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QUESTION

\begin{description}

\item[(a)]
A car rental company purchases replacement tyres from a supplier.
Usage for the tyres is steady, and the annual usage is 2000 tyres.
Storage costs are \pounds20 per tyre per annum.

The tyres can either be purchased from a small local supplier or
from a larger national supplier.  For the local supplier, the cost
of placing an order is \pounds50, and each tyre costs
\pounds28.50. For the national supplier, the cost of placing an
order is \pounds200 (which reflects the non-local transportation
cost), and the purchase price depends on the order size $Q$ as
follows:

$$\textrm{Cost per
tyre}=\left\{\begin{array}{ll}\pounds28&\textrm{for }Q<300\\
\pounds27&\textrm{for }300\leq Q<600\\ \pounds26&\textrm{for
}Q\geq600.\end{array}\right.$$

By analysing the relevant costs, determine whether the local or
the national supplier should be used, and what order size should
be chosen.

\item[(b)] Consider the planning of production over four time periods
for which the demand is given in the following table.

\begin{center}

\begin{tabular}{lcccc}
\hline Period&1&2&3&4\\ \hline Demand&40&50&20&40\\ \hline
\end{tabular}

\end{center}

The initial stock level is zero.  A set-up cost of \pounds120 is
incurred for each period in which there is production. The stock
holding cost is \pounds2 per unit of stock held at the end of each
period.  The objective is to schedule production so that demand is
met at minimum cost.

Formulate the problem of finding the minimum cost as a shortest
path problem in a network, and hence determine an optimal
production schedule.

Find by how much the set-up cost can increase and decrease,
without producing a change in the optimal production schedule.

\end{description}


ANSWER


\begin{description}

\item[(a)]
The annual cost (under the usual notation) is

$$K=\frac{sd}{Q}+\frac{1}{2}hQ+cd$$

$\frac{dK}{dQ}=0$ gives $-\frac{sd}{Q^2}+\frac{1}{2}h=0$ (if $c$
independent of $Q$), so $Q=\sqrt{\frac{2sd}{h}}$

\begin{itemize}

\item
For the local supplier, $Q=\sqrt{\frac{100.2000}{20}}=100$

$$K=\frac{50.2000}{100}+\frac{1}{2}20.100+28.5.2000=\pounds59000$$

\item
For the national supplier, $Q=\sqrt{\frac{400.2000}{20}}=200$

For $Q=200,$

$$K=\frac{200.2000}{200}+\frac{1}{2}20.200+28.2000=\pounds60000$$

For $Q=300$,

$$K=\frac{200.2000}{300}+\frac{1}{2}.20.300+27.2000=\pounds58333.33$$

For $Q=600$

$$K=\frac{200.2000}{600}+\frac{1}{2}.20.600+26.2000=\pounds58666.67$$

\end{itemize}

The minimum cost is \pounds58333.33, which is for the national
supplier with an order size of 300.

\item[(b)]
If $s$ denotes the set up cost, the network is as follows

DIAGRAM

For $s=120$, the shortest path length is 400, which is given by
path 0 - 1 - 3 - 4. The optimal production schedule is

\begin{center}

\begin{tabular}{lcccc}
\hline Period&1&2&3&4\\ \hline Production&40&70&0&40\\ \hline
\end{tabular}

\end{center}

For this solution to be optimal, the label at nodes 3 and 4 must
be $2s+40$ and $3s+40$, respectively. This gives

\begin{tabular}{ll}
$s+40\leq25$&$s\geq40$\\ $s+40\leq180$&$s\leq180$\\
\\
$2s+40\leq3s$&$s\geq40$\\ $2s+40\leq s+180$&$s\leq140$\\
$2s+40\leq420$&$s\leq190$
\end{tabular}

Thus, the required range is $40\leq s\leq140$.

\end{description}




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