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A single repairperson looks after both machines 1 and \(2 .\) Each time it is repaired, machine \(i\) stays up for an exponential time with rate \(\lambda_{i}, i=1,2 .\) When machine \(i\) fails, it requires an exponentially distributed amount of work with rate \(\mu_{i}\) to complete its repair. The repairperson will always service machine 1 when it is down. For instance, if machine 1 fails while 2 is being repaired, then the repairperson will immediately stop work on machine 2 and start on \(1 .\) What proportion of time is machine 2 down?

There are two machines, one of which is used as a spare. A working machine will function for an exponential time with rate \(\lambda\) and will then fail. Upon failure, it is immediately replaced by the other machine if that one is in working order, and it goes to the repair facility. The repair facility consists of a single person who takes an exponential time with rate \(\mu\) to repair a failed machine. At the repair facility, the newly failed machine enters service if the repairperson is free. If the repairperson is busy, it waits until the other machine is fixed; at that time, the newly repaired machine is put in service and repair begins on the other one. Starting with both machines in working condition, find (a) the expected value and (b) the variance of the time until both are in the repair facility. (c) In the long run, what proportion of time is there a working machine?

Consider a taxi station where taxis and customers arrive in accordance with Poisson processes with respective rates of one and two per minute. A taxi will wait no matter how many other taxis are present. However, an arriving customer that does not find a taxi waiting leaves. Find (a) the average number of taxis waiting, and (b) the proportion of arriving customers that get taxis.

If \(\\{X(t)\\}\) and \(\\{Y(t)\\}\) are independent continuous-time Markov chains, both of which are time reversible, show that the process \(\\{X(t), Y(t)\\}\) is also a time reversible Markov chain.

A total of \(N\) customers move about among \(r\) servers in the following manner. When a customer is served by server \(i\), he then goes over to server \(j, j \neq i\), with probability \(1 /(r-1)\). If the server he goes to is free, then the customer enters service; otherwise he joins the queue. The service times are all independent, with the service times at server \(i\) being exponential with rate \(\mu, i=1, \ldots, r .\) Let the state at any time be the vector \(\left(n_{1}, \ldots, n_{r}\right)\), where \(n_{i}\) is the number of customers presently at server \(i, i=1, \ldots, r, \sum_{i} n_{i}=N\) (a) Argue that if \(X(t)\) is the state at time \(t\), then \(\\{X(t), t \geqslant 0\\}\) is a continuous-time Markov chain. (b) Give the infinitesimal rates of this chain. (c) Show that this chain is time reversible, and find the limiting probabilities.

Consider the two-state continuous-time Markov chain. Starting in state 0 , find \(\operatorname{Cov}[X(s), X(t)]\)