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{\bf Question}

An electrician services $m$ machines which experiences random
breakdowns.  If a machine is working at time $t$ then the
probability that it will require attention in the time interval
$(t, t + \delta t]$ is $\lambda \delta t + o(\delta t)$ for each
machine.  The machines work independently and the electrician can
only service one machine at a time.  The service times are
independent random variables with cumulative distribution function
$1 - e^{-\mu x}$ (for $x\geq0$).  Let $X(t)$ denote the number of
machines working at time $t.$

Show that the limiting probabilities $$P_j = \lim_{t \to \infty}
p\{X(t) = j\} \hspace{.2in} j = 0, 1, 2,\ldots, m$$ satisfy the
equations $$j \lambda p_j - \mu p_{j-1}= (j+1)\lambda p_{j+1} -
\mu p_j \hspace{.2in} (j = 0, 1, 2,\ldots, m-1), $$ where
$p_{-1}=0$ and $m \lambda p_m = \mu p_{m-1}.$

By deducing that $j \lambda p_j - \mu p_{j-1} = 0$, or otherwise,
show that in the long run the expected  number of machines working
simultaneously is $(1 - p_m)\frac{\mu}{\lambda}$.



\vspace{.25in}

{\bf Answer}

Service times have c.d.f. $\ds 1 - e^{-\mu x}$ i.e. negative
exponential.  So they are competed according to a Poisson process
with rate $\mu$.

When $X(t) = j$ the change in $(t, t + \delta t]$ is
\begin{eqnarray*} + 1 & {\rm with\ probability} &
\mu \delta t + o(\delta t) \\ - 1 & {\rm with\ probability} &
\lambda j \delta t + o(\delta t) \\ 0 & {\rm with\ probability} &
 1 - \left( \mu  + \lambda j\right) \delta t +
o(\delta t) \end{eqnarray*}

When $X(t) = 0$ the change in $(t, t + \delta t]$ is
\begin{eqnarray*} + 1 & {\rm with\ probability} &
\mu \delta t + o(\delta t) \\ 0 & {\rm with\ probability} & 1 -
\mu  \delta t  + o(\delta t) \end{eqnarray*}

When $X(t) = m$ the change in $(t, t + \delta t]$ is
\begin{eqnarray*} - 1 & {\rm with\ probability} &
\lambda m \delta t + o(\delta t) \\ 0 & {\rm with\ probability} &
1 - \lambda m \delta t  + o(\delta t) \end{eqnarray*}

\newpage
\begin{eqnarray*} P(X(t + \delta t)=j) & = & P(X(t + \delta t) = j |
X(t) = j-1)P(X(t) = j-1) \\ & + & P(X(t + \delta t) = j | X(t) =
j+1)P(X(t) = j+1) \\ & + & P(X(t + \delta t) = j | X(t) = j)P(X(t)
= j) \\ \\ p_j (t + \delta t) & = & (\mu \delta t + o(\delta
t))p_{j-1}(t) \\ & + & (\lambda(j+1) + o(\delta t))p_{j+1}(t)
\\ & + & (1 - (\lambda j + \mu) \delta t +
o(\delta t)) p_j(t)\hspace{.2in} (0< j < m)  \end{eqnarray*}
giving $$p_j'(t) = \mu p_{j-1}(t) + \lambda(j+1)p_{j+1}(t) -
(\lambda j+ \mu)p_j(t) $$ Similarly
\begin{eqnarray*} p_0'(t) & = & - \mu p_0(t) +
\lambda p_1(t) \\ p_m'(t) & = & -\lambda m p_m(t) + \mu p_{m-1}(t)
\end{eqnarray*}

The equilibrium equations are therefore: \begin{eqnarray*} 0 & = &
-\mu p_0 + \lambda p_1 \\ 0 & = & -\lambda m p_m + \mu p_{m-1} \\
0 & = & \mu p_{j-1} + \lambda(j+1)p_{j+1} - (\lambda j + \mu)p_j
\hspace{.2in} (0< j < m) \end{eqnarray*}

So $\ds j\lambda p_j - \mu p_{j-1} = (j+1) \lambda p_{j+1} - \mu
p_j \hspace{.2in} (0<j<m)$

Since $\ds 1 \cdot \lambda p_1 - \mu p_0 = 0$ it follows by
induction that $\ds j \lambda p_j - \mu p_{j-1} = 0$ for $0 < j <
m.$  Also $\ds m\lambda p_m - \mu p_{m-1} = 0$.

The expected number of machines working is $$\sum_{j=1}^m j p_j =
\frac{\mu}{\lambda} \sum_{j=1}^m p_{j-1} = \frac{\mu}{\lambda}(1 -
p_m)$$


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