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hw10sln

# hw10sln - IEOR 161 Operations Research II University of...

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IEOR 161 Operations Research II University of California, Berkeley Spring 2008 Homework 10 Solution Chapter 4. 28. Let the states be the pair of outcomes of two consecutive trials, i.e. { ss, fs, sf, ff } . Therefore the transition probabilities are 3 / 4 0 1 / 4 0 2 / 3 0 1 / 3 0 0 2 / 3 0 1 / 3 0 1 / 2 0 1 / 2 Limiting probabilities are π ss = 1 / 2 , π fs = 3 / 16 , π sf = 3 / 16 , π ff = 1 / 8. Therefore the proportion of time that a trial is a success is π ss + π fs = 11 / 16. 29. π 1 = 0 . 7 π 1 + 0 . 2 π 2 + 0 . 1 π 3 π 2 = 0 . 2 π 1 + 0 . 6 π 2 + 0 . 4 π 3 π 3 = 0 . 1 π 1 + 0 . 2 π 2 + 0 . 5 π 3 π 1 + π 2 + π 3 = 1 π 1 = 6 / 17 , π 2 = 7 / 17 , π 3 = 4 / 17 30. Let the state X n be 0 if the nth vehicle is a car and let it be 1 if it is a truck. Then X n is a Markov chain and the transition probabilities are 4 / 5 1 / 5 3 / 4 1 / 4 π 0 = 4 / 5 π 0 + 1 / 5 π 1 π 1 = 1 / 5 π 0 + 1 / 4 π 1 π 0 + π 1 = 1 π 0 = 15 / 19 , π 1 = 4 / 19 Thus 4 out of every 19 vehicles are trucks. 31. Let the state X n on day n be 0 if sunny, 1 if cloudy, and 2 if rainy. The transition probability matrix is P = 0 1 / 2 1 / 2 1 / 4 1 / 2 1 / 4

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hw10sln - IEOR 161 Operations Research II University of...

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