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Unformatted text preview: 3.30 15,000 = 2000 + 2000(P/A,15%,3) + 1000(P/A,15%,3)(P/F,15%,3) + x(P/F,15%,7) 15,000 = 2000 + 2000(2.2832) + 1000(2.2832)(0.6575) + x(0.3759) x = $18,442 3.36 P for maintenance = [11,500(F/A,10%,2) + 11,500(P/A,10%,8) + 1000(P/G,10%,8)](P/F,10%,2) = [11,500(2.10) + 11,500(5.3349) + 1000(16.0287)](0.8264) = $83,904 P for accidents = 250,000(P/A,10%,10) = 250,000(6.1446) = $1,536,150 Total savings = 83,904 + 1,536,150 = $1,620,054 Build overpass 3.46 Find P in year –1 and then move to year 0. P (yr –1) = 15,000{[1 – (1 + 0.10) 5 /(1 + 0.16) 5 ]/(0.16 – 0.10)} = 15,000(3.8869) = $58,304 P = 58,304(F/P,16%,1) = 58,304(1.16) = $67,632 3.48 P = [2000(P/A,12%,6) – 200(P/G,12%,6)](F/P,12%,1) = [2000(4.1114) – 200(8.9302](1.12) = $7209.1...
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This note was uploaded on 12/06/2009 for the course ELECTRICAL GENG 360 taught by Professor Dr.tarekelmekkawy during the Fall '09 term at Qatar University.
 Fall '09
 Dr.TarekElMekkawy

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