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Summer CH6 Solutions

# Summer CH6 Solutions - MIME 310 ENGINEERING ECONOMY...

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1 M I M E 3 1 0 E N G I N E E R I N G E C O N O M Y SOLUTIONS TO PROBLEM EXERCISES – PROJECT EVALUATION CRITERIA 1. The benefits associated with the two types of power plants must be assumed identical, oth- erwise the comparison could not be made. Accordingly, the alternatives are compared on a cost basis. i) All elements for each type of plant are converted to equivalent annual values and added algebraically to determine the equivalent annual cost. Steam plant: 200 000 (A/P,15%,10) + 40 000 - 50 000 (A/F,15%,10) 200 000 (0.1993) + 40 000 - 50 000 (0.0493) = 77 395 Using a financial calculator with N=10, I/Y=15%, PV=200 000 and FV= -50 000, compute PMT to obtain -37 395. Change the sign and add 40 000 to obtain 77 395. Diesel plant: 300 000 (A/P,15%,10) + 30 000 - 50 000 (A/F,15%,10) 300 000 (0.1993) + 30 000 - 50 000 (0.0493) = 87 325 The steam plant is preferred because it has the lower equivalent annual cost. ii) All elements are converted to present worth equivalents and added algebraically. Steam plant: 200 000 + 40 000 (P/A,15%,10) - 50 000 (P/F,15%,10) 200 000 + 40 000 (5.0188) - 50 000 (0.2472) = \$388 392 Using a financial calculator with N=10, I/Y=15%, PMT=40 000 and FV= -50 000, compute PV to obtain -188 312. Change the sign and add 200 000 to obtain 388 392. Diesel plant: 300 000 + 30 000 (P/A,15%,10) - 50 000 (P/F,15%,10) 300 000 + 30 000 (5.0188) - 50 000 (0.2472) = \$438 204 Since the alternatives have the same service period, the present worth equivalents can be com- pared directly. The results suggest the same conclusion as in part i, i.e. the steam plant is pre- ferred. 2. All elements are converted to equivalent annual values and added algebraically to deter- mine the equivalent annual cost. Initial cost: 10 000 (A/P,12%,8) = 10 000 (0.2013) = 2013 Salvage value: 2000 (A/F,12%,8) = 2000 (0.0813) = 163 Maintenance costs: 200 Periodic overhauls: 600 [(P/F,12%,2) + (P/F,12%,4) + (P/F,12%,6)] (A/P,12%,8) 600 [0.7972 + 0.6355 + 0.5066] 0.2013 = 234

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