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Q. A brick manufacturer has two depots $A$ and $B$, with stocks of $30000$ and $20000$ bricks respectively. He receives orders from three builders $P, Q$ and $R$ for $15000, 20000$ and $15000$ bricks respectively. The cost (in $₹\,$) of transporting $1000$ bricks to the builders
from the depots are as given in the table :
image
The manufacturer wishes to find how to fulfil the order so that transportation cost is minimum.
Formulate the $ L P P$.

Linear Programming

Solution:

The given information can be expressed as given in the diagram:
In order to simplify, we assume that $1 $ unit $= 1000$ bricks
image
Suppose that depot $A$ supplies $x$ units to $P$ and $y$ units to $Q$, so that depot $A$ supplies $ (30 - x - y)$ bricks to builder $R$.
Now as $P$ requires a total of $15000$ bricks, it requires $(15 - x)$ units from depot $B$.
Similarly $Q$ requires $(20 - y)$ units from $B$ and $R$ requires
$15 - (30 - x - y) = x + y - 15$ units from $B$.
Using the transportation cost given in table, total transportation cost,
$ Z = 40 x + 20 y + 20(30 - x - y) + 20(15 - x) + 60(20 -y) + 40 (x + y- 15)$
$= 40 x - 20 y + 1500$
Obviously the constraints are that all quantities of bricks supplied from $A$ and $B$ to $P, Q, R$ are non-negative.
$\therefore x \ge 0$, $y \ge 0$, $30-x-y\ge 0$, $15-x \ge 0$, $20-y \ge 0$,
$x + y - 15 \ge 0$
Since, $1500$ is a constant, hence instead of minimizing
$Z = 40 x - 20 y + 1500$, we can minimize $Z = 40 x - 20 y$.
Hence, mathematical formulation of the given $L P P$ is
Minimize $Z = 40 x - 20 y$,
subject to the constraints :
$ x + y \ge 15, x + y \le 30, x \le 15, y \le 20, x \ge 0, y \ge 0$