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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2 Therefore, the equivalent annual cost of the machine tool is: 2013 + 200 + 234 - 163 = $2284 Using a financial calculator with N=8, I/Y=12%, PV=10 000 and FV= -2000, compute PMT to obtain -1850.42. Change the sign and add 200 to obtain 2050.42. Using the cash flow work- sheet with CF 1 =0, CF 2 =600, CF 3 =0, CF 4 =600, CF 5 =0 and CF 6 =600, compute NPV using I=12% to obtain 1163.61. Enter this value in PV (PV, 2 nd QUIT), clear FV (0, FV) and compute PMT to obtain -234.24. Change the sign and add to 2050.42 to obtain 2284.66. 3. i) The net present value is the algebraic sum of discounted cash flows. For instance, the net present values determined at a discount rate of 5 percent are: PROJECT A -800 + 380 (P/A,5%,4) = -800 + 380 (3.5460) = 547 Using the cash flow worksheet in the BA II Plus with CF 0 =-800, CF 1 =380 and F 1 =4, compute NPV using I=5% to obtain 547.46.
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This note was uploaded on 10/06/2009 for the course MIME 310 taught by Professor Bilido during the Summer '08 term at McGill.

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Summer CH6 Solutions - MIME 310 ENGINEERING ECONOMY...

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