hmwk3sol - = 1 2 . So for q = 17 we have 11 20 >...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 1 2 . So for q = 17 we have 11 20 &gt; 1 2 . For q &lt; 17 F ( q ) &lt; 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...
View Full Document

Ask a homework question - tutors are online