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Unformatted text preview: ISyE 3232 Stochastic Manufacturing and Service Systems Spring 2011 H. Ayhan and J. Dai Solutions to Homework 9 1. • Approach 1. Let S = { 1 ,...,N } be the state space. From π T P = π , we have π T P = π 1 ... π N p 11 ··· p 1 N . . . . . . . . . p N 1 ··· p NN = h ∑ N i =1 π i p i 1 ··· ∑ N i =1 π i p iN i = π 1 ... π N . Hence, for every state j ∈ S , we have N X i =1 π i p ij = π j = π j N X i =1 p ij , and N X i =1 ( π i π j ) p ij = 0 . Since p ij ≥ 0, π i π j = 0 for every i . Therefore, π = 1 N ··· 1 N . • Approach 2. Consider the following exemplary transition matrices: P 1 = 1 3 2 3 2 3 1 3 , and P 2 = 1 4 1 4 2 4 1 4 2 4 1 4 2 4 1 4 1 4 . Let π 1 and π 2 be the stationary distribution for each transition matrix. Then from the equation system, πP = π and ∑ N i =1 π i = 1, we have π 1 = 1 2 1 2 and π 2 = 1 3 1 3 1 3 . Now, we can guess that if the state space is { 1 ,...,N } , then the stationary distribution will be π = 1 N ··· 1 N . To show that this is correct, we check π T P = π . π T P = 1 N ... 1 N p 11 ··· p 1 N . . . . . . . . . p N 1 ··· p NN = h ∑ N i =1 1 N p i 1 ··· ∑ N i =1 1 N p iN i = h 1 N ∑ N i =1 p i 1 ··· 1 N ∑ N i =1 p iN i = 1 N ··· 1 N = π, where the fourth equality holds because the chain is doubly stochastic....
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 Spring '07
 Billings
 Exponential distribution, TA, stock level

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