COMPARING ALTERNATIVES CHAPTER 5 COMPARING ALTERNATIVES

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 )

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

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

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.

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?

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

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.

Study Period =Useful Life I II III IV Investment cost (I) $100,000 $152,000 $184,000 $220,000 Net Annual receipts 15,200 31,900 35,900 41,500 Salvage value (SV) 10,000 0 15,000 20,000 Useful life 10 10 10 10 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%) = +28.241, PWIII (12%) = +23,672, PWIV (12%)=+20,923 Select Alternative ___ to maximize PW.

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 1-7 -7,400 -12,100 -10,050 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.

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%) = +24.89 Using ROR Ranking _ is superior to _ Using a PW(10%) approach _ is superior to _ The two methods do not rank the same?

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 900% 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

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 = -9,900 + 14,000(P|F, i'%, 1) i' A-B = 41.4% > MARR

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

Example – Problem 5-2 on page 232 Given three MEAs and MARR = 15% per year 1 2 3 Investment (FC) -28,000 -16,000 -23,500 Net Cash Flow/year 5,500 3,300 4,800 Salvage Value 1,500 0 500 Useful Life 10 yrs 10 yrs 10 yrs Use the Incremental IRR procedure to choose the best alternative

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

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.

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.

Example: Problem 5-24a (pp. 236) Cost Alternatives; Study Period > Useful Life; MARR = 15%. A B Investment cost $14,000 $65,000 Annual costs 14,000 9,000 Useful life 5 20 SV (MKT value) 8,000 13,000 If the study period = 20 years, which alternative is preferred?

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

AW for Unequal Lives Consider the AW over the useful life of Alternative A: AWA = -14,000(A|P, 15, 5)- 14,000 + 8,000(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 + 8,000(A|F, 15, 5) = -16,990 Life 3: AW 11-15 = -16,990 Life 4: AW 16-20 = -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).

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

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).

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

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 $ 50,000 120,000 Useful life 20 yrs. 40 yrs. SV@end of useful life 10,000 20,000 Annual costs 9,000 6,000 Useful life of A = 20 years = study period Useful life of B = 40 years > study period Assume MV B @ EOY 20 = $50,000 The MARR is 10% per year.

Solution AWA (10%) = -14,700, AWB (10%) = -19,225 Select _ What would the market value of Boiler B @ 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.

Problem 5-24a Revisited with Cotermination If the study period = 10 years and the estimated market value for alternate B = $25,000 @ EOY 10, which alternative is preferred?

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

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