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Chapter 7 Rate-of-Return Analysis Concept of Rate of Return Note: Symbol convention---The symbol i* represents the breakeven interest rate that makes the NPW of the project equal to zero.. The symbol IRR represents the internal rate of return of the investment. For a simple (or pure) investment, IRR = i*. For a nonsimple investment, generally i* is not equal to IRR.
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Investment Classification and Calculation of i*
7.6 (a) Simple investment: Project A (Note: Project C is a simple borrowing) (b) Non simple investment: Project B and Project D (c)
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∴ * 54.05%i =
7.8 (a) Classification of investment projects:
• Simple projects: A,B, and E • Non simple projects: C and D
(b)
2
$60 $150$100 01 (1 )i i
− + + =+ +
Let 1(1 )
Xi
=+
, then,
2
1 2
$100 $60 $150 00.6406, 1.0406
X XX X− + + =
= = −
∴ * 56.09%i =
(c) Find by plotting the NPW as a function of interest rate: *i
Project *i
A 56.09% B 47.94% C 8.32% D 178.8% E 24.21%
7.9 (a) Classification of investment projects:
• Simple projects: A,B, and D • Non simple projects: C
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(c) Use the PW plot command provided in Cash Flow Analyzer, or you may use the Excel’s Chart Wizard
7.10 (a)
$50,000 ($25,000 $9,000)( / , ,8) $10,000( / , ,8) 0P A i P F i− + − + = Solving for yields i * 28.45%i =
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7.12 (a) Cash flow sign rules:
Projects Number of Sign Changes Possible Number of *iA 1 0, 1 B 1 0, 1 C 1 0, 1 D 1 0, 1 E 2 0, 1, 2 F 1 0, 1
(b) Use the PW plot command provided in Cash Flow Analyzer, or you may use
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(b) Investment classification:
• Project 1: simple and pure investment, 20%IRR = • Project 2: simple and pure investment, 18%IRR = • Project 3: non simple and mixed investment, 31.07%IRR RIC= = 0(31.07%,12%) $1,000PB = −
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7.19
(a) * *1 215.99%, 0%i i= =
(b) Since all projects pass the net investment test, all projects are pure
investments. (c) All projects are acceptable at MARR = 10%.
7.20 (a) Project A: , Project B: * *
1 210%, 100%i i= = * *1 2350.33%, 80.83%i i= = −
(b) Pure investment: C, mixed investments: A,B, D and E
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* 0% or 30%B Ai − = Since this is a mixed incremental investment, we need to find the RIC using an external interest rate of 15%.
B-A B-A 16.95% 15%RIC IRR= = >
∴ Project B is preferred (b) Use the PW plot command provided in Cash Flow Analyzer.
7.23 (a) Apply the net investment test using * 10%i =
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7.27 (a) Since IRR = 10% and PW(10%) = 0, we have,
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7.32
(a) Project A: IRR 11.71%= Project B: IRR 19.15%=
(b) Only project B is acceptable (c) Since project A is not acceptable, select project B.
7.33 Option 1:Buy a certificate, Option 2: Purchase a bond, and assume that MARR 9%=
Net Cash Flow n Option 1 Option 2 Option 1 – Option 2
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Since the incremental cash flow series portrays a nonsimple investment, we need to find the RIC at 10%, which is 2 1 7.36% 10%A ARIC − = < . So, select A1.
(b) We can verify the same result by applying the NPW criterion.
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n A – B 0 -$2,376 1 0 2 0 3 0 4 2,500
IRR 1.28%A B− =
∴ If MARR Model A is a preferred. 1.28%,<
7.38 IRR for Model A: 6.01%, IRR for Model B: 7.25%
∴ The best decision is “do-nothing.”
7.39 (a) The least common multiple project lives = 6 years → Analysis period 6 years
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Net Cash Flow n
Project C Project D C – D 0 -$4,000 -$2,000 -$2,000 1 2,410 1,400 1,010 2 2,930 1,720 1,210
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7.41
(a) • Project A vs. Project B
Net Cash Flow n Project A Project B B – A 0 -$1,000 -$1,000 0 1 900 600 -$300 2 500 500 0 3 100 500 400 4 50 100 50
∴ , select B. * IRR 21.27% 12%B A B Ai − −= = >
• Project B vs. Project C
Net Cash Flow n Project B Project C C – B 0 -$1,000 -$2,000 -$1,000 1 600 900 300 2 500 900 400 3 500 900 400 4 100 900 800
∴ , select C. * IRR 26.32% 12%C B C Bi − −= = >
(b)
$1,000 $300( / , , 4)7.71%
P A ii==
(c) Since BRR (borrowing rate of return) is less than MARR, project D is
acceptable.
(d) Net Cash Flow n Project C Project E C – E
0 -$2,000 -$1,200 -$800 1 900 400 500 2 900 400 500 3 900 400 500 4 900 400 500 ∴ , select C. * IRR 50.23% 12%C E C Ei − −= = >
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Net Cash Flow
n Project A Project B B – A 0 -$10,000 -$20,000 -$10,0001 5,500 0 -5,5002 5,500 0 -5,5003 5,500 40,000 34,500
∴ Since IRR 24.24% 15%,B A− = > select project B.
7.44 Select Model C. Note that all three projects would be acceptable individually,
as each project’s IRR exceeds the MARR. The incremental IRR of Model (C – B) is 40%, indicating that Model C is preferred over Model B at MARR = 12%. Similarly, the incremental IRR of Model (C – A) is 15% which exceeds the MARR. Therefore, Model C is again preferred.
7.45 All projects would be acceptable because individual ROR exceed the MARR. Based on the incremental analysis, we observe the following relationships:
2 1IRR 10% 15%A A− = < (Select A1)
3 1IRR 18% 15%A A− = > (Select A3)
3 2IRR 23% 15%A A− = > (Select A3)
∴ Therefore, A3 is the best alternative.
7.46 From the incremental rate of return table, we can deduce the following
relationships:
2 1IRR 9% 15%A A− = < (Select A1)
3 2IRR 42.8% 15%A A− = > (Select A3)
4 3IRR 0% 15%A A− = < (Select A3)
5 4IRR 20.2% 15%A A− = > (Select A5)
6 5IRR 36.3% 15%A A− = > (Select A6)
It is necessary to determine the preference relationship among A1, A3, and A6.
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∴ A6 is the best alternative.
7.47 For each power saw model, we need to determine the incremental cash flows
over the “by-hand” operation that will result over a 20-year service life.
Power Saw Category Model A Model B Model C Investment cost $4,000 $6,000 $7,000 Salvage value $400 $600 $700 Annual labor savings $1,296 $1,725 $1,944 Annual power cost $400 $420 $480 Net annual savings $896 $1,305 $1,464
Net Cash Flow
n Model A Model B Model C 0 -$4,000 -$6,000 -$7,000 1 896 1,305 1,464 2 896 1,305 1,464
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Net Cash Flow
n Project A Project B B – A 0 -$100 -$200 -$100 1 60 120 60 2 50 150-200 -100 3 50-100 120 170 4 60 150-200 -110 5 50 120 70 6 50 150 100
Since the incremental cash flow series is a nonsimple investment, we may abandon the IRR analysis, and use the PW decision rule.
(15%) $100 $60( / ,15%,1)$100( / ,15%,6)
$3.48
B APW P FP F
− = − ++ +=
Since or , select project B. (15%) 0,B APW − > (15%) (15%)BPW PW> A
Comments: Even though the incremental flow is a nonsimple, it has a unique rate of return. As shown in Problem 7.39, this incremental cash flow series will pass the net investment test, indicating that the incremental cash flow is a pure investment.
IRR 15.98% 15%B A− = >
7.49
(a) Since there is not much information given regarding the future replacement options and the required service period, we may assume that the required service period is indefinite and both projects can be repeated at the same cost in the future.
(b) The analysis period may be chosen as the least common multiple of project lives, which is 3 years.
n A2 – A1 0 -$5,0001 02 03 15,000
2 1IRR 44.195%A A− =
∴ The MARR must be less than 44.195% for Project A1 to be preferred.
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Short Case Studies
ST 7.1 (a) Analysis period of 40 years: • Without the “mothballing” cost:
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(a) Assumptions required:
• There is no information regarding the expected cash flows from the current operation if B&E Cooling decides to defer the introduction of the absorption technology for 3 years. Therefore, we need to make an explicit assumption of the expected cash flows for the first three years if B&E Cooling decides to defer the decision. Assume that the annual cash flow during this period would be X.
• Another assumption we have to make is about the analysis period. Assuming that the firm will be in business for an indefinite period, we also need to make an explicit assumption regarding the future cooling technology. Since there is no information about the future cooling technology options, we may assume that the best cooling technology will be the absorption technology that will be introduced 3 years from now. Therefore, if B&E Cooling decides to select Option1, we could assume that, at the end of 8 years, Option 2 (the best cooling technology at that time) will be adopted for an indefinite period.
(b) Investment decision:
• Present worth analysis: First, we will determine the equivalent present worth for each option:
Now we can determine the value X that makes the two options economically equivalent at an interest rate of 15%. In other words, if we evaluate the two present worth functions at 15%i = , we have
1
2
(15%) $41.31(15%) 2.2832 $13.28
PWPW X
== +
Letting and solving for X gives 1(15%) (15%)PW PW= 2
$12.28X =
As long as the current operation continues to generate annual net revenue of
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$12.28 millions for 3 years, Option 2 is a better choice.
• Rate of return analysis: The present worth analysis above indicates that, if , the break-even rate of return on incremental investment is
$12.28X =
*1 2 15%i − =
Therefore, the ultimate choice will depend on the level of annual revenues generated during the first 3 years when the advanced cooling technology is deferred. Clearly, if
$12.28X < , then , Option 1 is preferred. *1 2 15%i − >
ST 7.3 n Current Pump(A) Larger Pump(B) B-A0 $0 -$1,600,000 -$1,600,0001 $10,000,000 $20,000,000 $10,000,0002 $10,000,000 $0 -$10,000,000
IRR = 25%400%
The incremental cash flows result in multiple rates of return (25% and 400%), so we may abandon the rate of return analysis. Using the PW analysis,
(20%) $1.6 $10 ( / , 20%,1) $10 ( / , 20%, 2)
$0.21 0Reject the larger pump.
PW M M P F M P F= − + −= − <
Return on Invested Capital: Using the procedure outlined in Section 7.3.4,
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1If 525%, then ( , 20%) 0i PB i> < , no RIC exists. So, the RIC on the incremental
cash flows should be 4.17%, which indicates “Select a smaller pump.”
ST 7.4
(a) Whenever you need to compare a set of mutually exclusive projects based on the rate of return criterion, you should perform an incremental analysis. In our example, the incremental cash flows would look like the following:
n B - A 0 -$10,0001 +23,0002 -13,200
This is a nonsimple investment with two rates of return.
* 10% or 20%B Ai − =
We could abandon the IRR analysis and use either the PW analysis to rank the projects.
(b) If we plot the present worth as a function of interest rate, we will observe
the following:
• If MARR < 10% select project A. • If 10% ≤ MARR 18.1%, select project B. ≤• If MARR > 18.1%, do nothing.
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Return on Invested Capital: Using the procedure outlined in Section 7.3.4, the true rate of return can be found as a function of MARR.
Let i = IRR and assume 1.3i <
0
1
2
( , MARR) $10,000( , MARR) $10,000(1 ) $23,000
$13,000 10,000( , MARR) ($13,000 10,000 )(1 MARR)
$13, 2000
PB iPB i i
iPB i i
= −= − + += −= − +−=
(Note that, if, there will be no feasible solution.) Rearranging the terms in gives an expression of IRR as a function of MARR.
1.3,i >
2( ,MARR)PB i
1.32IRR 1.31 MARR
= −+
For example, at MARR 15%=
IRR 15.3% 15%B A− = >
∴ Project B (the higher cost investment) would be justified.
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