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Unformatted text preview: IEOR 4106: Spring 2011, Professor Whitt Topics for Discussion: Thursday, April 14 Alternating Renewal Processes and The Renewal Equation 1 An Example: Syed Khurshid’s Computer Syed’s computer has three critical parts, each of which is needed for the computer to work. The computer runs continuously as long as the three required parts are working. The three parts have mutually independent exponential lifetimes before they fail. The expected lifetime of parts 1, 2 and 3 are 10 weeks, 20 weeks and 30 weeks, respectively. When a part fails, the computer is shut down and an order is made for a new part of that type. When the computer is shut down (to order a replacement part), the remaining two working parts are not subject to failure. The time to fix part 1 is exponentially distributed with mean 1 week; the time to fix part 2 is uniformly distributed between 1 week and 3 weeks; and the time to fix part 3 has a gamma distribution with mean 3 weeks and standard deviation 10 weeks. (a) Assuming that all parts are initially working, what is the expected time until the first part fails? ——————————————————————— Let T be the time until the first failure. Then T is exponential with rate equal to the sum of the rates; see Chapter 5; i.e., ET = 1 λ 1 + λ 2 + λ 3 = 1 (1 / 10) + (1 / 20) + (1 / 30) = 1 (11 / 60) = 60 11 = 5 . 45 weeks . ——————————————————————— (b) What is the probability that part 1 is the first part to fail? ——————————————————————— Let N be the index of the first part to fail. Since the failure times are mutually independent exponential random variables (see Chapter 5), P ( N = 1) = λ 1 λ 1 + λ 2 + λ 3 = (1 / 10) (1 / 10) + (1 / 20) + (1 / 30) = 6 11 = 0 . 545 . ——————————————————————— (c) What is the long-run proportion of time that the computer is working? ——————————————————————— Now for the first time we need to consider the random times it takes to get the replacement parts. Actually these distributions beyond their means do not affect the answers to the ques- tions asked here. Only the means matter here. Use elementary renewal theory. The successive times that the computer is working and shut down form an alternating renewal process. Or, equivalently, apply renewal reward processes, as in Section 7.4: Look at the expected reward per cycle divided by the expected length of a cycle. Let T be a time until a failure (during which the computer is working) and let D be a down time. A cycle is T + D . Then the long-run proportion of time that the computer is working is ET/ ( ET + ED ). By part (a) above, ET = 60 / 11. It suffices to find ED . To find ED , we consider the three possibilities for the part that fails: ED = P ( N = 1) E [ D | N = 1] + P ( N = 2) E [ D | N = 2] + P ( N = 3) E [ D | N = 3] = (6 / 11)...
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This note was uploaded on 01/16/2012 for the course IEOR 4106 taught by Professor Whitward during the Spring '11 term at Columbia College.
- Spring '11