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Unformatted text preview: IEOR 6711: Stochastic Models I, Fall 2003, Professor Whitt Solutions to Final Exam: Thursday, December 18. Below are six questions with several parts. Do as much as you can. Show your work. 1. TwoPump Gas Station (12 points) Potential customers arrive at a fullservice, twopump gas station according to a Poisson process at a rate of 40 cars per hour. There are two service attendants to help customers, one for each pump. If the two pumps are busy, then arriving customers wait in a single queue, to be served in the order of arrival by the first available pump. However, customers will not enter the station to wait if there are already two customers waiting, in addition to the two in service. Suppose that the amount of time required to service a car is exponentially distributed with a mean of three minutes. ———————————————————————– This problem is a minor variant of Problem 15 in Chapter 6, on page 393, of the easier Ross book, Introduction to Probability Models . To answer the questions posed here, we need to find the steadystate distribution of the CTMC. In general, the steadystate distribution can be found by solving the equation αQ = 0, where α is the steadystate probability vector. This CTMC is a birthanddeath process, so we can find the steadystate solution directly, by solving the local balance equations α k λ k = α k +1 μ k +1 for all k . But, first, we need to read carefully to make sure we get the model right. We need to recognize that there are five states: k for 0 ≤ k ≤ 4. We thus can write down the steadystate probabilities in terms of an unknown x , and then normalize to find x : α = 1 * x, α 1 = λ μ 1 * x, α 2 = λ μ 1 λ 1 μ 2 * x , α 3 = λ μ 1 λ 1 μ 2 λ 2 μ 3 * x , α 4 = λ μ 1 λ 1 μ 2 λ 2 μ 3 λ 3 μ 4 * x . A main issue is to be careful and get the rates right, using the same units. We could either use cars per hour or cars per minute, but we should be consistent. We should realize that a mean service time of 3 minutes means one service per 3 minutes or 20 services per hour. So, here, inserting the birth rates and death rates, we get α = 1 * x, α 1 = 40 20 * x, α 2 = 40 20 40 40 * x , α 3 = 40 20 40 40 40 40 * x , α 4 = 40 20 40 40 40 40 40 40 * x or α = 1 * x, α 1 = 2 * x, α 2 = 2 * x , α 3 = 2 * x , α 4 = 2 * x , so that α = 1 / 9 , α 1 = α 2 = α 3 = α 4 = 2 / 9 . ———————————————————————– (a) In the longrun, what rate are cars served? ———————————————————————– Cars are served at rate 20 * α 1 + 40 * ( α 2 + α 3 + α 4 ) = 280 / 9 = 31 . 111 ... per hour ....
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This note was uploaded on 01/05/2010 for the course ECE 01 taught by Professor All during the Spring '09 term at Aarhus Universitet.
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