Strategic Cost Analysis : For Project Managers and Engineers [1 ed.] 9781781830437, 9781906574888

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STRATEGIC COST ANALYSIS For Project Managers and Engineers

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STRATEGIC COST ANALYSIS For Project Managers and Engineers

ROBERT C CREESE PhD, CCE

Professor of Industrial and Management Systems Engineering West Virginia University, USA

M ADITHAN PhD, FIE

Senior Professor of Mechanical Engineering and Director, Academic Staff College VIT University, Vellore Tamil Nadu, India

New Academic Science Limited NEW ACADEMIC SCIENCE

The Control Centre, 11A Little Mount Sion Tunbridge Wells, Kent TN1 1YS, UK www.newacademicscience.co.uk e-mail: [email protected]

Copyright © 2013 by New Academic Science Limited The Control Centre, 11 A Little Mount Sion, Tunbridge Wells, Kent TN1 1YS, UK www.newacademicscience.co.uk • e-mail: [email protected]

ISBN : 978 1 781830 43 7 All rights reserved. No part of this book may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the copyright owner. British Library Cataloguing in Publication Data A Catalogue record for this book is available from the British Library Every effort has been made to make the book error free. However, the author and publisher have no warranty of any kind, expressed or implied, with regard to the documentation contained in this book.

PREFACE

The basic principles and tools for Strategic Cost Analysis should be well understood by project managers and engineers who are responsible for planning and executing large scale engineering, construction and infrastructure projects. Strategic Cost Analysis involves a thorough understanding of cash flows, tax implications, financial statements, interest calculations, money values, inflation, depreciation accounting procedures, return-oninvestment, benefit-cost ratio analysis, risk analysis and overall profitability for the company in planning and executing the projects. All these aspects are lucidly explained through several worked out examples, problems and solutions. Various techniques for evaluating the worth of the projects and to judge the profitability of executing the projects are also presented in this book. The book will be a valuable resource and study material to all project managers, building and construction engineers, infrastructure developers, industry planners, cost accountants and financial executives in successfully executing the projects. Dr. Robert C. Creese Dr. M. Adithan

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ACKNOWLEDGEMENT

The contents of this book have been adopted from the course material developed by Dr. Robert C. Creese, Industrial Engineering Department, West Virginia University, U.S.A. The text was first used for a short course for AACEI (Association for the Advancement of Cost Engineering, International) and has been further developed as a textbook. Dr. Robert C. Creese Dr. M. Adithan

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CONTENTS

Preface Acknowledgement

v vii

Chapter 1 CASH FLOWS, PROFITS AND FINANCIAL STATEMENTS 1.1 1.2 1.3 1.4 1.5 1.6

Introduction Cash Flows versus Profits Cash Flow Diagrams Financial Statements and the Purcell Diagram References Evaluative Questions

1 2 4 5 9 9

Chapter 2 EARNED VALUE MANAGEMENT CONCEPTS 2.1 2.2 2.3 2.4 2.5

Introduction Example Problem Summary References Evaluative Questions

13 15 21 21 22

Chapter 3 INTEREST TYPES AND MONEY VALUES 3.1 3.2

History of Interest and its Significance Definitions and Terms Related to Interest Calculations

25 26

x // Contents //

3.3

Interest Rate Fundamentals and Basic Calculations

30

3.4

Types of Money Value and Interest Rates

34

3.5

Exchange Rate Impact

35

3.6

Mathematical Relationships for Engineering Economic Calculations

36

3.7

Evaluative Questions

38

Chapter 4 EXPRESSIONS USED IN ENGINEERING ECONOMICS—BASIC EXPRESSIONS 4.1

Introduction

39

4.2

Single Cash Flow Expressions

39

4.3

Uniform Payment Expressions

41

4.4

Continuous Expressions and Summary Table

45

4.5

Evaluative Questions

47

Chapter 5 GRADIENT EXPRESSIONS 5.1

Introduction

49

5.2

Uniform Gradient

49

5.3

Geometric Gradient

51

5.4

Escalation Gradient

54

5.5

Relationships between Factor Expressions

56

5.6

Evaluative Questions

58

Chapter 6 DEPRECIATION PARAMETERS AND METHODS 6.1

Introduction

59

6.2

Cash Flows

60

6.3

Depreciation

60

6.4

Methods of Depreciation

66

6.5

References

71

6.6

Evaluative Questions

71

// Contents // xi

Chapter 7 CASH FLOWS, PROFITS, TAXES, DEPRECIATION AND LOANS 7.1

Introduction

73

7.2

Loan Calculations

74

7.3

Examples in Cash Flow

77

7.4

Evaluative Questions

80

Chapter 8 PROJECT EVALUATION TECHNIQUES (BASIC METHODS) 8.1

Introduction

83

8.2

Evaluative Questions

92

Chapter 9 PROJECT EVALUATION TECHNIQUES (ADVANCED METHODS) 9.1 9.2 9.3 9.4 9.5 9.6

Introduction Internal Rate of Return Modified Internal Rate of Return (MIRR) Benefit/Cost Ratio Positive and Negative Project Balances Evaluative Questions

95 95 98 99 103 106

Chapter 10 RISK ANALYSIS 10.1 Introduction 10.2 Sensitivity Analysis 10.3 Optimistic-Pessimistic Analysis 10.4 Probabilistic Methods 10.5 Summary 10.6 References 10.7 Evaluative Questions Evaluative Questions and Answers (Problems and Solutions) Formula Sheets and Tables Index

109 110 113 114 120 121 121 125 161 181

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1

CASH FLOWS, PROFITS

AND FINANCIAL STATEMENTS 1.1 INTRODUCTION Strategic cost management requires an understanding of not only profits and financial statements, but also of engineering economics and the concepts of cash flows and interest. Engineering economics is also known as engineering economy or engineering economic analysis. Products and projects must not only be technically feasible, but also economically viable, especially in the global market place of today. Engineering Economics involves Engineering Economic Analysis. Engineering Economics can be defined as: 1.

The body of knowledge and techniques concerned with the evaluation of the worth of commodities and services relative to their costs and with the methods of estimating inputs.

2.

The application of economic or mathematical analysis and synthesis for taking engineering decisions.

3.

Engineering economics is the science which utilizes quantitative techniques for selecting a preferred alternative from amongst several technically feasible and economically viable alternatives.

4.

Engineering economy is the systematic evaluation of proposed engineering solutions to technical problems on an economic basis.

The field of engineering economics uses mathematical and economic techniques to systematically analyze situations in economic terms to resolve a given situation. It is not restricted only to engineering situations but also involves capital budgeting models, planning horizon models, and

2 // Strategic Cost Analysis for Project Managers and Engineers //

other higher level management concerns. It, therefore, also includes financial planning and analysis, inventory modeling, production planning, scheduling and control, statistical quality control, and any optimization models involving economics. It involves the economics of a product, process, or project over time by analyzing the magnitude and timing of the cash inflows and outflows over time for the production of a product, operation of a process, or completion of a project. The evaluation of engineering alternatives can be performed using different parameters and merits of performance and the basic tools/ techniques used are: 1.

Cash Flows Before Taxes,

2.

Cash Flows After Taxes,

3.

Profitability Analysis,

4.

Present Worth Analysis,

5.

Rate of Return,

6.

Benefit/Cost Analysis and

7.

Payback Period.

1.2 CASH FLOWS versus PROFITS Cash flows and profits are the two most widely used financial measures used to evaluate companies, products and projects. These two financial measures are desirable, but decisions must be made emphasizing one or the other as both are not maximums under the same conditions. This is similar to the conditions in optimization that maximum profits and minimum costs usually do not occur at the same level of production. Cash Flow 1.

Cash Flow is the actual monetary amount (Dollars, Euros, Pounds Sterling, Yen, Won, Rupees, etc.) passing into and out of a business financial venture or project being analyzed. Net Cash Flow

2.

Net Cash Flow is the difference between total cash receipts (inflows) and total cash disbursements (outflows) for a given period of time. Net Cash Flows shows the funds derived from the operation. Net Cash Flow represents the funds available and companies must have adequacy of funds to pay for expenses; otherwise it is necessary to borrow funds.

// Cash Flows, Profits and Financial Statements // 3

Profits 3.

Profits represent the net revenues minus the expenses and companies must be profitable to survive in the long-term. Shortterm periods of losses often occur during the start-up of businesses or during periods of recession, but long-term losses are not sustainable. Thus profits can be represented by: Profits = Net Revenues – Expenses

(1.1)

A basic relationship between cash flows and profits is: Cash Flows = Net Profits + Depreciation

(1.2)

Further, cash flows may take into consideration adjustments, as well. Cash Flows = Net Profits + Depreciation + Adjustments (1.3) (loans, inventory changes, changes in accounts receivable or payable, etc.) Cash flows have also been a better index rather than net profits for gauging a company’s performance as they depict the result of the various operations and activities of the company, without taking into consideration the higher receivables such as sales not yet realized into cash or inventory holdings.

Negative cash flows and Positive cash flows Generally infrastructure and construction projects tend to involve longer credit periods and high inventory levels with the result they suffer from negative operating cash flows. A longer working capital cycle often results in negative cash flows for short periods. For example, when a customer orders a part, the manufacturer has capital invested in the part and has a negative cash flow until he receives payment for the part. Hence, a reasonable amount of positive cash flows from operations always assume significance for two reasons: (i)

first, healthy cash flows can help a company meet its funding requirements internally in a high borrowing cost environment, and

(ii) second, a company’s ability to manage its debtors and inventory levels indicates the efficiency and strength of the business.

4 // Strategic Cost Analysis for Project Managers and Engineers //

With infrastructure and construction projects, Earned Value Management techniques involving cost performance and schedule performance are used and these are presented in detail in a latter Chapter. Cash Flows or Profits?

What should be the focus of economic studies; cash flows or profits? This is an interesting question with respect to the concept of accelerated depreciation. Accelerated depreciation will increase cash flows as one can reduce taxes, but it decreases profits from the increased expense. Thus one would appear to have to make a difficult choice when considering the use of accelerated depreciation; however, if one keeps two sets of books, one using accelerated depreciation to reduce taxes and make reports to the Government and a second set using straight line (nonaccelerated depreciation) to the stockholders to indicate higher earnings, the solution is to focus upon cash flows. The depreciation difference multiplied by the tax rate is under the item of deferred taxes which occurs in the reports to stockholders. Although this practice is legal, one does question the ethics of it. It does, however, make the decision easy as whether to focus upon cash flows or profits and this is a major reason as to why cash flows are the focus of engineering economics and project management. Another advantage of using cash flows is that one can focus only on costs whereas revenue inputs are required in addition to costs in order to focus on profits.

1.3 CASH FLOW DIAGRAMS A cash flow diagram is a plot of Cash Flows versus Time. The abscissa (x-axis) represents time and the ordinate (y-axis) represents the amount of cash flow. Cash flows are assumed to occur at the end of the period. There are three primary types of cash flow diagrams, but the net cash flow diagram is the most common. Consider an investment of $5 which results in a revenue of $3 per time period with an expense of $1 per time period.

// Cash Flows, Profits and Financial Statements // 5

A basic assumption in cash flow analysis is that the cash flows are assumed to occur at the end of the period; payments are assumed to be at the end of the period (instead of the beginning of the period). Adjustments can be to alter the results to consider beginning of period cash flows and this can be done relatively easily, but to avoid confusion only end of period cash flows will be presented.

1.4 FINANCIAL STATEMENTS AND THE PURCELL DIAGRAM The primary financial statements used are the Income Statement and the Balance Sheet. The Purcell diagram integrates these two statements into a cash flow diagram. The Purcell diagram indicates the relationship between the accounting financial statements and the overall cash flows for a company. The items such as long-term and short-term debts are not included to keep the problem easy to visualize and to reduce the complexity. The financial statements generally considered are the income statement and the balance sheet and they are presented in Figure 1.1 and the corresponding Purcell Diagram is shown in Figure 1.2. The income statement summarizes the revenues or sales, the expenses or costs, the taxes, and the profits. The balance sheet indicates the

6 // Strategic Cost Analysis for Project Managers and Engineers //

financial position of the company typically at the end of the accounting period. The basic equation of the balance sheet is that: (1.4)

Assets = Liabilities + Equities

The relationship between cash flows and profits is written as indicated by Eq. 1.3 as: Cash Flows = Net Profits + Depreciation + Adjustments This relationship is reduced in most engineering economic models to Eq. 1.2 as: Cash Flows = Net Profits + Depreciation All money values are in $ (Thousands) Income Statement RCC Corporation 2010 (End of Year) Sales Expenses Cost of Goods Sold Operating Costs Depreciation Profit Before Tax Taxes Net Profit

500 200 140 20

360 140 40 100

All money values are in $ (Thousands)

Fig. 1.1. Income and Financial Statements for RCC Corporation for 2010.

// Cash Flows, Profits and Financial Statements // 7

Fig. 1.2. Purcell Diagram for Financial Flows for RCC Corporation.

This can be derived from: Cash Flows (After Taxes) = Revenues – Costs – Taxes = Revenues – Costs – (Tax Rate)*(Revenues – Costs – Depreciation) = (1 – Tax Rate) * (Revenues – Costs – Depreciation) + Depreciation = Net Profits + Depreciation

(1.5)

8 // Strategic Cost Analysis for Project Managers and Engineers //

For accounting models, the cash flow after taxes expression must include adjustments such as changes in inventory, new equipment purchases, accounts payable, accounts receivable, dividends, stock sales, and other items not included in the profit calculations. The Purcell Diagram gives the cash flows in a much easier format and is recommended for use. The Purcell Diagram includes the information of the Income Statement and the Balance Sheet. The center section of the Purcell Diagram contains the Assets part of the balance sheet and the Accounts Payable and Stockholder items of the Purcell Diagram represent the Liabilities and Equities part of the Balance Sheet. It also includes additional information such as the amount spent on equipment and the total spent via suppliers and the total paid to suppliers which is not readily available on the financial statements. The Purcell Diagram as indicated in Figure 1.2 gives the cash flows in a much easier format and is recommended for use. If one rewrites Equation 1.3 to determine the end of period cash flows, one has: Cash Flows = Net Profits + Depreciation + Adjustments Cash Flows (end) – Cash Flows (start) = Net Profits + Depreciation + Adjustments Or Cash Flows (end) = Cash Flows (start) + Net Profits + Depreciation + Adjustments

(1.6)

From Figure 1.2, one can calculate the ending cash flows as: Cash Flows (end) = Cash Flows (start) + Net Profits 305 = 260 + 100 + Depreciation + Adjustments (–Purchase Equipment + 20 –50 + stock sales – dividends paid – inventory increase + 5 –10 –20 + Accounts Payable – Accounts Receivable) + 20 – 20 = 260 + 100 + 20 + (– 50 +5 – 10 – 20 + 20 – 20) = 260 + 120 (cash flow change) –75 (total adjustments) = 305

// Cash Flows, Profits and Financial Statements // 9

This can also be calculated in another manner by examining the particular activity. The end values need to be calculated and the general equation used is: End Value = Starting Value + Inputs to Activity – Outputs of Activity (1.7) If one examines the cash flow box of the Purcell Diagram, note that: 305 (end) = 260 (start) + 480 (receipts from customers) – 440 (outgoing funds) + 5 (stock sales) = 305

1.5 REFERENCES 1.

Purcell, W.R., Understanding a Company’s Finances – A Graphical Approach, Houghton Mifflin Company, Boston, 1981, pp. 143.

1.6 EVALUATIVE QUESTIONS 1. A company made an investment of $1,200,000 for a machine to manufacture a new product. The sale of the product produced is expected to provide a uniform annual revenue of $500,000 for 6 years. The annual operating, material, and maintenance expenses are $200,000 and the salvage value of the machine at the end of the 6 years is $400,000. Draw the cash flow diagram, the net cash flow diagram and the cumulative cash flow diagram. 2. Explain the difference between cash flows and profits. 3. Given the following data, draw the basic cash flow diagram, the net cash flow diagram and the cumulative cash flow diagram. Money values are in $ (thousands). Period

Investment

0 1 2 3 4 5 6

20

Revenue

Expenses

5 6 10 10 10 10

4 3 4 4 4 6

When is the break-even or payback period?

10 // Strategic Cost Analysis for Project Managers and Engineers //

4. Use the Income Statement and the Balance Sheet in Figure 1.3 to complete the Purcell Diagram for the Financial Flows for MA Corporation in Figure 1.4.

Fig. 1.3. Income Statement and Balance Sheet for MA Corporation for 2010.

// Cash Flows, Profits and Financial Statements // 11

Fig. 1.4. Purcell Diagram for Financial Flows for MA Corporation.

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// Earned Value Management Concepts // 13

2

EARNED VALUE MANAGEMENT CONCEPTS 2.1 INTRODUCTION Earned Value Management (EVM) concept is a management tool used to measure the cost and schedule performance on projects. The projects one typically thinks of are large construction projects such as highways, dams, bridges, etc., but earned value can be applied to any project such as developing new products, research projects, installation projects of new equipment and basically any project that involves a schedule and costs, which includes much of the work being done. It is a powerful tool and provides an early warning of performance problems as well as keeping the project focused upon successful completion. Earned value management tracks two main parameters in project performance, the cost and the timing. The following three elements are used to measure project performances in EVM: 1.

The Planned Value (PV) of the work scheduled, also known as the Budgeted Cost of Work Scheduled (BCWS);

2.

The Earned Value (EV) of the work accomplished, also known as the Budgeted Cost of Work Performed (BCWP); and

3.

The Actual Cost (AC) of the work accomplished, also known as the Actual Cost of Work Performed (ACWP).

The values are traditionally measured in monetary units, such as Dollars, but can also be measured in other units, such as work-hours. In construction projects, work-hours are often used instead of Dollars.

14 // Strategic Cost Analysis for Project Managers and Engineers //

The traditional performance measures derived from these elements are the Schedule Variance (SV), the Cost Variance (CV), the Schedule Performance Index (SPI), and the Cost Performance Index (CPI), which are presented in the following equation forms. The schedule values will be presented first SV = EV – PV

(2.1)

Where SV = Schedule Variance EV = Earned Value PV = Planned Value If the Schedule Variance is positive, the project is ahead of schedule whereas if the Schedule Variance is negative, the project is behind schedule. This same information can be obtained from the Schedule Performance Index where: SPI = EV/PV

(2.2)

Where SPI = Schedule Performance Index EV = Earned Value PV = Planned Value If the Schedule Performance Index is greater than unity (1.0), the project is ahead of schedule and if it is less than unity, the project is behind schedule. The cost performance equations are similar to the schedule performance equations, but focus upon the actual values rather than the scheduled values. Thus the equations are: CV = EV – AC

(2.3)

Where CV = Cost Variance EV = Earned Value AC = Actual Cost If the Cost Variance is positive, the project is under budget whereas if the Cost Variance is negative, the project is over budget. This same information can be obtained from the Cost Performance Index where:

// Earned Value Management Concepts // 15

CPI = EV/AC

(2.4)

Where CPI = Cost Performance Index EV = Earned Value AC = Actual Cost If the Cost Performance Index is greater than unity (1.0), the project is under budget and if it is less than unity, the project is over budget. Note that both performance measures indicate that values greater than one are desirable. The values can be kept on a periodic basis, such as weekly, and on a cumulative basis for monitoring the project performance and direction. An example problem from AACEI Skills and Knowledge is expanded for illustrative purposes.

2.2 EXAMPLE PROBLEM A construction project is executed with a delay of one week over the life of the project. The information is presented in Table 2.1. To illustrate how the values are calculated, Week 4 will be analyzed in detail for the variances and indexes. The values in the Table represent work hours, but could also be considered as monetary units. For the period values for Week 4 one obtains: SV = 85 – 80 = + 5 SPI = 85/80

= 1.06

CV = 85 – 78 = + 7 CPI = 85/78

= 1.09

For the cumulative values for Week 4 one obtains: SV c

= 195 – 200 = –5

SPIc = 195/200

= 0.98

CV c = 195 – 187 = + 8 CPIc = 195/187

= 1.04

30 45 45 80 80 50 25 30 30 25 0

1

2

3

4

5

6

7

8

9

10

11

14

28

33

30

30

51

66

78

62

31

16

15

25

30

25

25

55

70

85

65

30

15

Period AC EV PV (Week) (BCWS) (ACWP) (BCWP) (AT)

+15

0

0

–5

0

+5

– 10

+5

+ 20

–15

–15

SV

SPI

N/A

1.00

1.00

0.83

1.00

1.10

0.88

1.06

1.44

0.67

0.50

Period Values

+1.0

– 3.0

– 3.0

– 5.0

– 5.0

+4.0

+4.0

+7.0

+3.0

–1.0

–1.0

CV

1.07

0.89

0.91

0.83

0.83

1.08

1.06

1.09

1.05

0.97

0.94

CPI

440

440

415

385

355

330

280

200

120

75

30

439

425

397

364

334

304

253

187

109

47

16

440

425

400

370

345

320

265

195

110

45

15

PVC ACC EVC (BCWS) (ACWP) (BCWP)

1.00

0.97

0.96

0.96

0.97

0.97

0.95

0.98

0.92

0.60

0.50

SPIC

0

–15

–15

–15

–10

–10

–15

–5

–10

–30

–15

SVC

Cumulative Values

Table 2.1. Period and Cumulative Data for Project Delayed

1.00

1.00

1.01

1.02

1.03

1.05

1.05

1.04

1.01

0.96

0.94

CPIC

+1.0

0.0

+3.0

+6.0

+11.0

+16.0

+12.0

+8.0

+1.0

–2.0

–1.0

CVC

16 // Strategic Cost Analysis for Project Managers and Engineers //

// Earned Value Management Concepts // 17

The cumulative PV, AC and EV values are plotted in Figure 2.1.

Fig. 2.1. Planned Value, Actual Cost and Earned value.

The period values and cumulative values of the SPI and CPI can be quite different, and are indicated by the schedule parameters in Table 2.1. One problem with the schedule indices is that the SV tends to zero and the SPI goes to 1.0 regardless of whether or not the project is delayed. This occurs as the planned and earned values will be equal when the project is completed. This has led to a new calculation procedure for the Schedule Variance and the Schedule Performance Index on a time based analysis rather than on a cost based (or work-hour based) analysis. Lipke (2, 3) introduced the concept of Earned Schedule (ES), which is somewhat similar to the Earned Value. The earned schedule values are used for cumulative values, not period values. The equation for ES is mathematically expressed as: ESC = N + (EVC – PVC(N))/(PVC(N+1) – PVC(N))

(2.5)

18 // Strategic Cost Analysis for Project Managers and Engineers //

Where N = Period where the cumulative PV is less than the current cumulative EV EVC = Cumulative Earned Value at Actual Time, AT PVC(N) = Cumulative Planned Value at period, N PVC(N+1) = Cumulative Planned Value at next period, N+1 AT = Actual Time or current time period The time based schedule focuses on comparing the Earned Schedule (ES) and the Actual Time (AT). If, at the start of the project, the PV is greater than the EV, then the ES will be less than one and can be calculated by ESC = (PV1 – EV1)/PV1

(2.5a)

This calculation can be repeated for the initial periods until EVC exceeds PV1 and then Equation 2.5 can be used and this is when N = 1. The Schedule Variance, based upon time, is calculated by: SV(t) = ES – AT

(2.6)

Where SV(t) = Schedule Variance based upon time E S = Earned Schedule AT = Actual Time or current time period And the Schedule Performance Index based upon time is: SPI(t) = ES/AT

(2.7)

If one examines Period 4, then AT = 4 EVC = 195 The first cumulative value of PV below ES is 120 which occurs in period 3, thus N =3 PVC = 120 PVC(N + 1) = 200

// Earned Value Management Concepts // 19

And ES = 3 + (195 – 120)/(200 – 120) = 3 + 0.938 = 3.938 Thus the time based Schedule Variance using Equation 2.6 is: SV(t) = 3.938 – 4.00 = – 0.062 This implies that the project is 0.062 time periods behind schedule. The schedule performance index based upon time using Equation 2.7 is: SPI(t) = ES/AT = 3.938/4.00 = 0.9845 = 0.985 Now examining the data for the last period, that is period 11, note that AT = 11 EVC = 440 The first cumulative value of PV below ES is 415 which occurs in period 9, thus N =9 PVC = 415 PVC(N+1) = 440 And ES = 9 + (440 – 415)/(440 – 415) = 9 + 1 = 10 Thus, the time based Schedule Variance using Equation 2.6 is: SV(t) = 10 – 11 = –1 This implies that the project is 1.0 time periods behind schedule. The schedule performance index based upon time using Equation 2.7 is: SPI(t) = ES/AT = 10/11 = 0.909 The time based values are the actual values, that is a Schedule Variance of –1.0 and Schedule Performance Index of 0.909 instead of the traditional values of 0 and SPI of 1.0. Table 2.2 indicates the cumulative based and cumulative time based values. The major differences are in the Schedule Variance and the Schedule Performance Index. The Schedule Variance is based upon Dollars or work hours, whereas the Earned Schedule is based upon amount of work completed for that particular period. Figure 2.2 show the comparison of the SPI values based upon the cumulative Dollars (or work-hours) base versus the SPI values based upon the cumulative project time over the project life.

30

75

120

200

280

330

355

385

415

440

440

2

3

4

5

6

7

8

9

10

11

439

425

397

364

334

304

253

187

109

47

16

440

425

400

370

345

320

265

195

110

45

15

0

–15

–15

–15

–10

–10

–15

–5

–10

–30

–15

SVC

1.0

0.97

0.96

0.96

0.97

0.97

0.95

0.98

0.92

0.6

0.5

SPIC

Cumulative Values

AC EV PV (BCWS) (ACWP) (BCWP)

1

Period (Week) (AT)

1

0

3

6

11

16

12

8

1

–2

–1

CVC

1.0

1.0

1.01

1.02

1.03

1.05

1.05

1.04

1.01

0.96

0.94

CPIC

10

9

8

7

6

5

4

3

2

1

N/A

N

440

425

400

370

345

320

265

195

110

45

N/A

EV (AT)

440

415

385

355

330

280

200

120

75

NA

PV (N+1)

440

415

385

355

330

280

200

120

75

30

NA

PV (N)

10.0

9.40

8.50

7.50

6.60

5.80

4.81

3.94

2.78

1.33

0.5

ES

–1.00

–0.60

–0.50

–0.50

–0.40

–0.20

–0.19

–0.06

–0.22

–0.67

–0.5

SV(t)

Cumulative Time Based Values

Table 2.2. Time Based Calculations

0.91

0.94

0.94

0.94

0.94

0.97

0.96

0.98

0.93

0.67

0.5

SPI(t)

20 // Strategic Cost Analysis for Project Managers and Engineers //

// Earned Value Management Concepts // 21

Fig. 2.2. Comparison of cost based SPI($) (Dollars or work hour units) and time based SPI(t).

The major differences between the two SPI indices occur at the end of the project life. The initial differences are primarily due to the difference in units; that is Dollars versus the time estimate calculation. The time based calculation is the relative amount of work earned for the period versus the amount planned for the period. Table 2.3 is a combination of Tables 2.1 and 2.2 to indicate all the information in a single table.

2.3 SUMMARY The performance measures evaluated are the Schedule Variance, the Schedule Performance Index, the Cost Variance, the Cost Performance Index, the time based Schedule Variance and the time based Schedule Performance Index. The time based Schedule Variance and Schedule Performance Indices are more realistic than the traditional Schedule Variance and Schedule Performance Index and should be used whenever possible.

2.4 REFERENCES 1. Fang, Y., 2008, Estimate at Completion for Construction Projects, (M.S.) Problem Report, Industrial and Management Systems Engineering, West Virginia University. 2. Lipke, W., 2002, A Study of the Normality of Earned Value Management Indicators, Measurable News, 2002, pp. 1–16.

22 // Strategic Cost Analysis for Project Managers and Engineers //

3. Lipke, W., 2003, Schedule is Different, Measurable News, 2003 (March). 4. Vandevoorde, S., and Vanhoucke, M., 2006, A Comparison of Different Projects in Forecasting Method using Earned Value Metrics, International Journal of Project Management, Vol. 24, pp. 289–302. 5. Creese, R.C., and Fang, Yi, Time-Based Schedule Performance Index, Cost Engineering, Vol. 53, No. 3 March 2010, pp. 18–20. 6. AACE International, Skills and Knowledge of Cost Engineering, 4th Edition (2002), Chapter 12.

2.5 EVALUATIVE QUESTIONS 1. Use the data in Table 2.1 and calculate the values for SV, SPI, CV, and CPI for the period and cumulative values to three decimals and compare the values with those in Table 2.1 for periods 6 and 7. 2. Calculate the time based values for SV(t) and SPI(t) and compare the values with those in Table 2.2 for periods 6 and 7. 3. Use the data in Table 2.3 and calculate the period and cumulative values of SV, SPI, CV, CPI, and the time based values of SV(t) and SP I(t) for all the periods for the Project Manager of an U.S. Construction Company, Ohio. Table 2.3 Weekly Data of an U.S. Construction Company for a Windmill Power Generation Project. Money Values are in $(1000). Table 2.3 Period

PV

AC

EV

1

40

25

20

2

40

35

35

3

40

50

45

4

40

25

30

5

0

25

20

6

0

20

10

Discuss the differences between the SV and SV(t) and SPI and SPI(t) values. 4. Change the PV values (Table 2.1) for Periods 6, 7, and 8 from 50, 25, 30 to 30, 35, 40 and note the differences in the SV, SPI, CV, CPI, SPIC, SVC, CPIC, CVC, ES, SV(AT), and SPI(AT) values from those in Table 2.4.

Period Values

Cumulative Values

30

45

45

80

80

50

25

30

30

25

0

1

2

3

4

5

6

7

8

9

10

11

14

28

33

30

30

51

66

78

62

31

16

15

25

30

25

25

55

70

85

65

30

15

1.06 7 1.09

1

1.1 –5 0.83

4 1.08

1

1

15 NA

0

0

1 1.07

–3 0.89

–3 0.91

–5 0.83 –5 0.83

0

5

–10 0.88 4 1.06

5

20 1.44 3 1.05

–15 0.67 –1 0.97

–15 0.5 –1 0.94

16 47 109 187 253 304 334 364 397 425 439

30 75 120 200 280 330 355 385 415 440 440

440

425

400

370

345

320

265

195

110

45

15

1.04

1

0

0.97 –15

1.0

1.0

0.96 –15 1.01

0.96 –15 1.02

0.97 –10 1.03

0.97 –10 1.05

0.95 –15 1.05

0.98 –5

0.92 –10 1.01

0.6 –30 0.96

0.5 –15 0.94

1

0

3

6

11

16

12

8

1

–2 30

425 415

400 385

370 355

345 330

320 280

265 200

195 120

110 75

45

10 440 440

9

8

7

6

5

4

3

2

1

–1 NA NA NA

440

415

385

355

330

280

200

120

75

NA

–0.5

0.5

10.00 –1.00 0.91

9.40 –0.60 0.94

8.50 –0.50 0.94

7.50 –0.50 0.94

6.60 –0.40 0.94

5.80 –0.20 0.97

4.81 –0.19 0.96

3.94 –0.06 0.98

2.78 –0.22 0.93

1.33 –0.67 0.67

0.5

SV SPI (AT) (AT)

Cumulative Time Based Values

Period PV AC EV PVC ACC EVC EV PV PV ES SV SPI CV CPI SPIC SVC CPIC CVC N (AT) (N) (N+1) (AT) (Week) (BCWS) (ACWP) (BCWP) (BCWS) (ACWP) (BCWP)

AT

Table 2.4. Period and Cumulative Indices with Time Based Values

// Earned Value Management Concepts // 23

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// Interest Types and Money Values // 25

3

INTEREST TYPES AND MONEY VALUES 3.1 HISTORY OF INTEREST AND ITS SIGNIFICANCE The concept of interest has existed from the earliest records of human history. Historical documents and records reveal its existence in Babylon in 200 BC. In the earliest instances interest was paid in the form of grains or other goods for the use of commodities that were lent or borrowed. Many of the existing interest practices stem from early customs in the borrowing and repayment of grains and other crops. History also reveals that the idea of interest became so well established that a firm of international bankers existed even in 575 BC with Head Quarters in Babylon. In the Bible, the terms of money and interest are presented in the following references: 1.

“If you lend money to the poor of my people, do not act as the creditor to him; do not charge him interest” (Exodus 22: 25).

2.

“You must charge no interest on a loan made to your brother. On a loan to a foreigner, you must charge interest, but not from your brother. Charge no interest, so that the Lord your God may bless you in all your enterprises in the land of which you are taking possession” (Deuteronomy 23:19-20).

3.

“You should have placed my money with the bankers and on my return I would have gotten my capital with interest” – Parable of the talents (Matthew 25: 27).

4.

“Why did you not put my money in the bank so that when I come I might have gotten it with interest” (Luke 19: 23).

26 // Strategic Cost Analysis for Project Managers and Engineers //

5.

“For which of you, intending to build a tower, sit not down first, and count the cost, whether he has sufficient to finish it?” (Luke 14:28).

In modern times interest rates are decided by the Annual Monetary Policy of the Central Bank of the country, such as the Federal Reserve Bank in the USA or the Reserve Bank of India. The minimum interest rates must be theoretically equal to the inflation rate for overall financial systemic stability. High interest rates could impact the investment momentum and corporate cash flows and are likely to slow down the growth rate. Low interest rates discourage savings and encourage debt, leading to currency shortages and/or inflation. The trade off between growth momentum and inflation containment has always been a challenge to the economists and policy makers. Companies must develop appropriate business strategies so that they are in a position to combine long-term viable financing of investment projects with profitability in operations recognizing the reality of business cycles consistent with monetary policy responses of the Government and the regulatory agencies. Companies would experience a “positive free cash flow” by improving their market share. In a high interest rate regime, each project must be reviewed to rework its costs, if funds have not been adequately provided for initially. Industries must improve their efficiency and undertake cost-reduction measures under these conditions.

3.2 DEFINITIONS AND TERMS RELATED TO INTEREST CALCULATIONS 3.2.1 Interest There are various types of interest and several definitions of the meaning of interest in a financial sense. Some of the common definitions of interest are: 1.

The monetary return or other expectation which is necessary to divert money away from consumption and into long-term investments.

2.

The cost of the use of capital or the time value of money.

3.

In accounting and finance:

// Interest Types and Money Values // 27

(a) A financial share in a project or an enterprise (b) Periodic compensation for the lending of money The focus in this book is on the interest rate, which represents the return rate or charge rate for the investment of capital.

3.2.2 Interest Rate Interest rate is the rate at which interest is applied to the principal. The ratio of the interest amount accrued in a given period of time to the principal amount owed or invested at the start of that period. The two types of interest commonly used are the simple interest and the compound interest. 3.2.2.1 Simple Interest Simple interest is the interest rate that determines the interest amount only on the principal amount. Simple interest is defined as: 1.

Interest that is not compounded i.e., interest that is not added to the income producing investment or loan.

2.

Interest charges under the condition that interest in any time period is only charged on the principal. Simple Interest Calculations

The interest amount due at the end of a set of periods on an initial investment can be calculated by: I = niP where

(3.1)

I = total interest amount due n = number of periods i = interest rate (amount of interest per unit time, usually a year) P = principal or initial investment amount

To illustrate the calculations, an example will be presented. An amount P will be invested for n periods and the amount due at the end of n periods will be P plus the total interest rate, which is referred to as the Future Amount and is represented by F. Thus, F = Future worth = total amount at the end of period = P + I F = P + I = P + niP = P (1 + in)

28 // Strategic Cost Analysis for Project Managers and Engineers //

If P = $1,000, n = 2 years and i = 10%, then F = $1,000 (1 + 2 × 0.10) = $1,200 = $1,000 Principal

+ $100

+ $100

Interest for

Interest for

Year 1

Year 2

3.2.2.2 Compound Interest Compound interest is that which is most commonly used and will be used in most of the problems in this book. Compound interest includes not only interest on the principal amount, but also on the interest amount previously earned. Some definitions of compound interest are: 1. The type of interest that is periodically added (usually at quarterly intervals) to the amount of investment or loan so that subsequent interest is based on the cumulative amount. 2. The interest charges under the condition that interest is charged on any previous interest earned in any time period as well as on the principal. Compound Interest Calculations The table below indicates the calculations involved in compound interest and derivation of the formula to calculate the amount at the end of n periods. If P is the amount invested, i is the interest rate Year

Amount

Interest

Total Amount = F = Future worth

(Beginning of Year)

Amount

(Amount at End of Year)

1

P

iP

P(1 + i)

2

P(1 + i)

iP(1 + i)

P(1 + i)2

3

P(1 + i)2

iP(1 + i)2

P(1 + i)3

iP(1 + i)(n–1)

P(1 + i)n

and in general n

P(1 + i)(n–1)

Thus the expression for compound interest would be: F = P(1 + i)n

(3.2)

Using the data of the previous simple interest example and now applying continuous interest instead of simple interest, one has:

// Interest Types and Money Values // 29

If P = $1,000 and i = 10% and n = 2 Year

Amount

Interest

Total Amount = F = Future worth

(Beginning of Year)

Amount

(Amount at End of Year)

1

$1,000

$100

$1,100

2

$1,100

$110

$1,210

This can also be determined by using Equation 3.2 as: F = 1,000(1 + 0.10)2 = 1,210 Note that the total amount is $1,210 for compounding versus the $1,200 for simple interest compounding.

3.2.3 Rate of Return The rate of return for an investment is the interest rate earned by the investment. The rate of return is the interest rate considered by those making the investment and receiving the amount, whereas the interest rate is considered to be that paid for by those receiving the investment. For the examples considered in this book, the rate of return and the interest rate will be considered equivalent when only one rate is involved. Some definitions of rate of return are: 1.

The interest rate earned by an investment.

2.

The interest rate at which the present worth equation or the equivalent annual worth or future worth equations for the cash flows of a project becomes equal to zero.

3.2.4 Interest Amount Factors The interest rate includes many factors and the primary ones are presented. The interest considered is the market interest rate, and other interest rates are available, but not for the public or most investors. For example, the prime interest rate in the US was 0.25 percent, the interest rate given by banks to depositors was only 1.0 percent, the prime interest rate was 3.5 percent, the fixed mortgage rate for 30 years was 4.75 percent, the automobile interest rate was 6.75 percent, and the return rates desired by companies is typically 10 to 20 percent as they must make significant profit on the investment and the risk is typically high. The market interest rate is a rental amount charged (like parking fees for the amount invested) by financial institutions for the use of money which includes:

30 // Strategic Cost Analysis for Project Managers and Engineers //

(a) Risk of Loss (1–5%) (b) Administrative Expenses (1–5%) (c) Pure Gain or Profit (2–10%) (d) Risk of Inflation (1–200%, but typically 2–10%) The differences in the rates involve the high risk of loss and need for high profits, and for international companies, the risk of high inflation. Some countries have inflations rates exceeding 10 and even 20 percent, so international investors must be careful. Sometimes the inflation component is not considered and that is refered to as an “inflation free” interest rate. The most commonly used interest rate does include the effect of inflation.

3.2.5 Types of Payments and Types of Interests Considered in Engineering Economics There are two types of payments, either discrete or continuous, and two types of interest, discrete and continuous, used in engineering economic problems. The relative frequency of the problems with the different types of interest and payments can be expressed in Table 3.1. Note that most of the problems involve discrete interest and discrete payments and these will be emphasized. Table 3.1. Payment Types and Interest Types Considered in Engineering Economics Types of Interest Types of Payments

Discrete Continuous

Discrete

Continuous

> 95%

2–4%

0

< 1%

3.3 INTEREST RATE FUNDAMENTALS AND BASIC CALCULATIONS 3.3.1 Actual and Nominal Interest Actual interest rate (i), commonly called effective interest, is the interest rate per compounding period. Actual interest is the interest that is most commonly used in engineering economy calculations. Thus the actual interest value, i = interest rate (amount per compounding period), can be expressed as:

// Interest Types and Money Values // 31

3%/quarter;

1%/month;

12%/year;

0.03288%/day;

(Percent/unit time) Nominal interest rate(r) is the interest rate per year as obtained by multiplying the interest rate per period by the number of compounding periods per year. It is also known as the APR or Annual Percentage Rate. It is the interest rate used when considering continuous compounding situations. r = interest rate/year = APR(Annual Percentage Rate) = nominal interest If m = number of compounding periods per year i = interest rate/compounding period then r = mi

(3.3)

Some illustrations of various interest rates i being converted to nominal rates r are: i

r

3%/quarter

= 4 × 3%

= 12% per year

1%/month

= 12 × 1%

= 12% per year

12%/year

= 1 × 12%

= 12% per year

0.03288%/day

= 365 × 0.03288% = 12.00

= 12% per year (can be taken as 12% per year)

3.3.2 Effective Annual Interest Rate Effective annual interest is the actual interest based upon a one year period. It is the compound of the interest rate over the discrete number of compounding periods during the year. It is calculated by: ieff = (1 + r/m)m –1 where ieff = Effective Annual Interest r = nominal interest m = number of compounding periods per year

(3.4)

32 // Strategic Cost Analysis for Project Managers and Engineers //

If i = 1%/month, r = 12%, and m = 12 then ieff = (1 + 0.12/12)12 – 1 = 0.1268 = 12.68% instead of 12%(r) If i = 0.04%/day, r = 14.6%, and m = 365 then ieff = (1 + 0.146/365)365 –1 = 0.15716 = 15.72% versus 14.6%(r)

3.3.3 Continuous Compounding Effective annual interest can also be determined for continuous compounding situations. It is expressed as: ieff = limit [( 1 + r/m)m – 1 ] m→∞

= limit [ {1 + 1/(m/r)}m/r ]r –1 = er – 1 m→∞

or simply ieff = er –1

(3.5)

Some examples of effective interest when continuous compounding is used are: If i = 1%/month then m = 12 ieff = e0.12 – 1 = 1.127497 – 1 = 0.1275 = 12.75% compounded continuously Note the differences in the annual interest rates when using 1% per month, Nominal Interest

r = 12.00 %

Discrete Interest

ieff = 12.68 %

Continuous Interest

ieff = 12.75 %

// Interest Types and Money Values // 33

Most of the calculations will involve discrete interest, but one must be aware of the other types of interest rate calculations.

3.3.4 Inflation Free Interest Rate The problems which have been illustrated are in terms of actual currency. The actual currency is the amount associated with a cash flow at the point in time at which it occurs. For example, the amount of Dollars received or disbursed at any point in time are called actual Dollars. This is also referred to as current Dollars. Constant currency, are Dollars expressed in terms of the same purchasing power relative to a particular point in time. Constant Dollars represent the hypothetical purchasing power of future receipts and disbursements in terms of the purchasing power at some base year. Constant Dollars are often referred to as real Dollars or inflation free Dollars. Constant Dollars are often used in construction projects and in government projects. The relation between constant and actual Dollars is: Constant Dollars(C$) = Actual Dollars(A$)/(1+f)n where

(3.6)

f = inflation rate (as a decimal) over time period n years Bridge Example In the life cycle cost of a bridge, the deck would be expected to be replaced in 25 years. The cost to replace the deck today is $150,000 and this is the value one would use in constant Dollars if the replacement is not expected to change. However, if the inflation rate is 3 percent per year, if one uses actual Dollars, the deck replacement would be expected to cost $150,000 * (1 + 0.03)25 or $314,700. If one is using actual Dollars, then the return rate or interest rate (i) would be the market interest rate and would include the effects of inflation. This is the rate generally used by investors, bankers, and stockholders. If one is using constant Dollars, the inflation free interest rate (i′) would be used. This represents the earning power of money with the effects of inflation removed. If the inflation rate is f, the relation between the two interest rates is: (1 + i) = (1 + f) (1 + i′ )

(3.7)

34 // Strategic Cost Analysis for Project Managers and Engineers //

or i ′ = [(1 + i)/(1 + f)] –1 where

(3.8)

i = market interest rate i ′ = inflation free interest rate f = inflation rate

For most concerns in manufacturing, the actual Dollars and market interest rate (i) are used in the problems. For government projects and long-term civil engineering and construction projects, constant Dollars and the inflation free interest rate (i ′) would be used. Example If the interest rate is 26 percent and the inflation rate is 5 percent, what is the inflation free interest rate? Using Equation 3.7

(1 + 0.26) = (1 + i ′) (1 + 0.05)

or i ′ = [(1 + 0.26)/(1 + 0.05)] – 1.0 and solving i ′ = 0.20 = 20% In many long-term projects, such as the building of a nuclear reactor or a dam, it may take years to complete the project. In these instances, it is often difficult to predict the inflation rate far in the future, so one often uses an inflation free interest rate, which uses the symbol i′. The interest rates must be in decimal form when using Equations 3.7 and 3.8.

3.4 TYPES OF MONEY VALUE AND INTEREST RATES The types of money values used must be consistent with the interest rates applied. The two types of money values used are actual or current currency which includes the effects of inflation and real or constant currency which does not include the effects of inflation.

3.4.1 Actual Money Value Currency received or money spent at any point in time. Other names are current value of money or nominal currency. Actual currency values are used in most applications, especially in projects with relatively short

// Interest Types and Money Values // 35

investment life; that is less than 20 years. The interest rate used with actual currency is the effective or market interest rate which includes the effects of inflation.

3.4.2 Real Money Value Currency that represents the purchasing power of future receipts and disbursements in terms of purchasing currency with respect to some base year, which is normally time zero or the beginning of the investment. Other names are constant currency, real currency, inflation-free currency, and today’s currency. Real or constant currency values are typically used in Government projects involving life cycle costs where the maintenance, repair, and rehabilitation values are difficult to predict in the future. The investment life is more than 20 years for life of a building or equipment and frequently 50 or 100 years for the life of a highway bridge project or a river valley/irrigation canal project and inflation free interest rates are used.

3.5 EXCHANGE RATE IMPACT One of the problems of global projects is that of exchange rates in different countries. A project may be calculated in one currency, but performed in another country with a different currency. When long-term projects are involved, this can be a serious problem. The fluctuations in currency rates can be large due to different inflation rates in the two countries. The exchange rate is the amount of one country’s currency that one unit of another country’s currency would purchase. Example If in the year 2000 when one US Dollar purchased approximately 1.0 Euro, a US investor purchased 1,000 Euros. In 2010, the investor decided to convert the Euro back to US Dollars and the exchange rate was one US Dollar purchased 0.75 Euro. What did the investor receive? Currency (now) = Currency (original) [Currency Rate (now)/Currency Rate (original)] (3.9) Thus Currency (now) = $1,000 [1$/0.75 Euro/1$/1.0 Euro] = $1,000 [1/0.75] = $1,333

36 // Strategic Cost Analysis for Project Managers and Engineers //

Example If in the year 2000 when one US Dollar purchased approximately 9.5 Pesos, a US investor purchased $1,000 worth of Pesos. In 2010, the investor decided to convert the Pesos back to US Dollars and the exchange rate was one US Dollar purchased 13 Pesos. What did the investor receive? Currency (now) = $1,000 [1 $/13 pesos/1 $/9.5 pesos] = $1,000 [9.5/13] = $731 Thus, the effects of currency exchange rates can be rather large and must be considered in executing large international projects.

3.6 MATHEMATICAL RELATIONSHIPS FOR ENGINEERING ECONOMIC CALCULATIONS 3.6.1 Geometric Series The mathematics involved in the calculations for most of the formulas in engineering economics are not complex. The most commonly used expression is that of the geometric series when dealing with discrete interest relationships. The discrete relationships are those where there is a discrete payment at a discrete interest rate at a specific time. This discrete interest rate implies that it is compounded at a fixed time period, whereas a continuous interest implies that it is compounded continuously over time. The case of discrete payments and discrete interest rates is the most common of the engineering economic problems. The basic mathematical relationship for discrete interest problems is the geometric series and for continuous interest problems the infinite limits expression is used. The geometric series expression is: S = a + ar + ar2 + ar3 + ar4 + + ar(n–1) = a[(rn – 1)/(r – 1)] where

(3.10)

S = sum of the series a = constant which occurs in all terms r = ratio between terms n = number of terms in the sum (including the initial term without the ratio)

// Interest Types and Money Values // 37

Example Take the number 3 (a = 3) and double it (r = 2) for 4 additional periods and determine the sum. That is: S = 3 + (3 × 2) + (3 × 22) + (3 × 23) + (3 × 24) = 3 + 6 + 12 + 24 + 48 = 93 If one uses the geometric series expression, there are five terms (the initial term and the 4 additional periods), so n = 5 r = 2 a = 3 and thus S = 3 × [25 – 1]/[2 –1] = 3 × [32 – 1]/ [2 – 1] = 3 × [31]/[1] = 93 This becomes very useful when n is large. In most interest calculations, the ratio is r = (1 + i) which is the periodic compounding amount for continuous interest calculations. This will be used in the derivation of the compounding factors in the various engineering economic expressions.

3.6.2 Infinite Limit The infinite limit is used for continuous interest problems, that is when the interest is compounded continuously rather than over discrete time periods. The expression is based on:

Lim [ 1 + 1/k]k = e = 2.71828 k→ ∞ This can be illustrated by taking values of k such as 1, 2, 3, 100 and 1,000 and note that the value approaches the limit of e. Limit = [1 + 1/1], [1 + ½]2, [1 + 1/3]3,.....[1 + 1/100]100...[1 + 1/1000]1000 Limit =

2,

2.25,

2.37,

2.704,

2.716

The infinite limit was used in the derivation of continuous interest by letting the number of compounding periods go to infinity; that was: ieff = limit [(1 + r/m)m – 1] = limit [{1 + 1/(m/r)}m/r]r –1 = er – 1 m→∞

m→∞

38 // Strategic Cost Analysis for Project Managers and Engineers //

or simply ieff = er – 1 where ieff = interest rate (effective interest rate on an annual basis) r = nominal interest rate One can also express the relationship as: (ieff + 1) = e r

(3.11)

Equations 3.11 and 3.5 are frequently used to convert discrete compounding factors into continuous compounding factors.

3.7 EVALUATIVE QUESTIONS 1. What is the value of $100,000 invested at 10% simple interest per year for 10 years? 2. What is the value of $100,000 invested at 10% compound interest per year for 10 years? 3. The interest rate is 2% per month and the time period is 1 year. (a) What is the nominal interest rate? (b) What is the effective interest rate? (c) What is the continuous interest rate? 4. A person invests $50,000 at 9% simple interest for 4 years. At the end of the 4 years, the entire amount (principal and interest) is invested at 11% compounded annually for 10 years. What is the amount of the investment after 14 years? 5. The inflation rate is 4% and the inflation free interest rate is 8%, what is the market interest rate? 6. The market interest rate is 20% and the inflation rate is 5%, what is the inflation free interest rate? 7. A person earns $ 5 and the amount is doubled each period for the next 6 periods, what is the sum of all these amounts? 8. A person earns $ 10 and the amount is tripled each period for the next 4 periods, what is the sum of these 5 amounts? 9. If continuous interest is used and the nominal rate is 15 percent, what is the discrete annual effective interest rate? 10. If in the year 2000 one US Dollar purchased approximately 40 Rupees, a US investor purchased $1,000 worth of Rupees. In 2010, the investor decided to convert the Rupees back to US Dollars, when the exchange rate was Rupees 48 per US Dollar. What did the investor receive?

4

EXPRESSIONS USED IN ENGINEERING ECONOMICS–BASIC EXPRESSIONS 4.1 INTRODUCTION There are two categories of the basic engineering economic expressions –the single payment expressions and the uniform payment expressions. There are two single payment expressions and four uniform payment expressions and all six will be presented in this Chapter. The expressions will be developed for the discrete interest rate and discrete payment case. The developed expressions will be modified for the continuous interest rate and discrete payment case.

4.2 SINGLE CASH FLOW EXPRESSIONS The two single payment cases developed with discrete interest and discrete payments are the future worth expression and the present worth expression.

4.2.1 Future Worth Expression The future worth expression converts a present amount to a future worth for a given interest rate and given number of compounding periods. The notation used will be: F = Future worth (value after n compounding periods) P = Present worth (value at time zero, the beginning of the study period) i = interest rate per compounding period n = number of compounding periods

40 // Strategic Cost Analysis for Project Managers and Engineers //

A graph indicating where the Present Worth and Future Worth are with respect to time and Table 4.1 indicates the principal and interest earned.

0 P

1

2

3 4 5 6 Time (years)

F (n – 2) (n – 1) n

......

Table 4.1 Payment Illustration for Future worth Expression

Time (End of Period) 1

Principal (Start of Period) P

2

P × (1 + i)

+ iP × (1 + i)

= P × (1 + i)

3

P × (1 + i)

2

+ iP × (1 + i)

2

= P × (1 + i)

4

P × (1 + i)

3

+ iP × (1 + i)

3

= P × (1 + i)













n

n–1

P × (1 + i)

Interest (Earned During Period) + iP

n–1

+ iP × (1 + i)

Principal + Interest (End of Period) = P × (1 + i) 2 3 4

n

= P × (1 + i)

Thus from Table 4.1 it can be seen that the value at the end, the Future worth, is: F = P × (1 + i)n (4.1) Where P = Present worth (initial amount at time zero) i = Interest rate n = Number of compounding periods The conversion factor to convert the Present worth to the Future worth is called the single payment Future worth compound amount factor, is designated as (F/P, i, n), is stated as “F given P, i, n”, and is equal to (1 + i)n. Example If a single payment of $100,000 is invested at 15% interest compounded annually, the compound amount at the end of the 4th year would be: F = $100,000 (1 + 0.15)4 = $100,000 × 1.74901 = $174,900 Or F = $100,000 × (F/P, i = 15, n = 4) = $100,000 × 1.74901 = $174,900 The (F/P, i, n) factor is given in the Tables, in the Appendix but one can use the formulas to calculate the values as all the possible interest rates are not given in the tables.

// Expressions Used in Engineering Economics–Basic Expressions // 41

4.2.2 Present Worth Expression The Present worth expression is the inverse of the Future worth expression; that is the expression is: P = F/(1 + i)n

(4.2)

Where P = Present worth (initial amount at time zero) i = Interest rate n = Number of compounding periods The conversion factor to convert the Future worth to the Present worth is called the single payment. Present worth compound amount factor, is designated as (P/F, i, n), is stated as “P given F, i, n”, and is equal to (1 + i)–n or 1/(1+i)n. Example If $100,000 is desired at the end of 4 years and the interest is 15% compounded annually, what amount would need to be deposited initially? P = $100,000 /(1 + 0.15)4 = $100,000/1.74901 = $57,175 Or P = $100,000 × (P/F, i = 15, n = 4) = $100,000 × 0.57175 = $57,175 The (P/F, i, n) factor is also given in the tables at the end of most engineering economy books. Some basic values are given at the end of this text, but one can use the formulas to calculate the values as all the interest rates are not given in the tables (for example, 4.75% is not given in the tables, but it can easily be calculated using 1/(1 + i) n using i = 0.0475 for any particular n required).

4.3 UNIFORM PAYMENT EXPRESSIONS The four uniform payment factors to be developed are the compound amount factor(F/A, i, n), the sinking fund factor(A/F, i, n), the uniform series present worth factor (P/A, i, n) and the capital recovery factor (A/P, i, n). All of these factors involve the amount A payment which is made at the end of every period. In the development of the various engineering economic expressions, the assumption is made that the discrete payment is at the end of the payment period, not at the

42 // Strategic Cost Analysis for Project Managers and Engineers //

beginning. Modifications can be made for the beginning of payment periods.

4.3.1 Compound Amount Factor (F/A, i, n) (also called Uniform Series Compound Amount Factor) A series of payments of the amount A is made at the end of the period for n periods at an interest rate i will result in a Future worth of F at the end of the n periods. This results of a payment A in the final period to complete the Future worth value of F. A graphical representation of the payments and Future worth is:

The expression can be derived using the geometric series expression by: F = A (end of last period) + A(1 + i) + A(1+ i)2 + ... + A(1 + i)(n–1) (end of first period) = A + A (1 + i) + A(1 + i)2 + A(1 + i)3 + A(1 + i)4 + ... + A(1 + i)(n–1) = A (rn – 1)/(r – 1) = A [(1 + i)n – 1]/[(1 + i) – 1] = A[(1 + i)n – 1]/[i] ⇒ F = A(F/A, i, n) = A[(1 + i)n – 1]/[i]

(4.3)

Where (F/A, i, n) = [(1 + i)n – 1]/[i] Example What is the Future worth of $100,000 deposited at the end of the year each year for 10 years when the interest is 8%? F = $100,000 × [(1 + 0.08)10 – 1]/[0.08] = $100,000 × [1.1589/0.08] = $100,000 × [14.487] = $1,448,700 The compound amount factor [F/A, i = 8%, n = 10] = 14.487 may be listed in the tables, but if one had an interest rate of 8.2% it would not be available and thus one should be able to use the compound amount factor expression and thus: [F/A, i = 8.2%, n = 10] = [(1+ 0.082)10 – 1]/[0.082] = 14.625

// Expressions Used in Engineering Economics–Basic Expressions // 43

4.3.2 Sinking Fund Factor (A/F, i, n) The sinking fund factor involves the same terms as the compound amount factor and the graphical representation of the payments and Future worth is the same and is repeated as:

The Sinking Fund Factor is the inverse of the Compound Amount Factor and thus it would be: (A/F, i, n) = 1/(F/A, i, n) Therefore Sinking Fund Factor is: (A/F, i, n) = 1/[(1 + i)n –1]/[i] = i/[(1 + i)n –1] and A = F(A/F, i, n) = F/[(1 + i)n –1]/[i] or commonly expressed as A = F i/[(1 + i)n –1]

(4.4)

4.3.3 Sinking Fund Factor Example Yun wants to have $1,000,000 at the end of 10 years, so what amount would have to be saved at the end of each year if the interest rate over the 10 year period is 5% to obtain the $1,000,000? A = $1,000,000 [0.05/[(1 + 0.05)10 –1] = $1,000,000 × [0.05/(0.62889)] = $1,000,000 × (0.079505) = $79,505 Thus (A/F, i = 5%, n = 10) = 0.079505 Thus, Yun would need to deposit $79,505 at the end of each year for 10 years to have $1,000,000 at the end of the ten year period. His final payment is also necessary to make the total one million Dollars.

4.3.4 Uniform Series Present Worth Factor (P/A, i, n) The uniform series Present worth factor is used to convert a uniform series of n payments of the amount A at an interest rate i to a Present worth amount P. A graphical represent of the payments and the Present worth is presented as:

44 // Strategic Cost Analysis for Project Managers and Engineers //

To determine the present value P, each payment A must be discounted at the interest rate i by the number of periods it occurs in the future. The discount factor is (1 + i)–n for each period where n is the specific period for that particular payment. The expression can be derived using the geometric series expression by: P = A discounted 1 period + A discounted 2 periods +...+ A discounted n periods P = A/(1 + i) + A/(1 + i)2 + A/(1 + i)3 + A/(1 + i)4 + ... + A/(1 + i)n P = A/(1 + i)[1 + 1/(1 + i) + 1/(1 + i)2 + 1/(1 + i)3 + 1/(1 + i)4+...+ 1/(1+ i)(n–1)] Now using the geometric series equation (Equation 3.10) a = A/(1 + i) and r = (1 + i)–1 or 1/(1 + i) one obtains:

where

S = a [(rn – 1)/(r – 1)] P = A/(1 + i) × [(1/(1 + i))n –1] /[(1/(1 + i) –1)] = A/(1 + i) × [{1 – (1 + i)n }/(1 + i)n ] / [{1–(1 + i)} /(1 + i)] = A × [{1 – (1 + i)n }/(1 + i)n ]/[– i] = A × [{ (1 + i)n – 1}/{i(1 + i)n}]

(4.5)

The uniform series present worth factor is thus (P/A, i, n) = [{(1 + i)n – 1}/{i(1 + i)n}] Example Marcello won a lottery which promised $2,000,000 paid as $200,000 at the end of each year for 10 years. If the interest rate is 5%, what is the present worth of the Marcello’s lottery winnings? P = A × (P/A, i = 5%, n = 10) P = $200,000 × [(1.05)10 –1]/[0.05 × (1.05)10] = $200,000 × (7.7217) = $1,544,350

// Expressions Used in Engineering Economics–Basic Expressions // 45

Note that this amount is a lot less than the $2,000,000 prize amount. From the calculations, one notes that: (P/A, i = 5%, n = 10) = 7.7217

4.3.5 Capital Recovery Factor (P/A, i, n) The Capital Recovery Factor is the inverse of the Uniform Series Present Worth Factor and thus it has the same graphical representation as the Uniform Series Present Worth Factor and is presented as:

The solution is to determine the amount of the payment A based upon the values of P, i, and n. The Capital Recovery Factor is the inverse of the Uniform Series Present Worth Factor and is A = P × [i(1 + i)n]/[(1 + i)n – 1]

(4.6)

The capital recovery factor is thus: (A/P, i, n) = [i(1 + i)n] / [(1 + i)n – 1] Example Sucharitha has purchased a new motorcycle for $20,000 after a down payment of $3,000. Her loan is for $20,000 for a period of 3 years at an interest rate of 12% year. She wants to close the loan at the end of 3 years. a.

If annual payments are made, what is the payment? A = $20,000 × [0.12(1.12)3 ]/[1.123 – 1] = $8,326.97 = $8,327

b.

If monthly payments are made, what is the monthly payment?

The interest rate would be 1% per month and n = 36 A = $20,000 × [0.01(1.01)36]/[(1.01)36 –1] = $664 Note that $8,327/12 = $694 and not $664; why?

4.4 CONTINUOUS EXPRESSIONS AND SUMMARY TABLE As indicated in Chapter 3, Equation 3.5, the values of I and (1 + i) can be represented as:

n

Note: i = e –1 and (1+i) = e

rn

P

Present worth

r

F

Compound Amount

(continuous)

A

A

F

Capital Recovery

Sinking Fund

Future worth (compound amount)

P

Uniform Series

D. Uniform Payment or

(continuous)

Present worth

P

Present worth

C. Single Payment

F

Compound Amount

(discrete)

A

A

Capital Recovery

Sinking Fund

F

P

Find

Uniform Series

B. Uniform Payment or

Future worth (compound amount)

Present worth

A. Single Payment

(discrete)

Factor Name

A

A

P

F

P

F

A

A

P

F

P

F

Given

(P/A,r,n)

(F/A,r,n)

(A/P,r,n)

(A/F,r,n)

(F/P, r, n)

(P/F, r, n)

(P/A,i,n)

(F/A,i,n)

(A/P,i,n)

(A/F,i,n)

(F/P, i, n)

(P/F, i, n)

Symbol

n

n

n

r

rn

r

rn

r

[ (e

rn

rn

r

– 1)/(e (e –1))]

[ (e –1)/(e –1)]

rn

[ e (e –1)/(e –1)]

rn

[(e – 1)/ (e –1)]

e

rn

e

–r n

[(1 + i) – 1] / [i(1 + i) ]

n

[(1 + i) –1] / i

n

[i(1 + i) ] / [(1 + i) –1]

n

i / [(1 + i) –1 ]

(1 + i)

n

(1 + i)

–n

Formula

4.2. Discrete and Continuous Compounding Factors of Engineering Economic Expressions = Present worth A = uniform end-of-period payment i = i eff = effective annual interest = Future worth G = uniform gradient amount r = nominal annual interest = number of periods

Payment Type

Table Notation: P F n

46 // Strategic Cost Analysis for Project Managers and Engineers //

// Expressions Used in Engineering Economics–Basic Expressions // 47

ieff = er –1 and (1 + ieff) = and

er

(4.7)

(Where ieff = effective interest rate on an annual basis

r = nominal interest rate)

(4.8)

If these results are applied to Equations 4.1 to 4.6, the continuous expressions can be obtained and these results are included in the summary table, Table 4.2.

4.5 EVALUATIVE QUESTIONS 1. If the time period is 5 years and the interest rate is 10%, calculate the following values and compare them with the tables in the Appendix. a. (P/A, i = 10%, n = 5) b.

(F/A, i = 10%, n = 5)

c.

(A/P, i = 10%, n = 5)

d.

(A/F, i = 10%, n = 5)

e.

(P/F, i = 10%, n = 5)

f.

(F/P, i = 10%, n = 5)

2. If the time period is 5 years and the interest rate is 7.5%, find the following values: a. (P/A, i = 7.5%, n = 5) b.

(F/A, i = 7.5%, n = 5)

c.

(A/P, i = 7.5%, n = 5)

d.

(A/F, i = 7.5%, n = 5)

e.

(P/F, i = 7.5%, n = 5)

f.

(F/P, i = 7.5%, n = 5)

3. Engineer Frederick wants to retire in 20 years and would like to have 1 million dollar at that time. If the interest rate is expected to be 5% over the next 20 years, what annual amount would he need to save? 4. Mary has purchased a house with a loan of $500,000. If the loan interest rate is 5%, what will be her annual payments over the life of the 20 year loan? 5. John has decided to save for a trip in two years. If the monthly interest rate is 1% and he saves $2,500/month, how much will he have saved after two years?

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6. George has decided to purchase a new motor bike and the interest rate is ½ % per month and the purchase price is $20,000. The payments will be at the end of the month, what is her expected monthly payment over the 3 year period? 7.

William has purchased a new car and the interest rate is 2% per month and the purchase price is $100,000. a. What is the expected monthly payment over a 4 year period? b.

What is the total interest paid over the 4 year period?

8. Marlene won the US sweepstakes of $20 million. The prize is actually $2 million per year for 10 years. a.

If the payments are 10 end-of year payments and the interest rate is 5%, what is the annual amount she would receive each year?

b.

If the payments are 10 beginning-of-year payments and the interest rate is 5%, what is the annual amount she would receive each year?

9. If the time period is 5 years and the nominal interest rate is 10%, calculate the following values. a.

(P/A, r, = 10%, n = 5)

b.

(F/A, r = 10%, n = 5)

c.

(A/P, r = 10%, n = 5)

d.

(A/F, r = 10%, n = 5)

e.

(P/F, r = 10%, n = 5)

f.

(F/P, r = 10%, n = 5)

10. If the time period is 5 years and the APR (Annual Percentage Rate) interest rate is 7.5%, find the following values. a.

(P/A, r = 7.5%, n = 5)

b.

(F/A, r = 7.5%, n = 5)

c.

(A/P, r = 7.5%, n = 5)

d.

(A/F, r = 7.5%, n = 5)

5

GRADIENT EXPRESSIONS 5.1 INTRODUCTION Some of the more complex expressions are the gradient expressions and they can be expressed as a fixed amount for the increase, called a uniform gradient (which looks like a ramp function) or as a fixed rate, called a geometric gradient (which has a constant percentage amount applied to a specific amount). The geometric gradient fixed amount starts in the first period with the rate increases starting in the second period. Since it starts at the beginning of the first period, it does not appear graphically until the end of the first period, which is the start of the second period. The escalation expression is a special case of the geometric gradient where the rate increases also start in the initial time period and thus appears graphically at the end of the period zero, which is the start of period one. Thus in summary, the escalation gradients start the gradients one period sooner than either the uniform or the geometric gradients. The gradients that will be considered are the Uniform Gradient, the Geometric Gradient, and the Escalation Gradient. The Escalation Gradient can be considered as the Geometric Gradient starting one period earlier. The expressions of Present worth, Future worth, and Uniform Series will be developed for all three gradients, starting with the uniform gradient.

5.2 UNIFORM GRADIENT The uniform gradient can be represented in the sketch below. The gradient does not start until the second period. The Future worth of the gradient can be expressed by Equation 5.1. The amount of the gradient is G, the interest rate is i and the number of periods is n.

50 // Strategic Cost Analysis for Project Managers and Engineers //

F = (n – 1)G + (n – 2)G(1 + i) + (n – 3)G(1 + i)2 + ... + 2G(1 + i)(n–3) + G(1 + i)(n–2)

(5.1)

If one multiplies Equation 5.1 by (1 + i) and then subtracts Equation 5.1 from that, one obtains Equation 5.2. Rearranging the terms and using the geometric series, the expression become Equation 5.3. (1 + i)F = (n – 1)G(1 + i) + (n – 2)G(1 + i)2 + (n – 3)G(1 + i)3+ + 2G(1 + i)(n–2) + G(1 + i)(n–1) F = (n – 1)G + (n – 2)G(1 + i) + (n – 3)G(1 + i)2 + (n – 4)G(1 + i)3 + +2G(1+i)(n–3) + G(1+i)(n–2) iF = –(n – 1)G + G(1 + i) + G(1 + i)2+ G(1 + i)3 + G(1 + i)(n–2) + G(1 + i)(n–1) (5.2) = –nG + G[1 + (1 + i) + (1 + i)2 + (1 + i)3+ + (1 + i)(n–3)+ (1 + i)(n–2) + (1 + i)(n–1)] = – nG + G[(1 + i)n – 1]/[(1 + i) – 1] = –nG + G[(1 + i)n – 1]/[i] (5.3) = G{[(1 + i)n – 1 –ni]/[i]} Solving for F one obtains the geometric gradient expression for the Future worth. F = G{[(1 + i)n – 1 – ni]/[i2]}

(5.4)

The formula, represented by (F/G, i, n), is used to convert a gradient G to F and is what appears in Table 5.1 at the end of this chapter. (F/G, i, n) = {[(1 + i)n – 1 –ni]/[i2]}

(5.5)

The value for the present worth of the uniform gradient can be obtained by: P = F/(1 + i)n = G{[(1 + i)n – 1 –ni]/[i2 (1 + i)n]} = G{[(1 + i)n – 1 –ni]/[i2 (1 + i)n]} = G (P/G, i, n)

// Gradient Expressions // 51

Thus the conversion formula to convert a uniform gradient to a present worth is: (P/G, i, n) = {[(1 + i)n – 1 – ni]/[i2 (1 + i)n]}

(5.6)

The value for the uniform series of the uniform gradient can be obtained by A = Fi/[(1 + i)n –1] = G{[(1 + i)n – 1 – ni]/[i2]} × i/[(1 + i)n –1] = G {[(1 + i)n – 1 – ni]/[i((1 + i)n – 1)] = G (A/G, i, n) Thus the conversion formula to convert a uniform gradient to a uniform series is: (A/G, i, n) = {[(1 + i)n – 1 – ni]/[i((1 + i)n –1)]

(5.7)

5.2.1 Uniform Gradient Example What would be the value of a gradient of $100 per year for a period of 10 years? The first payment would be at the end of the second year and the last payment at the end of the 10th year. The interest rate is 10%. (a)

What is the final payment? Payment at 10 year = (n – 1) * $100 = $900

(b) What is the total payment, not including the interest? Total Payments Made = [n(n + 1)/2] × $100 = (9 × 10/2) × $100 = $4,500 (c)

What is the total value including the compounding of interest at the end of year 10?

Total Value is Future worth which is: F = G{[(1 + i)n – 1 – ni]/[i2]} = G{[(1 + i)n –1]/i2 – n/i} = $100 {[(1.05)10 – 1]/(0.05)2 – 10/0.05} = $100 {251.55785 – 200} = $5,156 Note the effect of compounding interest results in a total interest gain of $656.

5.3 GEOMETRIC GRADIENT The geometric gradient can be represented in the sketch below. The gradient does not start until the second period. The present worth of the

52 // Strategic Cost Analysis for Project Managers and Engineers //

gradient can be expressed by Equation 5.7. The initial amount is A1 starts in period 1 and has the gradient amount g applied each of the follow periods up to the end n.

P = A1/(1 + i) + A1[(1 + g)/(1 + i)2] + A1[(1 + g)2/(1 + i)3] + A1[(1 + g)3/(1 + i)4] + ... + A1[(1 + g)(n–1)/(1 + i)n]

(5.7)

Rearranging the terms and using the geometric series, this equation can be reduced to: P = [A1/(1 + i)][1 + (1 + g)/(1 + i) + {(1 + g)/(1 + i)}2 + ... + {(1 + g)/(1 + i)}n] = [A1/(1 + i)]{[(1 + g)/(1 + i)]n – 1]/[(1 + g)/(1 + i) –1] = [A1/(1 + i)]{[(1 + g)/(1 + i)]n – 1]/[[(1 + g)–(1 + i)]/(1 + i)] = A1{[(1 + g)/(1 + i)]n – 1]/ [(g – i)] } = A1{[1 – (1 + g)/(1 + i)]n ]/[(i – g)]}

(5.8)

Thus the geometric gradient Present worth factor is (P/A1, g, i, n) = {[1 – (1 + g)/(1+i)]n ]/[ (i – g)] } If

(5.9)

i =g P = A1/(1 + i) + A1[(1 + g)/(1 + i)2] + A1[(1 + g)2/(1 + i)3] + A1[(1 + g)3/(1 + i)4] + ... + A1[(1 + g)(n–1)/(1 + i)n] = [A1/(1 + i)][1 + (1 + g)/(1 + i) + {(1 + g)/(1 + i)}2 + ... + {(1 + g)/(1 + i)}(n–1) ] = [A1/(1 + i)][ 1 + 1 + 1 + ... + 1] = nA1/(1 + i)

(5.10)

Thus one has a rather simple expression when i = g and the geometric gradient Present worth factor expression becomes: (P/A1, i = g, n) = {n/(1 + i)}

(5.11)

// Gradient Expressions // 53

The value for the Future worth of the uniform gradient can be obtained by multiplying by (1+i)n to obtain: F = P(1 + i)n = A1{[1 – (1 + g)/(1 + i)]n/ [(i – g)]} (1 + i)n = A1{[(1 + i)n – (1 + g)n]/[i – g]}

(5.12)

Thus the geometric gradient future worth factor is: (F/A1, g, i, n) = {[(1+i)n – (1 + g)n]/[i – g]}

(5.13)

If i = g, then the Future Worth factor for the geometric gradient is the Present Worth factor for the geometric gradient multiplied by (1 + i)n which is: F = P(1 + i)n = (nA1/(1 + i)) (1 + i)n F = [nA1(1 + i)n–1]

(5.14)

Thus a simple expression occurs when i = g and the geometric gradient Future Worth factor expression becomes: (F/A1, i = g, n) = {n(1 + i)n–1}

(5.15)

The value for the uniform series of the uniform gradient can be obtained by: A = A1(F/A1g, i, n) × (A/F, i, n) = A1{[(1+i)n – (1 + g)n]/[i – g]} {i/[(1 + i)n –1]}

(5.16)

Thus the geometric gradient uniform series factor is: (A/A1, g, i, n) = {[(1 + i)n – (1 + g)n]/[i – g]} {i/[(1 + i)n –1]}

(5.17)

If i = g, then the uniform series factor for the geometric gradient is the Future worth factor for the geometric gradient multiplied by [i/[(1 + i)n –1]] which is: A = F[i/[(1 + i)n–1]] = (nA1(1 + i)n–1)[i/[(1 + i)n –1]] F = [niA1(1 + i)n–1]/[i/[(1 + i)n –1]]

(5.18)

Thus the expression occurs when i = g and the geometric gradient uniform series factor expression becomes: (A/A1, i = g, n) = [ni(1 + i)n–1]/[i/[(1 + i)n –1]]

(5.19)

Example What would be the value at the end of 10 years of a geometric gradient of 10% if the initial amount was $100 for a period of 10 years? The first payment would be at the end of the first year and the last payment at the end of the 10th year. The interest rate is 10%.

54 // Strategic Cost Analysis for Project Managers and Engineers //

(a)

What is the final payment? $100 (1 + 0.10)10 = $100 (2.5937) = $259.37 = $259

(b) What is the total payment including the compounding of interest? F = A1 {[(1 + i)n– (1 + g)n]/[i – g]} = $100 {[(1 + 0.05)10 – (1 + 0.10)]}/[0.05 – 0.10] = $100 [– .9648478]/[– .05] = $100 (19.2969) = $1,930

5.4 ESCALATION GRADIENT Escalation is used commonly in construction projects as the cost of materials and labour are expected to increase over time and this is done with the escalation rate. Since engineering economy uses end of period payments, the escalation must start in the first period and this is the difference between escalation and the geometric gradient series. The sketch of the escalation payments and Present worth is below and the Present worth of the escalation can be expressed by Equation 5.20.

P = A1(1 + g)/(1 + i)+A1[(1 + g)2/(1 + i)2] + A1[(1 + g)3/(1 + i)3] + A1[(1 + g)4/(1 + i)4] +... + A1[(1 + g)n/(1 + i)n]

(5.20)

= [A1(1 + g) /(1 + i)] [ 1 + (1 + g)/(1 + i) + {(1 + g)/(1 + i)}2 +...+ {(1 + g)/(1 + i)}n] = [A1(1 + g)/(1 + i)]{[(1 + g)/(1 + i)]n – 1]/[(1 + g)/(1 + i) –1] = [A1(1 + g)/(1 + i)]{[(1 + g)/(1 + i)]n – 1]/ [[(1 + g) – (1 + i)]/(1 + i)] = A1{[(1 + g)/(g–i)] [(1 + g)/(1 + i)]n – 1]}

// Gradient Expressions // 55

P = A1{[(1 + g)/(g–i)] [1–(1 + g)/(1 + i)]n]}

(5.21)

Equation 5.21 is the expression for the escalation gradient and g is the escalation amount, and often the symbol e is used instead of g. The present worth of an escalation gradient is: (P/A1, g = e, i, n) = {[(1 + g)/(g – i)] [1 – (1 + g)/(1 + i)]n]} where

(5.22)

g = e for escalation

If i = g Equations 5.21 and 5.22 will have denominators that are zero, so the equations must be derived initially using the case where i = g, or i = e. Doing this for the Present worth one obtains: P = A1(1 + g) /(1 + i) + A1[(1 + g)2/(1 + i)2] + A1[(1 + g)3/(1 + i)3] + ... + A1[(1 + g)n/(1 + i)n] = [A1(1 + g)/(1 + i)][1 + (1 + g)/(1 + i) + {(1 + g)/(1 + i)}2 + ... + {(1 + g)(n–1)/(1 + i)}(n-1) ] = [A1 × 1][1 + 1 + 1 +...+ 1] P = nA1

(5.23)

Thus for this special case; (P/A1, i = g = e, n) = n

(5.24)

The equation for the Future worth of the escalation gradient can be determined from the Present worth equation by: F = A1 × (P/A1, g = e, i, n) × (F/P, i, n) F = A1 × {[(1 + g)/(g – i)][1 – (1 + g)/(1 + i)]n ]} × (1 + i)n F = A1 {[(1 + g)/(g – i)][(1 + i)n – (1 + g)n ]} (5.25) This also results in (F/A1, g = e, i, n) = {[(1 + g)/(g – i)] [(1 + i)n – (1 + g)n ]} (5.26) For the special case where i = g, then F = A1 × (P/A1, i = g = e, n) × (F/P, i, n) = A1 n (1 + i)n

(5.27)

(F/A1 i = g = e, n) = n(1 + i)n

(5.28)

And thus The equation for the uniform series of the escalation gradient can be determined from the Present worth equation by: A = A1 × (P/A1, g = e, i, n) × (A/P, i, n) A = A1 × {[(1+g)/(g – i)][1 – (1+g)/(1 + i)]n]} × {[i((1 + i)n/[((1 + i)n –1)]} A = A1 {[i(1 + g)/(g – i)] [(1 + i)n – (1 + g)n]/[(1 + i)n –1]}

(5.29)

56 // Strategic Cost Analysis for Project Managers and Engineers //

This also results in (A/A1, g = e, i, n) = {[i(1 + g)/(g – i)][(1 + i)n – (1 + g)n]/ (5.30) [(1 + i)n – 1]} For the special case where i = g, then A = A1 × (P/A1, i = g = e, n) × (A/P, i, n) = A1 {ni(1 + i)n/[((1 + i)n – 1)]} (5.31) And thus (A/A1 i = g = e, n) = {ni(1 + i)n/[((1 + i)n –1)]}

(5.32)

Example A project is expected to have a material cost of $100,000 per year at the end of the year before escalation. The escalation rate is 5%, the interest rate is 10 percent, and the project life is 4 years. What is the annual endof-year equivalent of this amount? A

= A1 {[i(1 + g)[(1 + i)n –(1 + g)n]]/[(i – g){(1 + i)n –1}]} = $100,000{[0.10(1 + 0.05)[1.104 – 1.054 ]/[(0.10 – 0.05){(1.10)4 – 1}] = $100,000 {0.105}[0.24859]/[(0.05)(0.4641)] = $100,000 (1.1249) = $112,490

5.5 RELATIONSHIPS BETWEEN FACTOR EXPRESSIONS The various Capital Recovery Factors are related and some of the more useful relationships are listed below. These relationships are often useful in solving problems or in obtaining other expressions. (F/A, i, n) = (P/A, i, n) × (F/P, i, n) (P/G, i, n) = (F/G, i, n) × (P/F, i, n) (A/P, i, n) – (A/F, i, n) = i

Or

(A/F, i, n) = (A/P, i, n) – i

(P/F, i, n) = (P/A, i, n) × (A/F, i, n) = (P/A, i, n) × ((A/P, i, n) – i) = 1 – i(A/P, i, n) Table 5.1 gives the discrete compounding factors of the engineering economic expressions presented in Chapters 4 and 5. These expressions can be converted to the continuous expressions as indicated in Chapter 4.

P F A P F A

Uniform Gradient Present worth Uniform Gradient Future worth Uniform Gradient Uniform Series Geometric Gradient Present worth Geometric Gradient Future worth Geometric Gradient Uniform Series

F A

Escalation Gradient Future worth

Escalation Gradient Uniform Series

Find P

Find P F A A F P

Factor Name Present worth Future worth (compound amount) Sinking Fund Capital Recovery Compound Amount Present worth

D. Escalation Expressions (Escalation Gradient) Factor Name Escalation Gradient Present worth

Geometric Gradient

C. Gradient Expressions Uniform Gradient

B. Uniform Payment (Uniform Series)

Payment Type A. Single Payment

G = uniform gradient amount

(P/G, i, n) (F/G, i, n) (A/G, i, n) (P/A1, g, i, n) If i = g (F/A1, g, i, n) If i = g (A/A1, g , i, n) If i = g

Symbol (P/F, i, n) (F/P, i, n) (A/F, i, n) (A/P, i, n) (F/A, i, n) (P/A, i, n)

Given Symbol A1, g (P/A1,g,i,n) If i = g A1, g (F/A1, g, i, n) If i = g A1, g (A/A1, g, i, n) If i = g

A1,g

A1,g

G G G A1,g

Given F P F P A A 2

Formula n n {[(1+g)/(i – g)] [1 – ((1 + g) /(1 + i) )]} {n} n n {[(1 + g)/(i – g)][(1 + i) – (1+g) ]} n {n(1 + i) } n n n {[i(1 + g)[(1 + i) – (1+g) ]]/[(i – g)[(1+ i) –1]]} (n) n {[ni(1 + i) ] / [(1+i) – 1]}

[((1 + i) – 1 – ni)/(i (1 + i) )] n 2 {((1+i) – 1 – ni)/i ) } n n {[(1+i) – 1 – ni]/[i((1+i) – 1)]} n n {[1– ((1 + g) /(1 + i) )]/[I – g]} {n/(1 + i)} n n {[(1 + i ) – (1 + g) ]/[I – g]} (n–1) {n(1 + i) } n n n {[i[(1 + i) – (1 + g) ]]/[(I – g){(1 + i) –1}]} (n-1) n [ni(1 + i) ] / [(1+i) – 1]

n

Formula -–n (1 + i) n (1 + i) n i / [(1 + i) – 1 ] n n [i(1 + i) ] / [(1 + i) – 1] n [(1 + i) –1] / i n n [(1 + i) – 1] / [i(1 + i) ] n

i = ieff = effective annual interest

n = number of periods

F = Future worth

Compounding Factors – Discrete Payments and Discrete interest

r = nominal annual interest

A = uniform end-of-period payment

Table 5.1. Discrete Compounding Factors of Engineering Economic Expressions

Notation: P = Present worth

// Gradient Expressions // 57

58 // Strategic Cost Analysis for Project Managers and Engineers //

5.6 EVALUATIVE QUESTIONS 1. If the time period is 5 years and the interest rate is 10%, find the following values: (a) (P/G, i = 10%, n = 5) (b) (F/G, i = 10%, n = 5) (c) (A/G, i = 10%, n = 5) 2. If the time period is 5 years and the interest rate is 10 % and the gradient is 5%, find the following values: (a) (P/A1, g = 5%, i, = 10%, n = 5) (b) (F/A1, g = 5%, i, = 10%, n = 5) (c) (A/A1, g = 5%, i, = 10%, n = 5) 3. If the time period is 5 years and the interest rate is 10 % and the escalation rate is 5%, find the following values: (a) (P/A1, g = e = 5%, i, = 10%, n = 5) (b) (F/A1, g = e = 5%, i, = 10%, n = 5) (c) (A/A1, g = e = 5%, i, = 10%, n = 5) 4. The cost of materials for a project is $1,000,000 per year for the next 4 years. The interest rate (return rate) is expected to be 15% and the escalation rate is predicted to be 5%. What is the expected Present worth of the materials for the project? 5. A contractor is building a high rise apartment which will take 3 years to build. The contractor predicts that the labour costs will be $2,000,000 per year and the materials will be $3,000,000 per year. Consider these as end of year expenses. He expects the labour costs to escalate by 5% per year and the materials to escalate by 8% per year. His expected return rate is 20%. (a) What is the Present worth of this project to the contractor? (b) The contractor wants the owner to give him 3 payments, each at the beginning of the year. How much should the payments be? 6. Develop an equation for the geometric gradient Present worth function for continuous interest and continuous gradient. Evaluate it with n = 5, i = 10% and g = 5%. 7.

Develop an equation for the uniform gradient Future worth for continuous interest and evaluate it with n = 5 and r = 0.10.

6

DEPRECIATION PARAMETERS AND METHODS 6.1 INTRODUCTION There are various terms used in the analysis of cash flows and definitions of these terms are given here. Some of the terms are revenues, expenses, taxes, depreciation, cash flows before taxes, cash flows after taxes, profits, property life etc. The terms and the relationships between the terms are presented in this Chapter with an emphasis upon depreciation techniques. Depreciation considerations are important in analyzing profits and cash flows after taxes. The basic terms are: 1.

Revenues = Income generated from Sales of Products and Services, Consultancy, Royalties, etc.

2.

Expenses = Money consumed in production of Products and Services.

3.

Depreciation is the systematic allocation of the cost of a capital asset over the recovery period (useful life).

4.

Tax Rate = A percentage amount applied to determine the amount of taxes.

The amount of taxes and profits are determined from these terms by the relationships: Taxes = (Revenues – Expenses – Depreciation) × Tax Rate

(6.1)

Profits = Revenues – Expenses – Taxes

(6.2)

60 // Strategic Cost Analysis for Project Managers and Engineers //

6.2 CASH FLOWS Cash Flows Before Taxes (CFBT), considers only net cash flows. This analysis is used when taxes cannot be considered and one is considering only the expenses. In general, one can consider: CFBT (Cash Flows Before Taxes) = Revenues – Expenses

(6.3)

CFAT (Cash Flows After Taxes), considers taxes as an additional expense. Taxes are calculated based on a percentage of taxable income, and this is generally approximately 40 percent in the USA, but can be zero in some of the oil rich countries or 70 percent in some of the European countries. The federal government tax rate in the USA was as high as 90 percent in the 1950’s, but was reduced during the last half of the 20th century. The 40 percent represents a typical amount based upon the total of the federal and state taxes. The taxable income is based upon the net cash flows less the depreciation expenses. Thus, the amount allowed for depreciation must be determined in order to calculate the taxable income and taxes paid. CFAT (Cash Flows After Taxes) = Revenues – Expenses – Taxes (6.4)

6.3 DEPRECIATION Depreciation is the systematic allocation of the cost of a capital asset over the recovery period (useful life). The best reference for depreciation in the USA is IRS Publication 946 – How to Depreciate Property (IRS website : www.irs.gov). Depreciation can be applied to nearly everything, but one notable exception is land. There are a wide variety of terms applied in depreciation as to depreciable property, depreciation life and depreciation techniques. The terms used here apply to the USA system of depreciation, but other nations may have some differences in the definitions and methods of depreciation.

6.3.1 Depreciable Property There are two major classes of depreciable property and they are tangible property and intangible property. Some examples of each are: A. Tangible Property (1)

Personal Property – assets such as houses, machines, vehicles, equipment, furniture, etc.

// Depreciation Parameters and Methods // 61

(2)

Real (Estates) Property – land and buildings erected or agricultural produce growing on the land. The land is not depreciable, but buildings erected are depreciable.

B. Intangible Property This includes property that has value, but that you cannot see or touch. It includes items such as IPR’s (Intellectual Property Rights), computer software, copyrights, royalty, franchises, patents, trademarks, trade names and permits/licenses.

6.3.2 Depreciation Life Depreciable items must have a useful life of one year or more, be used in business or used to manufacture, to produce income and loses value via obsolescence, wear and tear, or natural causes, and is not inventory, stock in trade, or investment property. The useful life is the period during which the asset is kept in productive use in a trade or business. The useful life is not how long the asset will last, but how long the owner expects to be able to use it in a productive manner. The Internal Revenue Service (IRS) determines the depreciation life and lists the various values in IRS Publication, 946. Physical life The physical life of the asset is the life over which the asset can be used. Economic life The economic life is the life which reduces the total cost of the asset, that is, the costs of owning and operating the asset. The economic life is often less than that recovery period or useful life, especially if the ADS (Alternative Depreciation System) recovery period is used instead of the GDS (General Depreciation System). Recovery period Recovery period is the life used for depreciation calculations. This is normally the useful life and can be obtained from, standard tables such as Table 6.1 MACRS Property Classes and Types of Property and Table 6.2 Class Life Asset Depreciation Range (ADR) System.

6.3.3 Depreciation Systems The MACRS (Modified Accelerated Cost Recovery System) systems are the depreciation systems typically used in the USA for depreciation. These

62 // Strategic Cost Analysis for Project Managers and Engineers //

systems involve more than one method of depreciation and the methods of depreciation will be covered separately. The MACRS systems generally follow the declining balance method with a switchover to the straight line method. The salvage value in the MACRS system is taken as zero in the calculation of the depreciation amounts. The two MACRS Systems that will be discussed are: 1. 2.

GDS – General Depreciation System ADS – Alternative Depreciation System Table 6.1. MACRS (Modified Accelerated Cost Recovery System) Property Classes and Types of Property

Property Class

Example Types of Property

3-Year Property

Trailer units for use over the roadways; special tools for: manufacture of motor vehicles, fabricated metal products, glass products, and rubber products

5-Year Property

Automobiles, taxis, buses, light trucks; heavy duty trucks, information systems-computers and peripherals; construction; manufacture of apparel; cutting of timber; manufacture of chemicals and allied products; manufacture of electronic components, products and systems.

7-Year Property

Machine tools, Equipment for the manufacture of cement, glass products, primary ferrous metals, foundry products, fabricated metal products, aerospace products, electrical and non-electrical machinery, motor vehicles, ship and boat building; and office furniture and fixtures. (If no life is specified, a product is generally classified as 7 year property).

10-Year Property

Petroleum refining equipment; equipment for the manufacture of food grains and grain mill products; ship and boat building dry docks.

15-Year Property

Pipeline transportation, water transportation, telephone distribution equipment; electrical utility, nuclear power plant; municipal waste water treatment.

20-Year Property

Electrical utility thermal power plant, gas utility production plants; municipal sewer; water utilities.

Residential Rental Property (27.5)

Rental structures (depreciated over 27.5 years)

Nonresidential Property (31.5)

Buildings (depreciated over 31.5 years)

// Depreciation Parameters and Methods // 63

6.3.3.1

MACRS–GDS (Modified Accelerated Cost Recovery System – General Depreciation System)

The GDS (General Depreciation System) is generally used unless ADS (Alternative Depreciation System) is required by law. The GDS is a faster depreciation schedule; that is more depreciation occurs in the earlier years than in the ADS. The detailed Recovery Period and Property Classes for the GDS are listed in Table 6.1 with detailed examples. These represent the most frequent used classes and a brief summary of the recovery period and the eight property classes are: 1. 3 yr. property (200% generally) 2. 5 yr. property (200% generally) 3. 7 yr. property – any property that does not have a class life specified is generally considered to have a 7 year class life. (200% generally) 4. 10 yr. property (200% generally) 5. 15 yr. property (150% generally) 6. 20 yr. property (150% generally) 7. Residential Rental Property (27.5 yrs, straight line) 8. Non-Residential Real Property (31.5 yrs, straight line) There is also a separate category for Indian Reservation Property (Indigenous, natives) which can have shorter recovery periods. The percentage is the declining percentage amount used for the early depreciation years before the conversion to the straight line depreciation. The GDS does permit other systems and they are used in special instances when accelerated depreciation is not preferred. Accelerated depreciation leads to lower taxes paid, but also leads to lower profits or even losses which may not be desired. A list of all the possible ADS systems are: 1. 200% declining balance for 3, 5, 7, and 10 year property. 2. 150% declining balance over a GDS recovery period for all property used in farming businesses (except real estates property) and for all other property in the 15 and 20 year property classes. 3. straight line for 3, 5, 7, 10, 15, and 20 year property as well as the residential rental property and non-residential real property. 4. 150% declining balance over an ADS recovery period for property in the 3, 5, 7 and 10 year property classes.

64 // Strategic Cost Analysis for Project Managers and Engineers //

6.3.3.2 MACRS-ADS (Modified Accelerated Cost Recovery System– Alternative Depreciation System) The ADS system results in equal or longer recovery periods with one or two exceptions than the GDS. For example, the personal property without a specified class life is 12 years compared to the 7 years in the GDS system. Some of the differences are: 1.

Class Life Asset Depreciation Range (ADR) System (Table 6.2)

2.

Any personal property without a class life specified is 12 years

3.

Any real estate property without a class life specified is 40 years

There are certain instances where the ADS must be used instead of the GDS. The ADS (Alternative Depreciation System) is required for: 1.

Any tangible property predominantly used outside the USA

2.

Any tax-exempt use property (churches, non-profit organizations, etc.)

3.

Any property predominantly used in farming or agricultural business

4.

Any imported property covered by the executive order of the President of the USA.

The primary systems of ADS used in practice are: 1. 150% declining balance over the ADS recovery period 2. Straight line over the GDS Recovery Period (farming) 3. Straight line of the ADS Recovery Period

6.3.4 Primary Factors Governing Depreciation The primary factors governing the determination of the amount of depreciation that can be claimed are: 1.

Recovery Period (The values for typical assets are described in Tables 6.1 and 6.2) The recovery period has been established for various asset classes and the recovery periods listed are Class Life, GDS (General Depreciation System) and AD S (Alternative Depreciation System). The GDS is generally the shortest and the ADS is longer and typically equal to the class life.

2.

Method of Depreciation (The method of depreciation tends to be one of the accelerated depreciation methods, such as the 200%, 150% or straight line).

// Depreciation Parameters and Methods // 65 Table 6.2 Life Asset Depreciation Range (ADR) System

Asset Class

Asset Description

00.11 00.12

Recovery Period (Years) Class General Alternative Life (Years) (GDS)* (ADS)✸

Office Furniture 10 Information Systems, including 6 computers 00.13 Data Handling, except computers 6 00.22 Automobile, Taxis 3 00.23 Buses 9 00.24 Heavy General Purpose Trucks 6 01.21 Cattle, Breeding or Dairy 7 01.223 Racehorse, above 2 years age – 10.0 Mining 10 13.0 Offshore Drilling 7.5 13.3 Petroleum Refining 16 15.0 Construction 6 24.4 Manufacture of Wood Products 10 30.1 Manufacture of Rubber Products 14 30.2 Manufacture of Plastic Products 11 32.1 Manufacture of Glass Products 14 33.2 Manufacture of Primary 14 Nonferrous Metals 33.3 Manufacture of Foundry Products 14 33.4 Manufacture of Primary Steel 15 Mill Products 34.0 Manufacture of Fabricated 12 Metal Products 35.0 Manufacture of Electrical Machinery & Other Mechanical Products 10 36.0 Manufacture of Electronic 6 Components / Products 36.1 Semiconductor Manufacture 5 Equipment 37.11 Manufacture of Motor Vehicles 12 37.2 Manufacture of Aerospace Products 10 39.0 Manufacture of Athletic, Jewellery 12 and other goods 48.37 Satellite Space Segment Property 8 (Satellites) 80.0 Theme and Amusement Parks 12.5

*GDS = General Depreciation System ✸ ADS

= Alternative Depreciation System

7 5

12 5

5 5 5 5 5 3 7 5 10 5 7 7 7 7 7

6 5 9 6 7 12 10 7.5 16 6 10 14 11 14 14

7 7

14 15

7

12

7

10

5 5

6 5

7 7 7

12 10 12

5

8

7

12.5

66 // Strategic Cost Analysis for Project Managers and Engineers //

3. Depreciation Convention (full year, mid-year methods). The convention typically used is the mid year convention, which assumes the asset was purchased and disposed in the middle of the year. The other conventions used are the mid-quarter and mid-month. When investments do not occur uniformly throughout the year, but mainly in one quarter or month. 4. Book Value: The book value of an asset is the asset investment less the cumulative depreciation for the asset. The book value is used in the declining balance methods, but the details provided in the IRS (U.S.A.) tables gives the complete values and calculations are not necessary. However, in other countries the value may be needed to calculate the depreciation amounts.

6.4 METHODS OF DEPRECIATION There are numerous methods of depreciation and only a few of the more commonly used methods will be presented in detail. The straight line method is the traditional method which gives a uniform amount of depreciation over the life of the asset. The declining balance method gives a uniform percentage of depreciation over the investment life which means larger amounts initially, but which has the problem of value not reaching zero. The MACRS system is a combination of both systems, giving the higher initial amounts of depreciation via the declining balance method and switching to the straight line to reach zero value. The production based system is not used in the USA, but is used in other parts of the world and is based on the amount of use of the facility. There are special depreciation methods, and in the USA, the method “Section 179” permits allowing the entire purchase to be depreciated in one year but there are several restrictions. Each of the methods will be present in more detail. 1.

Straight Line (a constant amount of depreciation per year)

The expression for straight line depreciation is Dk = (B – SV)/N Where Dk = Depreciation amount for year k (k = 1, N)

(6.5)

// Depreciation Parameters and Methods // 67

B = investment (purchase cost + installation costs of asset) SV = salvage value at end of life of asset at N years N = depreciable life of asset, years k = year of interest The Salvage Value is often taken as zero when using the MACRS straight line schemes, and the net disposal value would be treated as a capital gain. 2.

Declining Balance (a constant percent of depreciation of the book value) The expression for declining balance depreciation, for full year depreciation, is: Dk = BR(1 – R)k–1

(6.6)

Where Dk = Depreciation amount for year k(k = 1, N) k = year of interest for depreciation B = investment R = depreciation rate = (usually 150% or 200%)/(N × 100) N = Life of asset, years If the life is 10 years and the depreciation rate of 200% is used, then R = 200/(10 × 100) = 0.2. where R is a decimal less than 1.0. Example 1 What is the depreciation using mid-year convention for an investment of $10,000 with an asset life of 4 years using straight line and double declining balance (200%)? N = 4 (since N = 4 and half-year is used, it will take 5 years to fully depreciate the item) Assume the asset is disposed of in the last year. Year 0 1 2 3 4 5

Straight Line Depreciation Book Value ($) ($) 0 10,000 1,250 8,750 2,500 6,250 2,500 3,750 2,500 1,250 1,250 0

Declining Balance Book Value ($) 0 10,000 2,500 7,500 3,750 3,750 1,875 1,875 938 938 234 704

Depreciation ($)

68 // Strategic Cost Analysis for Project Managers and Engineers //

SL (straight line) = $10,000/4 = $2,500 First & last year is $1,250 DDB (Double Declining Balance) = 2/4 = 50% Year Year Year Year

1 2 3 4

= = = =

½( ½ × 10,000) 0.50 × 7,500 0.50 × 3,750 0.50 × 1875

= = = =

$2,500 $3,750 $1,875 $938

Year 5 = ½ (½ × 938) = $234 Note that the initial depreciation is much more rapid with the declining balance in the early years, but it never gets to zero and this is the major problem with the declining balance method of depreciation. The MACRS was developed to give both the rapid early depreciation and also to force the book value to reach zero. 3.

MACRS (Modified Accelerated Cost Recovery System)

This is a declining balance method with a switchover to straight line. This is the most commonly used system. The MACRS assumes a zero salvage value for all cases. Table 6.3 gives the depreciation percentages for the various recovery periods. Note that there are only certain recovery periods; that is 3, 5, 7, 10 and 15 years. Table 6.3 Depreciation Percentages for MACRS Recovery Periods Recovery year 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 & 19 18 & 20 21

Recovery Period Double Declining (200%) 150% Declining Balance Balance 3 yrs 5 yrs 7 yrs 10 yrs 15 yrs 20 yrs 33.33 20.00 14.29 10.00 5.00 3.750 44.45 32.00 24.49 18.00 9.50 7.219 14.81 19.20 17.49 14.40 8.55 6.677 7.41 11.52 12.49 11.52 7.70 6.177 11.52 8.93 9.22 6.93 5.713 5.76 8.92 7.37 6.23 5.285 8.93 6.55 5.90 4.888 4.46 6.55 5.90 4.522 6.56 5.91 4.462 6.55 5.90 4.461 3.28 5.91 4.462 5.90 4.461 5.91 4.462 5.90 4.461 5.91 4.462 2.95 4.461 4.462 4.461 2.231

// Depreciation Parameters and Methods // 69

The MACRS method assumes the half-year convention (also referred to as mid-year convention) and this is typically used. Other conventions are the mid-quarter convention and mid-month convention and must be used depending on the investment. For example, if 40% or more of the purchases are in one quarter, the mid-quarter convention must be used. The mid-month is primarily for non-residential real estates property, residential property, e.g. railroad grading or tunnel bore. If an alternative MACRS method is selected such as straight line, the salvage value is usually taken as zero. However, it may have a non-zero value but the depreciation amounts will be lower and that is not usually desired. Example 2 Consider the case of a $10,000 investment, zero salvage value and determine the MACRS values for 200% GDS method for 5 year property. Using the mid-year convention it will take six years as the first and last years obtain only ½ of a years depreciation value. Also, when an asset is sold in a year, only ½ of the depreciation is applied for that year. Complete Table 6.4. R = 200/(5 × 100) = 0.40 That is 40% depreciation per year. (Double Declining Balance) Table 6.4 MACRS Depreciation Calculations from Declining Balance and Straight Line Year (End of Year)

Life Remaining (End of Year)

0 1

Depreciation ($)

MACRS Depreciation

Declining Balance (DDB)

Straight Line* (SL)

Amount Used ($) (Largest of DDB and SL)

End of Year Book Value ($)

4½

0 2,000

0 1,000

0 2,000

10,000 8,000

2

3½

3,200

1,778

3,200

4,800

3

2½

4

1½

5

½

6

0

5

* Straight line is determined by the book value of the previous year divided by the remaining life.

70 // Strategic Cost Analysis for Project Managers and Engineers //

Note: If one takes the Amount Used column and divides by 100, one obtains the percentages in Table 6.3 for the Double Declining (200%) Balance column for a recovery period of 5 years. 4. Production Based Depreciation This depreciation method is used in cases where the extent of use of equipment depends on the production quantity or volume of production. This system is similar to that used in the USA for depletion of assets, such as oil, natural gas and minerals equipment. However, the production based depreciation system is not used in the USA. Example 3 A truck has a capacity of hauling 500,000 tons of coal during its lifetime and the initial cost of the truck is $25,000. During the third year, the truck hauls 75,000 tons. What is the depreciation for the third year? Depreciation = (75,000/5,00,000) × $25,000 = $3,750. 5. Special Depreciation Method – (Section 179) This Special Depreciation Method is unique to the USA and is intended to give small companies a method of rapid depreciation. The maximum amount of depreciation that could be allowed in 2010 was $250,000. When the amount exceeds $800,000, the maximum limit is reduced Dollar for Dollar. The total amount of depreciation must be less than $1,050,000 during 2010 to use any of Section 179 Depreciation method. There are special limitations on using Section 179 Depreciation on passenger automobiles or Sport Utility Vehicles (SUV). The asset must be used 100% exclusively for business use. If less than 100%, only that percentage can be used for depreciation and one must use the GDS system to calculate yearly amounts. Examples 1.

Velan bought a tractor for $300,000 during the year and that was the only purchase he has made. He could have a Special Depreciation of $250,000 and the basis for the remaining depreciation of the tractor would then be $50,000.

2.

Jones bought a small tractor for $55,000 during the year and that was the only purchase he has made. He could have a Special Depreciation of $55,000 and the tractor would be fully depreciated.

// Depreciation Parameters and Methods // 71

3. American Farms Inc., bought a tractor for $450,000 during the year and had a total depreciation under all their purchases of $900,000 in 2010. Thus the maximum amount that could be used under Section 179 would be {$250,000 – (900,000 – 800,000)} or $150,000. The remaining depreciation of the tractor would be $450,000 – $150,000 or $300,000.

6.5 REFERENCES 1. IRS Publication 946 – How to Depreciate Property, www.irs.gov

6.6 EVALUATIVE QUESTIONS 1. An asset was purchased for $80,000 and it took $20,000 to prepare the site and install the equipment. The asset has a recovery period of 7 years and MACRS depreciation was used. (a) What is the depreciation amount for the first year? (b) What is the book value at the end of the third year? (c) What is the depreciation amount for the 5th year? (d) What is the book value after the 6th year? 2. An asset has a value of $100,000 and a recovery period of 3 years. Use MACRS depreciation and determine the depreciation amount and book value over the life of the investment. 3. Your company has purchased a large new tractor trailer truck (heavy duty truck). It has a basic cost of $180,000 and with additional options costing $20,000, the cost basis for depreciation purpose is $200,000. Its market value at the end of five years is estimated as $30,000 and will be depreciated under the GDS: (a) What is the cumulative depreciation through the end of year three? (b) What is the MACRS depreciation in the fourth year? (c) What is the book value at the end of year two? (d) What is the book value at the end of year five? 4. Your company has purchased a large new tractor trailer truck (heavy duty truck). It has a basic cost of $180,000 and with additional options costing $20,000, the cost basis for depreciation purpose is $200,000. Its market value at the end of five years is estimated as $30,000 and will be depreciated under the ADS with straight line depreciation.

72 // Strategic Cost Analysis for Project Managers and Engineers //

(a) What is three? (b) What is (c) What is (d) What is

the cumulative depreciation through the end of year the MACRS-ADS depreciation in the fourth year? the book value at the end of year two? the book value at the end of year five?

5. You are a private consultant and purchased a new computer system valued at $3,000. You decide to use Special Depreciation Method (Section 179) for the computer system you purchased. (a) What is the first year depreciation amount? (b) What is the second year depreciation amount? 6. You have started a consultancy company and purchased a new computer system valued at $3,000. You decided to use MACRS Depreciation for the computer. What is the first year depreciation amount? 7.

A mining equipment was purchased for $500,000 and is expected to mine 5,000,000 tons of coal during its life. The machine mined 400,000 tons the first year and 700,000 tons the second year. (a) What is the amount of depreciation using the production based system? Year 1______ Year 2______ (b) If the G DS system is used, what would be the amounts of depreciation for the first two years? Year 1_____ Year 2_____

// Cash Flows, Profits, Taxes, Depreciation and Loans // 73

7

CASH FLOWS, PROFITS, TAXES, DEPRECIATION AND LOANS 7.1 INTRODUCTION The basic relationships between revenues, expenses, cash flows, profits, taxes and depreciation are reviewed and discussed here. Depreciation is considered as an expense, but it is not paid to someone else and is considered in determination of the amount of taxes to be paid. Loans consist of the principal and interest, but only the interest is considered as an expense. The loan and the principal do effect the cash flows. Therefore, the basic relationships will be repeated and some detailed examples are presented indicating the cash flows before taxes, the depreciation amounts, the net profits and cash flows after taxes with and without consideration of loans. The basic terms used in the various relationships are: Table 7.1 Symbols and Meaning of Cash Flow Terms Symbol

Meaning

R E TR T D CFBT CFAT CFBTAL LA B I U

Revenue Expenses Tax Rate (decimal) Taxes (amount of taxes) Depreciation Cash Flows Before Taxes Cash Flows After Taxes Cash Flows Before Taxes After Loan Loan Amount Principal Payment of Loan Interest Payment of Loan Unpaid Balance

NP

Net Profits

74 // Strategic Cost Analysis for Project Managers and Engineers //

The basic expression for Cash Flows Before Taxes is: CFBT = Revenues – Expenses CFBT = R – E

(7.1)

The basic expression for Cash Flows After Taxes is: CFAT = CFBT – Taxes CFAT = R – E – T

(7.2)

The net profits can be determined by: NP = (Revenues – Expenses – Depreciation) * (1 – Tax Rate) NP = (R – E – D) *(1 – TR)

(7.3)

The amount of Taxes can be determined by: T = (Revenue – Expenses – Depreciation) * Tax Rate T = (R – E – D) * TR

(7.4)

If one combines Equations 7.2 and 7.4, one obtains: CFAT = (R – E) – (R – E – D) * TR = (R – E – D) *(1 – TR) + D CFAT = NP + D

(7.5) (7.6)

7.2 LOAN CALCULATIONS Loan payments include principal and interest and the interest portions of the payments are a deductible expense. Thus one needs to determine the interest payments for tax considerations and the unpaid balance when the loan is to be paid back early. The various details w.r.t. loan are considered in the following example. Example A loan is to be made for $20,000 (P) at a 10 (i) percent interest rate and is to be repaid in 5 end of year payments. The amount of the payments for the loan can be determined, using the expression from Table 5.1 for A, given P, by: A = P (A/P, i = 10%, n = 5) = $20,000 × [(i(1 + i)n)/((1 + i)n –1)] = $20,000 × [(0.10(1 + 0.10)5 )/((1 + 0.10)5 – 1)] = $20,000 × [0.26387] = $5,276

// Cash Flows, Profits, Taxes, Depreciation and Loans // 75

The various factors considered concerning the loan are presented in Table 7.2. Note that the interest payments decrease as t increases and the principal payment increases as t increases, but that the total payment remains constant at $5,276. Thus in considering loan problems, the amount of interest in each period need to be calculated. Table 7.2 Loan Payment Parameter Values for a $20,000 loan at 10% interest with a 5 year payment plan End of Loan Year Total Amount at Payment LA time = t A years 20,000

0

Unpaid Total Total Balance Principal Interest at time t paid to paid to U(t) time t time t TP(t) TI(t) 20,000

0

0

3,276

2,000

1

5,276

2,000

3,276

16,724

2

5,276

1,672

3,604

13,120

6,880

3,672

3

5,276

1,312

3,964

9,156

10,844

4,984

4

5,276

916

4,360

4,796

15,204

5,900

0

20,000

6,380

5 TOTAL

Interest Principal Payment Payment in time t in time t I(t) B(t)

5,276

480

4,796

26,380

6,380

20,000

Parameters I(t) = Interest on payment in time period t B(t) = Principal on payment in time period t U(t) = Total Unpaid balance at the end of time period t Note: The unpaid balance at the beginning of period t is the same as the unpaid balance at the end of the previous period, t – 1.

7.2.1 Loan Formula Development The approach to the formulas is based upon the extension of the individual payments to determine the unpaid principal. This starts with the remaining principal: U(t) = Amount of principal to be paid in the remaining life of (n – t) time periods U(t) = Present worth of n-t payments of A (A is the uniform end of-period payment) = A*[P/A, i, n – t]

(7.7)

76 // Strategic Cost Analysis for Project Managers and Engineers //

The interest for the period, I(t), is the product of the unpaid balance at the beginning of time period t (which is the unpaid balance at the end of period t – 1) multiplied by the interest rate for the period (i). Thus the Interest Payment in Table 7.2, I(t), is calculated by I(t) = U(t – 1) * i = A[P/A, i, {n – (t – 1)}] * i

(7.8)

where U(t – 1) = A * [P/A, i, n – (t – 1)] The principal for the period, B(t), is the amount of the payment minus the interest for the period t ; that is B(t) = A – I(t) = A – A[P/A, i, {n – (t – 1)}]*i = A[1 – (P/A, i, {n – (t – 1)}*i] but (P/F, i, n) = 1– (P/A, i, n) * i and thus B(t) = A*[P/F, i , n – (t – 1)] Example calculations for t = 4 B(4) = $5,276 * [P/F, 10, 5 – (4 – 1)] = $5,276 * (P/F, 10, 2) = 5,276 * (0.8265) = $4,360 I(4) = = = =

$5,276 $5,276 $5,276 $5,276

* * * *

[P/A, 10, 5 – (4 – 1)] * i [P/A, 10, 2] * 0.10 [(1 + i)n – 1]/[i(1 + I)n] * i [(1.10)2 – 1]/[(1.10)2]

= $916 U(4) = $5,276 * [P/A, 10, 5 – 4] = 5,276 * (P/A, 10, 1) = $5,276 * (.9091) = $4,796 TP(4) = Total Principal paid after 4 payments = loan — U(t) = $20,000 – $4,796 = $15,204 TI(4) = Total Interest paid after 4 payments = total payments – total principal paid = 4 * 5, 276 – 15, 204 = $5,900

(7.9)

// Cash Flows, Profits, Taxes, Depreciation and Loans // 77

7.3 EXAMPLES IN CASH FLOW Two examples are presented in detail, one with the investment from retained earnings and one using a loan to pay for the investment. The investment from retained earning is presented in Table 7.3. In this example the present worth of the CFAT is negative whereas the present worth of the profits is positive. A positive value implies that the project met the required return plus excess, whereas a negative value implies the project did not meet the required return rate of 15%. Thus since CFAT is the required criteria, the project would not be accepted. The second example in Table 7.4 considers the effect of interest on the cash flows where the investment is from a loan for the total amount which is to be repaid in 3 years at 16 percent interest. The use of the loan changes present worth of the cash flows of the project from negative $350 to a positive $997 at the required return rate of 15%. However, the cumulative CFAT at zero required return is $6,000 for the retained earnings versus $2,978 for the loan. The negative $350 is caused by the large negative of $– 15,000 in the CFAT at time zero whereas the loan analysis has a zero CFAT at time zero. However, the Net Profits are lower when the loan is taken rather than making the investment from retained earnings. The tables presented include the totals of the various columns and these present checks on the calculations. For the interest example of Table 7.2, the total of the payments must equal the sum of the total interest and the total principal. In Table 7.3 the total of the revenues less the total of the expenses equals the total of the CFBT. In Table 7.4, the totals of the cash flows before taxes after the loan (CFBTAL) and the Taxable Income are equal. In the two examples of Tables 7.3 and 7.4, if the item is fully depreciated, the total of the Net Profits (NP) and the Cumulative CFAT are equal even though the values of the net profits and CFAT are quite different in years in which depreciation occurs. The use of CFAT as the key performance measure will be illustrated in the remaining chapters and various methods for evaluating project performance will be presented. There are several methods used as no single method has been accepted as the “best” method. The improvement in computation abilities with the use of computers and the various software packages has reduced the need for some of the approximation methods of the past.

78 // Strategic Cost Analysis for Project Managers and Engineers // Table 7.3 Cash Flow Analysis for Investment from Retained Earnings .

Parameter

Values

Investment

$15,000

Depreciation

MACRS with 3 year class life and half year depreciation

Project Economic Life

5 years

Tax Rate

40% (0.40)

Annual Revenues

$15,000

Annual Expenses

$10,000

Required Return

15%

Year Revenues

Expenses

CFBT

D (Depre- Taxable T NP CFAT ciation) Income (Taxes) (Net Rate Amt. Paid Profits) %

0

–

1

15,000

15,000 –15,000 10,000

5,000

2

15,000

10,000

5,000

3

15,000

10,000

5,000

4

15,000

10,000

5,000

7.41

5

15,000

10,000

5,000

Total 75,000

65,000

Cumulative CFAT

0

0

0

0

33.33

5,000

–

–

–

–15,000 –15,000

44.45

6,667 –1,667

–667

–1,000

14.81

2,222

2,778 1,111

1,667

3,889

1,111

3,889 1,556

2,333

3,444 + 3,000

5,000 2,000

3,000

3,000 + 6,000

10,000 100.00 15,000 10,000 4,000

6,000

6,000

5,000 –10,000 5,667 – 4,333 – 443

Note: 1. CFAT = CFBT – T 2. CFAT = NP + D (except in year zero) PW(CFAT, 15%) = –15,000 + 5,000/1.15 + 5,667/1.152 + 3,889/1.153 + 3,444/1.154 + 3,000/1.155 = – $350 PW(Profits, 15%)

= 0 – 1,000/1.152 + 1,667/1.153 + 2,333/1.154 + 3,000/1.155 = $3,166

The assumption is that the negative taxes are offset by profits earned by other projects and some governments do permit companies to recover losses in future profitable years.

40% (0.40)

Tax Rate

Project Economic Life Revenues (yearly) Expenses (yearly)

5,000

5,000

5,000

5,000

5,000

10,000

1

2

3

4

5

Total

CFBT

Year End (t)

–15,000

(2)

0

(2)

(1)

(1)

–5,037

0

0

–6,679

–6,679

–6,679

15,000

Loan Cash Flow

(3)

(3)

4,963

5,000

5,000

–1,679

–1,679

–1,679

0

CFBTAL

(4) = (2)+(3)

(4)

5,037

921

1,716

2,400

0

Loan Interest

(5)

(5)

100.00

7.41

14.81

44.45

33.33

0

15,000

1,111

2,221

6,668

5,000

0

Depreciation Rate Amount %

(6)

(6)

(8)

(9)

4,963

5,000

3,889

1,858

–3,384

–2,400

0

Taxable Income

1,985

2,000

1,555

743

–1,353

–960

Taxes Paid

2,978

3,000

2,333

1,115

–2,030

–1,440

Net Profits

(11)

3,000

3,445

– 2,422

–326

–719

CFAT

2,978

–22

– 3,467

–1,045

– 719

Cumulative CFAT

(10)=(4)–(8)

(10)=(9)+(6)+(5)+(3)

(10)

5 yrs $15,000/yr $10,000/yr

(9)=(7)–(8)

(7)=(2)–(5)–(6) (8)=(7)×TR

(7)

= 15,000 [(0.16(1 + 0.16)3 )/((1 + 0.16)3 –1))] = $15,000 * (0.44526) = $6,679

A = $15,000 (A/P, i = 16, n = 3) = 15,000[(i(1 + i)n)/((1 + i)n –1))]

Loan for $15,000 at 16% interest and repayment in 3 years

$15,000 MACRS with 3 year class life and half year convention

Investment Depreciation

Table 7.4 Cash Flow Analysis for Investment from a Loan

// Cash Flows, Profits, Taxes, Depreciation and Loans // 79

80 // Strategic Cost Analysis for Project Managers and Engineers //

There are two methods for calculating the CFAT indicated by the column equations under column 10 and they are: CFAT(t = 1) = NP + D – B(t) = NP – D –(A – I(t = 1) = –1,440 + 5,000 – (6,679 – 2,400) = –1,440 + 5,000 – 4,279 = $ – 719 CFAT(t = 1) = CFBTAL(t = 1) – T(t = 1) = –1,679 – (– 960) = $ – 719 The loan interest for period 2 is calculated by: I(2) = Ai(P/A, i, n – (t – 1) = $6,679 (0.16) (P/A,16%, 2) = $6,679 (0.16) [[(1.16)2 – 1]/(0.16)(1.16)2] = $1,716 Or = $6,679 [(1 +

0.16)2

–1]/[(1 + 0.16)2] = $1,716

I(2) implies n–(t – 1) = 3 – (2 – 1) = 2 I(3) implies n–(t – 1) = 3 – (3 – 1) = 1 I(3) = 1069(.16)/(.16 × 1.16) = $921 PW(CFAT,15%) = 0 – 719/1.15 – 326/1.152 – 2,422/1.153 + 3,445/1.154 + 3,000/1.155 = $ + 997 (versus $ – 350 when paid from retained earnings) PW(Profits, 15%) = – 1,440/1.15 – 2,030/1.152 + 115/1.153 + 2,333/1.154 + 3,000/1.155 = $ +771 (versus $ + 3,166 when paid from retained earnings)

7.4 EVALUATIVE QUESTIONS 1. A loan is taken for a flat in an Housing Colony Seventh Heaven. The home is priced at $450,000 and the mortgage is for $400,000 at 6% APR for 30 years and the payments are made monthly.

// Cash Flows, Profits, Taxes, Depreciation and Loans // 81

(a)

What is the mortgage payment?

(b) What is the interest on the 125th payment? (c)

What is the principal on the 125th payment?

(d) What is the total interest paid on the loan during the 30 years? (e)

What is the remaining principal amount after the 125th payment is paid?

(f)

What is the total interest paid after the 125th payment is paid?

2. Company ABC has purchased a new grinder for $300,000. The annual income (savings) from this machine is expected to be $100,000 and the annual expenses are expected to be $30,000. MACRS depreciation is used, with the 5 year class life and the tax rate is 40%. (a) What is the amount of depreciation for year 4? (b) What is the book value after year 4? (c)

What would be the income taxes due for year 4 assuming this is the only machine?

(d) What would be the total cash flows for year 4? (e)

What would be the net profits for year 4?

3. Resolve the problem in Table 7.3 using an investment of $12,000 instead of $15,000 and calculate present worth of the cash flows after taxes as well as the profits. 4. Resolve the problem in Table 7.4 using an investment of $12,000 instead of $15,000 and calculate the present worth of the cash flows after taxes as well as the profits. 5. An investment of $100,000 is made for purchase of a new machine with attachments which are expected to generate an annual revenue of $50,000 with annual expenses of $10,000 and the machine parts will have a salvage value of $10,000 when the project is completed at the end of 7 years. The equipment with attachments have a class life of 5 years, MACRS depreciation will be used and the income taxes are 30%. Determine the CFAT for all years, including year 0. The MARR is 15%. Assume the capital gains tax is the same as the income tax. (You may use the Table given in the next page)

Total

7

7

6

5

4

3

2

1

0

Year

Revenues

Expenses

Net CFBT % Amount

Depreciation Taxable Income

Taxes Paid

Net Profits CFAT

Cumulative CFAT

End of Year Book Value

82 // Strategic Cost Analysis for Project Managers and Engineers //

Check the calculations and complete the table.

// Project Evaluation Techniques (Basic Methods) // 83

PROJECT EVALUATION TECHNIQUES (BASIC METHODS) 8.1 INTRODUCTION There is no single method used to evaluate projects and several methods are commonly used. The methods more or less come to the same conclusion, but in some cases the results will be different. The first six techniques are the methods that have been commonly used for project evaluation. The commonly used basic techniques are: 1. Payback Period 2. Present Worth (PW) Analysis 3. Future Worth (FW) Analysis 4. Average Annual Equivalent Analysis (Average Annual Cost) 5. Return on Original Investment (ROI) 6. Return on Average Investment (RAI) In addition, there are some advanced methods/techniques that are more complex and will be presented in the next Chapter. These techniques involve either more complex calculations or additional information. These techniques are: 7. Internal Rate of Return (IRR) 8. Modified Internal Rate of Return (MIRR) 9. Benefits /Cost (B/C) Ratio, and 10. Project Balance (PB) (Positive and Negative)

–15,000 0 0 0 0 –15,000

CFBT Cumulative Depreciation Taxable (Revenues- Cash Flows Percent Amount Income Expenses) Before (Gross Taxes Profits) –15,000 –15,000 0 0 0 +5,000 –10,000 33.33 5,000 0 +6,000 – 4,000 44.45 6,668 – 667 +5,000 + 1,000 14.81 2,221 + 2,778 +4,000 + 5,000 7.41 1,111 + 2,889 +5,000 100.00 15,000 + 5,000 0 0 – 267 +1,112 +1,155 +2,000

Taxes (40%) 0 0 – 400 +1,667 +1,733 +3,000

Net Profits –15,000 + 5,000 + 6,268 + 3,888 + 2,844 + 3,000

–15,000 –10,000 – 3,732 + 155 + 3,000

Cash Flows Cumulative After Taxes Cash Flows (CFAT) After Taxes

–20,000 –15,000 – 9,000 – 2,000 + 3,000 + 7,500

0 33.33 44.45 14.81 7.41 0 100.00

0 6,667 8,890 2,962 1,481 0 20,000

–20,000 + 5,000 + 6,000 + 7,000 + 5,000 + 4,500 + 7,500

0 1 2 3 4 5 Total

–20,000 0 0 0 0 0 –20,000

CFBT Cumulative Depreciation (Revenues- Cash Flows Percent Amount Expenses) Before Taxes

Year Investment

0 –1,667 – 2,890 + 4,038 + 3,519 + 4,500 +7,500

Taxable Income (Gross Profits)

0 –667 – 1,156 + 1,615 + 1,408 + 1,800 +3,000

Taxes (40%)

0 –1,000 – 1,734 + 2,423 + 2,111 + 2,700 +4,500

Net Profits

– 20,000 + 5,666 + 7,156 + 5,385 + 3,593 + 2,700 + 4,500

– 20,000 – 14,334 – 7,178 – 1,793 + 1,800 + 4,500

Cash Flows Cumulative After Taxes Cash Flows (CFAT) After Taxes

Project B (Manufacturing-Additive) : An investment of $20,000 is made which will give a total return of $27,500 over 5 years. MACRS depreciation is used and the investment has a three year recovery period. The tax rate is 40 percent.

0 1 2 3 4 Total

Year Investment

Project A (Research Computer): An investment of $15,000 is made which will give a total return of $20,000 over 4 years. MACRS (Modified Accelerated Cost Recovery System) depreciation is used and the investment has a three year recovery period. The tax rate is 40 percent.

The company is large and has numerous projects such that tax losses can be offset by profits from other projects.

Table 8.1 Cash Flows Before Taxes, Depreciation, Gross Profits, Taxes, Net Profits, Cash Flows After Taxes, and Cumulative Cash Flows for Projects A and B

84 // Strategic Cost Analysis for Project Managers and Engineers //

// Project Evaluation Techniques (Basic Methods) // 85

The first six techniques will be illustrated with a sample problem utilizing the data in Table 8.1. The techniques can be used on a cash flow before taxes or a cash flow after taxes basis and the results will be compared. In many instances, the CFBT (Cash Flow Before Taxes) is easier to perform and explain and the results are frequently equivalent with respect to the final decision. However, the analysis should be made on an after tax basis (CFAT) whenever possible as these are the preferred results. The two projects in Table 8.1 involve Project A, a Research Computer with software to run simulation models to make better products. Project B, Manufacturing-Additive, is to purchase a new system to make prototypes by additive manufacturing. There does not appear to be sufficient funds to support both projects, so a decision is to be made to select one of the two projects.

8.1.1 Payback Period 8.1.1.1 Traditional Payback Period The payback period occurs in the year when the cumulative cash flows becomes positive. The payback period is usually used for relatively small investments and when quick decisions need to be made. Frequently the limiting payback period may be 1 year, and occasionally 2 or 3 years. Longer periods are generally not advised as the time value of money is not considered. Comparing the two alternatives from the data in Table 8.1, the traditional payback periods are presented in Table 8.2 as: Table 8.2 Traditional Payback Period Alternative

Project A (yrs.)

Project B (yrs.)

CFBT Payback Period

3 years

4 years

CFAT Payback Period

3 years

4 years

Some problems with the payback period analysis are: (a) No consideration is given to the benefits after the payback period (b) The investments have different lives and thus the payback periods would be expected to be different. (c) The magnitude of the cumulative cash flows makes no difference, only the sign.

86 // Strategic Cost Analysis for Project Managers and Engineers //

Since Project A has a shorter payback period in both CFBT and CFAT, it would be the preferred project based upon the traditional payback period. 8.1.1.2 Discounted Payback Period The time value of money can be used to discount the future cash flows and a more realistic payback period can be determined. Consider the details of Project A discussed with data in Table 8.1. If the cash flows are discounted at a rate of 8 percent, the payback periods would become four years instead of three years as determined by the non-discounted cash flows. It is recommended that the reader perform the calculations for Project B. The discounting of the cash flows takes the cash flows to time period zero. Thus cash flows that occur further in the future are discounted more than cash flows near time zero, the start of the project. The selection of the appropriate discount rate, which is used as the interest rate in the calculations, is important. Often analysts will interpret and have a value of 3.4 years for the discounted CFBT of project A instead of 4 years. This is not advisable as the cash flows are taken as end-of-year cash flows, and not as uniform cash flows throughout the year. Table 8.3 Discounted Payback Data for Project A (a)

Discounted Payback Period – CFBT Basis

Year 0 1 2 3 4

CFBT –15,000 5,000 6,000 5,000 4,000

Discounted CFBT –15,000 4,630 5,144 3,969 2,940

Cumulative Discounted CFBT –15,000 –10,370 – 5,226 –1,257 +1,683

(b) Discounted Payback Period – CFAT Basis

Year 0 1 2 3 4

CFAT –15,000 5,000 6,267 3,889 2,844

Discounted CFAT –15,000 4,629 5,373 3,087 2,091

Cumulative Discounted CFAT –15,000 –10,371 – 4,998 –1,911 +180

// Project Evaluation Techniques (Basic Methods) // 87

8.1.2 Present Worth Analysis The present worth method discounts all the payments back to time zero. The present worth can be of the cash flows, the profits or any other item that one wants to compare. For profits and cash flows, one generally wants to maximize the value. The factor for discounting is (1 + i)–n where i is the discounting factor and n is the number of periods the amount is to be discounted. The present worth for the cash flows before taxes, profits, and the cash flows after taxes for Project A can be calculated using the following data from Table 8.1. Table 8.4 Project A Data for CFBT, Profits, and CFAT from Table 8.1

Year

CFBT

0

–15,000

0

–15,000

1

5,000

0

+ 5,000

2

6,000

– 400

+ 6,268

3

5,000

+1,667

+ 3,888

4,000

+1,733

+ 2,844

+ 5,000

+3,000

+ 3,000

4 Total

Profits

CFAT

PW(CFBT–8%) = –15,000 + 5,000/(1 + 0.08) + 6,000/(1 + 0.08)2 + 5,000/(1 + 0.08)3 + 4,000/(1 + 0.08)4 = –15,000 + 4,627 + 5,144 + 3,969 + 2,940 = $ + 1,683 (if > 0.0 accept if single alternative or if greatest value among alternatives) PW(Profits–8%) = 0/(1 + 0.08) – 400/(1 + 0.08)2 + 1,667/ (1 + 0.08)3 + 1,733/(1 + 0.08)4 = 0 – 343 + 1,323 + 1,274 = $ + 2,254 (if > 0.0 accept if single alternative or if greatest value among alternatives) PW(CFAT–8%) = –15,000 + 5,000/(1 + 0.08) + 6,268/(1 + 0.08)2 + 3,888/(1 + 0.08)3 + 2,844/(1 + 0.08)4 = –15,000 + 4,629 + 5,373 + 3,087 + 2,091 = $180 (if > 0.0 accept if single alternative or if greatest value among alternatives) If the discount rate had been zero percent, then the present worth of the cash flows after taxes would be $3,000. In general, as the discount rate increases, the present worth decreases.

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If one calculates the present worth of the cash flows after taxes for Project B the result would be: PW(CFAT–8%) = – 20,000 + 5,666/(1 + 0.08) + 7,156/(1 + 0.08)2 + 5,385/(1 + 0.08)3 + 3,593/(1 + 0.08)4 + 2700/ (1 + 0.08)5 = – 20,000 + 5,247 + 6,135 + 4,275 + 2,641 + 1,837 = $ + 135 (if > 0 accept if a single alternative or if the greatest value among alternatives) Note the similarity between the discounted cash flows of the payback period and the present worth analysis; the discounted cash flows is the same as the present worth up to the payback period and then the present worth method keeps track of the additional cash flows. The problem with the present worth method is that if the alternatives have different lives, the least common multiple of the lives must be used as the study period. For example, for the two alternatives of 4 and 5 years, the study period would need to be 20 years. Since the two alternatives have different lives, we cannot compare the present worth values of $180 and $135 as the live of the projects are different.

8.1.3 Future Worth Analysis The future worth measures the economic value of the project at the end of the project life. The future worth can be of the cash flows, the profits or any other item that one wants to compare. For profits and cash flows, one generally wants to maximize the value. The factor for inflating is (1 + i)n where i is the investment factor and n is the number of periods the amount is to be invested. The future worth for the cash flows before taxes, profits, and the cash flows after taxes for the Project A can be calculated as follows: Table 8.5 Project A Data for CFBT, Profits and CFAT from Table 8.1

Year 0 1 2 3 4 Total

CFBT –15,000 5,000 6,000 5,000 4,000 + 5,000

Profits 0 0 – 400 +1,667 +1,733 +3,000

CFAT –15,000 + 5,000 + 6,268 + 3,888 + 2,844 + 3,000

// Project Evaluation Techniques (Basic Methods) // 89

FW(CFBT–8%) = $ – 15,000(1 + 0.08)4 + 5,000(1 + 0.08)3 + 6,000(1 + 0.08)2 + 5,000(1 + 0.08) + 4,000 = – 20,407 + 6,298 + 6,998 + 5,400 + 4,000 = $2,289 (if > 0.0 accept if single alternative or if greatest value among alternatives) FW(Profits–8%) = $ + 0.0/(1 + 0.08)3 – 400(1 + 0.08)2 + 1,667(1 + 0.08)1 + 1,733 = 0 – 467 + 1,800 + 1,733 = $3,066 (if > 0.0 accept if single alternative or if greatest value among alternatives) FW(CFAT–8%) = $ –15,000 (1.08)4 + 5,000 (1 + 0.08)3 + 6,267 (1 + 0.08)2 + 3,888(1 + 0.08) + 2,844 = $ – 20,407 + 6,298 + 7,310 + 4,200 + 2,844 = $ 245 (if > 0.0 accept if single alternative or if greatest value among alternatives) If the investment rate had been zero percent, then the future worth of the cash flows after taxes would be $3,000. In general, as the discount rate increases, the present worth decreases. If one calculates the future worth of the cash flows after taxes for Project B the result would be: FW(CFAT–8%) = $ –20,000(1.08)5 + 5,666(1 + 0.08)4 + 7,156(1 + 0.08)3 + 5,384(1 + 0.08)2 + 3,592(1 + 0.08) + 2,700 = $ –29,386 + 7,709 + 9,015 + 6,280 + 3,880 + 2,700 = $ + 198 (if > 0 accept if a single alternative or if the greatest value among alternatives) The problem with the future worth method is the same as that with the present worth when the alternatives have different lives, the least common multiple of the project lives i.e. Project duration must be used as the study period. For example, for the two alternatives of 4 and 5 years, the study period would need to be 20 years. Since the two alternatives have different lives, we cannot compare the future worth values of $245 and $198 if the project lives (or duration) are different.

8.1.4 Average Annual Equivalent Analysis The advantage of the average annual equivalent (sometimes called average annual cost) method is that it considers the costs (or cash flows or profits) on an annual basis and this overcomes the problem of investments having different lives. It is rather easy to convert a present worth to an

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average annual equivalent using the capital recovery factor, which was presented in Chapter 4. For Project A Average Annual Equivalent (CFAT–8%) = $180 * (0.08) * (1.0 + 0.08)4/{ (1.0 + 0.08)4 – 1.0} = $180 * (0.3019) = $54 (This is the average annual equivalent of the present worth per year of this investment) For Project B Average Annual Equivalent (CFAT–8%) = $135 * (0.08) * (1.0 + 0.08)5/{(1.0 + 0.08)5 – 1.0} = $135 * (0.2505) = $34 (This is the average annual equivalent of the present worth per year of this investment) Since Average Annual Equivalent Project A has a greater value than that of Project B one would select Project A over Project B using an 8 percent discount factor. If the discount factor is zero percent, the present worth of the cash flows for Project A is $3,000 and the average value would be $750 per year. The present worth of Project B is $4,500 and the average value would be $900 per year. Thus, if a zero discount rate had been used, Project B would be selected over Project A and thus what discount rate is used is a very important decision.

8.1.5 Return on Original Investment (ROI) The return on original investment (often called return on investment) can be considered on a undiscounted (not discounted) or discounted basis. The return on investment is the percentage relationship on the average annual profit to the original investment. 8.1.5.1 ROI – Not Discounted The basic form does not include discounting the profits. The formula used to obtain the values on a percentage basis is: ROI = (Average Yearly Profit/Original Fixed Investment) * 100 (8.1) For Project A the ROI would be: ROI = {[(0 – 400 + 1,667 + 1,733)/4)]/15,000} × 100 = {[(3,000)/4]/15,000} × 100 = 5%

// Project Evaluation Techniques (Basic Methods) // 91

8.1.5.2

ROI – Discounted

If the profits are discounted, the profits would be the present worth of the profits which was calculated previously as $2,254 in Section 8.1.2 and thus the discounted ROI would be: ROI (discounted) = {[(2,254)/4]/15,000} × 100 = 3.76% Thus there is a large difference between the ROI values depending whether discounting is considered. In general, the undiscounted method is used, but the discounted ROI tends to give a lower limit.

8.1.6 Return on Average Investment (RAI) The return on average investment (RAI) can also be calculated on an undiscounted (not discounted) or discounted basis. The difference in these methods is that the average investment is used, not the total investment. 8.1.6.1 RAI – Not Discounted The return on average investment is the percentage relationship on the average annual profit to the average outstanding investment. The basic form does not include discounting the profits. The formula used is: RAI = (Average Yearly Profit/Average Outstanding Fixed Investment) * 100 (8.2) One must calculate the average outstanding investment per year and that is the original investment minus the depreciation, commonly referred to as the Book Value. Table 8.6 Net Profits, Depreciation, and Book Value for Project A Year End 1 2 3 4

Net Profits 0 – 400 1,667 1,733

Depreciation 5,000 6,777 2,221 111

Book Value During Year 15,000 10,000 3,223 111

Average outstanding investment = (15,000 + 10,000 + 3,223 + 111)/4 = 28,334/4 = $7,084 Return on Average Investment (RAI) = {[(3,000/4)]/7,084} ×100 = 10.59% (versus 5.0%) 8.1.6.2

RAI – Discounted

If the profits are discounted, the profits would be the present worth of the profits which was calculated previously as $2,254 in Section 8.1.2 and thus

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RAI(discounted) = {[(2,254/4)]/7,084} × 100 = 7.95% When one is dealing with return on investment, it is very important as to which method is being used (ROI, RAI, discounted, undiscounted) as RAI percentages are approximately twice that of the ROI percentages. The best method is probably the undiscounted RAI (Return on Average Investment) method as neither the investment nor the profits are discounted. The average outstanding investment is more realistic of the investment situation over the life of the investment.

8.2 EVALUATIVE QUESTIONS 1.

An investment of $100,000 is made for a new machine which is expected to generate an annual revenue of $50,000 with annual expenses of $10,000 and the machine will have a salvage value of $10,000 when the project is completed at the end of 7 years. The equipment has a class life of 5 years, MACRS depreciation will be used and the income taxes are 30%. Determine the CFAT for all years, including year 0. The required return MARR is 15% and assume the capital gains tax is the same as the income tax.

50,000

50,000

50,000

50,000

10,000

360,000

4

5

6

7

7

Total

Net CFBT

170,000

0

10,000

10,000

10,000

10,000

10,000

10,000

10,000

190,000

10,000

40,000

40,000

40,000

40,000

40,000

40,000

40,000

100,000 –100,000

Expenses

100.00

0

0

5.76

11.52

11.52

19.20

32.00

20.00

0

100,000

0

0

5,760

11,520

11,520

19,200

32,000

20,000

0

Depreciation % Amount

190,000

10,000

40,000

34,240

28,480

28,480

20,800

8,000

20,000

0

Taxable Income

7,000

28,000

23,968

19,936

19,936

14,560

5,600

14,000

0

Net Profits

57,000 133,000

3,000

12,000

10,272

8,544

8,544

6,240

2,400

6,000

0

Taxes Paid

133,000

7,000

28,000

29,728

31,456

31,456

33,760

37,600

34,000

–100,000

CFAT

133,000

126,000

98,000

68,272

36,816

+ 5,360

–28,400

–66,000

–100,000

Cumulative CFAT

0

0

0

5,760

17,280

28,800

48,000

80,000

100,000

End of Year Book Value

1. (a) Determine the Payback Period in years. (b) Determine the Discounted Payback Period in years. 2. Determine the Present Worth of the project CFAT. 3. Determine the Future Worth of the project CFAT. 4. Determine the Average Annual Worth of the Project CFAT. 5. Determine the Return on Investment (ROI). 6. Determine the Return on Average Investment (RAI).

Answer the following questions using the data from Table 8.7, which is the solution to Problem 5 in Section 7.4.

50,000

50,000

2

50,000

1

3

—

0

Year Revenues

Table 8.7 Problem Data for Question 1

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9

PROJECT EVALUATION TECHNIQUES (ADVANCED METHODS) 9.1 INTRODUCTION The next four techniques are the internal rate of return, the modified internal rate of return, benefit/cost analysis and project balance. The modified internal rate of return was developed as the internal rate of return was rather complicated to calculate, but with computers and calculators the internal rate of return is easily calculated and the need for the modified internal rate of return is reduced. The benefit/cost analysis gives similar results as the present worth method, but often involves parameters involving the value of life. For example in the evaluation of a new medical drug to reduce death due to a dreaded disease, what is the value and the number of the lives saved? These issues are typically not involved in the traditional present worth analysis. The project balance method indicates not only the value of the project, but also the timing of the cash flows and gives an indication of the risk of the project. These methods are presented and discussed in more detail in the following sections.

9.2 INTERNAL RATE OF RETURN The rate of return, also called the Internal Rate of Return (IRR), is compared to the firm’s Minimum Acceptable Rate of Return (MARR). Calculating the rate of return is an important task to be carried out by

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Project Managers and Engineers. The advantage of the IRR is that it is a single value that can be compared for projects that have different recovery periods and different investment amounts. Thus it is one of the preferred methods, but it is more difficult to perform. The rate of return method involves calculating the interest rate that makes the present worth of the project equal to zero. Typically the present worth is positive at a zero rate of return and is reduced as the rate of return increases and when it becomes zero, then the project would earn the desired rate of return as indicated in Figure 9.1. The disadvantages of the IRR method is that the solution requires an iterative procedure which is time consuming, but with computers and software packages, this problem is greatly reduced. In a few instances, there may be more than one rate of return and other methods can be used to determine which rate of return should be used. In most instances of multiple rates of return, the lowest is the most appropriate value. The rate of return is the rate which makes the present worth equal to zero. The present worth of the CFAT for Project A in Table 8.1 is: PW(CFAT–i%) = –15,000 + 5000/(1 + i) + 6,267/(1 + i)2 + 3,888/ (1 + i)3 + 2,844/(1 + i)4 The first estimate is i = 0% which results in: PW(CFAT–0%) = –15,000 + 5000/(1 + 0.0) + 6,268/(1 + 0.0)2 + 3,888/ (1 + 0.0)3 + 2,844/(1 + 0.0)4 = –15,000 + 5,000 + 6,268 + 3,888 + 2,844 = $3,000 Since this is a positive value, one then takes a higher estimate, for example 5% PW(CFAT–5%) = –15,000 + 5000(1 + 0.05) + 6,268/(1 + 0.05)2 + 3,888/(1 + 0.05)3 + 2,844/(1 + 0.05)4 = –15,000 + 4,762 + 5,684 + 3,358 + 2,340 = $1,144 This indicates it is greater than 5%, but since the value is less than half, it is most likely less than 10%. Thus one can use interpolation such as: i = 5% + 5% (1144 – 0)/(3000 – 1144) = 5% + 3.08% = 8.08%

// Project Evaluation Techniques (Advanced Methods) // 97

The interpolation will tend to give low results, but it will be somewhat close to 8.08%, so try 8.2% PW(CFAT–8.2%)= –15,000 + 5000/(1 + 0.082) + 6,268/(1 + 0.082)2 + 3,888 /(1 + 0.082)3 + 2,844/(1 + 0.082)4 = –15,000 + 4,621 + 5,353 + 3,069 + 2,075 = $118 Now the error is small and the solution is near 8.2%, but is slightly higher. If one interpolates again, i = 8.2% = 3.2(118 – 0)/(1144 – 118) = 8.2 + 0.369 = 8.569, so try 8.6 % PW(CFAT–8.6%)= 15,000 + 5000/(1 + 0.086) + 6,268/(1 + 0.086)2 + 3,888/ (1 + 0.086)3 + 2,844/(1 + 0.086)4 = –15,000 + 4,603 + 5,314 + 3.036 + 2,045 = $ – 2 (This means it is slightly less than 8.6%, but 8.5% is further away).

Fig. 9.1: Present Worth Versus Rate of Return to Determine Internal Rate of Return (IRR)

In general, one wants to select the project with the highest rate of return. In other cases, one wants to know what the rate of return will be if the project is essential to the business. For example, if the firm’s MARR is 5%, then a return of 8.6% would be acceptable for the project. If, however, the firms MARR is 10%, the return of 8.6% would not be acceptable and the project should be rejected. One advantage of the Internal Rate of Return is that it indicates the amount by which the rate of return is above or below the MARR. The previous methods do not give an indication of the amount above or below the desired MARR.

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The IRR can be calculated with software packages, such as Excel. Using the IRR function in Excel, the value obtained was 8.593%. The software greatly reduces the effort in calculating the IRR.

9.3 MODIFIED INTERNAL RATE OF RETURN (MIRR) The Modified Internal Rate of Return (MIRR), also called the External Rate of Return, (ERR) is compared to the MARR (Minimum Acceptable Rate of Return) in a manner similar to that of the IRR (Internal Rate of Return). The advantages of the MIRR (Modified Internal Rate of Return) over the IRR (Internal Rate of Return) are: 1. It can be solved directly without trial and error techniques or special software programs. 2. It does not have multiple rates of return, only a single rate of return. The disadvantages are: 1. It is not the actual value of the IRR (Internal Rate of Return). 2. It requires the MARR (Minimum Acceptable Rate of Return) to arrive at the solution. This method takes the Future worth of the net positive cash flows and the Present worth of the net negative cash flows to determine the MIRR (Modified Internal Rate of Return). It can be expressed as: Future worth of positive cash flows at MARR (Min. Acceptable Rate of Return) (1+MIRR)n = Present worth of negative cash flows at MARR (Min. Acceptable Rate of Return)

(9.1)

Example Determine the MIRR (Modified Internal Rate of Return) for Project A if the MARR (Min. Acceptable Rate of Return) is 8%. Using the data from Table 8.1. Table 9.1 Cash Flows After Taxes for Project A, Table 8.1

Year 0 1 2 3 4

CFAT (Cash Flows After Tax) –15,000 + 5,000 + 6,268 + 3,888 + 2,844

// Project Evaluation Techniques (Advanced Methods) // 99

(1 + MIRR)4 = [5000(1.08)3 + 6,268(1.08)2 + 3,888 (1.08) + 2,844]/ [15,000] = [6,298 + 7,310 + 4,200 + 2,844]/[15,000] = [20,652]/[15,000] = 1.3768 (1 + MIRR) = 1.083 Thus, MIRR = 8.3% Note this is less than the 8.6% for the IRR (Internal Rate of Return) Since the M IRR (8.3%) is greater than the MARR of 8%, the execution of the project is acceptable. This is not an indication of the actual IRR, but more of a comparison with the MARR. It compares the positive and negative cash flows, and thus has similarities with the Benefit/ Cost Ratio.

9.4 BENEFIT/COST RATIO 9.4.1 Conventional Benefit /Cost Ratio The benefit cost ratio is an extension of the present worth method. It is used in many application areas, especially in government projects where the benefits include effects upon society and the environment in addition to the direct costs and benefits. The conventional benefit-cost ratio is the present worth of the benefits divided by the Present worth of the costs. If the benefits exceed the costs, the ratio will be greater than unity and this is equivalent to the total Present worth being greater than zero. The ratio is also dependent upon the rate of return. The criteria for acceptance of a project by present worth is: PW (project value) ≥ 0

(9.2)

If one divides the project into benefits (positive values) and costs (negative values), Equation 9.2 can be rewritten as: PW (benefits) – PW (costs) ≥ 0 Or PW (benefits) ≥ PW (costs) Now if one divides by the PW (costs), one obtains: PW (benefits)/PW (costs) ≥ 1 Equation 9.3 is the standard form of the benefit/cost ratio.

(9.3)

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Example If the rate of return is zero percent, what is the benefit-cost ratio for the cash flows after taxes for Project A? The benefit/cost ratio would be evaluated, with a zero discount rate as: Present worth of cash flow after tax benefits

= 5,000 + 6,268 + 3,888 + 2,844 = $18,000

Present worth of the cash flow after tax costs = $15,000 Benefit-Cost Ratio

= 18,000/15,000 = 1.20 > 1.0

Therefore at a zero rate of return, the benefit/cost ratio exceeds one and the project is acceptable. If the rate of return is 8 percent, the benefit-cost ratio is: Present worth of cash flow after tax benefits

= 4,629 + 5,373 + 3,087 + 2,091 = $15,180

Present worth of the cash flow after tax costs = $15,000 Benefit/Cost Ratio

= 15,180/15,000 = 1.012

The conventional benefit/cost ratio method also has the same disadvantage as the present worth method in that to use the ratio for the comparison of different alternatives, the study period must be the same. If the study period is the same for the two alternatives, the one with the highest benefit/cost ratio is the one most desired. If however, the highest benefit/cost ratio is less than one, that indicates the project is not acceptable. Example Here is a project involving installation of traffic signal lights and lane expansion. Traffic lights and lane expansion are being proposed for a dangerous intersection where numerous accidents have occurred in the past and on an average one fatality and six serious injuries occur per year. The installation of the signal and controls will cost $300,000, the lane expansion will cost $1,000,000, and the maintenance of the signal will cost on an average $20,000 per year and the additional maintenance for

// Project Evaluation Techniques (Advanced Methods) // 101

the lane expansion at $40,000 per year. With the traffic signal, the safety engineers expect the fatalities to be reduced by 30% and the serious injuries to be reduced by 40%. The cost of a fatality is estimated at $500,000 and a serious injury at $80,000. The discount rate for such projects is 5% and the project life is estimated to be 20 years. What is the benefit cost ratio? Present worth Cost Values are: Investment in signal and controls and lane expansion = 300,000 + 1,000,000 = $1300,000 Traffic light Maintenance Costs = $20,000(P/A, i = 5%, n = 20) = $20,000[(1.05)20 –1}/{0.05(1.05)20] = $249,244 Lane Maintenance Costs = $40,000(P/A, i = 5, n = 20) = $498,488 Present worth Benefit Values are: Fatality Savings = (0.30) × 500,000)(P/A, i = 5%, n = 20) = $150,000 × 12.46221 = $1,869,332 Injury Savings = (0.40) × 80,000 × 12.4622 = $398,790 Conventional Benefit/Cost Ratio = (1,869,332 + 398,790) /( 249,244 + 498,488 + 1,300,000) = 2,268,112/2,047,732 = 1.107 > 1 (thus acceptable, but value is close to 1)

9.4.2 Modified Benefit /Cost Ratio The modified benefit/cost ratio, also known as the Net Benefit/Cost Ratio, considers the net benefits, that is the annual benefits less the annual operating costs divided by the investment. The calculations can be done on Present Worth Values over the life of the project or using average annual costs. The values will be higher than the conventional Benefit/ Cost ratio, but the relationship with respect to 1 as to being greater or less than one will not change. It results in a higher benefit/cost number, which makes the project appear better. Modified Benefit/Cost =

Benefits – Operating Costs Investment Costs

(9.4)

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Example If one takes the data for the Traffic Light and Lane Expansion Example one has: Present worth Cost Values are: Investment Costs Investment in signal and controls and lane expansion = 300,000 + 1,000,000 = $1,300,000 Operating Costs Traffic light Maintenance Costs = $20,000(P/A, i = 5%, n = 20) = $20,000[(1.05)20 – 1}/{0.05(1.05)20] = $249,244 Lane Maintenance Costs

= $40,000(P/A, i = 5, n = 20) = $498,488

Present worth Benefit Values are: Fatality Savings = (0.30) × 500,000)(P/A, i = 5%, n = 20) = $150,000 × 12.46221 = $1,869,332 Injury Savings = (0.40) × 80,000 × 12.4622 = $398,790 Modified Benefit/Cost = (1,869,332 + 398,790 – 249,244 – 498,488) / (Total Project Life) (1,300,000) = 1,520,390/1,300,000 = 1.1695 > 1 (thus acceptable, and larger than the conventional Benefit-Cost ratio) If one uses annual costs, the analysis must take the investment cost to an annual cost. Investment Costs Annual Investment in signal and controls and lane expansion = $1,300,000 * (A/P, i = 5%, n = 20) = $1,300,000[0.05(1 + 0.05)20]/[(1 + 0.05)20 –1] = $104,315 Operating Costs Annual Traffic light Maintenance Costs = $20,000 Annual Lane Maintenance Costs = $40,000

// Project Evaluation Techniques (Advanced Methods) // 103

Present Worth Benefit Values are: Fatality Savings = (0.30) × 500,000 = $150,000 Injury Savings = (0.40) × 80,000 = $32,000 Modified Benefit/Cost = ($150,000 + $32,000 – $20,000 (Annual Basis)

– $40,000)/ $104,315 = $122,000 / $104,315 = 1.1695 > 1 (thus acceptable, and larger than the conventional Benefit – Cost ratio)

Thus the modified benefit/cost analysis gives the same results, whether on a total life basis (life cycle cost) or on an annual basis.

9.5 POSITIVE AND NEGATIVE PROJECT BALANCES 9.5.1 Introduction The concept of Project Balance (PB) is to determine the project balance at the end of each period (year) and then sum up the negative and positive balances. This procedure often allows one to select between projects which may have similar present worth evaluations. This method is to be used for comparing projects which have the same study period. The project balance calculates the value of the project at the end of each period, so it is future worth approach. Example Project A If one takes Project A, from Table 8.1 the Future worth of the project as calculated in Section 8.3 would be: FW(CFAT–8%) = $ –15,000(1.08)4 + 5,000(1 + 0.08)3 + 6,267(1 + 0.08)2 + 3,888(1 + 0.08) + 2,844 = $ –20,407 + 6,298 + 7,310 + 4,200 + 2,844 = $ + 245 (if > 0.0 accept if single alternative or if greatest value among alternatives) The project balance approach calculates the project balance at the end of each period to determine the balance for each year and then check whether it is negative balance or positive balance. The shaded area is the project balance times the period length, which is one for all periods except for the last value which has a zero length.

104 // Strategic Cost Analysis for Project Managers and Engineers // Project Balance (PB)

Cumulative Balance Balance (–)

PB0 = –15,000

=

– 15,000

PB1 = –15,000* (1 + 0.08) + 5,000

=

– 11,200

PB2 = –11,200* (1 + 0.08) + 6267

=

– 5,829

PB3 = – 5,829* (1 + 0.08) + 3,888

=

– 2,407

PB4 = – 2,407*(1 + 0.08) + 2,845

=

Total

Balance (+)

Area Balance Negative

Positive

= –15,000 *1 = – 15,000 = –11,200 *1 = –11,200 = – 5,829 *1 = – 5,829 = – 2,407 *1 = – 2,407 = + 245

= + 245 = – 34, 436

= + 245

The project balance can be used as an indicator of the risk of the project in that large negatives imply that the returns do not come until late in the project life. A diagram of the project balance indicates the relative risk.

Fig. 9.2: Project Balance Diagram for Project A

The total area of the shaded region is $ –34,436 and there is no positive area, so the project is “risky” as the project doesn’t become profitable until the end of the last period. Example Project X Project (X) with an initial cost of $10,000 and an annual net cash flow after taxes of $3,074 per year for 4 years is to be considered. The return rate is 8 percent and the Present worth would be:

// Project Evaluation Techniques (Advanced Methods) // 105

PW(CFAT–8%) = –10,000 + 3,074/(1 + 0.08) + 3,074/(1 + 0.08)2 + 3,074 (1 + 0.08)3 + 3,074/(1 + 0.08)4 = –10,000 + 2,846 + 2,635 + 2,440 + 2,259 = $180 This Present worth is identical to Project A as calculated in Section 8.1.2. The project balance, however, is different, as indicated in the following calculations. The negative project balance for Project X, at $ –25,616 is much less than the negative project balance for Project A, at $ –34,436. Thus although the Present worth values are the same over the same project lives, the risk for loss is much greater for Project A than for Project X.

Project Balance (PB)

Cumulative Balance (–) Balance (+) Balance

Area Balance Negative

PB0 = –10,000

= –10,000

= –10,000 *1 = –10,000

PB1 = –10,000*(1 + 0.08) + 3,074

= – 7,726

= – 7,726 *1 = – 7,726

PB2 = – 7,726*(1 + 0.08) + 3,074

= – 5,271

= – 5,271 *1 = – 5,271

PB3 = – 5,271*(1 + 0.08) + 3,074

= – 2,619

= – 2,619 *1 = – 2,619

PB4 = – 2,691*(1 + 0.08) + 3,074 = + 245 Total

Positive

= + 245 = – 25,616

Fig. 9.3: Project Balance Diagram for Project X

= + 245

290,000

0 40,000 40,000 40,000 40,000 40,000 40,000 40,000 10,000

CFBT

Net

100.00

0 20.00 32.00 19.20 11.52 11.52 5.76 0 0

%

100,000

0 20,000 32,000 19,200 11,520 11,520 5,760 0 0

Amount

Depreciation

190,000

0 20,000 8,000 20,800 28,480 28,480 34,240 40,000 10,000

Income

Taxable

57,000

0 6,000 2,400 6,240 8,544 8,544 10,272 12,000 3,000

Paid

Taxes

133,000

0 14,000 5,600 14,560 19,936 19,936 23,968 28,000 7,000

Profits

Net

133,000

–100,000 34,000 37,600 33,760 31,456 31,456 29,728 28,000 7,000

CFAT

–100,000 – 66,000 – 28,400 +5,360 36,816 68,272 98,000 126,000 133,000

CFAT

Cumulative

100,000 80,000 48,000 28,800 17,280 5,760 0 0 0

Book Value

Year

End of

The required Minimum Acceptable Rate of Return (MARR) is 15%. (a) Determine the Internal Rate of Return (IRR) for the project. (Note that the Present worth of the cash flows after taxes with zero interest is $133,000. Combine the two cash flows for year seven into a single cash flow. If the MARR is given, it is usually good to take that value as the starting point.) (b) Determine the MIRR (Modified Internal Rate of Return) for the project. (Since MIRR is easier to calculate than the IRR, you may want to solve for the MIRR and use it as an estimate of the IRR).

170,000

Total 360,000

Expenses

100,000 10,000 10,000 10,000 10,000 10,000 10,000 10,000 0

nues

Reve-

— 50,000 50,000 50,000 50,000 50,000 50,000 50,000 10,000

0 1 2 3 4 5 6 7 7

Year

1. An investment of $100,000 is made for a new machine which is expected to generate an annual revenue of $50,000 with annual expenses of $10,000 and the machine will have a salvage value of $10,000 when the project is completed at the end of 7 years. The equipment has a class life of 5 years, MACRS (Modified Accelerated Cost Recovery System) depreciation will be used and the income taxes are 30%. Determine the CFAT (Cash Flow After Tax) for all years, including year 0. The MARR (Min. Acceptable Rate of Return) is 15% and the capital gains tax is the same as the income tax.

9.6 EVALUATIVE QUESTIONS 106 // Strategic Cost Analysis for Project Managers and Engineers //

0

0

0

0

2

3

4

5

–20,000

0

1

Total

– 20,000

Investment

0

Year

+ 7,500

+ 4,500

+ 5,000

+ 7,000

+ 6,000

+ 5,000

+ 7,500

+ 3,000

– 2,000

– 9,000

–15,000

– 20,000

Before Taxes

Expenses) – 20,000

Cumulative Cash Flows

CFBT (Revenues-

100.00

0

7.41

14.81

44.45

33.33

0

20,000

0

1,482

2,962

8,890

6,666

0

Depreciation % Amount

+7,500

+4,500

+3,518

+4,038

– 2,890

– 1,666

0 – 666

0

Taxes (40%)

+ 3,000

+1,800

+ 1,407

+ 1,615

– 1,156

(Gross Profits)

Taxable Income

+ 4,500

+ 2,700

+ 2,111

+ 2,423

– 1,734

– 1000

0

Net Profits

+ 4,500

+ 2,700

+ 3,593

+ 5,385

+ 7,156

+ 5,666

–20,000

(CFAT)

+ 4,500

+ 1,800

– 1.793

– 7,178

–14,334

– 20,000

After Taxes

Cash Flows Cumulative After Taxes Cash Flow

Project B: An investment of $20,000 is made which will give a return of $27,500 over 5 years. MACRS depreciation is used and the investment has a three year recovery period. The tax rate is 40 percent.

2. Two alternative public works projects are under consideration to prevent flooding. If the MARR is 5%, which of the projects should be selected? Use both the conventional and modified Benefit/Costs Ratios. Project R Project C Capital Investment $900,000 $750,000 Annual Operations & Maintenance $100,000 $80,000 Annual Benefit $180,000 $140,000 Useful Project Life (years) 40 30

// Project Evaluation Techniques (Advanced Methods) // 107

108 // Strategic Cost Analysis for Project Managers and Engineers //

3. Use the data given for Project B (Additive Manufacturing) and determine the following: (a) Payback Period based upon CFAT (b)

Present Worth of CFBT, CFAT, Profits, Depreciation using a 8% rate of return.

(c)

Average Annual Value of CFAT

(d)

Rate of Return on CFAT

(e)

Benefit/Cost Ratio on CFAT

(f)

Area of Negative Balances

Draw the picture showing the Project Balance at the end of each year.

10

RISK ANALYSIS 10.1 INTRODUCTION Projects involve making decisions about future events - such as cash flows, revenues, return rates, expenses (including taxes) but the future is inherently uncertain. In the previous discussions about economic analysis, it has been assumed that the cash flows were known with certainty. Although these types of analysis can provide a reasonable decision basis for most situations, one should also consider situations where the cash flows and other parameters are subject to some degree of uncertainty.

10.1.1 Risk and Uncertainty Risk refers to situations where cash flows and other economic parameters are not known with certainty, but which can be described by some array of outcomes whose probabilities can be estimated. These probabilities can be either discrete or continuous. Uncertainty, or decision making under uncertainty, implies the probabilities cannot be predicted. The techniques for considering decisions under conditions of uncertainty are more advanced. In most cases, decision under uncertainty implies decision under risk. Some of the approaches for considering risk are: 1. Sensitivity Analysis 2. Optimistic-Pessimistic Analysis (Scenario Analysis) 3. Probability Analysis (Discrete and Continuous) Each of the techniques will be discussed and will be illustrated by a basic example. The examples will emphasize the methodology adopted. The methodologies will be presented with the easier methods first and the more difficult (probabilistic) methods later.

110 // Strategic Cost Analysis for Project Managers and Engineers //

10.2 SENSITIVITY ANALYSIS Sensitivity analysis is a non-probabilistic technique to provide information about the potential impact of selected parameters in the estimate. It is often calculated as the change in Present worth as a function of a percentage change in one variable while the other variables are fixed. Some of the variables commonly used are selling price, capacity utilization, useful life, and escalation.

10.2.1 Example from Metal Casting A group of investors are planning to start a foundry in an area where a new automobile engine plant is being constructed. The investors believe that there will be a strong market for castings. The plant will cost $2.5 million Dollars and will have a capacity of 20,000 pounds of castings per day. The castings will vary in size from 1/2 to 100 pounds per casting. The plant will have 30 direct and indirect labor employees, which will cost about $30,000/employee in salary and benefits. The plant will also have 5 management and sales employees who will cost approximately $60,000 in salary, benefits, and other costs. The average selling price will be approximately $1.50/lb and the average total materials and energy costs will be $1.00 per pound. The plant will have an initial life of about 10 years and a return rate of 15% is desired. The expected capacity is 75% (that is 15,000 pounds of castings per day) and single shift operation is planned for 250 days/year. The data for analysis is in Table 10.1. Table 10.1: Metal Casting Investment Data

Investment Annual Revenue:

= $2,500,000

15,000 lb/day * 1.50 $/lb * 250 day/yr = $5,625,000 Total Annual Revenue $5,625,000 Annual Expenses: Labor - Direct & Indirect 30 employees * $30,000 /yr = $900,000 Management and Sales Staff 5 employees * $60,000 /yr

= $300,000

Materials and Energy Cost : 15,000 lb/day * 1.00 $/lb * 250 day/yr = $3,750,000 Total Annual Expenses

= $4,950,000

Total Annual Revenue – Total Annual Expenses = $675,000

// Risk Analysis // 111

The Present worth of the investment was: PW(Investment@15%) = $ – 2,500,000 + $675,000*[P/A, i = 15, n = 10] = $ – 2,500,000 + $675,000 * 5.0188 = $887,690 Therefore, the investment appears to be successful. The next steps are to investigate the sensitivity of the PW(15%) to changes in selling price, operating capacity, and investment life.

10.2.2 Sensitivity Analysis (Based on Selling Price Variation) The selling price will be evaluated at a 20% decrease, 10% decrease, 10% increase and 20% increase. The selling price must be made a variable in the evaluation of the PW(15%) expression. The expression for revenue becomes: Revenue = 15,000 lb/day * X $/lb * 250 day/yr = 3,750,000 X X is the selling price ($/pound) Total Revenue – Total Annual Expenses = 3,750,000 X – 4,950,000 and PW(15%) = – 2,500,000 + (3,750,000 X – 4,950,000) *[P/A, i = 10, n = 10] = – 2,500,000 + (3,750,000 X – 4,950,000) *[5.0188] = – 27,343,060 + 18,820,500 X Solving for various costs: Sales Price ($ / Ib)

Price Change (%)

PW (15%)

1.20

–20

–4,758,460

1.35

–10

–1,935,385

1.50

0

1.65

+10

3,710,765

1.80

+20

6,533,840

887,670

Thus one observes that the operation is very sensitive to Selling Price variation, as a 10 percent decrease in selling price converts the project from a successful operation to very unsuccessful operation.

10.2.3 Sensitivity Analysis (Based on Capacity Variation) The operating capacity will be evaluated similarly to the selling price changes at 20% decrease (60% capacity or 12,000 lb), 10% decrease (67.5%

112 // Strategic Cost Analysis for Project Managers and Engineers //

capacity or 13,500 lb), 10% increase (82.5% capacity or 16,500 lb) and 20% increase (90% capacity or 18,000 lb). The capacity is in pounds, so the revenue and material costs must be expressed in pounds. The variable Y will represent the capacity in 10,000 lb units. Revenue = Y lb/day * $1.50/lb * 250 day/yr = 375 Y($/yr) Materials and Energy Cost = Y lb/day * $1.00/lb * 250 day/yr = 250 Y($/yr) Total Revenue –Total Annual Expenses = 375Y – 250Y – 1,200,000 = 125Y – 1,200,000 Therefore PW(15%) = – 2,500,000 + [125Y – 1,200,000] *[P/A, i = 15, n = 10] = – 2,500,000 + [125Y – 1,200,000] * [5.0188] = 627.35Y – 8,522,560 Solving for various capacity levels: Capacity (lb) 12,000 13,500 15,000 16,500 18,000

Variation % – 20 – 10 0 10 20

PW (15%) – 994,360 –53,335 887,690 1,828,715 2,769,740

Thus one observes that the operation is sensitive to capacity variation, as a 10 percent decrease in capacity level converts the project from a successful operation to slightly unsuccessful operation.

10.2.4 Sensitivity Analysis (Based on Plant Life Variation) The useful plant life of the facility can be considered to vary from 8 to 12 years. This will affect the [P/A, i, n] expression. The PW(15%) expression would become: PW(15%) = – 2,500,000 + 675,000 * [P/A, i = 15, n = 8, 9, 10, 11, 12] Solving for various plant lives Useful Life 8 years 9 years 10 years 11 years 12 years

% Change –20 –10 0 10 20

PW (15%) 528,928 720,830 887,690 1,032,748 1,146,755

// Risk Analysis // 113

Thus one observes that the operation is not very sensitive to plant life variation as a 10 percent decrease in plant life reduces the present worth of the project by only about 20 percent and the project has a positive present worth over the entire project life range.

10.3 OPTIMISTIC-PESSIMISTIC ANALYSIS Optimistic-Pessimistic Analysis, also called Scenario Analysis, is used to evaluate variation in more than one variable. It usually considers three cases for each variable - a “worst-case” (pessimistic), “most-likely case”, and “best-case” (optimistic). Often more than one variable is analyzed in the study. The worst case is one which would be exceeded in 9 out of 10 (or 19 out of 20) cases and the best case would be exceeded in only 1 out of 10 (or 1 out of 20) cases. This is a simple and effective method for including risk in the analysis.

10.3.1 Example: CNC Machine Replacement A new machining center (CNC machine) is required for a R&D Laboratory to produce the parts required for conducting experiments and for research. The increased research capability is expected to increase the research funding. Two major concerns in the analysis are the expected life of the machine (before obsolescence) and the net revenue generated in research contracts from the use of the machine. The following three scenarios were developed and the desired rate of return is 10%. Table 10.2 CNC Machine Replacement Scenario Data

Scenarios Optimistic (O) Capital Investment ($) Investment Life (years) Net Annual Revenue ($)

200,000

Most-Likely (ML)

Pessimistic (P)

200,000

200,000

10

7

4

150,000

75,000

10,000

The present worth of the three scenarios were evaluated as: PW-O(10%) = – 200,000 + 150,000[P/A, i = 10, n = 10] = – 200,000 + 150,000[6.1445] = $721,675 PW-ML(10%) = – 200,000 + 75,000[P/A, i = 10, n = 7] = – 200,000 + 75,000[4.8684] = $165,130

114 // Strategic Cost Analysis for Project Managers and Engineers //

PW-P(10%) = – 200,000 + 10,000[P/A, i = 10, n = 4] = –200,000 + 10,000[ 3.1698 ] = $ – 168,302 These three scenarios give an average present worth of $239,501 and a range of $889,977. The question is, the problem of the short life or the low revenue, or both? A more detailed analysis evaluates the PW values for all nine combinations of Investment Life and Net Annual Revenue. The results are: Present worth ($) values for all 9 combinations Investment Life (years) 10 Net Annual Revenue ($1,000)

¸ ˝ ˛

7

4

150

721,675

530,260

275,350

75

260,888

165,130

37,675

10

– 138,555

– 151,316

–168,302

This analysis indicates that the low revenues are the primary cause and not the short life. Only when the revenues are low does the project lose money, and it does so at all three lives considered. The overall average is $170,312 on equal weight basis or $165,968 using 5% optimistic, 90% most likely, and 5% pessimistic basis. Although 1/3 of the values are negative, the probability of the low revenue is only about 5%, so the probability of a loss is approximately 10%.

10.4 PROBABILISTIC METHODS A random variable is a parameter (or variable) that can have more than one possible value (or outcome). The value of the random variable can take isolated (discrete) values or have values in a certain interval (continuous). The value of the random variable is unknown until the event occurs, but the probability of that event occurring is known in advance and can be described by its distribution (probability density function), which can be either continuous or discrete. The measures of mean and variance are used to describe the “expected value” and “variation” of the possible outcomes of the analysis. The expected values and variances for the discrete and continuous functions are given by:

// Risk Analysis // 115 N

E(X) = µ =

∑ i =1

pi xi (discrete case)

(10.1)

N = number of different outcomes pi = probability of ith event occurring E(X) = µ =

U

∫ = x * f(x) dx (continuous case)

(10.2)

L

U = upper limit, L = lower limit f(x) = distribution function Var(X) = σ2 = Σ [xi – µ]2 pi (discrete case) = Σ pi (xi)2 – [Σ pi xi]2 σ2 = E(x2) – [E(x)]2 Var(X) =

σ2

(10.3)

U

=

∫

[x – E(x)]2 f(x) dx (continuous case)

L

U

=

∫ x2 f(x) dx – [E(x)]2 L

σ2

= E(x2) – [E(x)]2

(10.4)

10.4.1 Discrete Probability Models Example The metal casting foundry example in Section 10.2 had the following capacity values and Present worth values. The probabilities for each capacity level are as assigned, so determine what would be the expected value and variance of the capacity and the expected value of the Present worth? Table 10.3 Metal Casting Foundry Discrete Probability Capacity (1000 lb units) y i

Probability Pi

12.0 13.5 15.0 16.5 18.0

.1 .3 .3 .2 .1

E(Y) = 14.85 E(Y2) = 223.425

PW (1,000 $ units) –944 –53 888 1,828 2,770

Expected Capacity P i yi 1.2 4.05 4.50 3.30 1.80

P i yi

2

14.4 54.675 67.5 54.45 32.4

116 // Strategic Cost Analysis for Project Managers and Engineers //

The expected value, or mean of the expected capacity is 14.85 or 14,850 pounds. The variance can be calculated from Equation 10.3 by: σ2 = E(Y2) – E(Y)2 = 223.425 – (14.85)2 = 223.425 – 220.5225 = 2.9025 The standard deviation(σ) is the square root of the variance and thus is: σ = (2.9025)1/2 = 1.704 The standard deviation is 1,704 pounds and the variance is 2,903,000. (The term 1,000 is squared in the variance calculations.) The expected value of the Present worth can be found from the expected value of the capacity or by calculating the expected value of the Present worth using the probability values. Using the expression: PW(15%) = $627.35 Y – 8,522,650 = $627.35 * 14,850 – 8,522,650 = $9,316,148 – 8,522,650 = $793,498 Since the mean is lower than the original capacity, the PW(15%) is also lower than the original PW value of $887,690. This occurs because in the original data, the values of capacity under 15,000 lbs had a higher total probability than those above 15,000 lbs. PW(15%) = 0 when Y = 13,585 lbs.

10.4.2 Continuous Probability Models There are numerous continuous probability distributions, but the two which will be considered are the normal distribution and the triangular distribution. The normal distribution is the most commonly used distribution of all the distributions, but the triangular distribution is commonly used in estimating and determining ranges of cost data. 10.4.2.1 Normal Distribution The normal distribution is frequently assumed for many problems either correctly or incorrectly. Some of the properties about means and variances of distributions are: N E(ΣX) = E(X1) + E(X2) + E(N) 1

(10.5)

// Risk Analysis // 117

E(aX1 + bX2) = aE(X1) + bE(X2)

(10.6)

N Variance(ΣX) = Variance(X1) + Variance(X2) +...+Variance(XN) (10.7) 1 σ = standard deviation = Square Root (variance) Standard Deviation (aX1 + bX1) = [a2 variance X1 + b2 variance X2]1/2 (10.8) Probability(X < b) = Φ[(b – µ)/σ] (10.9) Values of Φ(z) are in Table 10.4 Example (Normal Distribution) Use Equation 10.9 and Table 10.4 to answer the following questions to gain familiarity in obtaining probability values. Probability Calculations A. If the mean of a normal distribution is 10 and the standard deviation is 4, what is the probability that a random variable selected from that distribution is less than zero? B.

Prob(X < 0) = Φ[(0 – 10)/4] = Φ[– 2.5] = .006 What is the probability that the random variable selected is less than 6? Prob(X < 6) = Φ[(6 – 10)/4] = Φ[– 1] = .159

C. What is the probability that the random variable selected is greater than or equal to 6? Prob(X ≥ 6) = 1 – Prob(X < 6) = 1 – .159 = .841 Example (Cash Flow) The cash flows from a project are presented in the following table. The required rate of return is 15%. What is the Present worth of the expected cash flow on the project and what is the probability that the cash flow from the project has a loss? Year

Expected Cash Flow($) µ

Standard Deviation of Cash Flow σ

Variance of Cash Flow σ 2

0

–7,000

100

10,000

1

3,000

200

40,000

2

4,000

300

90,000

3

3,000

400

160,000

118 // Strategic Cost Analysis for Project Managers and Engineers //

The initial steps are to calculate the mean, variance and standard deviation of the Present worth of the cash flows at the required rate of return. This can be done by: PW(15%) = – 7,000 + 3,000[P/F, 15, 1] + 4,000[P/F, 15, 2] + 3,000[P/F, 15, 3] = –7,000 + 3,000 * 0.8695 + 4,000 * 0.7561 + 3,000 * 0.6575 = $605.40 µ = PW(15%) = $605.40 Table 10.4: Probability values for different Z values in increments of 0.05 Probability

Probability

Z value –3.0 –2.9 –2.8 –2.7 –2.6 –2.5 –2.4 –2.3 –2.2 –2.1

0.00 0.001 0.002 0.003 0.004 0.005 0.006 0.008 0.010 0.014 0.018

0.05 0.001 0.002 0.002 0.003 0.004 0.005 0.007 0.009 0.012 0.016

Z value 3.0 2.9 2.8 2.7 2.6 2.5 2.4 2.3 2.2 2.1

0.00 0.999 0.998 0.997 0.996 0.995 0.994 0.992 0.990 0.986 0.982

0.05 0.999 0.998 0.998 0.997 0.996 0.995 0.993 0.991 0.988 0.984

–2.0 –1.9 –1.8 –1.7 –1.6 –1.5 –1.4 –1.3 –1.2 –1.1

0.023 0.029 0.036 0.045 0.055 0.067 0.081 0.097 0.115 0.136

0.020 0.026 0.032 0.040 0.049 0.061 0.073 0.088 0.106 0.125

2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1

0.977 0.971 0.964 0.955 0.945 0.933 0.919 0.903 0.885 0.864

0.980 0.974 0.968 0.960 0.951 0.939 0.927 0.912 0.894 0.875

–1.0 –0.9 –0.8 –0.7 –0.6 –0.5 –0.4 –0.3 –0.2 –0.1 0

0.159 0.184 0.121 0.242 0.284 0.308 0.345 0.382 0.421 0.460 0.500

0.147 0.171 0.198 0.237 0.258 0.291 0.326 0.363 0.401 0.440 0.480

1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0.841 0.816 0.788 0.758 0.726 0.692 0.655 0.618 0.579 0.540 0.500

0.853 0.829 0.802 0.773 0.742 0.709 0.674 0.637 0.599 0.560 0.520

// Risk Analysis // 119

V(PW) = σ(PW)2 = 12 * 1002 + (.8695)2 * 2002 + (.7561)2 * 3002 + (.6575)2 * 4002 = 1 * 104 + 3.024 * 104 + 5.1451 * 104 + 6.9169 * 104 = 16.086 * 104 σ(PW) = 4.01 * 102 = 401 P(PW < 0) = Φ[(0 – 605.4)/401] = Φ[–1.509] ≈ 0.07 This indicates that although the mean value of the Present worth is $605.4, there is a 7% chance that the project will LOSE money; that is, it will have a present worth less than zero. 10.4.2.2 Modified Beta/Triangular Distribution The Beta distribution, is used in PERT (Programme Evaluation and Review Technique) and the triangular distribution uses a high (optimistic), low (pessimistic), and most likely values. The theory behind these distributions assume that the high and low values are known (or can be estimated) and are the end points of the distribution. The normal distribution has infinity (plus and minus) as its end points and this is not practical in most cost analysis situations. It is very difficult to estimate the end points, and Cooper and Davidson (1) have modified the parameters so the end points are 10 percent values; that is the low estimate implies there is only a 10 percent chance of having a value lower than this estimate and the high estimate implies there is only a 10 percent chance of having a value higher than this estimate. Using this definition, their expressions for the mean and standard deviation are: µ = [H + 2M + L]/4 σ = [H – L]/2.65

(10.10) (10.11)

where, H = High estimate L = Low estimate M = Most Likely Value The equations used in PERT are very similar, but they assume the high and low values are the actual high and low values. Those equations are: µ = [H + 4M + L]/6 σ = [H – L]/6

(10.12) (10.13)

120 // Strategic Cost Analysis for Project Managers and Engineers //

Example (Cash Flow Analysis) A present worth cash flow analysis, in million Dollar units, was performed using the Cooper and Davidson equations using the assumption that there is only a 10 percent chance that the value will be higher or a 10 percent chance that it will be lower. This represents a 80 percent confidence range and the following values were obtained: Most Likely

Range (%)

Low

High

µ

σ

Revenue

125

–20/+10

100

137.5

121.9

14.15

Expenses

40

–10/+20

36

48

41

4.52

Investment

70

–5/+5

66.5

73.5

70

2.64

Cash Flow

+15

µ = +10.9

σ2 = 14.152 + 4.522 + 2.642 = 227.62 σ = 227.621/2 = 15.087 = 15.1 P[CF