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QUESTION
\begin{description}
\item[(a)]
A retailer purchases animal feed from a supplier, which it then
sells to farms. The purchase price for the feed is \pounds5 per
kilogram. However, a discount of 5\% is given if an order of at
least $2\,000$ kilograms is placed with the supplier.
Alternatively, a discount of 10\% is given if an order of at least
3000 kilograms is placed.
Demand for the feed is steady, and the annual demand is $27\,000$
kilograms. The cost of placing an order with the supplier is
\pounds80, and the cost of holding the feed in stock is \pounds12
per kilogram per annum.
The retailer owns a storage facility that can store up to 500
kilograms of feed. Moreover, the retailer can hire additional
storage units at an annual cost of \pounds$1\,000$ per unit. Each
additional unit can hold up to 500 kilograms of feed.
Analyze the strategy that the retailer should follow regarding its
ordering of feed from the supplier and the hiring of additional
storage units.
\item[(a)]
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&60&40&20&30\\ \hline
\end{tabular}
\end{center}
The initial stock level is zero. A set-up cost of \pounds100 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 \textit{all}
optimal production schedules.
%\itemitem{} 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 cost per annum is
$$K=\frac{sd}{Q}+\frac{1}{2}hQ+cd\left(1-\frac{\textrm{discount}}{100}\right)+1000n$$
where $n$ is the number of storage units hired.
The EOQ value is
$$Q=\sqrt{\frac{2sd}{h}}=\sqrt{\frac{2.80.27000}{12}}=600$$
For $Q=600$
$$K=3600+3600+135000+1000=143200$$
It may be cheaper not to hire one of the storage units.
For $Q=500$
$$K=4320+3000+135000=142320$$
It is worthwhile to check if the 5\% or 10\% discount should be
obtained.
For $K=2000$
$$K=1080+12000+128250+3000=144330$$
For $K=3000$
$$K=720+18000+121500+5000=145220$$
Comparing the costs, the optimal strategy is to order 500 units
and not hire any additional storage.
\item[(b)]
We first evaluate the cost for the possible decisions.
\begin{tabular}{ccl}
$Q_1=$&60&100\\ &100&100+80\\ &120&100+120+40\\
&150&100+180+100+60\\ $Q_2=$&40&100\\ &60&100+40\\
&90&100+100+60\\ $Q_3=$&20&100\\ &50&100+60\\ $Q_4=$&30&100
\end{tabular}
The network is therefore as follows.
DIAGRAM
There are two shortest paths, 0 - 2 - 4 and 0 - 1 - 3 - 4, giving
production plans
\begin{tabular}{c|cccc}
$\delta$&1&2&3&4\\ \hline $Q_j$&100&0&50&0
\end{tabular}
\begin{tabular}{c|cccc}
$\delta$&1&2&3&4\\ $Q_j$&60&60&0&30
\end{tabular}
\end{description}
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