IEOR150F10_hw03_sol

# With each of these four values we may find the

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1. With each of these four values, we may find the corresponding order cycle time and the annual holding and setup cost: k F 1 = 2 T F 1 = 2 2 T = 0 . 4 G F 1 = K 1 T F 1 + h 1 λ 1 T F 1 2 = \$900 , k C 1 = 3 T C 1 = 2 3 T = 0 . 8 G C 1 = K 1 T C 1 + h 1 λ 1 T C 1 2 = \$1050 , k F 2 = - 2 T F 2 = 2 - 2 T = 0 . 025 G F 2 = K 2 T F 2 + h 2 λ 2 T F 2 2 = \$1175 , and k C 2 = - 1 T C 2 = 2 - 1 T = 0 . 05 G C 2 = K 2 T C 2 + h 2 λ 2 T C 2 2 = \$1150 . Since G F 1 < G C 1 and G C 2 < G F 2 , we know k * 1 = 2 and k * 2 = - 1 are the optimal k 1 and k 2 , respec- tively. (c) For retail store 1, we have G C 1 G * 1 = 900 894 . 43 = 1 . 0062 , which means the additional cost incurred by adopting the power-of-two policy is around 0 . 6% of the optimal total cost. For retail store 2, we have G C 2 G * 2 = 1150 1095 . 45 = 1 . 0498 , which means the additional cost incurred by adopting the power-of-two policy is around 5% of the optimal total cost. Our results show the effectiveness of the power-of-two policy. 3
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• Fall '09
• Trigraph, retail store, annual holding cost

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