HW1_Sol - a Y(100 =(1000(100.3219 = $227.09 b Y(10,000...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
*** Do Not Circulate *** HW1 Solutions 1.24 Although many firms should follow this pattern of evolution, it clearly does not make sense for all firms to do so. For example, the commercial printer shown in Figure 1.13 will never evolve beyond a jumbled flow shop and never beyond being a low volume one-of-a-kind producer. 1.25 a) I-I b) IV-IV c) II-II (could be III-III depending on the product volume and diversity of products produced) d) III-III e) I-I 1.29 Substituting u = 100,000 into the relationship Y(u) = 22.88u -.42276 gives Y(100,000) = .1761 hours 1.34 The curve is of the form Y(u) = au -b where a is the cost of producing the first unit and b = -ln(L)/ln(2). Here a = $1,000 and b = -ln(.80)/ln(2) = .3219
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 1-1...
View Full Document

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.

Ask a homework question - tutors are online