B formulate a ctmc describing the evolution of the

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———————————————————————- (b) Formulate a CTMC describing the evolution of the system. ———————————————————————- However, we can use 5 states with the states being: 0 for no copiers failed, 1 for copier 1 is failed (and copier 2 is working), 2 for copier 2 is failed (and copier 1 is working), (1 , 2) for both copiers down (failed) with copier 1 having failed first and being repaired, and (2 , 1) for both copiers down with copier 2 having failed first and being repaired. (Of course, these states could be relabelled 0, 1, 2, 3 and 4, but we do not do that.) From the problem specification, it is natural to work with transition rates, where these transition rates are obtained directly from the originally-specified failure rates and repair rates (the rates of the exponential random variables). In Figure 1 we display a rate diagram showing the possible transitions with these 5 states together with the appropriate rates. It can be helpful to construct such rate diagrams as part of the modelling process. From Figure 1, we see that there are 8 possible transitions. The 8 possible transitions should clearly have transition rates Q 0 , 1 = γ 1 , Q 0 , 2 = γ 2 , Q 1 , 0 = β 1 , Q 1 , (1 , 2) = γ 2 , Q 2 , 0 = β 2 , Q 2 , (2 , 1) = γ 1 , Q (1 , 2) , 2 = β 1 , Q (2 , 1) , 1 = β 2 . 3
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Rate Diagram (2,1) (1,2) 2 0 1 1 J 2 J 2 J 1 J 1 E 1 E 2 E 2 E = rate copier j fails, = rate copier j repaired j E j J Figure 1: A rate diagram showing the transition rates among the 5 states in Problem 2, involving copier breakdown and repair. We can thus define the transition-rate matrix Q . For this purpose, recall that the diagonal entries are minus the sum of the non-diagonal row elements, i.e., Q i,i = - X j,j 6 = i Q i,j for all i . That can be explained by the fact that the rate matrix Q is defined to be the derivative (from above) of the transition matrix P ( t ) at t = 0; see the CTMC notes. In other words, the rate matrix should be Q = 0 1 2 (1 , 2) (2 , 1) - ( γ 1 + γ 2 ) γ 1 γ 2 0 0 β 1 - ( γ 2 + β 1 ) 0 γ 2 0 β 2 0 - ( γ 1 + β 2 ) 0 γ 1 0 0 β 1 - β 1 0 0 β 2 0 0 - β 2 . ———————————————————————- (c) Suppose that γ 1 = 1, β 1 = 2, γ 2 = 3 and β 2 = 4. Find the stationary distribution. ———————————————————————- We first substitute the specified numbers for the rates γ i and β i in the rate matrix Q above, 4
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obtaining Q = 0 1 2 (1 , 2) (2 , 1) - 4 1 3 0 0 2 - 5 0 3 0 4 0 - 5 0 1 0 0 2 - 2 0 0 4 0 0 - 4 . Then we solve the system of linear equations αQ = 0 with α e = 1, which is easy to do with a computer and is not too hard by hand. Just as with DTMC’s, one of the equations in αQ = 0 is redundant, so that with the extra added equation α e = 1, there is a unique solution.
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