MS303_Assignment%205

MS303_Assignment%205 - Doublecheck: PW(15%) for the new...

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MS303 – ASSIGNMENT 5 6.22) List the alternatives in increasing order of initial cost: B, D, A, C. B is the base alternative (since its IRR=15.4% >MARR) In general, we have P = A (P/A, i′, 80) which becomes i′ = A / P when the MARR is 12% and N = 80 years, where i′ is the IRR of the project or the incremental cash flows. The incremental comparisons follow: B= 8/52 = 0.154 (15.4%), so select B. Δ(D−B) = 1/3 = 0.333 (33.3%), so select D. Δ(A−D) = 1/7 = 0.143 (14.3%), so select A. Δ(C−A) = 10/88 = 0.114 (11.4%), so keep A. Therefore, we recommend Alternative A even though it does not have the largest IRR or the largest ΔIRR. 6.23) Examine ∆(New Baghouse-New ESP): Incremental investment = $147,500 Incremental annual expenses = $42,300 per year for 10 years Therefore, by inspection, the extra investment required by the new baghouse is producing extra annual expenses, so the new ESP should be recommended.
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Unformatted text preview: Doublecheck: PW(15%) for the new baghouse = -$1,719,671 PW(15%) for the new ESP = -$1,359,876 The economic advantage of the ESP may not be sufficient enough to overcome is inability to meet certain design specifications under varying operating conditions. 6.26) Rank alternatives by increasing capital investment: A,B,C,D and E. Since the ERR on Equipment A > MARR, it is an acceptable base alternative. Therefore, ERRon Δ (B−A) must be examined. Δ (B−A) ($50,000 − $38,000)(F/P, i′, 10) = ($14,100 − $11,000)(F/A, 18%, 10) i′= 19.8% > MARR, therefore select B. Δ (C−B) $5,000 (F/P, i′, 10) = $2,200(F/A, 18%, 10) i′= 26.3% > MARR, therefore select C. Δ (D−C) $5,000 (F/P, i′, 10) = $500(F/A, 18%, 10) i′= 8.9% < MARR, therefore keep C. Δ (E−C) $15,000 (F/P, i′, 10) = $2,900(F/A, 18%, 10) i′= 16.4% < MARR, therefore keep C as best. Answer: Select C....
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MS303_Assignment%205 - Doublecheck: PW(15%) for the new...

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