\documentclass[a4paper,12pt]{article}
\begin{document}
\parindent=0pt
QUESTION
Describe briefly the transition diagram method of setting up the
equations relating the steady-state probabilities of the different
possible states of a queueing system.
Mass produced cars are `customised' in a two step process. Process
1 involves mechanical alterations, followed by Process 2 where the
trim is modified. Only one car at a time can be accepted by each
Process; no queueing is allow between the two Processes. Thus a
car finishing Process 1 must remain in it, blocking it, if there
is still a car undergoing Process 2. A car finishing Process 1
proceeds immediately to Process 2, if this is free. Processing
times are exponentially distributed with mean $1/\mu $ for both
Processes. Cars arrive at rate $\lambda ,$ but if Process 1 is not
free, cars do not wait but are turned away.
Set up equations relating the equilibrium probabilities of the
different states of the system, and hence find these
probabilities. Find also
\begin{description}
\item[(i)]
$L,$ the mean number of cars in the system,
\item[(ii)]
$F,$ the proportion of cars turned away.
\end{description}
Show that if $\rho =\lambda /\mu $ is small, then $L=2\rho +O(\rho
^{2})$ and $F=\rho +O(\rho ^{2}).$
If the cost of turning each car away is $C_{0},$ whilst the cost
per unit time of maintaining processing rate $\mu $ (for both
Processes) is $C_{1}\mu ,$ find an approximate expression for
$\rho^{*},$ the processing rate that minimises costs, assuming
this is small.
ANSWER
Description of transition diagram method should mention:
\begin{description}
\item[(i)]
Use of nodes to represent different states of system
\item[(ii)]
Links to represent possible transitions between states
\item[(iii)]
An equation for each node equating the long term balance of
transitions into and out of that node.
\end{description}
Car customizing problem:
Let $(m,n)$=(process 1 in state $m$, Process 2 in state $n$),
$m=0,1;\ n=0,1$.
Transition diagram:
DIAGRAM
Let $p_{mn}$=prob(system in state $(m,n)$). Then
$$\left.\begin{array}{cl}(0,0):&\lambda p_{00}=\mu p_{01}\\
(0,1):&(\mu+\lambda)p_{01}=\mu p_{10}\\ (1,0):&\mu p_{10}=\lambda
p_{00}+\mu
p_{11}\end{array}\right\}\Rightarrow\begin{array}{l}p_{01}=\rho
p_{00}\\ p_{10}=(1+\rho)p_{01}=\rho(1+\rho)p_{00}\\
p_{11}=p_{10}-\rho p_{00}=\rho^2 p_{00}\end{array}$$
Normalising condition gives:
$$p_{00}-[1+\rho+\rho(1+\rho)+\rho^2]^{-1}=[1+2\rho+2\rho^2]^{-1}$$
\begin{eqnarray*}
L&=&p_{01}+p_{10}+2p_{11}\\
&=&(2\rho+\rho^2+2\rho^2)(1+2\rho+2\rho^2)^{-1}\\
&=&2\rho(1+\frac{3}{2}\rho)(1+2\rho+2\rho^2)^{-1}\\
&\approx&2\rho(1+\frac{3}{2}\rho)(1-2\rho)+O(\rho^3)\\
&\approx&2\rho(1-\frac{\rho}{2})+O(\rho^3)
\end{eqnarray*}
$$F=p_{10}+p_{11}=(\rho+2\rho^2)(1-2\rho)+O(\rho^3)\approx\rho+O(\rho^3$$
The average unit time cost is $K=\frac{C_0\lambda}{\mu}+C_1\mu$,
assuming $\rho$ remains small.
$$\frac{dK}{d\mu}\equiv-\frac{C_0\lambda}{\mu^2}+C_1=0\textrm{ at
} \mu=\sqrt{\frac{C_0\lambda}{C_1}}$$
$$\frac{d^2K}{d\mu^2}=2\frac{C_0\lambda}{\mu^3}>0\textrm{ at
}\mu=\sqrt{\frac{C_0\lambda}{C_1}}$$
giving a minimum.
\end{document}