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QUESTION

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\item[(a)]
Consider the M/M/$1$/$n-2$ queue where $n \ge 2$. Let $p_i$ be the
steady state probability that there are $i$ customers in the
system. It is known that $$ p_{i+1} = {{\lambda}\over{\mu}} p_i
\quad {\rm for} \ i=0,1,2,\ldots,n-2. $$ Here, $\lambda^{-1}$ is
the mean inter-arrival time, and $\mu^{-1}$ is the mean service
time.

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\item[(i)] Show that $$ p_i = {{(1-\rho)\rho^i}\over{1-\rho^n}}, \quad
i=0,1,2,\ldots, n-1. $$ Here, $\rho = \lambda/\mu$.

\item[(ii)]
Find the expected number of customers in the system in terms of
$\rho$ and $n$.

\item[(iii)]
Let $L$ and $W$ be the expected number of customers and the
expected waiting time of customers in the system, respectively.
Write down an expression which demonstrates Little's queueing
formula.


\item[(iv)]
Supposing that $\lambda/\mu <1$, show that the expected waiting
time of a customer in the system is $(\mu-\lambda)^{-1}$ as $n \to
\infty$.

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A repairman is to be hired to repair machines which break down at
random at an average rate of $3$ per day. Non-productive time on
any machine costs the firm \pounds100 per day. The firm can hire a
slow cheap man charging \pounds50 per day who can repair $4$
machines per day on average. Alternatively, the firm can hire a
fast expensive man charging \pounds100 per day who can repair $6$
machines per day on average. In either case, service times are
exponentially distributed. Calculate the expected operation cost
in both cases, and determine which repairman is more economical to
hire.

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ANSWER

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\item[(a)]

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\item[(i)]
From $p_{i+1}=\left(\frac{\lambda}{\mu}\right)p_i$ we have
$p_i=\rho^ip_0$. Since $\sum_{i=0}^{n-1}p_i=1$, we have

$$p_0=\frac{1-\rho}{1-\rho^n}$$

\item[(ii)]
The expected number of customers in the system is given by

\begin{eqnarray*}
L&=&\sum_{i=0}^{n-1}ip_i\\ &=&\sum_{i=1}^{n-1}ip_0\rho^i\\
&=&\frac{\rho-n\rho^n-(n-1)\rho^{n+1}}{(1-\rho)(1-\rho^n)}
\end{eqnarray*}

\item[(iii)]
$L=\lambda W$ where $L$ is the expected number of customers in the
system and $W$ is the expected waiting time for the customers.

\item[(iv)]
$$L=\lim_{n\to\infty}\frac{\rho-n\rho^n-(n-1)\rho^{n+1}}{(1-\rho)(1-\rho^n)}=\frac{\rho}{1-\rho}.$$

Therefore as $n\to\infty$ we have

$$W=\frac{L}{\lambda}=\frac{1}{\mu(1-\rho)}=\frac{1}{(\mu-\lambda)}.$$

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\item[(b)]
One may regard the problem as an $M/M/1/\infty$ queuing problem
with the customers being the broken machines.

In the case of slower repairman, $\lambda=3$ and $\mu=4$,
therefore $\rho=0.75$ and the expected number of broken machines
in the system will be $\frac{\rho}{1-\rho}$. Therefore we have
$\frac{\rho}{1-\rho}=3$. Thus the operating cost will be

$$3\times100+50=350.$$

In the case of the faster repairman, $\lambda=3$ and $\mu=6$,
therefore $\rho=0.5$ and the expected number of broken machines in
the system will be $\frac{\rho}{1-\rho}=1$. Thus the operating
cost will be

$$1\times100+100=200.$$

Hence it is more economical to employ the faster repairman.

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