1. COMPARING ALTERNATIVES
CHAPTER 5
COMPARING ALTERNATIVES

2. Objective
To learn how to properly apply the profitability measures described in Chapter 4 to select the best alternative out of a set of mutually exclusive alternatives (MEA)
The cash-flow analysis methods (previously described) used in this process:
Present Worth ( PW )
Annual Worth ( AW )
Future Worth ( FW )
Internal Rate of Return ( IRR )

3. Alternatives
Organizations have the capability to generate potential beneficial projects for potential investment
The alternatives being considered may require different amounts of capital investment
The alternatives may have different useful lives
The subject of this section will help:
analyze and compare feasible alternatives
select the preferred alternative

4. Feasible Design Alternatives
Three types of investment categories
Mutually Exclusive Set
Independent Project Set
Contingent
Mutually exclusive set
The selection of one alternative excludes the consideration of any other alternative
Once selected, the remaining alternatives are excluded
Independent project set
Selecting the best possible combination of projects from the set that will optimize a given criteria
Subjects to constraints
The choice of the project is conditional on the choice of one or more other projects

5. Section 5.2 Fundamental Purpose of Capital Investment
To obtain at least the MARR for every dollar invested.
Basic Rule:
Spend the least amount of capital possible unless the extra capital can be justified by the extra savings or benefits.
In other words, any increment of capital spent (above the minimum) must be able to pay its own way.

6. Section 5.2 Two Types of Decisions
1. Investment Alternatives - each alternative has an initial investment producing positive cash flows resulting from increased revenues, reduced costs, or both.
"Do nothing" (DN) is usually an implicit investment alternative.
If positive cash flows > negative cash flows, then IRR>0.
If EW(MARR)>0, investment is profitable, or if EW(MARR)<0, do nothing (DN) is better, where EW refers to an equivalent worth method (e.g. PW)
2. Cost Alternatives - have all negative cash flows except for the salvage value (if applicable). These alternatives represent “must do” situations, and DN is not an option
IRR not defined for cost alternatives. Can you explain why?

7. Section 5.3. The Study Period
Must be appropriate for the decision being made
Study Period: The time interval over which service is needed to fulfill a specified function
Useful Life: The period over time during which an asset is kept in productive operation
Case 1: Study period = Useful life
Case 2: Study period ¹ Useful life
Fundamental Principle: Compare MEAs over the same period of time

8. Section 5.4.1 (EW) Methods: PW, AW, FW (Case 1)
Procedure for Selecting the Best MEA using the EW method:
1. Compute the equivalent worth of each alternative, using the MARR as the interest rate.
2. Investment Alternatives: Select the alternative having the greatest equivalent worth.
Note: If all equivalent worths are < 0 for investment alternatives, then "do nothing" is the best alternative.
3. Cost Alternatives: Select the alternative having the smallest equivalent cost (the one that is least negative).
All three equivalent worth methods (PW, AW, FW) will identify the same "best" alternative.

9. Study Period =Useful Life
I II III IV
Investment cost (I) $100, $152, $184, $220,000
Net Annual receipts , , , ,500
Salvage value (SV) , , ,000
Useful life
If the MARR is 12%, use the PW method to select the best alternative
PW(12%) = -I + A(P|A, 12%, 10) + SV(P|F, 12%, 10)
Solution
PWDN (12%) = 0, PWI (12%) = -10,897, PWII (12%) = , PWIII (12%) = +23,672, PWIV (12%)=+20,923
Select Alternative ___ to maximize PW.

10. Example – Problem 5-12a (p. 234)
Cost Alternatives
Study Period = Useful Life
A B C
Initial cost (I) -$85,600 -$63,200 -$71,800
Annual expenses
years , , ,050
Use the AW method to choose the best alternative (MARR = 12%)
AWA = -26,155
AWB = -25,947
AWC = -25,781
Assuming one must be chosen (i.e., DN is not an option), select alternative _ to minimize AW costs.

11. Rate-of-Return Analysis: Multiple Alt.s
Assume we have two or more mutually exclusive Alt.
Objective: Which, if any of the alternatives is preferred?
Two Investments A and B, Discount rate = 10%, Each investment requires $100 at t = 0, A is a 1-year investment, B is a 5- year investment.
i*A = 0.20 = 20%, i*B = 0.15 = 15%, PWA(10%) = +$9.09, PWB(10%) =
Using ROR Ranking _ is superior to _
Using a PW(10%) approach _ is superior to _
The two methods do not rank the same?

12. Using the IRR Method: Another Example
Why not select the investment opportunity that maximizes IRR?
Consider 2 alternatives: A B
Investment -$100 -$10,000
Lump-Sum Receipt $1,000 $15,000
IRR % 50%
If MARR = 20%, would you rather have A or B if comparable risk is involved?
If MARR = 20%, PWA = $733 and PWB = $2,500

13. Is It Worth It? Now the question is….
Is it worth spending an additional $9,900 to move from investment A to investment B?
NEVER simply select the MEA that MAXIMIZES the IRR
Never compare the IRR to anything except the MARR.
We don't maximize rate of return. Look at the increment
Answer: Compute the ROR or PW of the incremental investment to see!
IRR A-B : PW A-B = 0 = -9, ,000(P|F, i'%, 1)
i' A-B = 41.4% > MARR

14. Calculations of Incremental Cash Flows for ROR Analysis
Given two or more alternatives
Rank the investments based upon their initial time t = 0 investment requirements
Summarize the investments in a tabular format
Select the first investment to be the one with the lowest time t = 0 investment amount.
The next investment is to be the one with the largest investment at time t = 0

15. Example – Problem 5-2 on page 232
Given three MEAs and MARR = 15% per year
Investment (FC) , , ,500
Net Cash Flow/year 5, , ,800
Salvage Value ,
Useful Life 10 yrs yrs yrs
Use the Incremental IRR procedure to choose the best alternative

16. Incremental Investment Analysis Procedure
1. Order the feasible alternatives
2. Establish a base alternative
a. Cost alternatives -- The first alternative is the base
b. Investment alternatives - If the first alternative is acceptable, select as base. If the first alternative is not acceptable, choose the next alternative
3. Use iteration to evaluate differences (incremental cash flows) between alternatives until no more alternatives exist
a. If incremental cash flow between next alternative and current alternative is acceptable, choose the next
b. Repeat, and select as the preferred alternative the last one for which the incremental cash flow was acceptable

17. To Summarize
1. Each increment of capital must justify itself by producing a sufficient rate of return on that increment.
2. Compare a higher investment alternative against a lower investment alternative only when the latter is acceptable.
3. Select the alternative that requires the largest investment of capital as long as the incremental investment is justified by benefits that earn at least the MARR. This maximizes equivalent worth on total investment at i = MARR.

18. Section 5.5 Case 2: Study PeriodUseful Life
Up until now, study periods and useful lives have been the same length
The study period is frequently taken to be a common multiple of the alternatives’ lives when study period ¹ useful life
Repeatability Assumption (page 211)
Conditions:
1. Study period is either indefinitely long or equal to a common multiple of the lives of the alternative.
2. The cash flows associated with an alternative's initial life span are representative of what will happen in succeeding life spans.

19. Example: Problem 5-24a (pp. 236)
Cost Alternatives; Study Period > Useful Life; MARR = 15%. A B
Investment cost $14,000 $65,000
Annual costs , ,000
Useful life
SV (MKT value) 8, ,000
If the study period = 20 years, which alternative is preferred?

20. Different Lives Comparison must be made over equal time periods
Compare over the least common multiple, LCM, for their lives
Remember – if the lives of the alternatives are not equal, one must create or force a study period where the life is the same for all of the alternatives

21. AW for Unequal Lives
Consider the AW over the useful life of Alternative A:
AWA = -14,000(A|P, 15, 5)- 14, ,000(A|F, 15, 5)= -16,990
Life 1: AW 1-5 = -16,990
Life 2: AW 6-10 = -14,000(A|P, 15, 5)- 14, ,000(A|F, 15, 5) = -16,990
Life 3: AW = -16,990
Life 4: AW = -16,990
Shortcut: If the study period equals a common multiple of the alternatives' lives, simply compare AW computed over the respective useful lives (assuming repeatability is valid).

22. Problem?
What if the study period is not a common multiple of the alternatives' lives or repeatability is not applicable?
A finite and identical study period is used for all alternatives
This planning horizon, combined with appropriate adjustments to the estimated cash flows, puts the alternatives on a common and comparable basis
Used when repeatability assumption is not applicable
Frequently used in engineering practice

23. Use the Cotermination Assumption
Procedure: The cash flows of the alternatives need to be adjusted to terminate at the end of the study period.
Cost alternatives: Assuming repeatability, repeat part of the useful life of the original alternative, and then use an estimated MV to truncate it at the end of the study period
Without repeatability, we must purchase/lease the service/asset for the remaining years.
Investment alternatives: Assume all cash flows will be reinvested at the MARR to the end of the study period (i.e., calculate FW at end of useful life and move this to the end of the study period using the MARR).

24. Study Period < Useful Life
When the study period is explicitly stated to be shorter than the useful life, use the cotermination assumption
Procedure: The cash flows of the alternatives need to be adjusted to terminate at the end of the study period.
Truncate the alternative at the end of the study period using an estimated Market Value.
Alt-1: N = 5 yrs
Alt-2: N= 7 yrs

25. Example of Cotermination
Suppose the study period had been stated to be 20 years. Which boiler would you recommend? Boiler A Boiler B
Investment cost $ 50, ,000
Useful life yrs yrs.
of useful life 10, ,000
Annual costs 9, ,000
Useful life of A = 20 years = study period
Useful life of B = 40 years > study period
Assume MV EOY 20 = $50,000
The MARR is 10% per year.

26. Solution AWA (10%) = -14,700, AWB (10%) = -19,225 Select _
What would the market value of Boiler EOY 20 have to be in order to select Boiler B instead of A?
Set AWA = AWB
-14,700 = -6,000 - {120,000(A|P, 10, 20) - X(A|F,10, 20)}
X = $308,571 therefore, MVB > $308,571 to favor B
Such a value is very unlikely because X is more than the initial cost of Boiler B.

27. Problem 5-24a Revisited with Cotermination
If the study period = 10 years and the estimated market value for alternate B = EOY 10, which alternative is preferred?

28. Comparing Alternatives Using the Capitalized Worth Method
Capitalized Worth (CW) method -- Determining the present worth of all revenues and / or expenses over an infinite length of time
Capitalized cost -- Determining the present worth of expenses only over an infinite length of time
Capitalized worth or capitalized cost is a convenient basis for comparing mutually exclusive alternatives when a period of needed services is indefinitely long and the repeatability assumption is applicable

29. Capitalized Cost
CAPITALIZED COST- the present worth of a project which lasts forever.
Government Projects
Roads, Dams, Bridges, project that possess perpetual life
Infinite analysis period
Start with the closed form for the P/A factor
Next, let N approach infinity