University:
Michigan State University
Course:
EC 301 | Intermediate Microeconomics
Academic year:

2023
Views:

293

Pages:

19
Author:

Tyshawn Shaffer

Chapter 6: Comparison and Selection Among Alternatives Alternatives - Mutually exclusive: selection of one alternative excludes the others. - Independent: selection of one alternative does not exclude other alternatives. - Acceptable alternative with the least capital investment is called the base alternative. - If the incremental investment (over the base alternative) is justified by extra benefits ⇒ investment should be made. Investment and cost alternatives  Investment alternatives: involve initial (capital) investment and produce positive cash flows due to increased revenue or savings, reduced costs, or both. - Base? Selected?  Cost alternatives: all negative cash flows (decision involves the most economical way of conducting an activity/project). - Base? Selected? Study period Study period (or planning horizon): selected time period over which mutually exclusive alternatives are compared. Study period cases: - Useful lives of all mutually exclusive alternatives (MEAs) are the same and equal to the study period. ⇒ No cash flow adjustment. ⇒ MEAs are compared using equivalent worth methods (PW, FW, or AW) or rate of return methods (IRR or ERR). - Useful lives are unequal and at least one does not match the study period. ⇒ Repeatability assumption. ⇒ Coterminated assumption. Example N = 4 years • MARR = 10%. *** What if A and B were independent alternatives? Example • MARR = 10%. ⇒ Alternative C is the base alternative (lowest capital). ⇒ PW @ MARR < 0 for both alternatives … cost alternatives. Select alternative D (higher PW @ MARR) ⇒ additional capital investment in D is justified. Example Use a MARR of 12% to select one alternative among the four MEAs. All alternatives are acceptable (PW > 0) ⇒ C is the base alternative (lowest capital).  Select alternative A (highest PW) Repeat using AW and FW methods  Should get the same conclusion (not the same numbers though). Example A company is planning to install a new automated plastic-molding press. Four alternatives are considered as shown in the table. Assuming that each press has the same output capacity (120,000 units per year) with no market value at the end of its useful life, and using a 10% MARR, which machine will you select? Same output  same revenue  minimize cost Calculate the PW of all costs and select the one with the highest value. Alternatives A B C D Present worth -$142,273 -$140,818 -$145,098 -$138,734 Annual worth -$37,531 -$37,148 -$38,276 -$36,598 Future worth -$229,131 -$226,788 -$233,689 $-223,431 Example Three mutually exclusive design alternatives are being considered. The estimated cash flows for each alternative are given in the following table. At a MARR of 20% per year, which one will you select? PW (20%) A = -$28,000 + ($23,000 – $15,000) (P/A, 20%, 10) + $6,000 (P/F, 20%, 10) = $6,509. PW (20%) B = -$55,000 + ($28,000 – $13,000) (P/A, 20%, 10) + $8,000 (P/F, 20%, 10) = $9,180. PW (20%) C = -$40,000 + ($32,000 – $22,000) (P/A, 20%, 10) + $10,000 (P/F, 20%, 10) = $3,540. ⇒ Select B. *** selection based on maximum IRR is wrong *** Rate of Return Method N = 4 years • MARR = 10%. Calculate PW and IRR for both alternatives. Alternative IRR PW (10%) A 17.3% $9,738 B 16.3% $10,131 Both have positive PW, IRR > MARR … both acceptable alternatives. • Based on PW ⇒ select B (PW method is always correct). • Based on IRR ⇒ select A (misleading). • To solve using IRR, find the IRR of the incremental cash flow (B – A) IRR B-A = 11.4% > MARR … The incremental investment ($13,000) is justified. PW B-A = $393 > 0 … same conclusion. Select alternative B. Example – IRR method Six mutually exclusive alternatives with equal useful lives (10 years) are analyzed and compared using the IRR method. Assuming MARR = 10%, which alternative will you select? A B C D E F Capital investment $900 $1,500 $2,500 $4,000 $5,000 $7,000 Net annual income $150 $276 $400 $925 $1,125 $1,425 IRR 10.6% 13.0% 9.6% 19.1% 18.3% 15.6% - IRR is computed for each alternative … alternatives are ranked from lowest to highest capital. - IRR < MARR for alternative C … C is eliminated ⇒ The rest of alternatives (A, B, D, E, and F) are all acceptable alternatives. - Alternative A has the lowest capital among all acceptable alternatives … A is the base alternative. B becomes the new base (A is eliminated) ∆ Capital ∆ Annual income IRR ∆ Is increment justified? A B–A D–B E–D F–E $900 $600 $2,500 $1,000 $2,000 $150 $126 $649 $200 $300 10.6% 16.4% 22.6% 15.1% 8.1% - yes yes yes no Select E Example – IRR method Four alternatives are compared at MARR = 9%. C–A Alternative Capital IRR A $15,000 12% - B $20,000 15% C $25,000 D $30,000 A ∆ IRR B C - - 15% - - 10% 9% 10% - 20% 9.5% 12% 7% - If all alternatives are independent, which one will you select? All (all have IRR > MARR and the selection of one does not exclude others). - If all alternatives are mutually exclusive, which one will you select? Alternative C. Unequal useful lives – repeatability Two mutually exclusive alternatives with different useful lives. If MARR = 10% per year, and using the repeatability assumption, which alternative would you pick? Capital investment Annual net cash flow Useful lives (years) Market value at end of useful life A $3,500 $1,255 4 0 • The least common multiple of useful lives = 12 years. • A is repeated 3 times. • B is repeated 2 times. B $5,000 $1,480 6 0 Example – repeatability Alternative A $1,255 $1,255 $1,255 0 0 0 1 2 3 1 4 2 3 2 3 4 $3,500 $3,500 $3,500 1 4 Alternative B $1,480 $1,480 0 0 1 $5,000 2 3 4 5 6 $5,000 1 2 3 4 5 6 Select B Repeatability example – cont’d Solve using the annual worth method AW (10%)A = -$3,500 (A/P, 10%, 12) – $3,500 [(P/F, 10%, 4) + (P/F, 10%, 8)] (A/P, 10%, 12) + $1,255 = $151. Or AW (10%)A = PW (10%)A × (A/P, 10%, 12) = $151. AW (10%)B = -$5,000 (A/P, 10%, 12) – $5,000 (P/F, 10%, 6) (A/P, 10%, 12) + $1,480 = $332. Or AW (10%)A = PW (10%)B × (A/P, 10%, 12) = $332. Select B Calculate the annual worth of each alternative over one useful life cycle: AW (10%)A = -$3,500 (A/P, 10%, 4) + $1,255 = $151. AW (10%)B = -$5,000 (A/P, 10%, 6) + $1,480 = $332. ⇒ Same result ⇒ For repeatability assumption, we can calculate the AW for each alternative over its own useful (single) life and compare directly. Coterminated assumption 1) Useful life < study period a) Cost alternatives: - Contracting or leasing equipment/service for the remaining years. - Repeat part of the useful life and truncate at the end of the study period with an estimated market value. b) Investment alternatives: - Cash flows reinvested at MARR to the end of the study period. - Replace with another asset with possibly different cash flows over the remaining life. 2) Useful life > study period: truncate at the end of the study period with an estimated market value. Example – coterminated Two mutually exclusive alternatives with different useful lives. If MARR = 10% per year, and the study period is 6 years, which alternative would you pick? Capital investment Annual cash flow Useful lives (years) Market value at end of useful life A $3,500 $1,255 4 0 B $5,000 $1,480 6 0 • 6 years isn’t a multiple of both lives ⇒ repeatability isn’t applicable. • Coterminated assumption: Select B Example Two mutually exclusive alternatives with different useful lives. At 5% per year MARR: A B Capital investment $6,000 $14,000 Annual expenses $2,500 $2,400 Useful lives (years) 12 18 Market value at end of useful life 0 $2,800 • Determine which alternative to select assuming repeatability applies. • Determine which alternative to select if the repeatability does not apply, study period is 18 years, and a new system can be leased for $8,000 per year after the useful life of alternative A is over. Unequal useful lives – rate of return methods Capital investment Annual cash flow Useful lives (years) Market value at end of useful lives A $3,500 $1,255 4 0 B $5,000 $1,480 6 0 Which alternative should be selected assuming MARR = 20%? Use the IRR method. • Study period isn’t determined … repeatability. • Repeatability means finding AW. AW(𝑖𝑖*)A = AW(𝑖𝑖*)B -$3,500 (A/P, 𝑖𝑖*, 4) + $1,255 = -$5,000 (A/P, 𝑖𝑖*, 6) + $1480 𝑖𝑖* = 26% … increment is justified … Select B

Chapter 6 Comparison and Selection Among Alternatives

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