Book 2 - Whitman and Terry

YNTHESIS
YNTHESISL
L
LECTURES
ECTURES
ECTURESON
ON
ONE
E
ENGINEERING
NGINEERING
NGINEERING
SSSYNTHESIS
Series
SeriesEditor:
Editor:
Editor:Steven
Steven
StevenF.F.F.Barrett,
Barrett,
Barrett,University
University
UniversityofofofWyoming
Wyoming
Wyoming
Series
David
DavidL.
L.
L.Whitman,
Whitman,
Whitman,University
University
UniversityofofofWyoming
Wyoming
Wyoming
David
Ronald
RonaldE.
E.
E.Terry,
Terry,
Terry,Brigham
Brigham
BrighamYoung
Young
YoungUniversity
University
University
Ronald
The
Theauthors
authors
authorscover
cover
covertwo
two
twogeneral
general
generaltopics:
topics:
topics:basic
basic
basicengineering
engineering
engineeringeconomics
economics
economicsand
and
andrisk
risk
riskanalysis
analysis
analysisinininthis
this
thistext.
text.
text.Within
Within
Within
The
the
thetopic
topic
topicofofofengineering
engineering
engineeringeconomics
economics
economicsare
are
arediscussions
discussions
discussionson
on
onthe
the
thetime
time
timevalue
value
valueofofofmoney
money
moneyand
and
andinterest
interest
interestrelationships.
relationships.
relationships.
the
These
Theseinterest
interest
interestrelationships
relationships
relationshipsare
are
areused
used
usedto
to
todefine
define
definecertain
certain
certainproject
project
projectcriteria
criteria
criteriathat
that
thatare
are
areused
used
usedby
by
byengineers
engineers
engineersand
and
and
These
project
projectmanagers
managers
managerstototoselect
select
selectthe
the
thebest
best
besteconomic
economic
economicchoice
choice
choiceamong
among
amongseveral
several
severalalternatives.
alternatives.
alternatives.Projects
Projects
Projectsexamined
examined
examinedwill
will
will
project
include
includeboth
both
bothincome-and
income-and
income-andservice-producing
service-producing
service-producinginvestments.
investments.
investments.The
The
Theeffects
effects
effectsofofofescalation,
escalation,
escalation,inflation,
inflation,
inflation,and
and
andtaxes
taxes
taxes
include
on
onthe
the
theeconomic
economic
economicanalysis
analysis
analysisofofofalternatives
alternatives
alternativesare
are
arediscussed.
discussed.
discussed.Risk
Risk
Riskanalysis
analysis
analysisincorporates
incorporates
incorporatesthe
the
theconcepts
concepts
conceptsofofofprobability
probability
probability
on
and
andstatistics
statistics
statisticsinininthe
the
theevaluation
evaluation
evaluationofofofalternatives.
alternatives.
alternatives.This
This
Thisallows
allows
allowsmanagement
management
managementtototodetermine
determine
determinethe
the
theprobability
probability
probabilityofofof
and
success
successor
or
orfailure
failure
failureofofofthe
the
theproject.
project.
project.
Two
Twotypes
types
typesofofofsensitivity
sensitivity
sensitivityanalyses
analyses
analysesare
are
arepresented.The
presented.The
presented.Thefirst
first
firstisisisreferred
referred
referredtototoasasas
success
Two
the
therange
range
rangeapproach
approach
approachwhile
while
whilethe
the
thesecond
second
seconduses
uses
usesprobabilistic
probabilistic
probabilisticconcepts
concepts
conceptsto
to
todetermine
determine
determineaaameasure
measure
measureofofofthe
the
therisk
risk
risk
the
involved.
involved.
The
The
authors
authors
have
have
designed
designed
the
the
text
text
to
to
assist
assist
individuals
individuals
to
to
prepare
prepare
to
to
successfully
successfully
complete
complete
the
the
involved. The authors have designed the text to assist individuals to prepare to successfully complete the
economics
economicsportions
portions
portionsofofofthe
the
theFundamentals
Fundamentals
FundamentalsofofofEngineering
Engineering
EngineeringExam.
Exam.
Exam.
economics
FUNDAMENTALS OF
OF ENGINEERING ECONOMICS
ECONOMICS AND DECISION
DECISION ANALYSIS
FUNDAMENTALS
FUNDAMENTALS OF ENGINEERING
ENGINEERING ECONOMICS AND
AND DECISION ANALYSIS
ANALYSIS
Fundamentalsof
ofEngineering
Engineering
Fundamentals
of
Engineering
Fundamentals
Economicsand
andDecision
DecisionAnalysis
Analysis
Economics
and
Decision
Analysis
Economics
WHITMAN •• TERRY
TERRY
WHITMAN
WHITMAN • TERRY
Series
SeriesISSN:
ISSN:
ISSN:1939-5221
1939-5221
1939-5221
Series
M
M
Mor
Morgan
gan
gan&
Cl
Claypool
aypool
aypool Publishers
Publishers
Publishers
Mor
Cl
&
&
C
&C
&
Fundamentals of
of
Fundamentals
Engineering Economics
Economics
Engineering
and Decision
Decision Analysis
Analysis
and
David
DavidWhitman
Whitman
Whitman
David
Ronald
RonaldE.
E.
E.Terry
Terry
Terry
Ronald
About
AboutSYNTHESIs
SYNTHESIs
SYNTHESIs
About
&
&
&
Mor
Morgan
gan
gan Cl
Cl
Cl
aypool
aypoolPublishers
Publishers
Publishers
Mor
aypool
.mmooorrgrggaaannncccl lalaayyypppoooool l.l.c.cocoommm
wwwwwwwww. .m
ISBN:
ISBN:978-1-60845-864-6
978-1-60845-864-6
978-1-60845-864-6
ISBN:
90000
90000
90000
78
78
1608
608458646
458646
458646
99978
11608
Mor gan
gan & Cl
Cl aypool
Mor
Mor gan &
& Cl aypool
aypool
This
Thisvolume
volume
volumeisisisaaaprinted
printed
printedversion
version
versionofofofaaawork
work
workthat
that
thatappears
appears
appearsinininthe
the
theSynthesis
Synthesis
Synthesis
This
Digital
DigitalLibrary
Library
LibraryofofofEngineering
Engineering
Engineeringand
and
andComputer
Computer
ComputerScience.
Science.
Science.Synthesis
Synthesis
SynthesisLectures
Lectures
Lectures
Digital
provide
provideconcise,
concise,
concise,
original
originalpresentations
presentations
presentationsofofofimportant
important
importantresearch
research
researchand
and
anddevelopment
development
development
provide
original
topics,
topics,
published
publishedquickly,
quickly,
quickly,
digital
digitaland
and
andprint
print
printformats.
formats.
formats.
For
Formore
more
moreinformation
information
information
topics,
published
ininindigital
For
visit
visit
www.morganclaypool.com
www.morganclaypool.com
visit www.morganclaypool.com
YNTHESIS
YNTHESISL
L
LECTURES
ECTURES
ECTURESON
ON
ONE
E
ENGINEERING
NGINEERING
NGINEERING
SSSYNTHESIS
Steven
Barrett,
Series
Editor
StevenF.F.F.Barrett,
Barrett,Series
SeriesEditor
Editor
Steven
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
iii
Geometric Programming for Design and Cost Optimization (with Illustrative Case Study
Problems and Solutions)
Robert C. Creese
2009
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. 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
ISBN: 9781608458653
paperback
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
M
&C
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 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.
ix
Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii
1
2
3
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1
Engineering Economics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.1 Basic Engineering Economics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.2 Risk Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2
Decision Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.3
Fundamentals of Engineering Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Interest and the Time Value of Money . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.1
Time Value of Money . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.2
Sources of Capital . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.3
Interest Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.1 Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.2 Compound Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.3 Nominal, Effective, and Continuous Interest Rates . . . . . . . . . . . . . . . . . . . .
2.4
Cash Flow Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.5
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 The use of Financial Functions in Excel® . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.5.5 Example Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.6
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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4
4
4
5
Project Evaluation Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.2
Alternate Uses of Capital . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
x
4
5
3.3
Minimum Acceptable Rate of Return (MARR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.4
Equivalence Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.5
Net Present Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.1 Analysis of a Single Investment Opportunity . . . . . . . . . . . . . . . . . . . . . . . .
3.5.2 Do Nothing Project . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.3 Analysis of Multiple Investment Opportunities . . . . . . . . . . . . . . . . . . . . . .
27
27
29
30
3.6
Rate of Return Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.1 Internal Rate of Return (IRR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.2 Spreadsheet Formula for IRR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.3 External Rate of Return (ERR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.4 Spreadsheet Formula for ERR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
31
33
34
37
3.7
The Reinvestment Question in Rate of Return Calculations . . . . . . . . . . . . . . . . . .
3.7.1 Perception #1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.7.2 Perception #2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.7.3 Final Comments on ERR and IRR Relationships . . . . . . . . . . . . . . . . . . . .
37
39
40
41
3.8
Acceleration Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.9
Payout . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.10
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
Service Producing Investments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.2
Equal Life Alternatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.2.1 Equivalence Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.2.2 Rate of Return Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.3
Unequal Life Alternatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4.3.1 Least Common Multiple Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4.3.2 Common Study Period . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.4
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
Income Producing Investments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
5.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
5.2
Investment in a Single Project . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
5.3
Mutually Exclusive Alternatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.1 Equivalence Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.2 Rate of Return Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.3 Using Excel® . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
62
64
70
xi
6
7
5.4
Unequal Life Alternatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
5.5
Independent and Contingent Investments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5.1 Independent Investments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5.2 Contingent Investments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5.3 Limited Investment Capital . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.6
Ranking Alternatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
5.7
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
75
75
75
77
Determination of Project Cash Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
6.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
6.2
Escalation and Inflation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.1 Capital Investment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.2 Gross Revenue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.3 Operating Expenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.4 Before-Tax Profit Computation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.5 Before-Tax Cash Flow Computation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.6 Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.7 Taxable Income . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.8 State and Federal Income Tax . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.9 Net Profit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.10 Cash Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
106
106
107
107
107
107
108
109
109
110
110
Financial Leverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
7.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
7.2
Financial Leverage and Associated Risk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
7.3
Adjustment to Cash Flow Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
7.3.1 Leverage and Mutually Exclusive Projects . . . . . . . . . . . . . . . . . . . . . . . . . . 132
7.3.2 Excel® Spreadsheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
7.4
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
xii
8
Basic Statistics and Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
8.1
8.2
8.3
8.4
9
135
135
135
139
142
146
149
149
151
151
151
168
Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
9.1
9.2
A
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.1 Measures of Central Tendency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.2 Measures of Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.3 Frequency Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.4 Relative Frequency Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3.1 Classical Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3.2 Relative Frequency Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3.3 Subjective Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3.4 Probability Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.1.1 Range Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.1.2 Monte Carlo Simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
171
171
175
187
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. 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 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.
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
1
CHAPTER
1
Introduction
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 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, determining 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 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 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.
3
CHAPTER
2
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 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 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
INTEREST CONCEPTS
2.3.1
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 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
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. 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:
i m
ie = 1 +
−1
m
(2.2)
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 = e i − 1
(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 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.
5
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) 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.
2.4. CASH FLOW DIAGRAMS
0
1
2
3
…
n-2
n-1
n
CF0
CF1
CF2
CF3
…
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
-20,000
1
2
3
…
34
35
36
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.
7
8
2. INTEREST AND THE TIME VALUE OF MONEY
Present, P :
0
1
2
3
…
n-2
n-1
n
1
2
3
…
n-2
n-1
n
P
Future, F :
0
F
Annuity, A:
0
1
A
2
3
…
n-2
n-1
n
A
A
…
A
A
A
2
3
…
n-2
n-1
n
2G
…
(n-3)G
(n-2)G
Gradient, G:
0
1
G
2.5
(n-1)G
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
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.
2.5.1
SINGLE PAYMENTS
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 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,
(2.7)
P = F (P /F )i,n
9
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
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 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,
(2.11)
F = A(F /A)i,n
Rearranging Equation 2.10 and solving for A yields
A = F {i/[(1 + i)n − 1]}
(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)
2.5. 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
(2.15)
Rearranging Equation 2.14 and solving for A yields
A = P {[i(1 + i)n ]/[(1 + i)n − 1]}
(2.16)
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 equivalent value of A. That
is,
A = P (A/P )i,n
(2.17)
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
…
n-2
n-1
G
2G
…
(n-3)G
(n-2)G
n
(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]/[i 2 (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]/i 2 − 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 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
Symbol
Formula
Single Payment
Compound Amount
to F given P
(F / P) i,n
(1 + i) n
Single Payment Present
Worth
to P given F
(P / F) i,n
(1 + i) -n
Uniform Series
Compound Amount
to F given A
(F / A) i,n
(1 + i) n-1
i
Uniform Series Sinking
Fund
to A given F
(A / F) i,n
i
(1 + i) n-1
Uniform Series Present
Worth
to P given A
(P / A) i,n
(1 + i) n-1
i (1 + i) n
Capital Recovery
to A given P
(A/ P)
i,n
i (1 + i) n
(1 + i) n-1
Uniform Gradient
Present Worth
to P given G
(P/ G)
i,n
Uniform Gradient
Future Value
to F given G
(F / G) i,n
to A given G
(A / G)i,n
Uniform Gradient
Uniform Series
(1 + i) n-1
i 2(1 + i) n
n
i (1 + i) n
n
(1 + i) n-1
i2
i
1
i
n
(1 + i) n-1
2.5. INTEREST FORMULAS FOR DISCRETE COMPOUNDING
2.5.4
13
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:
• rate is the interest rate (as a fraction) per period
• nper is the number of interest bearing periods
• pmt is an annuity (A) sum of money
• pv is a present (P ) value sum of money (occurs at time = 0)
• fv is a future (F ) value sum of money (occurs at time = nper)
• 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
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
1
2
3
4
5
6
7
8
9
10
11
12
B
rate
nper
pmt(A)
pv (P)
fv (F)
type
F/A
F/P
P/A
P/F
A/P
A/F
0.1
10
-1
-1
-1
0
15.937
2.5937
6.1446
0.38554
0.16275
0.062745
1
2
3
4
5
6
7
8
9
10
11
12
A
rate
nper
pmt(A)
pv(P)
fv(F)
type
F/A
F/P
P/A
P/F
A/P
A/F
B
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:
• normal_rate = the nominal annual interest rate (as a fraction)
• 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 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 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% compounded 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 before computing
the correct formulas.
i m
0.08 12
ie = 1 +
−1= 1+
− 1 = 0.0830
m
12
F = A(F /A)i,n = 1, 000(F /A)8.30,10 = 1, 000{[(1 + 0.0830)10 − 1]/0.0830}
= 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
1000
2
1200
3
1400
4
1600
5
1800
6
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 = A(F /A)i,n = 1, 000(F /A)10,6 = 1, 000(7.7156) = $7, 715.60
FGradient = 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 to the $1,000 annuity and then
convert to the future.
AGradient = G(A/G)i,n = 200(A/G)10,6 = 200(2.2236) = $444.72
ATotal = 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
FCF 2
FCF 3
FCF 4
FCF 5
FCF 6
F
= P (F /P )i,n = 1, 000(F /P )10,5 = 1, 000(1.6105) = $1, 610.50
= P (F /P )i,n = 1, 200(F /P )10,4 = 1, 200(1.4641) = $1, 756.92
= P (F /P )i,n = 1, 400(F /P )10,3 = 1, 400(1.3310) = $1, 863.40
= P (F /P )i,n = 1, 600(F /P )10,2 = 1, 600(1.2100) = $1, 936.00
= P (F /P )i,n = 1, 800(F /P )10,1 = 1, 800(1.1000) = $1, 980.00
= P (F /P )i,n = 2, 000(F /P )10,0 = 2, 000(1.0000) = $2, 000.00
= 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
2000
2
1900
3
1800
…
…
8
1300
9
1200
10
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
2.6
19
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)]
2.6.2
(2.24)
(2.25)
(2.26)
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:
P̄ , F̄ , Ā = 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 /F̄ )i,n
(F /P̄ )i,n
(F /Ā)i,n
(P /Ā)i,n
= [i(1 + i)−n ]/[ln(1 + i)]
= [i(1 + i)n−1 ]/[ln(1 + i)]
= (ein − 1)/ i
= (ein − 1)/[i(ein )]
(2.27)
(2.28)
(2.29)
(2.30)
where, i is the nominal interest rate per period.
2.7
2.1.
PROBLEMS
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 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 $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
3000
2
3000
3
3000
4
3000
5
6
3300
3600
7
3900
8
4200
9
10
4500
4800
2.14. An engineer wishes to buy a house but can only afford monthly payments of $1500. 30year 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 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
0
X
2
1
2
X
X
3
4
100
120
5
6
140
160
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
1000
1
2
3
500
500
750
4
1000
5
6
800
7
8
600
400
9
10
2000
2.7. PROBLEMS
23
2.20. Calculate the future worth 5 years from now of a present sum of $2,000 if:
(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 continuously 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 it takes for an investment to
triple in value?
25
CHAPTER
3
Project Evaluation Methods
3.1
INTRODUCTION
In order to make informed decisions on one or more potential investments, methods must be developed 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.
A:
0
1
2
3
…
18
19
20
1,000,000
B:
0
1
100,000
2
100,000
3
100,000
…
18
19
…
100,000
100,000
20
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 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 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 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
3.4. EQUIVALENCE METHODS
27
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
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 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:
NP V =
Present Value of Cash Flows with i = MARR
(3.1)
NF V =
Future Value of Cash Flows with i = MARR
(3.2)
N AV =
Annuity Value of Cash Flows with i = MARR
(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 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
-1000
1
2
3
4
5
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 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
-1000
1
2
3
4
5
150
200
250
300
350
3.5. NET PRESENT VALUE
1
3
4
5
6
7
8
9
10
11
A
MARR =
B
10%
Year
0
1
2
3
4
5
CF
-1000
500
600
700
800
900
NPV =
1582
1
3
4
5
6
7
8
9
10
11
A
MARR =
29
B
0.1
Year
0
1
2
3
4
5
CF
-1000
500
600
700
800
900
NPV =
=B4+NPV(B1,B5:B9)
Figure 3.1: Demonstration of the use of the N P 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 acceptable 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 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 addition, 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
0
1
2
3
4
5
-800
100
100
100
100
900
Project B:
N 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
3.6
31
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.
3.6.1
INTERNAL RATE OF RETURN (IRR)
The I RR is defined as the interest rate which discounts a series of cash flows to an NP V value of
zero:
NP V = 0 =
Present Value of Cash Flows with the interest rate equal to I RR
(3.4)
The equation can also be written as:
NP V = 0 =
n
j =0
CFj (P /F )I RR,j =
n
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 NP 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
600
500
400
NPV, $
300
IRR = 0.125
200
100
0
-100 0
0.05
0.1
0.15
0.2
0.25
-200
-300
Interest Rate, fraĐƟon
Figure 3.2: General form of net present value as a function of interest rate. (Note, for this example, when
N P 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
3.6. RATE OF RETURN METHODS
33
Project B:
0
1
2
3
4
5
-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 +
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
0.0
100.0
10.0
75.8
-67.0
15.0
75.8−0
Interpolating for I RR: I RR = 10.0 + 75.8−(−67.0)
(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%.
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.
3.6.2
SPREADSHEET FORMULA FOR IRR
Excel® has a built-in function to calculate Internal Rate of Return.
34
3. PROJECT EVALUATION METHODS
The function is:
= I RR(values, guess)
where, 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.
1
3
4
5
6
7
8
9
10
11
12
A
MARR =
B
10%
Year
0
1
2
3
4
5
CF
-800
100
100
100
100
900
NPV =
IRR =
75.8
12.5%
1
3
4
5
6
7
8
9
10
11
12
A
MARR =
B
0.1
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)
Figure 3.3: Demonstration of the use of the N P 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.
⎡
⎤
n
n
Cj (P /F )MARR,j = ⎣
Ij (F /P )MARR,n−j ⎦ (P /F )ERR,n
(3.6)
j =0
j =0
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 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. The steps in the calculation procedure are:
n
PC = Cj (P /F )MARR,j (3.7)
j =0
n
Ij (F /P )MARR,n−j
(3.8)
FI =
j =0
ERR = (FI /PC )1/n − 1
(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. 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
-1000
1
2
3
4
5
500
500
-200
500
500
3.7. THE REINVESTMENT QUESTION IN RATE OF RETURN CALCULATIONS
37
NPV:
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
IRR:
Interest rate, %
10.0
20.0
30.0
25.0
NP V
369.4
90.2
-100.8
-13.8
90.2−0
Interpolating between 20% and 25%: I RR = 20.0 + 90.2−(−13.8)
(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.
3.6.4
SPREADSHEET FORMULA FOR ERR
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
1
3
4
5
6
7
8
9
10
11
12
13
A
MARR =
Year
0
1
2
3
4
5
NPV =
IRR =
ERR =
B
10%
CF
-1000
500
500
-200
500
500
369.5
24.3%
16.3%
1
3
4
5
6
7
8
9
10
11
12
13
A
MARR =
Year
0
1
2
3
4
5
NPV =
IRR =
ERR =
B
0.1
CF
-1000
500
500
-200
500
500
=B4+NPV(B1,B5:B9)
=IRR(B4:B9,0.1)
=MIRR(B4:B9,B1,B1)
Figure 3.4: Demonstration of the use of the N P 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 project’s I RR? If the MARR is 15%, what is the project’s ERR?
0
-5000
I RR :
1
2
1931.45
1931.45
3
1931.45
4
1931.45
NP V = −5000 + 1931.45(P /A)i,4
For NP V = 0, (P /A)I RR,4 = 2.5887
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 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 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 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
3.8. ACCELERATION PROJECTS
41
Table 3.1: Amortization table for a loan
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 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 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
500
200
500
500
+
−
+
+
=0
2
3
4
(1 + I RR) (1 + I RR)
(1 + I RR)
(1 + I RR)
(1 + I RR)5
I RR 5 + 4.5 I RR 3 + 7.5 I RR 3 + 5.7 I RR 2 + 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.
3.8. ACCELERATION PROJECTS
43
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%.
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 RR 6 + 5.1 I RR 5 + 9.3 I RR 4 + 5.4 I RR 3 − 2.7 I RR 2 − 3.1 I RR + 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 :
PC
FI
ERR
NP V :
= | − 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
-274
90
120
80
4
0
5
0
6
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 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%.
3.9. PAYOUT
0
1
2
3
4
-100
60
60
60
60
Year
Cash Flow
45
Undiscounted Payout:
0
1
2
о 100
60
60
CumulaƟve
Cash Flow
о 100
о 40
20
Interpolate between years 1 and 2 to find when the cumulative cash flow equals zero:
Payout = 1 +
−40 − 0
(2 − 1) = 1.67 years
−40 − (20)
Discounted Payout:
Year
0
1
2
3
Cash Flow
о 100
60
60
60
Discounted Cash Flow
100
60(P /F )
60(P /F )
60(P /F )
15.1
15.2
15.3
= 52.2
= 45.4
= 39.4
CumulaƟve Discounted
Cash Flow
о 100
о 47.8
о 2.4
37.0
Interpolate between years 2 and 3 to find when the cumulative cash flow equals zero:
Payout = 2 +
−2.4 − 0
(3 − 2) = 2.06 years
−2.4 − (37.0)
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
-2500
1
2
3
4
5
500
650
800
800
800
6
800
7
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 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
-2000
3.6.
1
-500
2
1000
3
1000
4
1000
5
1000
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.
3.10. PROBLEMS
3.7.
3.8.
3.9.
47
Determine the N P V , ERR, and modified I RR for the following cash flow diagram. Use
an MARR of 15%.
0
1
2
3
-50
100
100
-100
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
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
1
2
3
4
5
6
Net Revenue
1,000,000
800,000
600,000
400,000
200,000
100,000
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.
49
CHAPTER
4
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 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 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
-12000
Alternative B:
0
-20000
2
3
4
5
-3500
-3500
-3500
-3500
-3500
L = 5000
1
2
3
4
5
-1500
-1500
-1500
-1500
-1500
L = 10000
NPC:
N P CA = 12000 + 3500(P /A)15,5 − 5000(P /F )15,5 = $21, 250
N P CB = 20000 + 1500(P /A)15,5 − 10000(P /F )15,5 = $20, 060
NAC:
N ACA = 12000(A/P )15,5 + 3500 − 5000(A/F )15,5 = $6, 340
N ACB = 20000(A/P )15,5 + 1500 − 10000(A/F )15,5 = $5, 980
Both NPC and NAC analysis indicate that Alternative B is the best economic choice since it
has the lowest cost under these conditions.
4.2.2
RATE OF RETURN METHODS
Rate of return methods need to be altered since there are generally no positive cash flows in a service
producing investment except, perhaps, a salvage value. Under that scenario, the definitions of IRR
and ERR don’t make any sense and, in fact, generally do not result in positive values.
4.2. EQUAL LIFE ALTERNATIVES
51
When comparing two service producing investment alternatives, an incremental project rate
of return (either IRR or ERR) is determined and compared to the MARR. The cash flows for the
incremental project are found by taking the cash flows from the investment with the larger initial
capital cost and subtracting the cash flows from the investment with the lower initial capital cost. It
should be fairly obvious that if the alternative with the larger initial capital cost doesn’t have lower
annual costs than the alternative with the lower initial capital cost, it will never be the economic
choice. Therefore, one would expect the incremental project cash flow diagram to be represented by
a negative initial investment, followed by positive cash flows that represent the savings generated by
choosing the alternative with the larger initial capital cost over the alternative with the lower initial
capital cost. Thus, another name for this incremental project is the “savings project.”
The rate of return (either IRR or ERR) can now be calculated for the savings project. If the
rate of return is larger than the MARR, this indicates that the savings project is an acceptable project
which thereby insinuates that the correct economic choice would be the alternative with the larger
initial capital cost. The net savings that occur by choosing the alternative with the larger initial
capital cost more than offset its additional initial capital cost. If the IRR or ERR is less than the
MARR, the savings project is not an acceptable project and, therefore, the alternative with the lower
initial capital cost will be the economic choice.
If there are more than two alternatives, all of the alternatives should first be listed in descending
order of initial capital cost and the various pairings of alternatives would be evaluated using one of
the techniques above. For example, if there were three alternatives (A, B, C) in order of initial capital
costs (with A having the highest and C having the lowest), one would first compare A to B. If A is
the better choice, one would then compare A to C to determine the best overall choice. However, if
B were the better choice, the next comparison would be B to C to determine the best overall choice.
Example 4.2
Compare Alternatives A and B given in Example 4.1 and determine the best economic choice
using IRR and ERR techniques. Recall that the MARR is 15%.
Since Alternative B has the highest initial capital cost, the savings project would be created
by subtracting the cash flows of Alternative A from those of Alternative B:
Savings Project, B-A:
0
-8000
1
2000
2
2000
3
2000
4
2000
The N P V of this project is given by:
N P VB−A = −8000 + 2000(P /A)i,5 + 5000(P /F )i,5
5
2000
L = 5000
52
4. SERVICE PRODUCING INVESTMENTS
Interest Rate, %
15
20
NP V
1190.1
-9.4
Interpolation yields an I RR = 20%. Since I RR > MARR, B is the best economic choice.
ERR :
PC = | − 8000| = 8000
FI = 2000(F /A)15,5 + 5000 = 18485
ERR = (18485/8000)1/5 − 1 = 0.182 = 18.2%
Again, the ERR would indicate that Alternative B is the best economic choice.
Example 4.3
Given the 3 alternatives below that provide the same service over a 4 year period, develop
an Excel® spreadsheet that uses IRR analysis to determine which alternative is the best economic
choice. MARR is 10%.
Alternative A:
0
1
2
3
-300
-350
-400
0
1
2
3
-800
-320
-380
-440
-1000
4
-450
L = 200
Alternative B:
4
-500
L = 100
4.2. EQUAL LIFE ALTERNATIVES
53
Alternative C:
0
1
2
3
4
-700
-340
-410
-480
-550
L = 50
Spreadsheet and Results:
Incremental IRR calculations for Example 4.3:
A
B
C
D
E
F
G
1
2
3
4
5
6
7
8
9
10
11
12
The N P V and I RR functions are the same as presented in Chapter 3. The spreadsheet shows
the comparisons between all three projects. Since the initial investment of Alternative A is greater
than the initial investment of Alternative B and the initial investment of Alternative B is greater
than the initial investment of Alternative C, the alternatives are already correctly ordered by size
of initial investment. NPC analysis shows that Alternative B has the lowest net present cost and,
therefore, should be the alternative of choice.
The analysis of the incremental I RR calculations would be completed as follows:
1. Compare the first two alternatives.
2. Since the I RR of Incremental Project A-B (5.7%) is less than the MARR (10%), Alternative B
is a better choice than Alternative A.
54
4. SERVICE PRODUCING INVESTMENTS
3. Now compare Alternative B with Alternative C.
4. Since the I RR of Incremental Project B-C (23.5%) is greater than the MARR (10%), Alternative B is a better choice than Alternative C.
Therefore, Alternative B is the best economic choice (same as determined from the NPC
method).
Note that since Alternative B was a better choice than Alternative A, one never utilizes the incremental IRR that is calculated for the Incremental Project A-C. However, it is a necessary portion of the
Excel® spreadsheet since one does not know, ahead of time, which Alternatives will be eliminated
during the analysis of the results.
4.3
UNEQUAL LIFE ALTERNATIVES
The analysis of service producing investments that have alternatives which provide the same service
but have unequal project lives cannot be completed without modifications to the alternatives. A
common evaluation life for each alternative must be found before a proper economic decision can
be made. This is because the definition of two alternatives providing the same service includes the
assumption that they provide this service for the same length of time. For example, one cannot
compare an alternative to purchase a vehicle, keep it for 5 years, and then sell it for its salvage value
to a three-year lease option for the same vehicle. Both options are providing the service of a vehicle,
but the service is provided for different lengths of time.
There are, in general, two methods employed by evaluators to find common evaluation lives
in these situations. The first method requires the determination of a least common multiple of
service lives for the alternatives being considered. The second method involves the determination of
a common study period which will be either the life of the shortest or longest alternative. In both
methods, cost assumptions must be made that will impact the final analysis.
4.3.1
LEAST COMMON MULTIPLE METHOD
The least common multiple method of finding a common service life utilizes the same principles
that are involved in determining the common denominator when adding or subtracting fractions.
Consider the example of two alternatives having useful lives of 3 and 4 years. The least common
multiple in this case would be 12 since there is not a smaller number which is divisible by 3 and 4
without leaving partial years as a remainder. The alternative having a useful life of 3 years would be
repeated 4 times on a time line to reach the least common multiple of 12 years. The other alternative
would be repeated 3 times.
A couple of disadvantages of this method should immediately come to mind. First, costs do
not stay constant over time, so one would need to predict the future cost of each alternative. Cost
escalation will be discussed in Chapter 6, but even this approach requires a number of assumptions.
Secondly, one or more of the alternatives may be rendered obsolete by the development of new
4.3. UNEQUAL LIFE ALTERNATIVES
55
technology before the end of the time period that corresponds to the least common multiple is
reached.
4.3.2
COMMON STUDY PERIOD
The common study period method of finding a common service life utilizes either the life of the
shortest alternative or the life of the longest alternative as the common study period. To determine
which of these to use, the length of the common study period should be, if possible, the length of
time that the service is actually required.
If the life of the shortest alternative is used, the extra years of the longer life alternative are
neglected and a new salvage value is assigned at the end of the common study period. The new
salvage value will typically be larger than the original salvage value since it should reflect the value
of the extra years that are neglected.
If the life of the longest alternative is used, the shorter project needs to be extended via one
of two methods. The project can be extended by either estimating the cost involved to repair the
service to get additional years of service from it or by purchasing a new unit of service. Both of these
require some assumptions with regard to future cost.
Example 4.4
The cash flows shown below represent two alternatives which can provide the same service.
Assume that the MARR is 15%. Use both methods described above to determine which alternative
is the best economic choice. (Numbers are in $1,000.)
Alternative A:
0
1
2
3
-150
-3
-3
-3
0
1
2
3
-50
-18
-18
-18
……
9
10
-3
-3
L = 10
Alternative B:
4
-18
5
-18
L=8
56
4. SERVICE PRODUCING INVESTMENTS
Least Common Multiple Technique: The least common multiple of 5 and 10 is 10. Therefore,
one needs to extend Alternative B from 5 to 10 years. It will be assumed that there is no escalation
in the costs for Alternative B for the second 5 year period. In Chapter 6, we will consider this same
problem with cost escalation. Therefore, Alternative B extended to 10 years would be:
Alternative B (extended to 10 years):
0
1
2
-50
-18
-18
3
-18……
5
-18
L=8
-50
6
9
-18 …… -18
10
-18
L=8
N P C Analysis:
N P CA = 150 + 3(P /A)15,10 − 10(P /F )15,10 = $162.6
N P Cextended B = 50 + 18(P /A)15,10 + 42(P /F )15,5 − 8(P /F )15,10 = $159.2
N P C analysis indicates that Alternative B is the best economic choice under the assumptions
that were made (e.g., no increase in costs for the second 5 years). If costs increase or if technology
makes Alternative B obsolete, then this analysis will be inaccurate and one may need to consider
other non-economic factors in making this decision.
Common Study Period Technique: Let’s shorten Alternate A to 5 years by neglecting the costs
in the final 5 years and by increasing the salvage value that could be received at year 5 to $80,000.
Alternative A (shortened to 5 years):
0
1
2
3
4
5
-150
-3
-3
-3
-3
-3
L = 80
N P C Analysis:
N P Cshortened A = 150 + 3(P /A)15,5 − 80(P /F )15,5 = $120.3
NP CB = 50 + 18(P /A)15,5 − 8(P /F )15,5 = $106.4
4.4. PROBLEMS
57
N P C analysis indicates that Alternative B is the best economic choice under these set of
assumptions (e.g., the new estimated salvage value for Alternative A and the assumption that one
can actually “sell” Alternative A for salvage at the end of 5 years).
4.4
PROBLEMS
4.1.
A mining company is in need of four trucks. Suppliers will offer the options of purchasing
or leasing the trucks. The purchase price is $200,000. Maintenance, insurance, and general
operating costs (payable at the end of each year) will be $30,000 in year 1, $40,000 in year 2,
and $50,000 in year 3 with an expected salvage value of $70,000 at the end of year 3. The
lease price is $80,000 per year for the 3 years (payable at the beginning of each year). The
lease covers maintenance costs, but insurance and general operating costs will be $25,000
per year (payable at the end of each year). If the company’s MARR is 20%, determine the
best economic choice.
4.2.
A natural gas producing company is considering two engine systems for use in driving a
small compressor. System A can be purchased for $120,000 and is expected to have a life of 4
years. Annual diesel fuel consumption is estimated to be 60 gallons per day of use. System B
can be purchased for $150,000 and is expected to have a life of 4 years. Annual propane fuel
consumption is estimated to be 40 gallons per day of use. Both engines have salvage values
equal to 15% of initial cost and both will accomplish the needed requirements. Estimates
of fuel costs for each system and expected days of use each year are as follows:
Assume that MARR is 8% and that all other costs besides fuel will be the same for both
systems. Which system is the best economic choice?
58
4. SERVICE PRODUCING INVESTMENTS
4.3.
Use ERR analysis to determine which alternative would be the best economical choice.
Verify your decision with NPC analysis. Assume the MARR equals 10%.
Alternative A:
0
1
2
3
4
-500
-25
-25
-25
0
1
2
3
-300
-50
-50
-50
0
1
2
3
-250
-75
-60
-45
-30
L = 10
0
1
2
3
4
-450
-35
-35
-35
-25
L = 100
Alternative B:
4
-50
L = 25
Alternative C:
4
Alternative D:
-35
L = 100
4.4. PROBLEMS
4.4.
59
Consider the two service producing projects described below. They will provide the same
service but they do not have equal lives. Use NPC, IRR, and ERR analyses to determine
which alternative should be chosen. For the least common multiple method, assume no
increases in future costs for either project. For the common study period method, assume
that the salvage value for Alternative B will increase to $4,000 at the end of year 3. The
MARR is 10%.
Alternative A:
0
1
-15000
2
-1000
3
-1000
-1000
L=0
Alternative B:
0
-10000
1
-3000
2
-3000
3
-3000
4
-3000
L = 2000
4.5.
Use Excel® to solve Problem 4.1 for values of MARR of 10%, 15%, 20%, and 25%.
4.6.
Use Excel® to solve Problem 4.2 for values of MARR of 5%, 8%, and 12%.
4.7.
Use Excel® to determine what initial cost of Alternative A in Problem 4.2 would make the
two systems equal at an MARR of 8%.
4.8.
Use Excel® to solve Problem 4.3.
4.9.
Use Excel® to solve Problem 4.4.
61
CHAPTER
5
Income Producing Investments
5.1
INTRODUCTION
In the previous chapter, investments were considered that only provided a service of some kind for the
investor. In this chapter, investments that generate income (or profit) are discussed. The evaluation
techniques to be used will be identical to those introduced in Chapter 3. However, one additional
concept needs to be introduced when an investor is faced with making decisions between multiple
alternatives. This concept is the fact that income producing investment situations can be classified
as being either mutually exclusive, independent, or contingent as defined in later sections of this
chapter.
5.2
INVESTMENT IN A SINGLE PROJECT
If an investor is being offered the opportunity to invest in a single project (that is, without considering
any other alternatives other than the “do nothing” alternative), he needs to consider the following
two economic issues:
• Does he have enough money to invest in this project?
• Is the project profitable enough?
If one does not consider the option of the investor borrowing money from a lending institution, the
answer to the first question should be a clear “yes” or “no.” If the answer is “no,” then the investor
cannot invest in the project. Chapter 7 will cover financial leverage which will allow for the borrowing
of money.
If the answer to the first question is “yes,” then project profitability needs to be considered in
order to answer the second question. Utilizing the analysis techniques presented in Chapter 3, this
would mean one of the following:
• The N P V of the project, calculated at the investor’s MARR, is greater than zero. (Similarly,
N AV or N F V would be greater than zero.)
• The I RR of the project is greater than the investor’s MARR.
• The ERR of the project is greater than the investor’s MARR.
Of these three options, the authors strongly suggest the NPV method. This will become clearer as
this chapter proceeds.
62
5. INCOME PRODUCING INVESTMENTS
Example 5.1
An investor with MARR of 15% has been presented with the opportunity to invest in the
following income producing project. Assume that he has $20,000 to invest. Should he invest in this
project based on economic considerations?
0
1
-20000
7500
2
7500
3
7500
……
9
10
7500
7500
L = 10000
Using the NPV, IRR, and ERR techniques described in Chapter 3:
NP V :
I RR :
ERR :
N P V = −20000 + 7500(P /A)15,10 + 10000(P /F )15,10 = $20, 115
N P V = −20000 + 7500(P /A)I RR,10 + 10000(P /F )I RR,10 = 0
Trial and error solution yields I RR = 36.7%
PC = | − 20000| = $20, 000
FI = 7500(F /A)15,10 + 10000 = $162, 300
ERR = (162300/20000)1/10 − 1 = 0.233 = 23.3%
Since N P V > 0, I RR > MARR, and ERR > MARR, this project would be acceptable to
the investor.
5.3
MUTUALLY EXCLUSIVE ALTERNATIVES
When considering two or more alternatives in an economic analysis situation in which only one
alternative may be chosen, the alternatives are said to be mutually exclusive. Examples of mutually
exclusive alternatives would include the choice between two or more ways to develop a physical
property location (for example, build a gas station or a laundromat, but not both) or the choice
between two or more projects when faced with limited investment capital.
To evaluate choices in mutually exclusive situations, it is necessary to first determine if each
alternative is economically acceptable using the same questions as listed above. Any alternatives that
are not acceptable will be discarded. The remaining alternatives can then be ranked by a couple of
methods and the project at the top of the ranking is the best economic choice.
5.3.1
EQUIVALENCE TECHNIQUES
Equivalence techniques are those that use NPV, NAV, or NFV calculations. As explained earlier,
for a given project, if one of these values is greater than zero then the others will be also. Recall that
5.3. MUTUALLY EXCLUSIVE ALTERNATIVES
63
values greater than zero indicate that the alternative is an acceptable one. Obviously, if the value
is zero, the project earns exactly the MARR. Thus, the evaluation approach, using NP V as the
calculation choice, is as follows:
1. Calculate the N P V for each alternative.
2. Eliminate any alternative with NP V < 0.
3. If all alternatives have N P V < 0, then the investor’s decision should be the “do nothing”
alternative.
4. If one or more alternatives have NP V ≥ 0, the alternative with the largest positive NP V is
the best economic choice.
Example 5.2
In addition to the alternative given in Example 5.1, consider the situation where an investor
with an MARR of 15% has the choice between that alternative and the two additional ones given
below. Assume that the investor has $80,000 to invest. Also assume that the three alternatives are
mutually exclusive projects. This may occur because they represent alternatives in which only one
can actually be “built” or may occur because the investor only has $80,000 to invest so he only has
enough capital to invest in one.
Let’s call the project in Example 5.1 Alternative A. Thus, new alternatives are Alternative B
and Alternative C.
Alternative B:
0
-80000
1
20000
2
20000
3
20000
……
9
10
20000
20000
L = 25000
Alternative C:
0
-70000
1
17500
2
17500
3
17500
……
9
10
17500
17500
L = 21875
64
5. INCOME PRODUCING INVESTMENTS
Using the NPV technique described in Chapter 3:
N P V : N P VB = −80000 + 20000(P /A)15,10 + 25000(P /F )15,10 = $26, 560
N P VC = −70000 + 17500(P /A)15,10 + 21875(P /F )15,10 = $23, 240
Since N P VA , N P VB , and NP VC are all greater than zero, all three alternatives would be
acceptable to the investor. However, since these are mutually exclusive alternatives, Alternative B is
the overall best economic choice because its NP V is the largest.
One might think that the evaluator should directly compare any two projects (such as A
and B in the previous example) by using incremental NPV analysis. The following calculations will
demonstrate that this approach is not necessary because the NPV of an incremental project such as
B-A is governed by the following relationship:
NP V B−A = NP V B − NP V A
From Example 5.1, N P V A = $20, 115 and from Example 5.2, NP V B = $26, 560. Using the
relationship above, N P V B−A should be $6,445. The following cash flow diagram represents the
incremental project B-A:
Alternative B-A:
0
1
-60000
12500
2
12500
3
12500
……
9
10
12500
12500
L = 15000
NPV: N P V B−A = −60000 + 12500 (P A)15,10 + 15000 (P F )15,10 = $6, 445
Note that the NPV of the incremental project, B-A, is, indeed, numerically equal to the
difference between the NPV values of alternative B and A (B minus A).
5.3.2
RATE OF RETURN TECHNIQUES
One can use both the internal rate of return (IRR) and external rate of return (ERR) methods to
find the best alternative from a list of mutually exclusive alternatives. However, unlike N P V , it will
be shown that the alternative with the highest I RR or ERR is not necessarily the best economic
choice. One must be very careful not to simply rank the projects by I RR or ERR.
The process to determine the best alternative using I RR or ERR is as follows:
1. Calculate the I RR or ERR for each alternative.
5.3. MUTUALLY EXCLUSIVE ALTERNATIVES
65
2. Eliminate any alternative with I RR or ERR < MARR.
3. If all alternatives have I RR or ERR < MARR, then the investor’s decision should be the “do
nothing” alternative.
4. If one or more alternatives have I RR or ERR ≥ MARR, then those alternatives should be
rank ordered from the one with the highest initial investment to the one with the lowest initial
investment.
5. A comparison is made between the alternatives with the two largest initial investments. Create
an incremental project cash flow diagram by subtracting the cash flows of the lower initial
investment from those of the higher initial investment.
6. Calculate the I RR or ERR of the incremental project. If this I RR or ERR is ≥ MARR,
then the alternative with the larger initial investment is the better economic choice. Similarly,
if this I RR or ERR is < MARR, then the alternative with the lower initial investment is the
better economic choice. Keep the best alternative and discard the other one.
7. If additional alternatives are still available, return to step 5 and compare the alternative that
was kept from step 6 with the one with the next lower initial investment.
8. If no additional alternatives remain, the best economic choice is the alternative that was kept
from step 6.
Example 5.3
Consider the three alternatives A, B, and C introduced in the earlier example problems. Use
IRR and ERR analysis to determine the best economic choice. The MARR is 15%.
Alternative A:
0
1
-20000
7500
I RR :
2
7500
3
7500
……
9
10
7500
7500
L = 10000
N P V = 0 = −20000 + 7500(P /A)I RR,10 + 10000(P /F )I RR,10
Trial and error solution yields I RR = 36.7%
66
5. INCOME PRODUCING INVESTMENTS
ERR :
PC = | − 20000| = $20, 000
FI = 7500(F /A)15,10 + 10000 = $162, 300
ERR = (162300/20000)1/10 − 1 = 0.233 = 23.3%
Alternative B:
0
1
-80000
2
20000
I RR :
20000
3
20000
……
9
10
20000
20000
L = 25000
N P V = 0 = −80000 + 20000(P /A)I RR,10 + 25000(P /F )I RR,10
Trial and error solution yields I RR = 22.7%
ERR :
PC = | − 80000| = $80, 000
FI = 20000(F /A)15,10 + 25000 = $431, 100
ERR = (431100/80000)1/10 − 1 = 0.183 = 18.3%
Alternative C:
0
1
-70000
17500
I RR :
2
17500
3
17500
……
9
10
17500
17500
L = 21875
N P V = 0 = −70000 + 17500(P /A)I RR,10 + 21875(P /F )I RR,10
Trial and error solution yields I RR = 22.7%
5.3. MUTUALLY EXCLUSIVE ALTERNATIVES
ERR :
67
PC = | − 70000| = $70, 000
FI = 17500(F /A)15,10 + 21875 = $377, 200
ERR = (377200/70000)1/10 − 1 = 0.183 = 18.3%
As one can see, all three alternatives have I RR and ERR ≥ MARR. Therefore, all three
alternatives are acceptable. Putting them in ranked order by initial investment yields:
Alterna ve
B
C
A
Ini al
Investment
$80,000
$70,000
$20,000
I RR
22.7%
22.7%
36.7%
ERR
18.3%
18.3%
23.3%
At this point, one cannot simply choose the alternative with the highest I RR or ERR as the
best overall economic choice.
First, compare Alternative B to Alternative C:
Alternative B-C:
0
1
-10000
2
2500
2500
3
2500
……
9
10
2500
2500
L = 3125
Using the techniques described in Chapter 3:
I RR :
N P V = 0 = −10000 + 2500(P /A)I RR,10 + 3125(P /F )I RR,10
Trial and error solution yields I RR = 22.7%
ERR :
PC = | − 10000| = $10, 000
FI = 2500(F /A)15,10 + 3125 = $53, 900
ERR = (53900/10000)1/10 − 1 = 0.183 = 18.3%
Since both the I RR and ERR are greater than the MARR, this indicates that Alternative B
is better than Alternative C. Eliminate Alternative C from further consideration and compare
Alternative B to the next alternative.
68
5. INCOME PRODUCING INVESTMENTS
Comparing Alternative B to Alternative A:
Alternative B-A:
0
1
-60000
2
12500
12500
3
12500
……
9
10
12500
12500
L = 15000
Using the techniques described in Chapter 3:
I RR :
N P V = 0 = −60000 + 12500(P /A)I RR,10 + 15000(P /F )I RR,10
Trial and error solution yields I RR = 17.6%
ERR :
PC = | − 60000| = $60, 000
FI = 12500(F /A)15,10 + 15000 = $268, 800
ERR = (268800/60000)1/10 − 1 = 0.162 = 16.2%
Since both the I RR and ERR are greater than the MARR, this indicates that Alternative B
is better than Alternative A. Since the list of mutually exclusive alternatives has been exhausted,
Alternative B is the best overall economic choice.
In summary, one cannot use the values of the I RR and ERR from individual alternatives to
determine the best economic choice. If one were to do that, the results shown in the table for this
example would indicate that Alternative A is the best economic choice since it has the largest values
of I RR and ERR. However, both NP V and incremental rate of return analyses clearly show that
Alternative B is the best economic choice.
Example 5.4
To further reinforce the fact that one should not rank investments through the use of rate
of return, consider the following example. You are an investor with only $10 in your pocket. Two
friends offer you the following opportunities: Friend #1 needs $1 from you, but will give you $2 back
at the end of the day. Friend #2 needs all $10 of your money, but will give you $12 back at the end
of the day. Which opportunity is better for you from an economic point of view?
Examine this using NPV and incremental IRR approaches. Since the time frame is short (1
day), your daily MARR can be considered to be very close to 0%.
5.3. MUTUALLY EXCLUSIVE ALTERNATIVES
69
Friend #1 Alternative:
NP V :
0
1
-1
2
N P V = −1 + 2(P /F )0,1 = $1
I RR:
N P V = −1 + 2(P /F )I RR,1 = 0
Trial and error solution yields I RR = 100%
Friend #2 Alternative:
NP V :
0
1
-10
12
N P V = −10 + 12(P /F )0,1 = $2
I RR:
N P V = −10 + 12(P /F )I RR,1 = 0
Trial and error solution yields I RR = 20%
NPV analysis indicates that Friend#2 Alternative is the best economic choice, but IRR analysis
appears to indicate that Friend#1 Alternative is the best.
Friend #1 is offering a 100% rate of return and Friend #2 is offering a 20% rate of return. One
might think that Friend #1’s offer is the best. However, at the end of the day, you only have $11 in
your pocket if you invest with Friend #1, but $12 if you invest with Friend #2. It is clear, therefore,
that you should invest with Friend #2 even though that friend is offering a lower rate of return. The
reason that the higher rate of return option is not the best option in this case is that the other $9
in your pocket is earning 0% rate of return. Combining 0% rate of return on $9 and 100% rate of
return on $1 ends up yielding a 10% overall rate of return if you invest with Friend #1.
70
5. INCOME PRODUCING INVESTMENTS
Incremental Alternative of Friend#2 – Friend#1:
0
1
-9
10
I RR:
N P V = −9 + 10(P /F )I RR,1 = 0
Trial and error solution yields I RR = 11.1%
Since the incremental I RR is greater than your MARR, this indicates that Friend #2 Alternative is, indeed, the best economic choice.
This example also introduces the notion of risk in an investment. Obviously, the mathematical
analysis has shown that loaning Friend #2 is the better investment. But, it requires you, the lender,
to ‘give up’ all of your money. If there was a chance that neither friend could come through with
their repayment, then it might be better to keep the $9 in your pocket and invest in Friend #1. In
the event that neither friend could provide their repayment, at least you would still have $9 left of
your money. The concept of risk in investments will be discussed much more in Chapter 9.
5.3.3
USING EXCEL®
As shown in the previous chapters, Excel® can be used to choose the best alternative among a group
of mutually exclusive alternatives. Since Excel® offers the ability to quickly calculate incremental
rates of return, there is no need to manually choose the pairs of alternatives to be evaluated. However,
this requires that one needs to evaluate each possible pair of alternatives (starting with all alternatives ordered from highest to lowest initial investment) and then analyze the results table rather
than analyze specific pairs one at a time. For example, for Alternatives A, B, and C presented in
Examples 5.1 through 5.3, an Excel® spreadsheet might look like what is shown in Table 5.1. Recall
that within the original alternatives, Alternative B had the largest initial investment, C had the next
highest initial investment, and A had the lowest initial investment. Thus, the pairs of interest are
B-C, B-A, and C-A.
One would use Table 5.1 as follows:
If using NPV analysis:
1. Note that the values of NP V given in cells B17, C17, and D17 are all positive. This
indicates that all three alternatives are acceptable.
2. Note that Cell C17 contains the largest value of NP V . This would indicate that Alternative B is the best economic choice.
71
5.3. MUTUALLY EXCLUSIVE ALTERNATIVES
Table 5.1: Excel® solution of Examples 5.1 through 5.3
72
5. INCOME PRODUCING INVESTMENTS
If using IRR or ERR analysis (we will use IRR for this analysis):
1. Note that the values of I RR given in cells B18, C18, and D18 are all greater than the
MARR. This indicates that all three alternatives are acceptable.
2. Examine cell F18, which is the result of comparing the first pair of projects: B and C.
Since this value (22.7%) is larger than the MARR, this would indicate that Alternative B
is better than Alternative C. Alternative C is thus removed from further consideration
and the next viable pair would be B-A.
3. Examine cell G18, which is the result of comparing projects B and A. Since this value
(17.6%) is larger than the MARR, this would indicate that Alternative B is better than
Alternative A.
4. Since all necessary pairs have been examined, Alternative B is the best economic choice.
5. While column H is required to calculate the NP V , I RR, and ERR of the C-A pair, it
is not utilized in this example since Alternative C was removed from consideration after
its comparison against Alternative B. However, when developing this spreadsheet, one
does not know the result of the incremental analyses and, thus, all possible pairs must be
included. In addition, depending on the value of MARR, column H might be utilized
in other scenarios.
5.4
UNEQUAL LIFE ALTERNATIVES
Recall in Chapter 4 that if one was comparing service producing investments that have unequal
lives, one must choose one of two methods to force the projects to the same length of time. This is
because, to be comparable, the service must be offered for the same length of time.
In income producing investments, creating a common life is not required for NPV analysis.
However, for NAV, NFV, IRR, or ERR analysis, one must make the lives the same. Usually the life
of the longest alternative is used as the common evaluation life. If should be noted, however, that to
extend an income producing investment, one does not extend the positive cash flows. Instead, zero
cash flows are used to extend the life of the project. This is the case because one is assuming that the
cash flows from the income producing investment have already been estimated out to the full life of
the project and the project will be shut down at that time.
When conducting incremental rate of return analyses on unequal life alternatives, the evaluator
may find that the incremental project has multiple changes in sign of the yearly cash flows. This
was described in Chapter 3 as an acceleration project. Since the alternating signs may yield multiple
I RR values, either the modified IRR or ERR technique will need to be applied in the analysis.
Example 5.5
Use NPV, NAV, NFV, IRR, and ERR analyses to evaluate the unequal life alternatives below.
MARR is 12%.
5.4. UNEQUAL LIFE ALTERNATIVES
73
Alternative A:
0
1
2
3
4
-200
100
100
100
100
Alternative B:
0
1
2
3
4
5
6
-300
90
90
90
90
90
90
NPV analysis:
N P VA = −200 + 100(P /A)12,4 = $103.7
N P VB = −300 + 90(P /A)12,6 = $70.0
Alternative A is the best economic choice based on NPV analysis.
NAV, NFV, IRR, and ERR Analyses:
From the NPV analysis above, both alternatives are acceptable. Now, perform the incremental
analysis for NAV, NFV, IRR and ERR. Use six years as the common evaluation life by extending
Alternative A for two additional years with zero cash flows:
Alternative A extended to six years:
0
1
2
3
4
-200
100
100
100
100
5
0
6
0
74
5. INCOME PRODUCING INVESTMENTS
The incremental project is then:
Alternative B-A:
0
1
2
3
4
5
6
-100
-10
-10
-10
-10
90
90
N AV :
N AVB−A = −100(A/P )12,6 − 10(P /A)12,4 (A/P )12,6 + 90(F /A)12,2 (A/F )12,6
= $ − 8.20
Since the incremental NAV is less than zero, Alternative A is the best economic choice.
NF V :
N F VB−A = − 100(F /P )12,6 − 10(P /A)12,4 (F /P )12,6 + 90(F /A)12,2 = $ − 66.5
Since the incremental NF V is less than zero, Alternative A is the best economic choice.
I RR:
N P VB−A = 0 = −100 − 10(P /A)I RR,4 + 90(P /A)I RR,2 (P /F )I RR,4
Trial and error solution yields I RRB−A = 5.4%
Since I RRB−A is less than the MARR, Alternative A is the best economic choice.
ERR:
PCB−A =| − 100 − 10(P /A)12,4 | = $130.4
FIB−A =90(F /A)12,2 = $190.8
ERRB−A =(190.8/130.4)1/6 − 1 = 0.0655 = 6.55%
Since ERRB−A is less than the MARR, Alternative A is the best economic choice.
In summary, each of the analysis techniques of NPV, NAV, NFV, incremental IRR, and
incremental ERR, indicate that Alternative A is the best economic choice. However, of these five
options, the authors strongly suggest the NPV method because it usually involves the least amount
of calculations and never requires the use of incremental analyses.
5.5. INDEPENDENT AND CONTINGENT INVESTMENTS
5.5
INDEPENDENT AND CONTINGENT INVESTMENTS
5.5.1
INDEPENDENT INVESTMENTS
75
Consider the case when an investor is faced with the choice of investing in one or more projects (rather
than just one from a list of mutually exclusive alternatives) depending upon how much investment
capital is available. These alternatives are said to be independent alternatives. The final decision of
which projects to invest in will be based on maximizing the NP V for the given investment dollars.
This could mean that several combinations of projects will need to be evaluated.
5.5.2
CONTINGENT INVESTMENTS
A contingent project is a project that is conditional on the choice of one or more other projects. For
example, in the discipline of petroleum engineering, consider that the regional office of a large oil
company must make a decision to invest in one of the following projects for a particular producing
field: a series of well workovers to increase production from the existing wells; a polymer flood to
capture more oil from the field; or drilling a number of new wells within the field to expedite the oil
recovery from the field. Unfortunately, prior to investing in a full-scale polymer flood, the regional
office must also invest in a pilot polymer flood that will, most likely, not be an economic success
by itself. However, if the pilot is technically successful, then the full-scale polymer flood could be
considered. Therefore, the full-scale polymer flood would be considered a contingent project because
it could not be implemented without also choosing to invest in the pilot flood.
Example 5.6
Projects A, B, and C are being considered as investments. List the combinations that will need
to be considered under each of the following scenarios:
(a) The projects are mutually exclusive
(b) The projects are independent
(c) Projects A and B are mutually exclusive, but project C is contingent on project B.
(a) If the projects are already mutually exclusive, then the investor can only invest in one project.
Therefore, the list of combinations would be:
Mutually
Exclusive
AlternaƟve
Projects
Included
A B C
1
2
3
4
0
1
0
0
0
0
1
0
0
0
0
1
Possible
CombinaƟons
None
A
B
C
76
5. INCOME PRODUCING INVESTMENTS
(b) If the projects are independent, then the investor can invest in any or all projects. Therefore, the
list of combinations would be:
Mutually
Exclusive
AlternaƟve
Projects
Included
A B C
1
2
3
4
5
6
7
8
0
1
0
0
1
1
0
1
0
0
1
0
1
0
1
1
0
0
0
1
0
1
1
1
Possible
CombinaƟons
None
A
B
C
A,B
A,C
B,C
A,B,C
(c) For the contingencies given, the list of combinations would be:
Mutually
Exclusive
AlternaƟve
Projects
Included
A B C
1
2
3
4
0
1
0
0
0
0
1
1
0
0
0
1
Possible
CombinaƟons
None
A
B
B,C
Of the list from (b) in this example, the following combinations are missing for the following
reasons:
1. C only – Since C is contingent on project B, it cannot stand by itself
2. A,B – Since A and B are mutually exclusive, they cannot be combined together
3. A,C – Since C is contingent on project B and project B is not in this combination, A and C
cannot be combined
4. A,B,C – Since A and B are mutually exclusive, they cannot be combined together
5.5. INDEPENDENT AND CONTINGENT INVESTMENTS
5.5.3
77
LIMITED INVESTMENT CAPITAL
When investment capital is unlimited and more than one project may be chosen, the analysis simply
requires the determination of which project(s) will earn more than the MARR. This can be done
with any of the analysis techniques discussed previously. Once the list of acceptable alternatives has
been generated, the economic choice is to invest in all of them.
When investment capital is limited, the analysis approach is a bit more complicated. The
basic approach is to determine all possible combinations of projects in which the total investment
is within the capital constraints and then to analyze each of the combinations as being mutually
exclusive. The combination with the highest NPV will represent the set of projects in which one
should invest.
Example 5.7
The cash flow diagrams of six projects, A through F, are shown below. For these projects,
determine what combination of projects is the best economic choice using NPV analysis and a
MARR of 10%. Projects B, C, and E are mutually exclusive. Projects A and D are mutually exclusive
but both are contingent on the acceptance of C. Project F is contingent on the acceptance of either
B or E. Consider two separate scenarios:
(a) Assume unlimited capital
(b) Assume limited capital of $30,000
A:
B:
0
2
3
-5000
2500
2500
2500
0
1
2
3
-30000
C:
1
0
-15000
13500
1
10000
13500
2
10000
13500
3
10000
78
5. INCOME PRODUCING INVESTMENTS
D:
E:
0
2
3
-10000
6000
6000
6000
0
1
2
3
-20000
F:
1
0
-15000
10000
10000
1
2
11000
11000
10000
3
11000
It can be shown that the individual projects have the following NP V s:
Project
A
B
C
D
E
F
NP V
$1,220
$3,570
$9,870
$4,920
$4,870
$12,360
5.6. RANKING ALTERNATIVES
79
The table of mutually exclusive alternatives would be:
Mutuall y
Exclusive
Alterna ve
1
2
3
4
5
6
7
8
A
0
0
0
0
1
0
0
0
B
0
1
0
0
0
0
1
0
Projects
C D E
0 0 0
0 0 0
1 0 0
0 0 1
1 0 0
1 1 0
0 0 0
0 0 1
F
0
0
0
0
0
0
1
1
Possible
Combina ons
None
B
C
E
A,C
C,D
B,F
E,F
I n v es t m e nt
Capital
Needed
0
$30,000
$15,000
$20,000
$20,000
$25,000
$45,000
$35,000
NP V
0
$3,570
$9,870
$4,870
$11,090
$14,790
$14,930
$17,230
(a) There are eight mutually exclusive alternatives that result from the original six individual
projects and their interrelationships. When capital is unlimited, the correct economic choice is
the alternative that maximizes the NPV. In this case, alternative #8, which consists of investing
in projects E and F is the correct economic choice because it has the largest NPV.
(b) When capital is limited to $30,000, alternatives #7 and #8 are no longer considered. With
those removed, the correct economic choice will be alternative #6 since it will maximize the
NPV for those projects whose total investment is less than or equal to $30,000.
5.6
RANKING ALTERNATIVES
As mentioned earlier, one can always correctly rank alternatives according to their NP V values.
The combination of projects that is within any constraint of investment capital and has the highest
NP V will be the alternative of choice. However, one cannot correctly rank alternatives by I RR or
ERR values unless one utilizes incremental analyses. To illustrate this further, another example is
presented below.
Example 5.8
For the six projects listed in Example 5.7, use the IRR and ERR techniques to choose the
best mutually exclusive alternative.
It can be shown that mutually exclusive alternatives 1 through 8 have the following I RRs and
ERRs:
80
5. INCOME PRODUCING INVESTMENTS
AlternaƟve
1
2
3
4
5
6
7
8
I RR
10.0%
16.7%
44.6%
23.4%
39.5%
41.3%
29.2%
36.3%
ERR
10.0%
14.2%
30.2%
18.3%
27.4%
28.4%
21.7%
25.7%
Direct ranking by I RR or ERR would indicate that Alternative 3 (project C alone) would
be the best economic choice. This is, of course, inconsistent with the previous NPV analysis. To
overcome this inconsistency, the evaluator must perform incremental IRR or incremental ERR
analyses. For example, the incremental IRR technique is shown below:
1. Order the alternatives from the largest investment to the smallest:
AlternaƟve
Capital
Investment
$45,000
$35,000
$30,000
$25,000
$20,000
$20,000
$15,000
$0
7
8
2
6
4
5
3
1
Annual
Cash Flow
$24,500
$21,000
$13,500
$16,000
$10,000
$12,500
$10,000
$0
2. Calculate the I RR of the incremental project 7-8:
0
-10000
1
3500
2
3500
N P V7−8 = 0 = −10000 + 3500(P /A)I RR,3
Trial and error solution yields I RR7−8 = 2.5%
3
3500
5.6. RANKING ALTERNATIVES
81
Since the incremental I RR is less than the MARR, Alternative #8 is better than Alternative #7.
Keep Alternative #8, discard Alternative #7, and compare Alternative #8 with the next one on
the list (#2).
3. Calculate the I RR of the incremental project 8-2:
0
1
-5000
7500
2
3
7500
7500
N P V8−2 = 0 = −5000 + 7500(P /A)I RR,3
Trial and error solution yields I RR7−8 = 139.0%
Since the incremental I RR is greater than the MARR, Alternative #8 is better than Alternative #2. Keep Alternative #8, discard Alternative #2, and compare Alternative #8 with the
next one on the list (#6).
4. Continue in this manner (comparing the best choice with the next one on the list) until one
has exhausted all of the alternatives. Alternative #8 will be the last remaining alternative and,
thus, will be the best economic choice. This result is now consistent with the one from NPV
analysis. A similar method for ERR will yield the same ultimate results of Alternative #8 being
the best economic choice.
Another way to complete the incremental IRR technique is to compare each alternative with
all other alternatives that have a lower capital investment and compute the incremental IRR. This
would result in a table of incremental IRRs as given below:
Row #
1
2
3
4
5
6
7
Alterna ve with
higher capital
investment
Alterna ve with lower capital investment
8
2.5*
7
8
2
6
4
5
3
2
52.8
139
6
6.1
23.4
-1 7 4
4
33.8
52.8
2.5
106
5
20.7
32.1
-4 2 . 4
48.7
3
21.2
29.9
- 15.9
36.3
23.4
*IRR7-8
1
29.2
36.3
1 6.6
41.3
23.4
3 9 .5
44.6
82
5. INCOME PRODUCING INVESTMENTS
The use of this table would be as follows (see the arrows):
1. Start in row 1. Compare incremental alternative 7-8. Since the incremental I RR (2.5%) is less
than the MARR (10%), choose Alternative #8. Drop to row 2 (that belongs to Alternative #8).
2. Compare incremental alternative 8-2. Since the incremental I RR (139%) is greater than the
MARR (10%), choose Alternative #8. Stay in row 2.
3. Compare incremental alternative 8-6. Since the incremental I RR (23.4%) is greater than the
MARR (10%), choose Alternative #8. Stay in row 2.
4. Compare incremental alternative 8-4. Since the incremental I RR (52.8%) is greater than the
MARR (10%), choose Alternative #8. Stay in row 2.
5. Compare incremental alternative 8-5. Since the incremental I RR (32.1%) is greater than the
MARR (10%), choose Alternative #8. Stay in row 2.
6. Compare incremental alternative 8-3. Since the incremental I RR (29.9%) is greater than the
MARR (10%), choose Alternative #8. Stay in row 2.
7. Compare incremental alternative 8-1. Since the incremental I RR (36.3%) is greater than the
MARR (10%), choose Alternative #8. Since there are no more alternatives to be compared
with Alternative #8, then Alternative #8 is the best economic choice.
For the case of limited capital ($30,000), omit Alternatives #7 and #8 from the table. Follow
the arrows to show that Alternative #6 is the best economic choice.
Alterna ve with
higher capital
investment
2
6
4
5
3
Alterna ve with lower capital investment
6
-174
4
2.5
106
5
-42.4
48.7
-
3
-15.9
36.3
23.4
1
16.6
41.3
23.4
39.5
44.6
In summary, once the incremental I RR table has been created, start with the alternative
with the largest initial investment and compare it to the alternative with the second largest initial
investment. If the incremental I RR is less than the MARR, drop to the row of the lower initial
investment and proceed to compare with the next alternative. If the incremental I RR is greater than
the MARR, stay on the same row and proceed to compare with the next alternative. Eventually, one
will “exit” from the table on the best economic choice.
5.7. PROBLEMS
5.7
5.1.
PROBLEMS
Projects A and B below are mutually exclusive alternatives. The cash flow diagrams are
given. Determine which project is the best economic choice using NPV, IRR, and ERR
analyses. Use a value of 15% for MARR.
Project A:
0
-8000
1
2
3
5000
5000
5000
1
2
3
9
10
5000
5000
L = 8000
……
Project B:
0
-12000
5.2.
83
6000
6000
6000
……
9
10
6000
6000
L = 12000
Two mutually exclusive, but unequal life, investment projects A and B are shown below.
Project A:
0
1
2
3
4
5
-100
40
40
40
40
140
Project B:
0
1
2
-120
60
180
84
5. INCOME PRODUCING INVESTMENTS
(a) Determine the best economic choice using NPV, IRR, and ERR analyses. Use an
MARR of 20%.
(b) What value of MARR would reverse the ranking of projects A and B found in part
(a)?
For Problems 5.3 and 5.4.
The following projects are utilized in Problems 5.3 and 5.4. Projects A and B are independent. Projects C and D are mutually exclusive and both are dependent on the acceptance of
B. Project E is dependent on the acceptance of A.
-
B:
C:
-
D:
-
5.7. PROBLEMS
85
E:
5.3.
For the projects described above, do the following:
(a) List all mutually exclusive alternatives.
(b) Which alternative should be chosen if the MARR equals 10% and one has unlimited
capital?
(c) Which alternative should be chosen if the MARR equals 10% and investment capital
is limited to $80?
5.4.
For the projects described above, do the following:
(a) List all mutually exclusive alternatives.
(b) Develop the incremental I RR table.
(c) Use the table to determine which alternative should be chosen if the MARR equals
10% and one has unlimited capital.
(d) Use the table to determine which alternative should be chosen if the MARR equals
10% and investment capital is limited to $80.
5.5.
Use NPV and ERR analyses to determine which of the following two mutually exclusive
projects is the best economic choice. Use MARR of 15%.
Project A:
0
1
2
3
-500
200
200
200
4
200
L = 500
86
5. INCOME PRODUCING INVESTMENTS
Project B:
0
1
2
-200
100
100
L = 200
5.6.
Suppose you are considering two independent sets of two mutually exclusive projects each
plus a fifth project. The fifth project is contingent on two of the first four occurring. Make a
table that shows all of the mutually exclusive alternatives that are possible and the projects
that each alternative contains.
5.7.
Projects A through E are being considered by an investor. They all are ten-year projects and
the MARR is 10%. Projects A and B are mutually exclusive. Projects C and D are mutually
exclusive and contingent on the acceptance of B. Project E is contingent on the acceptance
of A.
Project
A
B
C
D
E
NP V
$5,000
ERR
8%
$20,000
$15,000
$10,000
Capital Investment
$20,000
$15,000
$30,000
$22,000
$15,000
(a) List all of the possible mutually exclusive alternatives.
(b) Which alternative is the best economic choice with unlimited capital?
(c) Which alternative is the best economic choice with a capital constraint of $40,000?
5.8.
Use Excel® to solve Problem 5.1 for values of MARR of 5%, 15%, 25%, 35%, and 45%.
5.9.
Use Excel® to solve Problem 5.2 for values of MARR of 10%, 20%, and 30%.
5.10. Use Excel® to solve Problem 5.3 for values of MARR of 10%, 20%, 25%, and 30%.
5.11. Use Excel® to develop the incremental I RR table for Problem 5.4. Use the table to determine which alternative should be chosen if the MARR equals 10% and one has unlimited
capital.
5.12. Use Excel® to solve Problem 5.5 for values of MARR of 5%, 15%, 25%, and 35%.
87
CHAPTER
6
Determination of Project Cash
Flow
6.1
INTRODUCTION
This chapter contains a discussion of escalation, depreciation, income taxes, and the subsequent
generation of cash flows when considering taxes. This chapter is not meant to be a detailed presentation on all of the ramifications of taxes. Most companies will use tax consultants and/or tax lawyers
instead of engineers to handle complicated tax questions. This chapter is meant to provide a basic
working knowledge of taxes so that the engineer can develop a stream of before- and after-tax cash
flows for a particular project.
6.2
ESCALATION AND INFLATION
When considering the effects of escalation on cash flows, it is necessary to define three types of
dollars with which evaluators work. The first is what is called today dollars. Today dollars simply refer
to the situation where all of the cash flows are calculated without any consideration for changes in
prices and costs as a function of time. This, of course, is not consistent with what actually occurs in
real life. A second type is escalated or actual dollars. When an evaluator attempts to estimate price
and cost changes and subsequently incorporates these changes into the cash flow calculations, then
the dollars are said to be escalated. The final type is constant dollars. When inflation is removed from
escalated dollars, then the resulting cash flows are said to be in constant dollars.
In order to more fully understand what is meant by these various types of dollars, the terms
inflation and escalation need to be defined.
Inflation refers to the general increase of prices with time due to an expanded money supply
with no hard assets to support the additional money. By definition, inflation affects prices of all
commodities by the same percentage amount. If the money supply decreases, there could be deflation
or the decrease in prices. There are as many causes of inflation as there are people who talk about it.
It is not the intent of the authors to discuss these causes. Using the Consumer Price Index (CPI)
that is published by the Bureau of Labor Statistics (http://www.bls.gov/data/), the average
inflation rate for 2000 to 2010 was 2.39% per year. The values of the CP I are shown in Figure 6.1
and Table 6.1 for various time periods. Published values are available back to 1913.
One can determine the average inflation rate for a given period of time by using the (F /P )i,n
formula. Consider the CP I values from two different years, n and m (with n > m). The CP I from
88
6. DETERMINATION OF PROJECT CASH FLOW
CPI Since 1960
250.0
200.0
150.0
CPI
100.0
50.0
0.0
1960
1970
1980
Year
1990
2000
2010
Figure 6.1: Consumer price index values for 1960-2010—from http://www.bls.gov/data/.
year n will be considered as a future value and the CP I from year m will be considered a present
value. Thus:
CP In = CP Im (1 + f )n−m which can be solved for the inflation rate, f , as
[1/(n−m)]
f = CP Iyr n /CP Iyr m
− 1 ∗ 100
(6.1)
For example, the average inflation rate between 1980 and 1990 was:
(130.7/82.4)[1/10] − 1 ∗ 100 = 4.72%
Similarly, the average inflation rate between 2009 and 2010 was:
(218.056/214.537)[1/1] − 1 ∗ 100 = 1.64%
Escalation, on the other hand, refers to the total change in the price of a specific commodity or
service over a period of time. Prices of individual commodities can change due to supply and demand,
as well as many other factors. While the inflation rate is a single numerical value for all commodities,
the escalation rate may be different for each commodity. For example, for 2000 to 2010, the price
of food increased an average of 2.69% per year (similar to inflation), the price of unleaded gasoline
increased an average of 6.32% per year, and the price of computers actually dropped an average of
6.2. ESCALATION AND INFLATION
89
Table 6.1: Consumer price index values for 1980-2010—from http://www.bls.gov/
data/
Year
1980
1981
1982
1983
1984
1985
1986
1987
1988
1989
** Star
CPI
82.4
90.9
96.5
99.6
103.9
107.6
109.6
113.6
118.3
124.0
Year
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
CPI
130.7
136.2
140.3
144.5
148.2
152.4
156.9
160.5
163.0
166.6
Year
CPI
2000
172.2
2001
177.1
2002
179.9
2003
184.0
2004
188.9
2005
195.3
2006
201.6
2007 207.342**
2008
215.303
2009
214.537
2010
218.056
ng in 2007, the Bureau of Labor Sta s cs began publishing the CPI with three
decimal places instead of one
16.4% per year over that same time frame. It should be pointed out that escalation includes the effect
of inflation. Figure 6.2 shows the price of unleaded gasoline ($/U.S. gallon) from 1976 to 2010.
One should notice that the CPI curve in Figure 6.1 is relatively smooth, but the price of any
one commodity may fluctuate significantly over the same time frame as shown in Figure 6.2. This
is due to the fact that the CPI “measures the average change in prices paid for a market basket of
goods and services.” (U.S. Department of Labor)
The escalated or actual dollar type of analysis referred to above includes both the effect of
inflation and escalation. This type of analysis attempts to predict the future prices of those elements
that are part of the cash flow calculation. One can either let all income and expenses rise at the
average inflation rate or one can attempt to isolate each commodity and use various escalation rates
for each income or expense item.
The constant dollar analysis reflects the purchasing power of money over the life of the project
by factoring out the effect of inflation. For example, in constant dollars, the price of unleaded gasoline
has increased an average of 3.84% per year and the price of food only increased an average 0.29%
per year from 2000 to 2010. This calculation will be shown later in this chapter.
The today dollar analysis simply uses the current prices for the commodities that are part of
the cash flow calculation for the project and maintains them at this level throughout the life of the
project. Thus, there is no consideration of the effects of inflation and escalation.
The authors believe that one should use either an escalated dollar analysis or a constant dollar
analysis when attempting to determine the economic viability of a project. Today dollar analyses
90
6. DETERMINATION OF PROJECT CASH FLOW
Price of Unleaded Gasoline
3.500
3.000
2.500
$/US gal
2.000
1.500
1.000
0.500
0.000
1970
1980
1990
2000
2010
2020
Year
Figure 6.2: Price of unleaded gasoline from 1976 to 2010 (Government Accounting Office analysis of
Bureau of Labor Statistics (BLS) data).
should only be used for projects that have short enough lives that the costs of the commodities that
are part of the cash flow calculation do not change substantially.
Two rules should be kept clearly in mind when incorporating the effects of inflation and
escalation. The first is that the dollar types, constant or escalated, should never be mixed within
a single cash flow diagram. The second is that the MARR that is used in the evaluation must be
consistent with the type of dollars used. A rate of return calculated from a set of cash flows that are
based on constant dollars should be compared with an MARR that is also based on constant dollars.
Similarly, consistency between escalated dollar cash flows and an MARR that is based on escalated
dollars is necessary. It should be noted that bank interest rates and investment bond interest rates
are based on escalated dollars. Thus, if an investor’s MARR is derived from those types of interest
rates, it should also be considered to be an escalated dollar MARR.
The relationship between interest rates in escalated and constant dollars can be obtained by
comparing the corresponding P /F factors:
(P /F )i,n = (P /F )f,n (P /F )ii,n
where, i = escalated dollar interest rate, fraction per period
f = inflation rate, fraction per period
ii = constant dollar interest rate, fraction per period
6.2. ESCALATION AND INFLATION
91
If one substitutes the definition of (P /F ) and does some basic algebra, one can show:
ii = (1 + i)/(1 + f ) − 1
(6.2)
This equation was utilized earlier to “factor out” the effect of inflation from escalation. For
example, for the period of 2000-2010, the inflation rate was 2.39% per year and the escalation rate
for unleaded gasoline was 6.32% per year. Using Equation 6.2, the constant dollar growth of this
commodity is:
ii = (1 + 0.0632)/(1 + 0.0239) − 1 = 0.0384 = 3.84%
Note that just subtracting the inflation rate from the escalation rate (a difference of 3.93% in this
example) is not the correct way to factor out inflation from escalation.
The relationships between today dollars, escalated dollars, and constant dollars are shown
below:
Escalated $ price at the end of the year n = (Today $) ∗ (1 + i)n
Constant $ price at the end of the year n = (Escalated $)/(1 + f )n
Constant $ price at the end of the year n = (Today $) ∗ ((1 + i)/(1 + f ))n
(6.3)
(6.4)
(6.5)
where, i = escalation rate, fraction per year
f = inflation rate, fraction per year
Example 6.1
Cash flow diagrams for projects A and B are shown below. Assume that the cash flows are
in escalated dollars and that the escalated dollar MARR is 15%. (a) Calculate the NP V of each
project as given and (b) calculate the NP V if one assumes a 5% inflation rate.
0
1
2
3
-100
55
60.5
66.55
0
1
2
3
-100
40
60
80
A:
B:
92
6. DETERMINATION OF PROJECT CASH FLOW
(a):
= −100 + 55(P /F )15,1 + 60.5(P /F )15,2 + 66.55(P /F )15,3
= $37.33
= −100 + 40(P /F )15,1 + 60(P /F )15,2 + 80(P /F )15,3
= $32.75
N P VA
N P VB
(b):
For part (b), one needs to factor out the effect of inflation from the escalated cash flows. In
addition, the MARR will have to be adjusted to a constant dollar basis. The cash flows are adjusted
by using the (P /F ) factor at 5% for the corresponding number of years. For example, the 55 (year 1
cash flow for Project A) is multiplied by (P /F )5,1 to yield 52.38. When this is done, the cash flows
become:
0
A:
-100
0
B:
-100
1
52.38
1
38.10
2
54.88
2
54.42
3
57.49
3
69.11
The constant dollar MARR will be:
ii = (1 + i)/(1 + f ) − 1 = (1 + 0.15)/(1 + 0.05) − 1 = 0.0952 = 9.52%
The N P V s then become:
N P VA = −100 + 52.38(P /F )9.52,1 + 54.88(P /F )9.52,2 + 57.49(P /F )9.52,3
= $37.34
N P VB = −100 + 38.10(P /F )9.52,1 + 54.42(P /F )9.52,2 + 69.11(P /F )9.52,3
= $32.66
Note that within numerical round off, the NP V s are the same for either escalated or constant dollar
analysis. This will always be the case.
6.2. ESCALATION AND INFLATION
93
Example 6.2
A five-year life project has an initial capital expenditure of $250,000 and annual operating
costs beginning at the end of the year 1 of $100,000. At the end of the years 3, 4, and 5 the project
receives $500,000 as income. Calculate the I RR for the following cases:
(a) Assume the cash flows given are in escalated dollars and the escalated dollar MARR is 25%.
(b) Assume the cash flows given are in today dollars and that incomes are escalated at 6% and
costs are escalated at 10%.
(c) Assume inflation is 4% and rework part (b) in terms of constant dollars.
(a) (numbers are in $1000):
0
1
2
3
4
5
-250
-100
-100
400
400
400
N P V = 0 = −250 − 100(P /A)I RR,2 + 400(P /A)I RR,3 (P /F )I RR,2
Trial and error solution yields I RR = 34.3%. This value would be compared to the escalated
MARR of 25% to indicate that it’s an economically acceptable project.
(b) (numbers are in $1000):
Use Equation 6.3 to convert today dollars to escalated dollars:
94
6. DETERMINATION OF PROJECT CASH FLOW
NP V = 0
= −250 − 110(P /F )I RR,1 − 121(P /F )I RR,2 + 463(P /F )I RR,3
+ 485(P /F )I RR,4 + 508(P /F )I RR,5
Trial and error solution yields I RR = 39.9%. This value would be compared to the escalated
MARR of 25% to indicate that it’s an economically acceptable project.
(c) (numbers are in $1000):
Use Equation 6.4 to convert the today dollars to constant dollars:
Year
0
1
2
3
4
5
Constant $ income
500 (1.06)3 / (1.04)3 = 529
500 (1.06)4 / (1.04)4 = 540
500 (1.06)5 / (1.04)5 = 550
Constant $ costs
-250 (1.10)0 / (1.04)0 = -250
-100 (1.10)1 / (1.04)1 = -106
-100 (1.10)2 / (1.04)2 = -112
-100 (1.10)3 / (1.04)3 = -118
-100 (1.10)4 / (1.04)4 = -125
-100 (1.10)5 / (1.04)5 = -132
Constant $ CF
-250
-106
-112
411
415
418
NP V = 0
= −250 − 106(P /F )I RR,1 − 112(P /F )I RR,2 + 411(P /F )I RR,3
+ 415(P /F )I RR,4 + 418(P /F )I RR,5
Trial and error solution yields I RR = 34.5%. This value would be compared to the constant
dollar MARR that is calculated according to Equation 6.1:
ii = (1 + 0.25)/(1 + 0.04) − 1 = .202 = 20.2%
This value would still indicate that it’s an economically acceptable project.
6.3
DEPRECIATION
Certain capital assets of a company lose their value with use and/or with time. A building or an
item of equipment are examples of such assets. These assets have an initial value that is equal to
the original cost of the asset. However, they may lose value over time due to physical deterioration,
development of improved facilities by technological advances, or different demands of their use. The
reduction in value is called depreciation.
6.3. DEPRECIATION
95
One also needs to recognize that most governments (including the United States) do not
allow companies, for tax purposes, to deduct the entire cost of an asset against their income in
the year that the asset is purchased. Since the asset retains at least some portion of its value over
its life, companies must prorate the deduction of the original asset cost over the usable life of the
asset. Governments will specify particular techniques for this proration. These techniques are called
depreciation methods.
Therefore, there are two interpretations of a depreciation account for a capital asset. Under the
first, a company would set aside actual cash in a depreciation account in order to have the necessary
funds to replace the asset at the end of its useful life. Under the second, rather than setting aside actual
cash in the depreciation account, the company would simply establish depreciation accounts for tax
purposes. That is, the depreciation account represents the allowable annual deduction of the asset
against the project’s income. The second interpretation represents reality. Thus, the depreciation
account that is maintained does not involve real dollars and depreciation expenses are known as
“paper” expenses in that they reduce the tax liability of the project but do not represent actual cash
expenditures. This chapter contains information on how to handle these paper expenses in the
calculation of after-tax cash flows for a project.
The most popular depreciation methods used in the United States are straight-line, sumof-the-years-digits, declining-balance, and the accelerated-cost-recovery-system. All four of these
methods will be discussed in this chapter.
In addition to the depreciation account, one also maintains a book value account that represents
the remaining value of the asset. Book value is simply the initial cost of the asset minus all accumulated
depreciation up to a specific point in time.
Depreciation calculations are based on the initial cost of the asset, P , any salvage value of the
asset at the end of its useful life, L, and the length of its useful life, N. The quantity P – L represents
the total allowable depreciation of the asset if it is held for the entire time period N.
6.3.1
STRAIGHT-LINE DEPRECIATION (SL)
When using the straight-line depreciation method, the yearly amount of depreciation is given by
Equation 6.6:
Dn = (P − L)/N
(6.6)
where, Dn = depreciation amount in year n, $
n = year of depreciation
P = initial cost of the asset, $
L = salvage value of the asset at the end of its useful life, $
N = length of the asset’s useful life, years
It should be evident that the depreciation is constant with time when using the straight-line
method.
96
6. DETERMINATION OF PROJECT CASH FLOW
At the end of any given year, the book value of the asset is given by Equation 6.7:
Bn = P − n(P − L)/N
(6.7)
where, Bn = book value of the asset at the end of year n, $
Excel® has a built-in function called SLN that computes straight-line depreciation:
= SLN(Initial_Cost, Salvage, Life)
where, Initial_Cost = initial cost of the asset (P )
Salvage = salvage value of the asset (L)
Life = asset’s useful life (N)
6.3.2
DECLINING-BALANCE DEPRECIATION
Unlike straight-line depreciation, the annual depreciation amount determined using the decliningbalance method is not constant with time. The declining-balance method provides for a larger
depreciation deduction in the early years of an asset’s life than when using straight-line depreciation.
In this method, the depreciation amount is a fixed percentage of the remaining book value of the
asset.
The equations to calculate the annual depreciation amount and the book value at the end of
each year are given in Equations 6.8 and 6.9:
Dn = f (1 − f )n−1 P
Bn = (1 − f )n P
(6.8)
(6.9)
where, f = a fixed percentage as a fraction
It should be noted that while the salvage value, L, is not utilized in the equations, one must
be careful that the total depreciation does not exceed the amount (P − L).
Limits have been placed on the value of f that can be used in the declining-balance method.
The value of f cannot exceed 2/N. When the value of 2/N is used, the method is referred to as the
double-declining-balance (DDB) method.
Excel® has a built-in function called DDB that computes double-declining balance depreciation:
= DDB(Initial_Cost, Salvage, Life, Period, Factor)
6.3. DEPRECIATION
97
where, Initial_Cost = initial cost of the asset (P )
Salvage = salvage value of the asset (L)
Life = asset’s useful life (N )
Period = the period of interest
Factor = 2 (or omitted) for double-declining balance
6.3.3
SUM-OF-THE-YEARS-DIGITS (SYD) DEPRECIATION
This method, like the declining-balance method, provides for an accelerated depreciation deduction
in the early years of the useful life of an asset.
The equations to calculate the annual depreciation amount and the book value at the end of
each year are given in Equations 6.10 and 6.11:
Dn = [N − (n − 1)](P − L)/S
n
Dj
Bn = P −
(6.10)
(6.11)
j =1
where,
S = sum of the digits of the useful life of the asset = N(N + 1)/2
Excel® has a built-in function called SYD that computes sum-of-the-years-digits depreciation:
= SYD(Initial_Cost, Salvage, Life, Period)
where, Initial_Cost = initial cost of the asset (P )
Salvage = salvage value of the asset (L)
Life = asset’s useful life (N )
Period = the period of interest
When calculating depreciation amounts for the determination of after-tax cash flows, it is
advantageous to use the most accelerated depreciation schedule possible. The sum-of-the-yearsdigits and the declining-balance methods give larger depreciation amounts in the early years of an
asset. The straight-line method may, however, be more advantageous in later years.
Example 6.3
A device costs $5000 and has a salvage value of $800 after its useful life of 7 years. Calculate
the depreciation deduction that can be taken each year and the book value at the end of each year
for the useful life of the asset. Use the following depreciation methods:
98
6. DETERMINATION OF PROJECT CASH FLOW
(a) Straight-Line (SL)
(b) Double-Declining-Balance (DDB)
(c) Sum-of-the-Years-Digits (SYD)
(a) For Straight-Line:
Dn = (P − L)/N = (5000 − 800)/7 = $600 which remains constant over the 7 years
Bn = P − n(P − L)/N = 5000 − 600 n
(b) For Double-Declining Balance:
f = 2/N = 2/7 = 0.28571
Dn = f (1 − f )n−1 P = 0.28571(0.71429)n−1 5000 = 1428.55(0.71429)n−1
Bn = (1 − f )n P = (0.71429)n 5000 = 5000(0.71429)n
0
1
2
3
4
5
6
7
1429
1020
729
521
372
129*
0**
5000
3571
2551
1822
1301
929
800
800
*D6 would have been calculated as $266, but it was limited to $129 because the book value
cannot go below the salvage value.
**D7 would have been calculated as $190, but it was limited to $0 because the book value had
already reached the salvage value at the end of year 6.
6.3. DEPRECIATION
99
(c) For Sum-of-the-Years-Digits:
S = N (N + 1)/2 = (7)(8)/2 = 28
Dn = [N − (n − 1)](P − L)/S = [7 − (n − 1)](5000 − 800)/28 = 150(8 − n)
n
n
Dj = 5000 −
Dj
Bn = P −
j =1
j =1
The depreciation and book values are shown in Figures 6.3 and 6.4 below to further demonstrate the differences between these three methods.
Solution with Excel® :
A
1
2
3
4
5
6
7
8
9
10
11
12
B
P=
L=
N=
5000
800
7
Year
1
2
3
4
5
6
7
SL
$600
$600
$600
$600
$600
$600
$600
C
DDB
$1,429
$1,020
$729
$521
$372
$130
$0
D
SYD
$1,050
$900
$750
$600
$450
$300
$150
1
2
3
4
5
6
7
8
9
10
11
12
5000
800
7
SL
=SLN($B$1,$B$2,$B$3)
=SLN($B$1,$B$2,$B$3)
=SLN($B$1,$B$2,$B$3)
=SLN($B$1,$B$2,$B$3)
=SLN($B$1,$B$2,$B$3)
=SLN($B$1,$B$2,$B$3)
=SLN($B$1,$B$2,$B$3)
Ye a r
1
2
3
4
5
6
7
B
P=
L=
N=
A
DDB
=DDB($B$1,$B$2,$B$3,A6)
=DDB($B$1,$B$2,$B$3,A7)
=DDB($B$1,$B$2,$B$3,A8)
=DDB($B$1,$B$2,$B$3,A9)
=DDB($B$1,$B$2,$B$3,A10)
=DDB($B$1,$B$2,$B$3,A11)
=DDB($B$1,$B$2,$B$3,A12)
C
SYD
=SYD($B$1,$B$2,$B$3,A6)
=SYD($B$1,$B$2,$B$3,A7)
=SYD($B$1,$B$2,$B$3,A8)
=SYD($B$1,$B$2,$B$3,A9)
=SYD($B$1,$B$2,$B$3,A10)
=SYD($B$1,$B$2,$B$3,A11)
=SYD($B$1,$B$2,$B$3,A12)
D
100
6. DETERMINATION OF PROJECT CASH FLOW
6.3. DEPRECIATION
101
DepreciaƟon Values
1600
1400
1200
1000
SL
D(n) $ 800
600
DDB
400
SYD
200
0
1
2
3
4
5
6
7
Year
Figure 6.3: Comparison of depreciation values for straight-line, double declining balance, and sum-ofthe-years-digits methods.
Book Values
5000
4500
4000
3500
3000
B(n) $ 2500
2000
1500
1000
500
0
SL
DDB
SYD
0
1
2
3
4
5
6
7
Year
Figure 6.4: Comparison of book values for straight-line, double declining balance, and sum-of-theyears-digits methods.
102
6. DETERMINATION OF PROJECT CASH FLOW
6.3.4
MODIFIED ACCELERATED COST RECOVERY SYSTEM (MACRS)
In 1981, the United States government passed the Economic Recovery Tax Act which made significant changes in depreciation calculations. The act was further modified in 1986 which led to
the Modified Accelerated Cost Recovery System (MACRS) for assets that were placed after 1980.
MACRS generally simplified the calculation of depreciation by (a) removing any reference to the
salvage value of the asset at the end of its useful life by assuming that L = 0 and (b) using various
combinations of the three previously presented depreciation methods to calculate annual depreciation values that are simply percentages of the original asset cost. As in the other methods, the asset’s
book value is the original cost minus all accumulated depreciation. That is:
Dn = P ∗ Depreciation Rate(depreciable life, n)
n
Dj
Bn = P −
(6.12)
(6.13)
j =1
To determine what depreciation rate to use, one must first determine the depreciable life of the
asset. The MACRS method created the following classifications: 3-year property, 5-year property,
7-year property, 10-year property, 15-year property, 20-year property, and 25-year property. IRS
publication 946 (http://www.irs.gov/pub/irs-pdf/p946.pdf) defines the types of assets that
fit in each classification. Table 6.2 shows a summary of this publication. One should note that any
property that doesn’t specifically fit in another category is automatically classified as 7-year property.
Table 6.3 shows the depreciation rates that are used for various classifications, assuming a
half-year convention (most common assumption). A half-year convention simply recognizes that
assets are put into service at various times during any one year. Rather than beginning to depreciate
the asset on the actual day that it is put into service, the U.S. government allows a half year of
depreciation in the first year of use, a full year of depreciation from year two until year N and a half
year of depreciation in year N+1. Thus, depreciation for assets that fit in the 7-year depreciation
category, are actually spread over a total of 8 years.
6.3. DEPRECIATION
Table 6.2: Various classifications of depreciable property—from http://www.irs.gov/
pub/irs-pdf/p946.pdf
Property
Classification
3-year property
5-year property
7-year property
10-year property
15-year property
20-year property
25-year property
Examples
Tractor units for over-the-road use
Qualified rent-to-own property
Automobiles, taxis, buses, and trucks
Computer and peripheral equipment
Office machinery
Certain geothermal, solar, and wind energy property
Office furniture and fixtures
Agricultural machinery and equipment
Any property that does not have a class life and has not been designated
by law as being in any other class
Any natural gas gathering line placed in service after April 11, 2005
Vessels, barges, tugs and similar water transportation equipment
Qualified small electric meter and qualified smart electric grid system
placed in service after Oct 3, 2008
Any municipal wastewater treatment plant
Any qualified restaurant property placed in service before Jan 1, 2012
Electric transmission property used in transmission at 69 or more kilovolts
of electricity placed in service after April 11, 2005
Any natural gas distribution line placed in service after April 11, 2005
Farm buildings
Municipal sewers
Water utility property that is not included as 20-year property
103
104
6. DETERMINATION OF PROJECT CASH FLOW
Table 6.3: Depreciation rates for various property lives—from http://www.irs.gov/pub/
irs-pdf/p946.pdf
6.3. DEPRECIATION
105
Although Excel® does not have a built-in function that can be used to directly compute
MACRS depreciation, one can use the VDB function if one recognizes that MACRS is defined as
DDB depreciation, using 1/2 year convention, and then switching to straight-line depreciation:
=VDB(Initial_Cost, Salvage, Life, Start_Period, End_Period, Factor,no_Switch)
where, Initial_Cost = initial cost of the asset (P )
Salvage = salvage value of the asset (L) – set to zero for MACRS
Life = asset’s useful life (N )
Start_Period to End_Period = the period of interest (which can be fractional time periods).
For 1/2 year convention in year 1, use Start_Period = 0 and End_Period = 0.5.
For full year convention for years 2 through N , use Start_Period = (n − 1.5) and
End_Period = (n − 0.5).
For 1/2 year convention in year N + 1, use Start_Period = (N − 0.5) and
End_Period =N .
Factor = 2 (or omitted for DDB)
no_Switch = FALSE (or omitted for automatic switching to straight-line)
For example, for 7-year property:
106
6. DETERMINATION OF PROJECT CASH FLOW
Example 6.4
Determine the yearly depreciation for the device described in Example 6.3 if it fits in the
7-year life category. Recall that P = $5000.
6.4
CASH FLOW COMPUTATION
As described in Chapter 2, cash flow is simply the net change (+ or -) in a company’s or individual’s
cash balance relative to a given project. That is, a positive project cash flow for a period would
indicate that the company had more cash due to that project at the end of that period than it did at
the beginning. A negative project cash flow would indicate just the opposite.
The following discussion lists the major considerations in determining cash flow for a project.
Cash flows can be calculated as before-tax or after-tax (where the tax is state and federal income tax).
It should be noted that the implications of a specific project on the company’s overall tax situation
will ultimately be determined by the company’s accountants and/or tax attorneys. Therefore, most
engineering economic analyses will be conducted on before-tax cash flows. However, sometimes it
is necessary or informational to evaluate after-tax cash flows. Therefore, both types are covered in
this discussion.
6.4.1
CAPITAL INVESTMENT
Capital investment is the cash that is expended by the company or the individual necessary to get
the project underway. That is, it is money used by the company or individual to purchase fixed assets
such as land, machinery, or buildings rather than money used for day-to-day operations. While
cash expenditures for fixed assets will generally occur over the length of a specific time period, it is
assumed that, for economic evaluation purposes, it all occurs at the beginning of that time period.
Thus, if it takes $500,000 over 6 months to construct a manufacturing facility, one would consider
all $500,000 to be spent at the beginning of year 1 (which, recall, is year 0 on the cash flow diagram).
Capital investment will include all costs associated with the fixed assets that are being purchased.
6.4. CASH FLOW COMPUTATION
107
For example, labor costs, materials, services, etc. that are part of the construction of a manufacturing
facility are considered capital investment.
6.4.2
GROSS REVENUE
Gross revenue is all revenue that is generated through the sale of a product or service. In most cases,
revenue for each product stream can be computed with Equation 6.14:
{Gross Revenue} = {#of items sold during a period} ∗ {price per item}
(6.14)
It should be noted that, for economic evaluation purposes, the period’s gross revenue will be assumed
to occur at the end of the particular time period in which it is generated.
6.4.3
OPERATING EXPENSES
Operating expenses are all cash outlays that are necessary to produce and sell the product or service.
These expenses may include, but are not limited to, items such as labor costs, building rent, utility
costs, raw materials, supplies, interest on loans, etc. Operating expenses are normally classified as
either fixed costs or variable costs. Fixed costs represent costs that are independent of the number of
units produced (for example building rent), whereas variable costs are proportional to the number
of units produced (for example raw materials). Equation 6.15 shows how to compute operating
expenses:
{Operating Expenses} ={Fixed Costs during a period} +
{#of items sold during a period} ∗ {variable cost per item}
(6.15)
It should be noted that, for economic evaluation purposes, the period’s operating expenses will be
assumed to all occur at the end of the particular time period in which it is spent. The assumption
that all capital investment will occur at the beginning of each year and that the income and operating
expenses will occur at the end of each year is known as end-of-year convention.
6.4.4
BEFORE-TAX PROFIT COMPUTATION
For the computation of before-tax profit, one only needs to consider gross revenues and operating
expenses:
{Before tax Profit} = {Gross Revenue} − {Operating Expenses}
6.4.5
(6.16)
BEFORE-TAX CASH FLOW COMPUTATION
For the computation of before-tax cash flows, one needs to have information on capital investment,
gross revenue, and operating expenses for each time period, n:
{Before tax cash flow} = {Gross Revenue} − {Operating Expenses}
− {Capital Investment}
(6.17)
108
6. DETERMINATION OF PROJECT CASH FLOW
Example 6.5
Create the cash flow diagram for the following project. $300,000 is to be expended over 6
months to build a bicycle manufacturing facility. It is assumed that the facility will build 500 bicycles
the first year and 1000 bicycles in years two through five.The bicycles will be sold for $500 in the first
year with an estimated 4% escalation rate in years two through five. In the first year, fixed operating
costs will be $20,000 and variable operating costs will be $100 per bicycle. Assume an estimated 3%
escalation rate in years two through five for both operating costs.
The table below shows the detailed cash flow calculations for each year that results in the
following cash flow diagram ($ in thousands):
Year
0
1
-300
180
0
1
2
Capital
Investment
300,000
0
0
3
0
4
0
5
0
6.4.6
2
396.4
3
4
413.5
431.3
Gross Revenue
Opera ng Expenses
0
500*500 = 250,000
1000*500*(1.04) =
520,000
1000*500*(1.04)2 =
540,800
1000*500*(1.04)3 =
562,400
1000*500*(1.04)4 =
584,900
0
20,000 + 500*100 = 70,000
(20,000 + 1000*100)*(1.03)
= 123,600
(20,000 + 1000*100)*(1.03)2
= 127,300
(20,000 + 1000*100)*(1.03)3
= 131,100
(20,000 + 1000*100)*(1.03)4
= 135,100
5
449.8
Before-tax cash
ow
-300,000
180,000
396,400
413,500
431,300
449,800
DEPRECIATION
As mentioned above, depreciation costs are “paper expenses” that result from the depreciation of a
capital item. That is, there is no actual cash expenditure for this category. The cost does, however,
reduce the company’s income tax burden as will be shown. One can pick any of the methods given
above to calculate the depreciation expenses.
6.4. CASH FLOW COMPUTATION
6.4.7
109
TAXABLE INCOME
Taxable income is the income (or sometimes called gross profit) that is subject to taxation by the
United States government:
{Taxable Income} = {Gross Revenue} − {Operating Expenses} − {Depreciation}
6.4.8
(6.18)
STATE AND FEDERAL INCOME TAX
As shown in Table 6.4, U.S. companies compute their U.S. federal income tax (FIT) as a percentage
of their taxable income. (United States Code: Title 26, Subtitle A, Chapter 1, Part II, § 11) Even
though the FIT rate varies as the taxable income increases, it is common for engineering economic
analyses to use a flat tax rate of 35% on all taxable income. In addition, many states in the U.S. have a
state income tax of a few percent (0 to 12% with a U.S. average of 6.56%). For engineering economic
calculations, it is sufficiently accurate to add the state and federal income tax rates together to arrive
at an effective tax rate.
Table 6.4: United States corporate income tax (FIT) rates—from United States Code: Title 26,
Subtitle A, Chapter 1, Part II, § 11.
Therefore,
{FIT} = {Taxable Income} ∗ {Tax Rate}
(6.19)
In some circumstances, FIT can be allowed to be a negative value. That is, if the taxable
income is negative (a “loss”), multiplying any tax rate by that taxable income would yield a negative
value for FIT. This would be the same as the government paying the project for losing money!!
However, this computation can be defensible if the project that is being evaluated is only one of
many for a large company. Since the company only pays taxes on its total taxable income (that is,
from all projects taken together), a loss from one project will reduce the taxes that would be paid by
a profitable project. Thus, the project that generates a negative taxable income does indeed yield a
negative tax. Allowing negative FIT values is known as a “corporate analysis.”
If the project is a “stand alone” project (that is, its profit or loss will not be combined with any
other project), then any negative values of FIT must be changed to zero for that year. However, the
loss in that year may be carried forward into the future to reduce taxes from a profitable year that
occurs later. This is an area where consultation with a corporate tax expert would be necessary.
110
6. DETERMINATION OF PROJECT CASH FLOW
6.4.9
NET PROFIT
Net Profit is computed as the taxable income minus the income tax:
{Net Profit} = {Taxable Income} − {FIT}
= {Taxable Income} ∗ (1 − Tax Rate)
(6.20)
6.4.10 CASH FLOW
The values defined above can now be combined in order to compute the cash flow (or net cash flow)
for a particular period:
{Cash Flow} = {Net Profit} + {Depreciation} − {Capital Investment}
(6.21)
As mentioned before, since depreciation is only a “paper” expense (that is, no actual cash
payment is made for depreciation), it must be added back into the cash flow calculation. Depreciation’s
only effect, therefore, is to reduce the income tax that is paid.
Any capital investment (cash spent on depreciable assets) made during the particular period
is subtracted after all other cash flow considerations are taken into account.
Example 6.6
Determine the after tax cash flows for the ten years of the following project’s life:
Initial capital investment: $1,000,000
Use 7-year MACRS depreciation
Total tax rate of 40%
Corporate tax analysis
Sales Schedule:
Year
1
2
3
4
5
6-10
# of units sold
5,000
5,000
7,000
7,000
10,000
10,000
Fixed Costs: $200,000 per year
Variable Costs: $30 per unit
Price per unit
$100
$110
$120
$120
$140
$140
6.4. CASH FLOW COMPUTATION
111
Solution:
Year 0
For evaluation purposes, assume that the initial capital investment occurs at the beginning of
year 1 (which, by definition, is year 0).
CF0 = −1, 000, 000
Year 1
Gross Revenue = 5,000 * 100 = $500,000
Operating Costs = 200,000 + 5000 * 30 = $350,000
Depreciation = 0.143 * 1,000,000 = $143,000
Taxable Income = 500,000 - 350,000 - 143,000 = $7,000
FIT = 0.40 * 7,000 = $2,800
CF 1 = 7,000 - 2,800 + 143,000 = $147,200
The remaining nine years are calculated in a similar manner and are shown in the following
cash flow table:
Year
0
1
2
3
4
5
6
7
8
9
10
Gross
Revenue
500,000
550,000
840,000
840,000
1,400,000
1,400,000
1,400,000
1,400,000
1,400,000
1,400,000
Opera ng
Costs Deprecia on
350,000
350,000
410,000
410,000
500,000
500,000
500,000
500,000
500,000
500,000
143,000
245,000
175,000
125,000
89,000
89,000
89,000
45,000
0
0
Taxable
Income
7,000
-45,000
255,000
305,000
811,000
811,000
811,000
855,000
900,000
900,000
Capital
FIT Investment
1,000,000
2,800
-18,000
102,000
122,000
324,000
324,000
324,000
342,000
360,000
360,000
Cash Flow
-1,000,000
147,200
218,000
327,000
308,000
576,000
576,000
576,000
558,000
540,000
540,000
At a value of MARR of 20%, the NP V of this project can be shown to be $518,000 (after
tax).
One might wish to generate an Excel® spreadsheet to allow additional analysis of this problem
if any or all of the given numerical values change. Such a spreadsheet is shown on the next page.
From the formulas it can be seen that key numerical values can be easily changed and the remainder
of the spreadsheet will change accordingly.
The formulas and/or values in each column are shown on the next pages.
112
6. DETERMINATION OF PROJECT CASH FLOW
6.4. CASH FLOW COMPUTATION
113
114
6. DETERMINATION OF PROJECT CASH FLOW
6.5. PROBLEMS
6.5
115
PROBLEMS
6.1.
Using the CP I , compute the average inflation rate from 1992 to 2009.
6.2.
Cash flow diagrams for projects A and B are shown below. Assume that the cash flows are
in escalated dollars and that the escalated dollar MARR is 10%.
(a) Calculate the N P V of each project as given.
(b) Calculate the N P V if one assumes a 5% inflation rate.
6.3.
0
1
2
3
-80
40
45
50
0
1
2
3
-120
100
80
60
An eight-year life project has an initial capital expenditure of $450,000, annual income of
$300,000 beginning at the end of year 1, and annual operating costs of $80,000 beginning
at the end of year 1. Calculate the I RR for the following cases:
(a) Assume the cash flows given are in escalated dollars and the escalated dollar MARR
is 20%.
(b) Assume the cash flows given are in today dollars and that incomes are escalated at 7%
and costs are escalated at 6%.
(c) Assume inflation is 4% and rework part (b) in terms of constant dollars.
116
6. DETERMINATION OF PROJECT CASH FLOW
6.4.
An investment related to developing a new product is estimated to have the following costs
and revenues in today dollars. Do not consider any tax issues.
0
1
Investment: 50,000
Income:
Oper Costs:
Salvage:
2
3
4
5
150,000
200,000
100,000
200,000
100,000
200,000
100,000
200,000
100,000
0
(a) Evaluate the project’s escalated dollar I RR if both capital costs and operating costs
are estimated to escalate at 15% per year from time zero and income is estimated to
escalate at 10% per year from time zero.
(b) Evaluate the project’s escalated dollar I RR assuming a “washout” of escalation of
income and operating costs with a 15% escalation of capital costs per year. “Washout”
means any operating cost escalation is offset by the same dollar escalation of revenue
(not the same percentage escalation) so that the before-tax profit remains uniform.
(c) Compute the constant dollar I RR of case (b) assuming that the rate of inflation will
be 10% per year.
6.5.
Determine the breakeven escalated dollar selling price per unit, X, required in each of years
1 and 2 to achieve a 15% constant dollar project I RR, assuming a 12% per year inflation
rate. All values are given in today dollars.
0
Investment: 100,000
Income:
Oper Costs:
1
1000(X)
50,000
2
1000(X)
50,000
Income escalation = 10% per year from time zero when selling price is $X per unit.
Operating Cost escalation = 15% per year from time zero.
1,000 units are to be produced and sold per year.
6.5. PROBLEMS
6.6.
117
Equipment has been purchased for $2,000,000 and put into service with an expected salvage
value at the end of 10 years of $200,000. Calculate the annual depreciation using:
(a) 10-year straight-line method
(b) 10-year double-declining balance method
(c) 10-year sum-of-the-years-digits method
(d) 10-year MACRS
6.7.
Consider a mining and processing project for an oil tar sands project. From the data given
below, calculate the after-tax cash flows for a 30-year life of the project and the NP V for
an MARR of 15%.
• Initial capital expenditures totaled $415.5 million and were distributed over four years
(10% in year 0, 30% in year 1, 40% in year 2, and 20% in year 3).
• Beginning in year 4:
– 17.666 million tons of ore will be mined per year
– Bitumen production rate will be 7.347 million barrels per year
– Product yield will be 0.841 barrels of oil per barrel of bitumen
– Product selling price will be $80 per barrel
– Operating costs:
∗ $10.47 per barrel of bitumen for plant and upgrading costs
∗ $9.02 per ton of ore for mining costs
– 10-year straight-line depreciation
– 40% tax rate (state and federal)
6.8.
The XYZ oil company owns several natural gas wells and is negotiating a 10-year contract
to sell the gas from these wells to another company. They are negotiating on the price of the
gas in the first year, in dollars per thousand cubic feet ($/MCF), including a 4% escalation
clause. XYZ expects the wells to produce 33,000 MCF the first year and to decline at the
rate of 15% every year thereafter. Operating costs are estimated to be $2/MCF and escalate
at 3% per year. XYZ has agreed to spend $500,000 now to lay pipelines from each well to
the second company’s processing plant. What should the minimum price be the first year
for this to be acceptable to XYZ? Assume an end-of-year convention and an MARR of
15%.
118
6. DETERMINATION OF PROJECT CASH FLOW
6.9.
An investment of $80,000 is projected to generate escalated dollar net revenues (income
minus costs) of $10,000 in year 1, $30,000 in year 2, and $40,000 in year 3 with a $40,000
salvage value at the end of year 3.
(a) Calculate the escalated dollar I RR for an escalated dollar MARR of 20%. Is this an
acceptable investment?
(b) Calculate the equivalent constant dollar I RR assuming that inflation will be 8% in
year 1, 10% in year 2, and 12% in year 3. Is this an acceptable investment?
6.10. The projected cost of the Alaskan oil pipeline was $900 million in 1969 dollars. The final
cost estimate was nearly $8.5 billion in 1977. What was the average yearly escalation rate
for the pipeline?
6.11. Boston’s “Big Dig” is one of the most expensive highway projects in the U.S. The project’s
original estimated cost was $2.6 billion in 1982 dollars. The costs in 2005 had risen to over
$14.6 billion.
(a) What is the value of the $14.6 billion in 1982 dollars?
(b) What was the average yearly escalation rate for the project?
6.12. Using Excel® and the CP I values given in Table 6.1, calculate the annual inflation rate for
each year from 1980 to 2010.
6.13. Use Excel® to solve Problem 6.2 for all 9 combinations of the following:
Values of MARR of 5%, 10%, and 15%
Inflation rates of 2%, 5%, and 7%
6.14. Use Excel® to solve the following problem. An eight-year life project has an initial capital
expenditure of $450,000, annual income of $300,000 beginning at the end of year 1, and
annual operating costs of $80,000 beginning at the end of year 1. Calculate the I RR for
the following cases:
(a) Assume the cash flows given are in escalated dollars and the escalated dollar MARR
is 10%, 20%, and 30%.
(b) Assume the cash flows given are in today dollars and pairs of escalation rates are:
a. Incomes are escalated at 7% and costs are escalated at 6%
b. Incomes are escalated at 3% and costs are escalated at 5%
c. Incomes are escalated at 4% and costs are escalated at 4%
(c) Assume inflation is 4% and rework all portions of part (b) in terms of constant dollars.
6.5. PROBLEMS
119
6.15. Use Excel® to solve Problem 6.6. Create a line graph that shows the values generated by
all four of the methods.
6.16. Use Excel® to solve Problem 6.7. The spreadsheet should allow for the user to easily change
any of the numerical values given.
6.17. Use Excel® to solve Problem 6.8. The spreadsheet should allow for the user to easily change
any of the numerical values given.
121
CHAPTER
7
Financial Leverage
7.1
INTRODUCTION
Earlier in this text, a brief description of the financial aspects involved in economic analyses was
presented. It was pointed out that one of the important financial aspects had to do with obtaining
the funds required to initiate the project. These funds are referred to as the investment capital. As
a source for this investment capital, a company could use its own internal funds (what is known as
equity funds), borrow funds from an external source (known as debt funds), or use a combination
of the two. The ratio of total borrowed funds to the total capital investment is called the financial
leverage factor. The ratio of borrowed funds to equity funds is called the debt to equity ratio. The
degree of financial leverage for any given project will affect the economic analysis of the project.
7.2
FINANCIAL LEVERAGE AND ASSOCIATED RISK
Under the correct conditions, financial leverage will allow an investor (company or individual) to
obtain a higher rate of return on its equity capital than it could achieve with no leverage. However,
there is often a good deal of added risk associated with leveraged projects. This additional risk is
due to the fact that when projects are financed with borrowed funds, those funds must be repaid to
the lender, independent of the ultimate success or failure of the project. If a leveraged project is only
marginally successful during any particular time period, the borrowed funds must be repaid to the
lender before any funds are used to pay a return on the equity portion of the investment.
7.3
ADJUSTMENT TO CASH FLOW EQUATIONS
Equations 6.14 and 6.15, as well as 6.18 through 6.21, allow the analyst to compute the after-tax cash
flows from a project. Some of these equations need to be modified for the case where the project is
leveraged. These modifications will account for the fact that (a) interest paid on the debt is a pre-tax
deduction while (b) the principal paid on the debt is not a pre-tax deduction.
Equation 6.18 is modified as follows:
{Taxable Income} ={Gross Revenue} − {Operational Expenses} −
{Depreciation} − {Interest paid on debt}
(7.1)
Equation 6.21 is modified as follows:
{Cash Flow} ={Net Profit} + {Depreciation} −
{Equity Investment} − {Principal paid on debt}
(7.2)
122
7. FINANCIAL LEVERAGE
It should be noted that the investor is allowed to compute depreciation on the total value of
each asset in the project independent of the source of funds. Despite the source of funds, the investor
owns the full value of the depreciable assets that it procures for the project.
Example 7.1
A company is considering a one year investment which will cost $1000.The company’s beforetax MARR is 10%. The $1000 will purchase assets that will be fully depreciated in the one year
of operation. There are three possible economic conditions that the company needs to investigate.
Details of these conditions are shown below. In addition, the company will consider three different
leverage factors: 0.0, 0.4, and 0.7. Interest on any borrowed funds will be 10% over the one year of
operation. Use a 40% corporate tax rate and determine the after-tax I RR on the equity funds for
each combination of the three economic conditions and the three leverage factors. Note that, for
economic condition A, the before-tax IRR on total assets (in this case $1000) is less than the interest
rate that will be charged on the loan. For economic condition B, the before-tax I RR on total assets
is equal to the interest rate to be charged on the loan and, for economic condition C, the before-tax
I RR is greater than the loan interest rate.
Economic Conditions
A
B
C
Revenue – Oper Costs
$1050 $1100 $1200
Depreciation
1000
1000
1000
Taxable income without leverage
50
100
200
I RR on total assets before taxes
5%
10%
20%
Before-tax cash flow diagrams for each economic condition:
A:
0
-1000
1
1050
B:
0
-1000
1
1100
C:
0
-1000
1
1200
Table 7.1 shows the cash flows and the computed after-tax I RRs for the 9 different combinations. Figure 7.1 shows the after-tax IRR on equity as a function of the leverage factor for the
three different economic conditions.
7.3. ADJUSTMENT TO CASH FLOW EQUATIONS
Table 7.1: Effect of leverage and economic conditions on the after-tax I RR on equity for
Example 7.1
123
124
7. FINANCIAL LEVERAGE
Economic Condition A
Economic Condition B
Economic Condition C
Figure 7.1: Effect of leverage factor for various economic conditions for Example 7.1.
From the results of Example 7.1, the following observations can be made:
1. Figure 7.1 shows that when the project’s before-tax I RR on assets is less than the interest rate
charged on the loan (economic condition A), the after-tax I RR on equity decreases as the
leverage factor increases. This makes sense because the project must pay the lender a higher
rate of interest than it will be able to pay the owner in rate of return.
2. Figure 7.1 shows that when the project’s before-tax I RR on assets is equal to the interest rate
charged on the loan (economic condition B), the after-tax I RR on equity is not affected as
the leverage factor increases. This makes sense because the project pays the lender the same
rate of interest as it will be able to pay the owner in rate of return.
3. Figure 7.1 shows that when the project’s before-tax I RR on assets is greater than the interest
rate charged on the loan (economic condition C), the after-tax I RR on equity increases as the
leverage factor increases. This makes sense because the project pays the lender a lower rate of
interest than it is able to pay the owner in rate of return.
4. There is more risk to equity capital when projects are leveraged with borrowed money. If the
economic conditions are poorer than originally predicted (such as condition A occurring when
condition C was predicted when the decision to invest was made), the after-tax I RR on equity
will decrease.
5. If enough equity capital exists, companies should not borrow money to fund a project unless
the interest rate paid on the debt is less than the before-tax I RR on the project’s total assets.
Leverage factors vary from company to company and even within a company from project
to project. In general, for most companies other than public utilities (who typically have very high
7.3. ADJUSTMENT TO CASH FLOW EQUATIONS
125
leverage factors of 0.6 or greater), leverage factors usually run from 0.3 to 0.5. A highly leveraged
project can do very well in a favorable economic climate, but may run into some hard times as
economic conditions go from good to bad. Many companies have used this principle to expand
rapidly during thriving business conditions.
Example 7.2
Consider the following five-year project with different methods of financing. A company has
the opportunity to invest in a five-year project that has an initial capital investment of $100,000.
The entire capital investment (total assets) will be depreciated over the five-year life of the project
using straight-line depreciation. Annual incomes and operating costs are expected to be $50,000
and $10,000, respectively. Interest on borrowed money will be 10% compounded annually. Calculate
the after-tax IRR on equity for the following cases and assuming a corporate tax rate of 40%. Use
an after-tax MARR of 12%.
(a) 100% equity.
(b) Leverage factor of 0.4. The principal payments will be constant for each of the five years and
the interest paid each year will be based on the outstanding debt balance.
(c) Leverage factor of 0.7. The principal payments will be constant for each of the five years and
the interest paid each year will be based on the outstanding debt balance.
(d) Leverage factor of 0.4. The principal and interest will be paid with a constant annual payment
as calculated according to: P &I payment = Debt ∗ (A/P )10%,5 .
(e) Leverage factor of 0.4. Interest payments are made each year but the principal is paid back
in one lump sum at the end of the project. This is known as yearly interest with a “balloon
payment” of the principal at the end.
126
7. FINANCIAL LEVERAGE
First, solve for the before-tax I RR on assets. This would be represented by a 0.0 leverage
factor and a 0% FIT rate.
Therefore, the before-tax I RR on assets for this project is 28.6%. Since the interest rate on
borrowed funds is less than this value, leveraging the project should increase the after-tax I RR on
equity.
7.3. ADJUSTMENT TO CASH FLOW EQUATIONS
127
(a) This solution will show the effect of the 40% FIT rate compared to the before-tax solution shown
previously.
The 40% FIT tax rate reduces the after-tax I RR on total assets to 18.0%.
128
7. FINANCIAL LEVERAGE
(b) This solution will show the effect of a leverage factor of 0.4.
0.4
40000
4000
3200
2400
1600
800
16000
6400
9600
16800
6720
10080
17600
7040
10560
18400
7360
11040
19200
7680
11520
8000
8000
8000
8000
8000
60000
-60000
21600
22080
22560
23040
23520
40000
32000
24000
16000
8000
25.1%
18691
The after-tax I RR on an equity investment of $60,000 has increased to 25.1%. This increase
is as expected. Also, the after-tax NP V has increased from a leverage factor of 0.0.
7.3. ADJUSTMENT TO CASH FLOW EQUATIONS
129
(c) This solution will show the effect of a leverage factor of 0.7.
Increasing the leverage factor to 0.7 further increases the after-tax I RR on equity and the
after-tax N P V .
130
7. FINANCIAL LEVERAGE
(d) This solution will show the effect of paying constant annual principal and interest payments.
Using a more conventional method to repay the debt, the after-tax I RR and after-tax NP V
both increase slightly from the first repayment method.
7.3. ADJUSTMENT TO CASH FLOW EQUATIONS
131
(e) This solution will show the effect of paying annual interest and then a balloon payment for the
principal.
** Modified I RR using an after-tax MARR of 12%.
The balloon repayment method further increases the after-tax I RR and after-tax NP V .
From the results of Example 7.2, the following observations can be made:
1. If the results from parts (a), (b), and (c) are compared, it is again found that, under these
economic conditions, when the amount of borrowed funds is increased, a higher rate of return
is obtained on the equity investment. It should be stressed that this higher rate of return is on
a smaller amount of equity dollars compared to financing the project with 100% equity funds.
2. It can also be seen that after-tax NP V increases as the leverage factor is increased. NPV
analysis would further emphasize that, under the economic conditions of the before-tax I RR
on assets being greater than the interest rate paid on the debt, the best option is to maximize
132
7. FINANCIAL LEVERAGE
the amount of leverage. By using maximum leverage on each project, a company can invest in
more projects and grow more rapidly.
3. Parts (b), (d), and (e) compare three different, but acceptable, methods of repaying the debt
portion of the investment. Since the interest on the borrowed money is less than the I RR on
assets, it is better to push the repayment of the principal as far forward in time as possible
in order to increase I RR on equity and NP V . The balloon payment technique provides the
highest I RR and N P V .
7.3.1
LEVERAGE AND MUTUALLY EXCLUSIVE PROJECTS
When applying leverage concepts to the evaluation of several projects to determine which one is
best, the leverage factor is an important variable. It has been shown in the example problems, that the
project I RR on equity is a function of the leverage factor. In order to compare projects, the degree
of leverage must be the same on all projects. Many companies have a policy that the comparison of
projects is done without considering any leverage for all of the projects. Once a project is chosen, then
various methods of financing, including different amounts of leverage and repayment techniques,
can be investigated as to their effect upon the project.
7.3.2
EXCEL® SPREADSHEET
As shown on the next page, the project spreadsheet generated for Example 6.6 can be easily modified
to include the effect of leverage. The only assumptions in the spreadsheet are that the loan will be
paid with constant principal payments over the first five years and the interest paid each year will be
based on the outstanding debt balance. This could be modified for other repayment options.
7.3. ADJUSTMENT TO CASH FLOW EQUATIONS
133
134
7. FINANCIAL LEVERAGE
7.4
7.1.
PROBLEMS
A used piece of heavy equipment is available for purchase at $300,000. A rental company is
deciding whether or not to purchase the equipment. The company estimates the equipment
will create annual incomes of $110,000 and have annual operating costs of $20,000. The
equipment can be depreciated in five years with straight-line depreciation. Based on the
results from part (a) below, should the rental company purchase the equipment if their
corporate tax rate is 35%? Consider a five-year life project and an after-tax MARR of 15%.
(a) Determine the return on equity for each of three different leverage factors of 0, 0.4,
and 0.7. Assume an interest rate on borrowed funds to be 10% compounded annually.
The principal payments will be constant for each of the five years and the interest paid
each year will be based on the outstanding debt balance.
(b) Assume two additional economic conditions: (i) annual income increases to $125,000
and (ii) annual income decreases to $95,000. Repeat part (a) for these two economic
conditions. Prepare a plot of the I RR on equity versus the leverage factor.
7.2.
A corporation’s tax rate is 40%. An outlay of $35,000 is being considered for a new asset.
Estimated annual revenues are $30,000 and estimated annual operating costs are $10,000.
The useful life of the asset is 5 years and has no salvage value. Use the SYD method of
depreciation. A lending institution has offered to loan the corporation 50% of the initial
investment cost at an annual interest rate of 12.5%. The principal and interest will be
paid with a constant annual payment as calculated according to: P &I payment = Debt ∗
(A/P )12.5%,5 . If the corporation’s after-tax MARR is 15%, should it accept the loan?
7.3.
Solve Problem 6.7 using a leverage factor of 0.2.
7.4.
Use Excel® to solve Problem 6.8 for leverage factors of 0.2 and 0.4.
135
CHAPTER
8
Basic Statistics and Probability
8.1
INTRODUCTION
In previous chapters of this text, it was assumed that all of the information needed to make an economic analysis was known without any uncertainty. In practice, this is a rare situation. Nearly always,
an evaluator will need to include a measure of the uncertainty pertaining to one or more variables
in the analysis. This uncertainty may, in turn, add significant uncertainty about the profitability of
an investment. For example, with one set of economic assumptions, the project’s NPV might be
greater than zero which would indicate an acceptable investment. However, with a different set of
economic assumptions, the project’s NPV might be negative, thereby indicating that the investor
should pass on this opportunity. This range of uncertainty about the project’s profitability is one
way to define the “risk” in a project. Having a basic understanding of statistics and probability will
allow an evaluator to incorporate various risk factors into the analyses that are to be completed for a
project. Some techniques that are available to incorporate uncertainty into project variables, and that
apply the ideas of statistics and probability presented in this chapter, will be presented in Chapter 9.
8.2
STATISTICS
8.2.1
MEASURES OF CENTRAL TENDENCY
Averages are often used to represent a set of data. Several different types of averages can be calculated.
These include the arithmetic mean, the median, the mode, and the geometric mean.These are known
as measures of central tendency as they tend to be centrally located within the data.
Arithmetic Mean
The arithmetic mean of a set of data is calculated with Equation 8.1. The arithmetic mean is also
known as the expected value of the data.
N xi /N
μ=
i=1
where, xi = the ith value of the data
N = total number of data points
μ = arithmetic mean
(8.1)
136
8. BASIC STATISTICS AND PROBABILITY
Excel® has a built-in function to calculate the arithmetic mean:
= AVERAGE(number1, number2,…)
where, number1, number2, ... = list of data points
Median
When a set of data is arranged in order of magnitude, the median of the set is found by taking the
middle value (when there is an odd number of values) or the arithmetic mean of the two middle
values (when there is an even number of values).
Excel® has a built-in function to calculate the median:
= MEDIAN(number1, number2,…)
Mode
The mode is the value which occurs with the greatest frequency. A set of data can have a single
mode, several modes, or no modes.
Excel® has two built-in functions to calculate the mode:
single mode: = MODE.SNGL(number1, number2, …)
multiple modes: = MODE.MULT(number1, number2, …)
Geometric Mean
The geometric mean of a set of data is calculated with Equation 8.2:
N
N
xi
G= i=1
where, xi = the ith value of the data
N = total number of data points
N
= (x1 )(x2 )(x3 ) . . . (xN −1 )(xN )
i=1
Excel® has a built-in function to calculate the geometric mean:
= GEOMEAN(number1, number2,…)
(8.2)
8.2. STATISTICS
137
Example 8.1
Consider 100 exam scores from a college-level class as shown below:
75
76
57
77
54
84
51
94
95
85
67
91
88
38
86
91
77
46
46
87
96
81
94
93
94
67
88
99
48
83
73
87
59
76
60
79
79
85
79
98
78
91
97
35
88
78
46
79
85
74
53
55
81
78
75
73
97
34
87
88
51
88
78
94
86
78
80
85
76
51
47
90
90
88
38
79
90
39
56
57
31
53
83
67
67
39
78
91
48
95
42
74
65
89
99
90
84
82
32
61
(a) Calculate the arithmetic mean:
μ = (75 + 67 + 96 + · · · + 95 + 61)/100 = 73.7
(b) Calculate the median:
First, order the 100 scores from high to low. Since there is an even number of values, the
median is the average of 50th (79) and 51st (78) values, or 78.5.
(c) Calculate the mode:
Again, order the 100 scores from high to low and find the value that occurs most often. In this
case, the value of 78 occurs six times. Therefore, the mode is 78.
(d) Calculate the geometric mean:
G=
√
75 ∗ 67 ∗ 96 ∗ · · · ∗ 95 ∗ 61 = 70.9
100
138
8. BASIC STATISTICS AND PROBABILITY
Using Excel® :
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
A
B
C
D
E
F
G
H
I
J
75
76
57
77
54
84
51
94
95
85
67
91
88
38
86
91
77
46
46
87
96
81
94
93
94
67
88
99
48
83
73
87
59
76
60
79
79
85
79
98
78
91
97
35
88
78
46
79
85
74
53
55
81
78
75
73
97
34
87
88
51
88
78
94
86
78
80
85
76
51
47
90
90
88
38
79
90
39
56
57
31
53
83
67
67
39
78
91
48
95
42
74
65
89
99
90
84
82
32
61
Average =
Median =
Mode =
Geo Mean =
73.7
78.5
78
70.9
A
B
C
D
E
F
G
H
I
J
75
76
57
77
54
84
51
94
95
85
67
91
88
38
86
91
77
46
46
87
96
81
94
93
94
67
88
99
48
83
73
87
59
76
60
79
79
85
79
98
78
91
97
35
88
78
46
79
85
74
53
55
81
78
75
73
97
34
87
88
51
88
78
94
86
78
80
85
76
51
47
90
90
88
38
79
90
39
56
57
31
53
83
67
67
39
78
91
48
95
42
74
65
89
99
90
84
82
32
61
Average =
Median =
Mode =
Geo Mean =
=AVERAGE(A1:J10)
=MEDIAN(A1:J10)
=MODE.SNGL(A1:J10)
=GEOMEAN(A1:J10)
8.2. STATISTICS
8.2.2
139
MEASURES OF DISPERSION
It is frequently desired to determine how a set of data is dispersed or spread about its average.
Measures of dispersion which will be discussed in this chapter include the range, the mean deviation,
the standard deviation, and the variance.
Range
The range of a set of data is simply the difference between the largest and the smallest values of the
data.
In order to compute the range of a set of data with Excel® , use the MAX and MIN functions:
=MAX(number1, number2,…) − MIN(number1, number2,…)
Mean Deviation
The mean deviation (or average deviation) is the mean of the distances between each value and the
mean. It is computed with Equation 8.3:
N
|xi − μ| /N
M.D. =
(8.3)
i=1
where, xi = the ith value of the data
μ = the arithmetic mean of the data
N = total number of data points
Excel® does not have a built-in function to calculate the mean deviation. To use Excel® , do
the following:
1. Place the data in a single column (for example, assume 10 data points in cells A1 through
A10).
2. Use the formula =AVERAGE(A1:A10) in cell A11 to compute the average of this column of
data. This is the mean of the data.
3. In the adjacent column B, use the formula =ABS(A1−$A$11) in cell B1.
4. Copy this formula to cells B2 through B10.
5. Use the formula =AVERAGE(B1:B10) in cell B11 to compute the average of this column of
data. This is the mean deviation of the data.
Standard Deviation
Standard deviation is another measure of the variability of a data set about its mean. Its origins are
associated with the normal distribution that is discussed later in this chapter, but it has meaning
140
8. BASIC STATISTICS AND PROBABILITY
for any set of data. A small value of standard deviation indicates that the data points are clustered
more closely to the mean than a larger value of standard deviation. If the entire population has been
sampled (that is, N equals the total possible number of data points in the population), the standard
deviation is calculated with Equation 8.4:
N
(8.4)
σ =
(x − μ)2 /N
i
i=1
th
where, xi = the i value of the data
μ = the arithmetic mean of the data
N = total number of data points
If one was calculating the standard deviation of 100 exam scores in a particular college-level
class with 100 students, then N would be 100 in Equation 8.4. However, if only a subset of the
population is being sampled, N should be replaced with N − 1. It can be noted that when N gets
larger than about 30, there is very little error introduced by using N instead of N − 1. As an example
of a sample, assume that one wanted to measure the mean and standard deviation of the age of
the population in a city of 20,000 people. It would be difficult to get the age of all 20,000 people,
so a subset of the population is sampled (perhaps 1,000 people). One would use Equation 8.1 to
determine the mean age of the population and Equation 8.4 (with N − 1 instead of N ) to determine
the standard deviation of the population’s age.
Excel® has two built-in functions to calculate the standard deviation:
=STDEV.P(number1, number2,…) for the entire population or
=STDEV.S(number1, number2,…) for a sample of the population.
Per Equation 8.4, STDEV.P contains a division by N , whereas STDEV.S contains a division
by N − 1.
Example 8.2
Consider 100 exam scores from a college-level class as shown below (same as Example 8.1):
75
76
57
77
54
84
51
94
95
85
67
91
88
38
86
91
77
46
46
87
96
81
94
93
94
67
88
99
48
83
73
87
59
76
60
79
79
85
79
98
78
91
97
35
88
78
46
79
85
74
53
55
81
78
75
73
97
34
87
88
51
88
78
94
86
78
80
85
76
51
47
90
90
88
38
79
90
39
56
57
31
53
83
67
67
39
78
91
48
95
42
74
65
89
99
90
84
82
32
61
8.2. STATISTICS
141
(a) Calculate the range:
Order the numbers from high to low. The range is then given by the highest value minus the
lowest value. Range = 99-31 = 68.
(b) Calculate the mean deviation:
M.D. = (|75 − 73.7| + |67 − 73.7| + |96 − 73.7| + · · · + |95 − 73.7| + |61 − 73.7|) /100
= 15.4
(c) Calculate the standard deviation:
(75 − 73.7)2 + (67 − 73.7)2 + · · · + (95 − 73.7)2 + (61 − 73.7)2
100
= 18.5
σ =
Using Excel® :
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
A
75
76
57
77
54
84
51
94
95
85
B
67
91
88
38
86
91
77
46
46
87
Average =
Median =
Mode =
Geo Mean =
Range =
Mean Dev =
StdDev =
C
96
81
94
93
94
67
88
99
48
83
73.7
78.5
78
70.9
68
15.4
18.5
D
73
87
59
76
60
79
79
85
79
98
E
78
91
97
35
88
78
46
79
85
74
F
53
55
81
78
75
73
97
34
87
88
G
51
88
78
94
86
78
80
85
76
51
H
47
90
90
88
38
79
90
39
56
57
I
31
53
83
67
67
39
78
91
48
95
J
42
74
65
89
99
90
84
82
32
61
142
8. BASIC STATISTICS AND PROBABILITY
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
A
75
76
57
77
54
84
51
94
95
85
B
67
91
88
38
86
91
77
46
46
87
Average =
Median =
Mode =
Geo Mean =
Range =
Mean Dev =
StdDev =
C
96
81
94
93
94
67
88
99
48
83
D
73
87
59
76
60
79
79
85
79
98
E
78
91
97
35
88
78
46
79
85
74
F
53
55
81
78
75
73
97
34
87
88
G
51
88
78
94
86
78
80
85
76
51
H
47
90
90
88
38
79
90
39
56
57
I
31
53
83
67
67
39
78
91
48
95
J
42
74
65
89
99
90
84
82
32
61
=AVERAGE(A1:J10)
=MEDIAN(A1:J10)
=MODE.SNGL(A1:J10)
=GEOMEAN(A1:J10)
=MAX(A1:J10)-MIN(A1:J10)
=B121**
=STDEV.P(A1:J10)
**This assumes that the 100 data points are copied to cells A21:A120 and the procedure listed
above under Mean Deviation is followed.
8.2.3
FREQUENCY DISTRIBUTIONS
The creation of a frequency distribution is another technique to summarize large numbers of raw data.
When the raw data are summarized, they begin to take on more meaning and utility. A frequency
distribution is made by grouping the raw data into classes and counting the number of items that
fall into each class. This number is referred to as the class frequency. A table is then formed which
contains a column for the class, a column for the class frequency, and a column for the cumulative
class frequency. The resulting table is the frequency distribution. The size and number of classes will
depend upon the particular application that is being considered. Typically, frequency distributions
contain five to ten classes, all of equal size. However, some data might lend themselves to classes of
unequal size or even classes that might be open ended (normally, the first class or the last class or
both).
The cumulative frequency distribution, Fi , is the summation of the frequency distribution.
8.2. STATISTICS
143
Example 8.3
Consider 100 exam scores from a college-level class as shown below (same as Example 8.1):
75
76
57
77
54
84
51
94
95
85
67
91
88
38
86
91
77
46
46
87
96
81
94
93
94
67
88
99
48
83
73
87
59
76
60
79
79
85
79
98
78
91
97
35
88
78
46
79
85
74
53
55
81
78
75
73
97
34
87
88
51
88
78
94
86
78
80
85
76
51
47
90
90
88
38
79
90
39
56
57
31
53
83
67
67
39
78
91
48
95
42
74
65
89
99
90
84
82
32
61
Using ten classes from 0–100, develop the frequency distribution for the data.
Solution:
Within the frequency distribution, the range of numbers that is used to define the class is
called a class interval. The smaller number is the lower class limit and the larger number is the upper
class limit. Note that in Example 8.3, the upper class limit of one class is the same as the lower class
limit of the next class. If a value is exactly equal to one of the class limits, one needs to decide in
which class it belongs. It doesn’t matter if it is placed in the higher range or the lower range as long
as the evaluator remains consistent. In Example 8.3, any value that is equal to a class limit is placed
in the lower range (e.g., a value of 90 is placed in the 80-90 class). If this convention is used, then
one can define true class limits for a range. In this case, the true class limits would be 90.5-100.5,
80.5-90.5, 70.5-80.5, etc. Excel® utilizes this convention as well.
There are two other terms that need to be defined for frequency distributions. The class size
is the difference between the upper true class limit and the lower true class limit. The true class mark
144
8. BASIC STATISTICS AND PROBABILITY
is the midpoint of each true class interval or the average between the upper true class limit and the
lower true class limit. In Example 8.3, the class size is ten for all ten classes and the true class marks
are 95.5, 85.5, 75.5, etc.
True Class
Limits
0.5-10.5
10.5-20.5
20.5-30.5
30.5-40.5
40.5-50.5
50.5-60.5
60.5-70.5
70.5-80.5
80.5-90.5
90.5-100.5
Total
True Class
Mark
5.5
15.5
25.5
35.5
45.5
55.5
65.5
75.5
85.5
95.5
Frequency
fi
0
0
0
8
7
12
6
23
27
17
100
Cumula ve
Frequency, Fi
0
0
0
8
15
27
33
56
83
100
Frequency distributions are often represented graphically. Graphical representations include
histograms, frequency polygons, and relative and cumulative relative frequency diagrams.
A histogram consists of a set of rectangles, where a rectangle is drawn for each class interval
with the width of each rectangle equal to the class size and the height of the rectangle is the class
frequency. The histogram is constructed so that the center of each rectangle lies at its true class mark.
Figure 8.1 is the histogram for the data presented in Example 8.3.
A frequency polygon can be generated by creating a line graph of the frequency of each class
as a function of the true class marks. Figure 8.2 is the frequency polygon for the data presented in
Example 8.3. The first and last points of the polygon are found on the x-axis at what would be the
true class marks associated with class intervals before the first actual class interval (located at −4.5)
and after the last actual class interval (located at 105.5).
8.2. STATISTICS
Histogram
30
25
20
Freq 15
10
5
0
5.5
15.5
25.5
35.5
45.5
55.5
65.5
75.5
85.5
95.5
Exam Scores
Figure 8.1: Histogram for Example 8.3.
Frequency Polygon
30
25
20
Freq 15
10
5
0
-4.5
5.5
15.5 25.5 35.5 45.5 55.5 65.5 75.5 85.5 95.5 105.5
Exam Scores
Figure 8.2: Frequency polygon for Example 8.3.
145
146
8. BASIC STATISTICS AND PROBABILITY
8.2.4
RELATIVE FREQUENCY DISTRIBUTION
The relative frequency distribution is constructed by dividing the number of occurrences in each
class interval by the total number of points in the data set. The following shows the relative frequency distribution for Example 8.3 while Figures 8.3 and 8.4 show the graphical versions of these
distributions.
True Class
Limits
True Class
Mark
0.5-10.5
10.5-20.5
20.5-30.5
30.5-40.5
40.5-50.5
50.5-60.5
60.5-70.5
70.5-80.5
80.5-90.5
90.5-100.5
5.5
15.5
25.5
35.5
45.5
55.5
65.5
75.5
85.5
95.5
Rela ve
Frequency,
fi
0.00
0.00
0.00
0.08
0.07
0.12
0.06
0.23
0.27
0.17
Cumula ve
Rela ve
Frequency, Fi
0.00
0.00
0.00
0.08
0.15
0.27
0.33
0.56
0.83
1.00
RelaƟve Frequency DistribuƟon
0.3
0.25
0.2
Rel Freq 0.15
0.1
0.05
0
-4.5
5.5 15.5 25.5 35.5 45.5 55.5 65.5 75.5 85.5 95.5 105.5
Exam Scores
Figure 8.3: Relative frequency distribution for Example 8.3.
8.2. STATISTICS
147
CumulaƟve RelaƟve Frequency
1
0.9
0.8
0.7
Cumul 0.6
RelaƟve 0.5
Freq 0.4
0.3
0.2
0.1
0
-4.5
5.5
15.5 25.5 35.5 45.5 55.5 65.5 75.5 85.5 95.5 105.5
Exam Scores
Figure 8.4: Cumulative relative frequency for Example 8.3.
If the data are presented in frequency distribution form, items such as the mean, mean deviation, and standard deviation can be determined from the following equations, respectively.
μ =
M.D. =
M
j =1
M
fj xj
(8.5)
fj xj − μ (8.6)
j =1
M
2
fj xj − μ
σ =
j =1
where, fj
xj
M
μ
M.D.
σ
= the relative frequency of the j th class
= the true class mark of the j th class
= total number of classes
= the arithmetic mean of the data based on the distribution
= the mean deviation of the data based on the distribution
= the standard deviation of the data based on the distribution
(8.7)
148
8. BASIC STATISTICS AND PROBABILITY
Example 8.4
Calculate the mean, mean deviation, and standard deviation for the data in Example 8.1, using
the relative frequency distributions found in Example 8.3:
Mean:
μ = 0.00 ∗ 5.5 + 0.00 ∗ 15.5 + · · · + 0.27 ∗ 85.5 + 0.17 ∗ 95.5 = 73.3
This value compares favorably to 73.7 computed with all 100 data points.
Mean Deviation:
M.D. =0.00 ∗ |5.5 − 73.3| + 0.00 ∗ |15.5 − 73.3| + · · · +
0.27 ∗ |85.5 − 73.3| + 0.17 ∗ |95.5 − 73.3|
=15.1
This value compares favorably to 15.4 computed with all 100 data points.
Standard Deviation:
0.00 ∗ (5.5 − 73.3)2 + 0.00 ∗ (15.5 − 73.3)2 + · · · +
0.27 ∗ (85.5 − 73.3)2 + 0.00 ∗ (95.5 − 73.3)2
= 18.3
σ =
This value compares favorably to 18.5 computed with all 100 data points.
8.3. PROBABILITY
8.3
PROBABILITY
8.3.1
CLASSICAL DEFINITION
149
The classical definition of probability involves an event occurring from a group or a set of equally
likely outcomes. That is, when a fair coin is tossed, the specific outcome of that event is either a
heads or a tails with each outcome equally likely to occur. Suppose that a particular event occurs
a certain number of times out of a total possible number of other events. The probability that the
desired event will occur is given by Equation 8.8:
P (A) = nA /n
(8.8)
where, P (A) = the probability of event A occuring
nA = the number of times event A could occur
n = the total number of possible events
Example 8.5
Consider the probability of drawing an ace from a fair deck of cards. Since there are four aces
in a total of 52 cards and the chances of drawing any specific card is the same, the probability of
drawing an ace would be:
P (Ace) = 4/52 = 1/13 = 0.0769 = 7.69%
Note that in the above example, the probability of 4/52 is only correct for any given attempt
to draw an ace if, when an undesired card is drawn (not an ace), it is returned to the deck before
the next card is drawn. This is referred to as sampling with replacement. If the undesired card is not
returned to the deck, known as sampling without replacement, the probability changes to 4/51, then
4/50, etc.
Example 8.6
Consider the probability of rolling two fair die and getting a total of 8. When rolling two fair
die, the possible outcomes are:
Die 1
1
1
1
1
1
1
Die 2
1
2
3
4
5
6
Total
2
3
4
5
6
7
Die 1
2
2
2
2
2
2
Die 2
1
2
3
4
5
6
Total
3
4
5
6
7
8
Die 1
3
3
3
3
3
3
Die 2
1
2
3
4
5
6
Total
4
5
6
7
8
9
150
8. BASIC STATISTICS AND PROBABILITY
Die 1
4
4
4
4
4
4
Die 2
1
2
3
4
5
6
Total
5
6
7
8
9
10
Die 1
5
5
5
5
5
5
Die 2
1
2
3
4
5
6
Total
6
7
8
9
10
11
Die 1
6
6
6
6
6
6
Die 2
1
2
3
4
5
6
Total
7
8
9
10
11
12
As shown in the table, there are 36 possible outcomes of the dice rolls, five of which have a
value of 8. Therefore, the probability of getting exactly 8 is:
P (8) = 5/36 = 0.139 = 13.9%
Example 8.7
Consider the flipping of a fair coin (50% probability of a head and 50% probability of a tail).
The coin will be flipped three times. What is the probability that 0, 1, 2, and 3 heads occurred in
the three flips?
First Flip
T
T
T
T
H
H
H
H
Second Flip
T
T
H
H
T
T
H
H
Third Flip
T
H
T
H
T
H
T
H
Total # of Heads
0
1
1
2
1
2
2
3
As shown in the table, there are eight possible outcomes of the three flips. Therefore,
P (0Heads) = 1/8 = 0.125 = 12.5%
P (1Heads) = 3/8 = 0.375 = 37.5%
P (2Heads) = 3/8 = 0.375 = 37.5%
P (3Heads) = 1/8 = 0.125 = 12.5%
8.3. PROBABILITY
8.3.2
151
RELATIVE FREQUENCY DEFINITION
The classical definition of probability uses the concept of equally likely outcomes to aid in the
definition. Since the words “equally likely” themselves imply some notion of probability, the definition
would appear to be a bit circular in nature. To get around this, the concept of relative frequency
probability was introduced. If an experiment or trial is going to be repeated a large number of times,
the probability that a particular event of the experiment will occur is given by the relative frequency
shown in Equation 8.9:
P (A) = lim (nA /n)
(8.9)
n→∞
Example 8.8
Consider the tossing of a coin. One would normally assume that the probability of getting a
head on any one toss is 50%. However, consider an experiment where a coin is tossed 100 times and
heads occurs 52 times and tails 48 times. For this case, the probability of getting a head would be
predicted to be 52%. If the coin is a fair one, as the number of experimental data points gets larger
and larger, the probability of getting a head will approach 50%.
In most cases throughout engineering, one does not have the luxury of performing an infinite
number of experiments in order to determine the true probability of an event occurring. For example,
if an engineer is doing failure tests on a particular manufactured component, one can only do a limited
number of failure experiments in order to determine the probability of failure.
8.3.3
SUBJECTIVE DEFINITION
A third type of probability definition is known as subjective probability. This type of probability
is not determined from theoretical or experimental work, but rather from the experience that an
individual or group of individuals has gained during their career in a particular area. This experience
is then used to predict the probability of future events. For example, a civil engineer who does road
design will gain, over time, a “feeling” or estimation of the probability that a road surface will begin
to fail within a certain time range based on weather conditions, quantity of traffic, type of surfacing
materials used, etc.
In summary, for economic evaluations, it is necessary that probabilities of certain outcomes be
assigned. This allows the evaluator to incorporate risk factors into economic analysis situations. The
difficulty is often in the assigning of the actual probabilities. One or more of the above definitions
may assist the evaluator in this task.
8.3.4
PROBABILITY DISTRIBUTIONS
For a given event or set of events, if probability or frequency distributions can be established, the
statistical concepts discussed earlier can be applied to calculate means and standard deviations for
each event.
152
8. BASIC STATISTICS AND PROBABILITY
Two different types of probability distributions, discrete and continuous, will be discussed
below.
Discrete Distribution
A discrete distribution is one which involves an experiment with a finite number of possibilities.
For example, as described earlier, when a fair die is thrown, a 1, 2, 3, 4, 5, or 6 will occur. Thus,
the outcome of a throw is a discrete value. Using the classical probability definition, each possible
outcome would have a probability of 1/6. Note that the sum of the probabilities of all possible
outcomes will always equal 1.0.
Example 8.9
Consider that a certain discrete random variable, x, has a discrete probability distribution as
follows:
(a) Graph the distribution
(b) Find the mean
(c) Find the standard deviation
(a) Graphically, the distribution is shown in Figure 8.5.
Discrete Probability DistribuƟon
0.3
0.25
0.2
P(x) 0.15
0.1
0.05
0
-3
-1
0
1
2
3
5
8
x
Figure 8.5: Probability distribution for the random discrete variable x in Example 8.9.
8.3. PROBABILITY
153
For this distribution, it would be useful to calculate the mean and standard deviation using
Equations 8.5 and 8.7:
(b) μ = 0.1 ∗ (−3) + 0.2 ∗ (−1) + 0.15 ∗ (0) + 0.25 ∗ (1) + 0.1 ∗ (2) + 0.1 ∗ (3)
+ 0.05 ∗ (5) + 0.05 ∗ (8) = 0.90
0.1 ∗ (−3 − 0.9)2 + 0.2 ∗ (−1 − 0.9)2 + 0.15 ∗ (0 − 0.9)2 + 0.25 ∗ (1 − 0.9)2
+ 0.1 ∗ (2 − 0.9)2 + 0.1 ∗ (3 − 0.9)2 + 0.05 ∗ (5 − 0.9)2 + 0.05 ∗ (8 − 0.9)2
= 2.51
(c) σ =
Solution using Excel® :
A
1
2
3
4
5
6
7
8
9
10
11
12
13
B
x
-3
-1
0
1
2
3
5
8
P(x)
0.1
0.2
0.15
0.25
0.1
0.1
0.05
0.05
Total =
Mean =
StdDev =
1.00
0.90
2.51
C
x*P(x)
-0.3
-0.2
0
0.25
0.2
0.3
0.25
0.4
D
E
(x-mu)^2 P(x)*(x-mu)^2
15.2
1.521
3.6
0.722
0.8
0.122
0.0
0.003
1.2
0.121
4.4
0.441
16.8
0.841
50.4
2.521
154
8. BASIC STATISTICS AND PROBABILITY
A
1
2
3
4
5
6
7
8
9
10
11
12
13
B
x
-3
-1
0
1
2
3
5
8
P(x)
0.1
0.2
0.15
0.25
0.1
0.1
0.05
0.05
C
x*P(x)
=A2*B2
=A3*B3
=A4*B4
=A5*B5
=A6*B6
=A7*B7
=A8*B8
=A9*B9
D
(x-mu)^2
=(A2-B$12)^2
=(A3-B$12)^2
=(A4-B$12)^2
=(A5-B$12)^2
=(A6-B$12)^2
=(A7-B$12)^2
=(A8-B$12)^2
=(A9-B$12)^2
E
P(x)*(x-mu)^2
=B2*D2
=B3*D3
=B4*D4
=B5*D5
=B6*D6
=B7*D7
=B8*D8
=B9*D9
Total =
=SUM(B2:B9)
Mean =
=SUM(C2:C9)
StdDev = =SQRT(SUM(E2:E9))
Binomial Distribution
The binomial distribution is a standard discrete distribution that accounts for the case where there
are two possible events and the probabilities of each event are not the same. Given a number of
independent trials, n, of an experiment that has two possible outcomes (call them success and failure),
the probability of a certain number of successes occurring in those n trials is given by Equation 8.10:
Pn (x) = Cxn p x q n−x
(8.10)
where, Pn (x) = the probability of x successes in n trials
Cxn = the number of combinations of n items taken x at a time = n!/(x!(n − x)!)
p = the probability of a success for any given trial
q = the probability of a failure for any given trial = 1 − p
In addition to the probability of exactly x successes in n trials, it is also common to determine
the probability of less than k successes, greater than k successes, or between l and k successes.
8.3. PROBABILITY
155
Specifically,
Pn (x < k) =
Pn (x > k) =
Pn (l < x < k) =
k−1
Pn (j )
j =0
n
j =k+1
k−1
(8.11)
Pn (j )
(8.12)
Pn (j )
(8.13)
j =l+1
Excel® has a built-in function to compute a binomial distribution:
=BINOM.DIST(#_successes, #_trials, prob_of_success, cumulative).
where, #_successes = number of successes in n trials (x)
#_trials = number of trials (n)
prob_of_success = probability of success (p)
cumulative = FALSE for probability distribution
= TRUE for cumulative probability distribution
Example 8.10
The probability that a fuse will be defective when first installed is 0.08. If six fuses are selected
at random, find each of the following:
(a) The probability that less than two fuses are defective
(b) The probability that four or more fuses are defective
(c) The probability that at least one is defective
Solution:
Define a success as a fuse that is defective. Therefore, p = 0.08 and q = 0.92.
P6 (0) = C60 (0.08)0 (0.92)6 =
= 0.606
P6 (1) = C61 (0.08)1 (0.92)5 =
= 0.316
P6 (2) = C62 (0.08)2 (0.92)4 =
= 0.0688
6!
(0.08)0 (0.92)6
0!6!
6!
(0.08)1 (0.92)5
1!5!
6!
(0.08)2 (0.92)4
2!4!
156
8. BASIC STATISTICS AND PROBABILITY
P6 (3) = C63 (0.08)3 (0.92)3 =
= 0.00797
P6 (4) = C64 (0.08)4 (0.92)2 =
= 0.000520
P6 (5) = C65 (0.08)5 (0.92)1 =
= 0.0000181
P6 (6) = C66 (0.08)6 (0.92)0 =
= 0.000000262
6!
(0.08)3 (0.92)3
3!3!
6!
(0.08)4 (0.92)2
4!2!
6!
(0.08)5 (0.92)1
5!1!
6!
(0.08)6 (0.92)0
6!0!
(a) P6 (x < 2) = P6 (0) + P6 (1) = 0.606 + 0.316 = 0.922
(b) P6 (x ≥ 4) = P6 (4) + P6 (5) + P6 (6) = 5.20 · 10−4 + 1.81 · 10−5 + 2.62 · 10−7 =
5.38 · 10−4
(c) P6 (x > 0) = P6 (1) + P6 (2) + P6 (3) + P6 (4) + P6 (5) + P6 (6) = 0.394
Alternatively, P6 (x > 0) = 1 − P6 (0) = 1 − 0.606 = 0.394
Using Excel® :
8.3. PROBABILITY
157
In graphical form, the discrete distribution for Example 8.10 can be shown as:
Binomial DistribuƟon
0.7
6.06E-01
0.6
0.5
P(x)
0.4
3.16E-01
0.3
0.2
6.88E-02
0.1
7.97E-03 5.20E-04 1.81E-05 2.62E-07
0
0
1
2
3
4
5
6
# of Successes
Continuous Distributions
When the value of the event, x, can take on a continuous set of probability values, rather than
a set of specific values, then a probability density function, p(x), exists. While there are a wide
variety of continuous distributions possible, the authors have chosen to present three continuous
distributions: the uniform distribution, the triangular distribution, and the normal or Gaussian
distribution. Figure 8.6 is a representation of a continuous distribution.
For continuous probability distributions, the following statements and equations are pertinent:
1. For a given x, p(x) is not the probability of that exact value occurring. Since there are an
infinite number of values for x, the probability of any one specific value of x would be zero.
2. The total area under the curve will equal the value of unity.
3. The mean of the data is calculated with Equation 8.14.
∞
μ=
xp(x)dx
−∞
(8.14)
158
8. BASIC STATISTICS AND PROBABILITY
ConƟnuous DistribuƟon
0.25
0.2
0.15
p(x)
0.1
0.05
0
0
2
4
6
8
10
12
14
16
x
Figure 8.6: Continuous distribution.
4. The standard deviation of the data is calculated with Equation 8.15:
σ =
∞
−∞
x 2 p(x)dx − μ2
5. The cumulative probability, F (x), is defined with Equation 8.16.
x
p(x)dx
F (x) =
−∞
(8.15)
(8.16)
6. F (x1 ) represents the probability that the value of x is less than or equal to x1 .
7. The quantity (1 − F (x1 )) represents the probability that the value of x is greater than or equal
to x1 .
Uniform or Rectangular Distribution
The uniform or rectangular distribution is represented in Figure 8.7. Each value of x has the same
probability of occurring.
Let a be the minimum value of x and b be the maximum value of x. Since the area under the
probability curve must be unity, the height of the uniform distribution will be given by Equation 8.17:
h = 1/(b − a)
(8.17)
8.3. PROBABILITY
159
Figure 8.7: Uniform or rectangular distribution.
The uniform distribution then has the following properties:
p(x) = h for a ≤ x ≤ b
p(x) = 0 for all other values of x
μ = (a + b)/2
√
σ = (b − a)/ 12
F (x) = (x − a)/(b − a)
(8.18)
(8.19)
(8.20)
Triangular Distribution
The triangular distribution is represented in Figure 8.8.
Let a be the minimum value of x, c be the maximum value of x, and b be the mode. P1 and P2
represent the areas from a to b and b to c, respectively. The triangular distribution has the following
properties:
h = 2/(c − a)
P1 = (b − a)/(c − a)
P2 = (c − b)/(c − a)
(8.21)
(8.22)
(8.23)
160
8. BASIC STATISTICS AND PROBABILITY
h
Figure 8.8: Triangular distribution.
μ = (a + b + c)/3
σ = (c − a) ∗ (1 − P1 P2 )/18
F (x) = P1 [(x − a)/(b − a)]2 for a ≤ x ≤ b
F (x) = 1 − P2 [(c − x)/(c − b)]2 for b ≤ x ≤ c
(8.24)
(8.25)
(8.26)
(8.27)
Normal Distribution
The Normal or Gaussian distribution is a continuous probability function that takes on the common
“bell-shaped curve” as represented in Figure 8.9.
The shape of this distribution is calculated with Equation 8.28:
1
−1
p(x) = √ e 2
σ 2π
x−μ 2
σ
where, μ = the mean of the data
σ = the standard deviation of the data
range of variable x: − ∞ ≤ x ≤ ∞
(8.28)
8.3. PROBABILITY
161
Figure 8.9: Representation of a unit normal distribution (μ = 0, σ = 1).
When μ = 0 and σ = 1, the distribution is called a unit normal distribution and Equation 8.28
simplifies to Equation 8.29:
x2
1
(8.29)
p(x) = √ e− 2
2π
One can convert any set of normally distributed data to a unit normal distribution through
the substitution of the variable Z, defined as:
Z = (x − μ)/σ
(8.30)
This allows one to then use Table 8.1 to determine values of p(Z) and F (Z) as defined above.
Since the unit normal distribution is symmetrical about Z = 0, one only needs the positive
portion of the table. If Z < 0, then use the following equations for p(Z) and F (z):
p(−Z) = p(Z)
F (−Z) = 1 − F (Z)
(8.31)
(8.32)
Excel® has a built-in function that calculates p(x) and F (x) given x, the mean, and the
standard deviation:
=NORM.DIST(x,Mean,Std_Dev,Cumulative)
x = value at which to find the value of either of p(x) of F (x)
Mean = mean of the distribution
Std_Dev = standard deviation of the distribution
Cumulative = FALSE for p(x) or =TRUE for F (x).
where,
162
8. BASIC STATISTICS AND PROBABILITY
Table 8.1: Values of p(Z) and F (Z) for the unit normal distribution
Z
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
p(Z)
0.39894
0.39844
0.39695
0.39448
0.39104
0.38667
0.38139
0.37524
0.36827
0.36053
0.35207
0.34294
0.33322
0.32297
0.31225
0.30114
0.28969
0.27798
0.26609
0.25406
0.24197
0.22988
0.21785
0.20594
0.19419
0.18265
0.17137
0.16038
0.14973
0.13943
0.12952
F(Z)
0.50000
0.51994
0.53983
0.55962
0.57926
0.59871
0.61791
0.63683
0.65542
0.67364
0.69146
0.70884
0.72575
0.74215
0.75804
0.77337
0.78814
0.80234
0.81594
0.82894
0.84134
0.85314
0.86433
0.87493
0.88493
0.89435
0.90320
0.91149
0.91924
0.92647
0.93319
Z
1.55
1.60
1.65
1.70
1.75
1.80
1.85
1.90
1.95
2.00
2.05
2.10
2.15
2.20
2.25
2.30
2.35
2.40
2.45
2.50
2.55
2.60
2.65
2.70
2.75
2.80
2.85
2.90
2.95
3.00
p(Z)
0.12001
0.11092
0.10226
0.09405
0.08628
0.07895
0.07206
0.06562
0.05959
0.05399
0.04879
0.04398
0.03955
0.03547
0.03174
0.02833
0.02522
0.02239
0.01984
0.01753
0.01545
0.01358
0.01191
0.01042
0.00909
0.00792
0.00687
0.00595
0.00514
0.00443
F(Z)
0.93943
0.94520
0.95053
0.95543
0.95994
0.96407
0.96784
0.97128
0.97441
0.97725
0.97982
0.98214
0.98422
0.98610
0.98778
0.98928
0.99061
0.99180
0.99286
0.99379
0.99461
0.99534
0.99598
0.99653
0.99702
0.99744
0.99781
0.99813
0.99841
0.99865
8.3. PROBABILITY
163
There is also a built-in function that calculates x given F (x), the mean, and the standard
deviation:
=NORM.INV(F(x),Mean,Std_Dev)
where,
F(x) = value of the cumulative distribution at which to find the value of x
Example 8.11
An engineer estimates that the selling price of a particular commodity will range from a low
of $5.00 per item to a high of $10.00 per item.
(a) If the distribution is assumed to be uniform, calculate the mean (or expected) value and the
standard deviation for the price of this commodity. Also, calculate the probability that the
price will be greater than $9.00.
(b) If the distribution is assumed to be triangular with a most likely value (mode) of $7.00 per
item, calculate the mean (or expected) value and the standard deviation for the price of this
commodity. Also, calculate the probability that the price will be greater than $9.00.
Solution (a):
The distribution would be:
The mean would be: μ = (a + b)/2 = (5 + 10)/2 = 7.50
√
√
The standard deviation would be: σ = (b − a)/ 12 = (10 − 5)/ 12 = 1.443
The probability of the price being greater than $9.00:
Prob(> $9) = 1 − F (9) = 1 − {(9 − 5)/(10 − 5)} = 0.20
164
8. BASIC STATISTICS AND PROBABILITY
Solution (b):
The distribution would be:
The mean would be: μ = (a + b + c)/3 = (5 + 7 + 10)/3 = 7.33
The areas would be: P1 = (b − a)/(c − a) = 2/5 = 0.4 and P2 = (c − b)/(c − a) =
3/5 = 0.6
The standard deviation would be:
σ = (c − a) ∗ (1 − P1 P2 )/18 = 5 ∗ (1 − (0.4)(0.6))/18 = 1.03
The probability of the price being greater than $9.00:
Prob(> $9) = 1 − F (9) = 1 − 1 − (0.6) [(10 − 9)/(10 − 7)]2 = 0.067
Example 8.12
300 ball bearings are tested for their diameters. The mean diameter was determined to be
0.452 cm and the standard deviation was determined to be 0.010 cm. Assume that the diameters
are normally distributed.
(a) How many ball bearings would be expected to be smaller than 0.4425 cm?
(b) Seventy percent of the ball bearings would be expected to have a diameter greater than what
value?
8.3. PROBABILITY
165
The distribution would be:
Solution for (a) using the F (Z) table:
Z = (x − μ)/σ = (0.4425 − 0.452)/0.010 = −0.95
F (−0.95) = 1 − F (0.95) = 1 − 0.82894 = 0.17106
# of ball bearings less than 0.4425 cm diameter = 0.17106(300) = 51
Solution for (a) using Excel® :
1
2
3
4
5
A
Mean =
StdDev =
x=
F(x) =
# bearings < x =
B
0.452
0.010
0.4425
0.17106
51
1
2
3
4
5
A
B
Mean =
0.452
St d D e v =
0.010
x=
0.4425
F(x) = =NORM.DIST(B3,B1,B2,TRUE)
# bearings < x =
=300*B4
Solution for (b) using the F (z) table:
One needs the value of Z that produces an F (Z) of 0.30 (for 70% greater than that value).
Since 0.30 is less than 0.5, one needs the negative side of the curve. F (−Z) = 1 − F (z) = 1 − .3 =
.7. The value of Z for F (0.7) lies between 0.50 and 0.55. Interpolating, Z = 0.524. Therefore,
Z for F (Z) of 0.30 is Z = −0.524. Solving for x in Equation 8.30 yields x = Z ∗ σ + μ =
−0.524(0.010) + 0.452 = 0.447 cm.Therefore, 70% of the ball bearings will have a diameter greater
than 0.447 cm.
166
8. BASIC STATISTICS AND PROBABILITY
Solution for (b) using Excel® :
1
2
3
4
A
Mean =
StdDev =
F(x) =
x=
B
0.452
0.010
0.3
0.44676
1
2
3
4
A
B
Mean =
0.452
StdDev =
0.010
F(x) =
0.3
x = =NORM.INV(B3,B1,B2)
Combined Distributions
In some applications, it will be necessary to work with more than one distribution to describe a
particular variable. In order to find the mean and standard deviation for the combined distributions,
the mean and standard deviation for each separate distribution are first determined. Equations 8.33
and 8.34 are then used to calculate the overall average and standard deviation:
μc =
(8.33)
Ai μi
σc =
Ai σi2 + (μi − μc )2
where, μc = mean of the combined distributions
σc = standard deviation of the combined distributions
Ai = probability area associated with each distribution
(8.34)
Ai = 1
μi = mean of each distribution
σi = standard deviation of each distribution
Example 8.13
An oil well has a 25% chance of being a “dry hole” (no oil found) and a 75% chance of finding
an oil reservoir that contains between 10,000 and 60,000 barrels as shown in the distribution below.
(a) Calculate the mean and standard deviation of the combined distributions.
(b) What is the probability that the reservoir will contain less than 40,000 barrels?
(c) What is the probability that the reservoir will contain at least 50,000 barrels?
(d) Sketch the cumulative probability distribution, F (x).
8.3. PROBABILITY
167
0.25
p(x)
0
barrels
10,000
60,000
Solution for (a):
Since the discrete probability at x = 0 is 0.25, the remaining area under the uniform distribution is then 0.75. The mean of the discrete probability distribution is 0 barrels and the mean
of the uniform distribution is 35,000 barrels.
The mean of the combined distribution: μc = 0.25(0) + 0.75(35, 000) = 26, 250 barrels
The standard deviation of the discrete probability distribution is 0 barrels and the standard
deviation of the uniform distribution is 14,434 barrels.
The standard deviation of the combined distribution is
σc = 0.25 ∗ 02 + (0 − 26, 250)2 + 0.75 ∗ 14, 4342 + (35, 000 − 26, 250)2
= 19, 645 barrels
Solution for (b):
P rob(< 40, 000) = F (40, 000). Recall that F (x) is the area under the probability curve.
Therefore, F (40, 000) = 0.25 + 0.75 ∗ [(40, 000 − 10, 000)/(60, 000 − 10, 000)] = 0.70.
There is a 70% probability that the reservoir will contain less than 40,000 barrels.
Solution for (c):
P rob(> 50, 000) = 1 − F (50, 000)
= 1 − [0.25 + 0.75 ∗ [(50, 000 − 10, 000)/(60, 000 − 10, 000)]] = 0.15.
There is a 15% probability that the reservoir will contain at least 50,000 barrels.
168
8. BASIC STATISTICS AND PROBABILITY
Solution for (d):
Again, F (x) is the area under the probability distribution. Therefore,
for
for
for
for
x<0
0 ≤ x ≤ 10, 000
10, 000 ≤ x ≤ 60, 000
x > 60, 000
F (x) = 0
F (x) = 0.25
F (x) = 0.25 + 0.75 [(x − 10, 000)/(60, 000 − 10, 000)]
F (x) = 1.0
CumulaƟve Probability
1
0.9
0.8
0.7
0.6
F(x) 0.5
0.4
0.3
0.2
0.1
0
-10000
0
10000 20000 30000 40000 50000 60000 70000
Barrels
8.4
8.1.
PROBLEMS
The following values of Young’s Modulus for a rubber compound (in 1000 lb/in 2 ) have
been measured. Determine the following:
(a) The frequency distribution using class boundaries of 4-6, 6-8, etc
(b) True class boundaries and true class marks
(c) The histogram and cumulative frequency diagrams
8
11
5
13
5
6
6
12
12
11
8
6
14
9
11
10
13
8
4
9
10
15
8
8
9
8
10
13
8.4. PROBLEMS
169
8.2.
For the data in Problem 8.1, calculate the mean, median, mode, and standard deviation.
Recalculate the mean and standard deviation using the frequency distribution determined
in Problem 8.1.
8.3.
A particular event has two possible outcomes of true and false. There is a 50% probability of
getting a true outcome. The event is repeated four times. Construct a table that contains all
possible combinations of results and determine the probabilities of getting 0 true outcomes,
1 true outcome, 2 true outcomes, 3 true outcomes, and 4 true outcomes.
8.4.
Fifteen castings of a certain type are produced per day in a foundry. The finished castings
are inspected and classified as defective or non-defective. Records indicate that of the last
500 castings inspected, 16 were defective. Based on this information, find the following:
(a) The probability of having at no defective castings in a day’s production
(b) The probability of having at least two defective castings in a day’s production
8.5.
The height of trucks on an interstate highway is approximately normally distributed with
mean of 10 ft and standard deviation of 1.5 ft. What is the height of an overpass if the
probability that a truck will clear it is 0.999?
8.6.
The average life of a certain type of compressor is 10 years with a standard deviation of 1
year. The lives of the compressors follow a normal distribution. The manufacturer replaces,
at no cost, all compressors that fail while under the guarantee. If the manufacturer is willing
to replace only 3% of all compressors sold, how long of a guarantee should they offer?
8.7.
A discrete distribution is given in the table below. Calculate the mean and standard deviation
of the distribution.
x
p(x)
1
0.2
2
0.3
3
0.1
4
0.4
170
8. BASIC STATISTICS AND PROBABILITY
8.8.
Determine the mean and standard deviation of the following combined distribution.
8.9.
A company is desirous of purchasing a service. The service will cost $10,000 and have a
probability of its life that can be described by a triangular distribution with values of a, b,
and c equal to 1, 3, and 6 years, respectively.
(a) Calculate the mean and the standard deviation of the life of the service
(b) What is the probability that the service will last at least 2 years?
(c) What is the probability that the service will last at least 5 years?
171
CHAPTER
9
Sensitivity Analysis
9.1
INTRODUCTION
A simple way of incorporating the elements of uncertainty into an economic analysis is to use
sensitivity analysis. As described earlier, an evaluator will normally need to include a measure of the
uncertainty pertaining to one or more variables in the analysis. This uncertainty may, in turn, add
significant uncertainty about the profitability of an investment. This range of uncertainty about the
project’s profitability is one way to define the risk in a project.
Uncertainty in any particular variable can occur for a number of reasons. For example, the
method of measuring a parameter may have a certain amount of inaccuracy, the parameter may have
to be predicted into the future, or there may be a limited amount of data for a certain parameter. In
any case, the best that can be done for a variable with an uncertain value is to choose a reasonable
range over which it may vary and, perhaps, the type of distribution that the variable might take on
over that range.
Two types of sensitivity analysis will be considered in this chapter. The first is called the
range approach and involves the systematic variation of key variables to determine their overall
effect on the profitability of the investment. The second approach uses the concepts of probability
and statistics and is referred to as Monte Carlo Simulation (MCS). MCS has also been called
probabilistic sensitivity analysis.
9.1.1
RANGE APPROACH
When applying the range approach, ranges of variations of key variables, defined by the evaluator, are
established. For example, the estimated sales price of a commodity to be sold could be allowed to vary
±10% from a base value during the analysis. The economic analysis is conducted by: (a) choosing an
evaluation criteria (normally NPV or IRR); (b) computing the value of this criteria for a base case
set of variable values; and (c) repeating the computations by varying each key parameter within the
specified range.
There are, in general, two ways of conducting the range approach sensitivity analysis.
The first is by identifying the most likely, most optimistic, and the most pessimistic cases
by varying all of the parameters simultaneously. The most likely case is defined as the case where
all variables are at their respective mean values. This method allows the evaluator to determine
the minimum and maximum values that could be obtained for the evaluation criteria. It does not,
however, allow the evaluator to study the effects of any one variable on the economic analysis.
172
9. SENSITIVITY ANALYSIS
The second way is to use the mean values for each key variable and calculate the corresponding
value for the evaluation criteria. This is then designated as the base case value. Each key parameter
is then varied about its mean value while the other parameters are held constant at their base case
values and the evaluation criteria is recalculated. The process is repeated until each parameter has
been varied. Typically, each parameter is varied plus or minus 10 to 20% about its mean value. When
the calculations have been completed, the results are usually summarized in a “spider plot.” The
spider plot, shown schematically in Figure 9.1, is constructed by plotting the evaluation criteria on
the vertical axis and the percent variation on the horizontal axis. Quick inspection of the spider
plot provides information to the evaluator on which parameter or parameters affect the economic
analysis to the greatest degree. The parameter which yields the line with the greatest slope (positive
or negative slope) on the spider plot has the most effect on the analysis. As illustrated in the sample
plot shown in Figure 9.1, Variable A has the greatest effect on the evaluation criteria.
Variable A
EvĂůƵĂƟŽŶ
Criteria
Variable B
Variable C
-X%
0
+X%
% variĂƟon from the base case
Figure 9.1: Sample spider plot.
Example 9.1
A ten-year life project has an initial investment of $87,500, annual operating expenses of
$7,500, and annual incomes of $30,000. It is desired to conduct a range approach sensitivity analysis
by the two methods described earlier:
9.1. INTRODUCTION
173
(a) Determine the most likely, the most optimistic, and the most pessimistic values for the I RR
by assuming the values given are the mean values for each parameter and that each parameter
has a range of ±20% from the mean value.
(b) Vary each parameter independently by ±20% from the mean values while holding the other
two constant at their mean values and develop a spider plot for the calculated I RR values.
The cash flow diagram for the most likely (or base case) is as follows:
0
-87,500
1
22,500
2
22,500
3
22,500
…
----
8
22,500
9
22,500
10
22,500
Solution for (a):
The I RR for the most likely case is computed from the cash flow above to be 22.3%.
The most pessimistic case would be the combination of these three variables that would have
the most negative influence on the project’s I RR. This would be a 20% increase in initial investment,
a 20% increase in operating expenses, and a 20% decrease in income.
The most optimistic case would be the combination of these three variables that would have
the most positive influence on the project’s IRR. This would be a 20% decrease in initial investment,
a 20% decrease in operating expenses, and a 20% increase in income.
Solution for (b):
In this analysis, start with the most likely case from (a) and denote that as the base case. Then
vary one parameter at a time by plus and minus 20% and recalculate I RR for each new cash flow
diagram.
174
9. SENSITIVITY ANALYSIS
Spider Plot
35
30
25
IRR
20
IniƟal Investment
15
OperaƟng Expenses
10
Annual Income
5
0
-25
-15
-5
5
15
25
% sĂƌŝĂƟon from Base Case
From the spider plot, it can be readily seen that the parameter which has the greatest effect on the
I RR is the annual income. A larger change in I RR is observed for the same percentage change in the
annual income than for initial investment or operating expenses. Changes in the operating expenses
have the least effect on the I RR. To minimize the risk that would be created by a ±20% uncertainty
in the annual income, it would be beneficial for the evaluator to do additional research into this
portion of the cash flow calculation and determine if the range in uncertainty can be reduced.
In the example problem, only two values for each variable have been used.To be more complete,
several values between −20% and +20% could have been used to generate additional values of I RR.
9.1. INTRODUCTION
175
The primary drawback on the spider plot approach is that it ignores the interactions that occur
when more than one variable is allowed to change at a time. Not only does one have to define many
more cases to account for all possible interactions, but it is also difficult to tabulate these results in a
meaningful way. Often, the results simply become a tabulated list of I RRs for each case evaluated.
For example, consider a problem that has four variables that have uncertain values and that three
numerical values for each of the variables are chosen in the range approach. This will result in 81
(3x3x3x3) individual solutions to the problem that must be presented to evaluate the effect of all
four variables. It may be very difficult for the evaluator to draw any conclusions from a long tabular
list of 81 results.
9.1.2
MONTE CARLO SIMULATION
Probabilistic sensitivity analysis or Monte Carlo Simulation (MCS) was introduced in the early
1960s. While it is a very powerful technique, it is often avoided due to the general lack of knowledge
among engineers and managers on how it works and the conclusions that can be drawn from it. This
section will attempt to explain the technique and its benefits.
Consider the example proposed above that contains four variables with uncertain values.
Instead of creating a tabular list of 81 evaluator-chosen cases, MCS allows each of the variables to vary
between minimum and maximum values according to some prescribed probability distribution and
then solves the problem for a large, randomized set of these input variables. The results of the MCS
are presented graphically as a cumulative probability distribution of the dependent variable (e.g.,
I RR, N P V , etc). This probability can then be interpreted using statistical methods to determine
the likelihood of a particular solution occurring or not occurring.
Distributions that are frequently used are the uniform, triangular, and normal distributions
presented in Chapter 8. These are fairly easy to describe mathematically and to input into an Excel®
spreadsheet or a computer program. The choice of the particular distribution for a certain variable
should be guided by the evaluator’s knowledge of that variable.
Within the MCS method, the selection of a value for an independent variable is accomplished
by using the fact that the cumulative probability distribution, F (x), will lie between zero and one
and will be monotonic in behavior. Thus, the selection of a random number between zero and one
will yield a distinct random value for the independent variable between the variable’s minimum and
maximum values. A different random number is chosen for each independent variable, resulting in
a truly random set of independent variable values. Once the values of all independent variables have
been determined, the dependent variable, i.e., an evaluation criteria such as I RR or NP V , can be
calculated. This process is then repeated a large number of times and the results of the dependent
variable calculations are grouped into class intervals and then the cumulative probability distribution
is constructed.
The following is a summary of the steps involved in the MCS method:
1. Select the independent variables that contain uncertainty in their values.
176
9. SENSITIVITY ANALYSIS
2. Estimate the minimum and maximum values for each independent variable.
3. Estimate the minimum and maximum values for the dependent variable and set up class
intervals in that range such that a probability distribution can be generated.
4. Select a probability distribution that best describes the behavior of each independent variable
between its minimum and maximum values.
5. Set up equations which will allow for the calculation of each of the independent variables.
This is done by determining expressions for the cumulative probability distributions, F (x),
for each independent variable and then solving this expression for the variable, x.
6. Generate a random number for each independent variable. A different random number is
determined for each independent variable. Random numbers are available from scientific
calculators, Excel® , or by using Table 9.1. One can enter this table at any random point and
then proceed through the table either by rows or columns. Excel® uses the =RAND() function
to generate a uniformly distributed random number between zero and one.
7. Use the random numbers to calculate the values for the independent variables using the
equations developed in step 5.
8. Calculate the dependent variable (or variables if necessary) for this set of independent variables
and increment a counter in the respective class interval.
9. Return to step 6 and repeat steps 6 through 8 a relatively large number of times. A large
number of trials might be 100, 1000, or 10,000 depending on the sensitivity of the dependent
variable to the independent variables.
10. Construct the cumulative probability distribution for the dependent variable.
177
9.1. INTRODUCTION
Table 9.1: Random numbers
178
9. SENSITIVITY ANALYSIS
Example 9.2
For the problem described in Example 9.1 and the distributions given below, conduct a Monte
Carlo Simulation and summarize the results in a cumulative probability distribution. Complete 10
cases using I RR as the dependent variable.
Initial Investment:
/ŶŝƟal Investment
0.00003
0.000025
0.00002
p(x) 0.000015
0.00001
0.000005
0
60000
70000
80000
90000
100000
110000
Dollars
Operating Expenses:
OperĂƟng Expenses
0.0007
0.0006
0.0005
p(x)
0.0004
0.0003
0.0002
0.0001
0
5000
6000
7000
8000
Dollars/Year
9000
10000
9.1. INTRODUCTION
179
Annual Income:
Annual Income
0.00009
0.00008
0.00007
0.00006
0.00005
p(x)
0.00004
0.00003
0.00002
0.00001
0
23000
25000
27000
29000
31000
33000
35000
37000
Dollars/Year
Solution:
Develop equations for the independent variables:
Let x1 be the value for the initial investment. Since it has a uniform distribution, its cumulative
probability (F1 ) is given by
F1 = (x1 − a)/(b − a) = (x1 − 70000)/(105000 − 70000) = (x1 − 70000)/35000
Solving this equation for x1 yields
x1 = 35000F1 + 70000
(9.1)
Let x2 be the value for the operating expenses. Since it has a triangular distribution, its
cumulative probability (F2 ) is given by
P1 = (b − a)/(c − a) = (7500 − 6000)/(9000 − 6000) = 0.5
P2 = 1 − P1 = 0.5
For 6000 ≤ x2 ≤ 7500 or F2 ≤ 0.5 (the value of P1 ),
F2 = P1 [(x2 − a)/(b − a)]2 = 0.5 [(x2 − 6000)/1500]2
Solving this equation for x2 yields
x2 = 6000 + 1500 2F2
For 7500 ≤ x2 ≤ 9000 or F2 ≥ 0.5 (the value of P1 ),
F2 = 1 − P2 [(c − x2 )/(c − b)]2 = 1 − 0.5 [(9000 − x2 )/1500]2
(9.2)
180
9. SENSITIVITY ANALYSIS
Solving this equation for x2 yields
x2 = 9000 − 1500 2(1 − F2 )
(9.3)
Let x3 be the value for the annual income. Since it has a uniform distribution, its cumulative
probability (F3 ) is given by
F3 = (x3 − a)/(b − a) = (x3 − 24000)/(36000 − 24000) = (x3 − 24000)/12000
Solving this equation for x3 yields
x3 = 12000F1 + 24000
(9.4)
First iteration:
Choose the first random number from the table. This will be the value for F1 . (F1 = 0.90535).
Use this number in Equation 9.1 to determine the value to be used for the initial investment:
x1 = 35000(0.90535) + 70000 = 101, 700
Choose the second random number from the table. This will be the value for F2 . (F2 =
0.86245). Since F2 ≥ 0.5, use this number in Equation 9.3 to determine the value to be used
for the operating expense:
x2 = 9000 − 1500 2(1 − 0.86245) = 8, 200
Choose the third random number from the table.This will be the value for F3 . (F3 = 0.32775).
Use this number in Equation 9.4 to determine the value to be used for the annual income:
x3 = 12000(0.32775) + 24000 = 27, 900
These values for initial investment, operating expense, and annual income yield the following
cash flow diagram for the project:
0
1
-101,700 19,700
2
19,700
3
19,700
…
----
8
19,700
9
19,700
10
19,700
I RR analysis yields a value of 14.3%. This value is tabulated in a list for further processing.
Second and successive iterations:
Follow the same procedure as listed for the first iteration. Three new random numbers are
used during each iteration. The results of the first ten iterations are shown in the table below.
9.1. INTRODUCTION
181
The cumulative probability distribution for I RR can be developed from information in the
following table:
182
9. SENSITIVITY ANALYSIS
Interval
10.5-12.5
12.5-14.5
14.5-16.5
16.5-18.5
18.5-20.5
20.5-22.5
22.5-24.5
24.5-26.5
26.5-28.5
28.5-30.5
30.5-32.5
32.5-34.5
Mid-point
11.5
13.5
15.5
17.5
19.5
21.5
23.5
25.5
27.5
29.5
31.5
33.5
Frequency
0
2
1
0
0
4
0
2
0
0
0
1
Prob
CumulProb
0.0
20.0
30.0
30.0
30.0
70.0
70.0
90.0
90.0
90.0
90.0
100.0
0.0
20.0
30.0
0.0
0.0
40.0
0.0
20.0
0.0
0.0
0.0
10.0
CumuůĂƟǀe Probability for IRR
100.0
90.0
80.0
70.0
60.0
Cumul Prob, % 50.0
40.0
30.0
20.0
10.0
0.0
10
15
20
25
30
35
IRR, %
With only ten iterations, this graph is too jagged to interpret correctly. The figure below
shows the same analysis after 100 iterations. One can see that the curve is much smoother. If even
more iterations are added, the curve will become smoother yet. However, the usefulness of the curve
may not increase proportionally to the number of iterations. One should only complete enough
iterations to get a reasonably smooth curve. Generally, this takes about 100 iterations, but this may
be a function of the actual problem being solved.This number of calculations can be easily completed
with Excel® .
9.1. INTRODUCTION
183
CumuůĂƟǀe Probability for IRR
100.0
90.0
80.0
70.0
60.0
F(x), % 50.0
40.0
30.0
20.0
10.0
0.0
10
15
20
25
30
35
40
IRR, %
As discussed in Chapter 8, the use of this graph is as follows. The value of F (x) at any I RR
is the probability that the project will attain that I RR or less. For example, there is approximately
an 18% probability that the project will earn an I RR value of less than 15%. Thus, if one uses
the investor’s MARR, F (x) provides the probability that the project will earn less than that value.
The quantity (100 − F (x)) would provide the probability that the project earns greater than that
MARR. For example, if the investor’s MARR is 20%, one would enter the horizontal axis at 20%
and read a cumulative probability of about 44%. The interpretation would be that there is a 56%
probability that the project will yield a 20% I RR or greater. This probability is then a direct measure
of risk associated with the project. If an evaluator feels that a 44% probability that the project will
not be economically viable is an unacceptable level of risk, then the project should be eliminated
from further consideration. However, if, in this example, the investor’s MARR is only 15%, there is
only a 18% probability that the project will not be economically viable. This investor would have a
more acceptable level of risk.
Example 9.3
A particular investment has three uncertain variables of initial cost, future value, and the
investment life.
The initial cost can be described by a uniform distribution from $100 to $200.
The future value can be described by a normal distribution with μ = $300 and σ = $30.
The investment life can be described by a discrete probability distribution with 40% probability
that n = 5 years, 30% probability that n = 6 years, 20% probability that n = 7 years, and 10%
probability that n = 8 years.
184
9. SENSITIVITY ANALYSIS
Based on this information,
(a) Calculate the minimum rate of return that can be earned, the maximum rate of return that
can be earned and the mean rate of return that will be earned. For the normal distribution,
assume that the minimum and maximum values of future value will be ±3σ from the mean
($210 and $390, respectively).
(b) Complete a Monte Carlo Simulation for this project to determine the probability that the
ROR will be at least 15%.
Solution for (a):
The mean values for the three distributions are:
Pmean = (100 + 200)/2 = 150
Fmean = 300
nmean = 0.4 ∗ 5 + 0.3 ∗ 6 + 0.2 ∗ 7 + 0.1 ∗ 8 = 6.0
The rate of return, ROR, can be calculated using the relationship, F = P ∗ (1 + i)n or
i = (F /P )1/n − 1
The three cases will be defined as follows:
Case
Minimum ROR
(Most PĞƐƐŝŵŝƐƟĐ
ĂƐĞͿ
DĞĂn ROR
DĂdžŝmum ROR
(Most KƉƟŵŝƐƟĐ
ĂƐĞͿ
/ŶŝƟĂl Cost, P
Future VĂůƵe, F
Life, n
ROR
$200
$210
8
0.61%
$150
$300
6
12.2%
$100
$390
5
31.3%
9.1. INTRODUCTION
185
Solution for (b):
Develop equations for the independent variables:
Let x1 be the value for the initial investment. Since it has a uniform distribution, its cumulative
probability (F1 ) is given by
F1 = (x1 − a)/(b − a) = (x1 − 100)/(200 − 100) = (x1 − 100)/100
Solving this equation for x1 yields
x1 = 100F1 + 100
Let x2 be the value for the future value. Since it has a normal distribution, its cumulative
probability (F2 ) is given by the values in Table 8.1. Once F2 (Z) is randomly chosen, the
appropriate value of Z is determined from Table 8.1 and then x2 = Z ∗ σ + μ.
Let x3 be the value for the project life. Since it has a discrete distribution, its cumulative
probability (F3 ) is given by
CumƵůĂƟve Probability for Project Life
5 ≤ x3 < 6
6 ≤ x3 < 7
7 ≤ x3 < 8
x3 ≥ 8
F3
F3
F3
F3
= 0.4
= 0.7
= 0.9
= 1.0
1
0.9
0.8
0.7
0.6
F(x) 0.5
0.4
0.3
0.2
0.1
0
4
5
6
7
n, years
Thus, once F3 is randomly chosen, the value for x3 will be:
F3 ≤ 0.4
0.4 < F3 ≤ 0.7
0.7 < F3 ≤ 0.9
0.9 < F3 ≤ 1.0
x3
x3
x3
x3
=5
=6
=7
=8
8
9
186
9. SENSITIVITY ANALYSIS
The dependent variable is ROR which is calculated by using: ROR = (x2 /x1 )1/x3 − 1.
Each iteration can be calculated using the following Excel® spreadsheet:
If one tabulates the result in cell B9 into another column of results (for example start in cell
A20), the spreadsheet will automatically select three new random numbers and a new result of
ROR will be calculated. This result would then be tabulated in cell A21. In order to generate
the histogram, this process is repeated manually 100 times (which would result in a column
of data from A20 to A119).
The results would be as follows:
CumuůĂƟǀe Probability for ROR
100.0
90.0
80.0
70.0
60.0
F(x), % 50.0
40.0
30.0
20.0
10.0
0.0
0
5
10
15
20
25
30
35
ROR, %
At ROR of 15%, F (x) is approximately 74%. This means that there is a 74% probability that
the ROR will be less than 15% or a 26% probability that the ROR will be at least 15%.
9.2. PROBLEMS
9.2
9.1.
187
PROBLEMS
You are to conduct an extensive sensitivity analysis on the problem described below. The
sensitivity analysis will consist of three parts:
(a) A range approach where the most optimistic, most likely, and most pessimistic values
of the dependent variables NP V and I RR are determined.
(b) A range approach where the mean value is determined for each independent variable
and then each variable is allowed to vary ±20% about that mean while all other
independent variables are held constant. Create spider plots for NP V and I RR using
the results.
(c) A probabilistic approach using Monte Carlo Simulation. Complete 10 iterations and
create cumulative probability curves for NP V and I RR.
The project life is 7 years and the MARR is 15%.
The initial investment is given by a uniform distribution between $200,000 and $300,000.
The annual profit is given by a triangular distribution that has a minimum value of
$55,000/year, a mode of $67,500/year, and a maximum value of $85,000/year.
The salvage value of the investment is given by a triangular distribution that has a minimum
value of $60,000, a mode of $75,000, and a maximum value of $85,000.
The cash flow diagram would be:
0
1
-Invest
ProĮt
2
ProĮt
3
ProĮt
…
----
6
ProĮt
7
ProĮt +
Salvage
9.2.
Complete Problem 9.1 using an Excel® spreadsheet to calculate 100 iterations. Create
cumulative probability curves for NP V and I RR.
9.3.
The following distributions are given for three independent variables, x1 , x2 , and x3 and the
relationship for the dependent variable, y. Calculate the largest, smallest, and mean values
of the dependent variable, y.
x1 : Uniform distribution between 35 and 50
x2 : Triangular distribution between 20 and 40 with a mode of 35
188
9. SENSITIVITY ANALYSIS
x3 : Discrete distribution with a 50% probability that the value will be 2 and 50%
probability that the value will be 4
y = (x1 )(x2 ) + x3
9.4.
Using the information given in Problem 9.3, use Monte Carlo Simulation to calculate 10
iterations of the dependent variable and create the cumulative probability diagram for the
dependent variable y.
9.5.
Complete Problem 9.4 using an Excel® spreadsheet and 100 iterations.
9.6.
Complete Problem 9.4 using an Excel® spreadsheet and assuming that independent variable
x2 has a normal distribution with a mean of 30 and a standard deviation of 3. Compute 100
iterations and create the cumulative probability diagram for the dependent variable y.
189
APPENDIX
A
Compound Interest Factors
190
APPENDIX A
i=
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
F/P
1.01000
1.02010
1.03030
1.04060
1.05101
1.06152
1.07214
1.08286
1.09369
1.10462
1.11567
1.12683
1.13809
1.14947
1.16097
1.17258
1.18430
1.19615
1.20811
1.22019
1.28243
1.34785
1.41660
1.48886
1.56481
1.64463
1.72852
1.81670
1.90937
2.00676
2.10913
2.21672
2.32979
2.44863
2.57354
2.70481
P/F
0.99010
0.98030
0.97059
0.96098
0.95147
0.94205
0.93272
0.92348
0.91434
0.90529
0.89632
0.88745
0.87866
0.86996
0.86135
0.85282
0.84438
0.83602
0.82774
0.81954
0.77977
0.74192
0.70591
0.67165
0.63905
0.60804
0.57853
0.55045
0.52373
0.49831
0.47413
0.45112
0.42922
0.40839
0.38857
0.36971
F/A
1.0000
2.0100
3.0301
4.0604
5.1010
6.1520
7.2135
8.2857
9.3685
10.4622
11.5668
12.6825
13.8093
14.9474
16.0969
17.2579
18.4304
19.6147
20.8109
22.0190
28.2432
34.7849
41.6603
48.8864
56.4811
64.4632
72.8525
81.6697
90.9366
100.6763
110.9128
121.6715
132.9790
144.8633
157.3538
170.4814
1%
A/F
1.00000
0.49751
0.33002
0.24628
0.19604
0.16255
0.13863
0.12069
0.10674
0.09558
0.08645
0.07885
0.07241
0.06690
0.06212
0.05794
0.05426
0.05098
0.04805
0.04542
0.03541
0.02875
0.02400
0.02046
0.01771
0.01551
0.01373
0.01224
0.01100
0.00993
0.00902
0.00822
0.00752
0.00690
0.00636
0.00587
P/A
0.9901
1.9704
2.9410
3.9020
4.8534
5.7955
6.7282
7.6517
8.5660
9.4713
10.3676
11.2551
12.1337
13.0037
13.8651
14.7179
15.5623
16.3983
17.2260
18.0456
22.0232
25.8077
29.4086
32.8347
36.0945
39.1961
42.1472
44.9550
47.6266
50.1685
52.5871
54.8882
57.0777
59.1609
61.1430
63.0289
A/P
1.01000
0.50751
0.34002
0.25628
0.20604
0.17255
0.14863
0.13069
0.11674
0.10558
0.09645
0.08885
0.08241
0.07690
0.07212
0.06794
0.06426
0.06098
0.05805
0.05542
0.04541
0.03875
0.03400
0.03046
0.02771
0.02551
0.02373
0.02224
0.02100
0.01993
0.01902
0.01822
0.01752
0.01690
0.01636
0.01587
A/G
0.00000
0.49751
0.99337
1.48756
1.98010
2.47098
2.96020
3.44777
3.93367
4.41792
4.90052
5.38145
5.86073
6.33836
6.81433
7.28865
7.76131
8.23231
8.70167
9.16937
11.48312
13.75566
15.98711
18.17761
20.32730
22.43635
24.50495
26.53331
28.52167
30.47026
32.37934
34.24920
36.08013
37.87245
39.62648
41.34257
COMPOUND INTEREST FACTORS
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
F/P
1.02000
1.04040
1.06121
1.08243
1.10408
1.12616
1.14869
1.17166
1.19509
1.21899
1.24337
1.26824
1.29361
1.31948
1.34587
1.37279
1.40024
1.42825
1.45681
1.48595
1.64061
1.81136
1.99989
2.20804
2.43785
2.69159
2.97173
3.28103
3.62252
3.99956
4.41584
4.87544
5.38288
5.94313
6.56170
7.24465
P/F
0.98039
0.96117
0.94232
0.92385
0.90573
0.88797
0.87056
0.85349
0.83676
0.82035
0.80426
0.78849
0.77303
0.75788
0.74301
0.72845
0.71416
0.70016
0.68643
0.67297
0.60953
0.55207
0.50003
0.45289
0.41020
0.37153
0.33650
0.30478
0.27605
0.25003
0.22646
0.20511
0.18577
0.16826
0.15240
0.13803
i=
2%
F/A
1.0000
2.0200
3.0604
4.1216
5.2040
6.3081
7.4343
8.5830
9.7546
10.9497
12.1687
13.4121
14.6803
15.9739
17.2934
18.6393
20.0121
21.4123
22.8406
24.2974
32.0303
40.5681
49.9945
60.4020
71.8927
84.5794
98.5865
114.0515
131.1262
149.9779
170.7918
193.7720
219.1439
247.1567
278.0850
312.2323
A/F
1.00000
0.49505
0.32675
0.24262
0.19216
0.15853
0.13451
0.11651
0.10252
0.09133
0.08218
0.07456
0.06812
0.06260
0.05783
0.05365
0.04997
0.04670
0.04378
0.04116
0.03122
0.02465
0.02000
0.01656
0.01391
0.01182
0.01014
0.00877
0.00763
0.00667
0.00586
0.00516
0.00456
0.00405
0.00360
0.00320
P/A
0.9804
1.9416
2.8839
3.8077
4.7135
5.6014
6.4720
7.3255
8.1622
8.9826
9.7868
10.5753
11.3484
12.1062
12.8493
13.5777
14.2919
14.9920
15.6785
16.3514
19.5235
22.3965
24.9986
27.3555
29.4902
31.4236
33.1748
34.7609
36.1975
37.4986
38.6771
39.7445
40.7113
41.5869
42.3800
43.0984
A/P
1.02000
0.51505
0.34675
0.26262
0.21216
0.17853
0.15451
0.13651
0.12252
0.11133
0.10218
0.09456
0.08812
0.08260
0.07783
0.07365
0.06997
0.06670
0.06378
0.06116
0.05122
0.04465
0.04000
0.03656
0.03391
0.03182
0.03014
0.02877
0.02763
0.02667
0.02586
0.02516
0.02456
0.02405
0.02360
0.02320
A/G
0.00000
0.49505
0.98680
1.47525
1.96040
2.44226
2.92082
3.39608
3.86805
4.33674
4.80213
5.26424
5.72307
6.17862
6.63090
7.07990
7.52564
7.96811
8.40732
8.84328
10.97445
13.02512
14.99613
16.88850
18.70336
20.44198
22.10572
23.69610
25.21471
26.66323
28.04344
29.35718
30.60635
31.79292
32.91889
33.98628
191
192
APPENDIX A
i=
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
F/P
1.03000
1.06090
1.09273
1.12551
1.15927
1.19405
1.22987
1.26677
1.30477
1.34392
1.38423
1.42576
1.46853
1.51259
1.55797
1.60471
1.65285
1.70243
1.75351
1.80611
2.09378
2.42726
2.81386
3.26204
3.78160
4.38391
5.08215
5.89160
6.82998
7.91782
9.17893
10.64089
12.33571
14.30047
16.57816
19.21863
P/F
0.97087
0.94260
0.91514
0.88849
0.86261
0.83748
0.81309
0.78941
0.76642
0.74409
0.72242
0.70138
0.68095
0.66112
0.64186
0.62317
0.60502
0.58739
0.57029
0.55368
0.47761
0.41199
0.35538
0.30656
0.26444
0.22811
0.19677
0.16973
0.14641
0.12630
0.10895
0.09398
0.08107
0.06993
0.06032
0.05203
F/A
1.0000
2.0300
3.0909
4.1836
5.3091
6.4684
7.6625
8.8923
10.1591
11.4639
12.8078
14.1920
15.6178
17.0863
18.5989
20.1569
21.7616
23.4144
25.1169
26.8704
36.4593
47.5754
60.4621
75.4013
92.7199
112.7969
136.0716
163.0534
194.3328
230.5941
272.6309
321.3630
377.8570
443.3489
519.2720
607.2877
3%
A/F
1.00000
0.49261
0.32353
0.23903
0.18835
0.15460
0.13051
0.11246
0.09843
0.08723
0.07808
0.07046
0.06403
0.05853
0.05377
0.04961
0.04595
0.04271
0.03981
0.03722
0.02743
0.02102
0.01654
0.01326
0.01079
0.00887
0.00735
0.00613
0.00515
0.00434
0.00367
0.00311
0.00265
0.00226
0.00193
0.00165
P/A
0.9709
1.9135
2.8286
3.7171
4.5797
5.4172
6.2303
7.0197
7.7861
8.5302
9.2526
9.9540
10.6350
11.2961
11.9379
12.5611
13.1661
13.7535
14.3238
14.8775
17.4131
19.6004
21.4872
23.1148
24.5187
25.7298
26.7744
27.6756
28.4529
29.1234
29.7018
30.2008
30.6312
31.0024
31.3227
31.5989
A/P
1.03000
0.52261
0.35353
0.26903
0.21835
0.18460
0.16051
0.14246
0.12843
0.11723
0.10808
0.10046
0.09403
0.08853
0.08377
0.07961
0.07595
0.07271
0.06981
0.06722
0.05743
0.05102
0.04654
0.04326
0.04079
0.03887
0.03735
0.03613
0.03515
0.03434
0.03367
0.03311
0.03265
0.03226
0.03193
0.03165
A/G
0.00000
0.49261
0.98030
1.46306
1.94090
2.41383
2.88185
3.34496
3.80318
4.25650
4.70494
5.14850
5.58720
6.02104
6.45004
6.87421
7.29357
7.70812
8.11788
8.52286
10.47677
12.31407
14.03749
15.65016
17.15557
18.55751
19.86004
21.06742
22.18407
23.21454
24.16342
25.03534
25.83490
26.56665
27.23505
27.84445
COMPOUND INTEREST FACTORS
i=
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
F/P
1.04000
1.08160
1.12486
1.16986
1.21665
1.26532
1.31593
1.36857
1.42331
1.48024
1.53945
1.60103
1.66507
1.73168
1.80094
1.87298
1.94790
2.02582
2.10685
2.19112
2.66584
3.24340
3.94609
4.80102
5.84118
7.10668
8.64637
10.51963
12.79874
15.57162
18.94525
23.04980
28.04360
34.11933
41.51139
50.50495
P/F
0.96154
0.92456
0.88900
0.85480
0.82193
0.79031
0.75992
0.73069
0.70259
0.67556
0.64958
0.62460
0.60057
0.57748
0.55526
0.53391
0.51337
0.49363
0.47464
0.45639
0.37512
0.30832
0.25342
0.20829
0.17120
0.14071
0.11566
0.09506
0.07813
0.06422
0.05278
0.04338
0.03566
0.02931
0.02409
0.01980
F/A
1.0000
2.0400
3.1216
4.2465
5.4163
6.6330
7.8983
9.2142
10.5828
12.0061
13.4864
15.0258
16.6268
18.2919
20.0236
21.8245
23.6975
25.6454
27.6712
29.7781
41.6459
56.0849
73.6522
95.0255
121.0294
152.6671
191.1592
237.9907
294.9684
364.2905
448.6314
551.2450
676.0901
827.9833
1012.7846
1237.6237
4%
A/F
1.00000
0.49020
0.32035
0.23549
0.18463
0.15076
0.12661
0.10853
0.09449
0.08329
0.07415
0.06655
0.06014
0.05467
0.04994
0.04582
0.04220
0.03899
0.03614
0.03358
0.02401
0.01783
0.01358
0.01052
0.00826
0.00655
0.00523
0.00420
0.00339
0.00275
0.00223
0.00181
0.00148
0.00121
0.00099
0.00081
P/A
0.9615
1.8861
2.7751
3.6299
4.4518
5.2421
6.0021
6.7327
7.4353
8.1109
8.7605
9.3851
9.9856
10.5631
11.1184
11.6523
12.1657
12.6593
13.1339
13.5903
15.6221
17.2920
18.6646
19.7928
20.7200
21.4822
22.1086
22.6235
23.0467
23.3945
23.6804
23.9154
24.1085
24.2673
24.3978
24.5050
A/P
1.04000
0.53020
0.36035
0.27549
0.22463
0.19076
0.16661
0.14853
0.13449
0.12329
0.11415
0.10655
0.10014
0.09467
0.08994
0.08582
0.08220
0.07899
0.07614
0.07358
0.06401
0.05783
0.05358
0.05052
0.04826
0.04655
0.04523
0.04420
0.04339
0.04275
0.04223
0.04181
0.04148
0.04121
0.04099
0.04081
A/G
0.00000
0.49020
0.97386
1.45100
1.92161
2.38571
2.84332
3.29443
3.73908
4.17726
4.60901
5.03435
5.45329
5.86586
6.27209
6.67200
7.06563
7.45300
7.83416
8.20912
9.99252
11.62743
13.11984
14.47651
15.70474
16.81225
17.80704
18.69723
19.49093
20.19614
20.82062
21.37185
21.85693
22.28255
22.65498
22.98000
193
194
APPENDIX A
i=
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
F/P
1.05000
1.10250
1.15763
1.21551
1.27628
1.34010
1.40710
1.47746
1.55133
1.62889
1.71034
1.79586
1.88565
1.97993
2.07893
2.18287
2.29202
2.40662
2.52695
2.65330
3.38635
4.32194
5.51602
7.03999
8.98501
11.46740
14.63563
18.67919
23.83990
30.42643
38.83269
49.56144
63.25435
80.73037
103.03468
131.50126
P/F
0.95238
0.90703
0.86384
0.82270
0.78353
0.74622
0.71068
0.67684
0.64461
0.61391
0.58468
0.55684
0.53032
0.50507
0.48102
0.45811
0.43630
0.41552
0.39573
0.37689
0.29530
0.23138
0.18129
0.14205
0.11130
0.08720
0.06833
0.05354
0.04195
0.03287
0.02575
0.02018
0.01581
0.01239
0.00971
0.00760
F/A
1.0000
2.0500
3.1525
4.3101
5.5256
6.8019
8.1420
9.5491
11.0266
12.5779
14.2068
15.9171
17.7130
19.5986
21.5786
23.6575
25.8404
28.1324
30.5390
33.0660
47.7271
66.4388
90.3203
120.7998
159.7002
209.3480
272.7126
353.5837
456.7980
588.5285
756.6537
971.2288
1245.0871
1594.6073
2040.6935
2610.0252
5%
A/F
1.00000
0.48780
0.31721
0.23201
0.18097
0.14702
0.12282
0.10472
0.09069
0.07950
0.07039
0.06283
0.05646
0.05102
0.04634
0.04227
0.03870
0.03555
0.03275
0.03024
0.02095
0.01505
0.01107
0.00828
0.00626
0.00478
0.00367
0.00283
0.00219
0.00170
0.00132
0.00103
0.00080
0.00063
0.00049
0.00038
P/A
0.9524
1.8594
2.7232
3.5460
4.3295
5.0757
5.7864
6.4632
7.1078
7.7217
8.3064
8.8633
9.3936
9.8986
10.3797
10.8378
11.2741
11.6896
12.0853
12.4622
14.0939
15.3725
16.3742
17.1591
17.7741
18.2559
18.6335
18.9293
19.1611
19.3427
19.4850
19.5965
19.6838
19.7523
19.8059
19.8479
A/P
1.05000
0.53780
0.36721
0.28201
0.23097
0.19702
0.17282
0.15472
0.14069
0.12950
0.12039
0.11283
0.10646
0.10102
0.09634
0.09227
0.08870
0.08555
0.08275
0.08024
0.07095
0.06505
0.06107
0.05828
0.05626
0.05478
0.05367
0.05283
0.05219
0.05170
0.05132
0.05103
0.05080
0.05063
0.05049
0.05038
A/G
0.00000
0.48780
0.96749
1.43905
1.90252
2.35790
2.80523
3.24451
3.67579
4.09909
4.51444
4.92190
5.32150
5.71329
6.09731
6.47363
6.84229
7.20336
7.55690
7.90297
9.52377
10.96914
12.24980
13.37747
14.36444
15.22326
15.96645
16.60618
17.15410
17.62119
18.01759
18.35260
18.63463
18.87120
19.06894
19.23372
COMPOUND INTEREST FACTORS
i=
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
F/P
1.06000
1.12360
1.19102
1.26248
1.33823
1.41852
1.50363
1.59385
1.68948
1.79085
1.89830
2.01220
2.13293
2.26090
2.39656
2.54035
2.69277
2.85434
3.02560
3.20714
4.29187
5.74349
7.68609
10.28572
13.76461
18.42015
24.65032
32.98769
44.14497
59.07593
79.05692
105.79599
141.57890
189.46451
253.54625
339.30208
P/F
0.94340
0.89000
0.83962
0.79209
0.74726
0.70496
0.66506
0.62741
0.59190
0.55839
0.52679
0.49697
0.46884
0.44230
0.41727
0.39365
0.37136
0.35034
0.33051
0.31180
0.23300
0.17411
0.13011
0.09722
0.07265
0.05429
0.04057
0.03031
0.02265
0.01693
0.01265
0.00945
0.00706
0.00528
0.00394
0.00295
F/A
1.0000
2.0600
3.1836
4.3746
5.6371
6.9753
8.3938
9.8975
11.4913
13.1808
14.9716
16.8699
18.8821
21.0151
23.2760
25.6725
28.2129
30.9057
33.7600
36.7856
54.8645
79.0582
111.4348
154.7620
212.7435
290.3359
394.1720
533.1282
719.0829
967.9322
1300.9487
1746.5999
2342.9817
3141.0752
4209.1042
5638.3681
6%
A/F
1.00000
0.48544
0.31411
0.22859
0.17740
0.14336
0.11914
0.10104
0.08702
0.07587
0.06679
0.05928
0.05296
0.04758
0.04296
0.03895
0.03544
0.03236
0.02962
0.02718
0.01823
0.01265
0.00897
0.00646
0.00470
0.00344
0.00254
0.00188
0.00139
0.00103
0.00077
0.00057
0.00043
0.00032
0.00024
0.00018
P/A
0.9434
1.8334
2.6730
3.4651
4.2124
4.9173
5.5824
6.2098
6.8017
7.3601
7.8869
8.3838
8.8527
9.2950
9.7122
10.1059
10.4773
10.8276
11.1581
11.4699
12.7834
13.7648
14.4982
15.0463
15.4558
15.7619
15.9905
16.1614
16.2891
16.3845
16.4558
16.5091
16.5489
16.5787
16.6009
16.6175
A/P
1.06000
0.54544
0.37411
0.28859
0.23740
0.20336
0.17914
0.16104
0.14702
0.13587
0.12679
0.11928
0.11296
0.10758
0.10296
0.09895
0.09544
0.09236
0.08962
0.08718
0.07823
0.07265
0.06897
0.06646
0.06470
0.06344
0.06254
0.06188
0.06139
0.06103
0.06077
0.06057
0.06043
0.06032
0.06024
0.06018
A/G
0.00000
0.48544
0.96118
1.42723
1.88363
2.33040
2.76758
3.19521
3.61333
4.02201
4.42129
4.81126
5.19198
5.56352
5.92598
6.27943
6.62397
6.95970
7.28673
7.60515
9.07220
10.34221
11.43192
12.35898
13.14129
13.79643
14.34112
14.79095
15.16012
15.46135
15.70583
15.90328
16.06202
16.18912
16.29050
16.37107
195
196
APPENDIX A
i=
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
F/P
1.07000
1.14490
1.22504
1.31080
1.40255
1.50073
1.60578
1.71819
1.83846
1.96715
2.10485
2.25219
2.40985
2.57853
2.75903
2.95216
3.15882
3.37993
3.61653
3.86968
5.42743
7.61226
10.67658
14.97446
21.00245
29.45703
41.31500
57.94643
81.27286
113.98939
159.87602
224.23439
314.50033
441.10298
618.66975
867.71633
P/F
0.93458
0.87344
0.81630
0.76290
0.71299
0.66634
0.62275
0.58201
0.54393
0.50835
0.47509
0.44401
0.41496
0.38782
0.36245
0.33873
0.31657
0.29586
0.27651
0.25842
0.18425
0.13137
0.09366
0.06678
0.04761
0.03395
0.02420
0.01726
0.01230
0.00877
0.00625
0.00446
0.00318
0.00227
0.00162
0.00115
F/A
1.0000
2.0700
3.2149
4.4399
5.7507
7.1533
8.6540
10.2598
11.9780
13.8164
15.7836
17.8885
20.1406
22.5505
25.1290
27.8881
30.8402
33.9990
37.3790
40.9955
63.2490
94.4608
138.2369
199.6351
285.7493
406.5289
575.9286
813.5204
1146.7552
1614.1342
2269.6574
3189.0627
4478.5761
6287.1854
8823.8535
12381.6618
7%
A/F
1.00000
0.48309
0.31105
0.22523
0.17389
0.13980
0.11555
0.09747
0.08349
0.07238
0.06336
0.05590
0.04965
0.04434
0.03979
0.03586
0.03243
0.02941
0.02675
0.02439
0.01581
0.01059
0.00723
0.00501
0.00350
0.00246
0.00174
0.00123
0.00087
0.00062
0.00044
0.00031
0.00022
0.00016
0.00011
0.00008
P/A
0.9346
1.8080
2.6243
3.3872
4.1002
4.7665
5.3893
5.9713
6.5152
7.0236
7.4987
7.9427
8.3577
8.7455
9.1079
9.4466
9.7632
10.0591
10.3356
10.5940
11.6536
12.4090
12.9477
13.3317
13.6055
13.8007
13.9399
14.0392
14.1099
14.1604
14.1964
14.2220
14.2403
14.2533
14.2626
14.2693
A/P
1.07000
0.55309
0.38105
0.29523
0.24389
0.20980
0.18555
0.16747
0.15349
0.14238
0.13336
0.12590
0.11965
0.11434
0.10979
0.10586
0.10243
0.09941
0.09675
0.09439
0.08581
0.08059
0.07723
0.07501
0.07350
0.07246
0.07174
0.07123
0.07087
0.07062
0.07044
0.07031
0.07022
0.07016
0.07011
0.07008
A/G
0.00000
0.48309
0.95493
1.41554
1.86495
2.30322
2.73039
3.14654
3.55174
3.94607
4.32963
4.70252
5.06484
5.41673
5.75829
6.08968
6.41102
6.72247
7.02418
7.31631
8.63910
9.74868
10.66873
11.42335
12.03599
12.52868
12.92146
13.23209
13.47598
13.66619
13.81365
13.92735
14.01458
14.08122
14.13191
14.17034
COMPOUND INTEREST FACTORS
i=
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
F/P
1.08000
1.16640
1.25971
1.36049
1.46933
1.58687
1.71382
1.85093
1.99900
2.15892
2.33164
2.51817
2.71962
2.93719
3.17217
3.42594
3.70002
3.99602
4.31570
4.66096
6.84848
10.06266
14.78534
21.72452
31.92045
46.90161
68.91386
101.25706
148.77985
218.60641
321.20453
471.95483
693.45649
1018.91509
1497.12055
2199.76126
P/F
0.92593
0.85734
0.79383
0.73503
0.68058
0.63017
0.58349
0.54027
0.50025
0.46319
0.42888
0.39711
0.36770
0.34046
0.31524
0.29189
0.27027
0.25025
0.23171
0.21455
0.14602
0.09938
0.06763
0.04603
0.03133
0.02132
0.01451
0.00988
0.00672
0.00457
0.00311
0.00212
0.00144
0.00098
0.00067
0.00045
F/A
1.0000
2.0800
3.2464
4.5061
5.8666
7.3359
8.9228
10.6366
12.4876
14.4866
16.6455
18.9771
21.4953
24.2149
27.1521
30.3243
33.7502
37.4502
41.4463
45.7620
73.1059
113.2832
172.3168
259.0565
386.5056
573.7702
848.9232
1253.2133
1847.2481
2720.0801
4002.5566
5886.9354
8655.7061
12723.9386
18701.5069
27484.5157
8%
A/F
1.00000
0.48077
0.30803
0.22192
0.17046
0.13632
0.11207
0.09401
0.08008
0.06903
0.06008
0.05270
0.04652
0.04130
0.03683
0.03298
0.02963
0.02670
0.02413
0.02185
0.01368
0.00883
0.00580
0.00386
0.00259
0.00174
0.00118
0.00080
0.00054
0.00037
0.00025
0.00017
0.00012
0.00008
0.00005
0.00004
P/A
0.9259
1.7833
2.5771
3.3121
3.9927
4.6229
5.2064
5.7466
6.2469
6.7101
7.1390
7.5361
7.9038
8.2442
8.5595
8.8514
9.1216
9.3719
9.6036
9.8181
10.6748
11.2578
11.6546
11.9246
12.1084
12.2335
12.3186
12.3766
12.4160
12.4428
12.4611
12.4735
12.4820
12.4877
12.4917
12.4943
A/P
1.08000
0.56077
0.38803
0.30192
0.25046
0.21632
0.19207
0.17401
0.16008
0.14903
0.14008
0.13270
0.12652
0.12130
0.11683
0.11298
0.10963
0.10670
0.10413
0.10185
0.09368
0.08883
0.08580
0.08386
0.08259
0.08174
0.08118
0.08080
0.08054
0.08037
0.08025
0.08017
0.08012
0.08008
0.08005
0.08004
A/G
0.00000
0.48077
0.94874
1.40396
1.84647
2.27635
2.69366
3.09852
3.49103
3.87131
4.23950
4.59575
4.94021
5.27305
5.59446
5.90463
6.20375
6.49203
6.76969
7.03695
8.22538
9.18971
9.96107
10.56992
11.04465
11.41071
11.69015
11.90154
12.06016
12.17832
12.26577
12.33013
12.37725
12.41158
12.43650
12.45452
197
198
APPENDIX A
i=
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
F/P
1.09000
1.18810
1.29503
1.41158
1.53862
1.67710
1.82804
1.99256
2.17189
2.36736
2.58043
2.81266
3.06580
3.34173
3.64248
3.97031
4.32763
4.71712
5.14166
5.60441
8.62308
13.26768
20.41397
31.40942
48.32729
74.35752
114.40826
176.03129
270.84596
416.73009
641.19089
986.55167
1517.93203
2335.52658
3593.49715
5529.04079
P/F
0.91743
0.84168
0.77218
0.70843
0.64993
0.59627
0.54703
0.50187
0.46043
0.42241
0.38753
0.35553
0.32618
0.29925
0.27454
0.25187
0.23107
0.21199
0.19449
0.17843
0.11597
0.07537
0.04899
0.03184
0.02069
0.01345
0.00874
0.00568
0.00369
0.00240
0.00156
0.00101
0.00066
0.00043
0.00028
0.00018
F/A
1.0000
2.0900
3.2781
4.5731
5.9847
7.5233
9.2004
11.0285
13.0210
15.1929
17.5603
20.1407
22.9534
26.0192
29.3609
33.0034
36.9737
41.3013
46.0185
51.1601
84.7009
136.3075
215.7108
337.8824
525.8587
815.0836
1260.0918
1944.7921
2998.2885
4619.2232
7113.2321
10950.5741
16854.8003
25939.1842
39916.6350
61422.6755
9%
A/F
1.00000
0.47847
0.30505
0.21867
0.16709
0.13292
0.10869
0.09067
0.07680
0.06582
0.05695
0.04965
0.04357
0.03843
0.03406
0.03030
0.02705
0.02421
0.02173
0.01955
0.01181
0.00734
0.00464
0.00296
0.00190
0.00123
0.00079
0.00051
0.00033
0.00022
0.00014
0.00009
0.00006
0.00004
0.00003
0.00002
P/A
0.9174
1.7591
2.5313
3.2397
3.8897
4.4859
5.0330
5.5348
5.9952
6.4177
6.8052
7.1607
7.4869
7.7862
8.0607
8.3126
8.5436
8.7556
8.9501
9.1285
9.8226
10.2737
10.5668
10.7574
10.8812
10.9617
11.0140
11.0480
11.0701
11.0844
11.0938
11.0998
11.1038
11.1064
11.1080
11.1091
A/P
1.09000
0.56847
0.39505
0.30867
0.25709
0.22292
0.19869
0.18067
0.16680
0.15582
0.14695
0.13965
0.13357
0.12843
0.12406
0.12030
0.11705
0.11421
0.11173
0.10955
0.10181
0.09734
0.09464
0.09296
0.09190
0.09123
0.09079
0.09051
0.09033
0.09022
0.09014
0.09009
0.09006
0.09004
0.09003
0.09002
A/G
0.00000
0.47847
0.94262
1.39250
1.82820
2.24979
2.65740
3.05117
3.43123
3.79777
4.15096
4.49102
4.81816
5.13262
5.43463
5.72446
6.00238
6.26865
6.52358
6.76745
7.83160
8.66566
9.30829
9.79573
10.16029
10.42952
10.62614
10.76832
10.87023
10.94273
10.99396
11.02994
11.05508
11.07256
11.08467
11.09302
COMPOUND INTEREST FACTORS
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
F/P
1.10000
1.21000
1.33100
1.46410
1.61051
1.77156
1.94872
2.14359
2.35795
2.59374
2.85312
3.13843
3.45227
3.79750
4.17725
4.59497
5.05447
5.55992
6.11591
6.72750
10.8347
17.4494
28.1024
45.2593
72.8905
117.391
189.059
304.482
490.371
789.747
P/F
0.90909
0.82645
0.75131
0.68301
0.62092
0.56447
0.51316
0.46651
0.42410
0.38554
0.35049
0.31863
0.28966
0.26333
0.23939
0.21763
0.19784
0.17986
0.16351
0.14864
0.09230
0.05731
0.03558
0.02209
0.01372
0.00852
0.00529
0.00328
0.00204
0.00127
i=
10%
F/A
1.0000
2.1000
3.3100
4.6410
6.1051
7.7156
9.4872
11.4359
13.5795
15.9374
18.5312
21.3843
24.5227
27.9750
31.7725
35.9497
40.5447
45.5992
51.1591
57.2750
98.3471
164.494
271.024
442.593
718.905
1163.91
1880.59
3034.82
4893.71
7887.47
A/F
1.00000
0.47619
0.30211
0.21547
0.16380
0.12961
0.10541
0.08744
0.07364
0.06275
0.05396
0.04676
0.04078
0.03575
0.03147
0.02782
0.02466
0.02193
0.01955
0.01746
0.01017
0.00608
0.00369
0.00226
0.00139
0.00086
0.00053
0.00033
0.00020
0.00013
P/A
0.9091
1.7355
2.4869
3.1699
3.7908
4.3553
4.8684
5.3349
5.7590
6.1446
6.4951
6.8137
7.1034
7.3667
7.6061
7.8237
8.0216
8.2014
8.3649
8.5136
9.0770
9.4269
9.6442
9.7791
9.8628
9.9148
9.9471
9.9672
9.9796
9.9873
A/P
1.10000
0.57619
0.40211
0.31547
0.26380
0.22961
0.20541
0.18744
0.17364
0.16275
0.15396
0.14676
0.14078
0.13575
0.13147
0.12782
0.12466
0.12193
0.11955
0.11746
0.11017
0.10608
0.10369
0.10226
0.10139
0.10086
0.10053
0.10033
0.10020
0.10013
A/G
0.00000
0.47619
0.93656
1.38117
1.81013
2.22356
2.62162
3.00448
3.37235
3.72546
4.06405
4.38840
4.69879
4.99553
5.27893
5.54934
5.80710
6.05256
6.28610
6.50808
7.45798
8.17623
8.70860
9.09623
9.37405
9.57041
9.70754
9.80229
9.86718
9.91125
199
200
APPENDIX A
i=
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
F/P
1.12000
1.25440
1.40493
1.57352
1.76234
1.97382
2.21068
2.47596
2.77308
3.10585
3.47855
3.89598
4.36349
4.88711
5.47357
6.13039
6.86604
7.68997
8.61276
9.64629
17.0001
29.9599
52.7996
93.0510
163.988
289.002
509.321
897.597
1581.87
2787.80
P/F
0.89286
0.79719
0.71178
0.63552
0.56743
0.50663
0.45235
0.40388
0.36061
0.32197
0.28748
0.25668
0.22917
0.20462
0.18270
0.16312
0.14564
0.13004
0.11611
0.10367
0.05882
0.03338
0.01894
0.01075
0.00610
0.00346
0.00196
0.00111
0.00063
0.00036
F/A
1.0000
2.1200
3.3744
4.7793
6.3528
8.1152
10.0890
12.2997
14.7757
17.5487
20.6546
24.1331
28.0291
32.3926
37.2797
42.7533
48.8837
55.7497
63.4397
72.0524
133.334
241.333
431.663
767.091
1358.23
2400.02
4236.01
7471.64
13173.9
23223.3
12%
A/F
1.00000
0.47170
0.29635
0.20923
0.15741
0.12323
0.09912
0.08130
0.06768
0.05698
0.04842
0.04144
0.03568
0.03087
0.02682
0.02339
0.02046
0.01794
0.01576
0.01388
0.00750
0.00414
0.00232
0.00130
0.00074
0.00042
0.00024
0.00013
0.00008
0.00004
P/A
0.8929
1.6901
2.4018
3.0373
3.6048
4.1114
4.5638
4.9676
5.3282
5.6502
5.9377
6.1944
6.4235
6.6282
6.8109
6.9740
7.1196
7.2497
7.3658
7.4694
7.8431
8.0552
8.1755
8.2438
8.2825
8.3045
8.3170
8.3240
8.3281
8.3303
A/P
1.12000
0.59170
0.41635
0.32923
0.27741
0.24323
0.21912
0.20130
0.18768
0.17698
0.16842
0.16144
0.15568
0.15087
0.14682
0.14339
0.14046
0.13794
0.13576
0.13388
0.12750
0.12414
0.12232
0.12130
0.12074
0.12042
0.12024
0.12013
0.12008
0.12004
A/G
0.00000
0.47170
0.92461
1.35885
1.77459
2.17205
2.55147
2.91314
3.25742
3.58465
3.89525
4.18965
4.46830
4.73169
4.98030
5.21466
5.43530
5.64274
5.83752
6.02020
6.77084
7.29742
7.65765
7.89879
8.05724
8.15972
8.22513
8.26641
8.29222
8.30821
COMPOUND INTEREST FACTORS
i=
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
F/P
1.15000
1.32250
1.52088
1.74901
2.01136
2.31306
2.66002
3.05902
3.51788
4.04556
4.65239
5.35025
6.15279
7.07571
8.13706
9.35762
10.7613
12.3755
14.2318
16.3665
32.9190
66.2118
133.176
267.864
538.769
1083.66
2179.62
4384.00
8817.79
17735.7
P/F
0.86957
0.75614
0.65752
0.57175
0.49718
0.43233
0.37594
0.32690
0.28426
0.24718
0.21494
0.18691
0.16253
0.14133
0.12289
0.10686
0.09293
0.08081
0.07027
0.06110
0.03038
0.01510
0.00751
0.00373
0.00186
0.00092
0.00046
0.00023
0.00011
0.00006
F/A
1.0000
2.1500
3.4725
4.9934
6.7424
8.7537
11.0668
13.7268
16.7858
20.3037
24.3493
29.0017
34.3519
40.5047
47.5804
55.7175
65.0751
75.8364
88.2118
102.444
212.793
434.745
881.170
1779.09
3585.13
7217.72
14524.1
29220.0
58778.6
118231
15%
A/F
1.00000
0.46512
0.28798
0.20027
0.14832
0.11424
0.09036
0.07285
0.05957
0.04925
0.04107
0.03448
0.02911
0.02469
0.02102
0.01795
0.01537
0.01319
0.01134
0.00976
0.00470
0.00230
0.00113
0.00056
0.00028
0.00014
0.00007
0.00003
0.00002
0.00001
P/A
0.8696
1.6257
2.2832
2.8550
3.3522
3.7845
4.1604
4.4873
4.7716
5.0188
5.2337
5.4206
5.5831
5.7245
5.8474
5.9542
6.0472
6.1280
6.1982
6.2593
6.4641
6.5660
6.6166
6.6418
6.6543
6.6605
6.6636
6.6651
6.6659
6.6663
A/P
1.15000
0.61512
0.43798
0.35027
0.29832
0.26424
0.24036
0.22285
0.20957
0.19925
0.19107
0.18448
0.17911
0.17469
0.17102
0.16795
0.16537
0.16319
0.16134
0.15976
0.15470
0.15230
0.15113
0.15056
0.15028
0.15014
0.15007
0.15003
0.15002
0.15001
A/G
0.00000
0.46512
0.90713
1.32626
1.72281
2.09719
2.44985
2.78133
3.09223
3.38320
3.65494
3.90820
4.14376
4.36241
4.56496
4.75225
4.92509
5.08431
5.23073
5.36514
5.88343
6.20663
6.40187
6.51678
6.58299
6.62048
6.64142
6.65298
6.65929
6.66272
201
202
APPENDIX A
i=
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
F/P
1.20000
1.44000
1.72800
2.07360
2.48832
2.98598
3.58318
4.29982
5.15978
6.19174
7.43008
8.91610
10.6993
12.8392
15.4070
18.4884
22.1861
26.6233
31.9480
38.3376
95.3962
237.376
590.668
1469.77
3657.26
9100.44
22644.8
56347.5
P/F
0.83333
0.69444
0.57870
0.48225
0.40188
0.33490
0.27908
0.23257
0.19381
0.16151
0.13459
0.11216
0.09346
0.07789
0.06491
0.05409
0.04507
0.03756
0.03130
0.02608
0.01048
0.00421
0.00169
0.00068
0.00027
0.00011
0.00004
0.00002
F/A
1.0000
2.2000
3.6400
5.3680
7.4416
9.9299
12.9159
16.4991
20.7989
25.9587
32.1504
39.5805
48.4966
59.1959
72.0351
87.4421
105.931
128.117
154.740
186.688
471.981
1181.88
2948.34
7343.86
18281.3
45497.2
113219
281733
20%
A/F
1.00000
0.45455
0.27473
0.18629
0.13438
0.10071
0.07742
0.06061
0.04808
0.03852
0.03110
0.02526
0.02062
0.01689
0.01388
0.01144
0.00944
0.00781
0.00646
0.00536
0.00212
0.00085
0.00034
0.00014
0.00005
0.00002
0.00001
0.00000
P/A
0.8333
1.5278
2.1065
2.5887
2.9906
3.3255
3.6046
3.8372
4.0310
4.1925
4.3271
4.4392
4.5327
4.6106
4.6755
4.7296
4.7746
4.8122
4.8435
4.8696
4.9476
4.9789
4.9915
4.9966
4.9986
4.9995
4.9998
4.9999
A/P
1.20000
0.65455
0.47473
0.38629
0.33438
0.30071
0.27742
0.26061
0.24808
0.23852
0.23110
0.22526
0.22062
0.21689
0.21388
0.21144
0.20944
0.20781
0.20646
0.20536
0.20212
0.20085
0.20034
0.20014
0.20005
0.20002
0.20001
0.20000
A/G
0.00000
0.45455
0.87912
1.27422
1.64051
1.97883
2.29016
2.57562
2.83642
3.07386
3.28929
3.48410
3.65970
3.81749
3.95884
4.08511
4.19759
4.29752
4.38607
4.46435
4.73516
4.87308
4.94064
4.97277
4.98769
4.99451
4.99757
4.99894
COMPOUND INTEREST FACTORS
i=
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
F/P
1.25000
1.56250
1.95313
2.44141
3.05176
3.81470
4.76837
5.96046
7.45058
9.31323
11.6415
14.5519
18.1899
22.7374
28.4217
35.5271
44.4089
55.5112
69.3889
86.7362
264.698
807.794
2465.19
7523.16
22958.87
70064.92
213821.2
652530.4
P/F
0.80000
0.64000
0.51200
0.40960
0.32768
0.26214
0.20972
0.16777
0.13422
0.10737
0.08590
0.06872
0.05498
0.04398
0.03518
0.02815
0.02252
0.01801
0.01441
0.01153
0.00378
0.00124
0.00041
0.00013
0.00004
0.00001
0.00000
0.00000
F/A
1.0000
2.2500
3.8125
5.7656
8.2070
11.2588
15.0735
19.8419
25.8023
33.2529
42.5661
54.2077
68.7596
86.9495
109.687
138.109
173.636
218.045
273.556
342.945
1054.79
3227.17
9856.76
30088.7
91831.5
280256
855281
2610118
25%
A/F
1.00000
0.44444
0.26230
0.17344
0.12185
0.08882
0.06634
0.05040
0.03876
0.03007
0.02349
0.01845
0.01454
0.01150
0.00912
0.00724
0.00576
0.00459
0.00366
0.00292
0.00095
0.00031
0.00010
0.00003
0.00001
0.00000
0.00000
0.00000
P/A
0.8000
1.4400
1.9520
2.3616
2.6893
2.9514
3.1611
3.3289
3.4631
3.5705
3.6564
3.7251
3.7801
3.8241
3.8593
3.8874
3.9099
3.9279
3.9424
3.9539
3.9849
3.9950
3.9984
3.9995
3.9998
3.9999
4.0000
4.0000
A/P
1.25000
0.69444
0.51230
0.42344
0.37185
0.33882
0.31634
0.30040
0.28876
0.28007
0.27349
0.26845
0.26454
0.26150
0.25912
0.25724
0.25576
0.25459
0.25366
0.25292
0.25095
0.25031
0.25010
0.25003
0.25001
0.25000
0.25000
0.25000
A/G
0.00000
0.44444
0.85246
1.22493
1.56307
1.86833
2.14243
2.38725
2.60478
2.79710
2.96631
3.11452
3.24374
3.35595
3.45299
3.53660
3.60838
3.66979
3.72218
3.76673
3.90519
3.96282
3.98580
3.99468
3.99804
3.99929
3.99974
3.99991
203
205
Authors’ Biographies
DAVID L. WHITMAN
David L. Whitman, P.E., Ph.D. received a B.S. degree (1975) in Electrical Engineering from the
University of Wyoming (UW). He also received a Ph.D. degree (1978) in Mineral Engineering
from the University of Wyoming. He worked in the synthetic fuels arena prior to becoming a faculty
member in Petroleum Engineering at the University of Wyoming in 1981. From 1989 to 2005,
he was the Associate Dean of Academics and since 2005 has been a professor of Electrical and
Computer Engineering. He received UW’s Ellbogen Outstanding Teacher Award in 1985, UW’s
College of Engineering Outstanding Undergraduate Teaching Award in 1990 and 2004 and the
ASEE Rocky Mountain Section Outstanding Teaching Award in 2001. He is a Past President
of the National Council of Examiners for Engineers and Surveyors (NCEES), chairman of the
IEEE-USA Licensure & Registration Committee, and an active member of ASEE.
RONALD E. TERRY
Ronald E. Terry, Ph.D. received a B.S. in Chemical Engineering from Oregon State University
(1971) and a Ph.D. from Brigham Young University (BYU) (1976). He worked for Phillips Petroleum
Company after graduate school and began his academic career in 1977 at the University of Kansas
in the Chemical and Petroleum Engineering Department. He taught in the Petroleum Engineering
Department at the University of Wyoming (1981-1987) and at BYU in the Chemical Engineering
Department (1987-2007) and in the Technology and Engineering Education Department (2007present). He has received teaching awards at the University of Kansas, University of Wyoming, and
at Brigham Young University.
Early in his career, his scholarship efforts involved researching methods to enhance the production of oil and gas. After joining BYU, his scholarship centered on pedagogy, student learning,
and engineering ethics. He has served as acting department chair, associate dean, and in BYU’s
central administration as an Associate in the Office of Planning and Assessment for five years
(2003-2008). He is past president of the Rocky Mountain Section of the American Society for
Engineering Education.