IEOR150F10_hw03_sol

# IEOR150F10_hw03_sol - IEOR 150, Fall 2010 Suggested...

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Unformatted text preview: IEOR 150, Fall 2010 Suggested Solution to Homework 3 1. Recall that in Problem 3 of Homework 1, we discussed how to combine orders between hex nuts and molly screws. We mentioned that the optimal policy among all possible policies has not been found in that problem. Now we are going to find the optimal policy. Because we have only two products, we have only two possibilities: Order them separately or order them together. The optimal way to order them separately has been found in Part 3(a). In this problem we shall focus on how to optimally order them together. Since we want to order them together, they share the same order cycle time. Denote this common cycle time by T . (a) The annual holding cost is 100 T . (b) The order quantity is Q H ( T ) = 20000 T . The annual holding cost is h H Q H ( T ) 2 = 375 T . (c) The order quantity is Q M ( T ) = 14000 T . The annual holding cost is h M Q M ( T ) 2 = 665 T . (d) The problem to solve is min T 100 T + 375 T + 665 T = min T 100 T + 1040 T. The first order condition leads to T * = r 100 1040 = 0 . 31 year . (e) The corresponding total annual cost is 100 . 31 + 2080 × . 31 = \$644 . 98 , which is lower than \$662, the best we did in Problem 1 of Homework 3....
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IEOR150F10_hw03_sol - IEOR 150, Fall 2010 Suggested...

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