SELECTED NOTES ON ENGINEERING ECONOMICS

by

A. M. Clausing Department of Mechanical and Industrial Engineering

University of Illinois at Urbana-Champaign Urbana, IL 61801

Copyright © 2002 by A. M. Clausing All Rights Reserved

Printed March 14, 2007

March 14, 2007 2

NOMENCLATURE a a uniform amount per interest

period, an annuity

BAL the balloon; the remaining balance at end of a series of payments

BV book value

c base amount in a geometric progression

D depreciation

EUAC equivalent uniform annual cost

EUAB equivalent uniform annual benefit

f future worth at end of the n-th interest period

FC first cost

g gradient: the periodic increase at end of each period in an arithmetic progression

i interest or discount rate per period expressed as a decimal

ic interest rate with continuous compounding; units: (time)-1

ie effective annual interest rate

in nominal interest rate

IRR internal rate of return

m number of interest compounding periods per year

n number of interest periods or life of asset

NPW net present worth

p present worth

r rate of change per period expressed as a decimal

RR rate of return

SV Salvage Value

t time

w defined by Eq. (29)

Subscripts

j j-th time (interest) period

Interest Factors

f/p single-payment compound amount factor (CAF)

p/f single-payment present worth factor (PWF)

f/a uniform series compound amount factor (SCAF)

a/f sinking fund factor (SFF)

p/a uniform series present worth factor (SPWF)

a/p capital recovery factor (CRF)

p/g gradient present worth factor (GPWF)

a/g gradient uniform series

p/c exponential present worth factor (EPWF)

March 14, 2007 3

1. INTRODUCTION The basis of many engineering decisions is an economic one, and engineering economics

provides the tools necessary to economically evaluate alternatives. For example, a proposed scheme for the utilization of solar energy must not only be technically feasible, but it must also be economically viable in order to be embraced. The question is how does one determine the economic viability of proposed engineering systems?

Systems like highly automatic manufacturing plants and nuclear power generation plants are capital intensive. Therefore, such plants generally would not be competitive if systems were selected on the basis of initial costs. However, making decisions based solely on initial costs is an approach used only in the simplest financial transactions. In general, the time value of money, the difference in annual operating costs, the escalating costs of conventional fuels, and other factors must be taken into account in order to make a rational economic comparison of alternatives. Life-cycle costing rather than initial-investment costing is a more appropriate way to determine if a given project is an economically viable one.

The economic analysis associated with proposed engineering and business ventures is complicated by unknowns like future product demand, future global competition, escalating energy costs due to fossil fuel depletion, future inflation rates, future environmental impacts, etc. Today's decisions should be strongly influenced by projections of these quantities. Unfortunately, such projections are very difficult to make with any degree of confidence.

Although engineering economics is concerned with the influences of economic factors, one should not lose sight of the importance of non-monetary attributes. Environmental, social, political, and aesthetic factors all play significant roles. Even national security should be considered. High sulfur coal is an inexpensive fuel for the production of power, yet US legislation resulting from environmental concerns severely restricts the use of this fuel. Likewise, few nuclear power plants are being built today in the United States due to social and safety concerns.

These notes are only concerned with making economic comparisons and decisions based solely on economic factors. It is assumed that the relevant discount rates, inflation rates, etc. are given, and the many variables that influence these economic factors are not of concern. The basic terminology, fundamental concepts, and mathematical tools required to perform economic comparisons will be presented. Examples are provided to illustrate the use of these tools and concepts.

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2. FUNDAMENTAL CONCEPTS and DEFINITIONS Engineering economics utilizes a collection of simple mathematical techniques to make

rational economic comparisons of various engineering alternatives. These mathematical techniques enable one to account for the time value of money in making comparisons. As a simple example, consider a $1,000 loan. If one borrows $1,000 today and uses this money for ten years, one clearly would be required to repay more than $1,000 at the end of ten years. Similarly, if the loan were repaid in ten equal annual payments, one would expect to pay more than $100 each year. Although the sum of the ten $100 payments is numerically equal to the original $1,000 loan, these ten payments are equivalent in economic value only if the time value of money is ignored. The time value of money means that a monetary amount on one date is not directly comparable with the same monetary amount on another date, even if inflationary influences are negligible. Interest charges (or payments) are a direct consequence of the time value of money.

Consider again the $1,000, ten-year loan with some positive rate of interest, i. The ability to determine the future worth, f, at the end of ten years of a present worth, p (the original $1,000 loan), is essential in economic analyses. Alternatively, if the loan is to be repaid with ten equal annual payments, a, the series of ten equal annual payments that is equal in economic value to the present worth, p, would be required. The concept of equivalence, the fact that different amounts of money at different times can be equal in economic value, is fundamental to engineering economy. Other examples of equivalence are:

(i) the present worth of a series of anticipated maintenance costs paid annually over the nyear life of a proposed piece of equipment,

(ii) the present worth of the anticipated fuel costs for heating a proposed factory over its n-year life with a fuel escalation rate of r percent per year, and

(iii) the present worth of the salvage value of a delivery truck at the end of its useful life.

The dependence of the economic value of a loan or investment on time is a consequence of interest. Interest may be thought of as money paid for the use of borrowed money or the return obtained by the productive investment of capital. Interest is defined as

interest = total amount accumulated - original amount (1) The interest rate, i, is by definition

!

i =interest accrued during unit time period

original amount (2)

March 14, 2007 5

The interest rate is usually expressed as a percent, that is, 100% times i; however, the fraction i as defined by Eq. (2) is the quantity to be used in all formulas that are presented. The unit time period or interest period is typically one year, but shorter interest periods such as one month are also used. The total number of interest periods is denoted by n, and the number of interest compounding periods per year is denoted by m.

If more than one interest period is involved, the type of interest, either simple or compound, must be considered. Simple interest ignores the time value of the interest accrued in preceding interest periods; therefore, it is almost never used in engineering economy. Specifically,

simple interest = p • n • i (3) where p is the original amount or principal. Compound interest is based on the total amount at the end of the previous period, that is, on the sum of the original amount (or principal) and the total interest accumulated during previous periods. This type of interest accounts for the time value of money and is the quantity used exclusively in these notes. Thus, when the word interest is used, compound interest is implied; hence, interest is to be added, that is, compounded, at the end of each interest period.

If interest is compounded over a period less than one year, the nominal interest rate, in, is defined as the interest rate times the number of compounding periods per year, m.

in =m • i (4) Nominal interest rates ignore the time value of accrued interest; hence, it would be better if nominal interest rates were never used. Unfortunately, this is not the case. Worse yet, the nominal rate of x% in loan transactions is often specified simply as x% or as x% per year. For example, an interest rate of 1.5% per month on the unpaid balance compounded monthly may be stated as 18% per annum. To be correct, the rate should be described as a nominal 18% per annum compounded monthly. When the value of accrued interest is taken into account, one arrives at the actual annual interest rate being charged. This rate is called the effective annual interest rate, ie, and is given by ie = 1 + i( )m !1 (5)

Hence, an interest rate, i, of 1.5% per month compounded monthly has an effective annual rate of

March 14, 2007 6

ie = 1 + 0.015( )12 !1 = 0.1956

or 19.56% not 18%. The influence of the compounding frequency on the effective annual rate is shown in Table 1. The influence of compounding frequency is greater, and the asymptotic value—the value with continuous compounding—is approached more slowly at higher rates of interest. Table 1 The Influence of Compounding Frequency on the Effective Annual Interest

Rate in Percent

Compounding Frequency, m

in = 6% i [%] ie [%]

in = 12% i [%] ie [%]

in = 18% i [%] ie [%]

1 (Annual) 6 6.000 12 12.000 18 18.000

2 (Semiannual) 3 6.090 6 12.360 9 18.810

4 (Quarterly) 3/2 6.136 3 12.551 9/2 19.252

12 (Monthly) 1/2 6.168 1 12.683 3/2 19.562

52 (Weekly) 3/26 6.180 3/13 12.734 9/26 19.685

∞ (Continuously) 0 6.184 0 12.750 0 19.722

The interest rate or rate of return, RR, expected on an investment would normally include a reasonable profit. The minimum attractive rate of return, MARR, would generally be larger on an investment involving risk than that expected from a bank, or another safe investment, before an investor would make a capital commitment. This is partly a consequence of the risk and partly a consequence of the limited amount of available investment capital or credit.

The life of a piece of equipment or property is the number of years the prudent user retains the property for its stated purpose. The salvage value, SV, is the net amount received, after disposal expenses are deducted, from the disposal of the property when it is retired from service. The salvage value may be a positive or negative amount.

Taxes and depreciation must also be considered in engineering economy. Three main types of taxes are of importance: property taxes, state income taxes, and federal income taxes. The property tax is usually a fixed percentage of the first cost of the property. Since state income taxes are presently deductible on federal tax returns, the effective-state-federal income tax rate is

March 14, 2007 7

effective rate = federal rate (1 - state) + state rate (6) The federal income rate for any but the smallest corporations is approximately 50 percent of the net profit. For individuals, it is dependent on the taxable income. State rates are dependent on the particular state as well as taxable income and tax status.

Depreciation is the decrease in value of property due to wear, deterioration, and obsolescence. The book value is the original cost minus the total amount of depreciation charged to date. The market value is the amount of money that can be obtained for the object on the open market. The market value and the book value normally are not equal. The major effect of depreciation in engineering economics is a reduction in income tax. Income taxes are paid on net income less depreciation; hence, the reduction in income tax allows the company to retain some of its income for replacement, maintenance, and investment.

Cash flow diagrams are often used in the analysis of the equivalent present worth or future worth of a series of receipts (income) and/or disbursements (costs), or in the calculation of the equivalent uniform annual amount, a. The cash flow is defined as the net receipt that results from the receipts and disbursements in the same interest period: cash flow = receipts - disbursements (7) The cash flow diagram is a graphic description of the cash flows. The horizontal line is the time axis. An upward arrow represents a positive cash flow, that is, a net receipt (a benefit). A downward arrow represents a negative cash flow: a net disbursement (a cost). The arrows are generally not drawn to scale.

Representative cash flow diagrams are given in Fig. 1. In this figure, several repayment plans of a $1,000 loan over a four-year period with an interest rate of 8 percent compounded annually are presented. The time interval on the horizontal scale is one year. The total amount repaid in Plans A, B, and C is, respectively, $1,207.68, $1,360.49, and $1,320.00; however, the economic value of all repayment plans has the same present worth: the original $1,000 loan.

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Plan A Equal annual payments (Amount repaid: $1,207.68)

1 2 3 4

$1000

$301.92 $301.92 $301.92 $301.92

i = 8 %

Plan B Repayment of loan with accrued interest at the end of the four year period

1 2 3 4

$1000

$1360.49

i = 8 %

Plan C Payment of interest charges each year with principal repayment after four year period

(Amount repaid: $1320)

1 2 3 4

$80

$1000

$1080$80 $80

i = 8 %

Figure 1 Cash Flow Diagrams for Several Repayment Plans of a $1,000 Loan with an 8%

Interest Rate Compounded Annually Note that if simple interest were charged in repayment Plan B, the interest payment would be p•n•i = ($1000)(4)(.08) = $320; whereas, annual compounding results in an interest payment of $360.49

Monetary computations in banks, insurance companies, and other financial institutions are generally required by law or business practice to be precise to the nearest penny. This may

March 14, 2007 9

require ten or more places of accuracy. On the other hand, many estimates are incorporated in economic analyses in engineering and such accuracy is usually neither justifiable nor necessary. The output of an analysis can be no more accurate than the input. Since illustrative points may be lost if quantities are rounded, numerical values that follow are given either to the nearest dollar or nearest penny. 3. INTEREST FORMULAS The mathematical formulas that permit the conversion of an amount at one period of time, for example, the present, to an equivalent amount at some other period of time, for example, 10 years in the future, are derived and presented in this section. These formulas enable one to account for the time value of money. The conversion of an arbitrary series of n annual cash flows to a present worth, a future worth, or a uniform annual series will also be addressed. Examples are given to illustrate the use of the conversion formulas. 3.1 Compound-Amount Factor (CAF), f/p, and Present-Worth Factor (PWF), p/f

If an amount p is allowed to accumulate for n time periods with an interest rate i, the interest and the total amount will grow with time as shown in Table 2.

Table 2 The Influence of the Number of Time Periods on the Interest and the Accumulated Amount—the Future Value, f

Period

Interest, (f - p)

Accumulated Amount, f

1

!

pi

!

p 1+ i( )

2

!

p 1+ i( ) i p 1+ i( )2

• • •

• • •

• • •

n p 1+ i( )n!1i p 1+ i( )

n

Hence, after n time periods, a present value p, will have grown to the future value f that is given

by:

f = p 1+ i( )

n (8) or

March 14, 2007 10

f

p= 1 + i( )

n= CAF( ) (9)

The factor f/p is referred to in engineering economy as the single-payment Compound-Amount Factor (CAF). Equation (8) also gives

p

f=

1

1 + i( )n= PWF( ) (10)

The factor p/f is called the single payment Present-Worth Factor (PWF). The most common notation used to denote these factors and simultaneously to specify the interest rate and the number of periods is (f/p, i%, n). This notation will be used throughout these notes.

The influence of n and i on the single-payment compound amount factor (CAF), f/p, is shown graphically in Fig. 2. Note that the limit of f/p as i approaches zero is one; that is, f = p if i = 0, an intuitively obvious result.

March 14, 2007 11

1

10

0 5 10 15 20 25 30

f/p

Number of Periods

i = 30% 24 18 15 12 10

9

8

7

6

5

4

3

2

i = 1%

Figure 2 The Influence of i and n on the Compound Amount Factor (CAF), f/p

If an investment earns 7% compounded annually, Fig. 2 shows that it will approximately double every 10-year period—a good number to remember. On the other hand, money sitting in one's shoe box would lose approximately half of its value in 14 years if the effective average rate of inflation were 5% per year. At an 18% inflation rate, it would take only four years for the money to lose half of its economic value. Clearly, a fixed income at high rates of inflation is not attractive.

A rule exists, the so-called 72 Rule, for quickly estimating the number of years required for an investment to double, nd. Specifically:

nd!72

i (%)

If p, f, and n or i are given and i or n is to be determined, Eq. (8) could be solved for i or n which gives, respectively:

March 14, 2007 12

i =f

p

!

" # $

%

1

n

&1 (11)

and

n =

lnf

p

!

" # $

%

ln(1+ i) (12)

Hence, given any three of four quantities p, f, i, and n, Fig. 2 or eqs. (9), (10), (11), or (12) can be used to obtain the fourth quantity. The effective interest rate (see Eq. (5), Section 2), follows from Eq. (9), that is,

!

1+ ie

( )1

= 1+ i( )m

or ie

= 1+ i( )m

"1

Since the present worth of a receipt or disbursement is always less than its future value (assuming that i > 0), the word discount is used in financial circles for the difference between the future worth and the present worth, that is:

!

discount{ } = f " p{ } = pf

p, i%, n

#

$ % &

' "1

( ) *

+ , -

Example 3.1: A savings deposit of $4,000 will accumulate to what value after ten years if the interest is left in the account and the account pays 6% interest compounded quarterly? Solution: The future value f, the accumulated value of the saving account at the end of the ten-year period, is obtained by multiplying the present value p, $4000, by f/p. Thus:

!

f = pf

p,1.5%, 40

"

# $ %

& = $4, 000 1.8140( )

or

!

f = $7,256.07 Example 3.2: You have won a million dollars in the Poor Man's Lottery. You later read in the fine print that all winnings are paid 25 years after the date on which the winning ticket is drawn. At 9 percent interest compounded annually, what is the present worth of your winnings?

March 14, 2007 13

Solution: In this example, the present worth, p, of the future value, f = $1,000,000 is determined by dividing f by f/p. Hence:

!

p =f

f

p, 9%, 25

"

# $

%

& '

=$1,000,000

8.6231= $115,968

Note: the discount is $884,032! Clearly, you will have to wait a while before you can call yourself a millionaire.

The conversion of the economic value of a present amount, p, to its equivalent economic

value in the future, f, is effected simply by multiplying by f/p or dividing by p/f. Analogously, the present value is obtained from the future value by dividing by f/p or multiplying by p/f. Once the appropriate factor is calculated, the conversions are no different than the conversion of 15 inches to 1.25 feet by multiplying 15 inches by the conversion factor: (1 ft)/(12 inches). Fifteen inches and 1.25 feet are equivalent lengths. Likewise, in Example 3.1, the present value of $4,000 has an equivalent future economic value of $7,256.07 for the conditions stated in that example. There are, however, some fundamental differences between engineering data and economic data, and between the conversion factors like (12 inches)/(ft), (0.4536 kg)/(lbm), etc., with which we are all familiar, and the economic factors f/p, f/a, etc. that are being presented. Specifically:

(i) While the numerical values of the three lengths, 12 inches, 1 ft., and 0.3048 m, are vastly different, an examination of their units readily shows that the three lengths are equivalent. The units associated with engineering data provide critical information that is used in converting from one quantity to another. The data being analyzed in engineering economy, on the other hand, typically all have the same monetary unit (the US$ in the United States). Hence, economic values would be indistinguishable if only the monetary unit were supplied; therefore, all monetary amounts must be appropriately qualified: the present value, p, the future value, f, etc. In the example presented earlier in this section, the three monetary amounts: the present value of $1,000, the four annual $301.92 payments, and future value of $1,360.49 are vastly different, yet they are economically equivalent quantities.

(ii) The factors used to convert economic values are all dimensionless quantities. Hence, they

also must be given names in order to identify them. Thus, the notations f/p, (f/p, i, n), and the name: single-payment Compound-Amount Factor (CAF) are introduced. Learning these names is not a trivial task. This task is compounded by the fact that the reciprocals of all economic factors are also named. For example, p/f is called the single-payment Present Worth Factor (PWF). In contrast, the conversion factor (1 in)/(25.4 mm) needs no name.

March 14, 2007 14

(iii) The economic conversion factors are not constants. For example, f/p is dependent on i and n; hence, the notation (f/p, i, n). A factor, p/c, will later be introduced that is also dependent on an escalation rate, r. The notation used for this quantity is (p/c, i, n)r.

Tabulations of the compound-amount factor f/p and, amazingly, its reciprocal p/f, are given in typical engineering economics books. With the advent of inexpensive programmable calculators and personal computers, such tabulations are no longer of sufficient value to warrant their inclusion in these notes. Some tips for programming interest formulas are given in Section 3.6. 3.2 Uniform Series Compound-Amount Factor, f/a, and Sinking Fund Factor, a/f

A retirement fund, a building fund, a fund for purchasing a home, etc., could be built up by depositing a dollars at the end of each time period for n periods in an account earning interest at a rate i per period. The future sum accrued at the end of the n periods is desired. The cash flow diagram is

1 2 n-1 n

f

aaaa It should be emphasized that the first payment is made at the end of the first time period, and succeeding payments are made at the end of succeeding time periods including the n-th period. The resulting progression is:

Period Accrued Amount at End of Respective Period 1 a 2 a 1+ (1 + i){ } 3 a 1+ (1 + i) + (1 + i)2{ }

• • • n

a 1+ (1 + i) + (1 + i)2 +K + (1+ i)n!1{ }

Hence, the future value, f, is

March 14, 2007 15

!

f = a 1+ (1+ i) + (1+ i)2 +K+ (1+ i)n"1{ } = a 1+ i( )j

j= 0

n"1

# (13)

The terms are seen to form a geometric progression with the common ratio

!

(1+ i) . The sum of the geometric series can be easily obtained by multiplying both sides of Eq. (13) by

!

(1+ i) and subtracting the resulting expression from Eq. (13). This gives f 1 ! (1 + i)[ ] = a 1 ! (1+ i)n[ ]

or

f

a=(1 + i)n !1

i= (SCAF) (14)

The factor f/a is called the Uniform Series Compound-Amount Factor (SCAF). Its reciprocal is called the Sinking Fund Factor (SFF); that is,

a

f=

i

(1 + i)n!1

= (SFF) (15)

The uniform-series compound-amount factor, f/a, is shown in Fig. 3. This figure is a log-log plot of f/a as a function of n for i equal to 1, 4, 8, 12, 24, and 48 percent. It is seen that f/a grows much more rapidly than f/p due to the uniform annual payment, a. For example, (f/p, 0.08, 20) = 4.661 whereas (f/a, 0.08, 20) = 45.76. Note also that the limit of f/a as i approaches zero is n; that is, f = n•a if i = 0, an intuitively obvious result.

March 14, 2007 16

1

10

100

1000

1 10

Number of Periods

f/a

i = 48% 24

12

8

4

1

Figure 3 The Influence of i and n on the Uniform-Series Compound-Amount Factor

(SCAF), f/a

Example 3.3: Equal annual deposits are to be made at the end of each year for the next ten years to build up a fund to purchase a home that will cost $40,000. If the deposits earn 8% interest compounded quarterly, what is the required annual deposit? Solution: Since the compounding period and the payment period are different, the effective annual interest rate must be utilized. It is (1 + 0.02)4 - 1 = 0.08243. The cash flow diagram is

1 2 9 10

$40,000

aaaa

Hence, the annual deposit, a, that must be made to accumulate the future amount, f = $40,000 is obtained by dividing this future value by f/a based on the effective annual interest rate of 0.08243. Thus:

March 14, 2007 17

a =$40,000

f /a, 8.234%, 10( )=

$40, 000

14.65496

or

!

a = $2729.45

3.3 Uniform Series Present Worth, p/a, and Capital Recovery Factor, a/p

An expression for determining the future worth, f, of a uniform series of amounts, a, was derived in Section 3.2. In transactions involving the recovery of invested capital, for example, loans and mortgages, the uniform series of amounts a that is equivalent to the present worth, p, is desired. The relevant cash flow diagram is

1 2 n-1 n

p

aaaa For example, p could represent the amount of a mortgage and a the periodic fixed payment that will fully amortize the debt at the end of n time periods with an interest charge on the unpaid balance of i percent per time period. The progression for this case is Period Value at End of Respective Period 1 p 1+ i( ) ! a 2 p 1+ i( )2 ! a 1+ (1+ i){ }

3 p 1+ i( )3 ! a 1 + (1 + i) + (1+ i)2{ }

• • • n

p 1+ i( )n ! a 1+ (1+ i) +K + (1 + i)n !1{ }

If the loan is to be fully amortized, the future value of the original loan

!

p(1i + i)n must be equal

to the future value of the periodic payments. Hence,

March 14, 2007 18

!

p

a=

1+ i( )j

j= 0

n"1

#

1+ i( )n =

(1+ i)n "1

i (1+ i)n =

f

a

f

p

The expression on the right-hand side of Eq. (16) reveals a more direct way of deriving an expression for p/a assuming expressions for f/a and f/p are known. The factor p/a is called the uniform Series Present Worth Factor (SPWF) or simply the Series Present Worth

p

a=(1 + i)n !1

i (1 + i)n = (SPWF) (17a)

Its reciprocal is called the Capital Recovery Factor.

a

p=

i (1 + i)n

(1 + i)n!1

= (CRF) (17b)

The uniform-series present worth factor, p/a, is shown in Fig. 4. This figure is a log-log plot of p/a as a function of n for i equal to 1, 4, 8, 12, 24, and 48 percent. It is seen that p/a grows more slowly than f/a since each annual payment, a, is being discounted. For example, (f/a, 0.08, 20) = 45.76 whereas (p/a, 0.08, 20) = 9.818. Since a is being discounted, p/a decreases with increasing i, whereas f/a increases with increasing i (see Fig. 3). As i approaches zero, both p/a and f/a approach n. That is, without interest, the present value and the future value of a series of payments is simply the total value of all of the payments: a•n. For a finite positive value of i, p/a is less than n, and f/a is greater than n.

March 14, 2007 19

1

10

1 10

Number of Periods

p/a

i = 1% 4

8

12

24

i = 48%

Figure 4 The Influence of i and n on Uniform-Series Present Worth Factor, p/a It can be seen from Eq. (16) as well as Fig. 4 that the asymptotic value of p/a as n approaches infinity is 1/i. This limit is of interest, for example, in working with endowments or with systems like roads that are to be maintained forever. If the uniform annual cost associated with maintaining such a system is a, the present worth, p, of these costs, p = a•(p/a) = a/i, is called the Capitalized Cost. Likewise, if an endowment p that supports, for example, scholarships is to be maintained in perpetuity, and this endowment is earning the interest rate, i, the institution may award the amount given by:

a = pa

p

!

" # $

% = p • i

each year without diminishing the value of the endowment. Again, this result is intuitively obvious. If one only spends the interest, p•i, that is earned each year, the principal, p, remains to earn the same amount in succeeding years.

March 14, 2007 20

Example 3.4: Determine the monthly payment required to fully amortize a 30-year, $60,000 mortgage on a home with a 9 percent rate of interest compounded monthly. Solution: The cash flow diagram is:

1 2 359 360

p = $60,000

aaaa

i = 9/12 = 0.75 %

Note that the monthly interest rate is 0.75%, and n is equal to 360, that is, there are 360 months in the 30-year period. The monthly payment, a, follows as:

!

a =$60,000

p

a, 0.75%, 360

" #

$ %

=$60,000

124.2819= $482.77

The value placed on capital is clearly evident by the difference between the amount borrowed, $60,000, and the amount paid back, $173,797. Payments are usually made at the end of the time periods (ordinary annuity) in direct reduction loans and mortgages, whereas payments are commonly made at the beginning of the time periods (annuity due) in leases. A sum can be shifted one time period to the right by multiplying it by

!

(1+ i) ; hence, if each a is multiplied by

!

(1+ i) , the cash flow diagram is shifted from the annuity due mode,

1 2 n-1 n

aaaa to the ordinary annuity mode,

1 2 n-1 n

a(1+i) a(1+i) a(1+i) a(1+i)

March 14, 2007 21

Hence, the same formulas are applicable if the periodic payment a is replaced by

!

a(1+ i) . Alternatively:

!

p

a

"

#

$

% annuity due

=p

a

"

#

$

% (i+ i) (18)

!

f

a

"

#

$

% annuity due

=f

a

"

#

$

% (i+ i) (19)

The present worth of an annuity due is easily calculated by dividing it into two parts: the first payment whose present worth is a, and the remaining ordinary annuity whose present value is a(p/a, i, n-1). This avoids introducing a new factor or the confusion that might otherwise arise in shifting the quantity, a, to the next time period.

Example 3.5: A rental agency is considering purchasing a fleet of large trucks priced at $80,000 each and leasing them to their customers for a ten-year period. They wish to achieve a yield of 1 percent per month, and they expect to be able to sell the trucks at the end of the lease for $30,000. What should they specify for the monthly payment in order to realize the desired yield?

Solution: The cash flow diagram for this 120 month period is:

$80,000

1 2 119 120

aaaa

i = 1 % $30,000

The monthly payment follows as:

!

a =

$80,000 "$30,000

f

p, 1%, 120

#

$ % &

'

p

a

(

)

*

+ annuity due

March 14, 2007 22

or

!

a =

$80,000 "$30,000

3.300387

p

a, 1%, 120

# $

% & (1+ 0.01)

=$70,910.16

(69.70052)(1.01)= $1007.28

Note that the present value of the $30,000 received for the fleet at the end of 10 years is only $9,089; hence, the salvage value of $30,000 is discounted by $20,910.

3.4 Arithmetic and Geometric Progressions

A series of equal amounts is common in dealing, for example, with loans, annuities, and leases. A series of increasing or decreasing amounts is also a common occurrence in engineering economy. For example, maintenance costs are expected to increase with the age of the property. Depreciation schedules, such as the sum-of-the-year's digits method, result in a rapidly decreasing series. Fuel costs and capital costs are often assumed to increase exponentially with time.

Most actual series of receipts or disbursements from items such as net profit or maintenance costs would be sufficiently complex that each year would have to be considered separately in order to determine the equivalent present worth, future worth, or uniform annual amount. However, sufficiently accurate approximations for engineering purposes can often be generated by simplified progressions. Two types of progressions, arithmetic progressions and geometric progressions, are considered in these notes. The results, used in combination with the previously derived factors, enable the treatment of relatively complex problems.

Consider first the constant gradient series, that is, a series of end-of-period disbursements increasing by a constant amount, g, per period. The growth for this case is linear, and the progression is arithmetic. The cash flow diagram is:

3 4 n-1 n

(n-1)g

(n-2)g

3g2g

1 2

g

March 14, 2007 23

Note that the first disbursement is made at the end of the second period, and the disbursement made at the end of the n-th period is

!

n "1( ) g . The present worth of this series, the sum of the

individual present values, is

p = g1

1 + i( )2+

2

1 + i( )3+

3

1 + i( )4+K +

n !1

1+ i( )n" # $

% & '

(20)

Multiplying both sides of Eq. (20) by

!

(1+ i) and subtracting Eq. (20) from the resulting equation, one obtains

p 1+ i( ) !1{ } = g1

1 + i( )+

1

1+ i( )2+ K+

1

1 + i( )n!

n

1+ i( )n" # $

% & '

or

p

g=1

i

1 + i( )n !1

i 1 + i( )n!

n

1+ i( )n" # $

% & '

(21)

p/g is called the Gradient Present Worth Factor (GPWF). The gradient present worth factor also can be written in terms p/a and p/f. Specifically,

p

g=1

i

p

a

! "

# $ % n

p

f

!

" & #

$

' ( )

* + ,

(22)

Or, if Eq. (22) is multiplied by f/p, one obtains

f

g=1

i

f

a

! "

# $ % n

& ' (

) * +

(23)

Likewise, a/g follows by multiplying Eq. (23) by a/f which gives:

a

g=1

i1 ! n

a

f

"

# $ %

&

' ( )

* + ,

(24)

The uniform series worth, a, of a gradient series would be used, for example, in investment analysis if all costs were being translated to a uniform stream of annual payments over the system lifetime. Equations (22), (23), and (24) can be used to determine the present worth, future

March 14, 2007 24

worth, and uniform annual amount, respectively, that are equivalent to the uniform gradient series.

The gradient series, p/g, is shown in Fig. 5, a log-log plot of p/g as a function of n for i equal to 1, 4, 8, 12, 24, and 48 percent. Since the disbursements are being discounted, p/g decreases with increasing i as is clearly seen in Fig. 5. A comparison of figs. 4 and 5 shows that p/g grows more rapidly than p/a because the end-of-period disbursements are increasing by a constant amount, g, each period in the gradient series. This is especially true at small values of i when these future payments are not discounted very much. For example, (p/g, 0.04, 20) = 111.56 whereas (p/a, 0.04, 20) = 13.59.

1

1 0

1 0 0

1 1 0

p / g

Number of Periods

i = 1%4

8

1 2

2 4

i = 48%

Figure 5 The Influence of i and n on the Gradient Present Worth Factor, p/g Example 3.6: The annual maintenance cost of a small office building is $1,200 for the first year and increases by $100 each year thereafter. The life of the building is assumed to be 25 years; the interest rate is 9 percent compounded annually; and the payments can be assumed to be made at the end of the time periods. What is the present worth of the maintenance costs?

Solution: The cash flow diagram is:

March 14, 2007 25

3 24 25

$3600

$3500

$1400$1300

1 2

$1200

i = 9 %

The given series is equivalent to the superposition of a uniform annual series with a = $1,200 and a gradient series with g = $100; hence, the present worth of the maintenance costs is:

!

p = $1,200p

a, 9%, 25

" #

$ %

+ $100p

g, 9%, 25

"

# & $

%

or p = $1,200(9.8226) + $100(76.9265) = $19,479.74

Consider next costs that grow at a constant rate, r, for example, fuel costs with a fuel escalation rate, r. If the base amount, the amount at the end of the first year, is denoted by c, the cash flow diagram is

3 4 n-1 n1 2

c (1+r)c (1+r)c

2

n-2

c (1+r)3

c (1+r)

n-1c (1+r)

The depicted growth, a constant rate of change per period, is an exponential growth, and the progression is geometric. In contrast, recall that the gradient series, an arithmetic progression, is appropriate for a constant amount of change per period. The present worth of each term in the geometric progression is

p = c1

1+ i( )+1+ r( )

1 + i( )2+1+ r( )2

1+ i( )3+K +

1+ r( )n! 2

1 + i( )n!1+1 + r( )n!1

1+ i( )n" # $

% & '

(25)

March 14, 2007 26

If

!

(1+ i)"1 is factored out of Eq. (25), the sum is seen to be

p =c

1 + i

1 + r

1 + i

! "

# $

n

%1

1+ r

1 + i%1

&

' (

) (

*

+ (

, (

(26)

Hence, p/c, the Exponential Present Worth Factor (EPWF) is:

p

c=

1+ r

1+ i

! "

# $

n

%1

r % i (27)

The exponential present worth factor, p/c, is shown in Fig. 6 for an escalation rate, r, of 10 percent. Figure 6 is a log-log plot of p/c as a function of n for i equal to 1, 4, 8, 12, 24, and 48 percent. On this log-log plot, the slope increases with increasing n for i < r and decreases with increasing n if i > r.

1

1 0

1 0 0

1 1 0

i = 48%

Number of Periods

p/c

2 4

1 2

8

4

i = 1%

r = 10 %

Figure 6 The Influence of n and i on the Exponential Present Worth Factor, p/c, for

an Escalation Rate, r, of 10 Percent

March 14, 2007 27

Alternatively, the amount at the end of the first year is sometimes denoted as

!

1+ r( ) c . In

this case, p/c becomes

!

p

c

"

#

$

% shift

=1+ r

r & i

' (

) * 1+ r

1+ i

' (

) *

n

&1+ , -

. / 0

Returning to Eq. (27), this expression is undefined if

!

r = i( ) . However, the limit as

!

r" i( ) can

be found by applying L' Hospital's rule, which gives:

!

p

c r= i

=n

1+ i=

n

1+ r (28)

The factor p/c can also be written in terms of f/a or p/a; specifically,

!

p

c=

f

a, w, n

" #

$ %

1+ i, where w =

1+ r

1+ i&1 (29)

and

!

p

c=

p

a, w' , n

" #

$ %

1+ r, where w '=

1+ i

1+ r&1 (30)

Also, it should be noted that

!

p

c

"

#

$

% shift

is simply

!

p a ,w', n( ) . The validity of Eqs. (29) and (30) is

easily established by substituting (1 + w) and (1 + w'), respectively, into Eq. (26). Since the limit

of both f/a and p/a obviously must approach n as i approaches zero, Eqs. (29) and (30) are valid

for all values of r and i including those that result in negative values of w and w'. Specifically, w

is positive only if r > i and w' is positive only if r < i. Hence, if conventional tables of f/a and p/a

are to be used, p/c would be determined from Eq. (28) for r = i, Eq. (29) for r > i, and Eq. (30)

for r < i. At the same time, all relations presented are valid for both positive and negative values

of i. Since p/c is a function of three variables, i, r, and n, its tabulation is illogical, especially in

the light of the simplicity of Eqs. (29) and (30).

Once the geometric series is converted to its equivalent present worth, its future worth and the equivalent uniform annual amount can be obtained by multiplying by

!

f p, i, n( ) and

!

a p, i, n( ) , respectively. The notation to be used for p/c is

!

p c, i, n( )r .

March 14, 2007 28

Example 3.7: The fuel cost for the next 12-month period for a small office building is projected to be $1,800. If the fuel escalation rate is 10 percent per year and the interest rate is 9 percent compounded annually; (a) What is the annual fuel cost for the 20th year (the estimated useful life of the facility)? (b) What is the present value of the total fuel costs for this 20-year period? (c) What is the equivalent uniform annual amount?

Solution: The cash flow diagram is:

2 19 20

$1800 (1.1)

$1800 (1.1)

$1800 (1.1)$1800

1

18

19

(a) The annual fuel cost for the 20th year is

!

cash flown= 2 0

= $1,800 1.1( )1 9

= $11,008.64

(b) The present value of the total fuel costs for this 20-year period is:

!

p = $1,800p

c, 9%, 20

" #

$ %

10%

!

p = $1,800

1.1

1.09

" #

$ %

20

&1

0.01= $1,800 20.03938( ) = $36,070.88

(c) The equivalent uniform annual amount is:

!

a =p

p

a, 9%, 20

" #

$ %

=$36,070.88

9.12855= $3,951.44

March 14, 2007 29

3.5 Continuous Compounding

Continuous compounding is being introduced for two reasons: (i) An understanding of continuous compounding clearly reveals the true nature of the interest

rate, i, and shows why a time period must be associated with any interest quotation, for example, i = 0.015 per month or in = 18% with monthly compounding.

(ii) Continuous compounding provides a useful and meaningful limit. This limit is the effective rate of interest, ie, as the number of compounding periods per year, m, approaches infinity.

It should be noted, however, that discrete payments and discrete interest formulas are used almost exclusively in practice. The quantity ic is used in these notes to denote the continuous rate of interest. The units of ic are [time]-1. The infinitesimal change in principal, the interest accumulated, during the infinitesimal time interval dt is:

!

dp = pic dt (31)

Equation (31) clearly reveals that ic is a rate of interest. If time is measured in years, the units of i would be (yr)-1. In order to determine the future value f to which the present value p has grown during the time t we need to separate variables and integrate. This gives:

!

dp

pp

f

" = ic dt0

t

" (32)

or

f

p= e

ic t (33)

Consider, for example, the calculation of f/p for a time period of 4 years with continuous compounding at an interest rate ic equal to 0.12/yr. Equation (33) gives for this case:

f

p= e

(0 .12/ yr )(4yr )= e

0.48= 1.6161

In comparison, an interest rate of 0.12 with annual compounding gives: (f/p, 0.12, 4) = 1.5735, and an interest rate of 0.12 with monthly compounding gives: (f/p, 0.01, 48) = 1.6122.

The effective annual interest rate with continuous compounding is ie = e

( ic )(1yr ) !1 (34)

March 14, 2007 30

Discrete interest formulas with discrete monthly, quarterly, yearly, etc. payments can be used with continuous compounding if i is replaced by the appropriate value of ie. For example, with quarterly payments and continuous compounding at an interest rate of ic equal to 0.12/yr, ie for this quarter would be:

!

ie = e(0.12/ yr)( 0.25yr)

"1= 0.0305 per quarter Thus, p/a for the 10 year period, assuming a series of 40 quarterly discrete payments, is: (p/a, 0.0305, 40) = 22.9459. 3.6 Discussion and Summary of Interest Formulas

A summary of key interest factors and their limiting values is given in Table 3. The names of all factors are listed in the NOMENCLATURE section. The ratios f/p, f/a, p/a, p/g, and p/c can be used singly or in combination to convert any of the quantities, p, f, a, g, and c, to its equivalent value in terms of any of the four remaining quantities. For example, a present value p could be converted to: (i) a future value f, (ii) a uniform series a, (iii) an arithmetic series g, or (iv) a geometric series c. All of the basic ratios used in the conversion must be based on the given values of: i, n, and, if c is involved, r. The monetary unit and economic value remain unchanged. Only the time(s), the magnitudes of the amount(s), and possibly the number of amounts change.

March 14, 2007 31

Table 3 Interest Factors and Their Limiting Values

Factor

!

n"#

i > 0( )

!

i" 0

any n( )

!

i "#

any n( )

f

p= 1 + i( )

n= CAF( ) ∞ 1 ∞

f

a=(1 + i)n !1

i= (SCAF) ∞ n ∞

p

a=(1 + i)n !1

i (1 + i)n = (SPWF)

!

1

i n 0

p

g=1

i

1 + i( )n !1

i 1 + i( )n!

n

1+ i( )n" # $

% & '

or

!

p

g=1

i

p

a

" # $ % & n

p

f

"

# ' $

%

( ) *

+ , -

=GPWF

!

1

i2

!

n2 " n

2

0

!

p

c=

1+ r

1+ i

" #

$ %

n

&1

r & i= EPWF

!

", r = i

1

i # r, r < i

!

f

a, r, n

" #

$ % 0

In previous decades, engineering economic calculations were often worked using printed tables of interest functions. However, personal computers and programmable calculators now provide a much easier way to perform the calculations. Most spreadsheet programs have the main interest formulas available as built-in functions. For example, consider Eq. (10), which relates the present value p and future value f with a constant payment a over n periods with interest rate i,

!

p

f=

1

1+ i( )n

Given the values of any four of the five quantities (p, f, a, n, i), one can solve for the fifth quantity. Microsoft Excel offers the following functions to do this: PV returns the present value p FV returns the future value f

March 14, 2007 32

PMT returns the payment a NPER returns the number of periods n RATE returns the interest rate i The arguments for each function are the other four values. In Excel, go to Help … Excel Help Contents; Function Reference; Financial to see information on these and other financial functions. If you want to program your own calculator, the following relationships between these factors are useful.

!

f

a=

f

p

"

# $ %

& '1

i (35)

!

p

a=

f

a

" #

$ %

f

p

"

# & $

%

(36)

and

!

p

g=1

i

p

a"

n

f

p

#

$ % &

'

(

) *

+ *

,

- *

. *

(37)

The quantities, p/a, p/g, and p/c are examples of factors that collapse special distributions of cash flows into a single number: the present value of the distribution. Any cash flow distribution could be converted to its present value—a unique number that summarizes the cash flow distribution for a given i and n. On the other hand, there are, for given values of i and n, an infinite number of possible cash flow distributions; hence, equal present values in no way imply equal or similar distributions of cash flows. 4. OTHER CONSIDERATIONS 4.1 Depreciation

Depreciation is the decrease in value of property due to wear, deterioration, and obsolescence. Since income taxes are paid on net income less depreciation, an increase in depreciation produces a reduction in income tax. An investment tax credit has been given in some years by the federal government to stimulate economic growth. This tax credit is similar to

March 14, 2007 33

depreciation in that it effects a reduction in income tax; however, it leaves the book value of the property unchanged. The investment tax credit is the fraction of an investment outlay in a given year that can be claimed as a credit against income tax due for that year. Hence, the investment tax credit (it has been as high as 10 percent) effects a direct reduction in income tax; whereas, depreciation effects a reduction in taxable income.

Several methods of depreciation are used. Two of these will be considered here: the straight-line method and the sum-of-the-year's-digits method. Straight-line depreciation derives its name from the fact that the book value of the asset decreases linearly with time. The annual depreciation, D, is

!

D =FC " SV

n (38)

where FC is the first cost of the asset including delivery and installation costs; SV is the salvage value; and n, the number of years of the tax life of the object. n is a dimensionless number. The Internal Revenue Service specifies restrictions on the tax life of a piece of property; hence, the tax life may not be equal to the economic life of the asset. The book value, BV, after j years of service with the straight-line method of depreciation is

!

BVj = FC " j •D (39) The straight-line depreciation results in a uniform annual series of deductions; thus, D is analogous to the quantity a for this method. Hence, the present value of the deductions over the tax life of the asset is:

!

p = Dp

a, i%, n

" #

$ %

The sum-of-the-year's-digits method is an accelerated depreciation schedule. Specifically, the depreciation for the j-th year is

!

Dj =No. of years remaining

sum - of - year' s - digitsfirst cost " salvage value( ) (40)

The number of years remaining, (n - j + 1), includes the year for which the depreciation is being determined. The sum-of-the-year's-digits, SYD, is the sum (1+ 2 + … + n) which is equal to [(n)(n +1)/2]. Hence, Eq. (40) can be written as

March 14, 2007 34

!

Dj =2 n " j +1( )n n +1( )

FC " SV( ) (41)

The accelerated nature of this method is seen by comparing the fraction

!

Dj / FC " SV( ) versus j.

The first year this fraction is n/SYD; whereas the n-th year, it is 1/SYD. In contrast, this fraction is equal to 1/n and is independent of j in the straight-line method. The book value with the SYD depreciation schedule for the j-th year is

!

BVj = FC "j 2n " j +1( )n n +1( )

FC " SV( ) (42)

The derivation of Eq. (42) is left as an exercise. It will also be left as an exercise to show that the present value of all of the annual depreciations for the tax life of the asset with the SYD depreciation schedule is

!

p =2 FC " SV( )n n +1( )

n "p

a, i%, n

# $

% &

' ( )

* + ,

i (43)

This quantity is useful, for example, in the determination of the present value of the income tax reduction effected through the depreciation of the asset.

The Accelerated Cost Recovery System (ACRS) and the Modified Accelerated Cost Recovery System (MACRS) are accelerated depreciation schedules that are to be used in the United States for property placed into service after 1981 and 1986, respectively (see Lindeburg, 1995). The life of the property, that is, the recovery period n, is specified by the government according to the type of property under consideration, and the salvage value of the property is specified to be zero. The depreciation for the j-th year of an asset's cost recovery period is given by: Dj = (CRj) (First Cost) = CRj•FC, j = 1, 2, . . ., n, (n+1) (44) where the factors CRj are specified in Table 4 for recovery periods of 3, 5, 7 and 10 years. Note that the sum of CRj for j = 1 to (n+1) is 1.000. The book value at the end of the j-th year is:

BVj = FC ! Dk

k=1

j

" (45)

March 14, 2007 35

Table 4 MACRS Depreciation Factors, CRj (see Lindeburg, 1995)

j: 1 2 3 4 5 6 7 8 9 10 11 n = 3 0.333 0.445 0.148 0.074 n = 5 0.200 0.320 0.192 0.115 0.115 0.058 n = 7 0.143 0.245 0.175 0.125 0.089 0.089 0.089 0.045

n = 10 0.100 0.180 0.144 0.115 0.092 0.074 0.066 0.066 0.065 0.065 0.033 4.2 Inflation and Fuel Escalation Rates

The role of inflation in investment analysis, and its effect on the evaluation of the economic viability of alternative energy systems, is not clear. Typical introductory books on engineering economy do not consider the subject of inflation. The prevailing opinion appears to be that since most quantities inflate at a similar rate, the net effect on the merit of an investment is small or negligible.

Interest rates are, of course, closely related to the rate of inflation. Specifically, the rate of interest charged by lending organizations is composed of three components that account for inflation, risk, and profit. The influence of inflation on the future worth f of a present value p is to decrease it by a factor (1 + ri) each year where ri is the rate of inflation. If the inflation and

interest rates are constants, then the future worth of an investment p, after n years, taking into account the influences of inflation, is:

!

f = p1+ i

1+ ri

"

# $ %

& ' n

(46)

Without inflation or deflation, the future value was previously shown to be:

!

f = p 1+ i( )n (47)

Hence, the real interest rate, ir, with inflation ri > 0, or deflation, ri < 0, follows from eqs. (46) and (47) as:

!

1+ ir

( ) =1+ i

1+ ri

or

March 14, 2007 36

!

ir

=i " r

i

1+ ri

(48)

If ri is small, the real interest rate, Eq. (48), reduces to:

!

ir" i # r

i (49)

For fixed investments, businesses expect a real rate of return of about 2 to 5 percent whereas individuals expect a real rate of return of approximately 1 to 3 percent.

Inflation would similarly reduce the future cost c of an item such as fuel. If it is assumed that fuel, for example, is escalating at a rate rf, the fuel escalation rate, then the real fuel escalation rate, rrf, is

!

rrf =rf " ri

1+ ri (50)

Likewise, if the inflation rate is small,

!

rrf " rf # ri (51)

If real interest and escalation rates are used in an economic analysis, future values are obtained in terms of constant base year dollars. On the other hand, future values include the influence of inflation if actual interest and escalation rates are used. Costs in base year dollars are easier to comprehend. On the other hand, including the effects of inflation enables the use of actual interest and escalation rates and results in future cash flows that are close to those that actually occur.

Since some quantities, for example, fuel and electrical power costs, may increase at a rate greater than the rate of inflation, it is necessary to include these influences in estimates of relevant costs, for example, energy costs. Unfortunately, the prediction of escalation rates is difficult. The use of a range of escalation rates, a sensitivity analysis, is recommended in order to provide insight on the influences of escalating costs on the outcome of the economic analysis. 4.3 Life-Cycle Costing

Solar energy systems, for example, require a large capital investment in order to gather the "free" energy. On the other hand, fossil-fueled systems require a smaller initial investment but are experiencing continually increasing primary energy costs. Hence, alternative systems must be compared over a period of time, preferably over the economic life of the equipment. Such

March 14, 2007 37

costing is called life-cycle costing. Life-cycle cost analysis is the conversion of expenses that occur over the n time periods of interest into a single cost number that can be used to compare alternative systems.

Several meaningful cost numbers could be used to represent the total costs (or economic value) over the n time periods. The two most common methods of life-cycle costing are the present value method and the annual cost method. Consider, for example, the following cash flow diagram that describes the expenses associated with a particular system.

2 3 n-1 n

aaaaa

1

k,o k,1 k,2 k,3 k,n-1ak,n

The cash flow ak,j† is that which occurs for system k at the end of the j-th time period. In the present value method, the cash flows are converted to the present, that is, all costs throughout the lifetime of the equipment are treated as if they could be paid at the present. Hence, the cost number is the present value which for system k is

!

pk = ak, jj= 0

n

" 1+ i( )# j

= ak , jj= 0

n

" p

f, i%, j

$

% & '

( (52)

Equation (52) shows that a greater value is assigned to expenditures made at the present than those made sometime in the future due to the time value of money, that is, future values are discounted. Systems with large initial costs, ak,o, have to overcome this disadvantage. The system with the smallest present value, pk, is the best of the alternatives based on the provided costs, ak,j, and the assumed discount rate, i. It should be stressed, however, that the results of the cost analyses are only as good as the input data. The ak,j's are future values if j is greater than zero; hence, they are predicted quantities based on assumed economic conditions. For example, the assumed fuel costs and the assumed fuel escalation rate drastically influence the relative ordering of systems like nuclear, solar, and conventional power plants.

In the annual cost method, the costs throughout the n years of interest are converted to a uniform annual amount, a; that is, the costs are treated as if they occurred in n equal annual payments. Hence, the cost number for system k is the uniform annual amount, ak, which is

† The cash flows ak,j could be negative (a net expense) or positive. For example, ak,n may be positive due to the salvage value of the system.

March 14, 2007 38

!

ak = pka

p, i%, n

"

# $ %

& = ak , j

j= 0

n

' 1+ i( )( j)

* +

, - .

a

p, i%, n

"

# $ %

& (53)

The annual cost method is especially meaningful, for example, in heating and cooling studies because the annual cost, ak, divided by the amount of energy supplied during the year, Qk, provides an estimate of the average energy cost in $/GJ (or $/MBTU). This energy cost facilitates comparisons between conventional and alternative energy sources.

The difficult part of life-cycle costing is the determination of the ak,j's. These quantities are not only dependent on the system but are also dependent on the interest rate and the economic conditions, specifically, the general rate of inflation and the escalation rates of fuel costs, maintenance costs, capital costs, and other costs. There are two general ways to handle the influences of interest and escalating costs. The first is to determine the "real" costs in terms of constant base year dollars. The base year would usually correspond to n = 0. The interest rate in this case is the "real" interest rate, which can be approximated by (See eqs. 47 and 48):

!

Real Interest Rate "Annual Nominal

Market Discount Rate

# $ %

& ' ( )

Expected General

Inflation Rate

# $ %

& ' (

(54)

The Annual Nominal Market Discount Rate is the current rate charged by banks on loans. With this method, the projected changes in unit fuel costs and hourly maintenance costs, for example, are only those that occur due to differences between the general rate of inflation, ri, and the escalation rates, rf and rm respectively. That is, these costs are to be expressed in "constant"

dollars; hence, only "real" price changes are included.

The other way to determine the costs is to use the actual interest rates and respective escalation rates. The resultant costs, in this case, include the effects of inflation. Since the influence of inflation is implicit in the interest rates charged by banks, in fuel costs, in maintenance charges, etc., the cash flows found in this manner are close to those that actually occur. 4.4 Rate of Return Analysis

The two common methods of life-cycle costing, the present value method and the annual cost method, considered in Section 4.3 both require a specified interest rate. Given i, these methods are used, for example, in comparing the costs associated with two alternatives to yield the present worth of these costs and the equivalent uniform annual cost (EUAC). If only the benefits are considered, the present worth of these benefits and the equivalent uniform annual

March 14, 2007 39

benefit (EUAB) would be calculated and compared. If both the costs and the benefits are specified, the net present worth (NPW) would be calculated where

!

Net Present Worth " PW of Benefits{ } # PW of Costs{ } (55)

In contrast to the present value method and the annual cost method, the interest rate is the calculated quantity in the rate of return analysis. Specifically, the rate of return (RR), also called, the internal rate of return (IRR), is defined as the interest rate at which the benefits are equal to the costs; hence:

!

Net Present Worth = 0 or Equivalent Uniform Annual Cost = Equivalent Uniform Annual Benefit The interest rate paid by the borrower on a loan is the rate of return the lender receives from the loan. If one invested in capital equipment, the rate of return on this investment is analogous to the interest the lender receives from a loan. Example 4.1: A company is considering purchasing a truck for $80,000. Assume this purchase saves the company $12,000 each year for the 10 year life of the truck, and the salvage value of the truck at the end of this ten year period is $30,000. What is their rate of return on this investment?

Solution: The cash flow diagram is:

$80,000

1 2 9 10

$12,000

i = ?

$42,000

$12,000 $12,000

Equating the present worth of the costs to the present worth of the benefits gives: $80,000 = $12,000 (p/a, i, 10) + $30,000 (p/f, i, 10)

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The variation of the present worth of the benefits with i is given in the following table:

i % 0 10 % 11 % 11.3 %

PW of benefits $150,000 $85,301 $81,236 $80,075

Thus, the rate of return on their investment is 11.3%. Clearly, the accuracy of the assumptions (n = 10, SV = $30,000) does not justify the determination of four significant digits.

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5. PROBLEMS 1. The interest rate paid on one year certificates of deposit is advertised as 6.5 percent,

compounded daily, with an annual yield of 6.81 percent. (a) Calculate the effective annual yield with continuous compounding using an interest rate, ic, of 0.065/yr. Show that the lower annual yield with continuous compounding is a consequence of banks basing the daily interest rate on a year length of 360 days. That is, determine the corresponding annual yield with daily compounding in a 365 day year.

Answers: (a) 6.716%; (b) 6.812% 2. The high initial cost of a solar heating system is not easily balanced by the high future fuel

costs that are incurred with conventional fossil fuel systems if the discount rate is high. For example, determine the present value of a $1,000 fuel cost that occurs 20 years from today if the discount rate is 12%. Note: It is difficult to make accurate estimates of salvage values in a life-cycle cost analysis. This problem shows that a 100% error in the salvage value has the same influence as a 10% error in a present value if I is 12% and n is 20 years.

Answer: $103.67 3. What yearly deposit would have to be made at the end of each year for 12 years to

accumulate a fund of $12,000 on the last payment date. Assume an interest rate of: (a) 10% compounded yearly, (b) 10% compounded weekly. Determine also the total of the 12 deposits for both parts (a) and (b).

Answers: (a) $561.16; $6733.92; (b) $544.31; $6531.70 4. A $20,000 automobile loan at 10% interest, compounded monthly, is to be repaid in 48

monthly payments. (a) What is the monthly payment and the total amount repaid? (b) What monthly deposit would have to be make at the end of each month for 48 months in order to accumulate a fund of $20,000 on the last deposit date with which to purchase this automobile? Assume an interest rate of 6% compounded monthly is being paid on all deposits. What is the total of the 48 deposits? Compare your results with those from Part (a).

Answers: (a) $507.25; $24,348.08; (b) $369.70; $17,745.63 5.1 (a) What is the interest charged the first month on a $150,000 home mortgage if the

effective annual interest rate is 10 percent? (b) What is the monthly payment if the term of the mortgage is 25 years? (c) What is the total amount paid when the loan is fully amortized?

Answers: (a) $1,196; (b) $1,318; (c) $395,323 5.2 Assume, for simplicity, the homeowner who took out the $150,000 mortgage described in

Problem 5.1 (i) paid no down payment, (ii) sold the house after 5 years for $150,000, and (iii) paid the realtor a 6% sales commission. Compare the amount the homeowner received from this sale with the unpaid principal on the $150,000 mortgage.

Answer: Unpaid principal: $140,689; Received from sale: $141,000

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5.3 Assume the homeowner who took out the $150,000 mortgage described in Problem 5.1 is in the 31% Federal tax bracket, pays $4000/yr in real estate taxes on this home, and can deduct from his Federal taxable income all of his mortgage interest and real estate taxes. What reduction in Federal income tax paid by this home owner results from the first 12 mortgage payments and one $4000 real estate tax payment? Answer: $5,669

6. An interest charge of $9.50 per $100 of loan means that one must repay $109.50 per $100 of

loan in 12 equal monthly payments. What is the effective annual interest rate being charged? Answer: 18.5% 7. You are interested in giving your alma mater a gift that will support eight $1,000

undergraduate scholarships in perpetuity. (a) Assuming the university's endowments earn a constant rate of return of 10% per year what must be the amount of your gift? (b) Assuming the scholarships are to be increased annually to account for inflation (for example, the fraction of the board paid by the $1000 is to remain constant) and the net rate of return after inflation is 6%, a constant, what must be the amount of your gift?

Answers: (a) $80,000; (b) $133,333 8. Consider two proposed manufacturing plants: a conventional plant and a highly automated

one. Both plants are designed to have a life of 20 years and to meet the demand of 100,000 units a year. Neither plant is estimated to have a significant salvage value at the end of its 20 year life. Due to reduced labor costs, warranty work, and other cost reductions, the automated plant results in a fixed profit per unit that is $150 higher, than the conventional plant, the first 10 years and $50 higher the next 10 years. Assume all profits are realized at the end of the respective year and the discount rate to be applied to these future profits is 18% per year. (a) How much more can the company afford to invest in the highly automated plant? (b) Repeat Part (a) assuming that the profit per unit is reversed, that is, $50 per unit for the first 10 years and $150 per unit for the last ten years. Compare the answers to parts (a) and (b). Do your answers make sense? Note that if the resulting higher quality product is needed in order to remain competitive, the company will have to invest in the automated plant regardless of the differences projected with the stated assumptions.

9. You intend to have $20,000 available 6 years from the present to replace your car. Assume

you earn 6% interest compounded quarterly on your savings. 9.1 What semiannual deposit would result in the desired amount? Assume the deposits are

made at the end of each six-month period including the period ending 6 years from present.

9.2 What sum of money would you have to deposit today to have the required amount available at the end of 6 years?

9.3 If you borrowed $20,000 at the same interest rate and amortized the loan in 6 years with equal payments made at the end of each three-month period, what would be the quarterly payment?

9.4 Compare the total amounts paid in parts 9.1, 9.2, and 9.3.

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10. Assume that $200 is deposited at the end of each three-month period and an interest rate of 6%, compounded monthly, is received on all deposits. The number of months required to accumulate $5,000 is to be determined. 10.1 Derive an expression for the number of payments, n. Make your expression

applicable regardless of the signs used for the cash flows, a and f. 10.2 Determine the number of payments required to accumulate $5,000 and the amount of

the final payment. 11. Trucks that cost $50,000 are to be leased at the rate of $1,000 per month. The payments are

made at the beginning of each month. A 12% effective annual return is to be achieved with monthly compounding. 11.1 Derive an expression for the number of periods required to recover the original

investment at the specified rate of return. Maintenance costs, insurance, taxes, and salvage value are to be neglected. Make your expression applicable regardless of the signs used for the cash flows, a and p.

11.2 How many months are required to recover the original investment with the desired annual return?

12. A solar water and space heating system costing $5,000 results in a reduction in fuel costs at

current prices which is just equal to the monthly payments on the 20-year mortgage. The interest rate is 8 percent compounded monthly. What is the average monthly savings in fuel costs over the 20-year period if fuel costs escalate at monthly rates of 1/2, 2/3, and 1%, respectively? What is the monthly cost of the fuel displaced by the solar heating system at the end of the 20th year?

13. A pumping station costs $70,000 and has an expected economic life of 25 years. The

maintenance cost of the system is $500 the first year and this cost is expected to escalate at the rate of 3 percent per year. The taxes and insurance on the system are $800 per year, a constant. All costs can be assumed to be paid at the end of the year. What is the uniform annual cost of this system if the interest rate is 12 percent compounded annually?

14. Derive Eq. (48). 15. Derive Eq. (49). 16. Assume income taxes are paid each year during the n-year tax life of an asset at a tax rate of

50 percent. Derive an expression for the present value of the income tax reduction effected by using the sum-of-the-year's-digits method instead of the straight line method. If an asset costs $100,000 and has a 20-year tax life, what is the savings in income tax effected by using the SYD method of depreciation? Assume an interest rate of 9 percent compounded annually and a salvage value of zero.

17. Yearly payments are to be made on an n-year mortgage. The initial amount of the mortgage

is Mo, and the annually compounded rate of interest being charged is i.

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17.1 Derive an expression for the remaining principal at the end of the k-th year, Mk. 17.2 If n, i, and Mo are 15 years, 8%, and $160,000, respectively, what is the amount of

interest paid the seventh year? 18. Consider a piece of equipment that has an installed cost, FC. The equipment is purchased

with a cash down-payment, DP, and with the balance, Mo, supplied by taking out an nm year mortgage which has an annually compounded interest charge im. Assume the discount rate id represents the true value of money. Derive an expression for the present value of the cost of this equipment over its economic life, n, where n > nm.

19. The effective annual interest rate, ie, is specified. If a series of payments is being made at a

frequency of m payments per year, the interest rate, i, based on the time period between payments is required for use in the interest formulas. Derive an expression for this interest rate. Is your expression valid if m is less than unity, for example, biannual payments, m = 0.5? Is your expression valid if m is greater than one but not an integer, for example, a payment period of five months, m = 2.4?

20. The dependence of the economic value of money on time disappears as the interest, i,

approaches zero. Utilizing this fact and the respective cash flow diagrams for f/p, f/a, p/a, p/g, and p/c, deduce the limits of these quantities as i approaches zero. Describe verbally the basis for your answers. Also determine these limits mathematically from the expressions for the respective factors given in Table 3.

21. Determine the present value of the fuel savings if a car's fuel consumption could be cut in

half. Assume: (i) an infinite car life, (ii) an interest rate of 10% per year, (iii) an annual fuel cost of $600: for example, (12,000 miles/yr) x (20 miles/gallon) x ($1.20/gallon) = $600/yr. Discuss your result.

22. Assume the addition of 6" of insulation in the attic of a home would cost $500. The

estimated annual reduction in the heating and cooling costs of this home with this change is $200. Assume the annual heating and cooling costs occur at the end of the year. (a) What is the pay back period for this investment at an effective annual interest rate of 10%? What did the person ignore who arrived at a pay back period of 2.5 yrs? (b) If this $200 savings persists over the estimated 30 years of the remaining life of this home, what is the homeowners rate-of-return on this investment? (c) Repeat parts (a) and (b) if the annual saving is only $100.

Answers: (a) 3.02 yrs; (b) 40.0%; (c) 7.27 yrs; 19.9%. 23. The annual maintenance cost of a small manufacturing facility with an assumed life of 25

years is $2,500 the first year and $1,600 the second year. If this cost increases by $100 in each succeeding year (a3 = $1,700, a4 = $1,800, ...), what is the present value of these costs? Assume all costs occur at the end of the year and an interest rate of 10%, compounded annually. What is the equivalent uniform annual cost?

Answers: p = $21,294.29; a = $2,345.95

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24.1 Determine the monthly payment required to fully amortize a $165,000 home mortgage and

the total amount of interest paid when the loan is fully amortized. Assume: (a) the rate of interest, compounded monthly, is 7.5 percent, and the term of the mortgage is 15 years; (b) an 8% interest rate and a 30 year term. Why is the rate of interest higher for the 30 year term?

24.2 Assume this is your home; you elected the 15 year, 7.5% mortgage; you are in the 31%

federal income tax bracket; the real estate taxes paid on this home during the first year are $2500; and your taxable income is decreased by the total of the interest and real estate taxes paid. (a) Generate a spreadsheet showing for each of the 180 payments: the monthly interest paid, the monthly principal paid, the remaining unpaid principal, and the cumulative interest and principal paid. (b) Determine the reduction in your federal tax liability during the first twelve-month period. (c) Determine the net, after tax, monthly payment including the prorated real estate tax.

24.3 Denoting the amount of the initial mortgage as Mo, the remaining unpaid principal at the

end of the k-th month as Mk, and the rate of interest per month as i, derive expressions for Mk and the total interest paid, Itot,k, in the first k payments.

25. Mary and John, two engineering graduates of identical age, both start investment plans (for

example, an IRA) for their retirement by making $2000 annual deposits on their birthdays. Mary starts on her 21st birthday and makes her last investment on her 40th birthday. John, a procrastinator, starts 5 years later on his 26th birthday and makes a deposit every year for 40 years. Assume that they both retire on their 65th birthday and that John adds $2000 to his account on the day he retires.

25.1 If their accounts earned a uniform annual rate of return of 11%, determine the value of

Mary’s and John’s accounts at retirement. Determine also the ratios of the final account value to the total savings invested.

Answers: Mary: $1,744,451; Ratio: 43.61 John: $1,163,652; Ratio: 14.55 25.2 Repeat 25.1 only assume a uniform annual rate of return of 6 percent. 6. REFERENCES

Lindeburg, M. R., 1995, EIT Review Manual, Professional Publications, Inc., Belmont, CA., pp. 13-1 and 13-2.