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Series ISSN: 1939-5221 Series ISSN: 1939-5221 Series ISSN: 1939-5221 SYNTHESIS LECTURES ON ENGINEERING SYNTHESIS LECTURES ON ENGINEERING SYNTHESIS LECTURES ON ENGINEERING Series Editor: Steven F. Barrett, University of Wyoming Series Editor: Steven F. Barrett, University of Wyoming Series Editor: Steven F. Barrett, University of Wyoming Fundamentals of Engineering Fundamentals of Engineering Fundamentals of Engineering Economics and Decision Analysis Economics and Decision Analysis Economics and Decision Analysis David L. Whitman, University of Wyoming David L. Whitman, University of Wyoming David L. Whitman, University of Wyoming Ronald E. Terry, Brigham Young University Ronald E. Terry, Brigham Young University Ronald E. Terry, Brigham Young University The authors cover two general topics: basic engineering economics and risk analysis in this text. Within The authors cover two general topics: basic engineering economics and risk analysis in this text. Within The authors cover two general topics: basic engineering economics and risk analysis in this text. Within the topic of engineering economics are discussions on the time value of money and interest relationships. the topic of engineering economics are discussions on the time value of money and interest relationships. the topic of engineering economics are discussions on the time value of money and interest relationships. These interest relationships are used to define certain project criteria that are used by engineers and These interest relationships are used to define certain project criteria that are used by engineers and These interest relationships are used to define certain project criteria that are used by engineers and project managers to select
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These interest relationships are used to define certain project criteria that are used by engineers and project managers to select the best economic choice among several alternatives. Projects examined will project managers to select the best economic choice among several alternatives. Projects examined will project managers to select the best economic choice among several alternatives. Projects examined will include both income-and service-producing investments. The effects of escalation, inflation, and taxes include both income-and service-producing investments. The effects of escalation, inflation, and taxes include both income-and service-producing investments. The effects of escalation, inflation, and taxes on the economic analysis of alternatives are discussed. Risk analysis incorporates the concepts of probability on the economic analysis of alternatives are discussed. Risk analysis incorporates the concepts of probability on the economic analysis of alternatives are discussed. Risk analysis incorporates the concepts of probability and statistics in the evaluation of alternatives. This allows management to determine the probability of and statistics in the evaluation of alternatives. This allows management to determine the probability of and statistics in the evaluation of alternatives. This allows management to determine the probability of success or failure of the project. Two types of sensitivity analyses are presented.The first is referred to as success or failure of the project. Two types of sensitivity analyses are presented.The first is referred to as success or failure of the project. Two types of sensitivity analyses are presented.The first is referred to as the range approach while the second uses probabilistic concepts to determine a
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analyses are presented.The first is referred to as the range approach while the second uses probabilistic concepts to determine a measure of the risk the range approach while the second uses probabilistic concepts to determine a measure of the risk the range approach while the second uses probabilistic concepts to determine a measure of the risk involved. The authors have designed the text to assist individuals to prepare to successfully complete the involved. The authors have designed the text to assist individuals to prepare to successfully complete the involved. The authors have designed the text to assist individuals to prepare to successfully complete the economics portions of the Fundamentals of Engineering Exam. economics portions of the Fundamentals of Engineering Exam. economics portions of the Fundamentals of Engineering Exam. About SYNTHESIs About SYNTHESIs About SYNTHESIs This volume is a printed version of a work that appears in the Synthesis This volume is a printed version of a work that appears in the Synthesis This volume is a printed version of a work that appears in the Synthesis Digital Library of Engineering and Computer Science. Synthesis Lectures Digital Library of Engineering and Computer Science. Synthesis Lectures Digital Library of Engineering and Computer Science. Synthesis Lectures provide concise, original presentations of important research and development provide concise, original presentations of important research and development provide concise, original presentations of important research and development topics, published quickly, in digital and print formats. For more information topics, published quickly, in digital and print formats. For
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topics, published quickly, in digital and print formats. For more information topics, published quickly, in digital and print formats. For more information topics, published quickly, in digital and print formats. For more information visit www.morganclaypool.com visit www.morganclaypool.com visit www.morganclaypool.com Mor gan Cl aypool Publishers Mor gan Cl aypool Publishers Mor gan Cl aypool Publishers & & & ISBN: 978-1-60845-864-6 ISBN: 978-1-60845-864-6 ISBN: 978-1-60845-864-6 90000 90000 90000 w w w . m o r g a n c l a y p o o l . c o m w w w . m o r g a n c l a y p o o l . c o m w w w . m o r g a n c l a y p o o l . c o m 9 78 1 608 458646 9 78 1 608 458646 9 78 1 608 458646 W W W H H H I I I T T T M M M A A A N N N • • • T E R R Y T E R R Y T E R R Y F F F U U U N N N D D D A A A M M M E E E N N N T T T A A A L L L S S S O O O F F F E E E N N N G G G I I I N N N E E E E E E R
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E N N N G G G I I I N N N E E E E E E R R R I I I N N N G G G E E E C C C O O O N N N O O O M M M I I I C C C S S S A A A N N N D D D D D D E E E C C C I I I S S S I I I O O O N N N A A A N N N A A A L L L Y Y Y S S S I I I S S S M M M o o o r r r g g g a a a n n n & & & C C C l l l a a a y y y p p p o o o o o o l l l CM& Mor gan Cl aypool Publishers CM& Mor gan Cl aypool Publishers CM& Mor gan Cl aypool Publishers & & & Fundamentals of Fundamentals of Fundamentals of Engineering Economics Engineering Economics Engineering Economics and Decision Analysis and Decision Analysis and Decision Analysis David Whitman David Whitman David Whitman Ronald E. Terry Ronald E. Terry Ronald E. Terry SYNTHESIS LECTURES ON ENGINEERING SYNTHESIS LECTURES ON ENGINEERING SYNTHESIS LECTURES ON ENGINEERING Steven F. Barrett, Series Editor Steven F. Barrett, Series Editor Steven F. Barrett, Series Editor Fundamentals of Engineering Economics and Decision
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F. Barrett, Series Editor Steven F. Barrett, Series Editor Steven F. Barrett, Series Editor Fundamentals of Engineering Economics and Decision Analysis Synthesis Lectures on Engineering Editor Steven S. Barrett, University of Wyoming Fundamentals of Engineering Economics and Decision Analysis David L. Whitman and Ronald E. Terry 2012 A Little Book on Teaching: A Beginner’s Guide for Educators of Engineering and Applied Science Steven F. Barrett 2012 Engineering Thermodynamics and 21st Century Energy Problems: A Textbook Companion for Student Engagement Donna Riley 2011 MATLAB for Engineering and the Life Sciences Joseph V. Tranquillo 2011 Systems Engineering: Building Successful Systems Howard Eisner 2011 Fin Shape Thermal Optimization Using Bejan’s Constructal Theory Giulio Lorenzini, Simone Moretti, and Alessandra Conti 2011 Geometric Programming for Design and Cost Optimization (with illustrative case study problems and solutions), Second Edition Robert C. Creese 2010 Survive and Thrive: A Guide for Untenured Faculty Wendy C. Crone 2010 Geometric Programming for Design and Cost Optimization (with Illustrative Case Study Problems and Solutions) Robert C. Creese 2009 iii Style and Ethics of Communication in Science and Engineering Jay D. Humphrey and Jeffrey W. Holmes 2008 Introduction to Engineering: A Starter’s Guide with Hands-On Analog Multimedia Explorations Lina J. Karam and Naji Mounsef 2008 Introduction to Engineering: A Starter’s Guide with Hands-On Digital Multimedia and Robotics Explorations Lina J. Karam and Naji Mounsef 2008 CAD/CAM of Sculptured Surfaces on Multi-Axis NC Machine: The DG/K-Based Approach Stephen P. Radzevich 2008 Tensor Properties of Solids, Part Two: Transport Properties of Solids Richard F.
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Machine: The DG/K-Based Approach Stephen P. Radzevich 2008 Tensor Properties of Solids, Part Two: Transport Properties of Solids Richard F. Tinder 2007 Tensor Properties of Solids, Part One: Equilibrium Tensor Properties of Solids Richard F. Tinder 2007 Essentials of Applied Mathematics for Scientists and Engineers Robert G. Watts 2007 Project Management for Engineering Design Charles Lessard and Joseph Lessard 2007 Relativistic Flight Mechanics and Space Travel Richard F. Tinder 2006 Copyright © 2012 by Morgan & Claypool All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means—electronic, mechanical, photocopy, recording, or any other except for brief quotations in printed reviews, without the prior permission of the publisher. Fundamentals of Engineering Economics and Decision Analysis David L. Whitman and Ronald E. Terry www.morganclaypool.com ISBN: 9781608458646 paperback ISBN: 9781608458653 ebook DOI 10.2200/S00410ED1V01Y201203ENG018 A Publication in the Morgan & Claypool Publishers series SYNTHESIS LECTURES ON ENGINEERING Lecture #18 Series Editor: Steven S. Barrett, University of Wyoming Series ISSN Synthesis Lectures on Engineering Print 1939-5221 Electronic 1939-523X Fundamentals of Engineering Economics and Decision Analysis David L. Whitman University of Wyoming Ronald E. Terry Brigham Young University SYNTHESIS LECTURES ON ENGINEERING #18 CM& Morgan & cLaypool publishers ABSTRACT The authors cover two general topics: basic engineering economics and risk analysis in this text. Within the topic of engineering economics are discussions on the time value of money and interest relationships. These interest relationships are used to define certain project criteria
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discussions on the time value of money and interest relationships. These interest relationships are used to define certain project criteria that are used by engineers and project managers to select the best economic choice among several alternatives. Projects examined will include both income- and service-producing investments. The effects of escalation, inflation, and taxes on the economic analysis of alternatives are discussed. Risk analysis incorporates the concepts of probability and statistics in the evaluation of alternatives. This allows management to determine the probability of success or failure of the project. Two types of sensitivity analyses are presented.The first is referred to as the range approach while the second uses probabilistic concepts to determine a measure of the risk involved. The authors have designed the text to assist individuals to prepare to successfully complete the economics portions of the Fundamentals of Engineering Exam. KEYWORDS engineering economics, time value of money, net present value, internal rate of return, cash flow analysis, probability, statistics, risk analysis vii To our parents, wives, children, and grandchildren with much love and gratitude for everything. Contents ix 1 2 Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii Introduction .
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. . . . . . . . . . . . . . . . . xiii Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Engineering Economics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1.1 Basic Engineering Economics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1.2 Risk Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Decision Analysis . . . . . . . . . . . . .
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. . . 1 1.2 Decision Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.3 Fundamentals of Engineering Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Interest and the Time Value of Money . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2.1 2.2 2.3 2.4 2.5 2.6 Time Value of Money . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Sources of Capital . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Interest Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.3.1 Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.3.2 Compound Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.3.3 Nominal, Effective, and Continuous Interest Rates . . . . . . . . . . . . . . . . . . . . 5 Cash Flow Diagrams . . . . . . . . . .
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. . . . . . 5 Cash Flow Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Interest Formulas for Discrete Compounding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.5.1 Single Payments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.5.2 Uniform Series (Annuities) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.5.3 Uniform Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.5.4
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. . . . . . . . . . . . . . . . . . 11 2.5.4 The use of Financial Functions in Excel® . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.5.5 Example Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Interest Formulas for Continuous Compounding . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.6.1 Continuous Compounding for Discrete Payments . . . . . . . . . . . . . . . . . . . . 19 2.6.2 Continuous Compounding for Continuous Payments . . . . . . . . . . . . . . . . . 19 2.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . 20 3 Project Evaluation Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 3.1 3.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Alternate Uses of Capital . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 x 4 5 3.3 Minimum Acceptable Rate of Return (MARR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.4 3.5 3.6 3.7 3.8 3.9 Equivalence Methods . . . . . . . . . . . . . .
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3.6 3.7 3.8 3.9 Equivalence Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Net Present Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 3.5.1 Analysis of a Single Investment Opportunity . . . . . . . . . . . . . . . . . . . . . . . . 27 3.5.2 Do Nothing Project . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.5.3 Analysis of Multiple Investment Opportunities . . . . . . . . . . . . . . . . . . . . . . 30 Rate of Return Methods . . . . . . . . . . . . . . . . . . .
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Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.6.1 Internal Rate of Return (IRR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.6.2 Spreadsheet Formula for IRR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 3.6.3 External Rate of Return (ERR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 3.6.4 Spreadsheet Formula for ERR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 The Reinvestment Question in Rate of Return Calculations . . . . . . . . . . . . . . . . . . 37
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Calculations . . . . . . . . . . . . . . . . . . 37 3.7.1 Perception #1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.7.2 Perception #2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 3.7.3 Final Comments on ERR and IRR Relationships . . . . . . . . . . . . . . . . . . . . 41 Acceleration Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Payout . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 3.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 Service Producing Investments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 4.1 4.2 4.3 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Equal Life Alternatives . . . . . . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 4.2.1 Equivalence Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 4.2.2 Rate of Return Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Unequal Life Alternatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 4.3.1 Least Common Multiple Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 4.3.2 Common Study Period . . . . . . . . . . . . . . . .
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4.3.2 Common Study Period . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 4.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Income Producing Investments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 5.1 5.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Investment in a Single Project . . . . . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 5.3 Mutually Exclusive Alternatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 5.3.1 Equivalence Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 5.3.2 Rate of Return Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 5.3.3 Using Excel® . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 5.4 5.5 5.6 5.7 Unequal Life Alternatives . . . . . . . . . .
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. . 70 5.4 5.5 5.6 5.7 Unequal Life Alternatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Independent and Contingent Investments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 5.5.1 Independent Investments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 5.5.2 Contingent Investments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 5.5.3 Limited Investment Capital . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 Ranking Alternatives . . . . . . . . . . . . . . . . .
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77 Ranking Alternatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 xi 6 Determination of Project Cash Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 6.1 6.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Escalation and Inflation . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 6.3 Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 6.3.1 Straight-Line Depreciation (SL) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 6.3.2 Declining-Balance Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 6.3.3 Sum-of-the-Years-Digits (SYD) Depreciation . . . . . . . . . . . . . . . . . . . . . . . 97 6.3.4 Modified Accelerated Cost Recovery System (MACRS) . . . . . . . . . . . . . 102 6.4 Cash Flow Computation . . . . . . . .
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. . . . . . . 102 6.4 Cash Flow Computation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 6.4.1 Capital Investment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 6.4.2 Gross Revenue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 6.4.3 Operating Expenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 6.4.4 Before-Tax Profit Computation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . 107 6.4.5 Before-Tax Cash Flow Computation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 6.4.6 Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 6.4.7 Taxable Income . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 6.4.8 State and Federal Income Tax . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 6.4.9 Net Profit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 6.4.10 Cash Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 6.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 7 Financial Leverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 7.1 7.2 7.3 7.4 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 Financial Leverage and Associated Risk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 Adjustment to Cash Flow Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 7.3.1 Leverage and Mutually Exclusive Projects . . . . . . . . . . . . . . . . . . . . . . . . . . 132 7.3.2 Excel® Spreadsheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 xii 8 Basic Statistics and Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 8.1 8.2 8.3 8.4 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 8.2.1 Measures of Central Tendency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . 135 8.2.2 Measures of Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 8.2.3 Frequency Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 8.2.4 Relative Frequency Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 8.3.1 Classical Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 8.3.2 Relative Frequency Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 8.3.3 Subjective Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 8.3.4 Probability Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 9 Sensitivity Analysis . . . . . . . . . . . . . . .
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. 168 9 Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 9.1 9.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 9.1.1 Range Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 9.1.2 Monte Carlo Simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 A Compound Interest Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 Authors’ Biographies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 xiii Preface Those individuals working on the development of an income-generating project, either for personal use or company use, are frequently called upon to determine if the endeavor will prove profitable if fully developed. By profitable, we simply mean that the project will provide a desirable rate of return on investment through the generation of revenue that offsets any capital and/or operating costs. The intent of this book is to provide individuals with the tools to evaluate projects to determine profitability.The subject has been called: Engineering Economics or Project Evaluation or Economic Evaluation or Decision Analysis.
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evaluate projects to determine profitability.The subject has been called: Engineering Economics or Project Evaluation or Economic Evaluation or Decision Analysis. Whatever one chooses to call it, the reader who studies this material and becomes proficient in its content, will be able to analyze project cash flows and make a decision as to the profitability of the project. The authors, mainly because of their engineering backgrounds, have chosen to refer to the subject matter as engineering economics. In addition to income-generating projects, this book will also assist those individuals who are analyzing two or more ways of doing a service-producing project. A service-producing project is one, that instead of generating income for the investor, provides a service at a cost to the investor. An example could be the renting versus purchasing of a vehicle to provide a needed service for a company. The authors cover two general topics: basic engineering economics and risk analysis in the text. Chapters 2-6 contain content relative to basic engineering economics and Chapters 7-9 present material on risk analysis. Within the topic of engineering economics are discussions on the time value of money and interest relationships. These interest relationships are used to define certain project criteria that are used by engineers and project managers to select the best economic choice among several alterna- tives. Projects examined will include both income and service producing investments. The effects of escalation, inflation, and taxes on the economic analysis of alternatives are discussed. There is always risk involved in undertaking
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of escalation, inflation, and taxes on the economic analysis of alternatives are discussed. There is always risk involved in undertaking a project. Risk analysis incorporates the concepts of probability and statistics in the evaluation of alternatives. This allows management to determine the probability of success or failure of the project. Two types of sensitivity analyses are presented. The first is referred to as the range approach while the second uses probabilistic concepts to determine a measure of the risk involved. The authors have designed the text to assist individuals to prepare to successfully complete the economics portions of the Fundamentals of Engineering Exam. xiv PREFACE The authors wish to thank Joel Claypool and his associates at Morgan & Claypool for their encouragement and excellent work on the preparation and production of this text. David L. Whitman and Ronald E. Terry May 2012 C H A P T E R 1 Introduction 1 1.1 ENGINEERING ECONOMICS Nearly all projects that are proposed to be undertaken by any engineering firm will be, at some point, subjected to close economic scrutiny. The results of this analysis will be a basis (perhaps one of many) for deciding whether or not to proceed with the project. The major emphasis of this text, therefore, is to provide the engineer with the tools necessary to make the aforementioned economic decision. There are two general topics which are included in this textbook: basic engineering economics and risk analysis. A very brief overview of each of these topics is
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included in this textbook: basic engineering economics and risk analysis. A very brief overview of each of these topics is presented in the following paragraphs. 1.1.1 BASIC ENGINEERING ECONOMICS Within this topic are discussions on the time value of money and interest relationships. These interest relationships are used to define certain project criteria that are used by engineers and project managers to select the best economic choice among several alternatives. Projects examined will include traditional projects that generate a profit for the company and service producing projects which do not provide income, but do provide a needed service. The effects of escalation, inflation, and taxes on the economic analysis of alternatives will be discussed. 1.1.2 RISK ANALYSIS There is always risk involved in undertaking a project. Management is interested in the quantification of that risk. Risk analysis incorporates the concepts of probability and statistics in the evaluation of alternatives. This allows management to determine the probability of success or failure of the project. While there are a variety of ways to incorporate risk analysis into the evaluation of a project, the authors will present two methods that utilize what is known as sensitivity analysis. That is, deter- mining the sensitivity of the economic viability of a project as the costs and/or incomes vary about estimated values. The first is referred to as the range approach, while the second uses probabilistic concepts to determine a measure of the risk involved. 1.2 DECISION ANALYSIS As described above, the overall objective of any
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concepts to determine a measure of the risk involved. 1.2 DECISION ANALYSIS As described above, the overall objective of any economic analysis is to provide a basis for making a sound decision regarding any particular project. For example, suppose that an engineer is given the assignment to implement a project for which there are multiple alternative methods that will achieve the goals of the project. The question is: which alternative should be chosen? The reader 2 1. INTRODUCTION should recognize that there is always a choice among two or more alternatives. However, if only one technical alternative is found, then that alternative must be compared with the “do nothing” case. The “do nothing” case represents the situation where a company keeps its money invested in other alternatives which earn some minimum rate of return. The minimum rate of return will be referred to as the minimum acceptable rate of return (MARR) and will be discussed in detail in Chapter 2. Thus, there are always at least two alternatives in any economic decision. This textbook will provide the engineer with the necessary tools to determine the “best” economic choice among the alternatives. However, one must realize that final decisions are not only made on the results of the economic evaluations. Other general areas of consideration could be classified as financial and intangible issues. The “best” economic choice will be made through the proper use of the time value of money formulas that will be presented. Financial aspects have to do with
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the proper use of the time value of money formulas that will be presented. Financial aspects have to do with the obtaining of funds required to initiate the project. There are several sources which may be considered, i.e., internal company funds, lending institutions, the issuing of bonds, or the issuing of new stock. The intangible area of a project is the most difficult to analyze. Included in the intangible aspects are environmental, social, and political impacts. These are the most difficult to quantify. The focus of this textbook will be on the economic aspects of a project and very little time will be devoted to the areas of financial and intangible aspects. However, they are alluded to from time to time in order to remind the engineer of their importance and the obligation to consider them in the final decision. 1.3 FUNDAMENTALS OF ENGINEERING EXAM It is envisioned that information found in this textbook will prepare students to successfully complete the economics portions of the Fundamentals of Engineering Exam. The specifications for this exam can be found at http://www.ncees.org/Exams/FE_exam.php. C H A P T E R 2 3 Interest and the Time Value of Money 2.1 TIME VALUE OF MONEY When an individual or a company desires to invest an amount of capital in a long-term project, the effect of time on the value of that capital needs to be considered. The effect of time on the value of money can be illustrated by the following examples. Consider a sum
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considered. The effect of time on the value of money can be illustrated by the following examples. Consider a sum of $1000 that an individual has accumulated. If the $1000 were buried in a can under a tree for some future need, the individual, one year later, would still have $1000. However, if the $1000 were placed in an insured savings account earning 3% interest for one year, the amount would have grown to $1030. Obviously, the length of time and the different investment opportunities (represented by different interest rates) lead to varying amounts of money that the $1000 can yield at some future date. A second example deals with the same $1000 and its purchasing power as a function of time. Suppose an individual has a choice of purchasing 1000 items now at a price of $1.00 per item or waiting until a future date to make the purchase. If, over the course of one year, the price increased to $1.03 per item, the $1000 will only be able to purchase 970 items. Thus, the value, in terms of purchasing power, has decreased with time. The longer the life of the project, the more important will be the considerations of the time value of money. Other factors that affect the outcome of investment projects are inflation, taxes, and risk. These will be discussed later in the text. 2.2 SOURCES OF CAPITAL There are, in general, two sources of capital needed to make an investment. Capital can be obtained either
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OF CAPITAL There are, in general, two sources of capital needed to make an investment. Capital can be obtained either from the investor’s own funds or from a lender. Wherever capital is obtained, there is a cost associated with the use of the funds. If they are obtained from a lender, the cost of capital is the interest rate at which the funds are loaned to the investor. This interest rate reflects the current state of the economy as a whole, the bank’s administrative costs, and, perhaps, the risk associated with the particular loan as viewed by the lender. If the investor chooses to use his own funds for the required capital, then the cost is called the opportunity cost of capital. The opportunity cost reflects the income that could be generated from other opportunities the investor might have for his funds. This opportunity cost is often referred to as the minimum acceptable rate of return 4 2. INTEREST AND THE TIME VALUE OF MONEY (MARR). This minimum acceptable rate of return could be the interest rate obtained by placing the funds in a certificate of deposit or savings account at a bank or it could be the rate of return on another investment opportunity.The MARR is an important concept in the evaluation of investment opportunities and will be discussed in Chapter 3. For now, the MARR will just be treated as an interest rate, i. 2.3 2.3.1 INTEREST CONCEPTS SIMPLE INTEREST The amount of interest earned by an investment
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treated as an interest rate, i. 2.3 2.3.1 INTEREST CONCEPTS SIMPLE INTEREST The amount of interest earned by an investment (for example, a single principal deposit in a savings account) is called simple interest when the interest is found by Equation 2.1: I = P in (2.1) where, I = total interest, dollars P = amount of principal, dollars i = interest rate per interest period, fraction n = number of interest periods. Consider the following example. Individual A agrees to loan individual B $1000 for a time period of 3 years. B agrees to pay A the $1000 at the end of the 3 years, plus an amount of interest determined by applying a simple interest rate of 10% per year. The total interest charge will be: I = (1000)(0.10)(3) = $300 Therefore, at the end of 3 years, B will pay a total of $1300 to A which would represent the $1000 initially borrowed plus $300 interest for the use of A’s money. 2.3.2 COMPOUND INTEREST Simple interest concepts are used infrequently in today’s business dealings, but they do provide the basis for compounded interest rate concepts that are utilized. Compounded interest is computed by applying the interest rate to the remaining unpaid principal plus any accumulated interest. One could consider it as “the interest earns interest.” Referring back to the example presented above, the total interest that B will pay A over 3 years would be calculated as the following: Iyr 1 = (1000)(0.1) = $100, which
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B will pay A over 3 years would be calculated as the following: Iyr 1 = (1000)(0.1) = $100, which would result in a balance at the end of year 1 of $1100 Iyr 2 = (1100)(0.1) = $110, which would result in a balance at the end of year 2 of $1210 Iyr 3 = (1210)(0.1) = $121, which would result in a balance at the end of year 3 of $1331 Therefore, at the end of 3 years, B will pay a total of $1331 to A which is $31 higher than for the simple interest case. This difference results from compounding the interest. One should note that the difference between these two methods will become larger as the interest rate and number of interest periods increase. 2.3. INTEREST CONCEPTS 5 2.3.3 NOMINAL, EFFECTIVE, AND CONTINUOUS INTEREST RATES The length of the interest period can and does vary from application to application. Common interest rate periods are annually, semi-annually, quarterly, monthly, daily, and in the limiting case, continuously. The amount of interest that is earned or charged to a principal will increase as the compounding period becomes smaller. Usually, a lending institution will quote a nominal annual percentage rate. However, payments on the loan are made more often than annually. For example, consider a loan that is quoted at 10% nominal with semi-annual compounding (and, thus, semi-annual payments). The 10% annual interest compounded semi-annually means that every one-half year, 5% interest is earned or charged to the principal.
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The 10% annual interest compounded semi-annually means that every one-half year, 5% interest is earned or charged to the principal. This leads to the concept of effective yearly interest rate. The effective yearly interest rate can be found by computing the value that the principal has grown to at the end of year one, F , subtracting the original principal, and then dividing by the principal: F = 1000 + 1000(.05) + (1000 + 1000(.05))(.05) = 1000(1.05)2 Therefore, the effective rate is: ie = (1000(1.05)2 − 1000)/1000 = 0.1025 or 10.25% per year In general, the effective rate can be found by: (cid:2) ie = (cid:3) m − 1 1 + i m where, m = number of interest periods per year i = yearly nominal interest rate, fraction ie = yearly effective interest rate, fraction. In the limiting case of continuous compounding, the effective rate is given by: ie = ei − 1 (2.2) (2.3) Table 2.1 lists the effective rates for various compounding time periods for a 10% nominal rate. As can be observed in Table 2.1, the difference between the effective rates generated by the various compounding periods is relatively small. The differences can become insignificant when considering the many uncertainties associated with analyzing most economic investments. One should be careful with the term Annual Percentage Rate (APR) when dealing with lending institutions. The APR is a yearly percentage rate that expresses the total finance charge on a loan over its entire term. The APR includes the
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yearly percentage rate that expresses the total finance charge on a loan over its entire term. The APR includes the nominal interest rate, fees, points, and mortgage insurance, and is therefore a more complete measure of a loan’s cost than the interest rate alone. The loan’s nominal interest rate, not its APR, is used to calculate the monthly principal and interest payment. 6 2. INTEREST AND THE TIME VALUE OF MONEY Table 2.1: Example of effective interest rates for compounding time periods 2.4 CASH FLOW DIAGRAMS The construction of a cash flow diagram, sometimes referred to as a time line, will greatly aid in the analysis of an investment opportunity. The cash flow diagram is a way of accounting for all cash incomes and outflows at their appropriate position in time. That is, in general terms, the cash flow for any particular period is the income received during that period minus the expenses incurred during that same period. A good analogy to a cash flow diagram is one’s checkbook. Deposits and checks are written at specific points in time. These transactions could be consolidated on a monthly basis to show the net cash flow in or out of the checkbook each month. Usually, once the cash flow diagram is constructed properly, the economic analysis becomes relatively easy to complete. There are several ways of constructing a cash flow diagram and the following method is utilized by the authors. A horizontal line is drawn which represents the length of time (life)
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the following method is utilized by the authors. A horizontal line is drawn which represents the length of time (life) of the investment opportunity (project). The interest periods are then marked off and labeled above the line. At the extreme left of the time line is time zero (or, as will be defined in the next section, the Present). Time zero represents the time when the first cash flow is made for this project. Time zero is, therefore, defined by each project and not by a specific calendar date. Time zero can also be interpreted as the beginning of time period 1. All cash flows are then placed beneath the time line, corresponding to the position in time (or interest periods) in which they occurred. Negative cash flows (expenses exceeding revenues) are given a minus sign. In the time line illustrated below, CF1, CF2, etc., represent the cash flows occurring at the end of interest period 1, 2, etc. The authors often use a break in the time line for brevity. When dealing with investments in engineering projects, the normal approach is to assume that all investments for a particular year are made at the beginning of the year, while all revenues and operating expenses occur at the end of the year. This will lead to a conservative evaluation of the project using the techniques presented in Chapter 3. 0 1 2 3 CF0 CF1 CF2 CF3 … … 2.4. CASH FLOW DIAGRAMS 7 n-2 n-1 n CFn-2 CFn-1 CFn
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1 2 3 CF0 CF1 CF2 CF3 … … 2.4. CASH FLOW DIAGRAMS 7 n-2 n-1 n CFn-2 CFn-1 CFn Example 2.1 Consider the example of a 3-year auto loan from the view of the lender. The lender provides $20,000 to the client (a negative cash flow for the lender) at month 0 at an interest rate of 0.5% per month. In exchange, the lender receives $608 per month from the client over the next 36 months. The resulting cash flow diagram would be: 0 1 2 3 … 34 35 36 -20,000 608 608 608 ... 608 608 608 Before equations can be developed that relate the time value of money, it is necessary to define a set of notations that will be used throughout the text. P = Present sum of money. The present (time zero) is defined as any point from which the analyst wishes to measure time. F = Future sum of money. The future is defined as any point n that is greater than time zero. A = Annuity. This is a uniform set of equal payments that occur at the end of each interest period from one to n. G = Uniform gradient. This is a series of payments that uniformly increase or decrease over the life of the project. i = Compound interest rate per period. n = Total number of compounding periods in the cash flow diagram. The cash flow diagrams that follow should help to define these sums of money. 8
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in the cash flow diagram. The cash flow diagrams that follow should help to define these sums of money. 8 2. INTEREST AND THE TIME VALUE OF MONEY Present, P : 0 1 2 3 … n-2 n-1 n P Future, F : 0 1 2 3 … n-2 n-1 n Annuity, A: 0 1 2 3 … n-2 n-1 n F A Gradient, G: 0 1 A 2 A 3 … … A A n-2 n-1 A n G 2G … (n-3)G (n-2)G (n-1)G 2.5 INTEREST FORMULAS FOR DISCRETE COMPOUNDING The following section contains the derivation and sample calculations for nine interest formulas used in most economic calculations. These formulas demonstrate the “equivalency” between the various 2.5. INTEREST FORMULAS FOR DISCRETE COMPOUNDING 9 sums of money described above at specific values of the interest rate, i, and the number of periods, n. For example, in the example of the 3-year car loan, the $608 monthly payment is “equivalent” to the $20,000 initial loan at an interest rate of 0.5% per month. These formulas are based on discrete compounding, i.e., the interest is compounded at the end of each finite interest period. Formulas used with continuous compounding will be presented later. SINGLE PAYMENTS 2.5.1 The first formula to be derived allows the calculation of the equivalent future amount F , of a present sum, P . Suppose P is placed in a bank account that earns i% interest per period. It will grow to a future amount, F , at
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in a bank account that earns i% interest per period. It will grow to a future amount, F , at the end of n interest periods according to: F = P (1 + i)n (2.4) The derivation of Equation 2.4 is given by: The factor (1 + i)n is frequently called the Single Payment Compound Amount Factor and is symbolized in this text by (F /P )i,n. If one is given the amount of P , one uses the (F /P )i,n factor to find the equivalent value of F . That is, F = P (F /P )i,n (2.5) Similarly, if a future amount, F , is known and it is desired to calculate the equivalent present amount, P , then Equation 2.4 can be arranged as: P = F (1 + i) −n (2.6) The factor (1 + i)−n is frequently called the Single Payment Present Worth Factor and is symbolized in this text by (P /F )i,n. If one is given the amount of F , one uses the (P /F )i,n factor to find the equivalent value of P . That is, P = F (P /F )i,n (2.7) 10 2. INTEREST AND THE TIME VALUE OF MONEY 2.5.2 UNIFORM SERIES (ANNUITIES) It is often necessary to know the amount of a uniform series payment, A, which would be equivalent to a present sum, P , or a future sum, F . In the following formulas that relate P , F , and A, it is
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or a future sum, F . In the following formulas that relate P , F , and A, it is imperative that the reader understands that: 1) P occurs one interest period before the first value of A; 2) A occurs at the end of each interest period; and 3) F occurs at the same time as the last A (at time n). These relationships were illustrated in the previous cash flow diagrams that originally defined each of them. The value of a future sum, F , of a series of uniform payments, each of value A, can be found by summing the future worth of each individual payment. That is, treat each A as a distinct present value (but with a different time zero) and use (F /P )i,n to calculate its contribution to the total F : F = A(1 + i)n−1 + A(1 + i)n−2 + A(1 + i)n−3 + . . . + A(1 + i)1 + A (2.8) Multiplying both sides of Equation 2.8 by (1 + i) yields F (1 + i) = A(1 + i)n + A(1 + i)n−1 + A(1 + i)n−2 + . . . + A(1 + i)2 + A(1 + i) (2.9) Subtracting Equation 2.8 from 2.9 yields F (1 + i) − F = A(1 + i)n − A Solving for F in terms of A results in: F = A{[(1 + i)n − 1]/i} (2.10) The term in the {} brackets is called the Uniform Series Compound
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F = A{[(1 + i)n − 1]/i} (2.10) The term in the {} brackets is called the Uniform Series Compound Amount Factor and is symbolized by (F /A)i,n. If one is given the amount of A, one uses the (F /A)i,n factor to find the equivalent value of F . That is, Rearranging Equation 2.10 and solving for A yields F = A(F /A)i,n A = F {i/[(1 + i)n − 1]} (2.11) (2.12) The term in the { } brackets is called the Sinking Fund Factor and is symbolized by (A/F )i,n. If one is given the amount of F , one uses the (A/F )i,n factor to find the equivalent value of A. That is, A = F (A/F )i,n (2.13) Substitution of Equation 2.10 into Equation 2.6 yields Equation 2.14 which contains the Uniform Series Present Worth Factor, (P /A)i,n in the {} brackets: P = A{[(1 + i)n − 1]/[i(1 + i)n]} (2.14) (2.15) (2.16) (2.17) 2.5. INTEREST FORMULAS FOR DISCRETE COMPOUNDING 11 If one is given the amount of A, one uses the (P /A)i,n factor to find the equivalent value of P . That is, P = A(P /A)i,n Rearranging Equation 2.14 and solving for A yields A = P {[i(1 + i)n]/[(1 + i)n − 1]} The term in the { } brackets is called the Capital Recovery Factor and is symbolized by (A/P )i,n. If one is given the amount of P , one uses the (A/P )i,n factor to find the
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(A/P )i,n. If one is given the amount of P , one uses the (A/P )i,n factor to find the equivalent value of A. That is, A = P (A/P )i,n 2.5.3 UNIFORM GRADIENT In some applications, a series of cash flows will be generated from a project analysis which uniformly increase or decrease from an initial value. The cash flow diagram is repeated here for clarity. 0 1 2 3 G 2G … … n-2 n-1 n (n-3)G (n-2)G (n-1)G Without derivation, Equations 2.18, 2.20, and 2.22 can be developed that relate the gradient, G, to an equivalent annuity, an equivalent present sum, and an equivalent future sum: A = G{1/i − n/[(1 + i)n − 1]} (2.18) The term in the { } brackets is symbolized by (A/G)i,n. If one is given the amount of G, one uses the (A/G)i,n factor to find the equivalent value of A. That is, A = G(A/G)i,n P = G{[(1 + i)n − 1]/[i2(1 + i)n] − n/[i(1 + i)n]} (2.19) (2.20) The term in the { } brackets is symbolized by (P /G)i,n. If one is given the amount of G, one uses the (P /G)i,n factor to find the equivalent value of P . That is, P = G(P /G)i,n F = G{[(1 + i)n − 1]/i2 − n/i} (2.21) (2.22) The term in the { } brackets is symbolized by (F /G)i,n. If one is given the amount of G, one uses the (F /G)i,n factor to find the
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by (F /G)i,n. If one is given the amount of G, one uses the (F /G)i,n factor to find the equivalent value of F . That is, F = G(F /G)i,n (2.23) 12 2. INTEREST AND THE TIME VALUE OF MONEY The equations for the nine factors are given in Table 2.2 and numerical values are tabulated in Appendix A for various values of interest rate, i, and number of periods, n so that the user can look them up rather than use the actual formulas. Rather than memorizing which factor is needed for a specific equivalency, think about the formulas in terms of “units conversion.” That is, if the input to a system has units of X and the output of that system has units of Y, the system provides a units conversion of (Y/X). Thus, if one is given A (input) and wants to find G (output), the correct formula to use would be (G/A). Knowing the value of the interest rate and the number of periods, one can look up or compute the value of the formula. Table 2.2: Formulas for discrete compounding Factor Name Converts Single Payment Compound Amount to F given P Symbol (F / P) i,n Single Payment Present Worth to P given F (P / F) i,n Uniform Series Compound Amount to F given A (F / A) i,n Uniform Series Sinking Fund Uniform Series Present Worth to A given F (A / F) i,n to P given A (P / A) i,n
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Uniform Series Present Worth to A given F (A / F) i,n to P given A (P / A) i,n Capital Recovery to A given P (A/ P) i,n Uniform Gradient Present Worth Uniform Gradient Future Value Uniform Gradient Uniform Series to P given G (P/ G) i,n to F given G (F / G) i,n to A given G (A / G)i,n Formula (1 + i) n (1 + i) -n -1n (1 + i) i i (1 + i) -1n (1 + i) i (1 + i) -1n n n i (1 + i) (1 + i) -1n (1 + i) 2 i (1 + i) -1n n - n (1 + i) i n -1n (1 + i) 2 i - n i 1 - i n (1 + i) -1n 2.5. INTEREST FORMULAS FOR DISCRETE COMPOUNDING 13 2.5.4 THE USE OF FINANCIAL FUNCTIONS IN EXCEL® Many cash flow situations can be simulated by using a spreadsheet such as Microsoft Excel®. This will become more evident in future chapters, but this chapter presents the following useful financial functions: Future Value: =FV(rate, nper, pmt, pv, type) Present Value: =PV(rate, nper, pmt, fv, type) Annuity: =PMT(rate, nper, pv, fv, type) Unfortunately, Excel® does not have a built-in function for gradient-type cash flows. That can, however, be overcome with functions that will be presented in later chapters. In each of these functions, the variables are as follows: (cid:129) rate is the interest rate (as a fraction) per period (cid:129) nper
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these functions, the variables are as follows: (cid:129) rate is the interest rate (as a fraction) per period (cid:129) nper is the number of interest bearing periods (cid:129) pmt is an annuity (A) sum of money (cid:129) pv is a present (P ) value sum of money (occurs at time = 0) (cid:129) fv is a future (F ) value sum of money (occurs at time = nper) (cid:129) type is 0 for end of period cash flows and 1 for beginning of period cash flows It should also be noted that in order to use these functions as equivalents for (P /A), (P /F ), etc., the values of pmt, pv, and fv need to be input as negative numbers. An example of a simple Excel® spreadsheet that computes the six functions given above is shown for 10% annual interest rate for 10 years. The actual formulas are shown as well. Recall that one needs to set “type” equal to zero to designate that the cash flows occur at the end of each period. An explanation of the values in the various Excel formulas may be necessary. For example, in the formula that computes F/A (cell B7), the values are as follows: “B1” is the interest rate as a fraction, “B2” is for 10 periods, “B3” is for an annual annuity payment of $1 per year, “0” represents the fact that there is no present value payment, and “B6” defines that the various payments are at the end of
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fact that there is no present value payment, and “B6” defines that the various payments are at the end of the period. Since the formula finds the future value of a $1 annuity, we have effectively computed (F/A). Some additional Excel® financial functions that might be of some interest at this point are: 14 2. INTEREST AND THE TIME VALUE OF MONEY A B A B rate 1 nper 2 pmt(A) 3 pv (P) 4 fv (F) 5 type 6 F/A 7 F/P 8 P/A 9 10 P/F 11 A/P 12 A/F 0.1 10 -1 -1 -1 0 15.937 2.5937 6.1446 0.38554 0.16275 0.062745 rate 1 nper 2 pmt(A) 3 pv(P) 4 fv(F) 5 type 6 F/A 7 F/P 8 P/A 9 10 P/F 11 A/P 12 A/F 0.1 10 -1 -1 -1 0 =FV(B1,B2,B3,0,B6) =FV(B1,B2,0,B4,B6) =PV(B1,B2,B3,0,B6) =PV(B1,B2,0,B5,B6) =PMT(B1,B2,B4,0,B6) =PMT(B1,B2,0,B5,B6) Effective Interest Rate: =EFFECT(normal_rate, npery) Number of periods: =NPER(rate, pmt, pv, fv, type) The new variables are defined as follows: (cid:129) normal_rate = the nominal annual interest rate (as a fraction) (cid:129) npery = the number of compounding periods per year The effective interest table for 10% nominal interest rate can be created in Excel® as follows (note that in the case of continuous compounding, npery=1,000,000 is close enough to give the answer to the desired number of significant digits). One can compare Table 2.3 with Table 2.1 to see consistency between the calculations in Excel® and those performed with the specific formula for ieff . The NPER function is
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consistency between the calculations in Excel® and those performed with the specific formula for ieff . The NPER function is useful for determining how many compounding periods are necessary to achieve a desired result. For example, one might want to determine how many years it will take for an original investment to double in value if the interest rate is varied from 1% per year to 25% per year. This is shown in Table 2.4. The explanation of the values in the NPER formulas in Table 2.4 is as follows: “A3/100” represents the interest rate as a fraction, “0” is for no annuity payment, “-1” is for a present value amount of $1, “2” is for a future value of $2, and “0” defines the amounts as end of year payments. One can also note that the product of the interest rate (as a percentage) and the # of periods to double the value of the investment varies from 70 to 75. This is commonly known as the “Rule of 2.5. INTEREST FORMULAS FOR DISCRETE COMPOUNDING 15 Table 2.3: Using Excel® to compute effective interest rates for a nominal 10% interest rate. Table 2.4: Using Excel to compute the number of years needed to double the value of an initial investment. 72.” If one takes 72 and divides by the interest rate (as a percentage), the resultant value is a close approximation of how long it will take for an investment to double. 2.5.5 EXAMPLE PROBLEMS At this point, it
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close approximation of how long it will take for an investment to double. 2.5.5 EXAMPLE PROBLEMS At this point, it would be beneficial to examine some of the practical applications of these formulas. Example 2.2 If $10,000 is invested in a fund earning 15% compounded annually, what will it grow to in 10 years? Solution: F = P (F /P )i,n = 10, 000(F /P )15,10 = 10, 000(4.0456) = $40, 456 16 2. INTEREST AND THE TIME VALUE OF MONEY Example 2.3 It is desired to accumulate $5,000 at the end of a 15-year period. What amount needs to be invested if the annual interest rate is 10% compounded semi-annually? Assume the given interest rate is a nominal rate and that the principal is compounded at 5% per period. Solution: P = F (P /F )i,n = 5, 000(P /F )5,30 = 5000(0.23138) = $1, 157 Example 2.4 What interest rate, compounded annually, will make a uniform series investment (at the end of each year) of $1,000 equivalent to a future sum of $7,442? The investment period is 5 years. Solution: F = A(F /A)i,n ⇒ 7, 442 = 1, 000(F /A)i,5 ⇒ (F /A)i,5 = 7.442 Searching the various interest tables in Appendix A for n = 5 yields i = 20% Example 2.5 An individual wishes to have $6,000 available after 8 years. If the interest rate is 7% com- pounded annually, what uniform amount must be deposited at the end of each year? Solution: A =
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is 7% com- pounded annually, what uniform amount must be deposited at the end of each year? Solution: A = F (A/F )i,n = 6, 000(A/F )7,8 = 6, 000(0.09747) = $585 Example 2.6 An individual wishes to place an amount of money in a savings account and, at the end of one month and for every month thereafter for 30 months, draw out $1,000. What amount must be placed in the account if the interest rate is 12% (nominal rate) compounded monthly? Solution: i(monthly) = 0.12/12 = 0.01(1%) P = A(P /A)i,n = 1, 000(P /A)1,30 = 1, 000(25.808) = $25, 808 Example 2.7 A principal of $50,000 is to be borrowed at an interest rate of 15% compounded monthly for 30 years. What will be the monthly payment to repay the loan? Solution: i (monthly) = 0.15/12 = 0.0125(1.25%). Since Appendix A does not contain a table for that interest rate, one must use the formulas. A = P (A/P )i,n = 50, 000(A/P )1.25,360 = 50, 000{[(0.0125)(1 + 0.0125)360]/[(1 + 0.0125)360 − 1]} = 50, 000(0.012644) = $632 2.5. INTEREST FORMULAS FOR DISCRETE COMPOUNDING 17 Example 2.8 An individual deposits $1,000 at the end of each year into an investment account that earns 8% per year compounded monthly. What is the balance in his account after 10 years? Solution: Since the time frame of the deposits (annually) does not match the time frame of the interest rate (monthly), one must convert to an effective annual interest rate
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does not match the time frame of the interest rate (monthly), one must convert to an effective annual interest rate before computing the correct formulas. (cid:2) 1 + i m (cid:3) m (cid:2) (cid:3) 12 ie = F = A(F /A)i,n = 1, 000(F /A)8.30,10 = 1, 000{[(1 + 0.0830)10 − 1]/0.0830} − 1 = 0.0830 − 1 = 1 + 0.08 12 = 1, 000(14.694) = $14, 694 Example 2.9 Calculate the future worth of the following 6-year cash diagram if the interest rate is 10% compounded annually. 0 1 2 3 4 5 6 1000 1200 1400 1600 1800 2000 There are a number of ways to solve this economic problem, which is the case for most cash flow evaluations. One technique might be shorter in terms of the number of formulas to look up or calculate, but all will result in the same answer. Solution 1: Note that this series of cash flows can be broken into an annuity of $1,000 per year and a gradient of $200 per year. One can compute the future value of each of these contributions separately and then add to get the final result. FAnnuity FGradient = A(F /A)i,n = 1, 000(F /A)10,6 = 1, 000(7.7156) = $7, 715.60 = G(F /G)i,n = 200(F /G)10,6 = 200(17.156) = $3, 431.22 F = 7, 715.60 + 3, 431.22 = $11, 147 18 2. INTEREST AND THE TIME VALUE OF MONEY Solution 2: Convert the gradient to an equivalent annuity, add this value
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2. INTEREST AND THE TIME VALUE OF MONEY Solution 2: Convert the gradient to an equivalent annuity, add this value to the $1,000 annuity and then convert to the future. AGradient ATotal = G(A/G)i,n = 200(A/G)10,6 = 200(2.2236) = $444.72 = 1, 000 + 444.72 = $1, 444.72 F = A(F /A)i,n = 1, 444.72(F /A)10,6 = 1, 444.72(7.7156) = $11, 147 Solution 3: Treat each cash flow as an individual, single payment, find the future value of each individual payment and then add to get the total future value. FCF 1 = P (F /P )i,n = 1, 000(F /P )10,5 = 1, 000(1.6105) = $1, 610.50 FCF 2 = P (F /P )i,n = 1, 200(F /P )10,4 = 1, 200(1.4641) = $1, 756.92 FCF 3 = P (F /P )i,n = 1, 400(F /P )10,3 = 1, 400(1.3310) = $1, 863.40 FCF 4 = P (F /P )i,n = 1, 600(F /P )10,2 = 1, 600(1.2100) = $1, 936.00 FCF 5 = P (F /P )i,n = 1, 800(F /P )10,1 = 1, 800(1.1000) = $1, 980.00 FCF 6 = P (F /P )i,n = 2, 000(F /P )10,0 = 2, 000(1.0000) = $2, 000.00 F = 1.610.50 + 1, 756.92 + 1, 863.40 + 1, 936.00 + 1, 980.00 + 2, 000.00 = $11, 147 Example 2.10 Calculate the present worth of the following 10-year cash flow diagram if the annual interest rate is 20% compounded annually. 0 1 2 3 … 8 9 10 2000
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flow diagram if the annual interest rate is 20% compounded annually. 0 1 2 3 … 8 9 10 2000 1900 1800 … 1300 1200 1100 Solution: Again, there are a variety of methods to solve this problem. One technique is to recognize that the cash flow is made up of an annuity of $2,000 and a gradient of −$100. ATotal = 2, 000 − 100(A/G)20,10 = 2, 000 − 100(3.0739) = $1, 692.61 P = A(P /A)i,n = 1, 692.61(P /A)20,10 = 1, 692.61(4.1925) = $7, 096 2.6. INTEREST FORMULAS FOR CONTINUOUS COMPOUNDING 19 2.6 INTEREST FORMULAS FOR CONTINUOUS COMPOUNDING In the last section, the assumption was made that money was received or dispersed and interest rates were compounded at the end of each discrete compounding period. In some projects (consider a banking institution for example), money is received and dispersed on a nearly continuous basis. If the evaluator wishes to consider the effect of continuous cash flow and/or continuous compounding of interest, one needs to utilize a slightly different set of formulas that relate P , F , and A. 2.6.1 CONTINUOUS COMPOUNDING FOR DISCRETE PAYMENTS The following formulas apply to the situation where payments (or withdrawals) to an account are made at discrete points in time, while the account accumulates interest on a continuous basis: (P /F )i,n = e−in (P /A)i,n = (ein − 1)/[ein(ei − 1)] (F /A)i,n = (ein − 1)/[(ei − 1)] (2.24) (2.25) (2.26) 2.6.2 CONTINUOUS COMPOUNDING FOR CONTINUOUS PAYMENTS The other
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1)] (F /A)i,n = (ein − 1)/[(ei − 1)] (2.24) (2.25) (2.26) 2.6.2 CONTINUOUS COMPOUNDING FOR CONTINUOUS PAYMENTS The other application of continuous compounding is the case where the deposits or withdrawals to an account are being made on a nearly continuous basis. One example of this situation would be a credit card company that receives charges and payments on millions of cards throughout each day. For this case, the following definitions need to be made: ¯P , ¯F , ¯A = the total amount of funds received over one period (present sum, future sum, or annuity, respectively). The following figures demonstrate these definitions: 20 2. INTEREST AND THE TIME VALUE OF MONEY The appropriate formulas are: (P / ¯F )i,n = [i(1 + i)−n]/[ln(1 + i)] (F / ¯P )i,n = [i(1 + i)n−1]/[ln(1 + i)] (F / ¯A)i,n = (ein − 1)/i (P / ¯A)i,n = (ein − 1)/[i(ein)] (2.27) (2.28) (2.29) (2.30) where, i is the nominal interest rate per period. 2.7 PROBLEMS 2.1. Given a nominal rate of 20%, what is the effective annual interest rate if the interest is compounded under each of the following scenarios: (a) Quarterly (b) Monthly (c) Daily (d) Continuously 2.2. What is the percentage difference between the effective rates determined by annual and continuous compounding for nominal interest rates of: (a) 10% (b) 20% (c) 30% 2.3. A company has decided to invest in a project to make a product. The initial investment cost will be $1,000,000 to be spread
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decided to invest in a project to make a product. The initial investment cost will be $1,000,000 to be spread over the first two years with $700,000 in the first year and $300,000 in the second. The plan calls for producing products at the following rates: 5,000 units in year 2; 10,000 in year 3; 30,000 in year 4; 30,000 in year 5; $10,000 in year 6; and $5,000 in year 7. Products will be sold for $50 each throughout the life of the project and cash operating expenses will be $60,000 per year for years 2 through 7. Construct a cash flow diagram for the project. 2.7. PROBLEMS 21 2.4. Example 2.1 presented a cash flow diagram for an automobile loan as seen through the eyes of the lender. Construct the corresponding cash flow diagram as seen through the eyes of the borrower. 2.5. A $1,000 investment has grown to $2,476 in 8 years. What interest rate (compounded annually) has it earned? 2.6. What present sum is equivalent to a future sum of $25,000 (after 5 years) at an interest rate of 8% compounded annually? 2.7. If $200 is placed at the end of each year for 10 years in an account earning 7% interest compounded annually, what amount will be accumulated at the end of 10 years? 2.8. What uniform series would be equivalent to a future sum of $10,000 if the series extends for 10 years and earns 12% interest compounded semi-annually? 2.9. An annual deposit of
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of $10,000 if the series extends for 10 years and earns 12% interest compounded semi-annually? 2.9. An annual deposit of $1,000 is placed in an account at the beginning of each year for 5 years. What is the present value of that series if interest is 12% compounded annually? What is the future value at the end of the 5th year? 2.10. What will be the future value, 10 years from the first payment, of the series of deposits in problem 2.9? 2.11. What monthly car payments for the next 30 months are required to amortize a loan of $4,000 if interest is 12% compounded monthly? 2.12. Payments of $1,000 are to be made at the end of each year for the next 3 years. What is the present worth of the three payments if interest is 12% compounded monthly? What series of monthly payments would be equivalent to the $1,000 year payments? 2.13. An individual agrees to lease a building to a firm with yearly payments shown on the cash flow diagram below.What is the future worth of the payments if interest is 15% compounded annually? 0 1 2 3 4 5 6 7 8 9 10 3000 3000 3000 3000 3300 3600 3900 4200 4500 4800 2.14. An engineer wishes to buy a house but can only afford monthly payments of $1500. 30- year loans are available at 5.75% interest compounded monthly. If the engineer can make a $20,000 down payment, what is the price of the most
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5.75% interest compounded monthly. If the engineer can make a $20,000 down payment, what is the price of the most expensive house that the engineer can afford to purchase? 22 2. INTEREST AND THE TIME VALUE OF MONEY 2.15. A young woman placed $200.00 in a savings account paying monthly interest. After one year, her balance has grown to $214.00. What was the effective annual interest rate? What was the nominal annual interest rate? 2.16. Find the value of cash flow X that will make the two cash flows equivalent. Interest is 10% compounded annually. Time on the diagram is given in years. 0 1 2 3 4 5 6 100 120 140 160 0 1 2 X X X 2.17. It takes a full $10,000 to put on a Festival of Laughingly Absurd Walks (FLAW) each year. Immediately before this year’s FLAW, the sponsoring committee finds that it has $40,000 in an account paying 15% interest compounded annually. After this year, how many more FLAWs can be sponsored without raising more money? 2.18. If $10,000 is borrowed at 12% interest compounded monthly, what would the monthly payments be if the loan is for 5 years? What would the annual payment be if the loan is for 5 years? Assume all payments occur at the end of a given period. 2.19. Calculate the value of the following cash flow diagram at the end of year 4. Interest is 10% per year compounded annually. 0 1 2 3 4 5 6
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at the end of year 4. Interest is 10% per year compounded annually. 0 1 2 3 4 5 6 7 8 9 10 1000 500 500 750 1000 800 600 400 2000 2.20. Calculate the future worth 5 years from now of a present sum of $2,000 if: 2.7. PROBLEMS 23 (a) Annual interest is 10% compounded annually (b) Annual interest is 10% compounded quarterly (c) Annual interest is 10% compounded continuously 2.21. Calculate the present value of 10 uniform $2,000 payments if: (a) Annual interest is 10% compounded continuously and payments are received at the end of each year (b) Annual interest is 10% compounded continuously and payments are received contin- uously over the year 2.22. A gas station sells $125,000 worth of gasoline over the course of a year. If this revenue is collected and deposited continuously into an account that earns 8% interest, compounded annually, how much money would the station have in its account at the end of the year? 2.23. Develop an Excel® spreadsheet that computes the six functions — (P /A), (P /F ), (F /A), (F /P ), (A/P ), (A/F ) — for a fixed interest rate and the number of periods ranging from 1 to 100. 2.24. Use the Excel® NPER function to determine how long it will take for an investment to triple in value at interest rates of 1%, 5%, 10%, 15%, 20%, and 25%. Can you determine an approximate “Rule” for how to quickly calculate how long
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1%, 5%, 10%, 15%, 20%, and 25%. Can you determine an approximate “Rule” for how to quickly calculate how long it takes for an investment to triple in value? C H A P T E R 3 25 Project Evaluation Methods INTRODUCTION 3.1 In order to make informed decisions on one or more potential investments, methods must be devel- oped that provide a numerical evaluation of a project. Both equivalence and rate of return methods will be developed in this chapter. Consider the following cash flow diagrams that contain income generating streams. 1 2 3 … 18 19 20 A: 0 1,000,000 B: 0 1 2 3 … 18 19 20 100,000 100,000 100,000 … 100,000 100,000 100,000 Since there are no cash flows for A after period 0, the present value of cash flow A is simply $1,000,000. For B, since the $100,000 occurs at the end of each period for 20 periods, multiplying the $100,000 by (P /A)i,20 will yield a present value for the interest rate used in the formula. For example, if the interest rate is 12% per year, the present value would be $746,944. If the question “which cash flow represents the largest present value?” is asked, the answer is obviously cash flow A. Now consider a different question. Suppose you have just won a lottery and you have a choice of receiving $1,000,000 now or receiving $100,000 at the end of each year for 20 years. If interest is expected to be constant at
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or receiving $100,000 at the end of each year for 20 years. If interest is expected to be constant at 12% for the next 20 years as in the previous paragraph, which set of payments would you prefer? Since this question is represented by the cash flow diagrams shown above and the interest rate of 12%, the choice can be made by analyzing the present values of the two cash flow diagrams. Since cash flow A yields a larger present value than cash flow B at an interest rate of 12%, the proper choice would be to accept option A. 26 3. PROJECT EVALUATION METHODS However, what if the interest rate is expected to be 0% over the 20 year period? What would the best choice be under that scenario? If interest is 0%, then money is worth the same no matter when it occurs. At 0% interest, the present value of cash flow B becomes $2,000,000 and cash flow B becomes the correct choice. The discussion in the previous two paragraphs infer that at some interest rate between 0% and 12%, the two cash flow diagrams are equivalent. A trial and error solution yields this interest rate to be about 7.75%. This discussion has just introduced two of the more popular techniques (equivalence methods and rate of return methods) used to evaluate the financial value of projects and help the evaluator choose between multiple projects. These will be discussed in more detail later in this chapter. 3.2 ALTERNATE USES
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the evaluator choose between multiple projects. These will be discussed in more detail later in this chapter. 3.2 ALTERNATE USES OF CAPITAL Investment analysis or project evaluation involves making a decision between alternative uses of capital. A cash flow diagram is constructed for each alternative according to the specific parameters of that alternative and evaluated using the concepts of time value of money that were discussed in Chapter 2. The results of the evaluations are then compared and a decision is made as to which alternative is the best option. Several evaluation methods can be used in analyzing investment opportunities. Two general types of calculations that will be introduced here are: (1) equivalence methods which involve the determination of an equivalent present, annual, or future worth of a cash flow diagram given a specific interest rate; and (2) rate of return methods which involve the determination of an interest rate produced by the cash flow diagram. 3.3 MINIMUM ACCEPTABLE RATE OF RETURN (MARR) When using either the equivalence method or the rate of return method for comparing alternatives, a minimum acceptable rate of return, MARR, needs to be defined. The value of MARR is set as the lower limit for investment which is acceptable to an individual or a company. The MARR may vary from individual to individual, company to company, and even within the structure of a specific company. The lower bound for the MARR is generally set at the cost of capital, which reflects the expense of obtaining
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The lower bound for the MARR is generally set at the cost of capital, which reflects the expense of obtaining funds for a given project. How much higher the MARR is above the cost of capital depends on a particular company’s or individual’s position and the particular project. For example, an individual who borrows money at 5% interest rate in order to invest in a profit-generating project would have an MARR of at least 5%, but would probably want to set the MARR at, say, 10% in order to generate a net increase in his/her personal worth based on the estimated profitability of the project. Similarly, if individuals are using their own funds to invest, their cost of capital would be the interest rate that their money is currently earning in a savings account, certificate of deposit, or other investments. A company’s MARR is usually set by the portfolio of projects in which the company can invest. That is, what is the minimum interest that a company can earn by investing its money in what it would consider to be a “guaranteed” success? For engineers performing economic evaluations for their companies, the MARR will be provided by upper management so that they will not have to make that determination. 3.4. EQUIVALENCE METHODS 27 3.4 EQUIVALENCE METHODS In the equivalence methods to determine either the acceptability of a single project or to choose the “best” project, the MARR is used as the interest rate in present, future, or annuity calculations. A
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to choose the “best” project, the MARR is used as the interest rate in present, future, or annuity calculations. A net present value, NPV (sometimes called the net present worth), net future value, NFV, or net annual value, NAV, is calculated by one of the following equations: (cid:4) N P V = N F V = N AV = Present Value of Cash Flows with Future Value of Cash Flows with (cid:4) (cid:4) Annuity Value of Cash Flows with i = MARR i = MARR i = MARR (3.1) (3.2) (3.3) Since N P V , N F V , and N AV are related by the interest formulas developed in Chapter 2, any one of the three calculations will yield the same conclusion (in terms of economic viability of the project) as the other two. Because of this fact, most analysts concentrate on the NP V method, as do the authors of this text. 3.5 NET PRESENT VALUE 3.5.1 ANALYSIS OF A SINGLE INVESTMENT OPPORTUNITY For a single investment opportunity, the NP V would be calculated using the MARR as the interest rate. A positive value for N P V indicates that the project which is represented by the cash flow diagram earns an actual interest rate greater than the MARR, a negative value for NP V indicates that it earns an actual interest rate less than the MARR, and an NP V value of zero indicates that it earns the MARR. Since the MARR represents the decision point
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and an NP V value of zero indicates that it earns the MARR. Since the MARR represents the decision point for determining the viability of a project for a particular investor, a positive NP V would indicate that the project is an acceptable one. Example 3.1 Consider the project represented by the following cash flow diagram. The project requires an initial investment of $1,000 that returns positive cash flows as shown. The MARR is 10%. 0 1 2 3 4 5 -1000 500 600 700 800 900 28 3. PROJECT EVALUATION METHODS N P V = −1000 + 500(P /A)10,5 + 100(P /G)10,5 = −1000 + 500(3.7908) + 100(6.8618) = $1582 Since the N P V is greater than zero, this project would be an acceptable one to the investor. An alternative method to calculate the NP V is to treat each individual cash flow as a future value at various values of n. While this technique might require more formulas than recognizing annuities and gradients in the cash flow diagram, it will always yield a correct value for NP V : N P V = − 1000 + 500(P /F )10,1 + 600(P /F )10,2 + 700(P /F )10,3 + 800(P /F )10,4 + 900(P /F )10,5 N P V = $1582 In Excel®, one can use the NP V function to make the same calculation. However, some caution is necessary. The function is: = NP V (rate, value1, value2, …). where, rate = interest rate per period (as
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is necessary. The function is: = NP V (rate, value1, value2, …). where, rate = interest rate per period (as a fraction). value1, value2, ... = cash flows that occur at the end of period 1, end of period 2, etc. One can see that the NPV function does not include the investment period 0. Therefore, in order to calculate the N P V of the entire cash flow diagram, one needs to include the initial investment. For example, the complete Excel formula to compute the NP V of a series of cash flows would be as shown in Figure 3.1: = CF0 + NP V (rate, value1, value2,…) One can see that the results from Excel match the NP V calculations from the other two methods. Example 3.2 Consider the project represented by the following cash flow diagram. The project requires an initial investment of $1,000 that returns positive cash flows as shown. The MARR is 10%. 0 1 2 3 4 5 -1000 150 200 250 300 350 A MARR = B 10% A B 1 MARR = 0.1 3.5. NET PRESENT VALUE 29 Year 0 1 2 3 4 5 CF -1000 500 600 700 800 900 NPV = 1582 3 4 5 6 7 8 9 10 11 Year 0 1 2 3 4 5 CF -1000 500 600 700 800 900 NPV = =B4+NPV(B1,B5:B9) 1 3 4 5 6 7 8 9 10 11 Figure 3.1: Demonstration of the use of the NP V function
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3 4 5 6 7 8 9 10 11 Figure 3.1: Demonstration of the use of the NP V function in Excel®. N P V = −1000 + 150(P /A)10,5 + 50(P /G)10,5 = −1000 + 150(3.7908) + 50(6.8618) = −$88 Since the N P V is negative, the project will not earn the MARR and, therefore, is not accept- able to this investor. Now a question arises: What does the investor do with the $1000? Since the time-line represents the only ‘new’ investment opportunity available to the investor and the NP V analysis suggests that it is not acceptable, the investor will choose to do nothing with the $1000. The concept of the “do nothing” project will be defined in the next section. 3.5.2 DO NOTHING PROJECT Example 3.2 indicates that there is always a choice to “do nothing” with investment funds. That is, even if a project, like the one described in Example 3.2, is the only new investment available and the financial analysis indicates that it is unacceptable, an investor can always choose to keep the proposed funds, $1000 in the case of Example 3.2, where they currently are and “do nothing” with those funds. The “do nothing” project does not mean that the investment funds are going to be buried in a can in the backyard where they earn nothing. The “do nothing” project means that the funds are already invested in a project that is earning the MARR. As mentioned before, for individuals, this could
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the funds are already invested in a project that is earning the MARR. As mentioned before, for individuals, this could mean leaving their funds in their savings accounts. By definition, the NP V of the “do 30 3. PROJECT EVALUATION METHODS nothing” project is zero.Thus, when a single investment opportunity is being evaluated, one is always comparing it against a second opportunity which is to leave the money in the “do nothing” project. 3.5.3 ANALYSIS OF MULTIPLE INVESTMENT OPPORTUNITIES For the purpose of this initial discussion of investing in multiple projects, assume that all of the prospective projects to be evaluated require the same initial investment, that the investor only has enough funds to invest in one of the projects, and that the decision will be based solely on NP V analysis. These assumptions will be removed in subsequent chapters and discussed further. In addi- tion, if at least one of the proposed projects has a positive NP V , then the “do nothing” project need not be considered. Example 3.3 Consider the following two investment opportunities. The investor’s MARR is 10% and the investor only has enough funds to invest in one of the projects. Which one should be chosen? Project A: 0 1 2 3 4 5 -800 215 215 215 215 215 Project B: 0 1 2 3 4 5 -800 100 100 100 100 900 N P V for Project A = −800 + 215(P /A)10,5 = $15.0 N P V for Project B = −800
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P V for Project A = −800 + 215(P /A)10,5 = $15.0 N P V for Project B = −800 + 100(P /A)10,5 + 800(P /F )10,5 = $75.8 Both projects show positive values of NP V . Therefore, both would be acceptable as long as the investor had at least $800 to invest. In addition, the “do nothing” alternative does not need to be considered. If the investor only has enough funds to invest in one of the projects, the NP V values indicate that Project B is the best economic choice. 3.6. RATE OF RETURN METHODS 31 3.6 RATE OF RETURN METHODS The second general type of project evaluation technique involves the determination of an unknown interest rate for a given cash flow diagram. This interest rate is usually referred to as a rate of return. There are several rates of return that can be calculated. Two will be presented in this chapter. The first is called the Internal Rate of Return (IRR) which is also known as the Discounted Cash Flow Rate of Return (DCFROR). The second is the External Rate of Return (ERR) which is also known as the Growth Rate of Return. The I RR is the rate of return earned by a particular individual’s or company’s investment.The ERR represents the overall growth of invested dollars for an individual or a company. The differences will become apparent in the following discussion and example problems. INTERNAL RATE OF RETURN (IRR) 3.6.1 The I RR is defined
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become apparent in the following discussion and example problems. INTERNAL RATE OF RETURN (IRR) 3.6.1 The I RR is defined as the interest rate which discounts a series of cash flows to an NP V value of zero: (cid:4) N P V = 0 = Present Value of Cash Flows with the interest rate equal to I RR (3.4) The equation can also be written as: N P V = 0 = n(cid:4) j =0 CFj (P /F )I RR,j = n(cid:4) j =0 CFj (1 + I RR)j (3.5) where, CFj = cash flow for period j j = period of cash flow n = total number of periods It should be noted that one cannot normally solve explicitly for the I RR from Equation 3.5. Therefore, a trial and error solution is usually required. Graphically, the relationship between N P V , interest rate, and I RR is demonstrated in Figure 3.2. Once the I RR is calculated, it is then compared with the MARR. If the I RR is greater than the MARR, the project is considered to be acceptable to the investor. 32 3. PROJECT EVALUATION METHODS NPV vs Interest Rate $ , V P N 600 500 400 300 200 100 0 -100 -200 -300 IRR = 0.125 0 0.05 0.1 0.15 0.2 0.25 Interest Rate, fra(cid:272)(cid:415)on Figure 3.2: General form of net present value as a function of interest rate. (Note, for this example, when NP V = 0, the interest rate, or I
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as a function of interest rate. (Note, for this example, when NP V = 0, the interest rate, or I RR, is 0.125.) Example 3.4 Consider the two investment opportunities examined in Example 3.3. The investor’s MARR is 10% and the investor only has enough funds to invest in one of the projects. What are the I RRs for each project? Project A: 0 1 2 3 4 5 -800 215 215 215 215 215 Project B: 0 1 2 3 4 5 3.6. RATE OF RETURN METHODS 33 -800 100 100 100 100 900 As noted, the calculation of I RR usually involves a trial and error approach. While the NP V versus interest rate curve is not a straight line, it is generally accurate enough to bracket the I RR solution within 5% and then linearly interpolate for the answer. Project A: N P V for Project A = −800 + 215(P /A)i,5 Interpolating for I RR: I RR = 10.0 + (cid:5) (cid:6) 15.0−0 15.0−(−79.3) (15.0 − 10.0) = 10.8% Project B: N P V for Project B = −800 + 100(P /A)i,5 + 800(P /F )i,5 Interest rate, % N P V 100.0 75.8 -67.0 (cid:6) 0.0 10.0 15.0 (cid:5) Interpolating for I RR: I RR = 10.0 + (15.0 − 10.0) = 12.6%. It should be noted that Figure 3.2 was generated with the cash flows from Project B. Thus, the “true” answer for I RR is 12.5% compared to the interpolated value of 12.6%.
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flows from Project B. Thus, the “true” answer for I RR is 12.5% compared to the interpolated value of 12.6%. 75.8−0 75.8−(−67.0) In this example, the I RRs of both projects are greater than the investor’s MARR, so both projects are acceptable. It would appear that since the I RR of Project B is greater than the I RR of Project A, then Project B is the best alternative. This is, indeed, the proper interpretation – but only because the initial investment values for both projects were the same. One must be very careful in ranking projects by I RR values as will be shown in Chapter 5. SPREADSHEET FORMULA FOR IRR 3.6.2 Excel® has a built-in function to calculate Internal Rate of Return. 34 3. PROJECT EVALUATION METHODS The function is: where, = I RR(values, guess) values = cash flows that occur for the project guess = initial estimate of the IRR (as a fraction) This function automatically takes care of the year 0 cash flow without having to include it as a separate term such as was necessary in the NP V calculation with Excel®. One can see that the cash flows in Figure 3.3 are the same as Project B in the previous example. A MARR = B 10% A B 1 MARR = 0.1 Year 0 1 2 3 4 5 CF -800 100 100 100 100 900 NPV = IRR = 75.8 12.5% 3 4 5 6 7 8 9 10 11 12 Year 0
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100 900 NPV = IRR = 75.8 12.5% 3 4 5 6 7 8 9 10 11 12 Year 0 1 2 3 4 5 CF -800 100 100 100 100 900 NPV = IRR = =B4+NPV(B1,B5:B9) =IRR(B4:B9,0.1) 1 3 4 5 6 7 8 9 10 11 12 Figure 3.3: Demonstration of the use of the NP V and I RR functions in Excel®. As in Figure 3.2, Excel provides the “true” value for I RR without the need for a trial and error solution and without interpolating. 3.6.3 EXTERNAL RATE OF RETURN (ERR) The External Rate of Return (ERR) or Growth Rate of Return is found by determining the interest rate that will satisfy the following equation. ⎡ ⎤ (cid:7) (cid:7) n(cid:4) (cid:7) (cid:7) (cid:7) (cid:7) j =0 (cid:7) (cid:7) (cid:7) (cid:7) (cid:7) (cid:7) Cj (P /F )MARR,j n(cid:4) ⎣ = j =0 Ij (F /P )MARR,n−j ⎦ (P /F )ERR,n (3.6) where, Cj = negative cash flow at period j Ij = positive cash flow at period j n = life of project 3.6. RATE OF RETURN METHODS 35 The equation states that positive cash flows (Ij s) derived from the project are reinvested at the MARR to generate a future value, which is called FI , at the end of the project life. All negative cash flows (investments) are brought back in time at the MARR to generate a present value, which is called PC, at year zero. The interest rate which will then discount FI
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to generate a present value, which is called PC, at year zero. The interest rate which will then discount FI to a value equal to the value of PC is determined to be the ERR. Another way of looking at the external rate of return is to set up a second project which is called the reinvestment project. The negative cash flows for the reinvestment project are the positive cash flows from the original project. A future value of the cash flows of the reinvestment project is determined using the MARR as the interest rate (FI ). The original project and reinvestment project are then added together to give a third project. The positive cash flows from the original project and the costs from the reinvestment project should have netted out to zero. The remaining cash flows for the third project will be the negative cash flow at year zero, any other negative cash flows from the original project at the year of occurrence, and the future value determined for the second project. All negative cash flows are brought back to time zero at the MARR to generate a present value (PC). The ERR is then determined by finding the interest rate which will bring the future value to a year zero value equal to the present value of the negative cash flows. The ERR method has a calculation advantage over the I RR method in that the ERR can be solved for directly without a trial and error procedure.
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over the I RR method in that the ERR can be solved for directly without a trial and error procedure. The steps in the calculation procedure are: Cj (P /F )MARR,j (cid:7) (cid:7) (cid:7) (cid:7) (cid:7) (cid:7) (cid:7) (cid:7) n(cid:4) (cid:7) (cid:7) (cid:7) (cid:7) j =0 n(cid:4) Ij (F /P )MARR,n−j PC = FI = j =0 ERR = (FI /PC)1/n − 1 (3.7) (3.8) (3.9) Example 3.5 Consider the two investment opportunities examined in Example 3.4. The investor’s MARR is 10% and the investor only has enough funds to invest in one of the projects. What are the ERRs for the projects? Project A: 0 1 2 3 4 5 -800 215 215 215 215 215 36 3. PROJECT EVALUATION METHODS Project B: 0 1 2 3 4 5 -800 100 100 100 100 900 Project A: PC = | − 800| = 800 FI = 215(F /A)10,5 = 1312.6 ERR = (1312.6/800)1/5 − 1 = 0.104 = 10.4% Project B: PC = | − 800| = 800 FI = 100(F /A)10,5 + 800 = 1410.5 ERR = (1410.5/800)1/5 − 1 = 0.120 = 12.0% In this example, the ERRs of both projects are greater than the investor’s MARR, so both projects are acceptable. It would appear that since the ERR of Project B is greater than the ERR of Project A, then Project B is the best alternative. This is, indeed, the proper interpretation – but only because the initial investment values for both projects were the same.
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This is, indeed, the proper interpretation – but only because the initial investment values for both projects were the same. Again, one must be very careful in ranking projects by ERR values as will be shown in Chapter 5. One additional observation can be made about the relationship between MARR, I RR, and ERR. The ERR will always lie between the MARR and the I RR. Thus, MARR ≤ ERR ≤ I RR or MARR ≥ ERR ≥ I RR Example 3.6 Consider the investment opportunity below. The investor’s MARR is 10%. What are the N P V , I RR, and ERR values for the project? Project A: 0 1 2 3 4 5 -1000 500 500 -200 500 500 3.7. THE REINVESTMENT QUESTION IN RATE OF RETURN CALCULATIONS 37 N P V = − 1000 + 500(P /F )10,1 + 500(P /F )10,2 − 200(P /F )10,3 + 500(P /F )10,4 + 500(P /F )10,5 N P V =$369.4 NPV: IRR: Interest rate, % 10.0 20.0 30.0 25.0 N P V 369.4 90.2 -100.8 -13.8 (cid:5) Interpolating between 20% and 25%: I RR = 20.0 + 90.2−0 90.2−(−13.8) (cid:6) (25.0 − 20.0) = 24.3% ERR: PC = | − 1000 − 200(P /F )10,3| = $1150.3 FI = 500(F /P )10,4 + 500(F /P )10,3 + 500(F /P )10,1 + 500 = $2447.6 ERR = (2447.6/1150.3)1/5 − 1 = 0.163 = 16.3% All three economic indicators show that this project is an acceptable one. SPREADSHEET FORMULA FOR ERR
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= 0.163 = 16.3% All three economic indicators show that this project is an acceptable one. SPREADSHEET FORMULA FOR ERR 3.6.4 Excel® has a built-in function that can be used to calculate the External Rate of Return. The function is: = MI RR(values, finance_rate, reinvestment_rate) where, values = cash flows that occur for the project finance_rate = interest rate for discounting the negative cash flows to year 0 (as a fraction) reinvestment_rate = interest rate for reinvesting the positive cash flows to year n (as a fraction) One needs to set both the finance_rate and the reinvestment_rate to MARR. As with I RR, this function automatically takes care of the year 0 cash flow without having to include it as a separate term. Figure 3.4 demonstrates this formula (along with NPV and IRR) for the cash flows given in Example 3.6. 3.7 THE REINVESTMENT QUESTION IN RATE OF RETURN CALCULATIONS The virtues of the I RR calculation have been argued for years by evaluators. When the I RR method was first introduced, it was met with a great deal of enthusiasm and is still one of the most popular 38 3. PROJECT EVALUATION METHODS A MARR = B 10% A B 1 MARR = 0.1 Year 0 1 2 3 4 5 CF -1000 500 500 -200 500 500 NPV = IRR = ERR = 369.5 24.3% 16.3% 3 4 5 6 7 8 9 10 11 12 13 Year 0 1 2 3 4 5 CF -1000 500 500
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5 6 7 8 9 10 11 12 13 Year 0 1 2 3 4 5 CF -1000 500 500 -200 500 500 NPV = IRR = ERR = =B4+NPV(B1,B5:B9) =IRR(B4:B9,0.1) =MIRR(B4:B9,B1,B1) 1 3 4 5 6 7 8 9 10 11 12 13 Figure 3.4: Demonstration of the use of the NP V , I RR, and MI RR(ERR) functions in Excel®. evaluation methods used. Surveys have indicated that a vast majority of the companies polled use I RR either by itself or in conjunction with other methods when evaluating projects. However, in spite of the popularity of the I RR method, many evaluators still question its meaning and validity. The basic question has to do with whether or not a reinvestment of incomes is implied in the calculation procedure. That is, one argument is that in order for the original project investment to “earn” the I RR, the positive cash flows generated by the project must be reinvested in another project that “earns” the same I RR. The other argument is that reinvestment is not necessary to “earn” the I RR. In fact, both arguments may be true depending on the evaluator’s perception of what is meant by the phrase “earning the IRR.” To begin the discussion of the reinvestment question, consider Example 3.7. 3.7. THE REINVESTMENT QUESTION IN RATE OF RETURN CALCULATIONS 39 Example 3.7 An investment of $5000 will yield $1931.45 at the end of each year for 4 years. What is the value of the
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of $5000 will yield $1931.45 at the end of each year for 4 years. What is the value of the project’s I RR? If the MARR is 15%, what is the project’s ERR? 0 1 2 3 4 -5000 1931.45 1931.45 1931.45 1931.45 I RR : NP V = −5000 + 1931.45(P /A)i,4 (P /A)I RR,4 = 2.5887 For NP V = 0, Examining the interest tables in Appendix A, one can determine that the I RR is 20.0%. ERR : PC = | − 5000| = $5000 FI = 1931.45(F /A)15,4 = $9644 ERR = (9644/5000)1/4 − 1 = 0.178 = 17.8% By definition, the calculation of ERR requires that the incomes be reinvested at the MARR of 15%. If the MARR had been higher, say 18%, the value of the ERR would have been higher. If the MARR were 20%, one can show that the ERR is now equal to 20% (same as the I RR). Thus, if the interest rate used for the reinvestment of incomes and for finding the present value of the costs (negative cash flows) is the I RR, then the values of MARR, I RR, and ERR will be identical. While not shown here, this can be demonstrated, mathematically, for any set of cash flows. Now, let’s expand on this example in order to determine the effect of different perceptions of an investment “earning” a particular interest rate. 3.7.1 PERCEPTION #1 The first perception of an investment “earning” a particular interest rate parallels
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“earning” a particular interest rate. 3.7.1 PERCEPTION #1 The first perception of an investment “earning” a particular interest rate parallels the concept of investing money in a savings account for a specified period of time. In this perception, “earning” means that an initial investment will yield a future value given by (F /P )i,n. Using the values from this example, a $5000 investment earning 20% (the I RR) for 4 years should result in a future sum of: F = 5000(F /P )20,4 = $10, 368 However, if the individual cash flows of $1931.45 (recall that these cash flows yielded an I RR of 20%) were buried in a can under a tree (thus earning no interest), the total future accumulated 40 3. PROJECT EVALUATION METHODS amount would be: F = (4)(1931.45) = $7, 725.80 Since the four individual cash flows yield a future sum significantly less than $10,368, the initial investment has not “earned” a 20% interest rate according to this perception of “earning.” In fact, the actual rate of return would be: i = (7725.80/5000)1/4 − 1 = 0.115 = 11.5%, not 20%! However, if the individual cash flows were reinvested in an account that earned 20% interest, the future sum accumulated in that account would be: F = 1931.45(F /A)20,4 = $10, 368 and the “earned” interest rate would indeed be 20%. Thus, in this perception, in order to “earn” the I RR (20%) interest rate on the entire initial investment ($5,000), any cash flows received before
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order to “earn” the I RR (20%) interest rate on the entire initial investment ($5,000), any cash flows received before the end of the project must be reinvested in another project that has the same I RR. 3.7.2 PERCEPTION #2 The second perception more closely parallels the concept of making a loan to a project and having that loan be paid back at some interest rate. In this perception, interest is “earned” only on the portion of the total loan that is still unpaid. The unpaid portion of the loan is also known as the unamortized portion. Again, consider the cash flows in Example 3.7. During the first year, interest is earned on the entire $5000 investment (or loan). The required interest amount at the calculated I RR of 20% would be: I1 = 5000(0.20) = $1000 This means that $931.45 can be used to “payback” a portion of the original investment, leaving an unamortized amount of $4,068.55. The required interest amount in the second year would then be: I2 = 4068.55(0.20) = $813.71 The reminder of that year’s cash flow, $1,117.74, would be used to further reduce the unamortized portion of the investment to $2,950.81. Table 3.1 summarizes this sequence for the entire project life. Note that the total interest “earned” is the same as would have been “earned” under perception #1 if the cash flows were not reinvested. However, banking institutions agree that this repayment scheme has indeed “earned” 20% on the original loan of $5,000. In the
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However, banking institutions agree that this repayment scheme has indeed “earned” 20% on the original loan of $5,000. In the opinion of the authors, the final conclusion is that the question of whether reinvestment of the cash flows at the I RR must occur or not is really more of an issue of perceiving what is meant by “earning a return.” Banking institutions readily “invest” in projects via loans to companies or Table 3.1: Amortization table for a loan 3.8. ACCELERATION PROJECTS 41 individuals and receive the I RR as defined in perception #2 without automatic reinvestment at that same rate. However, an individual or company that is expecting to generate a future sum of money based on earning the I RR on the original investment for the entire life of the project must depend on reinvestment of the cash flows at that specific I RR in order to actually have the desired future sum. It should be noted that, independent of the reinvestment question, I RR analysis still results in a powerful economic evaluation tool. 3.7.3 FINAL COMMENTS ON ERR AND IRR RELATIONSHIPS The ERR is a measure of the growth of the investment dollars. The I RR does not have the same meaning since it is a measure of the project profitability only. If a company wants a true measure of its growth based on a specific investment, then ERR analysis should be used. Both the I RR and ERR are valid investment analysis techniques and, if applied
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then ERR analysis should be used. Both the I RR and ERR are valid investment analysis techniques and, if applied correctly, will yield the same conclusion regarding the viability of an investment to the company or individual. It will be shown in the next section that the ERR method has some advantages in particular analysis situations. 3.8 ACCELERATION PROJECTS When a series of cash flows changes from a positive value to a negative value (or negative to positive) more than once, the cash flows may generate multiple positive real solutions to the I RR equation.The number of solutions is governed by Descartes’ rule. The rule states that if the terms of a polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is equal to the number of sign differences between consecutive nonzero 42 3. PROJECT EVALUATION METHODS coefficients. Since the I RR equation can be rearranged to form a polynomial of order n, this rule will apply since the coefficients will be related to the cash flows. A series of cash flows with more than one sign change is called an acceleration project.This type of project is created when a second capital investment must occur after one or more years of positive incomes. For example, consider a manufacturing facility that will require significant upgrading after several years. The multiple values of I RR rates calculated when there are multiple sign changes are difficult to interpret as to which might be
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of I RR rates calculated when there are multiple sign changes are difficult to interpret as to which might be the correct return on investment. Since the ERR equation does not form a polynomial, it always has a unique answer and, therefore, should be the rate of return technique of choice in acceleration projects. A modified I RR calculation can be made by finding the present value of all of the negative cash flows by discounting to year 0 at the MARR and then using the normal I RR equation. It should be noted that the investor can always use the equivalence methods (NP V specifically) in this situation without difficulty. In Example 3.6, a cash flow was presented that had sign changes between the 2nd and 3rd year and the 3rd and 4th year. In this case, the analyst should be aware that multiple positive values of I RR might exist. For that specific example, the nth order polynomial that is created by the NP V = 0 equation is developed as follows: − 1000 + 500(P /F )I RR,1 + 500(P /F )I RR,2 − 200(P /F )I RR,3 + 500(P /F )I RR,4 + 500(P /F )I RR,5 = 0 − 1000 + 500 (1 + I RR) + 500 (1 + I RR)2 − 200 (1 + I RR)3 + 500 (1 + I RR)4 + 500 (1 + I RR)5 = 0 I RR5 + 4.5 I RR3 + 7.5 I RR3 + 5.7 I
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500 (1 + I RR)5 = 0 I RR5 + 4.5 I RR3 + 7.5 I RR3 + 5.7 I RR2 + 1.4 I RR − 0.8 = 0 Since the 5th order polynomial only has one sign change, there is only one positive value of I RR for the cash flows in Example 3.6. Example 3.8 will demonstrate a situation where more than one positive value exists. Example 3.8 Given the following cash flow diagram, plot the NP V versus interest rate and determine the two positive values of I RR that would be predicted by Descartes’ rule. Assume an MARR of 5%. 3.8. ACCELERATION PROJECTS 43 Solution: From the plot of NP V versus interest rate and the Excel® spreadsheet, it can be seen that there are two values for I RR: 9.1% and 57.2%. One can use Excel® to find both rates of return by adjusting the initial guess. An initial guess of 10% will yield the 9.1% value and an initial guess of 50% will yield the 57.2% value. This creates an unfortunate situation in that one must have an idea of the value of the larger root in order to have Excel® compute it. The 6th order polynomial that could be developed is: I RR6 + 5.1 I RR5 + 9.3 I RR4 + 5.4 I RR3 − 2.7 I RR2 − 3.1 I RR + 0.3 = 0 One can see that there are two sign changes in the list of terms and, therefore,
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+ 0.3 = 0 One can see that there are two sign changes in the list of terms and, therefore, two positive values for I RR. As mentioned before, the multiple values of IRR cause difficulties in interpretation. With a total investment (without time value of money) of $320 and the total of the positive incomes (without time value of money) of $290, one would be hard pressed to accept that this project “earns” 9.1%, let alone 57.2%! Comparing 9.1% to the MARR of 5% would seem to indicate that this project is acceptable. 44 3. PROJECT EVALUATION METHODS Let’s examine the ERR, NP V , and modified I RR for this project: ERR : N P V : PC FI ERR = | − 100 − 90(P /F )5,4 − 80(P /F )5,5 − 50(P /F )5,6| = $274.0 = 90(F /P )5,5 + 120(F /P )5,4 + 80(F /P )5,3 = $353.3 = (353.3/274.0)1/6 − 1 = 0.043 = 4.3% . The table or the figure show that the NP V at an interest rate of 5% (the investor’s MARR) is -$10.4. The modified I RR would be calculated by replacing the negative cash flows with PC calculated above to create a new set of cash flows as follows: 0 1 2 3 4 5 6 -274 90 120 80 0 0 0 Using the trial and error solution technique or Excel®, the modified I RR is 2.9%. Thus, the ERR, the modified I RR, and the NPV
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solution technique or Excel®, the modified I RR is 2.9%. Thus, the ERR, the modified I RR, and the NPV indicate that this project is not an acceptable project for the investor. In summary, acceleration projects have the potential to add another level of complexity to the calculation of I RR in that multiple positive rates may exist. The authors strongly suggest that evaluators utilize N P V or ERR calculations to determine the economic viability of acceleration projects. 3.9 PAYOUT A supplementary evaluation technique that is frequently used is payout period or simply payout. Payout may be calculated with or without discounting although it is usually calculated without considering the time value of money. Payout refers to the time that it takes for a project to return its initial investment. Thus, it’s a quick measure of how long the investment is at risk. Although this time may be a very useful piece of information to compute for a particular project, payout analysis is limited in its use as an evaluation criterion. It does not serve as a useful screening criterion since it ignores any cash flows occurring past the payout period. Therefore, it must be used in conjunction with one of the evaluation techniques that have already been presented. Example 3.9 Given the following cash flow diagram, compute the undiscounted payout time and the discounted payout time if MARR is 15%. 0 1 2 3 4 3.9. PAYOUT 45 -100 60 60 60 60 Undiscounted Payout: Year Cash Flow
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is 15%. 0 1 2 3 4 3.9. PAYOUT 45 -100 60 60 60 60 Undiscounted Payout: Year Cash Flow 0 1 2 (cid:1086) 100 60 60 Cumula(cid:415)ve Cash Flow (cid:1086) 100 (cid:1086) 40 20 Interpolate between years 1 and 2 to find when the cumulative cash flow equals zero: Payout = 1 + (cid:3) (cid:2) −40 − 0 −40 − (20) (2 − 1) = 1.67 years Discounted Payout: Year Cash Flow Discounted Cash Flow 0 1 2 3 (cid:1086) 100 60 60 60 100 60(P /F ) 15.1 = 52.2 60(P /F ) 15.2 = 45.4 60(P /F ) 15.3 = 39.4 Cumula(cid:415)ve Discounted Cash Flow (cid:1086) 100 (cid:1086) 47.8 (cid:1086) 2.4 37.0 Interpolate between years 2 and 3 to find when the cumulative cash flow equals zero: Payout = 2 + (cid:3) (cid:2) −2.4 − 0 −2.4 − (37.0) (3 − 2) = 2.06 years Discounted payout measures the time for the project to return the initial investment and a 15% rate of return on that initial investment. 46 3. PROJECT EVALUATION METHODS 3.10 PROBLEMS 3.1. Calculate the present value and annual value of the following cash flow diagram. MARR is 15%. 0 1 2 3 4 5 6 7 -2500 500 650 800 800 800 800 800 3.2. Calculate the I RR and ERR for the cash flow diagram given in Problem 3.1. 3.3. An individual is considering the purchase of a property that he believes he can resell for $25,000 at the end of 10
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is considering the purchase of a property that he believes he can resell for $25,000 at the end of 10 years. The property will generate positive cash flows of $1,500 per year for the 10 years. What is the maximum that the individual should pay for the property if his MARR is 12%? 3.4. An investment of $10,000 will yield $33,000 at the end of 5 years with no other cash flows. What is the I RR of this investment? 3.5. Calculate the I RR for the following cash flow diagram. 0 1 2 3 4 5 -2000 -500 1000 1000 1000 1000 3.6. A company invests $30,650 in a project which yields an income (positive cash flow) of $10,000 in the first year, $9,000 in the second, $8,000 in the third, … etc … and $1,000 in the tenth, along with an extra $10,000 income at the end of year 10. The company’s MARR is 10%. Determine the I RR and ERR of this project. 3.7. Determine the N P V , ERR, and modified I RR for the following cash flow diagram. Use an MARR of 15%. 3.10. PROBLEMS 47 0 1 2 3 -50 100 100 -100 3.8. Determine the N P V , NAV , modified I RR, and ERR for the following cash flow diagram if the MARR is 10%. 0 1 2 3 4 -75 50 50 -30 200 3.9. You are a project engineer and you have to make a choice between two
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-75 50 50 -30 200 3.9. You are a project engineer and you have to make a choice between two contractors to perform some rebuilding work on a manufacturing facility. One contractor proposes that he will do the work for $1,300,000 payable immediately. The other contractor proposes that he will perform the same job for $1,400,000 payable in eight equal quarterly payments, starting 3 months after the job begins. A nominal rate of 14% should be used as the MARR. What equivalent annual interest rate is the second contractor offering? Which contractor’s offer would you accept? Repeat the analysis with the NP V technique. 3.10. John Q. Customer has received his bill for the next 6 months premium on his auto insurance. The bill allows him two methods to pay his premium of $189.00. He can either pay the entire amount now, or he can pay $99.00 now, which includes half of the premium plus a $4.50 prepaid “service charge” and $94.50 in two months, the other half of the premium. The insurance company is, implicitly, offering John a “loan.” What is the effective annual interest rate of the loan? Would you take the “loan?” Why or why not? 48 3. PROJECT EVALUATION METHODS 3.11. A project is expected to cost $2,000,000 and have the following net revenues: Year Net Revenue 1,000,000 800,000 600,000 400,000 200,000 100,000 1 2 3 4 5 6 Calculate the undiscounted and discounted payout periods. The MARR is 15%. 3.12. Engineer A retires at the
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4 5 6 Calculate the undiscounted and discounted payout periods. The MARR is 15%. 3.12. Engineer A retires at the age of 65 with a retirement account worth $500,000. At what interest rate would this amount need to be invested in order to withdraw $50,000 at the end of each of the next 15 years? 3.13. Develop an Excel® spreadsheet to compute NP V , NAV , NF V , I RR, and ERR for the cash flow diagram given in Problem 3.1. 3.14. Develop an Excel® spreadsheet to solve Problem 3.3 for MARR values of 5%, 10%, 12%, 15% and 20%. 3.15. Develop an Excel® spreadsheet to solve Problem 3.4 for initial investments of $5000, $10000 and $15000. 3.16. Develop an Excel® spreadsheet to solve Problem 3.5 for initial investments of $2000, $1500, and $1000. 3.17. Develop an Excel® spreadsheet to solve Problem 3.6. 3.18. Develop an Excel® spreadsheet to solve Problem 3.7. 3.19. Develop an Excel® spreadsheet to solve Problem 3.8. 3.20. Develop an Excel® spreadsheet to solve Problem 3.9. 3.21. Develop an Excel® spreadsheet to solve Problem 3.10. 3.22. Develop an Excel® spreadsheet to solve Problem 3.11 for MARR values of 5%, 10%, and 15%. 3.23. Develop an Excel® spreadsheet to solve Problem 3.12. C H A P T E R 4 49 Service Producing Investments 4.1 INTRODUCTION There are, in general, two types of investments—one which produces income and one which produces a service. A service producing investment is one that results in a cash flow
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produces income and one which produces a service. A service producing investment is one that results in a cash flow diagram that normally contains no positive cash flows with the exception of a possible salvage value of the service. Salvage value is the estimated value of an asset at the end of its useful life. It is assumed that the asset can be sold (as scrap metal for example) for this value as a positive cash flow to the project. The authors use the symbol L to represent the positive cash flow due to salvage value. An example of a service producing investment would be the consideration of either purchasing a new vehicle for a field office or leasing the vehicle. The vehicle provides a necessary service for the personnel in the field office but does not directly produce any income for the company. Generally, a leased vehicle would not have any salvage value since it is just returned to the leasing agency at the end of the lease period, while a purchased vehicle would have some salvage value since it could be sold to another owner. This chapter will discuss evaluation techniques for service producing investments for equal and unequal life alternatives. 4.2 EQUAL LIFE ALTERNATIVES Consider the following situation. An investment needs to be made by a company for a particular service that is necessary for the company to conduct its business. Two or more alternatives have been identified that provide the same service over the same time
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to conduct its business. Two or more alternatives have been identified that provide the same service over the same time period. These alternatives are known as equal life alternatives and they lend themselves to straight forward application of the evaluation methods that were presented in Chapter 3. 4.2.1 EQUIVALENCE TECHNIQUES The equivalence techniques, primarily NPV, are valid methods to choose the correct alternative. However, since service producing investments deal primarily with costs, NPV is replaced with Net Present Cost (NPC) which is the absolute value of the NPV. When the evaluator calculates NPC, the simplest approach is to change the signs of all of the project’s cash flows as will be demonstrated in Example 4.1. The alternative with the lowest NPC would be the best economic choice. Similarly, Net Annual Value (NAV) is replaced with Net Annual Cost (NAC). 50 4. SERVICE PRODUCING INVESTMENTS Example 4.1 Two alternatives are being considered which provide the same service and which have the same useful life of five years. Alternative A has an initial capital investment of $12,000, annual operating costs of $3,500, and a salvage value of $5,000. Alternative B has an initial capital investment of $20,000, annual operating costs of $1,500, and a salvage value of $10,000. If the company’s MARR is 15%, which alternative would be the best economic choice? Use NPC and NAC analysis. Alternative A: 0 1 2 3 4 5 -12000 -3500 -3500 -3500 -3500 -3500 L = 5000 Alternative B: 0 1 2 3 4 5
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