assignment9

Mmm 0 1 1 1 2 2 3 m 12

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Unformatted text preview: Pij �M k=0 πk m=0 Pkm Since the denominator is independent of i and j , and since the reversibility of the origi­ nal chain implies that πi Pij = πj Pji , we have the desired result. � π i Pij = �M Exercise 5.10: a) M/M/1: �� �� �� �0��� �� � �� �� �1��� λδ µδ 1−λδ λδ µδ 1−(λ+µ)δ �� � �� �� �2��� λδ µδ 1−(λ+µ)δ �� � �� �� �3��� λδ µδ 1−(λ+µ)δ �� � �� �� �4��� λδ � ... µδ 1−(λ+µ)δ From (5.40), we have πi = ρi (1 − ρ), for i ≥ 0 where ρ = λ/µ and ρ < 1 (positive recurrent). M/M/m: �� �� �� �0��� 1−λδ λδ µδ �� � �� �� �1��� 1−(λ+µ)δ λδ 2µδ �� � �� �� �2��� λδ � ... � 3µδ λδ mµδ 1−(λ+2µ)δ �� �m�� �� � ��� λδ mµδ 1−(λ+mµ)δ Using the steady-state equations (5.25) and defining ρ = λ/(mµ), we have πi /πi−1 = λ/(iµ), πi /πi−1 = ρ, for i < m; fo...
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This document was uploaded on 03/19/2014 for the course EECS 6.262 at MIT.

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