hw4 - solutions - f08

# hw4 - solutions - f08 - IE 336: Operations Research -...

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IE 336: Operations Research - Stochastic Models Homework Set #4 Solutions Fall 2008 1. (a) The transition diagram is; (b) It is impossible that p = 17 36 . Since a sum of a row of a transition matrix is 1, p + q 2 + r = 1 q 2 + r = 19 36 q = 19 18 - 2 r q 2 + 3 2 r + q 4 = 1 3 4 q + 3 2 r = 1 However, 3 4 q + 3 2 r = 19 24 - 3 2 r + 3 2 r is not equal to 1. (c) q = 2 3 Since, from (b), 3 4 q + 3 2 r = 1 ,r = 1 3 In the ﬁrst row of the transition matrix, p + q 2 + r = 1 p = 1 3 (d) P = 1 2 3 1 1 / 3 1 / 3 1 / 3 2 0 1 / 3 2 / 3 3 1 / 6 1 / 6 2 / 3 = 0 . 3333 0 . 3333 0 . 3333 0 0 . 3333 0 . 66667 0 . 1667 0 . 1667 0 . 6667 The two step and three step transition matrices are P 2 and P 3 respectively. 1

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Therefore, the two step matrix is; P 2 = P * P = 1 2 3 1 0 . 1667 0 . 2778 0 . 5556 2 0 . 1111 0 . 2222 0 . 6667 3 0 . 1667 0 . 2222 0 . 6111 The three step matrix is; P 3 = P * P * P = 1 2 3 1 0 . 1481 0 . 2407 0 . 6111 2 0 . 1481 0 . 2222 0 . 6296 3 0 . 1574 0 . 2315 0 . 6111 2. (a) P ( X 1 = aa | X 0 = ab ) = P ( X 1 = aa | X 0 = ab,father = aa ) P ( father = aa ) + P ( X 1 = aa | X 0 = ab,father = ab ) P ( father = ab ) + P ( X 1 = aa | X 0 = ab,father = bb ) P ( father = bb ) Since P ( X 1 = aa | X 0 = ab,father = aa ) = 1 2 ,P ( X 1 = aa | X 0 = ab,father = ab ) = 1 4 , P ( X 1 = aa | X 0 = ab,father = bb ) = 0 ,P ( father = aa ) = p 2 ,P ( father = ab ) = 2 p (1 - p ) , P ( X 1 = aa | X 0 = ab ) = 1 2 p 2 + 1 4 (2 p (1 - p )) = p 2 . P
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## This note was uploaded on 03/29/2010 for the course IE 383 taught by Professor Leyla,o during the Spring '08 term at Purdue University-West Lafayette.

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hw4 - solutions - f08 - IE 336: Operations Research -...

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