hmwk9sol - ISyE 3232 Stochastic Manufacturing and Service...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

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

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....
View Full Document

{[ snackBarMessage ]}

Page1 / 4

hmwk9sol - ISyE 3232 Stochastic Manufacturing and Service...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online