1. Comparing Alternatives? ?
Comparing Alternatives

2. Definitions
Investment Proposal: a single undertaking or project being considered as an investment possibility.
Independent Proposal: the acceptance from a set of alternatives has no effect on the acceptance of any other proposals in the set.
Dependent Proposals
Mutually exclusive proposal : the acceptance of one proposal precludes the acceptance of any of the others.
Contingent proposal: the acceptance of the proposal is dependent on the acceptance of some prerequisite proposal.
Dr. C.J. Su IEEM Dept. HKUST

3. Examples
Independent Proposals - the purchase of a CNC milling machine, a security system, office furniture, and fork lift trucks.
Dependent Proposals
Mutually exclusive proposal : Select a course from a set of courses that have the same time slot. Select different brand of equipment that perform the same functions.
Contingent proposal: the purchase of software is contingent on the purchase of hardware. The construction of the 3rd floor is contingent on the construction of 1st & 2nd floors.
Dr. C.J. Su IEEM Dept. HKUST

4. Comparing Alternatives with Equal Planning Horizon
A & B Are Mutually Exclusive
At i%= MARR
If PW(A) > PW(B) => Accept A
Else Accept B
If FW(A) > FW(B) => Accept A
If AW(A) > AW(B) => Accept A
Dr. C.J. Su IEEM Dept. HKUST

5. Example
Three mutually exclusive investment alternatives for implementing an office automation plan in a firm are being considered. The study period is 10 years, and the useful lives of all three alternatives are also 10 years. Market values of all alternatives are zero at the end of their useful lives. If the firm's MARR is 10% /year, which alternative should be selected ?
Alternative A B C
Investment
-390,000
-920,000
-660,000
Net Revenue
69,000
167,000
133,500
Dr. C.J. Su IEEM Dept. HKUST

6. Solution PW(10%)A = -$390,000 + $69,000(P/A, 10%, 10) = $33,977
PW(10%)B = -$920,000 + $167,000(P/A, 10%, 10) = $106,148
PW(10%)C = -$660,000 + $133,500(P/A, 10%,10) = $160,304
C > B > A, means C is preferred to B and B is preferred to A.
Dr. C.J. Su IEEM Dept. HKUST

7. IRR Method
If IRR of (A - B) > MARR => the incremental investment is justified; therefore proposal A should be selected
If A & B are mutually exclusive alternatives
If IRR(A - B) > MARR => Accept A, Reject B
Else Accept B and Reject A
Dr. C.J. Su IEEM Dept. HKUST

8. Example Alternative A B (B -A)
Capital investment , , ,000
Net Annual revenues , , ,225
N = 4 years, MARR = 10%
Alternative IRR PW (10%)
A % ,738
B % ,131
Dr. C.J. Su IEEM Dept. HKUST

9. Incremental Analysis Procedure
Arrange the alternatives on the order of increasing capital investment
Case 1: For Investment Alternatives:
Compute the IRR for each alternatives
If all IRR < MARR => Do nothing
If exactly one alternative’s IRR > MARR => Select this alternative
If more than one alternative’s IRR > MARR, use incremental criterion to select the best alternative.
Case 2: For Cost Alternatives:
Use incremental criterion to select the best alternative
Dr. C.J. Su IEEM Dept. HKUST

10. Investment Proposal Example
Suppose that we are analyzing the following six mutually exclusive alternatives for a project (arranged in ascending order of initial investment) using the IRR method. The useful life of each alternative is 10 years, and the MARR is 10% per year. Also, net annual revenues less expenses vary among all alternatives. If the study period is 10 years, and the salvage (market) values are 0, which alternative should be chosen?
A B C D E F
Capital investment , , , , ,000
Annual revenues
less expenses , ,425
Dr. C.J. Su IEEM Dept. HKUST

11. Solution Alternative C is unacceptable IRR(C) < MARR
A B C D E F
IRR 10.6% % 9.6% % % %
Alternative C is unacceptable IRR(C) < MARR
Select A as the base for comparison
Dr. C.J. Su IEEM Dept. HKUST

12. Cost Proposal Example Design Alternative D1 D2 D3 D4
Capital investment , , , ,000
Annual expenses - 29, , , ,100
Useful life (years)
Market value , , , ,000
Dr. C.J. Su IEEM Dept. HKUST

13. ERR Example
The analysis period is six years, and the MARR for capital investments at the plant is 20% per year before taxes. Using the ERR method, which alternative should be selected? ( = MARR.)
| - 640,000|(F/P, i '%,6)= 262,000(F/P,20%,5) ,000 = 2,853,535
Dr. C.J. Su IEEM Dept. HKUST

14. Parking Lot Construction Example
Capital Net Annual
Investment Income
P. Keep existing parking , ,000
lot, but improve
B1. Construct one-story ,000, ,000
building
B2. Construct two-story ,550, ,000
B3. Construct three-story - 7,500, ,000
MARR = 10%, N= 15 years, Salvage = Initial Investment
Dr. C.J. Su IEEM Dept. HKUST

15. Selection based on PW Method
PW(10%)p = -$200,000 + $22,000(P/A,10%,15) + $200,000(P/F,10%,15) = $15,214
PW(10%)B1 = - 4,000,000 + $600,000(P/A,10%.15) + $4,000,000(P/F,10%,15) = $1,521,260
PW(10%)B2 = - $5,550,000 + $720,000(P/A,10%,15) + $5,550,000(P/F,10%,15) = $1,255,062
PW(10%)B3 = - $7,500,000 + $960,000(P/A,10%,15) + $7,500,000(P/F,10%,15) = $1,597,356
From PW => Select B3
Dr. C.J. Su IEEM Dept. HKUST

16. Selection based on IRR Mutually Exclusive Alternatives
P B1 - P B2 - B B3 - B1
Capital , ,000, ,550, ,500,000
investment
Net annual , , , ,000
income
Residual 200, ,000, ,550, ,500,000
value
IRR % 15% % %
Dr. C.J. Su IEEM Dept. HKUST

17. Dr. C.J. Su IEEM Dept. HKUST

18. Conclusion
PW and IRR Incremental Analysis methods reach consistent selection for mutually exclusive alternatives.
Whenever possible try Not to use IRR to compare alternatives. Use PW, FW, or AW.
Dr. C.J. Su IEEM Dept. HKUST

19. Comparing Alternatives with Unequal Lives
When comparing alternatives with unequal lives, the principle that all alternatives under consideration must be compared over the same time span is basic to sound decision making.
Dr. C.J. Su IEEM Dept. HKUST

20. Method 1 Estimation of Required Cash Flow
When the required cash flow (salvage value) can be estimated, this method can be applied.
Case 1: Alternative’s Useful life > study period
the salvage value for the alternative extending
beyond the study period must be directly
estimated.
Dr. C.J. Su IEEM Dept. HKUST

21. Example Alternatives EOY A B 0 -15,000 - 20,000 1 - 6,000 - 2,000
0 -15, ,000
, ,000
, ,000
, ,000
, ____
, , ____
Dr. C.J. Su IEEM Dept. HKUST

22. Suppose study period (planning horizon) = 3
4,000
Estimated
3,000 1 2 3 4 5 A
- 6,000 /year
- 15,000 1 2 3 B
- 2,000 /year
- 20,000
Dr. C.J. Su IEEM Dept. HKUST

23. If MARR = 20%, Alternatives A & B “coterminated” at year 3
AW (A) = -15,000 (A/P, 20%, 3) - 6,000 +
4,000(A/F, 20%, 3) = -12,021
AW(B) = -20,000(A/P, 20%, 3) - 2,000 = -11,494
=> B > A
Using PW or FW will yield the same result.
Dr. C.J. Su IEEM Dept. HKUST

24. Case 2: Alternative’s Useful life < study period
the operational cost and/or revenue value for the alternative extending beyond the study period must be directly estimated.
Assuming that the study period = 5 and at year 4 & 5 will require costs $3,000 per year for the last two years of alternative B’s life.
Dr. C.J. Su IEEM Dept. HKUST

25. 3,000 1 2 3 4 5 A
- 6,000 /year
- 15,000
PW (A) = -15,000 (A/P, 20%,5) - 6, ,000(A/F, 20%, 5)
= -10,613 1 2 3 B
- 2,000 /year
- 3,000
- 3,000
- 20,000
Estimated
PW (B) = - 20,000(A/P, 20%, 5) - 2,000 -
1,000(F/A, 20%, 2) (A/F, 20%, 5) = -8,984
=> B > A
Dr. C.J. Su IEEM Dept. HKUST

26. Method 2
For alternatives that are repeatable (long term planning horizon). For example, public service facility.
The repeatability assumption assuming that the alternative will repeat identical cash flow pattern until the common study period is reached.
Dr. C.J. Su IEEM Dept. HKUST

27. Example A B Capital investment - $3,500 - $5,000
Two mutually exclusive investment alternatives, A and B, associated with a small engineering project for which revenues as well as expenses are involved. They have useful lives of 4 and 6 years, respectively. If the MARR = 10% per year, show which feasible alternative is more desirable by using equivalent worth methods. Use the repeatability assumption.
A B
Capital investment $3, $5,000
Annual revenue , ,500
Annual expenses ,020
Useful life (years)
Market value at end of useful life
Dr. C.J. Su IEEM Dept. HKUST

28. AW(10%)A = -3,500(A/P,10%,4) + (1, ) = 151
AW(10%)B = -5,000(A/P,10%,6) + (2, ,020) = 332
B > A
Dr. C.J. Su IEEM Dept. HKUST

29. Consistency in AW, PW, and FW
PW(10%)A = - 3, ,500[(P/F,10%,4) + (P/F,10%,8)] + (1, )(P/A,10%,12) = 1,028
PW(10%)B = - 5, ,000(P/F,10%,6) + (2, ,020)(P/A,10%,12) = 2,262
B > A
FW(10%)A = [- 3,500(F/P,10%,4) +
(1, )(F/A,10%,4)](F/P,10%,2) = 847
FW(10%)B = - 5,000(F/P,10%,6) +
(2, ,020) (F/A,10%,6) = 2,561
Dr. C.J. Su IEEM Dept. HKUST

30. Pump Model SP240 HEPS9 Capital investment - 33,200 - 47,600
Annual expenses:
Electrical energy - 2, ,720
Maintenance ,100 in year 1, in year 4, and increasing and increasing /yr thereafter
-500/yr thereafter
Useful life (years)
Salvage value ,000
The new processing facility is needed by your firm at least as far into the future as the strategic plan forecasts operating requirements. The MARR, before taxes, is 20% per year. Based on this information, which model slurry pump should you select?
Dr. C.J. Su IEEM Dept. HKUST

31. With the repeatability assumption
AW(20%)Sp240 = - 33,200(A/P,20%,5) - 2,165
- [1, (A/G,20%,5)] = -15,187
AW(20%)HEPS9 = - 47,600(A/P,20%,9) + 5,000(A/F,20%,9)
- 1,720 - [500(P/A,20%,6) + 100(P/G,20%,6)] x (P/F,20%,3) x (A/P,20%,9) = - 13,622
HEPS9 > SP240
Dr. C.J. Su IEEM Dept. HKUST

32. SP240 > HEPS9 AW(20%)HEPS9 = - 47,600(A/P,20%,5) +
Suppose that the estimated market value of pump model HEPS9 in five years is $15,000, and the firm's MARR remains 20% per year. Which pump model should be selected for this replacement action?
AW(20%)HEPS9 = - 47,600(A/P,20%,5) +
15,000(A/F,20%,5) - 1,720
- [$500(P/F,20%,4) + $600(P/F,20%,5)] x
(A/P,20%,5) = - 15,783
AW(20%)SP240 = - 15,187 (from previous example)
SP240 > HEPS9
Dr. C.J. Su IEEM Dept. HKUST

33. Imputed (Implied) Market Value Technique
When study period T < Useful Life
This technique estimates the value of the remaining life for an asset
The market value of an asset at time T, MVT
MVT = [EW at end of year T of remaining capital recovery amounts] + [EW at end of year T of original market value at end of useful life]
where EW means equivalent worth at i = MARR.
Dr. C.J. Su IEEM Dept. HKUST

34. Example
Suppose that the pump example is modified such that another market value for pump model HEPS9, at the end of year five, is developed using the imputed market value technique. The same question is again asked, which pump model (SP240 or HEPS9) should be selected for replacement of the current pump in the catalytic system? The MARR remains 20% per year and the study period remains five years.
Dr. C.J. Su IEEM Dept. HKUST

35. EWCR = [47,600(A/P,20%,9) - 5,000(A/F,20%,9)] x (P/A,20%,4) = 29,949
Compute the EW at end of year five, based on the original MV at end of useful life:
EWMV = 5,000(P/F,20%,4) = 2,412
Then, the new market value estimate at the end of year five is as follows:
MVs = EWCR + EWMV = 29, ,412 = 32,361
AW(20%)HEPS9 = - 47,600(A/P,20%,5) + 32,361 (A/F,20%,5) - 1,720 - [$500(P/F,20%,4) + 600(P/F,20%,5)]x (A/P,20%,5) = -13,449
AW(20%)SP240 = -15,187,
=> pump model HEPS9 > SP240
Dr. C.J. Su IEEM Dept. HKUST

36. Capitalized Worth (CW) Method
CW method involves in determining the present worth of all revenues and/or expenses over an infinite length of time (e.g., charity fund, scholarship, scientific foundation, etc.).
Suppose the end of period uniform payment = A
CW = PWN-> = A(A/P, i%, N)
= A {limN-> [(1+ i)N - 1]/[i [(1+ i)N]} = A/i
=> P * i = A = P(A/P, i, N) => i = (A/P, i, N)
Dr. C.J. Su IEEM Dept. HKUST

37. Example
A firm wishes to endow an advanced manufacturing processes laboratory at a university. The endowment principal will earn interest that averages 8% per year, which will be sufficient to cover all expenditures incurred in the establishment and maintenance of the laboratory for an indefinitely long period of time (forever). Cash requirements of the laboratory are estimated to be $100,000 now (to establish it), $30,000 per year indefinitely, and $20,000 at the end of every fourth year (forever) for equipment replacement.
(a) For this type of problem, what study period (N) is, practically speaking, defined to be "forever"?
(b) What amount of endowment principal is required to establish the laboratory and then earn enough interest to support the remaining cash requirements of this laboratory forever?
Dr. C.J. Su IEEM Dept. HKUST

38. For i = 8% , (A/P,8%,N) = 0.08 = i when N = 100.
As N-> , i = (A/P, i, N)
For i = 8% , (A/P,8%,N) = 0.08 = i when N = 100.
=> N = 100 is essentially forever
(b)
CW = - 100,000 - [30,000 + $20,000(A/F,8%,4)] / 0.08
= -$530,475
Dr. C.J. Su IEEM Dept. HKUST

39. A scholarship offers a student $15,000 a month
A scholarship offers a student $15,000 a month. What’s the deposit required if the bank’s annual interest rate is 12% nominal.
i= 12% / 12 = 1%
CW = A / i = 15,000 / 0.01 = 1,500,000
Dr. C.J. Su IEEM Dept. HKUST

40. Example
A selection is to be made between two structural designs. Because revenues do not exist (or can be assumed to be equal), only negative cash flow amounts (costs) and the market value at the end of useful life are estimated, as follows:
Structure M Structure N
Capital investment - $12,000 - $40,000
Market value ,000
Annual expenses , ,000
Useful life (years)
Using the repeatability assumption and the CW method of analysis, determine which structure is better if the MARR is 15% per year.
Dr. C.J. Su IEEM Dept. HKUST

41. AW(15%)M = -12,000(A/P,15%,10) - 2,200
= - 4,592
AW(15%)N = - 40,000(A/P,15%,25)
+ 10,000(A/F,15%,25)
- 1,000 = - 7,141
CW(15%)M = AWM / i = - 4,592 / 0.15 = - 30,613
CW(15%)N = AWN / i = -7,141 / 0.15 = - 47,607
M > N
Dr. C.J. Su IEEM Dept. HKUST

42. Forming mutually exclusive Alternatives
Independent Proposal: the acceptance from a set of alternatives has no effect on the acceptance of any other proposals in the set.
Dependent Proposals
Mutually exclusive proposal : the acceptance of one proposal precludes the acceptance of any of the others.
Contingent proposal: the acceptance of the proposal is dependent on the acceptance of some prerequisite proposal.
Dr. C.J. Su IEEM Dept. HKUST

43. For three mutually exclusive projects, the alternatives are:
If Xj = 1 => Accept Xj
Xj = 0 => Reject Xj
For three mutually exclusive projects, the alternatives are:
Dr. C.J. Su IEEM Dept. HKUST

44. If there are k independent proposals, then there are 2k possible selections of alternatives.
Dr. C.J. Su IEEM Dept. HKUST

45. A company is considering two independent sets of mutually exclusive projects. That is, projects A1 and A2 are mutually exclusive, as are B1 and B2. However, the selection of any project from the set of projects A1 and A2 is independent of the selection of any project from the set of projects B1 and B2
Dr. C.J. Su IEEM Dept. HKUST

46. Project B1 & B2 mutually exclusive and independent of C set
Five proposed projects are being considered. B1 and B2 are independent of C1 and C2. Also, certain projects are dependent on others that may be included in the final portfolio. Using the PW method and MARR = 10% per year, determine what combination of projects is best if the capital to be invested is (a) unlimited, and (b) limited to $48,000.
Project B1 & B2 mutually exclusive and independent of
C set
Project C1 & C2 mutually exclusive and dependent (contingent) on the acceptance of B2
Project D contingent on the acceptance of C1
Dr. C.J. Su IEEM Dept. HKUST

47. Dr. C.J. Su IEEM Dept. HKUST

48. (a) Alternative 6 is the best
(b) Alternatives 2 & 6 are excluded due to the budget limit $48,000 => alternative 5 is the best
Dr. C.J. Su IEEM Dept. HKUST