91Ó°ÊÓ

Potential customers arrive at a single-server station in accordance with a Poisson process with rate \(\lambda\). However, if the arrival finds \(n\) customers already in the station, then he will enter the system with probability \(\alpha_{n}\). Assuming an exponential service rate \(\mu\), set this up as a birth and death process and determine the birth and death rates.

Consider two machines, both of which have an exponential lifetime with mean \(1 / \lambda\). There is a single repairman that can service machines at an exponential rate \(\mu\). Set up the Kolmogorov backward equations; you need not solve them.

Consider a Yule process starting with a single individual- -that is, suppose \(X(0)=1\). Let \(T_{l}\) denote the time it takes the process to go from a population of size \(i\) to one of size \(i+1\). (a) Argue that \(T_{i}, i=1, \ldots, j\), are independent exponentials with respective rates \(i \lambda\). (b) Let \(X_{1}^{3}, \ldots, X_{j}\) denote independent exponential random variables each having rate \(\lambda\), and interpret \(X_{l}\) as the lifetime of component \(i\). Argue that \(\max \left(X_{1}, \ldots, X_{j}\right)\) can be expressed as $$ \max \left(X_{1}, \ldots, X_{j}\right)=\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{j} $$ where \(\varepsilon_{1}, \varepsilon_{2}, \ldots, 8_{j}\) are independent exponentials with respective rates \(j \lambda\), \((j-1) \lambda, \ldots, \lambda\) Hint: Interpret \(\varepsilon_{i}\) as the time between the \(i-1\) and the \(i\) th failure. (c) Using (a) and (b) argue that $$ P\left\\{T_{1}+\cdots+T_{j} \leqslant t\right)=\left(1-e^{-\lambda t}\right)^{j} $$ (d) Use (c) to obtain that $$ P_{1 j}(t)=\left(1-e^{-\lambda t}\right)^{j-1}-\left(1-e^{-\lambda t}\right)^{j}=e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-1} $$ and hence, given \(X(0)=1, X(t)\) has a geometric distribution with parameter \(p=e^{-\lambda t}\) (e) Now conclude that $$ P_{i j}(t)=\left(\begin{array}{c} j-1 \\ i-1 \end{array}\right) e^{-\lambda t i}\left(1-e^{-\lambda t}\right)^{j-i} $$

A small barbershop, operated by a single barber, has room for at most two customers. Potential customers arrive at a Poisson rate of three per hour, and the successive service times are independent exponential random variables with mean \(\frac{1}{4}\) hour. What is (a) the average number of customers in the shop? (b) the proportion of potential customers that enter the shop? (c) If the barber could work twice as fast, how much more business would he do?

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 waifing leaves. Find (a) the average number of taxis waiting, and (b) the proportion of arriving customers that get taxis.

In the \(M / M / s\) queue if you allow the service rate to depend on the number in the system (but in such a way so that it is ergodic), what can you say about the output process? What can you say when the service rate \(\mu\) remains unchanged but \(\lambda>s \mu ?\)

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.