MIME 310 - Solutions Chapter 6

MIME 310 - Solutions Chapter 6 - MIME 310 ENGINEERING...

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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 SET #6 – PROJECT EVALUATION CRITERIA 1. The present worth cost equivalent (PW) of each machine is determined at a discount rate of 10 percent. Under the repeated-projects assumption, the machines must be compared over a period of 18 years, the smallest common multiple of their lives. PW A = 9000 [1 + (P/F,10%,6) + (P/F,10%,12)] + 5000 (P/A,10%,18) = 9000 [1 + 0.5645 + 0.3186] + 5000 (8.2014) = $57 955 PW B = 16 000 + (16000 - 3000) (P/F,10%,9) - 3000 (P/F,10%,18) + 4000 (P/A,10%,18) = 16 000 + 13 000 (0.4241) - 3000 (0.1799) + 4000 (8.2014) = $53 779 Machine B is preferred because it has the lower present worth cost equivalent. 2. The maximum bid is the present worth equivalent (PW) of benefits over the expected period of ownership. The required return on investment is used as the discount rate. PW = (15 000 - 9000) (P/A,9%,12) + 150 000 (P/F,9%,12) = 6000 (7.1607) + 150 000 (0.3555) = $96 289 The bid should not exceed $96 289. 3. Since the study period is 50 years, the galvanized building (g) must be replaced after the first 30 years. The reinforced-concrete building (c) and the first galvanized building have no salvage value. The second galvanized structure, which has a remaining useful life of 10 years at the end of the 50-year study period, has a salvage value of: 310 000 (1/3) = 103 333 Present worth cost equivalent (PW) of each alternative: PW c = 440 000 + [19 200 + 440 000 (0.012)] (P/A,12%,50) = 440 000 + 24 480 (8.3045) = $643 294 PW g = 310 000 [1 + (P/F,12%,30)] + [30 600 + 310 000 (0.005)] (P/A,12%,50) - 103 333 (P/F,12%,50) = 310 000 [1 + 0.0334] + 32 150 (8.3045) - 103 333 (0.00346) = $586 986 The galvanized building is preferred because it has the lower present worth cost equivalent. 38
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4. The capitalized costs (CC) of the ditch-and-tunnel system (dt) and the wood-and-concrete flume (wc) are: CC dt = 500 000 + 4000 / 0.06 = $566 667 CC wc = 240 000 + 12 000 / 0.06 + [100 000 (A/F,6%,15)] / 0.06 = 240 000 + 200 000 + 100 000 (0.0430) / 0.06 = $511 667 The wood-and-concrete flume is preferred because it has the lower capitalized cost. 5. The lump-sum payment is equal to the present worth equivalent (PW) at time 3 of the six annual payments (due at the end of years 5 to 10). PW = 150 000 (P/A,7%,6) (P/F,7%,1) = 150 000 (4.7665) (0.9346) = $668 216 A lump-sum payment of $668 216 is required. 6. The discount rate used to compare the alternatives is 12%, the minimum acceptable return on investment. Cost of repairing machine today: $2000 Present worth equivalent (PW) of opportunity cost (i.e. spending more on operating expenses over the 5-year period if the machine is not repaired today): PW = [200 + 200 (A/G,12%,5)] (P/A,12%,5) = [200 + 200 (1.7746)] 3.6048 = $2000 The alternatives are equivalent because they have equal present worth equivalents. 7.
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MIME 310 - Solutions Chapter 6 - MIME 310 ENGINEERING...

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