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Unformatted text preview: IEOR 161  Introduction to Stochastic Processes Spring 2010 HW10 Solutions ** Note that the numbering is from Ross 9th Edition 4.36 (a) p P , + p 1 P , 1 = 0 . 4 p + 0 . 6 p 1 (b) p P 4 , + p 1 P 4 , 1 = 0 . 2512 p + 0 . 7488 p 1 (c) p π + p 1 π 1 = p / 4 + 3 p 1 / 4 (d) This is not a Markov Chain. 4.46 (i) Let the state be the number of umbrellas in his present location. The transition probabilities are P ,r = 1 P i,r i = 1 p, i = 1 ,...,r P i,r i +1 = p, i = 1 ,...,r (ii) The limiting probabilities should satisfy π r = π + pπ 1 π i = (1 p ) π r i + pπ r i +1 ,i = 1 ,...,r 1 π = (1 p ) π r r X i =0 π i = 1 and it is easy to verify that they are satisfied by the given solution (iii) The fraction of time our man gets wet is pπ = pq r + q . (iv) When r = 3, the result in (iii) is p (1 p ) 4 p . The first order condition implies p 2 8 p + 4 = 0, therefore p = 8 4 √ 3 2 ....
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This note was uploaded on 03/17/2011 for the course IEOR 161 taught by Professor Lim during the Spring '08 term at Berkeley.
 Spring '08
 Lim
 Operations Research

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