2. 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
3. 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.
4. 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.
5. 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
6. Example 1:
• MARR = 13%
Alternative Investment Annual
Income
Annual
Cost
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
7. 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.
8. 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:
9. Recall Incremental Analysis
1. 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.
10. Example 2: Comparing cost alternatives
Alternative Investment Operating Cost
A $5000 $240
B $3000 $875
C $4000 $500
D $2500 $1000
11. 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
12. 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.
13. 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.
14. 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
15. Example 3: Projects with Benefits
Alternative Investment Net Annual
Benefits
Life
A $100,000 $20,000 9
B $85,000 $24,000 5
C $60,000 $18,000 5
• Choose one of the three
16. Example 3 (cont’d)
• MARR = 13% (select one and only one)
• Ranked Projects: C, B, A
Alternative Investment Net Annual
Benefits
Life
C $60,000 $18,000 5
B $85,000 $24,000 5
A $100,000 $20,000 9
17. 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
18. 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!!
19. 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
20. 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
21. 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.
22. 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
23. 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
24. Example 4
• MARR = 13% (select one or none)
Alternative Investment Net
Income
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
25. 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.
26. 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