7298E105 5.8691E105 1.8472E106 5.8067E106 1.8241E107 5.6443E108 1.7190E111 1.5937E116 1.4775E121 1.3697E126

1.00000 2.16000 3.49960 5.04058 6.80711 8.82605 11.12717 13.74352 16.71172 20.07237 23.87045 28.15580 32.98357 38.41486 44.51725 51.36553 59.04244 67.63945 77.25774 88.00911 100.01716 113.41844 128.36382 145.01995 163.57088 184.21984 207.19121 232.73267 261.11763 292.64778 569.13571 874.33844 2015.58405 2474.26747 3033.65117 4110.22052 6787.34872 2.2230E104 2.9821E104 4.8565E104 7.1637E104 1.2809E105 2.2859E105 3.3603E105 7.2498E105 2.2905E106 7.2193E106 2.2719E107 7.0481E108 2.1484E111 1.9921E116 1.8468E121 1.7122E126

1.00000 2.20000 3.63000 5.32400 7.32050 9.66306 12.40093 15.58974 19.29230 23.57948 28.53117 34.23740 40.79957 48.33180 56.96248 66.83597 78.11454 90.98047 105.63843 122.31818 141.27750 162.80550 187.22632 214.90326 246.24332 281.70235 321.79077 367.07984 418.20881 475.89279 1011.68773 1645.79111 4233.47929 5335.94786 6714.75678 9452.95712 1.6608E104 6.2548E104 8.6720E104 1.4897E105 2.2902E105 4.3470E105 8.2144E105 1.2528E106 2.9003E106 1.0114E107 3.4915E107 1.1954E108 4.6192E109 1.8753E112 2.6078E117 3.2236E122 3.7357E127

1.00000 2.25000 3.79750 5.69812 8.01694 10.83000 14.22606 18.30868 23.19857 29.03631 35.98549 44.23643 54.01033 65.56415 79.19627 95.25296 114.13587 136.31073 162.31725 192.78075 228.42536 270.08942 318.74310 375.50887 441.68493 518.77238 608.50641 712.89237 834.24722 975.24739 2444.78343 4452.08581 1.4448E104 1.9325E104 2.5822E104 3.9811E104 8.1590E104 4.5000E105 6.8802E105 1.3940E106 2.4499E106 5.6992E106 1.3240E107 2.3211E107 7.1254E107 3.8253E108 2.0508E109 1.0985E110 1.6846E112 7.3876E115 1.4199E123 2.7289E130 5.2449E137

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

1.00000 2.14000 3.43560 4.90402 6.56428 8.43737 10.54642 12.91700 15.57726 18.55830 21.89438 25.62327 29.78663 34.43036 39.60508 45.36653 51.77616 58.90168 66.81766 75.60628 85.35803 96.17260 108.15978 121.44048 136.14783 152.42845 170.44376 190.37150 212.40736 236.76675 446.81247 674.30392 1507.44509 1838.06949 2239.27239 3006.87959 4899.36687 1.5646E104 2.0883E104 3.3756E104 4.9535E104 8.7982E104 1.5615E105 2.2884E105 4.9118E105 1.5433E106 4.8464E106 1.5215E107 4.7045E108 1.4325E111 1.3281E116 1.2312E121 1.1415E126

Appendix A 505

4%

5%

6%

10%

15%

n

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

To Find F Given A1 (F | A1 i %, j %,n)

1.00000 2.20000 3.63250 5.33500 7.35076 9.72965 12.52919 15.81567 19.66548 24.16663 29.42052 35.54394 42.67138 50.95774 60.58133 71.74746 84.69246 99.68834 117.04821 137.13240 160.35555 187.19485 218.19934 254.00076 295.32598 343.01123 398.01859 461.45483 534.59318 618.89830 1473.60036 2608.23558 8089.99442 1.0722E104 1.4205E104 2.1650E104 4.3653E104 2.3422E105 3.5634E105 7.1701E105 1.2543E106 2.9019E106 6.7131E106 1.1742E107 3.5921E107 1.9219E108 1.0283E109 5.5016E109 8.4258E111 3.6939E115 7.0994E122 1.3645E130 2.6224E137

1.00000 2.21000 3.66510 5.40588 7.47924 9.93935 12.84877 16.27972 20.31553 25.05233 30.60103 37.08948 44.66510 53.49780 63.78337 75.74744 89.64990 105.79016 124.51302 146.21558 171.35505 200.45787 234.13009 273.06935 318.07869 370.08236 430.14410 499.48806 579.52296 671.86979 1611.16222 2861.97587 8922.29822 1.1836E104 1.5694E104 2.3944E104 4.8345E104 2.5988E105 3.9549E105 7.9606E105 1.3929E106 3.2231E106 7.4572E106 1.3044E107 3.9908E107 2.1354E108 1.1425E109 6.1128E109 9.3620E111 4.1043E115 7.8882E122 1.5161E130 2.9138E137

1.00000 2.25000 3.79750 5.69812 8.01694 10.83000 14.22606 18.30868 23.19857 29.03631 35.98549 44.23643 54.01033 65.56415 79.19627 95.25296 114.13587 136.31073 162.31725 192.78075 228.42536 270.08942 318.74310 375.50887 441.68493 518.77238 608.50641 712.89237 834.24722 975.24739 2444.78343 4452.08581 1.4448E104 1.9325E104 2.5822E104 3.9811E104 8.1590E104 4.5000E105 6.8802E105 1.3940E106 2.4499E106 5.6992E106 1.3240E107 2.3211E107 7.1254E107 3.8253E108 2.0508E109 1.0985E110 1.6846E112 7.3876E115 1.4199E123 2.7289E130 5.2449E137

1.00000 2.30000 3.96750 6.08350 8.74503 12.06814 16.19143 21.28016 27.53121 35.17876 44.50114 55.82870 69.55325 86.13903 106.13559 130.19299 159.07955 193.70275 235.13362 284.63543 343.69729 414.07340 497.82915 597.39498 715.62940 855.89277 1022.13348 1218.98882 1451.90275 1727.26362 4794.31884 9316.99291 3.4201E104 4.7116E104 6.4803E104 1.0424E105 2.2873E105 1.4685E106 2.3265E106 4.9914E106 9.1664E106 2.2717E107 5.6049E107 1.0211E108 3.3736E108 2.0055E109 1.1803E110 6.8890E110 1.3188E113 7.7089E116 2.2224E124 5.6952E131 1.3682E139

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 36 40 48 50 52 55 60 72 75 80 84 90 96 100 108 120 132 144 180 240 360 480 600

1.00000 2.19000 3.60010 5.26498 7.22458 9.52492 12.21898 15.36776 19.04150 23.32103 28.29943 34.08380 40.79740 48.58208 57.60107 68.04218 80.12149 94.08761 110.22657 128.86740 150.38864 175.22570 203.87947 236.92611 275.02833 318.94842 369.56315 427.88099 495.06184 572.43977 1354.98109 2391.47751 7389.36531 9786.82507 1.2959E104 1.9736E104 3.9759E104 2.1308E105 3.2413E105 6.5207E105 1.1406E106 2.6385E106 6.1034E106 1.0675E107 3.2656E107 1.7472E108 9.3481E108 5.0015E109 7.6598E111 3.3581E115 6.4540E122 1.2404E130 2.3840E137

Time Value of Money Factors Geometric Series - Future Worth 15.00%

j

15.00%

TABLE A-c-5

APPENDIX B One purpose of this appendix is to briefly explore, from an engineering economy perspective, the world of accounting. Accounting is a broad and complex discipline. Our purpose here is not to cover these topics to the same depth and breadth that an accountant would; rather, our purpose is 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. A second purpose of this appendix 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.

B-1

INTRODUCTION

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

Cost Terminology Introductory Example Let us assume that the primary business of a small manufacturing firm is jobshop 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

506

OBTAINING AND ESTIMATING CASH FLOWS

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 percent 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 assume that 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 would 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 could 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 507

508 Appendix B Obtaining and Estimating Cash Flows

sources, such as company production records, accounting records, manufacturer’s 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 therefore be familiar with cost terminology, cost factors, and estimating techniques used by different specialists.

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COST TERMINOLOGY

In this section, cost definitions and concepts are considered from five 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 analysis’s purpose. 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.

Five Cost Viewpoints 1. 2. 3. 4. 5.

Life Cycle Viewpoint Past/Future Viewpoint Manufacturing Cost Structure Viewpoint Fixed/Variable Viewpoint Average/Marginal Viewpoint

B.2.1 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. 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

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(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 cost of testing. Supporting equipment costs may include computer-control hardware and software and a spare-parts inventory. 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 would consist of both preventive maintenance and maintenance on an as-needed basis. In any case, 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

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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 that 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. 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. B.2.2 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

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Cost Terminology

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.

Past Costs and Sunk Costs

EXAMPLE

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. 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 ($3500 2 $2,500 2 $85 2 $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 $2,500 2 $85 2 $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.

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

EXAMPLE

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512 Appendix B Obtaining and Estimating Cash Flows

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 over $2,000 at the end of 5 years, there is a $3,000 ($5,000 2 $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-yearold 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 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. 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 percent 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) 5 $50. The $50 amount is thus termed the opportunity cost associated with storing the $1,000 in the home safe. EXAMPLE

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 percent annual interest.

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Thus, the investment of $5,000 would yield (0.10)($5,000) 5 $500 each year. If, instead of investing the $5,000, the person purchases an automobile for the same amount, he will forgo the opportunity to earn $500 interest/ year. The $500 amount is again termed an annual opportunity cost associated with purchasing the automobile.

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 $600 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 percent if funded, then the $600 million tied up in inventories carries a $90 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 discussed 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. Some elaboration of bonds is therefore appropriate.

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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, since 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. B.2.3 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 B.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 you would see in 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. The second general comment about Figure B.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 does vary dramatically between sectors 1

Ostwald, P. F. and T. S. McLaren, Cost Analysis and Estimating for Engineering and Management, 4th Revised Edition, Prentice Hall, 2003.

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Direct Material Prime Cost Direct Labor

Cost Goods Manufactured

Indirect Material Conversion Costs

Indirect Labor

Factory Overhead Cost of Goods Sold

Selling Price

Fixed and Miscellaneous

General and Administrative

Nonfactory Overhead

Selling (Marketing)

Profit

FIGURE B.1

Manufacturing Cost Structure

(manufacturing, retail, etc.), between industries within sectors, and between 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 B.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

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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 B.1), the major cost elements can be defined as direct material, direct labor, and overhead costs. 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.

EXAMPLE

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 five sides, the unmachined surface is painted and air-dried, and then four 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, including indirect labor and indirect material. 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, departments

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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 appendix. B.2.4 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. Variable costs vary in proportion to quantity of output. These costs are usually for direct material and direct labor. For instance, in Example B.5, gray iron casting was used to produce a part. This cost was presented to illustrate direct material cost, since 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 since 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 purchase price of the castings. 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

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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 ‘‘What (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 TC1x2 5 FC 1 VC1x2

(B.1)

where x 5 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, since 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 TC1x2 5 $950 1 $0.15x is a valid relationship for a given year in question (the relevant range is restricted to a given year, since 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 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 B.2 graphically illustrates the total cost function.

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Total Cost

Cost

Variable Cost

Fixed Cost

Annual Mileage

FIGURE B.2

Total Annual Cost as a Function of Mileage

Now let us consider Figure B.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: TR1x2 5 $Rx

(B.2)

Adding this functional relationship to Figure B.2 and modifying the terminology for this example yields Figure B.3.

Revenue

Total Cost

Dollars

Break-even Point

Fixed Cost

Annual Production Volume (x in units)

FIGURE B.3

Revenue and Cost as a Function of Production Volume

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520 Appendix B Obtaining and Estimating Cash Flows

In Figure B.3, 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 B.3. TR1x* 2 5 TC1x* 2

5 FC 1 VC1x* 2

(B.3)

The break-even point is the 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) 2 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) 2 TC(x), evaluated for a particular value of x. It is generally desirable to have a low break-even value. For the general example of Figure B.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 breakeven point exists for two relationships—y 5 g(.) and y 5 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) 5 h(x) and solving for x*. The concept can be extended to more than two functions of a single variable, say y 5 h(x), y 5 g(x), and y 5 t(x). If these are all linear functions, then Figure B.4 depicts two of the possible results. In Figure B.4a, all three functions intersect at a single point; thus, a single value of x* can be determined (Point A). In Figure B.4b, there is no unique break-even value of x involving all three functional relationships. The linear equations y 5 h(x) and y 5 t(x) intersect at Point B, which is the

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h(x)

Values of y

Break-even Point A

g(x)

t(x) Values of x (a)

Break-even Point D

Values of y

Break-even Point B

Break-even Point C

h(x)

t(x) g(x)

Values of x (b)

F IGURE B.4

Break-even Points (a) Single, (b) Multiple

break-even value for these two relationships. Point C is the break-even value for y 5 h(x) and y 5 g(x). Point D is the break-even value for y 5 g(x) and y 5 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 appendix.

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522 Appendix B Obtaining and Estimating Cash Flows

EXAMPLE

Break-Even Point for Production 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. Letting x 5 the production volume in units, then R1x2 5 TC1x2 5 FC 1 VC1x2 and

$5.00x 5 $300 1 1$2.50 1 $1.002x

Simplifying and solving for x yields x* (the break-even value): x* 5 $300y$1.50 5 200 units For a production output of 1,000 units, the net profit, P, is calculated to be P 5 $511,000 units2 2 1$2.50 1 $1.002 11,000 units2 2 $300 5 $1,200

EXAMPLE

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 salt water 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 with 2

This example, with slight modification, is taken from Ferguson, E. J. and J. E. Shamblin, ‘‘Break-Even Analysis,’’ The Journal of Industrial Engineering, 18(8), August 1967 with permission of the publisher.

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TABLE B.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

1.00

Meter Calibration

0.40

Chemical pump power (1/4 hp)

0.32

Total Fixed Cost

$2.69

Fixed Cost per year 5 $2.69 3 365 5 $982 Variable Cost Pipeline pump (5 hp @ 50% utilization)

0.63

Chemical additives (7:5 qt/day)

3.75

Inhibitor (2 qt/day)

1.00

Gas (10.8 mcf/day 3 $0.0275/mcf)

0.30

Total Variable Cost

$5.68

Variable cost per barrel 5 0.01136/bbl @ 500 bbl/day

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 B.1 and B.2 and are considered valid for production quantities up to 1,000 barrels/day (365,000 barrels/yr). In addition to the fixed and variable costs given in Table B.1 for the automatic tank battery operation, other annual fixed costs are D1 5 annual cost of depreciation and interest 5 $3,082 M1 5 annual cost of maintenance, taxes, and labor 5 $5,485 TABLE B.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 5 $0.98 3 365 5 $358 Variable Cost Chemical additives (7.5 qt/day) (10.8 mcf/day 3 $0.0275/mcf) Total Variable Cost Variable cost per barrel 5 0.00810/bbl @ 500 bbl/day

3.75 0.30 $4.05

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524 Appendix B Obtaining and Estimating Cash Flows

Letting x 5 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 1x2 5 FC1 1 VC1 1x2 5 1$982 1 $3,082 1 $5,4852 1 $0.01136x 5 $9,549 1 $0.01136x

In addition to the fixed and variable costs given in Table B.2 for the manual tank battery operations, other annual fixed costs are D2 5 annual cost of depreciation and interest 5 $2,017 M2 5 annual cost of maintenance, taxes, and labor 5 $7,921 Then, the total annual cost, TC2(x), for the manual tank battery operation is given by TC2 1x2 5 FC2 1 VC2 1x2 5 1$358 1 $2,017 1 $7,9212 1 $0.00810x 5 $10,296 1 $0.00810x

By equating the two total cost functions, the break-even production volume can be determined as TC1 1x2 5 TC2 1x2 $9,549 1 $0.01136x 5 $10,296 1 $0.00810x x* 5 229,141 barrels/year

The interpretation of the break-even point in this example is that x* 5 229,141 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 5 200,000 barrels/year, then TC1(x) 5 $11,821 and TC2(x) 5 $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.

B.2.5 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

B-2

divided by the number of units produced while incurring those costs. The marginal cost is the incremental cost associated with increasing the output by one 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. The average cost of one unit of output (unit cost) is the ratio of total cost to the quantity of output (miles traveled, production volume, etc.); that is, AC1x2 5

TC1x2 x

(B.4)

where AC(x) 5 average cost per unit of x TC(x) 5 total cost for x units of output x 5 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 B.2.4, which had a total cost function of $(950 1 0.15x), the average cost, in dollars per mile, is given by AC1x2 5

950 950 1 0.15x 5 1 0.15 x x

If the automobile travels 10,000 miles/year, then the average operating cost is $(950y10,000 1 0.15) 5 $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. 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) 5 $950 1 0.15x, then dTC(x)/dx 5 $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 one unit.

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526 Appendix B Obtaining and Estimating Cash Flows

If the total cost function is discontinuous and defined only for discrete values of x (for example, x 5 1,2,3, . . .), then difference equations must be used to determine marginal costs. For example, TC(6) 2 TC(5) is the marginal cost of increasing the output quantity from x 5 5 to x 5 6. Thus, in the discrete case, marginal cost is always the cost required to increase the output quantity x by one 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. EXAMPLE

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) 5 60,000 1 30x. What is the marginal cost at x 5 10? x 5 20? 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 5 total cost at 11 2 total cost at 10 MC1102 5 TC1112 2 TC1102 MC1102 5 3$60,000 1 $301112 4 2 3$60,000 1 $301102 4

B-2

MC1102 5 $60,330 2 $60,300 MC1102 5 $30 Similarly, MC1202 MC1202 MC1202 MC1202

5 TC1212 2 TC1202 5 3$60,000 1 $301212 4 2 3$60,000 1 $301202 4 5 $60,630 2 $60,600 5 $30

Differentiation Marginal cost at 10 5 first derivative of the total cost function evaluated at 10 MC1x2 5 d/dx TC1x2 MC1x2 5 d/dx 3$60,000 1 $30x4 MC1x2 5 $30 therefore MC1102 5 $30 and MC1202 5 $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 one 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) 5 $950 1 $0.15x. The average cost is AC(x) 5 $950/x 1 $0.15 and the marginal cost is MC(x) 5 $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 B.3 and B.4 summarize the relationships between marginal cost and total cost (Table B.3) and between marginal cost and average cost (Table B.4). The following example illustrates some of these cost, revenue, and profit relationships.

Cost Terminology 527

528 Appendix B Obtaining and Estimating Cash Flows

Relationship Between Marginal Cost and Total Cost

TABLE B.3

EXAMPLE

Relationship Between Marginal Cost and Average Cost

TABLE B.4

If MC(x) . 0 then TC(x 1 1) . TC(x)

If MC(x) , AC(x) then AC(x 1 1) , AC(x)

If MC(x) 5 0 then TC(x 1 1) 5 TC(x)

If MC(x) 5 AC(x) then AC(x 1 1) 5 AC(x)

If MC(x) , 0 then TC(x 1 1) , TC(x)

If MC(x) . AC(x) then AC(x 1 1) . AC(x)

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). Let t 5 number of tons per season SP(t) 5 selling price in order to sell t tons 5 $1800 2 0.8t2 TR(t) 5 total revenue when t tons are sold at a particular selling price 5 selling price X demand 5 $(800 2 0.8t)t 5 $(800t 2 0.8t2) MR(t) 5 the marginal revenue at a sales volume of t tons 5 dTR1t2ydt 5 $1800 2 1.6t2 TC(t) 5 the total production cost for t tons 5 $110,000 1 400t2 TP(t) 5 total profit when t tons are sold 5 TR1t2 2 TC1t2 5 $1800t 2 0.8t2 2 2 $110,000 1 400t2 5 $120.8t2 1 400t 2 10,0002 AP(t) 5 average profit per ton when t tons are sold 5 TP1t2yt 5 $120.8t 1 400 2 10,000yt2 The equations apply for the range 0 # t , 1,000. Figure B.5 on the following page illustrates the total revenue, total cost, and total profit curves for this example.

B-2

500,000 400,000 300,000 200,000

Dollars

100,000 0 0

100

200

300

400

500

600

700

800

900

–100,000 –200,000 –300,000 –400,000 –500,000

Tons

Total Revenue

F IGURE B.5

Total Cost

Total Profit

Total Revenue, Total Cost, and Total Profit as a Function of Tons

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. dTR1t2ydt 5 800 2 210.82t 5 0 resulting in t* 5 500 tons The total revenue with a sales volume of 500 tons is TR15002 5 $80015002 2 $0.815002 2 5 $200,000 The marginal revenue at the output level of 500 tons is MR15002 5 $800 2 $1.615002 5 0 Thus, the rate of change in the TR(t) function with respect to t is 0 when TR(t) is evaluated at t 5 500. The TR(t) function is strictly concave with a

1000

Cost Terminology 529

530 Appendix B Obtaining and Estimating Cash Flows

unique maximum value at TR(500). For sales from t 5 1 to 500, the total revenue function is increasing at a decreasing rate. For sales volumes from t 5 500 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. dTP1t2ydt 5 2120.82t 1 400 5 0 resulting in t* 5 250 tons Thus, total profit will be maximized for a sales volume of 250 tons, and the maximum profit per season would be TP12502 5 2$0.812502 2 1 $40012502 2 $10,000 5 $40,000 The average profit per ton when 250 tons are sold is AP12502 5 20.812502 1 400 2 10,000y250 5 $160/ton Finally, we note that there are two break-even points in this example. By equating TR(t) 5 TC(t), or 800t 2 0.8t2 5 10,000 1 400t we obtain 20.8t2 1 400t 2 10,000 5 0 Solving for the positive roots of this quadratic equation yields t 5 26.39 and t 5 473.61 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.

B-3

COST ESTIMATION

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

B-3

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 that 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. Cost estimation is one of the most difficult challenges an engineering economist faces.

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 preciseness is exercised 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 5 C1M2 1 C1E2 where CT 5 the total cost of making the estimate in dollars C(M) 5 the functional cost of making the estimate in dollars C(E) 5 the functional cost of errors in the estimate in dollars. 3

Ibid.

(B.5)

Cost Estimation

531

532 Appendix B Obtaining and Estimating Cash Flows

Cost of Making Estimates

Cost

Optimal Amount of Detail

Total Cost of Estimating

Cost of Errors Resulting From Inaccurate Estimates Amount of Detail (increasing from left to right)

FIGURE B. 6

Cost of Detail in Estimating

As depicted in Figure B.6, the optimal amount of detail is that which minimizes the total cost of estimating. Since 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. B.3.1 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 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

B-3

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-ofmagnitude 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 FAA’s Life Cycle Cost Estimating Handbook4 suggests the following six-step process: 1. 2. 3. 4. 5. 6.

Plan the estimate Research, collect, and analyze data Develop the estimate structure Determine the estimating methodologies Compute the cost estimate 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 fourcategory classification scheme outlined earlier in this section (order-ofmagnitude, 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, since this step provides the framework for the remainder of the estimating process.

4

The FAA Life Cycle Cost Estimating Handbook is available at www.faa.gov.

Cost Estimation

533

534 Appendix B Obtaining and Estimating Cash Flows

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 is 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 is typically time-series data that represents or is 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 B.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 B.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.

B-3

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 decisionmaking 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. B.3.2 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

Cost Estimation

535

536 Appendix B Obtaining and Estimating Cash Flows

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. B.3.3 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.

B-4

General Accounting Principles

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 percent of the items survive the first 3 years of service. Then, 10 percent of the items fail in the fourth year, 20 percent fail in the fifth year, 50 percent in the sixth year, 15 percent in the seventh year, and the remaining 5 percent fail in the eighth year. A weighted-average time to failure for this equipment is [(0.10)(4) 1 (0.20)(5) 1 (0.50)(6) 1 (0.15)(7) 1 (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.

B-4

GENERAL ACCOUNTING PRINCIPLES

As already 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. 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 5

Stewart, R. D., J. D. Johannes, and R. M. Wyskida, Cost Estimator’s Reference Manual, 2nd ed., John Wiley & Sons, 2001.

537

538 Appendix B Obtaining and Estimating Cash Flows

Balance Sheet

FIGURE B. 7

Income Statement

Balance Sheet

Relationship Between Balance Sheets and Income Statements.

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 B.7. B.4.1 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. A balance sheet provides a statement of the financial condition of a firm at a point in time.

The fundamental accounting equation is defined as Assets 2 Liabilities 5 Net Worth Rewriting, we have the more common form of the fundamental equation of accounting: Assets 5 Liabilities 1 Net Worth (B.6) The Fundamental Equation of Accounting Assets 5 Liabilities 1 Net Worth

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 2006, shown in Figure B.8. Note in the figure that total assets

B-4

Fiscal Year Ended

General Accounting Principles

539

in thousand, except earnings per share Oct 1, 2006 Oct 1, 2005

ASSETS Current Assets Cash and Equivalents

$ 312,606

$ 173,809

Short Term Investments

141,038

133,227

Accounts Receivable

224,271

190,762

Inventories

636,222

546,299

Prepaid Expenses and other

126,874

94,429

88,777

70,808

1,529,788

1,209,334

Deferred Income Taxes Total Current Assets Long Term Investments Equity and other Investments Property, Plant, and Equipment Other Assets Other Intangible Assets Goodwill Total Assets

5,811

60,475

219,093

201,089

2,287,899

1,842,019

186,917

72,893

37,955

35,09

161,478

92,474

$4,428,941

$3,513,693

$ 340,937

$ 220,975

288,963

232,354

54,868

44,496

Liabilities and Shareholders’ Equity Current Liabilities Accounts Payable Accrued Compensation Accrued Occupancy Costs Accrued Taxes

94,010

78,293

Short-Term Borrowing

700,000

277,000

Other Accrued Expenses

224,154

198,082

Deferred Revenue

231,926

175,048

Current Portion of Long Term Total Current Liabilities Long Term Debt Other Long Term Liabilities Total Liabilities

762

748

1,935,620

1,226,996

1,958

2,870

262,857

193,565

2,200,435

1,423,431

Shareholders’ Equity Common Stock Additional Paid-In Capital Other Additional Paid-In Capital Retained Earnings Accumulated Other Comprehensive Income Total Shareholders’ Equity Total Liabilities and Shareholders’ Equity FIGURE B.8

Starbucks’ Condensed Consolidated Balance Sheet

756

767



90,201

39,393

39,393

2,151,084

1,938,987

37,273

20,914

2,228,506

2,090,262

$4,428,941

$3,513,693

540 Appendix B Obtaining and Estimating Cash Flows

($4,428,941 for the fiscal year ended October 1, 2006) equals the sum of total liabilities ($2,200,435) plus total net worth (called ‘‘shareholders equity’’ on Starbucks’s balance sheet) ($2,228,506). 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. 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 B.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 one 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 one 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 longterm 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 B.5. Of first importance, note that this sample balance sheet is ‘‘balanced’’ since assets 5 liabilities 1 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 was 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

B-4

TABLE B.5

General Accounting Principles

Example Balance Sheet BuiltRite Tool and Engineering Company Balance Sheet as of December 31, 2006

ASSETS Current Assets Cash

$ 25,000

Accounts Receivable (net)

115,000

Raw Materials

8,500

Work In Process

7,000

Finished Goods

3,000

Small Tools

12,500

Total Current Assets

$171,000

Fixed Assets Land Building Less depreciation reserve Equipment Less depreciation reserve Office Equipment

$ 30,000 $200,000 50,000

150,000

$750,000 150,000

600,000 10,000

Total Fixed Assets

$790,000

Total Assets

$961,000

LIABILITIES AND CAPITAL Current Liabilities Accounts Payable

32,000

Taxes Payable

15,000

Total Current Liabilities

47,000

Fixed Liabilities Mortgage loan payable

$130,000

Equipment loan payable

350,000

Total Fixed Liabilities

$480,000

Total Liabilities

$ 527,000

Capital Common Stock

$325,000

Retained Earnings

80,000

Earned Surplus (current year)

29,000

Total Capital Total Liabilities and Capital

$434,000 $961,000

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.

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542 Appendix B Obtaining and Estimating Cash Flows

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 B.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 B.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. B.4.2

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 B.9 shows a condensed version of Starbucks’s income statement (referred to as a statement of earnings).

An income statement provides a statement of the revenues and expenses incurred by a firm during a period of time.

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

B-4

Fiscal Year Ended

General Accounting Principles

in thousand, except earnings per share Oct 1, 2006 Oct 1, 2005

Net Revenues Company Operated Retail

$6,583,098

$5,391,927

Specialty Licensing

860,676

673,015

Foodservice and other

343,168

304,358

1,203,844

977,373

Total Net Revenues

7,786,942

6,369,300

Cost of Sales including Occupancy

3,178,791

2,605,212

Store Operating Expenses

2,687,815

2,165,911

Other Operating Expenses

260,087

197,024

387,211

340,169

473,023

357,114

6,986,927

5,665,430

Total Specialty

Depreciation and Amortization Expenses General and Administrative Expenses Subtotal Operating Expenses Income from Equity Investments

93,937

76,648

893,952

780,518

12,291

15,829

Earnings before Income Taxes

906,243

796,347

Income Taxes

324,770

301,977

Earnings before effect of accounting change

581,473

494,370

Operating Income Interest and Other Income

Cumulative effect of accounting change Net Earnings

17,214



$ 564,259

$ 494,370

Net Earnings—basic

$

0.74

$

0.63

Net Earnings—diluted

$

0.71

$

0.61

Per Common Share

Weighted Average Shares Outstanding Basic

766,114

789,570

Diluted

792,556

815,417

F IGURE B.9

Starbucks’ Condensed Consolidated Statement of Earnings

goods. For a manufacturing company (Table B.6), the cost of goods section usually includes the costs incurred in producing the product. For a retail company (Table B.7 and Figure B.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 B.6 and B.7 show a gross profit line, but Starbucks (Figure B.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

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544 Appendix B Obtaining and Estimating Cash Flows

TABLE B.6

Example Income Statement—Manufacturing BuiltRite Tool and Engineering Company Income Statement For Year Ended December 31, 2006

Net Sales

$1,200,000

Less Cost of Goods Manufactured Direct Labor

$420,000

Direct Materials

302,000

Indirect Labor

112,000

Depreciation

98,000

Repairs and Maintenance

41,500

Utilities

11,500

Gross Profit

985,000 $215,000

Other Expenses Marketing

$49,000

General and Administrative

76,000

Interest payments

35,000

Net Income before Taxes Less Income Taxes

26,000

Net Income (posted to Earned Surplus)

TABLE B.7

160,000 $55,000 $29,000

Example Income Statement—Retail BuiltRite Tool and Engineering Company Income Statement For Year Ended December 31, 2006

Net Sales

$1,200,000

Less Cost of Goods Sold Inventory, December 31, 20x5 Plus purchases Total Less Inventory, December 31, 2006

$26,000 432,000 $458,000 44,000

Gross Profit

414,000 $786,000

Less Expenses Direct Labor

$420,000

Depreciation—Building

10,000

Depreciation—Equipment

30,000

Repairs and Maintenance

41,500

Indirect Labor

218,000

Utilities

9,800

Supplies

1,700

Net Income before Taxes Less Income Taxes Net Income (posted to Earned Surplus)

731,000 $55,000 26,000 $29,000

B-4

General Accounting Principles

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 B.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 B.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. 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. Starbucks’s financial statements were extracted from their December 14, 2006, Form 10-K, which is a required disclosure statement that publicly held companies file with the SEC (Securities and Exchange Commission). Starbucks’s balance sheet (Figure B.8) and income statement (Figure B.9) are both referred to as comparative statements, since they show more than 1 fiscal year and therefore provide an opportunity for the reader to make year-to-year comparisons. B.4.3 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 appendix. 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. Ratios are only rough guides to interpreting financial statements; they are not mathematical conclusions. Ratios are generally used in one of three ways to draw conclusions about a company’s financial health: (1) comparison with a

545

546 Appendix B Obtaining and Estimating Cash Flows

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.

Uses of Ratio Analysis 1.

Comparison to a company’s historic values to spot trends and/or changes in performance

2. Comparison to expected or desired benchmarks 3. Comparison to competitor’s performance

Generally, the analysis of financial statements focus 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 B.8 contains comparative balance sheets (two balance sheets displayed side by side) for the years 2005 and 2006. Table B.9 contains comparative income statements for the years 2005 and 2006.

Areas of Focus of Ratio Analysis 1. Earning power 2. Short-term liability obligations 3. Long-term liability obligations

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 5 Return on Owner’s Equity 5

Net Income Average Total Assets

Net Income Average Owner’s Equity

(B.7) (B.8)

B-4

TABLE B.8

General Accounting Principles

Carson’s Cutlery Company Comparative Balance Sheet

Carson’s Cutlery Comparative Balance Sheet as of December 31, 2005 and 2006 ASSETS

2006

2005

Current Assets Cash

$61,750

$83,520

Accounts Receivable (net)

195,000

130,500

65,000

50,000

Inventory Prepaid Expenses Total Current Assets

22,750

31,900

$344,500

$295,920

Fixed Assets Machinery

$208,000

$187,830

Furniture

74,750

72,500

Other

22,750

23,750

$650,000

$580,000

Total Assets LIABILITIES AND CAPITAL Current Liabilities Notes Payable

$92,950

$87,000

Accounts Payable

147,212

109,653

Taxes Payable Total Current Liabilities

69,438

64,920

$309,600

$261,573

Fixed Liabilities Loans Total Liabilities

$100,000

$90,000

$409,600

$351,573

$100,000

$100,000

88,427

77,397

Capital Stock Retained Earnings Earned Surplus Total Capital Total Liabilities and Capital

51,973

51,030

$240,400

$228,427

$650,000

$580,000

The calculation of earning power ratios for Carson’s Cutlery for 2006 are shown below: Return on Assets Employed 5

$51,973 5 0.0845 or 8.45% 1$580,000 1 650,0002y2

Return on Owner’s Equity 5

$51,973 5 0.2217 or 22.17% 1$228,427 1 $240,4002y2

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

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548 Appendix B Obtaining and Estimating Cash Flows

TABLE B.9

Carson’s Cutlery Company Comparative Income Statement Carson’s Cutlery Comparative Income Statement For Years Ended December 31, 2005 and 2006

Net Sales

2006

2005

$1,625,450

$1,450,000

Cost of Goods Sold Beginning Inventory

$50,000

Direct Materials

406,000

350,000

Direct Labor

801,500

700,000

Factory Overhead Total Cost of Goods Sold

40,000

94,603

90,000

$1,352,103

$1,180,000

65,000

50,000

Less: Ending Inventory Gross Profit

$

$ 1,287,103

$1,130,000

$ 338,347

$ 320,000

$

$

Other Operating Expenses Selling Expenses Depreciation & Amortization

43,980 58,122

General and Administrative

122,484

Total Other Operating Expenses

37,200 53,791 120,580

$ 224,586

$

Net Operating Income

$ 113,761

$ 108,430

Less: Interest Expenses

$

$

Less: Income Taxes

21,600 40,188

Net Income

$

51,973

211,570 18,000 39,400

$

51,030

ability to meet its current obligations. The following ratios are commonly used to measure liquidity: Current Ratio ⫽

Current Assets Current Liabilities

(B.9)

Acid Test Ratio Cash 1 Accounts Receivable 1 Short Term Investments 5 (B.10) Current Liabilities Accounts Receivable Turnover ⫽

Net Sales (B.11) Average Accounts Receivable

Inventory Turnover 5

Cost of Goods Sold Average Inventory

(B.12)

B-4

General Accounting Principles

The calculation of liquidity ratios for Carson’s Cutlery for 2006 is shown below. Current Ratio ⫽ Acid Test Ratio 5 Accounts Receivable Turnover 5 Inventory Turnover 5

$344,500 ⫽ 1.112 $309,600 $61,750 1 $195,000 1 $0 5 0.829 $309,600 $1,625,450 5 9.98 1$130,500 1 $195,0002y2 $1,287,103 5 22.38 1$50,000 1 $65,0002y2

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: Debt to Equity Ratio ⫽

Total Liabilities Total Capital Worth

(B.13)

Times Interest Earned Ratio Net Income Before Taxes and Interest (B.14) 5 Interest Charges Operating Income to Total Assets Ratio ⫽

Net Operating Income (B.15) Total Assets

The calculation of solvency ratios for Carson’s Cutlery for 2006 is shown below: Debt to Equity Ratio 5

$409,600 5 1.70 $240,400

Times Interest Earned Ratio 5

$113,761 5 5.27 $21,600

Operating Income to Total Assets Ratio 5

$113,761 5 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

549

550 Appendix B Obtaining and Estimating Cash Flows

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 5 Net Income 1 Income Taxes 1 Interest Expense

(B.16)

EBITDA 5 Net Income 1 Income Taxes 1 Interest Expense 1 Depreciation 1 Amortization (B.17) The calculation of these measures for Carson’s Cutlery for 2006 is shown below: EBIT 5 $51,973 1 $40,188 1 $21,600 5 $113,761 EBITDA 5 $51,973 1 $40,188 1 $21,600 1 $58,122 5 $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.

B-5

COST ACCOUNTING PRINCIPLES

The balance sheet and the income statement in Section B.4 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.

B-5

B.5.1 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 on labor time cards, 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 B.10, which follows the step-bystep procedures, illustrates their application.

Traditional methods for allocating overhead 1. Allocation based on direct labor hours 2. Allocation based on direct labor dollars 3. Allocation based on prime dollars

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: Rate 5

overhead cost direct labor hours

(B.18)

Cost Accounting Principles

551

552 Appendix B Obtaining and Estimating Cash Flows

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 5 rate 3 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: Ratio 5

overhead cost 3 100% direct labor dollars

(B.19)

3. Determine (or estimate) the direct labor dollars required by the par-

ticular job for which overhead cost is being estimated 4. Calculate the overhead cost for the job: Estimated overhead 5 ratio 3 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: Ratio 5

overhead cost 3 100% (B.20) direct labor dollars 1 direct material dollars

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 5 ratio 3 1estimated direct labor dollars 1 estimated direct material dollars2

EXAMPLE

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

B-5

are to be based on the following cost totals collected during the previous accounting period. Total direct labor hours Total direct labor dollars Total direct material dollars Total overhead costs

48,000 $480,000 $600,000 $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. Direct labor hours Step 1: Previous period direct labor hours 5 48,000 Previous period overhead cost 5 $360,000 Step 2: Rate per direct labor hour 5 $360,000y48,000 5 $7.50/hour Step 3: Estimated direct labor hours for job 5 40 Step 4: Estimate overhead 5 $7.50/hour 3 40 hours 5 $300 b. Direct labor dollars Step 1: Previous period direct labor dollars 5 $480,000 Previous period overhead cost 5 $360,000 Step 2: Percentage ratio per direct labor dollar 5 ($360,000y $480,000) 3 100% 5 75% Step 3: Estimated direct labor dollars for job 5 40 hours 3 $12.50/hour 5 $500 Step 4: Estimate overhead 5 75% 3 $500 5 $375 c. Direct labor dollars 1 direct material dollars Step 1: Previous period direct labor dollars 5 $480,000 Previous period direct material dollars 5 $600,000 Previous period overhead cost 5 $360,000 Step 2: Percentage ratio per (direct labor dollar 1 direct material dollar) 5 [$360,000y($480,000 1 $600,000)] 3 100% 5 33.33% Step 3: Estimated direct labor dollars 1 direct material dollars for job 5 (40 hours 3 $12.50/hour) 1 $850 5 $1,350 Step 4: Estimate overhead 5 33.33% 3 $1,350 5 $450 a.

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 Accounting Principles

553

554 Appendix B Obtaining and Estimating Cash Flows

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. Since 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

Overhead Allocation to Departments TABLE B.10

Data for Example B.6 Department A

Department B

Direct Material Cost

$720,000

$240,000

$960,000

Direct Labor Cost

$260,000

$140,000

$400,000

25,200

16,200

41,400

14

9

23

$250,000

$200,000

$450,000

$25,000

$20,000

Direct Labor Hours Number of Employees First Cost of Equipment Annual Depreciation

Total

$45,000

Other Factory Overhead

$150,000

General Overhead

$350,000

The information in Table B.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

B-5

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. 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 percent of selling price? For Department A, the total overhead allocation is determined as follows: Annual depreciation Other factory overhead (14y23)($150,000) General overhead (14y23)($350,000) TOTAL

$25,000 $91,000 $213,043 $329,347

Thus, the overhead rate for Department A per direct labor hour is Rate 5

$329,347 5 $13.07 per direct labor hour 25,200

For Department B, the total overhead allocation is calculated as follows: Annual depreciation Other factory overhead (9y23)($150,000) General overhead (9y23)($350,000) TOTAL

$20,000 $58,696 $136,957 $215,653

Thus, the overhead rate for Department B per direct labor hour is Rate 5

$215,653 5 $13.31 per direct labor hour 16,200

The estimated total cost for Job Order D is then computed as Direct material cost Direct labor cost A ($260,000y25,200)(30 hours) Overhead cost A ($13.07)(30 hours) Direct labor B ($140,000y16,200)(100 hours) Overhead cost B ($13.31)(100 hours) TOTAL If x 5 the selling price of Job Order D, then x 5 total cost 1 profit

x 5 $3,796.82 1 10.252x

$900 $309.52 $392.10 $864.20 $1,331 $3,796.82

Cost Accounting Principles

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556 Appendix B Obtaining and Estimating Cash Flows

solving for x yields x5

$3,796.82 0.75

thus, x 5 $5,062.43

B.5.2 Activity Based Costing

A relatively new entrant in the field of cost accounting is activity-based costing (ABC). This has emerged due to the dramatic changes that have occurred, and continue to occur, 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 B.5.1). In many cases today, 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. 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.

Activity based costing is designed to associate costs with the activities which drive them.

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

B-5

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 most 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 B.5.3 Standard Costs

Although the first task of cost accounting is to determine per-item or perorder 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. Since 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 6

Kaplan, R. S., and S. R. Anderson, Time-Driven Activity Based Costing, Harvard Business School Press, 2007. 7 Cokins, G., Activity-Based Cost Management, Making It Work, McGraw-Hill, 1996.

Cost Accounting Principles

557

558 Appendix B Obtaining and Estimating Cash Flows

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. B.5.4 Economic Value Added

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. 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).

EVA is a management tool that examines the difference between the net operating profit after taxes and the cost of capital.

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

8

Stern, J. M., J. S. Shiely, and I. Ross, The EVA Challenge, Wiley, 2001.

B-5 Cost Accounting Principles

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: 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. 1.

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 R&D and in training, they should not be overlooked in the quest for value. The three examples below illustrate how EVA is calculated and how it can be used to improve economic decision making.

Economic Value Added—Example 1 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 $70 million, respectively. Taxes equal 40 percent of operating profit. Consequently, the annual net incomes for the firms are $24 million and $42 million, respectively. Dividing the net income by equity yields return on equity of 24 percent and 21 percent, respectively. The cost of capital is 12 percent. Which firm is best from a shareholder value point of view? Some would choose A, because it has the greatest ROE (return on equity) or ROA (return on assets). However, B maximizes value for shareholders. To determine the EVA, subtract the cost of capital from net operating profit after taxes (NOPAT). For A, this yields $24 million 2 0.12($100 million) 5 $12 million. For B, the EVA equals $42 million 2 0.12($200 million) 5 $18 million. (As was the case with the IRR method, one cannot choose the alternative that has the greatest return on equity; instead, incremental net operating profits should be compared with the cost of capital.)

EXAMPLE

559

560 Appendix B Obtaining and Estimating Cash Flows

It has long been known that maximizing return on investment is not the right objective. In 1924, Donaldson Brown, chief financial officer of General Motors Corporation, noted, ‘‘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.’’9

EXAMPLE

Economic Value Added—Example 2 Consider two firms, C and D, each with a 12 percent cost of capital. The financial data for the two firms (in $M’s) are

Equity Annual Operating Profit

Firm C

Firm D

$100

$200

$25

$35

Taxes (40 percent)

$10

$14

Net Income

$15

$21

On the basis of net income generated, one might conclude that D is the betterperforming company. However, for C there is a capital cost of $12 and an EVA of $3; whereas for D there is a capital cost of $24 and an EVA of 2$3. Therefore, firm C is superior to firm D in adding value for the shareholder.

EXAMPLE

Economic Value Added—Example 3 There are several ways to compute EVA. The method used in the previous examples was Firm C

Firm D

Capital Invested

$100

$200

Operating Profit

$25

$35

Taxes (40 percent)

$10

$14

Net Operating Profit After Taxes (NOPAT)

$15

$21

Cost of Capital (12 percent)

$12

$24

1$3

2$3

EVA 9

Viewgraphs used in EVA seminar conducted by Stern Stewart & Co.

Summary 561

Alternatively, EVA can be computed as follows: Firm C

Firm D

Return on Assets

15%

10.5%

Cost of Capital

12%

12%

Difference

13%

21.5%

EVA 5 difference 3 assets

1$3

⫺$3

In this appendix, we provided an introduction to the language and techniques of accountants, financial analysts, and managers. The ‘‘takeaway’’ messages from this appendix can be summarized as follows: 1.

2.

3.

4.

5.

6. 7.

8.

To be successful in developing and selling engineering designs and projects, the engineer must learn to communicate effectively in the language of accounting. Cost information is best understood and interpreted by considering it from one or more of five cost viewpoints: life cycle, past/future, manufacturing cost structure, fixed/variable, and average/marginal. Cost estimation is a complex and challenging task that must incorporate the trade-off between the cost of making the estimate and the consequences of an inaccurate estimate. Cost estimation can be thought of hierarchically. Order of magnitude estimates can be broken down into preliminary estimates that can be further broken down into detailed estimates. The two most important accounting documents are the balance sheet and the income statement. The balance sheet shows what a company’s assets, liabilities, and net worth are at a point in time. The income statement shows the revenues and expenses a company incurred over a period of time. Ratio analysis can be used to meaningfully interpret the values found on balance sheets and income statements. Cost accounting focuses on allocating overhead costs to specific products or orders. Traditional methods include allocation by direct labor hours, direct labor dollars, and prime dollars. Activity-based costing is an emerging area of accounting that more accurately connects costs with the activities that drive them.

SUMMARY

562 Appendix B Obtaining and Estimating Cash Flows

9. Economic value added is an emerging technique to focus manage-

ment’s attention to shareholder value by connecting profit with cost of capital.

Problem available in WileyPLUS GO Tutorial Tutoring Problem available in WileyPLUS Video Solution Video Solution available in WileyPLUS

FE-LIKE PROBLEMS 1.

On a balance sheet, a corporation’s economic obligations to nonowners are called a. owners’ equity. c. assets. b. liabilities. d. retained earnings.

2.

The information below has been extracted from the books of the Shelley Company. Which of the following represents Shelley’s current ratio? Current Assets Cash Accounts receivable Inventory Total

a. 0.76 b. 1.48

Accounts payable Wages payable Taxes payable Total

$78 $68 $28 $174

c. 2.05 d. 2.51

a. b. c. d.

The fundamental equation used within an accounting balance sheet is net profit 5 gross profit 2 expenses 2 taxes. assets 1 liabilities 5 net worth. assets 1 liabilities 1 net worth 5 0. assets 5 liabilities 1 net worth.

a. b. c. d.

Marginal cost is any cost occurring after “time now.” the ratio of total cost to total quantity of output. the market value of an asset at the end of its life less its disposal costs. the incremental cost of producing one more unit of output.

3.

4.

5.

Current Liabilities $86 $130 $140 $356

The three major categories that comprise cost of goods manufactured are a. material, labor, and overhead. b. average, marginal, and instantaneous.

Summary 563

c. past, present, and future. d. initial, operating, and salvage. 6.

The total cost equation for producing X widgets is given by TC 5 $1,000 1 $6X. The average cost per widget for producing 500 widgets is closest to which of the following? a. $1,000 c. $8 b. $6 d. $4,000

7.

The total cost equation for producing X widgets is given by TC 5 $1,000 1 $6X. The variable cost per widget is closest to which of the following? a. $1,000 c. $8 b. $6 d. $4,000

8.

The total cost equation for producing X widgets is given by TC 5 $1,000 1 $6X. The marginal cost per widget at a production level of 300 units is closest to which of the following? a. $2,800 c. $8 b. $6 d. $4,000

9.

Fixed cost is any cost that does not vary with the quantity of output. the ratio of total cost to total quantity of output. the market value of an asset at the end of its life less its disposal costs. the incremental cost of producing one more unit of output.

a. b. c. d. 10.

When an organization considers its work-in-process, how should it be classified on a balance sheet? a. Asset c. Net worth b. Liability d. Expense

PROBLEMS Section B.1 1. Reconsider the scenario presented in Example B.1, which details a firm con-

sidering 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?

564 Appendix B Obtaining and Estimating Cash Flows

Section B.2 2. Match the terms in the first column with the appropriate definition in the sec-

ond column. Terms (a) First cost (b) Average cost (c) Fixed cost (d) Overhead cost (e) Marginal cost (f) Opportunity cost (g) Salvage value (h) Sunk cost (i) Variable cost (j) Past cost (k) Maintenance costs (l) Future cost (m) Cost of debt capital

3.

Definitions (1) Any cost occurring after ‘‘time now’’ (2) The total initial investment required to get an asset ready for service (3) The ratio of total cost to total quantity of output (4) The market value of an asset at the end of its life less its disposal costs (5) The cost of obtaining fund through debt obligations (6) The incremental cost of producing one more unit of output (7) Any past cost or portion of past cost that is not recovered (8) The recurring costs that are necessary to operate and maintain an asset (9) Any cost that varies with the quantity of output (10) Any cost occurring prior to ‘‘time now’’ (11) Any cost that does not vary with the quantity of output (12) The cost of forgoing an opportunity to earn interest on funds (13) All cost in manufacturing other than direct material and direct labor

A firm has the capacity to produce 1,000,000 units of a product each year. At present, it is operating at 70 percent 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 percent 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?

4. 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? 5.

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 would 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?

Summary 565

6. 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 5 0.04n3 2 700n 1 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 7.

GO Tutorial The cost curve for producing widgets passes through the following points and is piecewise linear in between. Units Produced

Costs

0 200 400 600

$600 $1,200 $1,600 $1,800

a. b. c. d. e. f. g.

What is the fixed cost of producing 600 widgets? What is the variable cost of producing 600 widgets? What is the cost per unit if only 400 widgets are produced? What is the incremental cost of producing the 100th widget? What is the incremental cost of producing the 500th widget? What is the fixed cost per unit for producing 1,000 widgets? What is the variable cost per unit for producing 1,000 widgets?

8. 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 each. Annual fixed costs for the manufacturer are $3,500,000. a. If the plant is operating at 50 percent 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? 9. 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: TC 5 1200 1 80X 1 2X2 2A

where X 5 number of floors and A 5 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.

566 Appendix B Obtaining and Estimating Cash Flows

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.) 10.

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 M1 M2 M3

Fixed Cost/Order

Variable Cost/Unit

$300 $750 $500

$9 $3 $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? 11.

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?

12. The variable cost of producing specialty value-added stainless-steel 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

Summary 567

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 B.3 13. 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? 14. 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 5 63.63 1 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 r 2 of 0.9824. Interpret the meaning of this value. 15. 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 16. 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 this 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 r 2. b. Which driver do you recommend?

568 Appendix B Obtaining and Estimating Cash Flows

c. What percentage of the data does your driver and regression equation

explain? Week

Circuit Board Assembly Support Cost

Direct Labor Hours

Number of Boards Completed

Average Cycle Time

$66,402 $56,943 $60,337 $50,096 $64,241 $60,846 $43,119 $63,412 $59,283 $60,070 $53,345 $65,027 $58,220 $65,406 $35,268 $46,394 $71,877 $61,903 $50,009 $49,327 $44,703 $45,582 $43,818 $62,122 $52,403

7,619 7,678 7,816 7,659 7,646 7,765 7,685 7,962 7,793 7,732 7,771 7,842 7,940 7,750 7,954 7,768 7,764 7,635 7,849 7,869 7,576 7,557 7,569 7,672 7,653

2,983 2,830 2,413 2,221 2,701 2,656 2,495 2,128 2,127 2,127 2,338 2,685 2,602 2,029 2,136 2,046 2,786 2,822 2,178 2,244 2,195 2,370 2,016 2,515 2,942

186.44 139.14 151.13 138.30 158.63 148.71 105.85 174.02 155.30 162.20 142.97 176.08 150.19 194.06 100.51 137.47 197.44 164.69 141.95 123.37 128.25 106.16 131.41 154.88 140.07

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Section B.4 17.

Determine the missing values below to produce a valid balance sheet. BALANCE SHEET Year Ending December 31, 2004 ASSETS Cash Accounts receivable Raw materials inventory Work-in-process inventory Finished goods inventory Land Building Less: Depreciation reserve Equipment Less: Depreciation reserve TOTAL ASSETS

(a) $8,000 $10,000 $15,000 $18,500 $30,000 $80,000 $8,000 (c) $4,000

(b) $36,000 $283,727

Summary 569

LIABILITIES Notes payable Accounts payable Declared dividends TOTAL LIABILITIES NET WORTH Common stock Retained earnings TOTAL NET WORTH TOTAL LIABILITIES & NET WORTH

$25,000 (d) $20,000 $51,000 $200,000 (e) (f) (g)

18. Following is a list of balance sheet accounts for Branman Co. as of December 31,

2005. a. Classify each of the following accounts as an asset, liability, or net worth. Account

Balance

Account

Balance

Cash Accounts receivable Notes payable Raw materials inventory Common stock Work-in-process inventory

$45,000 $8,000 $25,000 $10,000 $100,000 $15,000

Declared dividends payable Finished goods inventory Land Net building (less depreciation) Net equipment (less depreciation) Retained earnings

$20,000 $18,500 $30,000 $72,000 $36,000 ??????

Accounts payable

$6,000

b. What is the balance in the Retained Earnings account if this data forms a

valid balance sheet? 19.

The Scott Company shows the following alphabetical list of account balances as of December 31, 2006. Account

Balance

Account

Balance

Accounts payable Accounts receivable Accrued taxes Buildings (net value) Cash Dividends payable

$50,000 $80,000 $20,000 $205,000 $40,000 $20,000

Finished goods Land value Long-term mortgages Notes payable Raw material inventory Stockholder’s equity

$30,000 $115,000 $390,000 $60,000 $40,000 $150,000

Equipment (net value)

$180,000

a. Construct a balance sheet for Scott Company as of December 31, 2006. b. Demonstrate that the fundamental accounting equation holds by stating

the dollar values of the terms in the equation. 20. In addition to the account balances in the previous problem, Scott Company

recorded the following summary transactions for the year ending December 31, 2006.

570 Appendix B Obtaining and Estimating Cash Flows

Category

Amount

Administrative expenses Factory depreciation Direct labor charges Factory overhead General interest payments Raw material expenses Sales Taxes paid

$30,000 $30,000 $70,000 $35,000 $40,000 $90,000 $400,000 $20,000

Construct an income statement for Scott Company for the period 1/1/2006 to 12/31/2006. 21.

Betty Smith is the owner of Accurate Tax Service. For the year ending April 30, 2007, 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, 2007. b. Prepare a balance sheet as of April 30, 2007. 22. 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.

Summary 571

Accounts payable Accounts receivable Advertising expense Cash Common stock Cost of goods sold Depreciation expense Inventory

$12,000 $21,000 $600 $22,950 $54,000 $15,000 $300 $14,400

Notes payable Prepaid rent Property, plant, and equipment Rent expense Retained earnings Sales revenue Wages expense Wages payable

$6,000 $1,050 $15,300 $2,100 $1,500 $27,000 $7,500 $1,200

a. Prepare a balance sheet for JanCo. b. Prepare an income statement for JanCo. 23.

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

? $8,739 $187,506 $92,827 $402 ? $9,474 $11,195

Other assets Other liabilities Property, plant, and equipment Retained earnings Total assets Total liabilities Total liabilities and net worth Total net worth

a. Determine the values for the missing items. b. Prepare a balance sheet for Packers. 24. The annual report from a major energy company contains the following: Net

income represented 9 percent return on average total assets of approximately $27 billion and a 19 percent 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 25.

Consider the comparative balance sheet and income statement for Starbucks provided in Figures B.8 and B.9. Based on these financial statements, determine the following for the year ending October 1, 2006: 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

$30,350 $57,786 $103,684 $35,097 $378,045 ? ? ?

572 Appendix B Obtaining and Estimating Cash Flows

26. 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 appendix. Section B.5 27.

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 49 3 49 sheet of aluminum that costs $3.50 per square foot. Overhead costs to produce a lot of size 10,000 are estimated to be $304,000.

Sequence 1 2 3 4 5

Dept

Avg. Labor ($/hr)

Avg. Prod. Time (min/part)

Floor Space (sq. ft.)

Cutting Stamping Trimming Assemble Package

$20.00 $18.20 $17.20 $16.00 $14.40

3.0 1.0 2.0 5.0 0.5

1,300 1,600 1,200 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? 28. 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 A B

Direct Labor Hours

Overhead Costs

900 1,350

$13,500 $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

Summary 573

• Overhead costs in each department are allocated based on an overhead rate per direct labor hour • Profit is to be 25 percent of total cost 29. 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.) 30. 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? GO Tutorial Consider the following table (shown in two parts):

31.

EOY 8 9 10

EOY 8 9 10

Revenue

Operating Expenses

Depreciation

Net Operating Profit

Taxes

$2,765,000 $2,765,000 $2,765,000

$2,379,200 $2,379,200 $2,379,200

$38,400 $23,040 $13,824

$347,400 $362,760 $371,976

$138,960 $145,104 $148,790

NOPAT

Capital

Cost of Capital

Capital Charge

EVA (ECON Profit)

$208,440 $217,656 $223,186

$57,600 $34,560 $20,736

0.1 0.1 0.1

$9,600 XXX

XXX

a. What is the EVA in year 8? b. What is the capital charge in year 9? 32. Consider the following table (shown in two parts): a. What is the EVA in year 5? b. What is the capital charge in year 6?

574 Appendix B

Obtaining and Estimating Cash Flows

EOY 5 6 7

Revenue

Operating Expenses

Depreciation

Net Operating Profit

Taxes

$2,765,000 $2,765,000 $2,765,000

$2,379,200 $2,379,200 $2,379,200

$49,536 $49,536 $48,768

$336,264 $336,264 $337,032

$134,506 $134,506 $134,813

EOY 5 6 7

33.

NOPAT

Capital

Cost of Capital

Capital Charge

EVA (ECON Profit)

$201,758 $201,758 $202,219

$74,304 $144,768 $96,000

0.1 0.1 0.1

$12,384 XXX

XXX

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 tax rate is 30 percent. MARR is 10 percent. What is the EVA for year 3?

34. Given the following information for Live Wire Electronics, construct an in-

come statement for the month ending June 30. Sales Cost of goods sold Administration cost Advertising expense Taxes

35.

$500,000 $170,000 $75,000 $25,000 $70,000

GO Tutorial 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 tax rate is 40 percent. MARR is 10 percent. What is the EVA for year 3?

36. Cost Center D within the Welding Department of the Mizer Corporation col-

lected the following data during 2004. 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 2005 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

Summary 575

d. Mizer Corp. has adopted the direct labor hours method of overhead alloca-

tion. 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? 37. 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? 38. 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? 39.

Video Solution Skrunchy Company produces two products, Lower and Upper. The following two tables give pertinent information about these two products. Category

Lower

Direct labor Direct material Direct labor hours Quantity produced Total overhead

Upper

$700,000 $1,000,000 $2,000,000 $1,500,000 28,000 25,000 20,000 10,000 $2,500,000

ABC Activity

Driver

Cost

Lower

Upper

Rework Material handling

Hours Moves

$1,800,000 $700,000

4,000 hours 40,000 moves

42,500 hours 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? 40. Reconsider Problem 39. 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? 41. LockTite Company produces two products, Pretty Safe (PS) and Virtually

Impenetrable (VI). The following two tables give pertinent information about these products.

576 Appendix B

Obtaining and Estimating Cash Flows

Category

PS

Number produced Direct material cost Direct labor hours Direct labor cost Total overhead

3,000 $350,000 5,000 hours $100,000

VI 2,000 $150,000 10,000 hours $200,000 $750,000

ABC Activity

Driver

Cost

PS

VI

Production Engineering

Machine hours Engineering hours

$500,000 $250,000

1,000 hours 2,000 hours

4,000 hours 3,000 hours

a. What is the cost per unit of Pretty Safe if LockTite uses traditional over-

head 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? 42. Reconsider Problem 41. a. What is the cost per unit of Virtually Impenetrable if LockTite uses tradi-

tional 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?

ANSWERS TO SELECTED EVEN NUMBERED PROBLEMS Chapter 1 FE2. b

FE4. a

2. (a) Yes

FE6. d

(b) $1,600 to $2,400

4. You are lender $450–$600 interest 6. (a) $1,900

(b) $6,400

(c) WSI—Future spending is less

8. (a) None; C; D; CD; BC; BD; BCD; AD; ABD

(b) None

(d) 2, 10

(e) 1, 4, 5, 7

(c) 1

10. Q; P; PR 12. (c) 2, 3, 4, 5, 9

(d) 1, 4, 5, 6, 7

14. (a) likely between 4 and 10 years, others acceptable

(b) 2, 3, 8

(c) 1, 2, 5, 7

Chapter 2 FE2. b

FE4. d

2. (a) $971.70

FE6. b (b)

FE8. c

FE10. a

FE12. b

$1030.00

4. (1) Money has a time value. (2) Money cannot be added or subtracted unless it occurs at the same point(s) in time. (3) To move money forward one time unit, multiply times one plus the discount or interest rate. (4) To move money backward one time unit, divide by one plus the discount or interest rate. 6.

$14,000

$14,000

1

2

$14,000

$14,000

$14,000

$14,000

4

5

6

$3,000

$4,000

(+) 0 (–)

3

$2,000

$5,000

$23,000 $450

8. (+) 0

1

2

3

4

5

6

7

8

9

10

11

(–)

12 $150

$550+150 $550+150 $550+150 $550+150 $550+150 $550+150 $550+150 $550+150 $550+150 $550+150 $550+150 $550+450

577

578

10.

Answers to Selected Even Numbered Problems

$150,000

(+) 0

1

2

3

(–)

12.

4

5

$20,000

$20,000

6 $X

7

$X+8000

8

9

$X+16,000

$X+24,000

$10,000

$4,000

(+) 0

1

(–)

2

3

4

5

6

7

8

9

10

11

13

$2,880 $2,880 $2,880 $2,880 $2,880 $2,880 $2,880 $2,880 $2,880 $2,880

14. $49,178.78 16. 4.8641% 18. 12.9116% 20. (a) 7.1773%

12

(b) 11.6123%

(c) 14.8698%

24. (a) 12 years

(b) 19 years

(c) 24 years

28. (a) $7,462.15

(b) $4,777.69

(c) $1,689.70

30. (a) $5,863.80

(b) $5,863.87

(c) $5,863.87

32. (a) $12,348.70 (d) $550.77

(b) $1,551.33 (e) $2,135.97

(c) $15.27

38. (a) $150.76

(b) $97.02

(c) $97.02

40. (a) $1,711.50

(b) $3,167.86

(c) $1,996.29

22. 18.5631%

34. $12,000.00 36. $2,714.10

(d) $8,382.44

Answers to Selected Even Numbered Problems

42. (a)

$8,000

0

1

2

3

4

5

6

7

$X

$X+100

$X+400

8

$2X

(b) $2,272.61

(c) 5 5 $2,272.61

6 5 $2,372.61

7 5 $2,672.61

8 5 $4,545.23

44. Payment method #2 is best. 46. (a) $3,939.68

(b) $4,962.86

48. $315.47 50. $46,573.45 52. (a) $2,818.55

(b) $3,771.85

54. Prefer to receive your money quicker at $10K for each of 5 years. Prefer to pay your money over a longer period at $5K for each of 10 years. 56. $6,097.27 58. (a) $5,746.64

(b) $10,636.63

(c) $13,399.09

60. $6,352.85 62. $25,180.40 64. (a) 6.367

(b) 9.196%

66. (a) $17,908.48

(b) $25,937.42

68. $11,194.61 70. (a) $43,157.13

(b) $63,856.54

(c) $95,010.53

72. (a) $402,532.54

(b) $39,181.05

(c) $36,227.93

74. $929.42 76. $7,169.73 78. (a) B*(F ZP i%,1)*(F ZA i%,n)

(b) B*(11i)*((11i)n⫺1)/i

82. $4,368.76 84. $65,199.82 86. (a) $3,154.12

(b) 2$719.29

88. $15,173.55 90. $13,725.46 92. $314.13 94. $107.87 96. $12,771.80 98. (a) $313,890.35 100. (a) $8,471.54 102. $4,622.20

(b) $2,354.17 (b) $53,357.69

(c) $47,356.75

(c) $6,715.61

579

580

Answers to Selected Even Numbered Problems

106. $14,129.51 108. (a) (F Z G i%,n) 5 (P Z G i%,n)*(F Z P i%,n) 110. (a) $8,528.80

(b) (F Z G i%,n) 5 ((11i)^n2(11n*i))/(i^2)

(c) $11,051.00

(b) $2,690.58

112. $3,774.14 114. $12,075.69 116. $926,130.53 120. (a) $7,783.80

(b) $2,455.56

122. $286,447.62 124. $8,933.81 126. (a) $12,544.41

(b) $18,366.27

128. $50,116.76 130. (c) $7,066,155.64 132. $2,317,332.23 134. $25,656.82 136. 35 138. (a) $17.99

(b) $19.10

140. (a) $477,120.00

(b) $549,899.47

(c) $19,867.87

142. $8,473.13 144. $8,197.46 146. $413.55 148. (a) $4,551.84

(b) $2,937.37

152. (a) 5.0625%

(b) 5.0838%

(c) 5.0945%

(d) 5.1053%

(e) 5.1162%

154. (a) $12,548.25 (f ) $12,376.16

(b) $12,463.34 (g) $12,375.69

(c) $12,419.86 (h) $12,375.67

(d) $12,390.48 (i) $12,375.67

(e) $12,379.09

156. (a) 0.911%

(b) 10.932%

(c) 11.497%

158. (a) $970.00

(b) 3.093%

(c) 160.825%

(d) 387.382%

160. 19.4 quarters 162. $965.95

Chapter 3 FE2. d

FE4. a

FE6. b

FE8. a

2. $2,545.50 with interest of 4% per quarter 4. (a) Pay $6500.00 outright 6. Q 5 $656.97

(b) 17.27%

R 5 $1,511.34

8. X 5 $798.12 Y 5 $849.73 10. (a) $658.35

(b) $150.07

(c) $6,825.49

12. (a) $2,921.71

(b) $151,127.13

(c) $0 to $8,719.49

14. (a) $40,015.30

(b) $940,550.79

(c) $500,000.00

16. $4,105.23 18. $6,056.31 or less 20. 10.00%

(d) $4,473.77 (d) $423,175.45

Answers to Selected Even Numbered Problems

$0.08*X $500+X

22. (a)

$0.08*X

$0.08*X

$0.08*X

$0.08*X

1

2

3

4

0

PAR=

5

$9,000.00

–$8,628.12

(b) $9,000.00/8 24. You should buy this bond since it earns a nominal rate of 8.4964%. 26. $85,718.21 28. 7.873% 30. $18,250.74 32. True 34. $38,653.15 36. $6,739.94 38. $3,596.16 or $3,989.96 40. second alternative 42. (a) $3,059.20

(b) $2,299.18

(c) $918.79

Chapter 4 FE2. d

FE4. c

FE6. c

FE8. b

FE10. d

FE12. d

FE14. b

2. the project 5 a, d, e, f, h, i do nothing 5 b, c, g, j, k, l (m) The uninvested funds can be invested elsewhere to earn at least MARR (n) No (o) Yes (p) The PW must be less than zero for the project (q) The IRR must be less than MARR for the project 4. (a) $837.93

(b) If PW . 0, Accept; Otherwise reject

(c) Yes

6. (a) 2$31,203.42

(b) If PW . 0, Accept; Otherwise reject

(c) No

8. (a) 2$4,870.44 (b) If PW . 0, Accept; Otherwise reject reducing emissions is a non-economic factor to be considered

(c) No

(d) The environmental issue of

10. $79,426.86 12. (a) $89.32

(b) If PW . 0, Accept; Otherwise reject

(c) Yes

14. (a) $3.87

(b) If PW . 0, Accept; Otherwise reject

(c) Yes

16. (a) $342.76

(b) If PW . 0, Accept; Otherwise reject

(c) Yes

18. (a) $1,479.98

(b) If PW . 0, Accept; Otherwise reject

(c) Yes

20. 1 inch insulation 22. (a) Lagrange 5 $1,341,897.12 Auburn 5 $1,222,453.79 Anniston 5 $1,198,721.88 with the highest PW (c) Lagrange

(b) Select the project

581

582

Answers to Selected Even Numbered Problems

24. (a) Machine 7745 5 2$11,649.63 (c) Machine A37Y 26. (a) Structure Y 5 2$20,804.41 (c) Structure Z 28. (a) Polisher 1 5 $2,408.80 30. (a) Telescope T2

Machine A37Y 5 2$10,208.03

Structure Z 5 2$19,376.90

Polisher 2 5 2$1,116.78

(b) Select the project with the highest PW

(b) Select the project with the highest PW

Polisher 3 5 $2,967.19

(b) Polisher 3

(b) Telescope T1

32. Yes 34. Option 1 36. Yes 38. (a) No

(b) $3.94

40. (a) 1.09

(b) $2,800.00

42. Route B 44. (a) SA

(b) SA

(c) SB

46. $2,082,210.98 48. (a) 3.65

(b) If DPBP ,5 3, Accept; Otherwise, Reject

50. (a) Model 127B 5 7.2

(c) No

Model 334A 5 Does not pay back

(b) Model 127B

52. $40,334,375.95 54. $2,909,493.70 56. (a) $36,790.88

(b) $28,915.05

(c) $23,275.10

(d) $20,000.00

58. $210.00

Chapter 5 FE2. c

FE4. a

2. (a) $198.93

FE6. c

FE8. d

FE10. b

(b) If AW .5 0, Accept; Otherwise, Reject

(c) Yes

(b) If AW .5 0, Accept; Otherwise, Reject

(c) Yes

4. $79,426.86 6. (a) $14.18 8. (a) $0.68

(b) If AW .5 0, Accept; Otherwise, Reject

(c) Yes

10. (a) $90.42

(b) If AW .5 0, Accept; Otherwise, Reject

(c) Yes

12. (a) $746.96

(b) If AW .5 0, Accept; Otherwise, Reject

(c) Yes

14. 1 inch insulation 16. (a) Lagrange 5 $247,555.20 Auburn 5 $225,520.00 Anniston 5 $221,142.40 the highest AW (c) Lagrange 18. (a) Machine 7745 5 2$1,582.83, Machine A37Y 5 2$1,386.96 highest AW (c) Machine A37Y 20. (a) Project A 5 $806.00,

Project B 5 2$41.00

(b) Select the project with

(b) Select the project with the

(b) Project A

22. Purchased 24. Yes 26. (a) $4,052.10

(b) $3,670.38

(c) (i) $2,284.87

28. (a) $15,985.04

(b) $1,449.68

(c) $568.21

30. (a) $1,075,273.98

(b) $141,370.33

(c) $126,301.28

(ii) $10,084.87 (d) $118,460.86

(d) (i) $2,384.55

(ii) $15,319.48

Answers to Selected Even Numbered Problems

32.

FIRST DEPOSIT

ANNUAL DEPOSIT

25 30 35 40 45 50 55 60 65

$2,049.80 $3,343.06 $5,496.21 $9,159.04 $15,624.39 $27,816.62 $53,963.14 $129,607.38 $1,000,000.00

34. (a) $7,892.60

(b) If FW .5 0, Accept; Otherwise, Reject

(c) Yes

36. (a) $1,121.34

(b) If FW .5 0, Accept; Otherwise, Reject

(c) Yes

38. (a) $596.85

(b) If FW .5 0, Accept; Otherwise, Reject

(c) Yes

40. (a) $12.01

(b) If FW . 0, Accept; Otherwise, Reject

(c) Yes

42. (a) Vendor A 5 $23,767.00, Vendor B 5 $30,947.00

(b) Select the project with the highest FW

(c) Vendor B

44. Approximately 7,765 Units 46. (a) Polisher 1 5 $9,744.91,

Polisher 2 5 2$4,518.02,

Polisher 3 5 $12,003.92

(b) Polisher 3

48. Savings Account 50. (a) No

(b) Yes

(c) X 5 379.98 or Approximately 380 Units/YR

52. Yes 54. Design 1 56. (a) $40,233.07

$37,634.00

$39,459.08

(b) They produce a consistent decision

Chapter 6 FE2. d

583

FE4. d

FE6. c

FE8. a

FE10. c

FE12. c

2. (a) Dollars

(b) Percentage

(c) Dollars

(d) Percentage

(e) Dollars

4. (a) $7,447.88

(b) $10,427.13

(c) $20,346.05

(d) $3,713.02

(e) 2$3,000.00

6. (a) 7.42%

(b) If IRR . MARR (10%), Accept; Otherwise, Reject

(c) No

8. (a) 15.70%

(b) If IRR .5 MARR (18%), Accept; Otherwise, Reject

(c) No

10. (a) 7.45%

(b) If IRR .5 MARR (8%), Accept; Otherwise, Reject

(c) No

12. (a) 36.35%

(b) If IRR .5 MARR (10%), then Attractive; Otherwise, Unattractive

14. (a) 17.91%

(b) If IRR .5 MARR (15%), Accept; Otherwise, Reject

(c) Yes

(c) Yes

16. IRR 5 16.417% This is a desirable venture 18. IRR 5 24.204% Condenser should be purchased 20. (a) IRR . MARR (e) IRR 5 MARR

(b) IRR 5 MARR (f ) IRR , MARR

(c) IRR , MARR (g) IRR . MARR

(d) IRR . MARR (h) IRR 5 MARR

(i) IRR , MARR

22. Yes 24. (a) 16.52%

(b) If IRR .5 MARR (18%), Accept; Otherwise, Reject

(c) No

26. No 28. (a) One sign change, therefore, one positive real root (c) 12.98% (d) Yes

(B) One sign change, therefore, unique positive real root

584

Answers to Selected Even Numbered Problems

30. (a) Four sign changes, therefore, either 4, 2, or 0 positive real roots (b) Multiple sign changes, therefore, no conclusion (c) 8.08% 37.43% 32. (a) 9.82%

25.67%

57.45%

(d) Yes

(b) No

34. Lagrange 36. Proposal B 38. (a) Prefer Null

(b) Prefer A

(c) Prefer D

(d) Prefer F

(e) Prefer G

40. Light Bar 42. (a) Null, X1 Only, X2 Only, X1&Y1, X2&Y2

(b) X2&Y2

44. Alternative 6 46. (a) 2$130,000.00; $80,000.00; $80,000.00; $80,000.00; $80,000.00; $80,000.00; $80,000.00; $80,000.00; $80,000.00; $80,000.00; $85,000.00 (b) 61.03% (c) Prefer A (d) If MARR 5 40%, PW(C) , 0 Since IRR(C) 5 39% If MARR 5 40%, PW(D) , 0 Since IRR(D) 5 36% If MARR 5 40%, PW(A) . 0 Since IRR(A) 5 44% If MARR 5 40%, PW(B) 5 0 Since IRR(B) 5 40% 48. Vendor B 50. C003 52. Yes 54. Purchased 56. (a) 11.57%

(b) Yes

58. (a) 9.9997%

(b) No

60. (a) 10.35%

(b) Yes

62. ERR 5 14.179% Recommend 64. ERR 5 19.850% Recommend 66. (a) 14.07%

(b) If ERR .5 MARR (13.5%), then Accept; Otherwise, Reject

68. (a) 16.68%

(b) If ERR .5 MARR, then Accept; Otherwise, Reject

(c) Yes

(c) YES

70. Gate 3 72. Project A 74. Proposal B 76. Alternative 6

Chapter 7 FE2. d

FE4. c

FE6. a

4. (a) Replace

(b) Replace

6. (a) $17,000.00

(b) Replace

FE8. a

FE10. b

(c) $17,000.00

(d) Replace

8. Trade in existing and purchase new 10. (b) Replace with new turret lathe

(d) Replace with new turret lathe

12. (b) Keep existing unit and buy additional unit 14. (a) Trade in existing system

(b) Trade in existing system

16. Retain old lathe 18. (a) Replace Furnace 20. Keep the old machine

(d) Keep existing unit and buy additional unit

(b) Replace Furnace

Answers to Selected Even Numbered Problems

22. Replace the present equipment 24. Trade in existing equipment 26. Trade in existing equipment 28. Overhaul the existing equipment 30. Year 8 32. Year 17 34. Purchase new machine immediately 36. (a) Replace every 10 years

(b) Replace at its maximum life of 12 years

38. 7 years $58,028.16 40. 6 years $126,696.35 42. 7 years $293,243.13 44. 5 years $276,608.18 46. True 48. True 50. True

Chapter 8 FE2. c

FE4. d

FE6. d

FE8. b

FE10. b

4. b, d 6. (a) Tangible, Personal, Depreciable (b) Tangible, Real, Nondepreciable (c) Intangible, Personal, Nondepreciable (d) Tangible, Real, Nondepreciable (e) Tangible, Personal, Depreciable 8. (a) Tangible, Personal, Nondepreciable (b) Tangible, Personal, Depreciable Depreciable (d) Tangible, Real, Depreciable 10. (a) Year 4: $24,000.00

(b) Year 4: $0.00

12. (a) Year 2: $300,000.00

(b) Year 2: $391,238.40

14. (a) $38,000.00 16. (a) $1,300.00 (c) $2,666.67 18. a, b, c 26. EOY

0 1 2 3 4 5

28.

EOY

0 1 2 3 4

(b) $1,500.00 $1,300.00 $1,185.19

(c) Year 4: $10,368.00

$1,300.00 $351.17

dt

Bt

$749.00 $3,589.60 $3,589.60 $3,589.60 $2,243.50

$140,000.00 $139,251.00 $135,661.40 $132,071.80 $129,231.20 $126,987.70

dt

Bt

$1,166.90 $2,545.20 $2,545.20 $1,378.65

$70,000.00 $68,833.10 $66,287.90 $63,742.70 $62,364.05

(b) $3,672.00 (d) $2,666.67

(d) Year 4: $12,960.00

(c) Year 2: $384,000.00

(c) i 5 $14,300.00

ii 5 $2,781.79 $1,074.72 $1,185.19

(c) Tangible, Personal,

(d) Year 2: $384,000.00

iii 5 $6,708.00

$170.17 $690.12

iv 5 $6,708.00

585

586

Answers to Selected Even Numbered Problems

30. (a) MACRS 5 20, 150% DBSLH (d) MACRS 5 5, 200% DBSLH

(b) NA (c) MACRS 5 5, 200% DBSLH (e) MACRS 5 20, 150% DBSLH

32. (a) MACRS 5 3, 200% DBSLH (d) MACRS 5 27.5, SLM

(b) NA

34. (a) $800,000.00

(c) MACRS 5 5, 200% DBSLH

(b) $449,820.00 using MACRS-GDS (7)

36. (a) MACRS-GDS(5) (b)

EOY

0 1 2 3 4 5 6 7 8

dt

Bt

$70,000.00 $112,000.00 $67,200.00 $40,320.00 $40,320.00 $20,160.00 $0.00 $0.00

$350,000.00 $280,000.00 $168,000.00 $100,800.00 $60,480.00 $20,160.00 $0.00 $0.00 $0.00

38. (a) MACRS-GDS(10) (b) EOY

0 1 2 3 4 5 6

dt

Bt

MACRSGDS(10)

$25,000.00 $45,000.00 $36,000.00 $28,800.00 $23,050.00 $9,212.50

$250,000.00 $225,000.00 $180,000.00 $144,000.00 $115,200.00 $92,150.00 $82,937.50

10.00% 18.00% 14.40% 11.52% 9.22% 3.685%

dt

Bt

MACRSGDS(10)

$25,000.00 $45,000.00 $36,000.00 $28,800.00 $23,050.00 $18,425.00 $8,187.50

$250,000.00 $225,000.00 $180,000.00 $144,000.00 $115,200.00 $92,150.00 $73,725.00 $65,537.50

10.00% 18.00% 14.40% 11.52% 9.22% 7.37% 3.275%

(c) EOY

0 1 2 3 4 5 6 7

40. (a) d1 5 $4,400.00 d3 5 $ 4,224.00 d6 5 $1,267.20 (b) d1 5 $4,400.00 d3 5 $ 4,224.00 d4 5 $1,267.20

Answers to Selected Even Numbered Problems

42. (a) MACRS-GDS(10) (b) EOY

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

dt

Bt

$270,000.00 $486,000.00 $388,800.00 $311,040.00 $248,940.00 $198,990.00 $176,850.00 $176,850.00 $177,120.00 $176,850.00 $88,560.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00

$2,700,000.00 $2,430,000.00 $1,944,000.00 $1,555,200.00 $1,244,160.00 $995,220.00 $796,230.00 $619,380.00 $442,530.00 $265,410.00 $88,560.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00

MACRSGDS(10)

10.00% 18.00% 14.40% 11.52% 9.22% 7.37% 6.55% 6.55% 6.56% 6.55% 3.28%

44. (a) MACRS-GDS(5) (b) EOY

0 1 2 3 4 5 6

dt

Bt

MACRSGDS(5)

$320,000.00 $512,000.00 $307,200.00 $184,320.00 $184,320.00 $92,160.00

$1,600,000.00 $1,280,000.00 $768,000.00 $460,800.00 $276,480.00 $92,160.00 $0.00

20.00% 32.00% 19.20% 11.52% 11.52% 5.76%

dt

Bt

MACRSGDS(5)

$320,000.00 $512,000.00 $153,600.00

$1,600,000.00 $1,280,000.00 $768,000.00 $614,400.00

20.00% 32.00% 9.600%

(c) EOY

0 1 2 3

587

588

Answers to Selected Even Numbered Problems

46. (a) MACRS-GDS(3) (b) dt

Bt

MACRSGDS(3)

$29,997.00 $40,005.00 $13,329.00 $6,669.00 $0.00 $0.00

$90,000.00 $60,003.00 $19,998.00 $6,669.00 $0.00 $0.00 $0.00

33.33% 44.45% 14.81% 7.41% 0.00% 0.00%

dt

Bt

MACRSGDS(3)

$29,997.00 $40,005.00 $13,329.00 $3,334.50

$90,000.00 $60,003.00 $19,998.00 $6,669.00 $3,334.50

33.33% 44.45% 14.81% 3.705%

dt

Bt

MACRSGDS(3)

$29,997.00 $40,005.00 $6,664.50

$90,000.00 $60,003.00 $19,998.00 $13,333.50

33.33% 44.45% 7.405%

EOY

0 1 2 3 4 5 6

(c) EOY

0 1 2 3 4

(d) EOY

0 1 2 3

50. 25.00%, 37.50%, 18.75%, 12.50%, 6.25% 52. (a) MACRS-GDS(27.5) (b and c) EOY

0 1 2 3 4 5

dt

Bt

MACRSGDS(27.5)

$1,932.05 $3,090.60 $3,090.60 $3,090.60 $2,189.18

$85,000.00 $83,067.95 $79,977.35 $76,886.75 $73,796.15 $71,606.98

2.273% 3.636% 3.636% 3.636% 2.576%

Answers to Selected Even Numbered Problems

589

54. (a) MACRS-GDS(27.5) (b and c) dt

Bt

MACRSGDS(27.5)

$16,864.60 $19,270.80 $19,270.80 $19,270.80 $19,270.80 $19,270.80 $4,014.75

$530,000.00 $513,135.40 $493,864.60 $474,593.80 $455,323.00 $436,052.20 $416,781.40 $412,766.65

3.182% 3.636% 3.636% 3.636% 3.636% 3.636% 0.758%

EOY

0 1 2 3 4 5 6 7

56. (a) MACRS-GDS(39) (b)

(c)

d15 d45 d75

$15,996.50 $16,666.00 $16,666.00

B15 B45 B75

$634,003.50 $584,005.50 $534,007.50

d15 d45 d75

$15,996.50 $16,666.00 $9,027.42

B15 B45 B75

$634,003.50 $584,005.50 $541,646.08

Chapter 9 FE2. b

FE4. b

FE6. b

FE8. b

FE10. b

2. (a) Gross Income—A company’s income before taking deductions or taxes into account (b) Expenses—Decreases in economic benefits during the accounting period (c) Depreciation—method of attributing cost of asset across its useful life (d) Interest—The “rent” paid to borrow money (e) Property Disposition—the offering or selling of property for tender 6. (a) $3,750 (b) $12,500 (c) $20,550 (g) $5,910,000 (h) $8,750,000

(d) $61,250

(e) $340,000

(f) $4,100,000

8. (a) $1,800.00; 15.00%; 15.00% (b) $11,250.00; 17.31%; 25.00% (c) $69,050.00; 31.39%; 39.00% (d) $340,000.00; 34.00%; 34.00% (e) $6,754,999.99; 35.00%; 35.00% 10. (a) $100,001.00

(b) $335,000.00

(c) $18,333,333.33

12. (a) 15.00%

(b) 31.25%

(c) 35.05%

(d) 39.00%

14. (a) 35.00%

(b) 35.00%

(c) 35.00%

(d) 35.00%

16. (a) $70,000.00

(b) $18,850.00

18. (a) $7,325,000.00

(b) $4,210,000.00

20. (a) 2$30,000.00, $12,399.60, $13,734.00, $10,177.20, $9,289.20, $8,400.00 (b) 2$30,000.00, $11,400.00, $11,400.00, $1,400.00, $11,400.00, $10,400.00 22. (a) (b) (c)

M(5)

ATCF(2)

ATCF(7)

PW

AW

IRR

ERR

$300,000.00 $419,200.00 $345,000.00

$300,000.00 $240,000.00 $259,667.97

$293,779.34 $369,389.01 $343,590.85

$55,067.18 $69,239.76 $64,404.05

15.24% 17.34% 16.61%

12.65% 13.27% 13.06%

590

Answers to Selected Even Numbered Problems

24. (a)

ATCF(2)

ATCF(8)

PW

AW

IRR

ERR

$57,000.00

$75,000.00

$50,425.36

$10,150.77

17.63%

14.70%

(b) PW here is $50,425.36 versus only $8,520.80 where MACRS-GDS(5) is used. Expensing the entire investment in year 0 is far more attractive than depreciating it over years 1–6. 26. (a) (b) (c)

28. (a)

M(7)

ATCF(2)

ATCF(4)

PW

AW

IRR

ERR

$225,000.00 $312,246.00 $255,000.00

$1,125,000.00 $956,142.00 $1,032,500.00

$16,519.66 $38,021.09 $31,038.34

$5,099.10 $11,735.92 $9,580.56

9.42% 10.03% 9.83%

9.33% 9.76% 9.62%

ATCF(2)

ATCF(12)

PW

AW

IRR

ERR

$27,000.00

$30,000.00

$70,925.57

$13,294.57

21.44%

14.53%

(b) Both investments are desirable. The metrics for the design portfolio are substantially better, due to the expensing of the portfolio vs. depreciation of the router over 8 years. 30. (a)

ATCF(2)

ATCF(4)

PW

AW

IRR

ERR

$22,200.00

$25,800.00

$18,829.86

$5,940.27

24.90%

18.54%

(b) Both investments are desirable. The metrics for the 100-day short-term project are substantially better, due to the expensing of the project vs. depreciation of the tractor over 4 years. 32. PW 5 $224,131.60, AW 5 $56,135.21, IRR 5 16.267%, ERR 5 13.469% 34. PW 5 2$116,270.10, AW 5 2$29,120.60, IRR 5 0.692%, ERR 5 2.492% 36. (a) and (b) Select the virtual mold apparatus costing $87,500 38. (a) Select investment a

(b) $104,039.51

40. $22,476.43 42. When we use BTCF, that means we have not yet deducted taxes from the cash flow. (Also, there is no cash flow due to a loan considered.) When we use BTLCF, that means we have not yet deducted taxes, loan interest, or loan principal from the cash flow. In both cases, BTCF or BTLCF is considered gross income less expense deductions, with no consideration of either taxes or loans. 44. (a) $148,020.00

(b) $59,208.00

(c) $260,292.00

46. (a) $183,408.00

(b) $6,000.00

(c) $77,592.00

48. (a) $178,188.73 (b) $190,620.64 (c) Loan payment plan 4 is preferred when the lender’s interest rate is less than the borrower’s MARRAT /(1 2 itr). Here, the lender’s interest rate is 12% and the borrower’s MARRAT /(1 2 itr) is 16.67%. Therefore, plan 4 is preferred to plan 2. 50. (a) (b) (c) (d)

PW

FW

AW

IRR

ERR

$470,030.09 $451,256.92 $453,286.63 $464,294.97

$936,564.39 $899,157.68 $903,202.00 $925,136.80

$84,922.39 $81,530.56 $81,897.28 $83,886.20

29.39% 24.04% 24.54% 31.54%

16.11% 16.43% 16.45% 15.73%

Answers to Selected Even Numbered Problems

52. (a) $220,209.06 54. (a) (b) (c) (d) (e) (f)

(b) $215,428.05

(c) $216,512.27

(d) $216,872.68

PW

AW

IRR

ERR

$37,475.30 2$38,519.14 2$25,820.80 $105,148.18 $29,153.73 $41,852.08

$5,839.40 2$6,002.06 2$4,023.40 $16,384.20 $4,542.74 $6,521.40

9.50% 8.64% 8.74% 10.52% 9.29% 9.44%

9.25% 8.72% 8.81% 9.69% 9.21% 9.30%

56. (a) MARRS(7) (b) 12.00% (c) $5,000.00 (d) 15.00% TI(3) 5 2$16,811.82; ATCF(3) 5 $33,921.66 plus more 58. (a) Select the virtual mold apparatus costing $87,500 costing $87,500 60. (a) Select investment A costing $70,000 to be equivalent

(e) Interest payment (3) 5 2$1,713.82;

(b) Select the virtual mold apparatus

(b) The cost of investment B must be $104,498.15 for the two

62. $19,431.19 64. (a) 2$109,854.04 (MACRS_GDS(39)) (d) $158,764.29

(b) 2$90,480.00

(c) 2$97,221.43

Chapter 10 FE2. c

FE4. d

FE6. c

FE8. d

2. Gasoline, copper, mechanic labor, tuition 4. Homes and land 6. The Consumer Price Index (CPI) Measures price changes from the perspective of the Purchaser The Producers Price Index (PPI) Measures price changes from the perspective of the Seller 8. Not irrelevant. CPI does not describe any individual’s purchasing habits. Also, the economy is very “connected.” 10. (a) 101.6,

104.0,

106.0,

109.5,

113.2

(b) 2.51744%

12. (a) 141.3,

160.7,

176.9,

210.3,

249.8

(b) 15.31554%

14. 2.1155% 16. 18.5396% 18. $262,268.73 20. (a) 3.18%

(b) $19,005.50

22. (a) $35,236.17

(c) $666.01

(b) $242,628.93

24. (a) $944,160.00; $1,273,483.01; $1,717,673.88; $2,316,798.53; $3,124,897.86 (b) $840,000.00; $1,008,000.00; $1,209,600.00; $1,451,520.00; $1,741,824.00 (c) $3,879,152.31 (d) $3,879,152.31 26. (a) $17,599,971.78

(b) $67,938.73

28. (a) Defer the payment. It is worth $1,085,748.75

(b) 5.24%

(c) Take it now and run.

591

592

Answers to Selected Even Numbered Problems

30. EOY

1 2 3 4

THENCURRENT

CONSTANT WORTH

$10,000.00 $12,000.00 $15,000.00 $17,500.00

$10,000.00 $11,428.57 $13,605.44 $15,117.16

32. 11.24% 34. (a) $52,347.07

(b) $52,347.07

36. (a) $111,292.23

(b) $111,292.23

38. 7.97%; 3.618% 40. $3,454,391,189 42. False 44. $1,316.60 46. (a) 2$9,000,000.00; $4,292,307.69; $4,065,088.76; $3,014,474.28; $2,154,504.39; $1,540,869.61; $1,183,879.62 (b) $1,615,078.51 (c) $461,767.31 (d) $4,359,991.90 (e) 26.68% (f) 21.29% (g) 31.74% (h) 26.14% 48. (a) 2$190,000.00; $37,452.74; $44,241.35; $38,851.05; $35,145.43; $32,605.15; $32,388.71; $32,192.26; $29,495.88; $27,000.00; $27,000.00; $27,000.00; $30,000.00 (b) $43,495.36 (c) $6,383.52 (d) $136,507.08 (e) 14.92% (f) 11.91% (g) 19.40% (h) 16.27% 50. (a) 2$1,400,000.00; $347,485.60; $405,045.07; $335,035.55; $294,722.97; $292,517.24; $265,200.21; $240,000.00; $360,000.00 (b) $326,090.32 (c) $61,123.68 (d) $699,003.57 (e) 16.49% (f) 12.92% (g) 21.38% (h) 17.66% 52. (a) 2$190,000.00; $37,860.40; $45,612.40; $40,292.40; $36,492.40; $33,786.80; $33,779.20; $33,786.80; $30,389.60; $27,000.00; $27,000.00; $27,000.00; $30,000.00 (b) $7,483.92 (c) $1,358.31 (d) $38,481.69 (e) 15.61% (f) 14.99% (g) 10.95% (h) 10.35% 54. (a) 2$600,000.00; $197,662.14; $220,556.13; $152,527.80; $135,800.85; $240,000.00 (b) $118,341.96 (c) 17.529% 56. (a) 2$800,000.00; $135,952.38; $182,879.82; $158,074.18; $150,328.10; $163,883.91; $163,754.24; $225,000.00 (b) $5,827.60 (c) 10.210% 58. (a) 2$450,000.00; $92,307.69; $136,464.50; $137,403.28; $145,945.85; $200,063.63 (b) $104,976.05 (c) 15.562% 60. 4; 4; 4; 4; all same; 2; 2; 2; 2; 2; 2 62. 4; 4; 4; 4; 4; 4; 4; 4; 4; 4; all same; 2; 2; 2 64. (a) 2$5,000,000.00; $4,015,384.62; $3,798,816.57; $2,758,443.33; $1,908,320.79; $1,304,154.60; 2$2,204,989.07 (b) $3,547,544.14 (c) $1,014,278.80 (d) $9,576,787.51 (e) 59.09% (f) 29.03% (g) 65.45% (h) 34.19% 66. (a) 2$90,000.00; $29,945.52; $37,015.92; $31,896.84; $28,452.25; $26,163.21; 2$53,300.36; $32,192.26; $29,495.88; $27,000.00; $27,000.00; $27,000.00; $30,000.00 (b) $68,533.67 (c) $10,058.23 (d) $215,088.00 (e) 26.13% (f) 14.77% (g) 31.05% (h) 19.25%

Answers to Selected Even Numbered Problems

68.

ATCF(2)

(a) (b) (c) (d)

(a) (b) (c) (d)

$4,108,800.00 $3,490,133.33 $3,592,237.79 $4,396.800.00

ATCF(6)

PW

FW

2$2,790,014.60 $783,318.73 $566,778.36 2$4,839,189.05

$3,547,544.14 $2,929,877.79 $3,021,558.55 $3,759,823.63

$12,117,691.37 $10,007,868.38 $10,321,031.27 $12,842,795.08

AW

IRRc

ERRc

IRRr

ERRr

$1,139,640.14 $941,216.29 $970,668.52 $1,207,834.43

65.45% 52.96% 54.95% 71.45%

34.19% 32.53% 32.78% 34.74%

59.09% 47.08% 48.99% 64.86%

29.03% 27.43% 27.67% 29.56%

(e) Method 4 70. 2$37,784.17

Chapter 11 FE2. a

FE4. c

FE6. b

FE8. c

FE10. b

2. a 5 3, b 5 1, c 5 2 4. (a) False

(b) True

6. (a) 6137

(b) Plan 1

8. (a) $67,584.00

(c) False

(b) 307,692 Units

(c) $132,480.00

10. $5,170.15 12. 426 Holes per Year 14. (a) 2$500,000.00

(b) 58.33%

16. (a) $25,000.00 annual loss (d) 132,000 pallets per year

(b) 220,000 pallets per year

(c) $125,000.00 annual profit

18. 7,883.24 hours 20. (a) 25.46%

(b) 41.37%

(c) 50.42 %

(d) 60.03%

22. between 3 and 4 years 24. Choose Motor X for more than 3176 hours. Choose Motor Y for 2823 to 3176 hours. Choose Motor Z for less than 2823 hours 26. $971.33 28. (a) AW 5 $7,120.24 Investment is attractive (b)

% CHANGE

280.00% 260.00% 240.00% 220.00% 0.00% 20.00% 40.00%

AW

2$3,279.76 2$679.76 $1,920.24 $4,520.24 $7,120.24 $9,720.24 $12,320.24

(c) 254.8% 34. (a) Not Attractive

(b) Attractive

(c) Attractive

(d) Attractive

593

594

Answers to Selected Even Numbered Problems

38. 0.55 40. 0.50 42. 0.682 44. 0.953 46. 0.281 48. $4,182.73 50. (a) 2$1,031,532.00

$835,870.00

(b) 0.109 or 10.9%

52. (a) 0.229 or 22.9 %

(b) 0.232 or 23.2 %

54. (a) 0.437 or 43.7%

(b) 0.437 or 43.7%

56. (a) $825,709 $649,192

(b) 0.102 or 10.2%

(c) 2$1,032,209

(c) 0.394 or 39.4%

$827,269

0.107 or 10.7%

(d) 0.104 or 10.4%

Chapter 12 FE2. a

FE4. c

2. (a) 2, 4, 5 (g) 2, 4, 5 (h) 2, 4, 5

FE8. c

FE10. c

(b) $43,617.79 (c) 16.94% (d) 2, 4, 5 (e) $43,617.79 $43,617.79 2, 3, 4, 5 $49,000.00 2, 3, 4, 5 $33,507.05 $43,617.79 2, 4, 5 $21,150.07 2, 4, 5 $68,860.59

4. (a) 1, 2, 3, 4 (g) 1, 2, 3, 4 (h) 1, 2, 3, 4 6. (a) 1, 3, 4 (g) 1, 3, 4 (h) 1, 3, 4 8. (a) 2, 4 (g) 2, 4 (h) 2, 4

FE6. d

(b) $113,550.45 (c) 24.60% (d) 1, 2, 4, 5 (e) $111,387.58 $113,550.45 2, 3, 4, 5 $130,492.90 2, 3, 5 $94,445.14 $113,550.45 1, 2, 3, 5 $86,844.50 1, 2, 3, 4 $143,619.14 (b) $90,978.88 (c) 12.00% (d) 1, 2, 3 (e) $79,606.52 $90,978.88 1, 2, 3, 4 $113,723.60 1, 2, 3 $79,606.52 $90,978.88 1, 3 $10,814.33 1, 3, 4 $191,650.08

(b) $530,720.51 (c) 21.99% $530,720.51 1, 2, 4 $741,563.04 $530,720.51 2, 4 $411,546.84

10. (a) 2, 4, 5, 6

(f) 16.94%

(b) $9,950.82

(d) Optimum Portfolio

(d) 2, 5 (e) $522,889.02 1, 4 $507,829.17 2, 5 $668,318.72

(c) 11.42% 1, 2, 3, 4, 6 $7,676.34 11.13%

Present Worth Portfolio IRR

(e) Optimum Portfolio

2, 4, 5, 6 $9,950.82 11.42%

Present Worth Portfolio IRR

(f ) Optimum Portfolio

2, 4, 5, 6 $9,950.82 11.42%

Present Worth Portfolio IRR

(g) Optimum Portfolio

1, 2, 4, 6 $7,486.80 11.32%

Present Worth Portfolio IRR

(h) 2, 4, 5, 6

$9,950.82

1, 2, 3, 4, 5, 6

(i) 2, 4, 5, 6

$9,950.82

None

$0.00

$10,519.43 2, 4, 5, 6

4, 5, 6 $25,253.89

$9,003.12

(f) 25.43%

(f) 12.33%

(f) 20.75%

Answers to Selected Even Numbered Problems

12. $11,372.36 12.00% 14. (a) Invest in M, O, P, R (b) Invest in M, O, P, Q, R 16. (a) H11 (d)

(e)

(b) B9:G9

(c) B9:G9 5 Binary H10 ,5 H12

DETERMINE X4 1 X5 THEN, SET X4 1 X5 ,5 1 SET THE CONSTRAINT X6 ,5 X5

(f ) DETERMINE THE SUM OF THE X’S SET THIS SUM 5 3

18. True 20. (a) 3, 5, 2, 4(60%) (b) CAP CONSTRAINT

(c) MARR

12.00% 15.00% 18.00%

(d) 3, 5, 2, 4(60%)

$1,000,000.00 $1,250,000.00 $1,500,000.00 3, 5, 2, 4(60%) 3, 5, 2, 4(60%) 3, 5, 2, 4(60%)

PORTFOLIO

PW

3, 5, 2, 4(10%) 3, 5, 2, 4(60%) 3, 5, 2, 4, 1(14.29%)

$522,258.61 $626,331.89 $729,864.52

$834,615.69 $626,331.89 $447,693.13

$626,331.89

22. (a) 2, 1, 4(43.75%) (b) CAP CONSTRAINT

(c) MARR

(d)

9.60% 12.00% 14.40%

$280,000.00 $350,000.00 $420,000.00 2, 1, 4(43.75%) 2, 1, 4(43.75%) 2, 1, 4(43.75%)

PORTFOLIO

PW

2, 1 2, 1, 4(43.75%) 2, 1, 4(87.5%)

$56,590.58 $68,171.11 $79,751.65

$115,128.65 $68,171.11 $27,695.48

Optimal Portfolio Percent Total PW Total Investment

2 100% $66,630.85 $350,000.00

4 100%

24. (a) Optimal Portfolio

5 100% $20,883.82 $250,000.00

4 100%

Percent Total PW Total Investment

1 40%

2 100.00%

3 17.65%

595

596

Answers to Selected Even Numbered Problems

(b)

(c)

(d)

26. (a)

CAP CONSTRAINT

$200,000.00 $250,000.00 $300,000.00

MARR

5, 4, 2, 3(17.65%) 5, 4, 2, 3(17.65%) 5

8.00% 10.00% 12.00%

Optimal Portfolio Percent Total PW Total Investment Optimal Portfolio Percent Total PW Total Investment

5 100% $20,883.82 $250,000.00 6 100% $1,058,608.60 $2,500,000.00

PORTFOLIO

PW

5, 4, 2(41.67%) 5, 4, 2, 3(17.65%) 5, 4, 2, 3(76.47%)

$18,775.04 $20,883.82 $21,816.13

$31,588.24 $20,883.82 $6,074.70 4 100%

1 100%

(b) CAP CONSTRAINT

(c) MARR

8.00% 10.00% 12.00%

(d) Optimal Portfolio Percent Total PW Total Investment

(e) Optimal Portfolio Percent Total PW Total Investment

(f ) Optimal Portfolio Percent Total PW Total Investment

$2,000,000.00 $2,500,000.00 $3,000,000.00

6 100% $1,058,608.60 $2,500,000.00 6 100% $1,028,578.56 $2,500,000.00

7 100%

3 17.65%

9 100%

4 100%

5 100%

PORTFOLIO

PW

6, 1, 7, 9, 4, 5(42.86%) 6, 1, 7, 9, 4, 5, 10(50%) 6, 1, 7, 9, 4, 5, 10, 3(88.89%)

$884,659.61 $1,058,608.60 $1,210,595.46

6, 1, 7, 9, 4, 5, 10(50%) 6, 1, 7, 9, 4, 5, 10(50%) 6, 1, 7, 9, 4, 5, 10(50%) 6 100% $1,043,918.42 $2,500,000.00

2 100%

1 100%

1 100%

1 100%

10 50%

$1,400,672.13 $1,058,608.60 $761,080.10 7 100%

7 100%

7 100%

9 100%

4 100%

10 100%

5 14.29%

9 100%

4 100%

5 100%

10 50%

9 100%

4 100%

2 100%

10 75%

Answers to Selected Even Numbered Problems

28. (a) Optimal Portfolio

4 100% $623,633.06 $500,000.00 $175,000.00

Percent Total PW Total Investment

1 100%

(c) MARR

$400,000.00 $500,000.00 $600,000.00

8.00% 10.00% 12.00%

PORTFOLIO

PW

1, 2, 4(90%) 4, 1, 2, 3(37.5%) 4, 1, 2, 3(87.5%)

$531,518.25 $623,633.06 $714,431.01

4, 1, 2, 3(37.5%) 4, 1, 2, 3(37.5%) 4, 1, 2, 3(37.5%)

(d) Optimal Portfolio

2 100% $581,080.90 $500,000.00 $125,000.00

Percent Total PW Total Investment

3 37.5%

YEAR 0 YEAR 1

(b) CAP CONSTRAINT

2 100%

$752,584.83 $623,633.06 $511,830.60 4 100%

1 50%

3 50%

YEAR 0 YEAR 1

30. (a) Invest in 100% of M, N, O, and R; Invest in 25% of P (b) Invest fully in N, P, and R (c) Invest fully (100%) in M, N, O, and R. Invest partially (26.32%) in Q

Appendix B FE2. c 2. (a) 2 (k) 8

FE4. d (b) 3 (l) 1

FE6. c (c) 11 (m) 5

FE8. b (d) 13

FE10. a (e) 6

(f) 12

(g) 4

(h) 7

(i) 9

( j) 10

4. for X ,5 7.5, prefer Method II; otherwise, prefer Method I 6. (a) 100 units

(b) $30,000

8. (a) 2$500,000 (loss)

(c) $50,000

(b) approximately 583,333 tools

(c) $15/tool

10. (a) graph required; for Order ,5 50, use M1, 50 , order ,5 125, use M3, order . 125 use M2 (b) use M3 for $875 (c) M2 is preferred, penalty for using M3 is $70, penalty for using M1 is $510 12. (a) $180,000

(b) $600/ton

14. (a) 63.63 (b) 0.279 (c) weight (d) cost (e) $231.03 (f ) 98.24% of the changes in cost can be explained by changes in weight 16. (a) graphs required, regression results for labor hours y 5 5.96x 1 9466:9, r 2 5 0.0059; for boards completed y 5 13.95x 1 21811.0, r 2 5 0.222; for average cycle time y 5 330.5x 1 6572.8, r 2 5 85.41 (b) average cycle time (c) 85.41% 18. (a) Assets: cash, accounts receivable, raw materials inventory, work-in-process inventory, finished goods inventory, land, net buildings, net equipment; Liabilities: notes payable, accounts payable, declared dividends payable; Net Worth: common stock, retained earnings (b) $318,000

597

598

Answers to Selected Even Numbered Problems

20. Net Income 5 $85,000 22. Balance Sheet balancing total 5 $74,700; Net Income 5 $1,500 24. (a) $2.43B

(b) $60.75B

(c) $12.79

(d) 69.43B

26. (a) many sites provide valuable information; among them are www.investopedia.com, www.fool.com, www.business.com (b) investopedia.com (c) investopedia.com has a well organized and easy to navigate tutorial on financial ratios, (d) there are many including cash conversion cycle, return on capital employed, interest coverage ratio, cash flow to debt ratio, fixed asset turn-over, price to sales ratio. 28. $4,062.50 30. (a) made

(b) units should be made unless overhead rate exceeds $10.8333 per hour

32. (a) $189,374

(b) $7,430.40

34. Net Income 5 $160,000 36. (a) 42.86%

(b) 33.80%

(c) $1.7143/hr

(d) $214

38. (a) many sites provide valuable information, among them are www.investopedia.com, www.valuebasedmanagement.net, www.valuationresources.com (b) investopedia.com (c) investopedia.com has a well organized and easy to navigate tutorial on economic value added 40. (a) Lower 5 $201.04/unit, Upper 5 $367.92/unit 42. (a) PS 5 $300/unit, VI 5 $325/unit

(b) Lower 5 $166.08/unit, Upper 5 $437.85/unit

(b) PS 5 $217/unit, VI 5 $450/unit

INDEX A

Abbott, Wallace C., 436 Abbott Laboratories, 436–437 ABC. See Activity-based costing (ABC) Abundance of riches, 439 Accounting, 537–550. See also Cost accounting balance sheet, 538–542 income statement, 542–545 ratio analysis, 545–550 Activity-based costing (ABC), 556–557 Adjusted basis. See Book value Adjusted cost basis. See Book value ADS. See Alternative Depreciation System (ADS) After-tax analysis before-tax vs., 128–129 borrowed capital, 330–336 inflation, 367–373 leasing vs. purchasing equipment, 336–338 retained earnings (no borrowing), 322–330 After-tax cash flow (ATCF), 322. See also After-tax analysis Alternative Depreciation System (ADS) defined, 297 MACRS-ADS, 301 Amortization, 289 Analogy, cost estimation by, 535–536 Annual percentage rate (APR), 111–112 additive approach, 112 defined, 112 subtractive approach, 112 Annual worth (AW) defined, 180 multiple alternatives, 183–184 single alternative, 180–182 APR. See Annual percentage rate (APR)

Association for the Advancement of Cost Engineering International, 13, 530–531 Autocorrelated, 407 Average cost, 524–530 AW. See Annual worth (AW)

Brown, Greg, 212 BTCF. See Before-tax cash flow (BTCF) BT&LCF. See Before-tax-and-loan cash flow (BT&LCF) C

B

Balance sheet, 538–542 B/C. See Benefit-cost ratio (B/C) Before-tax-and-loan cash flow (BT&LCF), 322. See also After-tax analysis Before-tax cash flow (BTCF), 322. See also After-tax analysis Before-tax vs. after-tax analysis, 128–129 Benefit-cost analysis, 137–143 multiple alternatives, 139–143 single alternative, 137–139 Benefit-cost ratio (B/C), 138 Best, Daniel, 358 Binary linear programming (BLP), 441 BLP. See Binary linear programming (BLP) Bond(s), 106–109, 514 defined, 106 face/par value, 106 maturity, 106 purchase price for, 108 rate of return for, 109 redeem, 106 selling price for, 107–108 Bond rate, 106–107 Book value, 290, 510 Borrowed capital, after-tax analysis, 330–336 Bottom-up estimating, 536 BP, 390–391 Break-even analysis, 393–398, 520–524 Break-even chart, 395 Break-even point, 395 Break-even value, 393 Brown, Donald, 9

CAP EX, 439 Capital budgeting problem defined, 439 divisible investments, 446–449 indivisible investment, 440–446 practical considerations, 449–450 Capitalized worth (CW), 152–157 defined, 152 multiple alternatives, 156–157 single alternative, 153–156 Capital rationing problem. See Capital budgeting problem Capital recovery cost, 265 Cash flow, 3 beginning-of-period, 26 end-of-period, 26 estimation. See Cost estimation future worth calculations (F Z P), 26–33 geometric series, 51–55 gradient series, 46–50 irregular, 35–39 patterns, 26 present worth calculations (P Z F), 33–35 uniform series, 39–46 Cash flow diagrams (CFD), 24–26 as communication tools, 16 defined, 24 Caterpillar, 358 CFD. See Cash flow diagrams (CFD) Challengers, 259 Coca-Cola Company, 558 Combined rate, 365 Comparing economic alternatives, 126–130. See also Economic worth 599

600 Index

Compounding, 26 Compounding and cash flow frequencies, 61–62 Compound interest, 26, 27 ConocoPhillips, 124–125 Constant dollars, 364 Consumer price index (CPI), 361–364 Contributed capital, 542 Corporate income tax, 319–321 computing, 319–320 effective tax rate, 320–321 incremental tax rate, 320–321 marginal tax rate, 320–321 Cost accounting, 550–561 activity-based costing (ABC), 556–557 economic value added (EVA), 558–561 overhead costs, 551–556 standard costs, 557–558 Cost basis, 289. See also Book value Cost estimation, 13, 530–537 methodologies, 535–536 project estimation, 532–535 sources of data, 536–537 Cost of capital, 513 Cost of goods sold, 515–516 Costs average, 524–530 fixed, 517–524 future, 512 manufacturing cost structure, 514–517 marginal, 524–530 operating and maintenance, 509–510 opportunity, 512–513 past, 511 sunk. See Sunk costs variable, 517–524 Cost viewpoints, 508–512 average and marginal viewpoint, 524–530 fixed and variable viewpoint, 517–524 manufacturing cost structure viewpoint, 514–517 past/future, 510–512 CPI. See Consumer price index (CPI) Cumulative depreciation charge, 290. See also Book value

D

DB. See Declining balance (DB) depreciation DCF. See Discounted cash flows (DCF) DDB. See Double declining balance (DDB) depreciation Debenture bonds, 514 Declining balance (DB) depreciation, 291–294 Defender, 259 Deferred payment loans, 104–106. See also Loan(s) defined, 104 interest and equity in, 104–106 Depreciable property, 288–289 intangible, 289 MACRS, 289, 296–301 tangible, 289 Depreciation DDB, 292–294 declining balance (DB), 291–294 defined, 286 depreciation allowances, 288 economic analysis, 287–288 MACRS, 289, 296–301 overview, 286–287 property. See Depreciable property SLN, 289–291 Depreciation allowances, 288 Descartes’ rule of signs, 217. See also Internal rate of return (IRR) Discounted cash flows (DCF), 6–7 defined, 6 rules, 6–7 Discounted payback period (DPBP) defined, 144 multiple alternatives, 149–152 single alternative, 144–149 Divisible investments, 446–449 Double declining balance (DDB) depreciation, 292–294 DPBP. See Discounted payback period (DPBP) E

Earned capital, 542 Earnings before interest, taxes, depreciation and amortization (EBITDA), 550

Earnings before interest and taxes (EBIT), 549–550 EBIT. See Earnings before interest and taxes (EBIT) EBITDA. See Earnings before interest, taxes, depreciation and amortization (EBITDA) Economic obsolescence, 259 Economic value added (EVA), 558–561 Economic worth before-tax vs. after-tax analysis, 128–129 comparing, methods of, 127 equivalence of methods, 128 incremental methods, 128 ranking methods, 128 single alternative. See Single alternative Economies of scope, 525–526 EDI. See Electronic data interchange (EDI) Effective annual interest rate, 58–61 Effective tax rate computing, 320–321 defined, 320 Electronic data interchange (EDI), 2 Engineering economic analysis defined, 3 principles, 7–10 Engineering economy, 3 Equal vs. unequal lives, 129 Equipment, leasing vs. purchasing, 336–338 Equity payments, 101 deferred payment loans, 104–106 Equivalence, 94–100 defined, 94 determining gradient step, 96–97 determining interest rate, 98–100 uniform series equivalency of decreasing gradient series, 95–96 Equivalent uniform annual cost (EUAC), 262 ERR. See External rate of return (ERR) EUAC. See Equivalent uniform annual cost (EUAC) EVA. See Economic value added (EVA)

Index

Excel after-tax effects of borrowed funds, 332–336 after-tax impact of inflation, 367–369, 370–373 break-even analysis, 396–398 MARR, 182–183 Monte Carlo simulation, 413–418 Excel® DDB, 292–294 Excel® EFFECT, 58–59 Excel® GOAL SEEK, 31–33 DPBP, 147 equivalent gradient step, 97 equivalent interest rate, 100 Excel® IRR function, 333, 334, 370, 415 Excel® MIRR function, 333, 334, 370 Excel® NORMSDIST, 408, 412 Excel® NPER, 31, 398 DPBP, 145–147, 151 Excel® NPV, 38–39, 99, 416 Excel® PMT, 397 Excel® RATE, 398 Excel® SOLVER, 31–33 capital budgeting problem, 441–446 DPBP, 147–149, 151 equivalent gradient step, 97 equivalent interest rate, 99–100 MARR, 183 optimum investment portfolio when investments are divisible, 446–448 Excel® VDB, 292, 293–295, 297 External rate of return (ERR) defined, 224 multiple alternatives, 228–230 single alternative, 224–228 F

Face/par value bond, 106. See also Bond Federal tax rate (ftr), 323 Fixed assets, 540–542 Fixed costs, 517–524 Functional obsolescence, 259 Future costs, 512 Future value. See Future worth (FW)

Future worth (FW), 28–29. See also Cash flow calculation (F Z P), 26–33 defined, 185 financial goal, 186–189 geometric series, 54–55 multiple alternatives, 189–190 multiple compounding periods per year, 56–57 portfolio analysis, 192–194 retirement planning, 190–192 series of cash flows, 37–38 single alternative, 185–186 uniform series of cash flows, 44–45 G

GDS. See General Depreciation System (GDS) General Depreciation System (GDS), 297 MACRS-GDS, 297–301 General Motors, 9 Geometric series of cash flows, 51–55 GIGO (garbage-in, garbage-out), 399 Gradient series of cash flows, 46–50 H

Half-year convention, 297 HEPI. See Higher Education Price Index (HEPI) Higher Education Price Index (HEPI), 364 Holt, Benjamin, 358 Hurricane Sandy, 392 I

Immediate payment loans, 101–103. See also Loan(s) Income statement, 542–545 Income taxes after-tax analysis. See After-tax analysis corporate, 319–321 depreciation, 288 leasing vs. purchasing equipment, 336–338 overview, 317–318 Income tax rate (itr), 323 Incremental methods, 128

601

Incremental tax rate computing, 320–321 defined, 320 Indivisible investment, 440–446 Inflation after-tax analysis, 367–373 before-tax analysis, 364–367 CPI, 361–364 defined, 360 HEPI, 364 PPI, 364 Inflation rate, 365 Intangible property, 289 Intel, 316 Interest payments, 101 deferred payment loans, 104–106 immediate payment loan, 102 Interest rate, 26 effective annual, 58–61 equivalence, 98–100 nominal annual, 55–56 period, 56–58 Internal rate of return (IRR), 215–223 defined, 215 Descartes’ rule of signs, 217 multiple alternatives, 220–223 multiple roots, 217–219 mutually exclusive alternatives, 220–222 Norstrom’s criterion, 218 single alternative, 215–217 Investment(s) divisible, 446–449 economically justified, 8 indivisible, 440–446 multiple. See Multiple alternatives mutually exclusive, 8 single. See Single alternative Irregular cash flow, 35–39 J

J. B. Hunt Transport Services Inc. (JBHT), 256–257 JBHT. See J. B. Hunt Transport Services Inc. (JBHT) L

LCF. See Loan cash flow (LCF) Leasing vs. purchasing equipment, 336–338

602

Index

Life Cycle Cost Estimating Handbook, 533 Lives, equal vs. unequal, 129 Loan(s) deferred payment, 104–106 immediate payment, 101–103 principal payments, 100–101 Loan cash flow (LCF), 331 M

MACRS. See Modified Accelerated Cost Recovery System (MACRS) Marginal cost, 8–9 Marginal costs, 524–530 Marginal revenue, 8–9 Marginal tax rate computing, 320–321 defined, 320 Market basket rate, 361. See also Consumer price index (CPI) MARR. See Minimum attractive rate of return (MARR) Maturity, bond, 106. See also Bond Microsoft’s VBA (Visual Basic for Applications) software, 413 Minimum attractive rate of return (MARR), 13–14 Modified Accelerated Cost Recovery System–Alternative Depreciation System (MACRS-ADS), 301 Modified Accelerated Cost Recovery System–General Depreciation System (MACRS-GDS), 297–301 Modified Accelerated Cost Recovery System (MACRS), 289, 296–301 defined, 296 MACRS-ADS, 301 MACRS-GDS, 297–301 Monte Carlo simulation, 413 Excel, 413–418 @RISK software, 413–418 Motorola Solutions, Inc., 212–213 Multiple alternatives annual worth (AW), 183–184 benefit-cost analysis, 139–143 capitalized worth (CW), 156–157

discounted payback period (DPBP), 149–152 external rate of return (ERR), 228–230 future worth (FW), 189–190 internal rate of return (IRR), 220–223 present worth (PW), 132–135 sensitivity analyses, 401–403 N

Net present value (NPV). See Present worth (PW) Nominal annual interest rate, 55–56 Norstrom’s criterion, 218. See also Internal rate of return (IRR) O

Oberhelman, Douglas R., 358 Obsolescence defined, 259 economic, 259 functional, 259 technological, 259 One-shot investments defined, 135 present worth (PW), 135–136 Operating and maintenance costs, 509–510 Opportunity cost, 261 Opportunity cost of money, 6 Opportunity costs, 512–513 Optimum replacement interval (ORI), 264–268 computing, 266–267 planning horizon, 267–268 ORI. See Optimum replacement interval (ORI) Overhead costs, 516–517, 551–556 P

Pallisade Corporation, 413 Parametric estimation, 535 Payoff, 102 Period interest rate, 56–58 Personal property, 289 MACRS-GDS, 298 Planning horizon, 11–12, 129 approaches used to determine, 12 defined, 11

Portfolio analysis, 192–194 future worth (FW), 192–194 PPI. See Producer price index (PPI) Present value. See Present worth (PW) Present worth (PW). See also Cash flow calculation, 33–35 defined, 26, 130 delayed uniform series of cash flows, 41–42 geometric series, 52–53 gradient series, 49–50 multiple alternatives, 132–135 one-shot investments, 135–136 series of cash flows, 36 single alternative, 130–132 uniform series of cash flows, 40 Principal payments, 100–101. See also Loan(s) Probability distributions, 405–407 Producer price index (PPI), 364 Project estimation, 532–535 Property. See specific properties Purchasing power of money, 361 Purchasing vs. leasing equipment, 336–338 Pure discount rate, 365 R

Radio frequency identifi cation (RFID), 2 Ranking methods, 128 Rate of return. See also Minimum attractive rate of return (MARR) bond, 109 external, 224–230 internal. See Internal rate of return (IRR) Ratio analysis, 545–550 areas of focus, 546 uses, 546 Real interest rate, 365 calculating, 365–366 Real property, 289 MACRS-GDS, 298 Redeem, 106. See also Bond Replacement analysis, 258–261 cash flow approach to, 261, 262–263 obsolescence. See Obsolescence

Index

opportunity cost approach to, 261, 263–264 ORI, 264–268 overview, 258 Risk analysis, 405–407 analytical solutions, 407–412 defined, 405 probability distributions, 405–407 simulation solutions, 412–418 @RISK software, 413–418 S

Salvage value, 261, 510 Schneider National, 284–285 Scrap value, 510 SEAT. See Seven-step systematic economic analysis technique (SEAT) Sensitivity analyses, 398–405 defined, 398 multiparameter, 404–405 multiple alternatives, 401–403 single alternative, 399–401 Seven-step systematic economic analysis technique (SEAT), 10–15 cash flow estimates, 14 comparing alternatives, 14 investment alternatives, 11 minimum attractive rate of return, 13–14 planning horizon, 11–12 selecting preferred investment, 15 supplementary analyses, 14–15 Single alternative, 129–130 annual worth (AW), 180–182 benefit-cost analysis, 137–139 capitalized worth (CW), 153–156

discounted payback period (DPBP), 144–149 external rate of return (ERR), 224–228 future worth (FW), 185–186 internal rate of return (IRR), 215–217 present worth (PW), 130–132 sensitivity analyses, 399–401 Single cash flow future worth calculations (F Z P), 26–33 present worth calculations (P Z F), 33–34 Single sum, future worth factor, 28 Sinking fund factor, 45 SLN. See Straight-line depreciation (SLN) Solvency, 549 Standard costs, 557–558. See also Cost accounting State tax rate (str), 323 Straight-line depreciation (SLN), 289–291 Sunk costs, 259–260, 511–512 Supplementary analyses, 14–15 Svanberg, Carl-Henric, 390

603

concept, 4–5 DCF, 6–7 illustration, 5, 25 inflation, 6 terms used to express, 6 TVOM. See Time value of money (TVOM) U

unequal vs. equal lives, 129 Uniform series of cash flows, 26, 39–45 future worth, 44–45 United States Securities and Exchange Commission, 285 Unrecovered investment. See Book value U.S. Department of Labor’s Bureau of Labor Statistics, 361 V

Variable costs, 517–524 Variable declining balance (VDB), 292–293, 294–295 Variable interest rates, 110–111 cash flow series with, 110–111 VDB. See Variable declining balance (VDB)

T

Tangible property defined, 289 personal, 289 real, 289 Tax. See Income taxes Technological obsolescence, 259 Technology followers, 261 Then-current dollars, 364 Time value of money (TVOM), 4–7, 8. See also Cash flow

W

WACC. See Weighted average cost of capital (WACC) Walmart Stores, Inc. (WMT), 2–3 Weighted average cost of capital (WACC), 13–14 defined, 13 formula, 13 White, Miles D., 436 WMT. See Walmart Stores, Inc. (WMT)

Interest Rate Equations 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.

Period Interest Rate

Effective Annual Interest Rate

Effective Rate per Cash Flow Period

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 year

r ⫽ nominal annual interest rate

r ⫽ nominal annual interest rate

m ⫽ number of compounding/interest periods per year

m ⫽ number of compounding/ interest periods per year k ⫽ number of cash flows per year

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 (MARK), 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:

best estimate of IRR, if needed

nper:

number of compound interest periods

finance_rate:

pv:

net present value

interest rate paid on borrowed investment capital (for our purposes, this is generally zero or left blank)

fv:

net future value

r:

nominal interest rate, usually per year

pmt:

magnitude of uniform series of cash flows

m:

number of compounding periods, usually in a year

type:

0 or omitted denotes end-of-period cash flow 1 denotes beginning-of-period cash flow

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