Unformatted text preview: a) Y(100) = (1000)(100).3219 = $227.09 b) Y(10,000) = (1000)(10,000).3219 = $51.57 1.38 a) From Figure 1.14, a value of a = .58 gives a value of u very close to 1.0. Solving x = u/r = 1/.12 = 8.33 years. The optimal size of each addition is xD = (8.33)(3,000) = 25,000 tons. b) f(y) = .0205y .58 = (.0205)(25,000) .58 = 7.29. Hence, the cost of each new addition is $7.29 million. c) At 12% interest the effective annual discount rate is 1/(1+0.12) = .8929. However replacements are made once every 8.33 years giving a discount factor per replacement of (.8929) 8.33 = .3891. Hence, the present cost of the next four replacements is 7.29[.39 + (.39) 2 + (.39) 3 + (.39) 4 ] = 7.29[.6245] = $4.55 million 11...
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This note was uploaded on 04/18/2010 for the course ISEN 315 taught by Professor Den during the Spring '10 term at Texas A&M University, Corpus Christi.
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