Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 7 - Section 7.7 - Markov Systems - Exercises - Page 532: 33

Answer

$v_\infty=\begin{bmatrix} \dfrac{1}{3}&\dfrac{1}{2}&\dfrac{1}{6} \end{bmatrix}$

Work Step by Step

The steady-state distribution vector $v_\infty$ can be written as: $v_\infty P=v_∞$ where, $v_\infty=[x~~y~~z]$ This gives: $[x~~y~~z] \begin{bmatrix} 0 & 1&0 \\ 1/3& 1/3&0\\1&0&0 \\ \end{bmatrix} =[x~~y~~z]$ We can have the following equations: $$1/3y+z=x\\ x+1/3 y=y\\1/3 y=z$$ or, $$-x+1/3y+z=0\\ x-2/3y=0\\1/3y-z=0$$ Also, we have: $x+y+z=1$ So, the new system of equations are: $x+y+z=1\\ x-2/3y=0 \\ 1/3y- z=0 $ After solving the above equations, we get: $x=1/3; y=1/2 ; z=1/6$ Thus, the required steady-state distribution vector $v_\infty$ $v_\infty=\begin{bmatrix} \dfrac{1}{3}&\dfrac{1}{2}&\dfrac{1}{6} \end{bmatrix}$
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