IEOR
IEOR150F10_hw04_sol

# Follows a normal distribution with mean 280 and

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follows a normal distribution with mean 280 and standard deviation 77, let μ = 280 × 5 = 1400 units be the mean of the demand and σ = 5 × 77 2 = 172 . 18 units be the standard deviation of the demand during a lead time. Finally, let λ = 280 × 12 = 3360 units be the annual demand size. (a) We iteratively compute the order quantity Q and the reorder point R by the following formulas: Q i = r 2[ K + pn ( R i - 1 )] λ h , R i = F - 1 ˆ 1 - Q i h ! , z i = R i - μ σ , and n ( R i ) = σL ( z i ) . Note that we initiate this process by setting n ( R - 1 ) = 0 so that Q 0 = q 2 h , the regular EOQ formula. Also note that we use Table A-4 in the textbook to find the value of L ( z i ) [you may also use softwares to find L ( z i )]. For this problem, the iterative calculation is summarized below. Iteration i 0 1 2 Q i 1264.91 1322.16 1326.15 R i 1786.71 1783.76 1783.56 z i 2.25 2.23 2.23 n ( R i ) 0.72 0.77 0.77 We stop at the end of iteration 2 since R 2 = R 1 and conclude that Q * = 1326.15 units and R * = 1783.56 units . The safety stock is R - λτ = 383.56 units . (b) The expected annual holding cost is h ( Q * 2 + R * - λτ ) = \$439 . 59 . The expected annual setup cost is Q * = \$253 . 36 . The expected annual stock-out penalty cost is λn ( R * ) Q * = \$25 . 13 . (c) According to Part 3b, the expected annual total cost is thus 439 . 59+253 . 36+25 . 13 = \$718 . 08. Now suppose there is no demand uncertainty, we may use the basic EOQ formula to find the annual total cost as 2 Kλh = \$531 . 28. The cost of uncertainty is thus 718 . 08 - 531 . 28 = \$186 . 82 . You may want to verify that the optimal order quantity and the optimal reorder point with no uncertainty are both lower than Q * and R * . 2
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• Fall '09
• Cumulative distribution function, \$1.5, \$0.05, \$0.42, \$0.27

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