homework5soln - ORIE 361/523 Homework 5 Solutions...

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ORIE 361/523 – Homework 5 Solutions Instructor: Mark E. Lewis due February 27, 2008 1. We introduce the states 1, 2, . .., m as the life of a light bulb in units of days. Define a Markov chain X n as the life of the current bulb at day n . The transition matrix looks like: P = 1 2 3 . . . m - 1 m p 1 - p 0 . . . 0 0 p 0 1 - p . . . 0 0 p 0 0 . . . 0 0 . . . . . . . . . . . . . . . . . . p 0 0 . . . 0 1 - p 1 0 0 . . . 0 0 . The stationary distribution π is given by the balance equation: π 1 = π 1 p + π 2 p + . . . + π m - 1 p + π m (0.1) π 2 = π 1 (1 - p ) (0.2) π 3 = π 2 (1 - p ) (0.3) . . . (0.4) π m = π m - 1 (1 - p ) (0.5) These equations together with the normalization equation i π i = 1 have a unique solution. Solving out π 1 = p 1 - (1 - p ) m . This is the probability that the current bulb will be new (with life 1) in the long run. So this is exactly the fraction of time that a student will find a newly-installed fresh light bulb in the lounge. 2. (a) There are 6 states in the state space:
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homework5soln - ORIE 361/523 Homework 5 Solutions...

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