Mutually Exclusive Alternative_Engineering Economy, Lecture notes of Engineering Economy

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MUTUALLY EXCLUSIVE

ALTERNATIVES

Mutually Exclusive Alternative

• (^) Only one on the viable projects can be selected because only one is needed. Each viable project is an alternative and these alternatives are compared against each other.
• (^) If no alternative is economically justifiable, “do nothing” is the default selection.
• (^) Selecting one project excludes other projects from investments.

Present Worth Method of

Analysis

Criteria: -If at least one alternative has a PW at the MARR that is positive, choose the alternative with the highest PW. Otherwise choose the do- nothing alternative.

• (^) One alternative : If PW ≥ 0, the requested MARR is met or exceeded and the alternative is economically justified.
• (^) Two or more alternatives : Select the alternative with the PW that is numerically largest, that is, less negative or more positive. This indicates a lower PW of cost for cost alternatives or a larger PW of net cash flows for revenue alternatives.

Formula: PW = -PC (Present Cost) A(P/A, i, n) F(P/F, i, n) General Formula: PW = -PC A() F()

• (^) Gain Money (+)
• (^) Lost Money (-)

Alternative A B First Cost $100000 $ Annual Saving $25000 $ Example 2: i = MARR = 15%, n = 10, which alternative is to be selected? Solution: PWA = -100000 + 25000(P/A, 15%, 10) = -100000 + 25000() = $25, PWB = -150000 + 34000(P/A, 15%, 10) = -150000 +34000() = $20, PWA>PWB, therefore select PWA

EQUIVALENT ANNUAL WORTH

METHOD OF ANALYSIS

Criteria: -If at least one alternative has a EAW at the MARR that is positive, choose the alternative with the highest EAW. Otherwise choose the do-nothing alternative.

• (^) One alternative: AW > or = to 0, MARR is met or exceeded.
• (^) Two or more alternatives: Choose the lowest cost or highest income AW value using the MARR value.

Example 1: Machine A Machine B First Cost, $ 26,000 36, Annual maintenance cost, $ 800 300 Annual labor cost, $ 11,000 7, Extra annual income taxes, $ - 2, Salvage value, $ 2,000 3, Life, years 6 10 i = 15%, which machine should be slected? Solution: Machine A: AWA = -26,000(A/P,15%,6) + 2,000 (A/F,15%,6) - 11, = -26,000() + 2,000() - 11, = -26,000(0.26424) + 2,000(0.11424) – 11, = -$18, Machine B: AWB = -36,000(A/P,15%,10) + 3,000 (A/F,15%,10) - 9, = -36,000() + 3,000() - 9, = -36,000(0.19925) + 3,000(0.04925) – 9, = -$16, Select machine B since AWB > AWA

Example 2: Assume the company in previous example is planning to exit the tomato canning business in 4 years. At that time, the company expects to sell machine A for $12,000 or machine B for $15,000. All other costs are expected to remain the same. Which machine should the company purchase under these conditions? Solution: AWA = -26,000(A/P,15%,4) +12,000 (A/F,15%,4) – 11, = -26,000() + 12,000() - 11, = -26,000(0.35027) + 12,000(0.20027) – 11, = -$18, AWB = -36,000(A/P,15%,4) +15,000 (A/F,15%,4) – 9, = -36,000() + 15,000() - 9, = -36,000(0.35027) + 15,000(0.20027) – 9, = -$19, Select machine A as AWA > AWB.

• (^) Equivalent annual cost (EAC) is used for a variety of purposes, including capital budgeting. But it is used most often to analyze two or more possible projects with different lifespans, where costs are the most relevant variable.
• (^) Other uses of EAC include calculating the optimal life of an asset, determining if leasing or purchasing an asset is the better option, determining the magnitude of which maintenance costs will impact an asset, determining the necessary cost savings to support purchasing a new asset, and determining the cost of keeping existing equipment.

Formula: EAC = (P-SV)(A/P, i, n) + SV(i) General Formula: EAC = (P-SV)() + SV(i) P = Purchase Cost SV = Salvage Value i = Rate n = Life

(P-SV)( ) =(P-SV)(A/P, r, n)

INCREMENTAL RATE OF RETURN

METHOD OF ANALYSIS

Criteria: -Perform incremental IRR analysis by pairwise comparison in a defender/challenge approach. At each comparison, choose the higher cost alternative (challenger) if the incremental IRR exceeds the MARR. Otherwise choose the lower cost alternative (defender). Continue until all alternatives have been considered. Summary:

• (^) If the ∆IRR is ≥ MARR then choose the higher cost alternative.
• (^) If the ∆IRR is ≤ MARR then choose the lower cost alternative.

• (^) When two alternatives are compared, the IRR for each one is calculate and each alternative should first satisfies the MARR
• (^) Alternative with the highest IRR not necessary being the best
• (^) If the cash flows for one or all alternatives contain expenditures only, in this case we couldn’t compute the IRR and accordingly we couldn’t compare these alternatives using the IRR
• (^) In this case, the incremental rate of return (∆IRR) is computed on the difference between the alternatives.

Finding i* for LATHE 1 4 = (P/A, i, 10) 4.054 = (P/A, 21%, 10) 3.923 = (P/A, 22%, 10) i = x* = x1 + (x2 – x1) = 0.21 + (0.22 – 0.21) i* = 21.4%

LATHE 2: (150000-100000) = (34000-25000)(P/A,i,10) 50000 = 9000(P/A, i, 10) 5.556 = (P/A, i, 10) Finding i for LATHE 2 5.650 = (P/A, 12%, 10) 5.426 = (P/A, 13%, 10) i* = x* = x1 + (x2 – x1) = 0.12 + (0.13 – 0.12) i* = 12.4%