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```									       Chapter 8
Rate of Return Analysis: Multiple
Alternatives
Alternative Comparison (Single Alternative)

For a single alternative/project

NPW >0 select the project
FW >0 select the project
AW >0 select the project
IRR >MARR select the project.
Multiple Projects
• NPWA > NPWB select A
• FWA > FWB select A
• AWA > AWB select A

• IRRA > IRRB select A??
Example
•   Alternative/Year         A     B
•     0                    -100   -200
•     1                     110    226
•   For A, NPW=0
•   -100 + 110(P/F i, 1)=0
•   IRRA=10%
•   For B
•   -200 + 226(P/F, i, 1) =0
•   IRRB = 13%
•   A or B?
The Challenge
• Assume with additional \$100 you can invest in a
project with MARR=20%.

•   Alternative/Year     A 1 A2      A    B
•     0                -100 -100    -200 -200
•     1                 110 120      230 226
•   Now
•   RORA = 15%
•   RORB = 13%
Example
• Suppose we have \$100,000 to spend and we have two
mutually exclusive investment alternatives both of
which yield returns greater than MARR = 15%.
• Alternative/Year      A        B
•      0            -50,000 -90,000
•      1             60,000 106,200
• RORA = 20%
• RORB = 18%
• Select A?
Challenge
•   Now calculate PW:
•   PWA = -50,000 + 60,000(P/F, 15%, 1)
•        = -50,000 + 60,000 (0.8696)
•        = \$2176
•   PWB = -90,000 + 106,200(P/F, 15%, 1)
•        = -90,000 + 106,200 (0.8696)
•        = \$ 2351.52
•   Select B?
Example
Remember, we have \$100,000 available in funds so we
could spend an additional \$50,000 above alternative A or
an additional \$10,000 above alternative B. If we assume
we can make MARR or 15% return on our money, then
if we invest in A, we have an extra \$50,000 which
can be invested at MARR (15%).
Example
• Year         A     A1    B    B1     A+A1 B+B1
• 0           -50   -50 -90 -10        -100 -100
• 1            60    57.5 106.2 11.5    117.5 117.7

• RORA = 17.5%
• RORB = 17.7%

• Select B.
Incremental Rate of Return
• The result of IRR analysis for each project is not
consistent with Present Worth or Annual Worth
analysis. To resolve that, the Incremental Rate of
Return technique has been developed.
• Define B – A as new cash flow
• Usually B is the alternative with greater initial
investment (absolute value).
• If B and A are defined over different years, find the
Least Common Multiplier (LCM) of A and B and
define the projects over LCM.
Incremental Rate of Return
•   Year A          B       A – B (Incremental)
•    0 -300 -200 -300 – (-200) = - 100
•    1   400 300          400 – 300 = 100
•    2   500 600          500 – 600 = -100
•    3   400 500          400 – 500 = -100
•   NPWA – B (i*) = -100 +100(P/F, i*,1) - 100(P/F, i*,2)
•                   - 100 (P/F, i*,3)
Decision Rule
•   Form cash flow for B – A.
•   Calculate internal interest rate for B – A (i*).
•   If i* < MARR, select A.
•   If i* > MARR, select B.

• Note that the internal rate of return for each project
should be greater than MARR. If ROR of an
alternative is smaller than MARR, do not select that
alternative.
Example
• A solid-waste recycling plant is considering two types of
storage bins. Determine which should be selected on the
basis of rate of return. Assume the MARR is 20% per
year.
• Alternative                    P           Q
• First cost,\$                 -18,000     -35,000
• Annual operating cost,\$       -4,000      -3,600
• Salvage value, \$               3,000       2,700
• Life, years                       3         9
Solution
Year         P                        Q              Q -P
0        -18,000                  - 35,000          -17,000
1         -4,000                    -3,600             400
2         -4,000                    -3,600             400
3 -4,000 -18,000+3,000               -3,600   15,000+400
4         -4,000                    -3,600             400
5         -4,000                    -3,600             400
6 -4,000-18,000+3,000                -3,600   15,000+400
7          -4,000                   -3,600             400
8          -4,000                   -3,600             400
9     -4,000 + 3000              -3,600+2700    -300+400
NPW(i*) = -17,000 + 400 (P/A,i*,9) + 15,000 (P/F,i*,3) +
15,000(P/F,i*,6) – 300 (P/F,i*,9)=0

Solve with trial and error i*=16.9%
iQ-P =16.9% < MARR =20%            Select P.

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