11.Sol.07.Fall - ISyE 3232 Stochastic Manufacturing and...

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Unformatted text preview: ISyE 3232 Stochastic Manufacturing and Service Systems Fall 2007 Professors Hayriye Ayhan and Jim Dai Solutions to Homework 11 1. (a) (b) P (0 . 2) = . 2299 . 1805 . 5896 . 2383 . 3989 . 3628 . 1411 . 0508 . 8081 . (c) P (1 . 0) = . 1673 . 1131 . 7196 . 1688 . 1176 . 7137 . 1662 . 1097 . 7241 . (d) P (1 . 2) = P (1 . 0) P (0 . 2) = . 1669 . 1119 . 7212 . 1675 . 1136 . 7189 . 1665 . 1105 . 7230 . and so P { X (1 . 2) = C | X (0) = A } = 0 . 7212. P (2 . 0) = P (1 . 0) P (1 . 0) = . 1667 . 1111 . 7222 . 1667 . 1112 . 7221 . 1667 . 1111 . 7222 . and so P { X (3 . 0) = A | X (1 . 0) = B } = 0 . 1667. (e) P (5 . 0) = . 1667 . 1111 . 7222 . 1667 . 1111 . 7222 . 1667 . 1111 . 7222 . The Markov Chain has reached steady-state by time 5. 1 2. (a) The arrival rate is = 5 / 3 (per min); the service rate is = 1 / 4 and the reneging rate is = 1 / 8. The state space is S = { (0 , 0) , (0 , 1) , (0 , 2) , (1 , 0) ,...
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This note was uploaded on 11/26/2009 for the course ISYE 3232 taught by Professor Billings during the Fall '07 term at Georgia Institute of Technology.

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11.Sol.07.Fall - ISyE 3232 Stochastic Manufacturing and...

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