Tow godowns, *A* and *B*, have grain storage capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, *D*, *E* and *F*, whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:

Transportation cost per quintal(in Rs.) | ||

From-> | A | B |

To | ||

D | 6.00 | 4.00 |

E | 3.00 | 2.00 |

F | 2.50 | 3.00 |

How should the supplies be transported in order that the transportation cost is minimum?

#### Solution

Let godown *A* supply *x* quintals and *y* quintals of grain to the shops *D* and *E *respectively.

Then, (100 − *x* − *y*) will be supplied to shop F.

The requirement at shop *D* is 60 quintals since, *x* quintals are transported from godown *A*.

Therefore, the remaining (60 − *x*) quintals will be transported from godown *B*.

Similarly, (50 − *y*) quintals and 40 − (100 − *x *− *y*) i.e. (*x* +* y* − 60) quintals will be transported from godown *B* to shop *E* and *F* respectively.

The given problem can be represented diagrammatically as follows.

Quantity of the grain cannot be negative.Therefore,

x ≥ 0 , y ≥ 0 and 100 - x - y ≥ 0

⇒ x ≥ 0 , y ≥ 0 , and x + y ≤ 100

60 - x ≥ 0 , 50 - y ≥ 0 , and x + y - 60 ≥ 0

⇒ x ≤ 60 , y ≤ 50 , and x + y ≥ 60

Total transportation cost Z is given by,

\[Z = 6x + 3y + 2 . 5\left( 100 - x - y \right) + 4\left( 60 - x \right) + 2\left( 50 - y \right) + 3\left( x + y - 60 \right)\]

\[ = 6x + 3y + 250 - 2 . 5x - 2 . 5y + 240 - 4x + 100 - 2y + 3x + 3y - 180\]

\[ = 2 . 5x + 1 . 5y + 410\]

The given problem can be formulated as:

Minimize Z = 2.5*x* + 1.5*y* + 410

subject to the constraints,

\[x + y \leq 100\]

\[x \leq 60\]

\[y \leq 50\]

\[x + y \geq 60\]

\[x, y \geq 0\]

First we will convert inequations into equations as follows:*x* + *y* = 100, *x* = 60, *y* = 50, *x *+ *y* =60, *x* = 0 and *y* = 0

Region represented by *x* + *y* ≤ 100:

The line *x* + *y* = 100 meets the coordinate axes at *A*_{1}(100, 0) and *B*_{1}(0, 100) respectively. By joining these points we obtain the line* x* + *y* = 100. Clearly (0,0) satisfies the *x* + *y* = 100. So, the region which contains the origin represents the solution set of the inequation* x* + *y* ≤ 100.

Region represented by *x* ≤ 60:*x* = 60 is the line that passes (60, 0) and is parallel to the *Y* axis.The region to the left of the line *x *= 60 will satisfy the inequation *x* ≤ 60.

Region represented by *y* ≤ 50:*y* = 50 is the line that passes (0, 50) and is parallel to the *X* axis.The region below the line *y *= 50 will satisfy the inequation *y* ≤ 50.

Region represented by *x** *+ *y* ≥ 60:

The line *x* + *y* = 60 meets the coordinate axes at *C*_{1}(60, 0) and \[D_1 \left( 0, 60 \right)\] respectively. By joining these points we obtain the line* x* + *y* = 60. Clearly (0,0) does not satisfies the inequation *x** *+ *y* ≥ 60. So,the region which does not contain the origin represents the solution set of the inequation *x** *+ *y* ≥ 60.

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* ≤ 100, *x* ≤ 60, *y* ≤ 50, *x** *+ *y* ≥ 60, *x* ≥ 0 and *y* ≥ 0 are as follows.

The corner points are *C*_{1}(60, 0), *G*_{1}(60, 40), *F*_{1}(50, 50), and *E*_{1}(10, 50).

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

Corner point | Z = 2.5x + 1.5y + 410 |

C_{1}(60, 0) |
560 |

G_{1}(60, 40) |
620 |

F_{1}(50, 50) |
610 |

E_{1}(10, 50) |
510 |

The minimum value of Z is 510 at *E*_{1}(10, 50).

Thus, the amount of grain transported from *A* to *D*,* E*, and *F* is 10 quintals, 50 quintals, and 40 quintals respectively and from *B* to *D*, *E*, and *F* is 50 quintals, 0 quintals, and 0 quintals respectively.

The minimum cost is Rs 510.