0866_c17

Vista previa del material en texto

17
 
Engineering Economics
 
17.1 Engi
 
Decisions made 
(some say 85%) 
century must hav
examines the mo
 
Chan S. Park*
 
Auburn University
 
Donald D. Tip
 
University of Alaba
 
* Department o
17.6 based on 
 
Co
 
Company, Reading
 
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© 2005 by CRC Pre
and Project
Management
17.1 Engineering Economic Decisions
17.2 Establishing Economic Equivalence
Interest: The Cost of Money • The Elements of Transactions 
Involving Interest • Equivalence Calculations • Interest 
Formulas • Nominal and Effective Interest Rates • Loss of 
Purchasing Power
17.3 Measures of Project Worth
Describing Project Cash Flows • Present Worth Analysis • 
Annual Equivalent Method • Rate of Return Analysis • 
Accept/Reject Decision Rules • Mutually Exclusive Alternatives
17.4 Cash Flow Projections 
Operating Profit — Net Income • Accounting Depreciation • 
Corporate Income Taxes • Tax Treatment of Gains or Losses 
for Depreciable Assets • After-Tax Cash Flow Analysis • Effects 
of Inflation on Project Cash Flows
17.5 Sensitivity and Risk Analysis 
Project Risk • Sensitivity Analysis • Scenario Analysis • Risk 
Analysis • Procedure for Developing an NPW Distribution • 
Expected Value and Variance • Decision Rule
17.6 Design Economics
Capital Costs vs. Operating Costs • Minimum-Cost Function
17.7 Project Management
Engineers, Projects, and Project Management • Project 
Planning • Project Scheduling • Staffing and Organizing • Team 
Building • Project Control • Estimating and Contracting
neering Economic Decisions
during the engineering design phase of product development determine the majority
of the costs of manufacturing that product. Thus, a competent engineer in the 21st
e an understanding of the principles of economics as well as engineering. This chapter
st important economic concepts that should be understood by engineers.
pett
ma in Huntsville
f Industrial & Systems Engineering, Auburn University, Auburn, AL 36849. Sections 17.1 through
ntemporary Engineering Economics, 2nd edition, by Chan S. Park, Addison-Wesley Publishing
, MA, 1997.
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17
 
-2
 
Chapter 17
 
Engineers participate in a variety of decision-making processes, ranging from manufacturing to mar-
keting to financing decisions. They must make decisions involving materials, plant facilities, the in-house
capabilities of company personnel, and the effective use of capital assets such as buildings and machinery.
One of the engin
enable the firm 
 
economic decision
 
17.2 Esta
 
A typical enginee
the investment, w
analytical purpos
to result from th
In such a fixed
the future. In the
principal. This is
 
form of cash gen
with the capital e
tenance costs, an
 
the project cash fl
a money series o
cash flow series, 
 
Interest: The
 
Money left in a s
the deposits. In t
and sold, money
 
percentage that i
specified length o
use of the lender
from providing a
 
for use.
The operation
depend on lengt
year. This princip
of a sum depend
work, earning mo
at some future ti
The changes in
with large amou
annual interest r
receive $1 millio
must take into ac
of different amou
 
The Element
 
Many types of tr
machinery on cr
 
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© 2005 by CRC Pre
eer’s primary tasks is to plan for the acquisition of equipment (fixed asset) that will
to design and produce products economically. These decisions are called engineering
s.
blishing Economic Equivalence
ring economic decision involves two dissimilar types of dollar amounts. First, there is
hich is usually made in a lump sum at the beginning of the project, a time that for
es is called today, or time 0. Second, there is a stream of cash benefits that are expected
is investment over a period of future years.
 asset investment funds are committed today in the expectation of earning a return in
 case of a bank loan, the future return takes the form of interest plus repayment of the
 known as the loan cash flow. In the case of the fixed asset, the future return takes the
erated by productive use of the asset. The representation of these future earnings along
xpenditures and annual expenses (such as wages, raw materials, operating costs, main-
d income taxes) is the project cash flow. This similarity between the loan cash flow and
ow brings us an important conclusion—that is, first we need to find a way to evaluate
ccurring at different points in time. Second, if we understand how to evaluate a loan
we can use the same concept to evaluate the project cash flow series.
 Cost of Money
avings account earns interest so that the balance over time is greater than the sum of
he financial world, money itself is a commodity, and like other goods that are bought
 costs money. The cost of money is established and measured by an interest rate, a
s periodically applied and added to an amount (or varying amounts) of money over a
f time. When money is borrowed, the interest paid is the charge to the brrower for the
’s property; when money is loaned or invested, the interest earned is the lender’s gain
 good to another. Interest, then, may be defined as the cost of having money available
 of interest reflects the fact that money has a time value. This is why amounts of interest
hs of time; interest rates, for example, are typically given in terms of a percentage per
le of the time value of money can be formally defined as follows: the economic value
s on when it is received. Because money has earning power over time (it can be put to
re money for its owner), a dollar received today has a greater value than a dollar received
me.
 the value of a sum of money over time can become extremely significant when we deal
nts of money, long periods of time, or high interest rates. For example, at a current
ate of 10%, $1 million will earn $100,000 in interest in a year; thus, waiting a year to
n clearly involves a significant sacrifice. In deciding among alternative proposals, we
count the operation of interest and the time value of money to make valid comparisons
nts at various times.
s of Transactions Involving Interest
ansactions involve interest — for example, borrowing or investing money, purchasing
edit — but certain elements are common to all of them:
ss LLC
 
Engineering Economics and Project Management
 
17
 
-3
 
1. Some initial amount of money, called the 
 
principal
 
 (
 
P
 
) in transactions of debt or investment
2. The 
 
interest rate
 
 (
 
i
 
), which measures the cost or price of money, expressed as a percentage per
period of 
3. A period o
 
interest is 
4. The specifi
certain 
 
nu
 
5. A 
 
plan for
 
of time (fo
6. A 
 
future a
 
number o
 
Cash Flow Dia
 
It is convenient t
flow diagram (se
of interest period
upward arrows fo
 
End-of-Period C
 
In practice, cash 
any point in time
 
end-of-period con
 
interest period. T
within an interes
 
Compound Int
 
Under the comp
the end of the pr
interest that has 
 
FIGURE 17.2.1
 
A 
equal annual instal
is $19,800. The bo
 
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© 2005 by CRC Pre
time
f time, called the interest period (or compounding period), that determines how frequently
calculated
ed length of time that marks the duration of the transaction and thereby establishes a
mber of interest periods (N)
 receipts or disbursements (An) that yields a particular cash flow pattern over the length
r example, we might have a series of equal monthly payments [A] that repay a loan)
mount of money (F) that resultsfrom the cumulative effects of the interest rate over a
f interest periods
grams
o represent problems involving the time value of money in graphic form with a cash
e Figure 17.2.1), which represents time by a horizontal line marked off with the number
s specified. The cash flows over time are represented by arrows at the relevant periods:
r positive flows (receipts) and downward arrows for negative flows (disbursements).
onvention
flows can occur at the beginning or in the middle of an interest period, or at practically
. One of the simplifying assumptions we make in engineering economic analysis is the
vention, which is the practice of placing all cash flow transactions at the end of an
his assumption relieves us of the responsibility of dealing with the effects of interest
t period, which would greatly complicate our calculations.
erest
ound interest scheme, the interest in each period is based on the total amount owed at
evious period. This total amount includes the original principal plus the accumulated
been left in the account. In this case, you are in effect increasing the deposit amount by
cash flow diagram for a loan transaction — borrow $20,000 now and pay off the loan with five
lments of $5,141.85. After paying $200 for the loan origination fee, the net amount of financing
rrowing interest rate is 9%.
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17
 
-4
 
Chapter 17
 
the amount of interest earned. In general, if you deposited (invested) P dollars at interest rate i, you
would have P + iP = P(1 + i) dollars at the end of one period. With the entire amount (principal and
interest) reinveste
This interest-earn
grow to
 
Equivalence 
 
Economic equiva
payments — can
 
sequence of cash 
Equivalence c
developed in Equ
 
expresses the equ
rate, i, and a num
$1000 will grow 
Thus at 8% in
trade $1000 now
application of th
 
Example 17.2.1
 
Suppose you are 
There is no quest
you would depos
indifferent in you
now?
 
Discussion
 
Our job is to det
the investment po
your option of u
you refers to eco
rate, you could t
 
Solution
 
From Equation (
Rearranging to s
 
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© 2005 by CRC Pre
d at the same rate i for another period, you would have, at the end of the second period,
ing process repeats, and after N periods, the total accumulated value (balance) F will
(17.2.1)
Calculations
lence refers to the fact that a cash flow — whether it is a single payment or a series of
 be said to be converted to an equivalent cash flow at any point in time; thus, for any
flows, we can find an equivalent single cash flow at a given interest rate and a given time.
alculations can be viewed as an application of the compound interest relationships
ation 17.2.1. The formula developed for calculating compound interest, F = P(1 + i)N,
ivalence between some present amount, P, and a future amount, F, for a given interest
ber of interest periods, N. Therefore, at the end of a 3-year investment period at 8%,
to
terest, $1000 received now is equivalent to $1,25l9.71 received in 3 years and we could
 for the promise of receiving $1259.71 in 3 years. Example 17.2.1 demonstrates the
is basic technique.
 — Equivalence
offered the alternative of receiving either $3000 at the end of 5 years or P dollars today.
ion that the $3000 will be paid in full (no risk). Having no current need for the money,
it the P dollars in an account that pays 8% interest. What value of P would make you
r choice between P dollars today and the promise of $3000 at the end of 5 years from
ermine the present amount that is economically equivalent to $3000 in 5 years, given
tential of 8% per year. Note that the problem statement assumes that you would exercise
sing the earning power of your money by depositing it. The “indifference” ascribed to
nomic indifference; that is, within a marketplace where 8% is the applicable interest
rade one cash flow for the other.
17.2.1), we establish
olve for P,
P i i P i P i i
P i
1 1 1 1
1 2
+( ) + +( )[ ] = +( ) +( )
= +( )
F P i N= +( )1
$ . $ .1000 1 0 08 1259 713+( ) =
$ .3000 1 0 08 5= +( )P
P = +( ) =$ . $3000 1 0 08 20425
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Engineering Economics and Project Management
 
17
 
-5
 
Comments
 
In this example, it is clear that if P is anything less than $2042, you would prefer the promise of $3000
in 5 years to P d
a lower interest r
P = $2466.
 
Interest Form
 
In this section is
flows. It classifies
and presents wor
 
Single Cash Fl
 
We begin our co
Given a present s
at the end of the
encountered in d
Because of its 
given P, i, and N”
 
this factor is one
important intere
process. (Note th
preceding examp
 
evaluate the equa
1.2597 in the F/P
Finding prese
 
discounting proce
 
sum F, we simply
The factor 1/(
 
Tables
 
*
 
 have been
the P/F factor ar
 
interest tables is 
the formulas der
 
A Stream of C
 
A common cash 
as car loans and in
bills typically inv
payment amoun
 
* All standard 
 
Addison Wesley, 19
Wide Web site at h
 
neering Economics.
 
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© 2005 by CRC Pre
ollars today; if P were greater than $2042, you would prefer P. It is less obvious that at
ate, P must be higher to be equivalent to the future amount. For example, at i = 4%,
ulas
 developed a series of interest formulas for use in more complex comparisons of cash
 four major categories of cash flow transactions, develops interest formulas for them,
king examples of each type.
ow Formulas
verage of interest formulas by considering the simplest cash flows: single cash flows.
um P invested for N interest periods at interest rate i, what sum will have accumulated
 N periods? You probably noticed quickly that this description matches the case we first
escribing compound interest. To solve for F (the future sum) we use Equation (17.2.1):
origin in compound interest calculation, the factor (F/P, i, N), which is read as “find F,
 is known as the single payment compound amount factor. Like the concept of equivalence,
 of the foundations of engineering economic analysis. Given this factor, all the other
st formulas can be derived. This process of finding F is often called the compounding
e time-scale convention. The first period begins at n = 0 and ends at n = 1.) Thus, in the
le, where we had F = $1000(1.08)3, we can write F = $1000(F/P, 8%, 3). We can directly
tion or locate the factor value by using the 8% interest table* and finding the factor of
 column for N = 3.
nt worth of a future sum is simply the reverse of compounding and is known as
ss. In Equation (17.2.1), we can see that if we were to find a present sum P, given a future
 solve for P.
(17.2.2)
1 + i)N is known as the single payment present worth factor and is designated (P/F, i, N).
 constructed for the P/F factors for various values of i and N. The interest rate i and
e also referred to as discount rate and discounting factor, respectively. Because using the
often the easiest way to solve an equation, this factor notation is included for each of
ived in the following sections.
ash Flow Series
flow transaction involves a series of disbursements or receipts. Familiar situations such
surance payments are examples of series payments. Payments of car loans and insurance
olve identical sums paid at regular intervals. However, when there is no clear pattern of
ts over a series, one calls the transaction an uneven cash-flow series.
engineering economy textbooks (such as Contemporary Engineering Economics by C. S. Park,
97) provide extensive sets of interest tables. Or you can obtain such interest tables on a World
ttp://www.eng.auburn.edu/-park/cee.html,which is a textbook web site for Contemporary Engi-
F P i P F P i NN= +( ) = ( )1 , ,
P F
i
F P F i NN= +( )




= ( )1
1
, ,
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17
 
-6
 
Chapter 17
 
The present worth of any stream of payments can be found by calculating the present value of each
individual payment and summing the results. Once the present worth is found, one can make other
equivalence calcu
the previous sect
 
Example 17.2.2
into Single Pay
 
Wilson Technolo
4 years in automa
and they wish to
Year 1: $25,00
Year 2: $3000 
Year 3: No ex
Year 4: $5000
How much mon
 
Discussion
 
This problem is e
P dollars today a
uneven series of 
the present value
17.2.2.
 
FIGURE 17.2.2
 
D
sition allows us to 
 
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© 2005 by CRC Pre
lations, such as calculating the future worth by using the interest factors developed in
ion.
 — Present Value of an Uneven Series by Decomposition 
ments
gy, a growing machine shop, wishes to set aside money now to invest over the next
ting their customer service department. They now earn 10% on a lump sum deposited,
 withdraw the money in the following increments:
0 to purchase a computer and data base software designed for customer service use
to purchase additional hardware to accommodate anticipated growth in use of the system
penses
 to purchase software upgrades
ey must be deposited now to cover the anticipated payments over the next 4 years?
quivalent to asking what value of P would make you indifferent in your choice between
nd the future expense stream of ($25,000, $3000, $0, $5000). One way to deal with an
cash flows is to calculate the equivalent present value of each single cash flow and sum
s to find P. In other words, the cash flow is broken into three parts as shown in Figure
ecomposition of uneven cash flow series into three single-payment transactions. This decompo-
use the single-payment present worth factor.
2
0
P4
1 3 4 
Years
$5,000
P2
P1 
1
0 0
1 2
$3,000
$25,000
P = P1 + P2 + P4
1 2 3 4
Years
$5,000
$3,000
0
$25,000
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Engineering Economics and Project Management
 
17
 
-7
 
Solution
 
Cash Flow Ser
 
Whenever one ca
expressions for c
classify cash flow
and (3) geometr
the following no
1. Uniform S
equal cash
 
17.2.3a). T
which arr
formulas d
in the seri
2. Linear Gra
always un
when each
5-year loa
by $50 ea
 
diagram p
formulas u
3. Geometric
is determi
 
For examp
be budget
series sugg
 
such serie
Table 17.2.1 sum
For example, the
equivalent lump-
rate i. Note that 
the same as the p
The next two exa
cash flow.
 
Example 17.2.3
 
Suppose you ma
10 years. If your 
10 years? (See Fi
 
Solution
 
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© 2005 by CRC Pre
ies with a Special Pattern
n identify patterns in cash flow transactions, one may use them in developing concise
omputing either the present or future worth of the series. For this purpose, we will
 transactions into three categories: (1) equal cash flow series, (2) linear gradient series,
ic gradient series. To simplify the description of various interest formulas, we will use
tation:
eries: Probably the most familiar category includes transactions arranged as a series of
 flows at regular intervals, known as an equal-payment series (or uniform series) (Figure
his describes the cash flows, for example, of the common installment loan contract,
anges for the repayment of a loan in equal periodic installments. The equal cash flow
eal with the equivalence relations of P, F, and A, the constant amount of the cash flows
es.
dient Series: While many transactions involve series of cash flows, the amounts are not
iform: yet they may vary in some regular way. One common pattern of variation occurs
 cash flow in a series increases (or decreases) by a fixed amount (Figure 17.2.3b). A
n repayment plan might specify, for example, a series of annual payments that increased
ch year. We call such a cash flow pattern a linear gradient series because its cash flow
roduces an ascending (or descending) straight line. In addition to P, F, and A, the
sed in such problems involve the constant amount, G, of the change in each cash flow.
 Gradient Series: Another kind of gradient series is formed when the series in cash flow
ned, not by some fixed amount like $50, but by some fixed rate, expressed as a percentage.
le, in a 5-year financial plan for a project, the cost of a particular raw material might
ed to increase at a rate of 4% per year. The curving gradient in the diagram of such a
ests its name: a geometric gradient series (Figure 17.2.3c). In the formulas dealing with
s, the rate of change is represented by a lowercase g.
marizes the interest formulas and the cash flow situations in which they should be used.
 factor notation (F/A, i, N) represents the situation where you want to calculate the
sum future worth (F) for a given uniform payment series (A) over N period at interest
these interest formulas are applicable only when the interest (compounding) period is
ayment period. Also in this table we present some useful interest factor relationships.
mples illustrate how one might use these interest factors to determine the equivalent
 — Uniform Series: Find F, Given i, A, N
ke an annual contribution of $3000 to your savings account at the end of each year for
savings account earns 7% interest annually, how much can be withdrawn at the end of
gure 17.2.4.)
P P F P F P F= ( ) + ( ) + ( )
=
$ , , %, $ , %, $ , %,
$ ,
25 000 10 1 3000 10 2 5000 10 4
28 622
F F A= ( )
= ( )
=
$ , %,
$ .
$ , .
3000 7 10
3000 13 8164
41 449 20
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17
 
-8
 
Chapter 17
 
Example 17.2.4
 
Ansell Inc., a me
the machines to 
changed layouts 
delivery system t
unused old pipe 
and fraught with
70% of the time 
of electricity at a 
to operate the cu
next 5 years due
compressed air r
now, it will cost 
23% less (or 70%
Ansell’s interest r
 
FIGURE 17.2.3
 
Fiv
gradient series.
 
FIGURE 17.2.4
 
Ca
$100 $100 $100 $100 $100
 
0866_book.fm Page 8 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
 — Geometric Gradient: Find P, Given A1, g, i, N
dical device manufacturer, uses compressed air in solenoids and pressure switches in
control the various mechanical movements. Over the years the manufacturing floor has
numerous times. With each new layout more piping was added to the compressed air
o accommodate the new locations of the manufacturing machines. None of the extra,
was capped or removed; thus the current compressed air delivery system is inefficient
 leaks. Because of the leaks in the current system, the compressor is expected to run
that the plant is in operation during the upcoming year, which will require 260 kW/hr
rate of $0.05/kW-hr. (The plant runs 250 days a year for 24 hr a day.) If Ansell continues
rrent air delivery system, the compressor run time will increase by 7% per year for the
 to ever-deteriorating leaks. (After 5 years, the current system cannot meet the plant’s
equirement, so it has to be replaced.) If Ansell decides to replace all of the old piping
$28,570. The compressor will still run the same number of days; however, it will run
 (1 – 0.23) = 53.9% usage during the day) because of the reduced air pressure loss. If
ate is 12%, is it worth fixing now?
e types of cash flows: (a) equal (uniform) payment series; (b) linear gradient series; and (c) geometric
sh flow diagram (Example 17.2.3).
0 1 2 3 4 5
(a) Equal (uniform) payment
series at regular Intervals
(b) Linear gradient series
where each cash flow in a
series increases or decreases
by a fixed amount,G.
$50 + 4G
$50 + 3G
$50 + 2G
$50 + G$50
0 1 2 3 4 5
$50(1 + g)4
$50(1 + g)3
$50(1 + g)2$50(1 + g)
$50
0 1 2 3 4 5
(c) Geometric gradient series
where each cash flow in a
series increases or decreases
by a fixed rate (percentage), g.
0 1 2 3 4 5 6 7 8 9
A = $3,000
F
10 Years
l = 7%
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E
n
gin
eerin
g E
con
om
ics an
d
 P
roject M
an
agem
en
t
 
17
 -9
 
TABLE 17.2.1 Summary of Discrete Compounding Formulas with Discrete Payments
 
Flow Type Factor Notation Formula Cash Flow Diagram Factor Relationship
 
Single
 
(
 
F/A, i, N
 
) + 1
 
 – (
 
P/A, i N
 
)
 
i
 
Equal Paym
 
A/P, i, N
 
) – i
 
Gradient S
 
(
 
P/G, i, N
 
)(
 
F/P, i, N
 
)
 
(
 
P/G, i, N
 
)(
 
A/P, i, N
 
)
 
 = (P/A1, g, i, N)(F/P, i N)
Adapted factors and you can obtain
such intere ineering Economics.
( / , , )
i
P F i N−1
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 Page 9 Thursday, A
ugust 5, 2004 3:37 PM
© 2005 by CRC 
Compound amount (F/P, i, N)
Present worth (P/F, i, N)
F = P(1 + i)N
P = F(1 + i)–N
(F/P, i, N) = i
(P/F, i, N) = 1
ent Series
Compound amount (F/A, i, N)
(A/F, i, N) = (
Sinking fund (A/F, i, N)
Present worth (P/A, i, N)
Capital recovery (A/P, i, N)
eries
Uniform gradient
Present worth (P/G, i, N)
(F/G, i, N) = 
(A/G, i, N) = 
Geometric gradient
Present worth (P/A1, g, i, N)
(F/A1, g, i, N)
 from Park, C.S. 1997. Contemporary Engineering Economics. Addison-Wesley, Reading, MA. Tables are constructed for various interest
st tables on a World Wide Web site at http://www.eng.auburn.edu/~park/cee.html, which is a textbook web site for Contemporary Eng
F A
i
i
N
=
+( ) −





1 1
A F
i
i
N
=
+( ) −





1 1
P A
i
i i
N
N
=
+( ) −
+( )






1 1
1
( / , , )A P i N =
A P
i i
i
N
N
=
+( )
+( ) −






1
1 1
P C
i iN
i i
N
N
=
+( ) − −
+( )






1 1
12
P
A
g i
i g
NA
i
i g
N N
=
− +( ) +( )
−






+
=(
−
1
1
1 1 1
1
if ))



Press LLC
17-10 Chapter 17
Solution
• Step 1. Ca
The powe
• Step 2. Ea
power cos
The equiv
• Step 3. If 
cost will b
equivalent
• Step 4. Th
Since the 
FIGURE 17.2.5 Ex
(Example 17.2.4).
a geometric gradient series
g = 7%
power c
0866_book.fm Page 10 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
lculate the cost of power consumption of the current piping system during the first year.
r consumption is equal to:
ch year the annual power cost will increase at the rate of 7% over the previous year’s
t. Then the anticipated power cost over the 5-year period is summarized in Figure 17.2.5.
alent present lump-sum cost at 12% for this geometric gradient series is
Ansell replaces the current compressed air system with the new one, the annual power
e 23% less during the first year and will remain at that level over the next 5 years. The
 present lump-sum cost at 12% is
e net cost for not replacing the old system now is $71,175 (= $222,283 – $151,108).
new system costs only $28,570, the replacement should be made now.
pected power expenditure over the next 5 years due to deteriorating leaks if no repair is performed
0 Years1 2 3 4 5
$54,440 $58,251
$62,328
$66,691
$71,360
ost = % of day operating days operating per year hours per day
kW hr kW-hr
 days year hr day 260 kW hr kW-hr
× ×
× ×
= ( ) × ( ) × ( ) × ( ) × ( )
=
$
% $ .
$ ,
70 250 24 0 05
54 440
P P AOld = ( )
=
− +( ) +( )
−




=
−
$ , , %, %,
$ , . .
. .
$ ,
54 440 7 12 5
54 440 1 1 0 07 1 0 12
0 12 0 07
222 283
1
5 5
P P ANew = −( )( )
= ( )
=
$ , . , %,
$ , . .
$ ,
54 440 1 0 23 12 5
41 918 80 3 6048
151 108
ss LLC
Engineering Economics and Project Management 17-11
Nominal and Effective Interest Rates
In all our examp
year, or annually
matters and engin
payments and da
different compou
led to the develo
Nominal Inter
Even if a financia
in calculating int
example, state th
say 18% is the no
monthly (12). To
per month. Ther
unpaid balance f
Although the a
to customers, wh
precisely the amo
compounding on
Effective Annu
The effective inter
time period. For
balance at the en
monthly basis, it
balance.
Suppose you p
entire amount w
would grow to
This implies that
the principal and
In terms of an ef
the principal am
Thus, the effectiv
Clearly, compo
nominal interest
duration. As you
(transaction) per
monthly. This qu
this, we may defi
0866_book.fm Page 11 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
les in the previous section, we implicitly assumed that payments are received once a
. However, some of the most familiar financial transactions in both personal financial
eering economic analysis involve nonannual payments; for example, monthly mortgage
ily earnings on savings accounts. Thus, if we are to compare different cash flows with
nding periods, we need to address them on a common basis. The need to do this has
pment of the concepts of nominal interest rate and effective interest rate.
est Rate
l institution uses a unit of time other than a year — a month or quarter, for instance —
erest payments, it usually quotes the interest rate on an annual basis. Many banks, for
e interest arrangement for credit cards in this way: “18% compounded monthly.” We
minal interest rate or annual percentage rate (APR), and the compounding frequency is
 obtain the interest rate per compounding period, we divide 18% by 12 to obtain 1.5%
efore, the credit card statement above means that the bank will charge 1.5% interest on
or each month.
nnual percentage rate, or APR, is commonly used by financial institutions and is familiar
en compounding takes place more frequently than annually, the APR does not explain
unt of interest that will accumulate in a year. To explain the true effect of more frequent
 annual interest amounts, we need to introduce the term effective interest rate.
al Interest Rate
est rate is the only one that truly represents the interest earned in a year or some other
 instance, in our credit card example, the bank will charge 1.5% interest on unpaid
d of each month. Therefore, the 1.5% rate represents an effective interest rate — on a
 is the rate that predicts the actual interest payment on your outstanding credit card
urchase an appliance on credit at 18% compounded monthly. Unless you pay off the
ithin a grace period (let’s say, a month), any unpaid balance (P) left for a year period
 for each dollar borrowed for 1 year, you owe $1.1956 at the end of the year, including
 interest. For each dollar borrowed, you pay an equivalent annual interest of 19.56 cents.
fective annual interest rate (ia), we can rewrite the interest payment as a percentage of
ount
e annual interest rate is 19.56%.
unding more frequently increases the amount of interest paid for the year at the same
 rate. We can generalize the result to compute the effective interest rate for any time
 will see later, we normally compute the effective interest rate based on payment
iod. For example, cash flow transactions occur quarterly but interest rate is compounded
arterly conversion allows us to use the interest formulas in Table 17.2.1. To consider
ne the effective interest rate for a given payment period as
F P i
P
P
N
= +( )
= +( )
=
1
1 0 015
1 1956
12
.
.
i
a
= +( ) − =1 0 015 1 0 1956 19 5612. . , . %orss LLC
17-12 Chapter 17
(17.2.3)
where
M = th
C = th
K = th
When M = 1,
(17.2.3), we find
interest is equal t
interest, we were
Example 17.2.5
Suppose you wan
with 8.5% annua
so the net amoun
Solution
The ad does not s
periods are almo
situation, we can
Example 17.2.6
Suppose you ma
compounded mo
Solution
We follow the pr
1. Identify th
* For an extrem
(M) becomes very
infinity and r/M ap
To calculate t
As an exampl
is ia = e0.12 – 1 = 12
i r M C= +( ) −1 1
0866_book.fm Page 12 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
e number of interest periods per year
e number of interest periods per payment period
e number of payment periods per year
 we have the special case of annual compounding. Substituting M = 1 into Equation
 it reduces to ia = r. That is, when compounding takes place once annually,* effective
o nominal interest. Thus, in all our earlier examples, where we considered only annual
 by definition using effective interest rates.
 — Calculating Auto Loan Payments
t to buy a car priced $22,678.95. The car dealer is willing to offer a financing package
l percentage rate over 48 months. You can afford to make a down payment of $2678.95,
t to be financed is $20,000. What would be the monthly payment?
pecify a compounding period, but in automobile financing the interest and the payment
st always both monthly. Thus, the 8.5% APR means 8.5% compounded monthly. In this
 easily compute the monthly payment using the capital recovery factor in Table 17.2.1:
 — Compounding More Frequent Than Payments
ke equal quarterly deposits of $1000 into a fund that pays interest at a rate of 12%
nthly. Find the balance at the end of year 2 (Figure 17.2.6).
ocedure for noncomparable compounding and payment periods described above.
e parameter values for M, K, and C:
e case, we could consider a continuous compounding. As the number of compounding periods
 large, the interest rate per compounding period (r/M) becomes very small. As M approaches
proaches zero, we approximate the situation of continuous compounding.
he effective annual interest rate for continuous compounding, we set K equal to 1, resulting in:
e, the effective annual interest rate for a nominal interest rate of 12% compounded continuously
.7497%.
r CK C= +( ) −1 1
i er K= −1
i e
a
r
= −1
i
N
A A P
= =
( )( ) =
( ) =
8 5 12 0 7083
4 48
0 7083 48 492 97
. % . %
, . %, $ .
 per month
= 12 months
= $20,000
M
K
C
=
=
=
12
4
3
compounding periods per year
payment periods per year
interest periods per payment period
ss LLC
Engineering Economics and Project Management 17-13
2. Use Equat
3. Find the t
4. Use i and 
Loss of Purc
It is important t
and the effects of
earning potentia
the general econ
purchasing powe
time or, to put it
opposite of infla
amount gains in 
of purchasing po
The Average In
To account for th
a single rate that
on the pervious y
the average inflat
the second year’s
• Step 1. We
FIGURE 17.2.6 Q
0
F
0866_book.fm Page 13 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
ion (17.2.3) to compute effective interest per quarter.
otal number of payment periods, N.
N in the (F/A, i, N) factor from Table 17.2.1:
hasing Power
o differentiate between the time value of money as we used it in the previous section
 inflation. The notion that a sum is worth more the earlier it is received can refer to its
l over time, to decreases in its value due to inflation over time, or to both. Historically,
omy has usually fluctuated in such a way that it experiences inflation, a loss in the
r of money over time. Inflation means that the cost of an item tends to increase over
 another way, the same dollar amount buys less of an item over time. Deflation is the
tion (negative inflation), in that prices decrease over time and hence a specified dollar
purchasing power. In economic equivalence calculations, we need to consider the change
wer along with the earning power.
flation Rate
e effect of varying yearly inflation rates over a period of several years, we can compute
 represents an average inflation rate. Since each individual year’s inflation rate is based
ear’s rate, they have a compounding effect. As an example, suppose we want to calculate
ion rate for a 2-year period for a typical item. The first year’s inflation rate is 4% and
 is 8%, using a base price index of 100.
 find the price at the end of the second year by the process of compounding:
uarterly deposits with monthly compounding (Example 17.2.6).
l = 3.030% per quarter (Step 2)
12% compounded monthly
41 2 3 5 6 7 8 9 10 11 12
Months
$1,000$1,000$1,000$1,000
i = 1+ 0.12 12
per quarter
( ) −
=
3 1
3 030. %
N K= ( ) = ( ) =number of years quarters4 2 8
F F A= ( ) =$ , . %, $ .1000 3 030 8 8901 81
 
100 1 0 04 1 0 08 112 32+( ) +( ) =. . .� ��� ���
� ����� �����
ss LLC
17-14 Chapter 17
• Step 2. To find the average inflation rate f over a 2-year period, we establish the following
equivalence equation:
Solving fo
We can say that t
of 5.98% per yea
previous year’s d
it is known as a 
Actual vs. Con
To introduce the 
terms.*
• Actual (cu
any antici
amounts a
• Constant 
of time. In
those estim
readily acc
we specify
Equivalence C
In previous sectio
of money — that
we may use eithe
the same solution
There are two 
inflation-free int
estimating the ca
• Market in
value of c
Virtually a
interest ra
[discount 
• Inflation-f
effects hav
if the mar
In calculating an
three common c
Case 1. All cas
value of a
* Based on the 
Economist. Vol. 33(
0866_book.fm Page 14 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
r f yields
f = 5.98%
he price increases in the last 2 years are equivalent to an average annual percentage rate
r. In other words, our purchasing power decreases at the annual rate of 5.98% over the
ollars. If the average inflation rate is calculated based on the consumer price index (CPI),
general inflation rate
stant Dollars
effect of inflation into our economic analysis, we need to define several inflation-related
rrent) dollars (An): Estimates of future cash flows for year n which take into account
pated changes in amount due to inflationary or deflationary effects. Usually these
re determined by applying an inflation rate to base year dollar estimates.
(real) dollars Dollars of constant purchasing power independent of the passage
 situations where inflationary effects have been assumed when cash flows were estimated,
ates can be converted to constant dollars (base year dollars) by adjustment using some
epted general inflation rate. We will assume that the base year is always time 0 unless
 otherwise.
alculation under Inflation
ns, our equivalence analyses have taken into consideration changes in the earning power
 is, interest effects. To factor in changes in purchasing power as well — that is, inflation —
r (1) constant dollar analysis or (2) actual dollar analysis. Either method will produce
; however, each uses a different interest rate and procedure.
types of interest rate for equivalence calculation: (1) the market interest rate and (2) the
erest rate. The interest rate that is applicable depends on the assumptions used in
sh flow.
terest rate (i): This interest rate takes into account the combined effects of the earning
apital (earning power) and any anticipated inflation or deflation (purchasing power).
ll interest rates stated by financial institutions for loans and savings accounts are market
tes. Most firms use a market interest rate (also known as inflation-adjusted rate of return
rate]) in evaluating their investment projects.
ree interest rate (i′): An estimate of the true earning power of money when inflation
e been removed.This rate is commonly known as real interest rate and can be computed
ket interest rate and inflation rate are known.
y cash flow equivalence, we need to identify the nature of project cash flows. There are
ases:
h flow elements are estimated in constant dollars. Then, to find the equivalent present
 cash flow series in constant dollars, use the inflation-free interest rate.
ANSI Z94 Standards Committee on Industrial Engineering Terminology. 1988. The Engineering
2): 145–171.
100 1 112 32 100 2 112 322+( ) = ← ( ) =f F P f. , , .
( ).f
(A ):
n′
ss LLC
Engineering Economics and Project Management 17-15
Case 2. All cash flow elements are estimated in actual dollars. Then, use the market interest rate to
find the equivalent worth of the cash flow series in actual dollars.
Case 3. Some 
actual dol
constant o
actual-dol
Removing the eff
of these constant
of the adjusted-d
we can combine 
This implies t
inflationary effec
increases. For exa
climb, because len
demand higher r
3%, you might b
beat inflation. If
insist instead on 
are stable, lender
to lend at lower 
17.3 Mea
This section show
economic standp
present worth, (2
of future cash flow
Annual worth is
basis. The third m
Describing P
When a compan
commits funds t
investment is sim
flow consists of i
form of cash flow
only with those c
or incremental ca
of the investmen
We must also 
estimation of the
viewed as before
organizations (e.
can be a valid ba
area of concern, t
tax net cash flow
0866_book.fm Page 15 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
of the cash flow elements are estimated in constant dollars and others are estimated in
lars. In this situation, we simply convert all cash flow elements into one type — either
r actual dollars. Then we proceed with either constant-dollar analysis for case 1 or
lar analysis for case 2.
ect of inflation by deflating the actual dollars with and finding the equivalent worth
 dollars by using the inflation-free interest rate can be greatly streamlined by the efficiency
iscount method, which performs deflation and discounting in one step. Mathematically
this two-step procedure into one by
(17.2.4)
hat the market interest rate is a function of two terms, i′, . Note that if there is no
t, the two interest rates are the same ( = 0 → i = i′). As either i′ or increases, i also
mple, we can easily observe that when prices are increasing due to inflation, bond rates
ders (that is anyone who invests in a money-market fund, bond, or certificate of deposit)
ates to protect themselves against the eroding value of their dollars. If inflation were at
e satisfied with an interest rate of 7% on a bond because your return would more than
 inflation were running at 10%, however, you would not buy a 7% bond; you might
a return of at least 14%. On the other hand, when prices are coming down, or at least
s do not fear the loss of purchasing power with the loans they make, so they are satisfied
interest rates.
sures of Project Worth
s how to compare alternatives on equal basis and select the wisest alternative from an
oint. The three common measures based on cash flow equivalence are (1) equivalent
) equivalent annual worth, and (3) rate of return. The present worth represents a measure
 relative to the time point “now” with provisions that account for earning opportunities.
 a measure of the cash flow in terms of the equivalent equal payments on an annual
easure is based on yield or percentage.
roject Cash Flows
y purchases a fixed asset such as equipment, it makes an investment. The company
oday in the expectation of earning a return on those funds in the future. Such an
ilar to that made by a bank when it lends money. For the bank loan, the future cash
nterest plus repayment of the principal. For the fixed asset, the future return is in the
s from the profitable use of the asset. In evaluating a capital investment, we are concerned
ash flows that result directly from the investment. These cash flows, called differential
sh flows, represent the change in the firm’s total cash flow that occurs as a direct result
t.
recognize that one of the most important parts of the capital budgeting process is the
 relevant cash flows. For all examples in this section, however, net cash flows can be
-tax values or after-tax values for which tax effects have been recalculated. Since some
g., governments and nonprofit organizations) are not taxable, the before-tax situation
se for that type of economic evaluation. This view will allow us to focus on our main
he economic evaluation of an investment project. The procedures for determining after-
s in taxable situations are developed in Section 17.4.
f
i i f i f= ′ + + ′
f
f f
ss LLC
17-16 Chapter 17
Example 17.3.1 — Identifying Project Cash Flows
Merco Inc., a machinery builder in Louisville, KY, is considering making an investment of $1,250,000 in
a complete struct
of the drilling sy
calculating produ
• Increased 
• Average sa
• Labor rate
• Tons of st
• Cost of st
• Number o
• Additiona
With the cost 
producing a ton 
contribution to o
increased produc
tion has been est
Since the drill
can run the syste
operator for load
three workers w
centerpunching, 
labor savings in t
can last for 15 ye
income taxes wou
Determine the n
Solution
The net investme
• Cash inflo
Increas
Project
Project
• Cash outfl
Project
Project
Project
Now we are read
The project’s cas
Assuming the
ahead for installa
0866_book.fm Page 16 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
ural-beam-fabrication system. The increased productivity resulting from the installation
stem is central to the justification. Merco estimates the following figures as a basis for
ctivity:
fabricated steel production: 2000 tons/year
les price/ton fabricated steel: $2566.50/ton
: $10.50/hr
eel produced in a year: 15,000 tons
eel per ton (2205 lb): $1950/ton
f workers on layout, holemaking, sawing, and material handling: 17
l maintenance cost: $128,500 per year
of steel at $1950/ton and the direct labor cost of fabricating 1 lb at 10 cents, the cost of
of fabricated steel is about $2170.50. With a selling price of $2566.50/ton, the resulting
verhead and profit becomes $396/ton. Assuming that Merco will be able to sustain an
tion of 2000 tons per year by purchasing the system, the projected additional contribu-
imated to be 2000 tons × $396 = $792,000.
ing system has the capacity to fabricate the full range of structural steel, two workers
m, one on the saw and the other on the drill. A third operator is required as a crane
ing and unloading materials. Merco estimates that to do the equivalent work of these
ith conventional manufacture requires, on the average, an additional 14 people for
holemaking with radial or magnetic drill, and material handling. This translates into a
he amount of $294,000 per year ($10.50 × 40 hr/week × 50 weeks/year × 14). The system
ars with an estimated after-tax salvage value of $80,000. The expected annual corporate
ld amount to $226,000. Determine the net cash flow from undertaking the investment.
et cash flows from the project over the service life.
nt cost as well as savings are as follows:
ws:
ed annual revenue: $792,000
ed annual net savings in labor: $294,000
ed after-tax salvage value at the end of year 15: $80,000
ows:
 investment cost: $1,250,000
ed increase in annual maintenance cost: $128,500
ed increase in corporate income taxes: $226,000
y to summarize a cash flow table as follows:
h flow diagram is shown in Figure 17.3.1.
se cost savings and cash flow estimates are correct, should management give the go-
tion of the system? If management has decided not to install the fabrication system,
Year Cash Inflows Cash Outflows Net Cash Flows
0 0 $1,250,000 –$1,250,000
1 1,086,000 354,500 731,500
2 1,086,000 354,500 731,500
� � � �15 1,086,000 + 80,000 354,500 811,500
ss LLC
Engineering Economics and Project Management 17-17
what do they do 
$1,250,000 of Tre
the company com
ering? This is an 
involves a compa
you answer.
Present Wort
Until the 1950s, 
flows in this met
project evaluatio
into account the
capital investmen
a proposed proj
criterion, we will
convenient point
the present worth
associated with a
called the net pre
When two or mo
project by compa
We will first s
investment proje
• Determine
interest ra
to this inte
Usually th
to change 
NPW.
FIGURE 17.3.1 Ca
* One of the pr
be recovered. The p
outlays. A commo
considered unless i
the acceptable rang
remember that pay
ceptable investmen
payback method of
method is its failu
measuring how lo
power of a project
$731,500 $80,000
0866_book.fm Page 17 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
with the $1,250,000 (assuming they have it in the first place)? The company could buy
asury bonds. Or it could invest the amount in other cost-saving projects. How would
pare cash flows that differ both in timing and amount for the alternatives it is consid-
extremely important question because virtually every engineering investment decision
rison of alternatives. These are the types of questions this section is designed to help
h Analysis
the payback method* was widely used as a means of making investment decisions. As
hod were recognized, however, business people began to search for methods to improve
ns. This led to the development of discounted cash flow techniques (DCF), which take
 time value of money. One of the DCFs is the net present worth method (NPW). A
t problem is essentially one of determining whether the anticipated cash inflows from
ect are sufficiently attractive to invest funds in the project. In developing the NPW
 use the concept of cash flow equivalence discussed in Section 17.2. Usually, the most
 at which to calculate the equivalent values is often time 0. Under the NPW criterion,
 of all cash inflows is compared against the present worth of all cash outflows that are
n investment project. The difference between the present worth of these cash flows,
sent worth (NPW), determines whether or not the project is an acceptable investment.
re projects are under consideration. NPW analysis further allows us to select the best
ring their NPW figures.
ummarize the basic procedure for applying the present worth criterion to a typical
ct.
 the interest rate that the firm wishes to earn on its investments. This represents an
te at which the firm can always invest the money in its investment pool. We often refer
rest rate as either a required rate of return or a minimum attractive rate of return (MARR).
is selection will be a policy decision by top management. It is possible for the MARR
over the life of a project, but for now we will use a single rate of interest in calculating
sh flow diagram (Example 17.3.1).
imary concerns of most business people is whether and when the money invested in a project can
ayback method screens projects on the basis of how long it takes for net receipts to equal investment
n standard used in determining whether or not to pursue a project is that no project may be
ts payback period is shorter than some specified period of time. If the payback period is within
e, a formal project evaluation (such as the present worth analysis) may begin. It is important to
back screening is not an end itself, but rather a method of screening out certain obvious unac-
t alternatives before progressing to an analysis of potentially acceptable ones. But the much-used
 equipment screening has a number of serious drawbacks. The principal objection to the payback
re to measure profitability; that is, there is no “profit” made during the payback period. Simply
ng it will take to recover the initial investment outlay contributes little to gauging the earning
.
0
Years
$1,250,000
15105
ss LLC
17-18 Chapter 17
• Estimate the service life of the project.*
• Determine the net cash flows (net cash flow = cash inflow – cash outflow).
• Find the p
their sum
• Here, a po
of outflow
project sh
Note that the dec
as costs associate
same revenues, 
minimizing costs
or least negative,
Example 17.3.2
Consider the inv
the firm’s MARR
Solution
Since the fabrica
equal annual savi
the NPW as follo
* Another speci
horizon is extrem
capitalization of pr
be invested today t
of i. Observe the li
Thus, it follow
Another way 
by this in perpetuit
the principal, whic
0866_book.fm Page 18 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
resent worth of each net cash flow at the MARR. Add up these present worth figures;
 is defined as the project’s NPW.
sitive NPW means the equivalent worth of inflows are greater than the equivalent worth
s, so project makes a profit. Therefore, if the PW(i) is positive for a single project, the
ould be accepted; if negative, it should be rejected. The decision rule is
ision rule is for a single project evaluation where you can estimate the revenues as well
d with the project. As you will find later, when you are comparing alternatives with the
you can compare them based on the cost only. In this situation (because you are
, rather than maximizing profits), you should accept the project that results in smallest,
 NPW.
 — Net Present Worth
estment cash flows associated with the metal fabrication project in Example 17.3.1. If
 is 15%, compute the NPW of this project. Is this project acceptable?
tion project requires an initial investment of $1,250,000 at n = 0 followed by the 15
ngs of $731,500, and $80,000 salvage value at the end of 15 years, we can easily determine
ws:
al case of the PW criterion is useful when the life of a proposed project is perpetual or the planning
ely long. The process of computing the PW cost for this infinite series is referred to as the
oject cost. The cost is known as the capitalized cost. It represents the amount of money that must
o yield a certain return A at the end of each and every period forever, assuming an interest rate
mit of the uniform series present worth factor as N approaches infinity:
s that
(17.5)
of looking at this, PW(i) dollars today, is to ask what constant income stream could be generated
y. Clearly, the answer is A = iPW(i). If withdrawals were greater than A, they could be eating into
h would eventually reduce to 0.
lim lim
N N
N
NP A i N
i
i i i→∞ →∞
( ) = +( ) −
+( )





 =, ,
1 1
1
1
PW i A P A i N A
i
( ) = → ∞( ) =, ,
If accept the investment
If remain indifferent
If reject the investment
PW i
PW i
PW i
( ) >
( ) =
( ) <
0
0
0
,
,
,
PW
outflow15 1 250 000% $ , ,( ) =
PW P A P F15 731 500 15 15 80 000 15 15
4 284 259
% $ , , %, $ , , %,
$ , ,
( ) = ( ) + ( )
=
inflow
ss LLC
Engineering Economics and Project Management 17-19
Then, the NPW of the project is
Since PW (15%)
Annual Equi
The annual equi
equal payments 
series of equal an
NPW by the cap
The accept-reject
Notice that the
value will be posi
AE(i) value is equ
should provide a
As with the pr
revenues are the 
alternative with t
Unit Profit/Co
There are many 
general procedur
• Determine
• Identify th
• Calculate 
determine
• Divide the
each year. 
into equiv
To illustrate th
be useful in estim
Example 17.3.3
Tiger Machine T
The required ini
over the 3-year p
Compute the equ
0866_book.fm Page 19 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
 > 0, the project would be acceptable.
valent Method
valent worth (AE) criterion is a basis for measuring investment worth by determining
on an annual basis. Knowing that we can convert any lump-sum cash amount into a
nual payments, we may first find the NPW for the original seriesand then multiply the
ital recovery factor:
(17.3.1)
 decision rule for a single revenue project is
 factor (A/P, i, N) in Table 17.2.1 is positive for –1 < i < ∞. This indicates that the AE(i)
tive if and only if PW(i) is positive. In other words, accepting a project that has a positive
ivalent to accepting a project that has a positive PW(i) value. Therefore, the AE criterion
 basis for evaluating a project that is consistent with the NPW criterion.
esent worth analysis, when you are comparing mutually exclusive service projects whose
same, you may compare them based on cost only. In this situation, you will select the
he minimum annual equivalent cost (or least negative annual equivalent worth).
st Calculation
situations in which we want to know the unit profit (or cost) of operating an asset. A
e to obtain such a unit profit or cost figure involves the following two steps:
 the number of units to be produced (or serviced) each year over the life of the asset.
e cash flow series associated with the production or service over the life of the asset.
the net present worth of the project cash flow series at a given interest rate and then
 the equivalent annual worth.
 equivalent annual worth by the number of units to be produced or serviced during
When you have the number of units varying each year, you may need to convert them
alent annual units.
e procedure, we will consider Example 17.3.3, where the annual equivalent concept can
ating the savings per machine hour for a proposed machine acquisition.
 — Unit Profit per Machine Hour
ool Company is considering the proposed acquisition of a new metal-cutting machine.
tial investment of $75,000 and the projected cash benefits and annual operating hours
roject life are as follows.
ivalent savings per machine hour at i = 15%.
PW PW PWin flow out flow15 15 15
4 284 259 1 250 000
3 034 259
% % %
$ , , $ , ,
$ , ,
( ) = ( ) − ( )
= −
=
AE i PW i A P i N( ) = ( )( ), ,
If accept the investment
If remain indifferent
If reject the investment
AE i
AE i
AE i
( ) >
( ) =
( ) <
0
0
0
,
,
,
ss LLC
17-20 Chapter 17
Solution
Bringing each flo
Since the project
annual equivalen
we obtain the AE
With an annual u
Comments
Note that we can
over the 3-year p
worth for each ho
Once we have th
period is 1 year. 
for the compoun
Rate of Retu
Along with the N
as rate of return.
financial manage
appealing to ana
Internal Rate o
Many different t
used in bond va
will define intern
project’s cash ou
Note that the NP
End of Year Net Cash Flow Operating Hours
0 –$75,000
PW 15%( ) =
=
0866_book.fm Page 20 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
w to its equivalent at time zero, we find
 results in a surplus of $3553, the project would be acceptable. We first compute the
t savings from the use of the machine. Since we already know the NPW of the project,
 by
sage of 2000 hr, the equivalent savings per machine hour would be
not simply divide the NPW amount ($3553) by the total number of machine hours
eriod (6000 hr), or $0.59/hr. This $0.59 figure represents the instant savings in present
urly use of the equipment, but does not consider the time over which the savings occur.
e annual equivalent worth, we can divide by the desired time unit if the compounding
If the compounding period is shorter, then the equivalent worth should be calculated
ding period.
rn Analysis
PW and AE, the third primary measure of investment worth is based on yield, known
 The NPW measure is easy to calculate and apply. Nevertheless, many engineers and
rs prefer rate of return analysis to the NPW method because they find it intuitively more
lyze investments in terms of percentage rates of return than in dollars of NPW.
f Return
erms refer to rate of return, including yield (that is, the yield to maturity, commonly
luation), internal rate of return, and marginal efficiency of capital. In this section, we
al rate of return as the break-even interest rate, i*, which equates the present worth of a
tflows to the present worth of its cash inflows, or
W expression is equivalent to
(17.3.2)
1 24,400 2,000
2 27,340 2,000
3 55,760 2,000
P F P F P F75 000 24 400 15 1 27 340 15 2 55 760 15 3
3553
$ , $ , , %, $ , , %, $ , , %,
$
− + ( ) + ( ) + ( )
AE A P15 3553 15 3 1556% $ , %, $( ) = ( ) =
Savings per machine hour hr hr= =$ $ $ .1556 2000 0 78
PW i PW PW*( ) = −
=
cash inflows cash outflows
0
 
PW i
A
i
A
i
A
i
N
N
*
* * *
( ) =
+( ) + +( ) + + +( ) =
0
0
1
1
1 1 1
0�
ss LLC
Engineering Economics and Project Management 17-21
Here we know the value of An for all n, but not the value of i*. Since it is the only unknown, we can solve
for i*. There will inevitably be N values of i* that satisfy this equation. In most project cash flows you
would be able to
some cash flow t
the NPW functio
a certain type of 
Finding the IR
We don’t need lab
for calculating i*
solve Equation (1
keyboard or by r
to locate the brea
Accept/Rejec
Why are we inter
worth of its rece
notice two impo
varying interest 
would be accepta
i > i*, indicating 
even interest rate
is consistent with
At the MARR 
which to judge p
Note that this 
compare mutual
see in a later sect
Example 17.3.4
Reconsider the fa
fabrication inves
* When applied
project), the i* prov
occur, none of the
a high priority on d
multiple i*s is to ge
In addition to
for predicting mult
our analysis when 
return (also known
Addison-Wesley, 1
must at a minimum
method such as NP
0866_book.fm Page 21 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
 find a unique positive i* that satisfies Equation (17.3.2). However, you may encounter
hat cannot be solved for a single rate of return greater than –100%. By the nature of
n in Equation (17.3.2), it is certainly possible to have more than one rate of return for
cash flow.* (For some cash flows, we may not find any rate of return at all.)
R
orious manual calculations to find i*. Many financial calculators have built-in functions
. It is also worth noting here that many spreadsheet packages have i* functions, which
7.3.2) very rapidly. This is normally done by entering the cash flows through a computer
eading a cash flow data file. As an alternative, you could try the trial-and-error method
k-even interest that makes the net present worth equal to zero.
t Decision Rules
ested in finding the particular interest rate that equates a project’s cost with the present
ipts? Again, we may easily answer this by examining Figure 17.3.2. In this figure, we
rtant characteristics of the NPW profile. First, as we compute the project’s PW(i) at a
rate (i), we see that the NPW becomes positive for i < i*, indicating that the project
ble under the PW analysis for those values of i. Second, the NPW becomes negative for
that the project is unacceptable for those values of i. Therefore, the i* serves as a break-
. By knowing this break-even rate, we will be able to make an accept/reject decision that
 the NPW analysis.
the company will more than break even. Thus, the IRR becomes a useful gauge against
roject acceptability, and the decision rule for a simple project is
decision rule is designed to be applied for a single project evaluation. When we have to
ly exclusive investment projects, we need to apply the incremental analysis, as we shall
ion.
 — Rate of Return Analysis
brication investment project in Example 17.3.1. (a) What is the projected IRR on this
tment? (b) If Merco’s MARR is known to be 15%, is this investment justifiable?
 to projects that require investments at the outset followed by a series of cash inflows (or a simple
ides an unambiguous criterion for measuring profitability. However, when multiple rates of return
m is an accurateportrayal of project acceptability or profitability. Clearly, then, we should place
iscovering this situation early in our analysis of a project’s cash flows. The quickest way to predict
nerate a NPW profile and check to see if it crosses the horizontal axis more than once.
 the NPW profile, there are good — although somewhat more complex — analytical methods
iple i*s. Perhaps more importantly, there is a good method, which uses a cost of capital, of refining
we do discover multiple i*s. Use of a cost of capital allows us to calculate a single accurate rate of
 as return on invested capital); it is covered in Contemporary Engineering Economics, C.S. Park,
997. If you choose to avoid these more complex applications of rate-of-return techniques, you
 be able to predict multiple i*s via the NPW profile and, when they occur, select an alternative
W or AE analysis for determining project acceptability.
If IRR MARR accept the project
If IRR MARR remain indifferent
If IRR MARR reject the project
>
=
<
,
,
,
ss LLC
17-22 Chapter 17
Solution
• (a) The ne
Using Exc
will recove
• (b) If Mer
pool and 
indicating
ment belie
of its fabri
deviations
Mutually Ex
Until now, we ha
were determinin
MARR requirem
In the real wor
of projects for a
alternatives will 
excluded.
Analysis Perio
The analysis peri
The analysis peri
period may be d
policy, or it may
these situations, w
period is stated a
the alternative in
of useful life of t
FIGURE 17.3.2 A 
The project breaks
Merco's fabrication system
project described in Example 17.8
0866_book.fm Page 22 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
t present worth expression as a function of interest rate (i) is
el’s financial function (IRR), we find the IRR to be 58.43%. (See Figure 17.3.2.) Merco
r the initial investment fully and also earn 58.43% interest on its invested capital.
co does not undertake the project, the $1,250,000 would remain in the firm’s investment
continue to earn only 15% interest. The IRR figure far exceeds the Merco’s MARR,
 that the fabrication system project is an economically attractive one. Merco’s manage-
ves that, over a broad base of structural products, there is no doubt that the installation
cating system would result in a significant savings, even after considering some potential
 from the estimates used in the analysis.
clusive Alternatives
ve considered situations in which only one project was under consideration, and we
g whether to pursue it, based on whether its present worth or rate of return met our
ents. We were making an accept or reject decision about a single project.
ld of engineering practice, however, it is more typical for us to have two or more choices
ccomplishing a business objective. Mutually exclusive means that any one of several
fulfill the same need and that selecting one alternative means that the others will be
d
od is the time span over which the economic effects of an investment will be evaluated.
od may also be called the study period or planning horizon. The length of the analysis
etermined in several ways: it may be a predetermined amount of time set by company
 be either implied or explicit in the need the company is trying to fulfill. In either of
e consider the analysis period to be a required service period. When no required service
t the outset, the analyst must choose an appropriate analysis period over which to study
vestment projects. In such a case, one convenient choice of analysis period is the period
he investment project.
net present worth profile for the cash flow series given in Figure 17.3.1 at varying interest rates.
 even at 58.43% so that the NPW will be positive as long as the discount rate is less than 58.43%.
NPW > 0 NPW < 0
accept reject
58.43%
PW(I)
$9,802,500
l
r
PW i P A i P F i( ) = − + ( ) + ( )
=
$ , , $ , , , $ , , ,1 250 000 731 500 15 80 000 15
0
ss LLC
Engineering Economics and Project Management 17-23
When useful life of the investment project does not match the analysis or required service period, we
must make adjustments in our analysis. A further complication, when we are considering two or more
mutually exclusiv
compare projects
ments in our ana
Analysis Perio
Let’s begin our a
this case, we com
or least negative 
Example 17.3.5
A pilot wants to
(formerly the U.
business, she dec
an old aircraft (A
revenues as well 
because of comp
are given in thou
Assuming that 
Solution
Since the require
analysis period c
equivalent NPW
• For A1:
• For A2:
Clearly, A2 is t
Project Lives D
Often project liv
example, two ma
both of them las
with some unuse
value is the amo
measure of its re
PW 15%(
PW 15%(
0866_book.fm Page 23 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
e projects, is that the investments themselves may have differing useful lives. We must
 with different useful lives over an equal time span, which may require further adjust-
lysis.
d Equals Project Lives
nalysis with the simplest situation where the project lives equal the analysis period. In
pute the NPW for each project and select the one with highest NPW for revenue projects
NPW for service projects. Example 17.3.5 will illustrate this point.
 — Two Mutually Exclusive Alternatives
 start her own company to airlift goods to the Commonwealth of Independent States
S.S.R.) during their transition to a free-market economy. To economize the start-up
ides to purchase only one plane and fly it herself. She has two mutually exclusive options:
1) or a new jet (A2) with which she expects to have higher purchase costs, but higher
because of its larger payload. In either case, she expects to fold up business in 3 years
etition from larger companies. The cash flows for the two mutually exclusive alternatives
sand dollars:
there is no do-nothing alternative, which project would she select at MARR = 10%?
d service period is 3 years, we should select the analysis period of 3 years. Since the
oincides with the project lives, we simply compute the NPW value for each option. The
 figures at i = 10% would be as follows:
he most economical option.
iffer from a Specified Analysis Period
es do not match the required analysis period and/or do not match each other. For
chines may perform exactly the same function, but one lasts longer than the other and
t longer than the analysis period for which they are being considered. We are then left
d portion of the equipment, which we include as salvage value in our analysis. Salvage
unt of money for which the equipment could be sold after its service or the dollar
maining usefulness.
n A1 A2
0 –3,000 –$12,000
1 1,350 4,200
2 1,800 6,225
3 1,500 6,330
P F P F P FA 3000 1350 10 1 1800 10 2 1500 10 3
842
1 $ $ , %, $ , %, $ , %,
$
) = − + ( ) + ( ) + ( )
=
P F P F P FA 12 000 4200 10 1 6225 10 2 6330 10 3
1719
2 $ , $ , %, $ , %, $ , %,
$
) = − + ( ) + ( ) + ( )
=
ss LLC
17-24 Chapter 17
When project lives are shorter than the required service period, we must consider how, at the end of the
project lives, we will satisfy the rest of the required service period. Replacement projects — additional projects
to be implemente
case. Sufficient re
To simplify ou
as the initial pro
necessary. For ex
technology — in
we select exactly 
likely to have so
required service p
to lease the nece
period. In this cas
Example 17.3.6
Analysis Perio
The Smith Novel
product announ
The two machine
$12,500 semiauto
model B will cos
two machines in
As business grow
of year 5. If that 
increased busine
Solution
Since both mode
explicit assumpti
leasing comparab
required service 
Here the bold fig
$5000 to operate
alternatives now 
0866_book.fm Page 24 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
d when the initial projecthas reached the limits of its useful life — are needed in such a
placement projects must be analyzed to match or exceed the required service period.
r analysis, we sometimes assume that the replacement project will be exactly the same
ject, with the same corresponding costs and benefits. However, this assumption is not
ample, depending on our forecasting skills, we may decide that a different kind of
 the form of equipment, materials, or processes — is a preferable replacement. Whether
the same alternative or a new technology as the replacement project, we are ultimately
me unused portion of the equipment to consider as salvage value at the end of the
eriod. On the other hand, if a required service period is relatively short, we may decide
ssary equipment or subcontract the remaining work for the duration of the analysis
e, we can probably exactly match our analysis period and not worry about salvage values.
 — Present Worth Comparison — Project Lives Shorter Than 
d
ty Company, a mail-order firm, wants to install an automatic mailing system to handle
cements and invoices. The firm has a choice between two different types of machines.
s are designed differently but have identical capacities and do exactly the same job. The
matic model A will last 3 years with a salvage value of $2000, while the fully automatic
t $15,000 and last 4 years with a salvage value of $1500. The expected cash flows for the
cluding maintenance, salvage value, and tax effects are as follows:
s to a certain level, neither of the models can handle the expanded volume at the end
happens, a fully computerized mail-order system will need to be installed to handle the
ss volume. With this scenario, which model should the firm select at MARR = 15%?
ls have a shorter life than the required service period (5 years), we need to make an
on of how the service requirement is to be met. Suppose that the company considers
le equipment that has an annual lease payment of $6000 (after taxes) for the remaining
period. In this case, the cash flow would look like Figure 17.3.3.
ures represent the annual lease payments. (It costs $6000 to lease the equipment and
 anually. Other maintenance will be paid by the leasing company.) Note that both
have the same required service period of 5 years. Therefore, we can use NPW analysis.
n Model A Model B
0 –12,500 –$15,000
1 –5,000 –4,000
2 –5,000 –4,000
3 –5,000 + 2,000 –4,000
4 –4,000 + 1,500
5
n Model A Model B
0 –12,500 –$15,000
1 –5,000 –4,000
2 –5,000 –4,000
3 –3,000 –4,000
4 –6,000 –5,000 –2,500
5 –6,000 –5,000 –6,000 –5,000
ss LLC
Engineering Economics and Project Management 17-25
Since these are se
Flaws in Project
Under NPW, the
the analogy does 
alternative. To illu
you have two mu
of $1000 with a re
the IRRs and NP
Would you prefe
FIGURE 17.3.3 Co
project life (Examp
Model A
PW 15%( )
0866_book.fm Page 25 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
rvice projects, model B is the better choice.
 Ranking by IRR
 mutually exclusive project with the highest worth figure was preferred. Unfortunately,
not carry over to IRR analysis. The project with the highest IRR may not be the preferred
strate the flaws of comparing IRRs to choose from mutually exclusive projects, suppose
tually exclusive alternatives, each with a 1-year service life; one requires an investment
turn of $2000 and the other requires $5000 with a return of $7000. You already obtained
Ws at MARR = 10% as follows:
r the first project simply because you expect a higher rate of return?
mparison for unequal-lived projects when the required service period is longer then the individual
le 17.3.6.)
n A1 A2
0 –$1000 –$5000
1 222000 7000
IRR 100% 40%
PW(10%) $818 $1364
0
$11,000
$11,000
1 2 3 4 5
5
$3,000$5,000$5,000
Year
Remaining
service
requirement
met by
leasing an
assetModel B
$12,500
$15,000
$4,000
10 2
$4,000
3 4
$2,500
$4,000
Required
service
period.
PW P A P F
P A P F
A15 12 500 5000 15 2 3000 15 3
11 000 15 2 15 3
34 359
% $ , $ , %, $ , %,
$ , , %, , %,
$ ,
( ) = − − ( ) + ( )
− ( )( )
= −
P A P F P FB 15 000 4000 15 3 2500 15 4 11 000 15 5
32 747
$ , $ , %, $ , %, $ , , %,
$ ,
= − − ( ) − ( ) − ( )
= −
ss LLC
17-26 Chapter 17
We can see that A2 is preferred over A1 by the NPW measure. On the other hand, the IRR measure
gives a numerically higher rating for A1. This inconsistency in ranking is due to the fact that the NPW
is an absolute (do
cannot be applied
the answer is no;
NPW. The NPW
project higher. A
Rate of Return
In our previous 
at an incrementa
interested in kno
value implies tha
1 year for $4000 i
make an addition
investment can b
Now we can g
mutually exclusiv
on incremental in
ment, we compu
the lower investm
decision rule is s
Example 17.3.7
Reconsider the tw
Since B1 is the lo
we compute the 
We obtain 
Comments
Why did we cho
investment durin
lower initial inve
flow. Ignoring th
cash flow and ha
If we erroneously
17.4 Cash
With the purchas
cash flows) that 
iB2-B1
*
0866_book.fm Page 26 Thursday, August 5, 2004 3:37 PM
© 2005 by CRC Pre
llar) measure of investment worth, while the IRR is a relative (percentage) measure and
 in the same way. That is, the IRR measure ignores the scale of the investment. Therefore,
 instead, you would prefer the second project with the lower rate of return, but higher
 measure would lead to that choice, but comparison of IRRs would rank the smaller
nother approach, called incremental analysis, is needed.
 on Incremental Investment
ranking example, the more costly option requires an incremental investment of $4000
l return of $5000. If you decide to take the more costly option, certainly you would be
wing that this additional investment can be justified at the MARR. The 10% of MARR
t you can always earn that rate from other investment sources — $4400 at the end of
nvestment. However, by investing the additional $4000 in the second option, you would
al $5000, which is equivalent to earning at the rate of 25%. Therefore, the incremental
e justified.
eneralize the decision rule for comparing mutually exclusive projects. For a pair of
e projects (A, B), rate of return analysis is done by computing the internal rate of return
vestment (IRR∆) between the projects. Since we want to consider increments of invest-
te the cash flow for the difference between the projects by subtracting the cash flow for
ent-cost project (A) from that of the higher investment-cost project (B). Then, the
elect B, if IRRB-A > MARR. Otherwise select A.
 — IRR on Incremental Investment: Two Alternatives
o mutually exclusive projects in Example 17.3.5.
wer cost investment project, we compute the incremental cash flow for B2–B1. Then
IRR on this increment of investment by solving
= 15%. Since IRRB2-B1 > MARR, we select B2, which is consistent with the NPW analysis.
ose to look at the increment B2-B1 instead of B1-B2? We want the increment to have
g at least some part of the time span so that we can calculate an IRR. Subtracting the
stment project from the higher guarantees that the first increment will be investment
e investment ranking, we might end up with an increment which involves borrowing
s no internal rate of return. This is the case for B1-B2. is also 15%, not –15%.)
 compare this i* with MARR, we might have accepted project B1 over B2.
 Flow Projections
e of any fixed asset such as equipment, we need to estimate the profits (more precisely,
the asset will generate during its service period. An inaccurate estimate of asset needs
n B1 B2 B2–B1
0 –$3,000 –$12,000 –$9,000
1 1,350 4,200 2,850
2 1,800 6,225 4,425
3 1,500 6,330 4,830
IRR 25% 17.43%
− + ( ) + ( ) + ( ) =$ $ , , $ , , $ , ,9000 2850 1 4425 2 4830 3 0P F i P F i P F i