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Unformatted text preview: IEOR 3106, Fall 2007, Professor Whitt Solutions for Visit to the Post Office. The Exponential Distribution 1. The LackOfMemory Property A nonnegative random variable X is said to have the lack of memory (LOM) property of P ( X > x + y  X > y ) = P ( X > x ) for all x > , y > . Theorem 0.1 A nonnegative random variable has the LOM property if and only if it has an exponential distribution. 2. A Trip to the Post Office Here is a variant of the Post Office Problem with different names: Five students – Ruxian, Shiqian, Yori, Yotam and Yu Hang – simultaneously enter an empty post office with three clerks. Ruxian, Shiqian and Yori begin to receive service immediately, while Yotam and Yu Hang wait in a single line, ready to be served by the first free clerk, with Yotam at the head of the line (to be served first when a server becomes free). Suppose that the service times of the three clerks are independent exponential random variables, each with mean 2 minutes. (a) What is the expected time (from the moment the students enter the post office) until Yotam completes his service? ———————————————————— Yotam must wait until one of the first three finish service. We use the fact that a minimum of independent exponential random variables is again exponentially distributed with a rate equal to the sum of the component rates. Here the minimum has a rate equal to three times an individual service rate. Since the mean is the reciprocal of the rate, the mean time for the first service completion is 2 / 3 minute. We must add to that Yotam’s own expected service time. Hence, the expected time until Yotam finishes service is (2 / 3) + 2 = (8 / 3) = 2 . 33 minutes. ———————————————————— (b) What is the expected time (again since entering the post office) until all five students finish service? ———————————————————— All three clerks are working for the first three service completions; for the fourth service completion two clerks are working; for the fifth (last) service completion one clerk is working. Hence the overall expected time is: (2 / 3) + (2 / 3) + (2 / 3) + (2 / 2) + (2 / 1) = 5 minutes. We use the lack of memory property; at each service completion, the remaining service time of each service not completed is exponential, just as if the service began at that point.service not completed is exponential, just as if the service began at that point....
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 Spring '11
 WhitWard
 Variance, Laplace, Probability theory, Yotam

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