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EGG 307 HW 8

# EGG 307 HW 8 - \$478.63 \$1,342.8/n A = \$2,164 Solve for =...

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EGG 307: Homework No. 8 Sample Solutions Problem 6-4 MARR = 20% Incremental Analysis A B C C vs A B vs C Investment -\$30,000 -\$60,000 -\$40,000 -\$10,000 -\$20,000 units/year 15000 20000 18000 Price \$3.10 \$4.40 \$3.70 Annual revenue \$46,500 \$88,000 \$66,600 \$20,100 \$21,400 Variable cost \$1.00 \$1.40 \$0.90 Annual fixed cost -\$15,000 -\$30,000 -\$25,000 Total cost -\$30,000 -\$58,000 -\$41,200 -\$11,200 -\$16,800 Market value \$10,000 \$10,000 \$10,000 \$0 \$0 Useful life 10 10 10 10 10 IRR 54.5% 49.2% 63.1% 88.8% 18.9% PW \$145,000 \$250,000 \$224,000 > MARR < MARR AW \$14,500 \$25,000 \$22,400 Select C Keep C Payback period 1.82 2.00 1.57 Problem 11-6 (b) Let n A = efficiency of Pump A At 2,000 hrs/year of operation EUAC(Pump A) = \$2,410(A/P,12%,8) - 80(A/F,12%,8) + \$0.06x2,000x15x0.746/n A = \$478.63 + \$1,342.8/n A EUAC(Pump B) = \$4,820(A/P,12%,8) + \$0.06x2,000x10x0.746/0.75 = \$2,164 To find the breakeven efficiency of Pump A EUAC(Pump A) = EUAC(Pump B)

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Unformatted text preview: \$478.63 + \$1,342.8/n A = \$2,164 Solve for = 0.7967 ==> 79.76% Pump A will be prefered over Pump B if its efficincey is over 79.67% Graph: Next page… Graph n A EUAC(Pump A) EUAC(Pump B) 60% \$2,717 \$2,164 65% \$2,544 \$2,164 70% \$2,397 \$2,164 75% \$2,269 \$2,164 80% \$2,157 \$2,164 85% \$2,058 \$2,164 90% \$1,971 \$2,164 Problem 12-2 Option 1: Build a 4-lane bridge now PW = -\$3,500,000 Option 2: Build a 2-lane bidge now, add two lanes later: Compute expected PW based on probabilities and costs of when the additional two lanes will be required (see table below) Decision: Build a 4-lane bridge now. \$1,000 \$1,500 \$2,000 \$2,500 \$3,000 50% 60% 70% 80% 90% 100% EUAC(Pump A) EUAC(Pump B) Pump A More Economical Pump B more economical Efficiency of Pump A (n A )...
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EGG 307 HW 8 - \$478.63 \$1,342.8/n A = \$2,164 Solve for =...

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