But we do not need the q matrix to solve for the

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But we do not need the Q matrix to solve for the steady-state distribution. We can use the SMP representation. If we do define Q , then we can alternatively obtain the steady-state probability vector α above by solving the equation αQ = 0, where the elements α i are required to sum to 1. Note that this is just a system of linear equations, just like π = πP . You should work to understand why we here have 0 instead of α for the vector. ———————————————————————- (b) What is the average number of trips Pooh makes per day from tree B to tree A ? ———————————————————————- The long-run fraction of time spent at B is 1 / 4, by part (a). Thus, on average, Pooh spends 6 hours per day at tree B . When at tree B , the rate of trips from B is 1 / 5 per hour (the reciprocal of 5 hours), and thus, on average, Pooh makes 6 / 5 = 1 . 2 trips per day from tree B . However, 3 / 4 of the trips from tree B are to tree A , so the average number of trips per day from B to A is (6 / 5) × (3 / 4) = (18 / 20) = 0 . 9. ———————————————————————- 2
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2. Copier Breakdown and Repair. Consider two copier machines that are maintained by a single repairman. Machine i func- tions for an exponentially distributed amount of time with mean 1 i , and thus rate γ i , before it breaks down. The repair times for copier i are exponential with mean 1 i , and thus rate β i , but the repairman can only work on one machine at a time. Assume that the machines are repaired in the order in which they fail. Suppose that we wish to construct a CTMC model of this system, with the goal of finding the long-run proportions of time that each copier is working and the repairman is busy. How can we proceed? (a) Let { X ( t ) : t 0 } be a stochastic process, where X ( t ) represents the number of working machines at time t . Is { X ( t ) : t 0 } a Markov process? (b) Formulate a CTMC describing the evolution of the system. (c) Suppose that γ 1 = 1, β 1 = 2, γ 2 = 3 and β 2 = 4. Find the stationary distribution. (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. ANSWERS: (a) Let { X ( t ) : t 0 } be a stochastic process, where X ( t ) represents the number of working machines at time t . Is { X ( t ) : t 0 } a Markov process? ———————————————————————- This process is not a Markov process. To be a Markov process, we need the conditional distribution of a future state, given a present state and past states to depend only upon the present state; i.e., we need P ( X ( t ) = j | X ( s ) = i, X ( u ) , 0 u s ) = P ( X ( t ) = j | X ( s ) = i ) for all s and t with 0 s < t , and for all i and j . Here, however, the Markov property does not hold: When both machines are down, the next transition depends on which of the two machines failed first.
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