lect0303

# Performing the calculation we see that the limiting

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Performing the calculation, we see that the limiting probability vector is α ( α 0 , α 1 , α 2 , α (1 , 2) , α (2 , 1) ) = 44 129 , 16 129 , 36 129 , 24 129 , 9 129 . Thus, the long-run proportion of time that copier 1 is working is α 0 + α 2 = 80 / 129 0 . 62, while the long-run proportion of time that copier 2 is working is α 0 + α 1 = 60 / 129 0 . 47. The long-run proportion of time that the repairman is busy is α 1 + α 2 + α (1 , 2) + α (2 , 1) = 1 - α 0 = 85 / 129 0 . 659, ———————————————————————- (d) Now suppose, instead, that machine 1 is much more important than machine 2, so that the repairman will always service machine 1 if it is down, regardless of the state of machine 2. Formulate a CTMC for this modified problem and find the stationary distribution. ———————————————————————- With this alternative repair strategy, we can revise the state space. Now it does suffice to use 4 states, letting the state correspond to the set of failed copiers, because now we know what the repairman will do when both copiers are down; he will always work on copier 1. Thus it suffices to use the single state (1 , 2) to indicate that both machines have failed. There now is only one possible transition from state (1 , 2): Q (1 , 2) , 2 = μ 1 . We display the revised rate diagram in Figure 2 below. The associated rate matrix is now Q = 0 1 2 (1 , 2) - ( γ 1 + γ 2 ) γ 1 γ 2 0 β 1 - ( γ 2 + β 1 ) 0 γ 2 β 2 0 - ( γ 1 + β 2 ) γ 1 0 0 β 1 - β 1 or, with the numbers assigned to the parameters, Q = 0 1 2 (1 , 2) - 4 1 3 0 2 - 5 0 3 4 0 - 5 1 0 0 2 - 2 . Just as before, we obtain the limiting probabilities by solving αQ = 0 with α e = 1. Now we obtain α ( α 0 , α 1 , α 2 , α (1 , 2) ) = 20 57 , 4 57 , 18 57 , 15 57 . Thus, the long-run proportion of time that copier 1 is working is α 0 + α 2 = 38 / 57 = 2 / 3 0 . 67, while the long-run proportion of time that copier 2 is working is α 0 + α 1 = 24 / 57 0 . 42. The 5

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Revised Rate Diagram (1,2) 2 0 1 1 J 2 J 2 J 1 J 1 E 1 E 2 E = rate copier j fails, = rate copier j repaired j E j J Figure 2: A revised rate diagram showing the transition rates among the 4 states in Problem 2, where the repairman always works on copier 1 first when both have failed. new strategy has increased the long-run proportion of time copier 1 is working from 0 . 62 to 0 . 67, at the expense of decreasing the long-run proportion of time copier 2 is working from 0 . 47 to 0 . 42. The long-run proportion of time the repairman is busy is 1 - α 0 = 37 / 57 0 . 649, which is very slightly less than before. We conclude by making some further commentary . We might think that the revised strategy is wasteful, because the repairman quits working on copier 2 when copier 1 fails after copier 2 previously failed. By shifting to work on copier 1, we might think that the repairman is being inefficient, “wasting” his expended effort working on copier 2, making it more likely that both copiers will remain failed. In practice, under other assumptions, that might indeed be true, but here because of the lack-of-memory property of the exponential distribution, the expended work on copier 2 has no influence on the remaining required repair times. From a
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• Spring '08
• Whitt
• Operations Research, Markov chain, long-run proportion, copier, CTMC, A. Pooh

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