05.2comp ror
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TRANSCRIPT
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Comparing Alternatives with Rate of Return
Use Incremental Analysis
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Topics
In this section, we • recall the definition of ROR,• discuss decision situations for multiple
alternatives, and• discuss the appropriate decision
methodology for each situation
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Rate of Return
• Recall that the ROR of an investment is the interest rate that makes
• NPW = 0• NAW = 0• PW Benefits = PW Costs• AW Benefits = AW Costs• For a single project, accept if ROR ≥
MARR.
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Decisions Involving Multiple Projects
Given a set of possible projects,• any subset of the projects may be
selected.• only one may be selected, but one must
be chosen.• only one may be selected, but it is OK to
choose none.
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Non-Mutually Exclusive Alternatives
• Suppose we can select any subset of the projects.
• Solution Methodology:– Compute the ROR of each alternative– Select each alternative for which ROR ≥ MARR
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Example 1:
• MARR = 13%
Alternative Investment AnnualIncome
AnnualCost
Life
A $100,000 $50,000 $30,000 9
B $85,000 $44,000 $20,000 5
C $60,000 $30,000 $12,000 5
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Analysis
– ROR of A = 13.7%– ROR of B = 12.7%– ROR of C = 15.3%
• Both A and C have ROR larger than our MARR. Therefore, select both A and C.
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Mutually Exclusive Alternatives
• Suppose we must select one, only one, but at least one, alternative.
• Solution Methodology:– Each increment of investment must yield the
MARR.– Perform Incremental Analysis:
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Recall Incremental Analysis1. Rank the alternatives in increasing order of
investment.2. Select as the defender the alternative with
the smallest investment.3. Let the challenger be the alternative with
the next higher investment.4. Accept or reject the challenger on the basis
of the return on the extra investment. The winner becomes the defender.
5. If the highest level of investment has been reached, stop. Otherwise return to step 3.
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Example 2: Comparing cost alternativesAlternative Investment Operating Cost
A $5000 $240
B $3000 $875
C $4000 $500
D $2500 $1000
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Ranked Alternatives
• MARR = 15%• Alternatives: All have five-year life and no
salvage• Ranked Alternatives : Rank by order of
increasing investment: D, B, C, A.Alternative Investment Operating Cost
D $2500 $1000
B $3000 $875
C $4000 $500
A $5000 $240
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Example 2: Incremental Analysis
• Defender: Project D.• Next greater investment (Challenger):
Project B. • Is the extra investment in B over D
justified?– Incremental Investment: -$500– Incremental Benefit: $125– NAW= -500 (A/P, i, 5) + 125 => ROR = 8% <
MARR– Decision: Reject Challenger, Project B.
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Example 2: Incremental Analysis (cont’d)
• Defender: Project D.• Next greater investment (Challenger):
Project C. • Is the extra investment in C over D
justified?– Incremental Investment: -$1500– Incremental Benefit: $500– NAW= -1500 (A/P, i, 5) + 500 => ROR = 20%
> MARR– Decision: Accept Challenger, Project C.
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Example 2: Incremental Analysis (cont’d)
• Defender: Project C.• Next greater investment (Challenger):
Project A. • Is the extra investment in A over C justified?
– Incremental Investment: -$1000– Incremental Benefit: $260– NAW= -1000 (A/P, i, 5) + 260 => ROR = 9% <
MARR– Decision: Reject Challenger, – Keep Defender, Project C
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Example 3: Projects with Benefits
Alternative Investment Net AnnualBenefits
Life
A $100,000 $20,000 9
B $85,000 $24,000 5
C $60,000 $18,000 5
• Choose one of the three
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Example 3 (cont’d)
• MARR = 13% (select one and only one)• Ranked Projects: C, B, A
Alternative Investment Net AnnualBenefits
Life
C $60,000 $18,000 5
B $85,000 $24,000 5
A $100,000 $20,000 9
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Example 3: Incremental Analysis
• Defender: Project C• Challenger: Project B• Is the extra investment in B over C
justified?– Incremental Investment: -$25,000– Incremental Benefit: $6000– NAW= -25 (A/P, i, 5) + 6 => ROR = 6.4% <
MARR– Decision: Reject Challenger, Keep Project C
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Example 3: Incremental Analysis (cont’d)
• Defender: Project C• Challenger: Project A• Is the extra investment in A over C
justified?– Incremental Investment: -$40,000– Incremental Benefit: $2000– Useful Life: Project lives are different!!
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Example 3: Incremental Analysis (cont’d)
• Select a common study period, say 45 years.
A
…
10 20 30 40
C
100,000
20,000
18,000
60,000
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Example 3: Incremental Analysis (cont’d)
• Compute the incremental cash flows.
• The cash flows of A-C represent a non-simple investment.
A-C 10 20 30 40
40,000
2,000
60,000
60,000
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Example 3: Incremental Analysis (cont’d)
• We must verify graphically (or with another method) that the interest found correspond to a rate of return. Luckily, it does. The rate of return of the incremental cash flows is 11.6%.
• Therefore reject the challenger, project A and accept C.
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An Easier way for Different Lives
• Find ROR of A - C• NAW(A - C) = NAW(A) - NAW(C) = 0• -100(A/P, i, 9) + 20 - [-60(A/P, i, 5) + 18]• -100(A/P, i, 9) + 60(A/P, i, 5) + 2 = 0• i NAW(A - C)• 0% 2889• 5% 1789• 10% 463• 15% -1058• 12% -123• ROR of A - C is 11.6%. Reject A - C and choose C
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Mutually Exclusive Alternatives with a Do-Nothing Alternative
• Suppose we can select one alternative or no alternative
• Solution Methodology: – Compute ROR of each alternative– Reject any alternatives that do not yield
the MARR– If only one remains, choose that
alternative– If more than one remains, do incremental
analysis to select the best
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Example 4
• MARR = 13% (select one or none)
Alternative Investment NetIncome
Life ROR Life
A $100,000 $20,000 9 13.7% 9
B $85,000 $24,000 5 12.7% 5
C $60,000 $18,000 5 15.3% 5
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Example 4 (cont’d)
• Since B does not return 13% it can be discarded.
• We most compare A and C, by computing the rate of return of the extra investment of A over C.
• The rate of return of A over C is 11.6%. Since the return on the extra investment is less than 13% (the MARR) reject the extra investment.
• We should choose option C.
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Conclusion:
• Measure of merit is the ROR (a percentage)
• For a single alternative, accept if ROR ≥ MARR
• For multiple alternatives: Accept an increment if the ROR for the increment≥ MARR