An oil company has two depots, *A* and *B*, with capacities of 7000 litres and 4000 litres respectively. The company is to supply oil to three petrol pumps, *D*, *E*, *F* whose requirements are 4500, 3000 and 3500 litres respectively. The distance (in km) between the depots and petrol pumps is given in the following table:

Figure

Assuming that the transportation cost per km is Rs 1.00 per litre, how should the delivery be scheduled in order that the transportation cost is minimum?

#### Solution

Let *x* and *y* litres of oil be supplied from *A* to the petrol pumps *D* and* E.* Then, (7000 − *x *− *y*) L will be supplied from* A *to petrol pump *F*.

The requirement at petrol pump *D* is 4500 L. Since, *x *L are transported from depot *A*, the remaining (4500 −*x*) L will be transported from petrol pump *B*.

Similarly, (3000 − *y*) L and [3500 − (7000 − *x *− *y*)] L i.e. (*x* +* y *− 3500) L will be transported from depot *B* to petrol pump *E* and *F*. respectively.

The given problem can be represented diagrammatically as follows.

Since, quantity of oil are non-negative quantities.Therefore,

x ≥0 , y ≥ 0, and (7000 - x - y ) ≥ 0

⇒ x ≥ 0 , y ≥ 0, and x + y ≥ 7000

4500 - x ≥ 0 , 3000 - y ≥ 0 , and x + y -3500 ≥ 0

⇒ x ≤ 4500, y ≤ 3000, and x + y ≥ 3500

Cost of transporting 10 L of petrol = Re 1

Cost of transporting 1 L of petrol = \[Rs \frac{1}{10}\]

Therefore, total transportation cost is given by,

`z = 7/10 xx x+6/10 y + 3/10 (7000 - x -y ) + 3/10 (4500 - x) + 4/10 ( 3000 - y ) + 2/10( x + y - 3500)`

= 0.3x + 0.1y + 3950

The problem can be formulated as follows.

Minimize Z = 0.3*x* + 0.1*y* + 3950

subject to the constraints,

\[x + y \leq 7000\]

\[x \leq 4500\]

\[y \leq 3000\]

\[x + y \geq 3500\]

\[x, y \geq 0\]

First we will convert inequations into equations as follows:*x* + *y* = 7000, *x* = 4500, *y* = 3000, *x *+ *y* = 3500, *x* = 0 and *y* = 0

Region represented by *x* + *y* ≤ 7000:

The line *x* + *y* = 7000 meets the coordinate axes at *A*_{1}(7000, 0) and \[B_1 \left( 0, 7000 \right)\] respectively. By joining these points we obtain the line *x* + *y* = 7000 . Clearly (0,0) satisfies the *x* + *y* = 7000 . So, the region which contains the origin represents the solution set of the inequation *x* + *y* ≤ 7000.

Region represented by *x* ≤ 4500:

The line *x* = 4500 is the line passes through *C*_{1}(4500, 0) and is parallel to *Y* axis. The region to the left of the line x = 4500 will satisfy the inequation*x* ≤ 4500.

Region represented by *y* ≤ 3000:

The line *y* = 3000 is the line passes through *D*_{1}(0, 3000) and is parallel to *X* axis. The region below the the line *y* = 3000 will satisfy the inequation*y* ≤ 3000.

Region represented by *x* + *y* ≥ 3500:

The line *x* + *y* = 7000 meets the coordinate axes at *E*_{1}(3500, 0) and \[F_1 \left( 0, 3500 \right)\] respectively. By joining these points we obtain the line*x* + *y* = 3500 . Clearly (0,0) satisfies the *x* + *y* = 3500. So, the region which contains the origin represents the solution set of the inequation *x* + *y* ≥ 3500.

Region represented by *x *≥ 0 and* y* ≥ 0:

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations *x* ≥ 0, and *y* ≥ 0.

The feasible region determined by the system of constraints *x* + *y* ≤ 7000,* x* ≤ 4500, *y* ≤ 3000, *x* + *y* ≥ 3500, *x* ≥ 0 and *y* ≥ 0 are as follows.

The corner points of the feasible region are *E*_{1}(3500, 0), *C*_{1}(4500, 0), *I*_{1}(4500, 2500), *H*_{1}(4000, 3000), and *G*_{1}(500, 3000).

The values of Z at these corner points are as follows.

Corner point | Z = 0.3x + 0.1y + 3950 |

E_{1}(3500, 0) |
5000 |

C_{1}(4500, 0) |
5300 |

I_{1}(4500, 2500) |
5550 |

H_{1}(4000, 3000) |
5450 |

G_{1}(500, 3000) |
4400 |

The minimum value of Z is 4400 at *G*_{1}(500, 3000).

Thus, the oil supplied from depot *A* is 500 L, 3000 L, and 3500 L and from depot *B* is 4000 L, 0 L, and 0 L to petrol pumps *D*, *E*, and *F* respectively.

The minimum transportation cost is Rs 4400.