hmwk3solution

# hmwk3solution - q such that F ( q ) ≥ p-c v p = 1 2 . So...

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ISyE 3232 Stochastic Manufacturing and Service Systems Fall 2011 H. Ayhan 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 } 2 10 1 10 4 10 2 10 1 10 (a) E ( D )= 11.9 (b) Since E [ min { D, 11 } ] = 10.8 and E [ max { 11 - D, 0 } ]= 0.2 it follows that expected proﬁt under the policy q = 11 is: g (11)= 15 E [ min { D, 11 } ] - 2 E [ max { 11 - D, 0 } ]= 161.6. (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.The objective function is the expectation of cost, which is minimized or the expectation of proﬁt which is maximized. (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 6 20 4 20 5 20 2 20 2 20 The appropriate decision rule is to choose the smallest
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Unformatted text preview: q such that F ( q ) ≥ p-c v p = 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 , hE [ max { 12-D, } ] = 12.5 and E [ C ( Q )] = 480 × 12+650+12.5 = 6422.5. • (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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## This note was uploaded on 12/07/2011 for the course ISYE 3232 taught by Professor Billings during the Fall '07 term at Georgia Tech.

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