Chapter 6 - Comparison and Selection among
Alternatives
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Comparing Alternative Projects
In Chapter 5, we learned how to evaluate the economic
profitability (acceptability) of a single alternative using various
measures (PW, AW, FW, IRR, ERR).
In Chapter 6, we will learn how to use these profitability
measures to evaluate and compare multiple alternatives.
If we have mutually exclusive alternatives, then we need to
compare the alternatives against each other, and we can select
at most one alternative. (Do-nothing may be an implicit
alternative.)
If we have independent alternatives, then we need to compare
each alternative separately against the do-nothing alternative
and decide whether it is acceptable or not. We can select more
than one alternative.
We will focus on mutually exclusive alternatives.
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Example 6.1: A company is considering two investment
alternatives (A and B) with the following cash flows:
Investment
Revenue next year
A
10,000
12,500
B
50,000
61,500
Suppose that the company uses an MARR of 20% and there is
no budget restriction. If the alternatives are mutually exclusive
what should the company do?
Let’s find the PW and IRR of each alternative.
IRR
PW
A
25%
416.67
B
23%
1,250
IRRA =25%≥ 20%, IRRB =23%≥ 20%, both alternatives are
profitable. A has higher return.
PW(MARR)B ≥ PW(MARR)A ≥ 0, both alternatives are
profitable. B has higher present worth. Which one to select?
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Remember the fundamental goal of capital investment: You
should invest an additional dollar only if it can obtain at least the
MARR. So, if the investment of the additional dollar has a
PW(MARR) greater than or equal to 0, then it is acceptable.
Alternative B requires higher investment and produces higher
revenue. Does this additional (incremental) next year’s revenue
of $61,500-$12,500=$49,000 economically justify the additional
(incremental) investment of $50,000-$10,000=$40,000 now?
Find the PW(MARR) of the difference (∆(B − A)):
PW(MARR)∆(B−A) =-40,000+49,000(P/F,20%,1)=833.3≥ 0. So,
it is justified. The company should select alternative B.
If we made the decision based on maximizing IRR, then the
conclusion would have been wrong! We will learn how to
correctly use the IRR method to identify the best alternative
later on.
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Types of decisions:
1
Investment alternatives
Each investment alternative requires initial investment and
produces positive cash flows (revenues, savings, etc).
Unless specified, we will assume there is an option of not
choosing any alternative, i.e., “do nothing".
2
Cost alternatives
Each cost alternative has only negative cash flows except
possibly the salvage value. You must choose one of the
alternatives, i.e., no “do nothing" option exists.
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Study period and useful lives of alternatives:
Study period is the time period over which the comparison of
alternatives is made.
Case 1. Study period=useful life (of all alternatives)
Case 2. Study period6=useful life (of at least one
alternative)
Rule: Compare mutually exclusive alternatives over the same
period of time.
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Case 1: Study period=useful life
We will use two different methods: Equivalent-Worth (EW)
methods (PW, AW, FW) and Rate-of-Return methods (IRR).
Equivalent-Worth methods:
Procedure to find the best alternative using the
equivalent-worth methods:
1
Find the equivalent worth (EW) of each alternative at
i%=MARR.
2
Investment alternatives: select the alternative with the
greatest positive equivalent worth.
3
Cost alternatives: select the alternative with the least
negative equivalent worth.
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Remark: PW, AW, FW are equivalent measures of project
acceptability (Chapter 5) because of the equivalency
relationships:
AW=PW(A/P,i%,N), AW=FW(A/F,i%,N), FW=PW(F/P,i%,N), etc.
If PW(i%)A <PW(i%)B , then
PW(i%)A (A/P,i%,N)<PW(i%)B (A/P,i%,N)⇒ AW(i%)A <AW(i%)B
and
PW(i%)A (F/P,i%,N)<PW(i%)B (F/P,i%,N)⇒ FW(i%)A <FW(i%)B .
Therefore, all EW methods will rank the alternatives in the
same order and they will find the same alternative as the best
one. (Consistency of alternative selection.)
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Example 6.2 (investment alternatives): Which one of the
following four mutually exclusive alternatives should be
selected if the planning horizon and the useful lives of all
alternatives are 10 years, and the MARR is 12% per year?
Capital investment
Annual cost savings
Salvage value
A
50,000
7,600
5,000
B
110,000
20,750
10,000
C
92,000
17,950
7,500
D
76,000
15,950
0
PW(12%)A =
-50,000+7,600(P/A,12%,10)+5,000(P/F,12%,10)=-$5,448
PW(12%)B =$10,461.65
PW(12%)C =$11,836.1
PW(12%)D =$14,129.7
PW(12%)do nothing = 0.
PW(12%)D > PW(12%)C > PW(12%)B >PW(12%)do nothing >
PW(12%)A .
Select Alternative D.
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Example 6.3 (cost alternatives): Which one of the following
four machines should be selected if the planning horizon and
the useful lives of all machines are 8 years, and the MARR is
12% per year?
Initial cost
Annual expenses
I
65,000
14,600
II
70,000
12,600
III
87,000
9,500
IV
68,000
16,950
Eliminate Machine IV since it has higher initial cost and higher
annual expenses than Machine I. We can use any EW method.
AW(12%)I =-65,000(A/P,12%,8)-14,600=-$27,684.5
AW(12%)II =-$26,691.2
AW(12%)III =-$27,013.4
Select the alternative with the least negative AW: Alternative II.
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Rate-of-Return methods:
Question: Can we select the alternative with the maximum IRR
as the best alternative? Answer: No! Look at Example 6.1:
Investment
Revenue next year
IRR
PW
A
10,000
12,500
25%
416.67
B
50,000
61,500
23%
1,250
If we maximize the IRR, then Alternative A is preferred. But, the
EW method selects Alternative B. Here, rate-of-return method
is not correctly used. How to correct it?
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Rules:
Compare an IRR only to MARR. Never compare the IRRs
of alternatives.
Never simply select the alternative that maximizes IRR.
Guidelines:
Each additional (incremental) capital must justify itself (i.e.,
must produce a rate of return greater than or equal to
MARR). We need to look at the increments; incremental
investment and incremental benefits.
Compare a higher investment alternative against a lower
“acceptable" investment alternative and check whether the
incremental investment is justified. To do this, we can
analyze the difference between the two alternatives, which
is usually an investment alternative by itself.
Select the alternative that requires largest capital
investment as long as the incremental investment is
justified. This will maximize EW(MARR).
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Back to Example 6.1:
Investment
Revenue next year
IRR
PW(20%)
A
10,000
12,500
25%
416.67
B
50,000
61,500
23%
1,250
∆(B-A)
40,000
49,000
22.5%
833.3
Alternative A is acceptable since 25%>MARR, and Alternative
B is a higher investment alternative. Analyze the difference
between the two alternatives, ∆(B-A). IRR of the difference
(incremental cash flow) is 22.5% >MARR. Additional $40,000
is justified and Alternative B is the preferred alternative.
This is consistent with the EW method conclusion:
PW(20%)∆(B−A) = 833.3 ≥ 0, therefore Alternative B is
preferred.
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If PW(MARR)∆ ≥ 0 then IRR∆ ≥ MARR. It is because PW(i)∆
is an decreasing function of i. So, the EW method and the IRR
method give consistent rankings of alternatives if we compare
the IRR of the incremental cash flow against the MARR.
Generalize this idea to a procedure.
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The incremental analysis procedure
Step 1. Order the alternatives by increasing capital
investment.
Step 2. Find a base alternative (current best alternative)
Cost alternatives: the first alternative in the ordered list (the
one with the least capital investment).
Investment alternatives: the first acceptable alternative in
the ordered list (EW≥ 0, or IRR≥MARR). The IRR of the
DN is assumed to be the MARR.
Step 3. Evaluate the difference between the next
alternative and the current best alternative.
If the incremental cash flow is acceptable, choose the next
alternative as the current best alternative.
Otherwise, keep the current best alternative and drop the
next alternative from further consideration.
Step 4. Repeat Step 3 until the last alternative is
considered. Select the current best alternative as the
preferred one.
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Example 6.4: Four mutually exclusive alternatives are
considered for constructing an office building near a parking lot.
The company determined a study period of 15 years. If the
MARR is 10%, which alternative should the company select?
Use the incremental analysis with the IRR method.
Investment
Annual income
Salvage value
P
200,000
22,000
100,000
B1
4,000,000
600,000
2,000,000
B2
5,500,000
720,000
2,775,000
B3
7,500,000
960,000
3,750,000
Step 1. DN-P-B1-B2-B3. (Alternatives are in the order of
increasing investment.)
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Step 2.
We have investment alternatives. IRR of DN alternative is 10%,
so it is the first acceptable alternative.
Step 3.
IRR of ∆(P-DN) is 9.3%, so it is unacceptable.
IRR of ∆(B1-DN) is 13.8%, so B1 is the current best alternative.
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Step 3 (cont.). Compare alternatives B2 and B1.
∆ Investment
∆ Annual income
∆ Salvage value
IRR∆
∆(B2-B1)
5,550,000-(4,000,000)=1,550,000
720,000-600,000=120,000
2,775,000-2,000,000=775,000
5.5%
(IRR∆ is the i’% such that
-1,550,000+120,000(P/A,i’%,15)+775,000(P/F,i’%,15)=0.)
Is the incremental cash flow acceptable?
No, because 5.5%<10%. Keep Alternative B1 as the current
best alternative and drop Alternative B2 from further
consideration. Go back to Step 3 and compare alternatives B3
and B1.
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∆ Investment
∆ Annual income
∆ Salvage value
IRR∆
∆(B3-B1)
7,500,000-(4,000,000)=3,500,000
960,000-600,000=360,000
3,750,000-2,000,000=1,750,000
8.5%
Is the incremental cash flow is acceptable?
No, because 8.5%<10%. Keep Alternative B1 as the current
best alternative and drop Alternative B3 from further
consideration.
We considered all alternatives, therefore stop.
Select Alternative B1, which is the current best alternative.
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Perform the PW analysis (Example 6.4) :
PW(10%)P =
-200,000+22,000(P/A,10%,15)+100,000(P/F,10%,15)=-$8,726.
PW(10%)B1 =$1,042,460
PW(10%)B2 =$590,727
PW(10%)B3 =$699,606
Select Alternative B1.
PW method is computationally easier than IRR method.
We can perform the incremental analysis using the PW method
to decide whether the incremental investment is justified:
PW(10%)∆(B2−B1) = −$451, 733 < 0, increment is not justified.
PW(10%)∆(B3−B1) = −$342, 854 < 0, increment is not justified.
Select Alternative B1.
Note: When there are multiple IRRs for some alternatives
(more than one sign changes in the cash flow), we can perform
the incremental analysis using the ERR method.
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Example (6.5)
Useful life of each alternative is 10 years and the MARR=10%.
Capital Investment
Annual net income
A
900
150
B
1,500
276
C
2,500
400
D
4,000
925
E
5,000
1,125
F
7,000
1,425
Which project should be selected?
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Solution
The incremental analysis proceeds as follows:
Find the IRRs:
IRR
A
10.6%
B
13%
C
9.6%
D
19.1%
E
18.3%
F
15.6%
The alternatives are already ranked in the order of increasing
capital investment. We can eliminate alternative C
(9.6%<10%). The incremental analysis starts with the current
best alternative as Alternative A. (The first acceptable project.)
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Compare A and B: Consider ∆(B-A). Incremental investment
for alternative B is justified. Current best alternative is B.
Compare D and B: Consider ∆(D-B). Incremental investment
for alternative D is justified. Current best alternative is D.
Compare E and D: Consider ∆(E-D). Incremental investment
for alternative E is justified. Current best alternative is E.
Compare F and E: Consider ∆(F-E). Incremental investment for
alternative F is not justified. Current best alternative is E.
Select the current best alternative: Alternative E.
∆ Capital Investment
∆ Annual net income
IRR∆
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∆(B-A)
600
126
16.4%
∆(D-B)
2,500
649
22.6%
∆(E-D)
1,000
200
15.1%
∆(F-E)
2,000
300
8.1%
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How to perform the incremental analysis for cost alternatives?
(All cash flows are negative except possibly the salvage
values.) Very similar to the analysis of investment alternatives.
Example (6.6): We consider four design alternatives each with
a useful life of 5 years. The MARR is 20%.
Capital Investment
Annual expenses
Salvage value
D1
100,000
29,000
10,000
D2
140,600
16,900
14,000
D3
148,200
14,800
25,600
D4
122,000
22,100
14,000
Increasing capital investment leads to decreasing annual
expenses. Which project should be selected?
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Solution
Start the incremental analysis:
D1-D4-D2-D3. Choose the first alternative, D1, as the current
best alternative.
Compare D4 and D1.
∆ Capital Investment
∆ Annual expenses (savings)
∆ Salvage value
IRR∆
∆(D4-D1)
122,000-(100,000)=22,000
22,100-(29,000)=-6,900
14,000-10,000=4,000
20.5%
∆(D4-D1) is an investment alternative. (An initial investment
produces positive cash flows.) The question is whether the
incremental capital investment can produce enough cost
savings to justify itself, i.e., lead to a return equal to or greater
than MARR.
The answer is yes since 20.5%>20%.
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The incremental analysis proceeds as follows:
Compare D2 and D4. IRR∆(D2−D4) = 12.3% < 20%. Increment
is not justified.
Compare D3 and D4. IRR∆(D3−D4) = 20.4% > 20%. Increment
is justified. Select Alternative D3.
∆ Capital Investment
∆ Annual expenses (savings)
∆ Salvage value
IRR∆
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∆(D2-D4)
18,600
-5,200
0
12.3%
∆(D3-D4)
26,200
-7,300
11,600
20.4%
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Case 2: Study period6= useful life (of at least one
alternative)
So far, we looked at situations where study periods are equal to
the useful lives of the alternatives. When the alternatives have
different useful lives, we need to make some modifications to
compare the alternatives over the same period of time, which is
needed for a fair comparison.
Example 6.7:
Investment
Annual revenues
Useful life (years)
A
5,250
1,882.5
4
B
7,500
2,220
6
We want to find the best alternative when the MARR=10%.
Let’s assume that we can repeat the same cash flow of each
alternative during its initial useful life span in its succeeding life
spans. This can be done by identical replacements of
alternatives at the end of their useful life spans.
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Based on this assumption, we can use a study period of 12
years (the least common multiple (LCM) of 4 and 6) where
Alternative A repeats itself 3 times and Alternative B repeats
itself twice during the entire period.
This assumption is called “repeatability", and it is useful for
situations where
the study period is very long (can be assumed infinity)
the study period is equal to a common multiple of the
alternatives’ lives.
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Let’s use the PW method to find the best alternative: (study
period=12 years)
PW(10%)A =-5,2505,250[(P/F,10%,4)+(P/F,10%,8)]+1,882.5(P/A,10%,12)=$1,542.
PW(10%)B =-7,5007,500(P/F,10%,6)+2,220(P/A,10%,12)=$3,393.
Select Alternative B.
Let’s find the AW(MARR) of each alternative using the
PW(MARR):
AW(10%)A =1,542(A/P,10%,12)=$226.5
AW(10%)B =3,393(A/P,10%,12)=$498. (Select Alternative B.
Conclusion is consistent with PW method as it should be.)
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What is the AW(MARR) of each alternative over its single
useful life span?
AW(10%)A =-5,250(A/P,10%,4)+1,882.5= $226.5
AW(10%)B =-7,500(A/P,10%,6)+2,220=$498
The AW of each alternative over its single life cycle is equal to
its AW over the entire study period. (The detail mathematical
proof is omitted.)
When repeatability assumption holds and the study period is a
common multiple of the alternatives’ lives, it is sufficient to
compare the AW values of the alternatives over their own useful
life.
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In Example 6.7, what if the study period of 12 years is not
acceptable and/or the repeatability assumption does not hold in
practice?
Suppose we are given a study period of 6 years. (Study period
> Useful life of A)
We need to adjust the cash flow of Alternative A to terminate at
the end of 6 years. Let’s assume that all cash flows of
Alternative A can be reinvested at the MARR from year 4 to
year 6.
Let’s use the FW method:
FW(10%)A =
[-5,250(F/P,10%,4)+1,882.5(F/A,10%,4)](F/P,10%,2)= $1,270.5
FW(10%)B =-7,500(F/P,10%,6)+2,200(F/A,10%,6)= $3,841.5
Alternative B is still preferred.
This was an investment alternatives situation. What if we have
cost alternatives?
Contract for the service or lease the equipment (at some cost)
for the remaining years.
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So far, we looked at alternatives with useful lives < study
period.
What if the useful life of an alternative is longer than the study
period?
The strategy is to truncate the alternative at the end of the
study period. We need to find the market value of the
alternative at the end of the study period, and we use the
imputed market value technique to do this.
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Let T denote the study period (T< useful life), and let MVT
denote the market value at time T. Then,
MVT =
PW at the end of year T of remaining capital recovery amount +
PW at the end of year T of original market (salvage) value at
end of useful life.
Formally, let I be the initial investment and S be the salvage
value of a project with a useful life of U time periods where
U>T. Let i%=MARR,
MVT =[I(A/P,i%,U)-S(A/F,i%,U)](P/A,i%,U-T)+S(P/F,i%,U-T).
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S
MVT
CR
CR
CR
U–1
U
…
0
T
T+1
CR = I(A/P, i%, U) – S(A/F, i%, U)
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Example 6.8: Consider a project with a capital investment of
$47,600, useful life of 9 years, annual expenses of $1,720, and
market (salvage) value of $5,000 at the end of its useful life.
What is the estimated market value of this project at the end of
year 5? The MARR is 20%.
MV5 =[47,600(A/P,20%,9)5,000(A/F,20%,9)](P/A,20%,4)+5,000(P/F,20%,4)=$32,361.
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How do we use the imputed market value in the comparison of
alternatives?
Example 6.9: Which of the following alternatives should be
selected if the study period is 5 years and the MARR is 20%?
Investment
Annual expenses
Useful life (years)
Salvage value
A
33,200
2,165
5
0
B
47,600
1,720
9
5,000
Alternative B is identical to the project in Example 6.8. The
imputed market value of Alternative B at the end of year 5 was
found to be $32,361. We can use the AW method to compare
the alternatives:
AW(20%)A = -33,200(A/P,20%,5)-2,165=-$13,267.
AW(20%)B =
-47,600(A/P,20%,5)-1,720+32,361(A/F,20%,5)=-$13,288.
Select Alternative A.
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