Practice Midterm4

Practice Midterm4 - IEOR 4106 Introduction to Operations Research Stochastic Models Spring 2005 Professor Whitt Second Midterm Exam Chapters 5-6 in

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IEOR 4106: Introduction to Operations Research: Stochastic Models Spring 2005, Professor Whitt, Second Midterm Exam Chapters 5-6 in Ross, Thursday, March 31, 11:00am-1:00pm Open Book: but only the Ross textbook plus one 8 × 11 page of notes Justify your answers; show your work. 1. The IEOR Department Ricoh Printer ( 30 points) The Columbia IEOR Department has a versatile Ricoh printer that can rapidly print one- sided and two-sided copies, but unfortunately it often goes down. Ricoh is alternately up (working) and down (waiting for repair or under repair). The average up time (time until breakdown) is 4 days, while the average down time (time until repair) is 3 days. Assume continuous operation. Let X ( t ) = 1 if the Ricoh is working at time t , and let X ( t ) = 0 otherwise. (a) What do we need to assume about the successive up and down times in order to make the stochastic process { X ( t ) : t 0 } a continuous-time Markov chain (CTMC)? —————————————————————————— We need the successive up times and down times to be mutually independent random variables. In addition, these random variables should have exponential distributions. The lack-of-memory property of the exponential distribution is critical for getting the Markov property for the stochastic process { X ( t ) : t 0 } . The exponential distribution has a single parameter, which can be taken to be its mean. Since the means are already specified, nothing more needs to be assumed, beyond assuming that the means agree with the specified averages. —————————————————————————— Henceforth assume that these extra assumptions are in place, so that indeed the stochastic process { X ( t ) : t 0 } is a CTMC. (b) Construct the CTMC; i.e., specify the model. —————————————————————————— A CTMC can be specified by its rate matrix , Q . That takes a simple form here because there are only two states: 0 and 1. The rate matrix here, ordering the two states 0 and 1, is Q = ± - 1 / 3 1 / 3 1 / 4 - 1 / 4 where the rates are expressed per day. That is, the repair rate is 1 / 3 per day, while the failure rate is 1 / 4 per day. In this case we have a birth-and-death (BD) process, so that λ 0 = Q 0 , 1 = 1 / 3, while μ 1 = Q 1 , 0 = 1 / 4. That is an equivalent specification; i.e., we say that it is BD and we specify these single λ i and μ i values. —————————————————————————— (c) Assuming that Ricoh has been working continuously for 7 days, what is the probability that it will remain working at least 8 more days? ——————————————————————————
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The holding time in each state is exponentially distributed. The exponential up time has mean 4 days, and thus rate 1 / 4, as indicated above. Let T be the failure time. Then P ( T > 8 + 7 | T = 7) = P ( T > 8) = e - (1 / 4)8
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This note was uploaded on 04/08/2008 for the course IEOR 4106 taught by Professor Whitt during the Spring '08 term at Columbia.

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Practice Midterm4 - IEOR 4106 Introduction to Operations Research Stochastic Models Spring 2005 Professor Whitt Second Midterm Exam Chapters 5-6 in

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