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84 Chapter 4 Present Worth Analysis

4.2 PRESENT WORTH ANALYSIS OF EQUAL-LIFE

ALTERNATIVES
In present worth analysis, the P value, now called PW, is calculated at the MARR
for each alternative. This converts all future cash flows into present dollar equiva-
lents. This makes it easy to determine the economic advantage of one alternative
over another.
The PW comparison of alternatives with equal lives is straightforward. If both
alternatives are used in identical capacities for the same time period, they are
termed equal-service alternatives.
For mutually exclusive alternatives the following guidelines are applied:
One alternative: Calculate PW at the MARR. If PW ⱖ 0, the
alternative is financially viable.
Two or more alternatives: Calculate the PW of each alternative at the
MARR. Select the alternative with the PW value that is numerically
largest, that is, less negative or more positive.
The second guideline uses the criterion of numerically largest to indicate a lower
PW of costs only or larger PW of net cash flows. Numerically largest is not the
absolute value because the sign matters here. The selections below correctly apply
this guideline.

Selected
PW1 PW2 Alternative

$⫺1500 $⫺500 2
⫺500 ⫹1000 2
⫹2500 ⫺500 1
⫹2500 ⫹1500 1

EXAMPLE 4.1 Perform a present worth analysis of equal-service machines with the costs
shown below, if the MARR is 10% per year. Revenues for all three alternatives
are expected to be the same.

Electric- Gas- Solar-

Powered Powered Powered

First cost, $ ⫺2500 ⫺3500 ⫺6000

Annual operating cost (AOC), $/year ⫺900 ⫺700 ⫺50
Salvage value, $ 200 350 100
Life, years 5 5 5
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4.2 Present Worth Analysis of Equal-Life Alternatives 85

Solution
These are cost alternatives. The salvage values are considered a “negative” cost,
so a  sign precedes them. The PW of each machine is calculated at i  10%
for n  5 years. Use subscripts E, G, and S.
PWE  2500  900(P兾A,10%,5)  200(P兾F,10%,5)  $5788
PWG  3500  700(P兾A,10%,5)  350(P兾F,10%,5)  $5936
PWS  6000  50(P兾A,10%,5)  100(P兾F,10%,5)  $6127
The electric-powered machine is selected since the PW of its costs is the low-
est; it has the numerically largest PW value.

Often a corporation or government obtains investment capital for projects by

selling bonds. A good application of the PW method is the evaluation of a bond
purchase alternative. If PW  0 at the MARR, the do-nothing alternative is selected.
A bond is like an IOU for time periods such as 5, 10, 20, or more years. Each bond
has a face value V of $100, $1000, $5000 or more that is fully returned to the pur-
chaser when the bond maturity is reached. Additionally, bonds provide the purchaser
with periodic interest payments I (also called bond dividends) using the bond coupon
(or interest) rate b, and c, the number of payment periods per year.
(bond face value)(bond coupon rate) Vb
I  [4.1]
number of payments per year c
At the time of purchase, the bond may sell for more or less than the face value,
depending upon the financial reputation of the issuer. A purchase discount is more
attractive financially to the purchaser; a premium is better for the issuer. For exam-
ple, suppose a person is offered a 2% discount for an 8% $10,000 20-year bond
that pays the dividend quarterly. He will pay $9800 now, and, according to Equa-
tion [4.1], he will receive quarterly dividends of I  $200, plus the $10,000 face
value after 20 years.
To evaluate a proposed bond purchase, determine the PW at the MARR of all cash
flows—initial payment and receipts of periodic dividends and the bond’s face value
at the maturity date. Then apply the guideline for one alternative, that is, if PW  0,
the bond is financially viable. It is important to use the effective MARR rate in the
PW relation that matches the time period of the payments. The simplest method is
the procedure in Section 3.4 for PP  CP, as illustrated in the next example.

Marcie has some extra money that she wants to place into a relatively safe EXAMPLE 4.2
investment. Her employer is offering to employees a generous 5% discount for
10-year $5,000 bonds that carry a coupon rate of 6% paid semiannually. The
expectation is to match her return on other safe investments, which have aver-
aged 6.7% per year compounded semiannually. (This is an effective rate of
6.81% per year.) Should she buy the bond?
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4.3 Present Worth Analysis of Different-Life Alternatives 87

4.3.2 Least Common Multiple (LCM)

This approach can result in unrealistic assumptions since equal service compari-
son is achieved by assuming:
ɀ The same service is needed for the LCM number of years. For example, the
LCM of 5- and 9-year lives presumes the same need for 45 years!
ɀ Cash flow estimates are initially expected to remain the same over each life
cycle, which is correct only when changes in future cash flows exactly match the
inflation or deflation rate.
ɀ Each alternative is available for multiple life cycles, something that is usually
not true.
Present worth analysis using the LCM method, as illustrated in Example 4.3, is
correct, but it is not advocated. The same correct conclusion is easier to reach
using each alternative’s life and an annual worth (AW) computation as discussed
in Chapter 5.

A project engineer with EnvironCare is assigned to start up a new office in a city EXAMPLE 4.3
where a 6-year contract has been finalized to collect and analyze ozone-level
readings. Two lease options are available, each with a first cost, annual lease cost,
and deposit-return estimates shown below. The MARR is 15% per year.

Location A Location B

First cost, $ 15,000 18,000

Annual lease cost, $ per year 3,500 3,100
Deposit return, $ 1,000 2,000
Lease term, years 6 9

a. EnvironCare has a practice of evaluating all projects over a 5-year period.

If the deposit returns are not expected to change, which location should be
selected?
b. Perform the analysis using an 8-year planning horizon.
c. Determine which lease option should be selected on the basis of a present
worth comparison using the LCM.
Solution
a. For a 5-year study period, use the estimated deposit returns as positive cash
flows in year 5.
PWA  15,000  3500(P兾A,15%,5)  1000(P兾F,15%,5)
 $26,236
PWB  18,000  3100(P兾A,15%,5)  2000(P兾F,15%,5)
 $27,397
Location A is the better economic choice.
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88 Chapter 4 Present Worth Analysis

b. For an 8-year study period, the deposit return for B remains at $2000 in
year 8. For A, an estimate for equivalent service for the additional 2 years is
needed. Assume this is expected to be relatively expensive at $6000 per year.
PWA  15,000  3500(P兾A,15%,6)  1000(P兾F,15%,6)
 6000(P兾A,15%,2)(P兾F,15%,6)
 $32,030
PWB  18,000  3100(P兾A,15%,8)  2000(P兾F,15%,8)
 $31,257
Location B has an economic advantage for this longer study period.
c. Since the leases have different terms, compare them over the LCM of
18 years. For life cycles after the first, the first cost is repeated at the begin-
ning (year 0) of each new cycle, which is the last year of the previous cycle.
These are years 6 and 12 for location A and year 9 for B. The cash flow dia-
gram is in Figure 4.2.
PWA  15,000  15,000(P兾F,15%,6)  1000(P兾F,15%,6)
 15,000(P兾F,15%,12)  1000(P兾F,15%,12)
 1000(P兾F,15%,18)  3500(P兾A,15%,18)
 $45,036
PWB  18,000  18,000(P兾F,15%,9)  2000(P兾F,15%,9)
 2000(P兾F,15%,18)  3100(P兾A,15%,18)
 $41,384
Location B is selected.

FIGURE 4.2 PWA = ?

Cash flow diagram
for different-life $1000 $1000 $1000
alternatives, 1 2 6 12 16 17 18
Example 4.3c.

$3500

$15,000 $15,000 $15,000

Location A

PWB = ?

$2000 $2000
1 2 9 16 17 18

$3100

$18,000 $18,000
Location B
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4.4 Capitalized Cost Analysis 91

three county judges agreed to purchase the software system. If the new system
will be used for the indefinite future, find the equivalent value (a) now and
(b) for each year hereafter.
The system has an installed cost of $150,000 and an additional cost of
$50,000 after 10 years. The annual software maintenance contract cost is $5000
for the first 4 years and $8000 thereafter. In addition, there is expected to be a
recurring major upgrade cost of $15,000 every 13 years. Assume that i  5%
per year for county funds.

Solution
a. The detailed procedure is applied.
1. Draw a cash flow diagram for two cycles (Figure 4.3).
2. Find the present worth of the nonrecurring costs of $150,000 now and
$50,000 in year 10 at i  5%. Label this CC1.
CC1  150,000  50,000(P兾F,5%,10)  $180,695
3. Convert the recurring cost of $15,000 every 13 years into an annual
worth A1 for the first 13 years.
A1  15,000(A兾F,5%,13)  $847
The same value, A1  $847, applies to all the other 13-year periods
as well.
4. The capitalized cost for the two annual maintenance cost series may be
determined in either of two ways: (1) consider a series of $5000 from
now to infinity plus a series of $3000 from year 5 on; or (2) a series
of $5000 for 4 years followed by a series of $8000 from year 5 to
infinity. Using the first method, the annual cost (A2) is $5000 forever.

i = 5% per year

0 2 4 6 8 10 12 14 20 26 Year

$5000
$8000

$15,000 $15,000
$50,000

$150,000

FIGURE 4.3 Cash flows for two cycles of recurring costs and all nonrecurring amounts,
Example 4.4.
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92 Chapter 4 Present Worth Analysis

The capitalized cost CC2 of $3000 from year 5 to infinity is found

using Equation [4.2] times the P兾F factor.
3000
CC2  (PF,5%,4)  $49,362
0.05
The CC2 value is calculated using n  4 because the present
worth of the annual $3000 cost is located in year 4, one period ahead
of the first A. The two annual cost series are converted into a capital-
ized cost CC3.
A1  A2 847  (5000)
CC3    $116,940
i 0.05
5. The total capitalized cost CCT is obtained by adding the three CC
values.
CCT  180,695  49,362  116,940  $346,997
b. Equation [4.3] determines the A value forever.
A  CCT (i)  $346,997(0.05)  $17,350
Correctly interpreted, this means Marin County officials have committed
the equivalent of $17,350 forever to operate and maintain the property
appraisal software.

The CC evaluation of two or more alternatives compares them for the same
number of years—infinity. The alternative with the smaller capitalized cost is the
more economical one.

EXAMPLE 4.5 Two sites are currently under consideration for a bridge over a small river. The
north site requires a suspension bridge. The south site has a much shorter span,
allowing for a truss bridge, but it would require new road construction.
The suspension bridge will cost $500 million with annual inspection and
maintenance costs of $350,000. In addition, the concrete deck would have to be
resurfaced every 10 years at a cost of $1,000,000. The truss bridge and approach
roads are expected to cost $250 million and have annual maintenance costs of
$200,000. This bridge would have to be painted every 3 years at a cost of
$400,000. In addition, the bridge would have to be sandblasted every 10 years
at a cost of $1,900,000. The cost of purchasing right-of-way is expected to be
$20 million for the suspension bridge and $150 million for the truss bridge.
Compare the alternatives on the basis of their capitalized cost if the interest rate
is 6% per year.
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5.1 AW Value Calculations 109

5.1 AW VALUE CALCULATIONS

The annual worth (AW) method is commonly used for comparing alternatives. All
cash flows are converted to an equivalent uniform annual amount over one life
cycle of the alternative. The AW value is easily understood by all since it is stated
in terms of dollars per year. The major advantage over all other methods is that
the equal service requirement is met without using the least common multiple
(LCM) of alternative lives. The AW value is calculated over one life cycle and is
assumed to be exactly the same for any succeeding cycles, provided all cash flows
change with the rate of inflation or deflation. If this cannot be reasonably assumed,
a study period and specific cash flow estimates are needed for the analysis. The
repeatability of the AW value over multiple cycles is demonstrated in Example 5.1.

New digital scanning graphics equipment is expected to cost $20,000, to be used EXAMPLE 5.1
for 3 years, and to have an annual operating cost (AOC) of $8000. Determine
the AW values for one and two life cycles at i  22% per year.
Solution
First use the cash flows for one life cycle (Figure 5.1) to determine AW.
AW  20,000(A兾P,22%,3)  8000  $17,793
For two life cycles, calculate AW over 6 years. Note that the purchase for the
second cycle occurs at the end of year 3, which is year zero for the second life
cycle (Figure 5.1).
AW  20,000(A兾P,22%,6)  20,000(P兾F,22%,3)(A兾P,22%,6)  8000
 $17,793
The same AW value can be obtained for any number of life cycles, thus demon-
strating that the AW value for one cycle represents the equivalent annual worth
of the alternative for every cycle.
FIGURE 5.1
0 1 2 3 4 5 6 Year
Cash flows over
two life cycles of
an alternative.

$8000 $8000

$20,000
Life Life
cycle 1 cycle 2

$20,000
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7.2 Benefit/Cost Analysis of a Single Project 167

The conventional B/C ratio is the most widely used. It subtracts disbenefits
from benefits.
benefits  disbenefits BD
B/C   [7.2]
costs C
The B/C value would change considerably were disbenefits added to costs. For
example, if the numbers 10, 8, and 5 are used to represent the PW of benefits, dis-
benefits, and costs, respectively, Equation [7.2] results in B/C  (10  8)兾5 
0.40. The incorrect placement of disbenefits in the denominator results in B/C 
10兾(8  5)  0.77, which is approximately twice the correct B/C value. Clearly,
then, the method by which disbenefits are handled affects the magnitude of the B/C
ratio. However, no matter whether disbenefits are (correctly) subtracted from the
numerator or (incorrectly) added to costs in the denominator, a B/C ratio of less
than 1.0 by the first method will always yield a B/C ratio less than 1.0 by the sec-
ond method, and vice versa.
The modified B/C ratio places benefits (including income and savings), dis-
benefits, and maintenance and operation (M&O) costs in the numerator. The
denominator includes only the equivalent PW, AW, or FW of the initial investment.
benefits  disbenefits  M&O costs
Modified B/C  [7.3]
initial investment
Salvage value is included in the denominator with a negative sign. The modified
B/C ratio will obviously yield a different value than the conventional B/C method.
However, as discussed above with disbenefits, the modified procedure can change
the magnitude of the ratio but not the decision to accept or reject the project.
The benefit and cost difference measure of worth, which does not involve a
ratio, is based on the difference between the PW, AW, or FW of benefits (includ-
ing income and savings) and costs, that is, B  C. If (B  C)  0, the project is
acceptable. This method has the advantage of eliminating the discrepancies noted
above when disbenefits are regarded as costs, because B represents net benefits.
Thus, for the numbers 10, 8, and 5 the same result is obtained regardless of how
disbenefits are treated.
Subtracting disbenefits from benefits: B  C  (10  8)  5  3
Adding disbenefits to costs: B  C  10  (8  5)  3

The Ford Foundation expects to award $15 million in grants to public high schools EXAMPLE 7.2
to develop new ways to teach the fundamentals of engineering that prepare stu-
dents for university-level material. The grants will extend over a 10-year period
and will create an estimated savings of $1.5 million per year in faculty salaries
and student-related expenses. The Foundation uses a discount rate of 6% per year.
This grants program will share Foundation funding with ongoing activities,
so an estimated $200,000 per year will be removed from other program funding.
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168 Chapter 7 Benefit/Cost Analysis and Public Sector Projects

To make this program successful, a $500,000 per year operating cost will be
incurred from the regular M&O budget. Use the B/C method to determine if the
grants program is economically justified.
Solution
Use annual worth as the common monetary equivalent. For illustration only, all
three B/C models are applied.
AW of investment cost. $15,000,000(A兾P,6%,10) ⫽ $2,038,050 per year
AW of M&O cost. $500,000 per year
AW of benefit. $1,500,000 per year
AW of disbenefit. $200,000 per year
Use Equation [7.2] for conventional B/C analysis, where M&O is placed in the
denominator as an annual cost. The project is not justified, since B/C ⬍ 1.0.
1,500,000 ⫺ 200,000 1,300,000
B/C ⫽ ⫽ ⫽ 0.51
2,038,050 ⫹ 500,000 2,538,050
By Equation [7.3] the modified B/C ratio treats the M&O cost as a reduc-
tion to benefits.
1,500,000 ⫺ 200,000 ⫺ 500,000
Modified B/C ⫽ ⫽ 0.39
2,038,050
For the (B ⫺ C) model, B is the net benefit, and the annual M&O cost is
included with costs.
B ⫺ C ⫽ (1,500,000 ⫺ 200,000) ⫺ (2,038,050 ⫺ 500,000)
⫽ $⫺1.24 million

7.3 INCREMENTAL B/C EVALUATION OF TWO OR

MORE ALTERNATIVES
Incremental B/C analysis of two or more alternatives is very similar to that for
incremental ROR analysis in Chapter 6. The incremental B/C ratio, ⌬B/C, between
two alternatives is based upon the PW, AW, or FW equivalency of costs and ben-
efits. Selection of the survivor of pairwise comparison is made using the follow-
ing guideline:
If ⌬B/C ⱖ 1, select the larger-cost alternative.
Otherwise, select the lower-cost alternative.
Note that the decision is based upon incrementally justified total costs, not incre-
mentally justified initial cost.
There are several special considerations for B/C analysis of multiple alternatives
that make it slightly different from ROR analysis. As mentioned earlier, all costs have
a positive sign in the B/C ratio. Also, the ordering of alternatives is done on the basis
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170 Chapter 7 Benefit/Cost Analysis and Public Sector Projects

EXAMPLE 7.3 The city of Garden Ridge, Florida has received two designs for a new wing to
the municipal hospital. The costs and benefits are the same in most categories,
but the city financial manager decided that the following estimates should be
considered to determine which design to recommend at the city council meet-
ing next week.

Design 1 Design 2

Construction cost, $ 10,000,000 15,000,000

Building maintenance cost, $/year 35,000 55,000
Patient benefits, $/year 800,000 1,050,000

The patient benefit is an estimate of the amount paid by an insurance carrier,

not the patient, to occupy a hospital room with the features included in the
design of each room. The discount rate is 5% per year, and the life of the addi-
tion is estimated at 30 years.
a. Use conventional B/C ratio analysis to select design 1 or 2.
b. Once the two designs were publicized, the privately owned hospital in the
adjacent city of Forest Glen lodged a complaint that design 1 will reduce its
own municipal hospital’s income by an estimated $600,000 per year because
some of the day-surgery features of design 1 duplicate its services. Subse-
quently, the Garden Ridge merchants’ association argued that design 2 could
reduce its annual revenue by an estimated $400,000 because it will eliminate
an entire parking lot used for short-term parking. The city financial manager
stated that these concerns would be entered into the evaluation. Redo the B/C
analysis to determine if the economic decision is still the same.
Solution
a. Apply the incremental B/C procedure with no disbenefits included and
direct benefits estimated.
1. Since most of the cash flows are already annualized, ⌬B/C is based on
AW values. The AW of costs is the sum of construction and mainte-
nance costs.
AW1 ⫽ 10,000,000(AⲐP,5%,30) ⫹ 35,000 ⫽ $685,500
AW2 ⫽ 15,000,000(AⲐP,5%,30) ⫹ 55,000 ⫽ $1,030,750
2. Since the alternatives have direct benefits estimated, the DN option is
added as the first alternative with AW of costs and benefits of $0. The
comparison order is DN, 1, 2.
3. The comparison 1–to–DN has incremental costs and benefits exactly
equal to those of alternative 1.
4. Calculate the incremental B/C ratio.
¢B/C ⫽ 800,000Ⲑ685,500 ⫽ 1.17
5. Since 1.17 ⬎ 1.0, design 1 is the survivor over DN.
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8.1 Breakeven Analysis for a Single Project 187

$ FIGURE 8.3
TC Breakeven points and
maximum profit point
for a nonlinear
Profit R analysis.
maximized

Profit range

Loss Loss

QBE QP QBE
Units, Q

If nonlinear R or TC models are used, there may be more than one breakeven
point. Figure 8.3 presents this situation for two breakeven points. The maximum
profit occurs at QP where the distance between the R and TC curves is greatest.

Nicholea Water LLC dispenses its product Nature’s Pure Water via vending EXAMPLE 8.1
machines with most current locations at food markets and pharmacy or chemist
stores. The average monthly fixed cost per site is $900, while each gallon costs
18¢ to purify and sells for 30¢. (a) Determine the monthly sales volume needed
to break even. (b) Nicholea’s president is negotiating for a sole-source contract
with a municipal government where several sites will dispense larger amounts.
The fixed cost and purification costs will be the same, but the sales price per
gallon will be 30¢ for the first 5000 gallons per month and 20¢ for all above
this threshold level. Determine the monthly breakeven volume at each site.
Solution
a. Use Equation [8.2] to determine the monthly breakeven quantity of
7500 gallons.
900
QBE   7500
0.30  0.18
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9.3 Performing a Replacement Study 221

Replacement study FIGURE 9.2

Overview of
replacement study
approaches.

No study period Study period

specified specified

Find AWD and Calculate PW, AW,

AWC using ESL or FW for D and C
over study period

Select better AW Select better option

determined in the ESL analysis to apply the following replacement study

procedure. This assumes the services provided by the defender could be obtained
at the AWD amount.
New replacement study:
1. On the basis of the better AWC or AWD value, select the challenger alterna-
tive (C) or defender alternative (D). When the challenger is selected, replace
the defender now, and expect to keep the challenger for nC years. This replace-
ment study is complete. If the defender is selected, plan to retain it for up
to nD more years, but next year, perform the following analysis.
One-year-later analysis:
2. Are all estimates still current for both alternatives, especially first cost, mar-
ket value, and AOC? If no, proceed to step 3. If yes and this is year nD, replace
the defender. If this is not year nD, retain the defender for another year and
repeat this same step. This step may be repeated several times.
3. Whenever the estimates have changed, update them, perform new ESL analy-
ses, and determine new AWC and AWD values. Initiate a new replacement
study (step 1).
If the defender is selected initially (step 1), estimates may need updating after
1 year of retention (step 2). Possibly there is a new best challenger to compare
with D. Either significant changes in defender estimates or availability of a new
challenger indicates that a new replacement study is to be performed. In actuality,
a replacement study can be performed each year to determine the advisability of
replacing or retaining any defender, provided a competitive challenger is available.

Two years ago, Toshiba Electronics made a $15 million investment in new EXAMPLE 9.3
assembly line machinery. It purchased approximately 200 units at $70,000 each
and placed them in plants in 10 different countries. The equipment sorts, tests,
and performs insertion-order kitting on electronic components in preparation for
special-purpose printed circuit boards. A new international industry standard
requires a $16,000 additional cost next year (year 1 of retention) on each unit
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222 Chapter 9 Replacement and Retention Decisions

in addition to the expected operating cost. Due to the new standards, coupled
with rapidly changing technology, a new system is challenging these 2-year-old
machines. The chief engineer at Toshiba USA has asked that a replacement
study be performed this year and each year in the future, if need be. At i ⫽ 10%
and with the estimates below, do the following:
a. Determine the AW values and economic service lives necessary to perform
the replacement study.
Challenger: First cost: $50,000
Future market values: decreasing by 20% per year
Estimated retention period: no more than 5 years
AOC estimates: $5000 in year 1 with increases of
$2000 per year thereafter
Defender: Current international market value: $15,000
Future market values: decreasing by 20% per year
Estimated retention period: no more than 3 more years
AOC estimates: $4000 next year, increasing by $4000
per year thereafter, plus the extra $16,000 next year
b. Perform the replacement study now.
c. After 1 year, it is time to perform the follow-up analysis. The challenger is
making large inroads into the market for electronic components assembly
equipment, especially with the new international standards features built in.
The expected market value for the defender is still $12,000 this year, but it
is expected to drop to virtually nothing in the future—$2000 next year on
the worldwide market and zero after that. Also, this prematurely outdated
equipment is more costly to keep serviced, so the estimated AOC next year
has been increased from $8000 to $12,000 and to $16,000 two years out.
Perform the follow-up replacement study analysis.
Solution
a. The results of the ESL analysis, shown in Table 9.2, include all the market
values and AOC estimates for the challenger in the top of the table. Note
that P ⫽ $50,000 is also the market value in year 0. The total AW of costs
is shown by year, should the challenger be placed into service for that num-
ber of years. As an example, if the challenger is kept for 4 years, AW4 is
Total AW4 ⫽ ⫺50,000(A兾P,10%,4) ⫹ 20,480(A兾F,10%,4)
⫺[5000 ⫹ 2000 (A兾G,10%,4)]
⫽ $⫺19,123
The defender costs are analyzed in the same way in Table 9.2 (bottom) up
to the maximum retention period of 3 years.
The lowest AW cost (numerically largest) values for the replacement
study are
Challenger: AWC ⫽ $⫺19,123 for nC ⫽ 4 years
Defender: AWD ⫽ $⫺17,307 for nD ⫽ 3 years