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IEOR150F10_hw07_sol

IEOR150F10_hw07_sol - IEOR 150 Fall 2010 Homework 6 1...

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IEOR 150, Fall 2010 Homework 6 1. Because D U (0 , 1), we have its pdf f ( x ) = 1 and cdf F ( x ) = x for all x [0 , 1]. We also have the inverse cdf F - 1 ( y ) = y for all y [0 , 1]. (a) i. The unit overage cost is c = 1 and the unit underage cost is r - c = 1. Therefore, we have q D = F - 1 ( 1 1 + 1 ) = 1 2 . ii. The expected sales quantity under an inventory level q [0 , 1] is min { D, q } = Z q 0 xf ( x ) dx + Z 1 q qf ( x ) dx = Z q 0 xdx + q Z 1 q dx = 1 2 q 2 + q (1 - q ) = q - 1 2 q 2 . Therefore, we have M D = r h q D - 1 2 ( q D ) 2 i - cq D = 1 4 . (b) i. The unit overage cost is w and the unit underage cost is r - w = 1. Therefore, we have q I ( w ) = F - 1 ( r - w r - w + w ) = r - w r . ii. We have M I = max w ( w - c ) r - w r · = max w - w 2 + ( r + c ) w - cr r . Take first order derivative and set it to 0, we get - 2 w * + ( r + c ) = 0 w * = r + c 2 = 3 2 . Note that w c > c = 1. iii. q I = q I ( w * ) = r - w * r = 1 4 . iv. M I = ( w * - c ) q I = 1 8 , R I = r h q I - 1 2 ( q I ) 2 i - w * q I = 1 16 , and S I = M I + R I = 3 16 . (c) q I = 1 4 < 1 2 = q D and M D = 1 4 > 3 16 = S I . 2. Let t i , d i , F i , L i , and T i , be the processing time, due time, flow time, lateness, and tardiness of job i .
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