PS3_W11_Sol - MS&E 261 Winter 2010-11 Prof. Warren H....

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MS&E 261 Prof. Warren H. Hausman Winter 2010-11 SOLUTIONS TO PROBLEM SET #3 Note: The stated solutions iterate to find the optimal Q values. As we said in class, use of the EOQ value for Q is a close approximation and is suggested for all calculations both for homework and exams. 1. Type 2 service of 95%. Requires iterative solution. Q 0 = EOQ = 1265 n(R 1 ) = (1 - β )Q = (.05)(1265) = 63.25 L(z 1 ) = = .3673 z 1 = .065, 1 - F(R 1 ) = .474 Q 1 = = = 1405 n(R 2 ) = (1 - β )Q 1 = (.05)(1405) = 70.25 L(z 2 ) = = .4080 z 2 -.02 1 - F(R 2 ) = .508 R 2 = σ z + μ = 1397 Q 2 = = 1411 n(R 3 ) = (.05)(1411) = 70.54 L(z 3 ) = .4097, z 3 -.02, R 3 = 1397 Same value. Stop. (Q,R) = (1411,1397)
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Imputed p = = $ .35 2. Weekly demand has mean 38 and standard deviation LTD has μ = 38 x 3 = 114 and σ = = 19.75 d) Use the relationship n(R) = (1 - β )Q to determine R and the fact that n(R) = σ sL(z). Hence, L(z) = = .10026 From Table B-4, z .90. Since R = σ z + μ we obtain R = (19.75)(.90) + 114 = 132. The imputed shortage cost is:
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PS3_W11_Sol - MS&E 261 Winter 2010-11 Prof. Warren H....

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