\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

QUESTION

\begin{description}

\item[(a)]
Two products are obtained from the same supplier.  For each
product, the price of an item, the cost of placing an order, the
holding cost per item per annum and the annual demand are shown in
the following table.  Demand for both products is steady, and
shortages must not occur.

\begin{tabular}{ccccc}
\hline Product&Price(\pounds)&Order cost (\pounds)&Holding cost
(\pounds)&Demand\\ \hline 1&4&10&5&900\\ 2&15&19&8&304\\ \hline
\end{tabular}


The supplier gives a discount of 2.5\% whenever a buyer spends at
least \pounds500.  Two types of ordering policy are possible:
either order both products independently of each other, or always
place an order for both products at the same time. Perform a cost
analysis of the two policies, and hence recommend a policy to be
adopted.


\item[(b)]
In a batch production system, $n$ products are manufactured on a
single machine which can produce only one type of product at a
time.  The machine can produce any desired quantity of product
after the necessary set-up procedure.

The following data are available for each product $j$
($j=1,\ldots,n$).

Demand = $d_j$ per month.

Production rate = $r_j$ per month.

Stock holding cost = $h_j$ per month.

Set-up time = $t_j$ months.

Set-up cost = $s_j$.

Assume that demand is steady. A common cycle approach is to be
adopted. Derive an expression for the monthly cost of this method,
and hence find the optimal cycle length.

\end{description}


ANSWER


\begin{description}

\item[(a)]
For a single product, the annual cost is

$$K(Q)=\frac{sd}{Q}+\frac{1}{2}hQ+cd(1-\textrm{discount})$$

Ignoring the discount, $K$ is minimized when

$$\frac{dK}{dQ}=-\frac{sd}{Q^2}+\frac{1}{2}h=0\
Q*=\sqrt{\frac{2sd}{h}}$$

For product 1, $Q*=\sqrt{\frac{2.10.900}{5}}=60$

$$K_1(60)=\frac{10.900}{60}+\frac{1}{2}.5.60+4.900=3900$$

To obtain discount $4Q_1=1200, Q_1=300$

$$K_1(300)=\frac{10.900}{300}+\frac{1}{2}5.300+4.900.0.975=4290$$

Thus, $Q_1=60$ is optimal order quantity for product 1. $K_1=3900$

For product 2, $Q_2*=\sqrt{\frac{2.19.304}{8}}=38$

$$K_2(38)=\frac{19.304}{38}+\frac{1}{2}.8.38+15.304=4864$$

To obtain discount $15Q_2=1200\ Q_2=80$

$$K_2(80)=\frac{19.304}{80}+\frac{1}{2}.8.80+15.304.0.975=4838.2$$

Thus $Q_2=80$ is optimal order quantity for product 2.
$K_2=4838.2$

Suppose both products ordered together every time $T$ time units.

$$K=\sum_{i=1}^2\left(\frac{S_i}{T}+\frac{1}{2}h_id_iT\right)$$

Ignoring the discount, $K$ is minimized when

$$\frac{dK}{dT}=-\frac{1}{T^2}\sum_{i=1}^2S_i+\frac{1}{2}\sum_{i=1}^2h_id_i=0\
T*=\sqrt{\frac{2\sum S_i}{\sum h_id_i}}$$

$T*=\sqrt{\frac{2(10+19)}{5.900+8.304}}=0.09147$

To obtain the discount $Q_1C_1+Q_2C_2\geq1200$

$$T(c_1d_1+c_2d_2)\geq1200\ T\geq0.14706$$

For $T=0.14706$,

$$K=\frac{19}{0.14706}+\frac{1}{2}(5.900+8.304)0.14706+(4.900+15.304)0.975=8594.91$$

For individual ordering, the total cost is 8738.2. Thus, order
quantities 132.4 and 44.7 respectively at the same time.


\item[(b)]
Let $Q_j$ denote the production quantity of product $j$, and let
$T$ denote the cycle length.

Since $T_1=\frac{Q_j}{r_j}$, the maximum stack level is
$T_1(r_j-d_j)=Q_j\left(1-\frac{d_j}{r_j}\right)$

\setlength{\unitlength}{.3in}

\begin{picture}(15,9)

\put(2,3){\line(1,0){12}} \put(2,3){\line(2,3){4}}
\put(6,3){\line(0,1){6}} \put(6,9){\line(4,-3){8}}
\put(2,2.8){\vector(1,0){4}} \put(6,2.8){\vector(-1,0){4}}
\put(2,1.5){\vector(1,0){12}} \put(14,1.5){\vector(-1,0){12}}
\put(0,7){slope $r_j-d_j$} \put(10,7){slope $-d_j$}
\put(3.5,2){$T_1$} \put(8,0){$T$}

\end{picture}

Thus, the monthly cost is

$$K=\sum_{j=1}^n\left(\frac{S_jd_j}{Q_j}+\frac{1}{2}h_jQ_j\left(1-\frac{d_j}{r_j}\right)\right)$$

For the common cycle method, $Q_j=d_jT$. Thus

$$K=\sum_{j=1}^n\left(\frac{S_j}{T}+\frac{1}{2}h_jQ_j\left(1-\frac{d_j}{r_j}\right)\right)$$

$\frac{dK}{dT}=0$ gives

$$-\sum_{j=1}^n\frac{s_j}{T^2}+\sum_{j=1}^n\frac{1}{2}h_jd_j\left(1-\frac{d_j}{r_j}\right)=0$$

$$T*=\sqrt{\frac{2\sum_{j=1}^nS_j}{\sum_{j=1}^nh_jd_j\left(1-\frac{d_j}{r_j}\right)}}$$

There is a lower bound $T^{LB}$ on $T$ due to feasibility. In a
unit time interval, set-ups and production must be done. Thus,

$$\sum_{j=1}^n\frac{t_j}{T}+\sum_{j=1}^n\frac{d_j}{r_j}\leq1$$

to give

$$T^{LB}=\sum_{j=1}^n\frac{t_j}{\left(1-\sum_{j=1}^n\frac{d_j}{r_j}\right)}$$

Choose $T=max\{T*, T^{LB}\}$

\end{description}



\end{document}