hmwk3sol

hmwk3sol - = 1 2 So for q = 17 we> 1 2 For q< 17 F q<...

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ISyE 3232 Stochastic Manufacturing and Service Systems Spring 2011 J. Dai Solutions to Homework 3 1. Data: p = 18, c v = 3, s v = 1, and the distribution of demand is: D = d 10 11 12 13 14 P { D = d } 1 10 2 10 4 10 2 10 1 10 (a) E ( D )= 12 (b) Since E [ min { D, 11 } ] = 10.9 and E [ max { 11 - D, 0 } ]= 0.1 it follows that expected proﬁt under the policy q = 11 is: g (11)= 15 E [ min { D, 11 } ] - 2 E [ max { 11 - D, 0 } ]= 163. (c) The smallest q satisfying F ( q ) p - c v p - s v = 15 17 is q = 13. (d) This is optimal provided that the environment does not change and that the game is played many times. (e) From tables of the Gaussian (i.e., normal) distribution or by using a calculator: F ( q ) = P { D < q } = 15 17 q = 1016 . 78 2. Data: p = 20, c v = 10, and the demand distribution is: b 15 16 17 18 19 20 P { D = d } 1 20 4 20 6 20 5 20 2 20 2 20 The appropriate decision rule is to choose the smallest q such that F ( q ) p - c v p
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Unformatted text preview: = 1 2 . So for q = 17 we have 11 20 > 1 2 . For q < 17 F ( q ) < p-c v p = 1 2 . 3. Data: unit retail price is 600, unit delivery charge is 50; consequently the unit penalty cost is p = 650, unit variable cost is c v = 480, unit inventory holding cost is h = 25, and the distribution of demand is: D = d 10 11 12 13 14 15 P { D = d } 1 6 1 6 1 6 1 6 1 6 1 6 • (a) pE [ max { D-12 , } ] = p = 650 and hE [ max { 12-D, } ] = 12.5 • (b) q is smallest number such that F ( q ) ≥ p-c v p + h = 170 675 = 34 135 ⇒ q = 11. • (c) Now q solves F ( q ) = p-c v p + h = 170 675 = 34 135 ⇒ q = 933 . 13; so order 933 cameras. 1...
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