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 = 2.5x + 1.5y + 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 A1(100, 0) and B1(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 C1(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 C1(60, 0), G1(60, 40), F1(50, 50), and E1(10, 50).
The values of Z at these corner points are as follows.
| Corner point | Z = 2.5x + 1.5y + 410 |
| C1(60, 0) | 560 |
| G1(60, 40) | 620 |
| F1(50, 50) | 610 |
| E1(10, 50) | 510 |
The minimum value of Z is 510 at E1(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.
