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QUESTION

The local supermarket in a small town buys and sells turkeys for
Christmas. Each turkey costs the supermarket \pounds5.00 and has
to be ordered well in advance. They are sold to customers for
\pounds7.50. A week after Christmas, turkeys have to go on sale at
half price. Based on previous experience, the owners of the
supermarket know that the demand for turkeys over the Christmas
period has the following distribution:

\begin{center}

\begin{tabular}{cc}
\hline Demand&Probability\\ \hline
100&0.10\\150&0.20\\200&0.35\\250&0.20\\300&0.10\\350&0.05\\
\hline
\end{tabular}

\end{center}

\begin{description}

\item[(i)]
Find an expression for the expected profit for a given order
quantity.

\item[(ii)]
Hence derive a method of finding the optimal order quantity.

\item[(iii)]
Use your answer in part (ii) to determine how many turkeys the
supermarket should order.

\end{description}

\bigskip

ANSWER


Inventories

Let $c=\pounds5.00,\ s=\pounds7.50,\ v=\pounds3.75,\ d$= demand
and $q$= quantity to be ordered. Then,

\begin{description}

\item[(i)]

\begin{eqnarray*}
\textrm{Profit}(d,q)=sd+v(q-d)-cq\textrm{ if }d\leq q\\
\textrm{Profit}(d,q)=sq-cq\textrm{ if }d> q\\
\end{eqnarray*}

$$E(\textrm{Profit})=\sum_{d=0}^q(sd+v(q-d)-cq)\textrm{Prob}(D=d)+
\sum_{d=q+1}^\infty(sq-cq)\textrm{Prob}(D=d)$$

\item[(ii)]
We then have

\begin{eqnarray}
\textrm{Profit}(d,q)&=&(v-c)q+(s-v)d\textrm{ for }d\leq q\\
\textrm{Profit}(d,q)&=&(s-c)q\textrm{ for }d>q
\end{eqnarray}

For the situation represented by (1), ordering $q+1$ turkeys
instead of $q$ would decrease profit by an amount ($v-c$). The
probability of case (1) is Prob$(D\leq q)$

In the situation represented by (2), ordering $q+1$ turkeys
instead of $q$ would increase profit by an amount of $(s-c)$. The
happens with Prob$(D>q)$

If we call $E(q)$ the expected profit when ordering $q$ turkeys,
we see that

\begin{eqnarray*}
E(q+1)-E(q)&=&(v-c)\textrm{Prob}(D\leq
q)+(s-c)\textrm{Prob}(D>q)\\ &=&(v-c)\textrm{Prob}(D\leq
q)+(s-c)[1-\textrm{Prob}(D\leq q)]\\
&=&(s-c)+(v-s)\textrm{Prob}(D\leq q)
\end{eqnarray*}

Now, $E(q+1)-E(q)\geq0$ if $(s-c)+(v-s)\textrm{Prob}(D\leq
q)\geq0$ therefore ,
\begin{equation}
\textrm{Prob}(D\leq q)\leq\frac{s-c}{s-v}
\end{equation}

and we look for the largest value of $q$ such that (3) holds.

\item[(iii)]
We need
$\ds\sum_{d=1}^q\textrm{Prob}(D=d)\leq\frac{7.50-5.00}{7.50-3.75}=0.67$

$$\textrm{Prob}(D=100)+\textrm{Prob}(D=150)+\textrm{Prob}(D=200)=0.65$$

The supermarket should order 200 turkeys.

\end{description}



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