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HW6_soln

# HW6_soln - Homework Set#6 Solutions IE 336 Spring 2011 1(a...

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Homework Set #6 Solutions IE 336 Spring 2011 1. (a) p = 1 / 4, q = 1 / 8, r = 1 / 4. Transition diagram omitted. P = 2 q r 2 p 3 r p 2 q 6 q 2 p 3 r 3 = 1 / 4 1 / 4 1 / 2 3 / 4 1 / 8 1 / 8 3 / 4 1 / 6 1 / 12 (b) Given p (4) = 0 . 2 0 . 6 0 . 2 , the vector at time step 8 is: p (8) = p (4) P 4 = 0 . 4813 0 . 2034 0 . 3153 (c) The steady-state probabilities are uniquely determined by the dependent set of linear equations ( P T - I ) π T = 0 and the additional normalizing equation π 1 = 1: π 1 π 2 π 3 = - 3 / 4 3 / 4 3 / 4 1 / 4 - 7 / 8 1 / 6 1 1 1 - 1 0 0 1 = 1 / 2 1 / 5 3 / 10 2. (a) Mean sojourn time in state 2. E ( N 2 ) = 1 1 - p 22 = 1 1 - 1 / 8 = 8 7 = 1 . 1429 (b) If the process begins in state 3, the probability that it will be in state 2 for the first time four steps later is f (4) 32 : f (1) 32 = p 32 = 1 6 f (2) 32 = p (2) 32 - f (1) 32 p (1) 22 = 2 9 - 1 6 · 1 8 = 29 144 f (3) 32 = p (3) 32 - 2 X k =1 f ( k ) 32 p (3 - k ) 22 = 163 864 - 1 6 · 43 192 - 29 144 · 1 8 = 109 864 f (4) 32 = p (4) 32 - 3 X k =1 f ( k ) 32 p (3 - k ) 22 = 210 1021 - 1 6 · 76 403 - 29 144 · 43 192 - 109 864 · 1 8 = 189 1667 = 0 . 1134 (c) Mean first passage time from state 1 state 2.

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