Question

There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below:
Figure

How many units should be transported from each factory to each depot in order that the transportation cost is minimum. What will be the minimum transportation cost?

Answer 1

Let the factory at P transports x units of commodity to depot at A and y units to depot at B. Then , 8 − (x + y) units of commodity would be transported to depot C .
The requirement at depot A is 5 units since, x units are transported from factory P.
Therefore, the remaining (5 − x) units are transported from factory Q.

Similarly, (5 − y) units and 4 − [8 − (x + y)] i.e. (x + y − 4) units will be transported from factory Q to depot B and C respectively.



Units of commodity cannot be negative.Therefore,

$$\begin{gathered} x\ge 0,y\ge 0\mathrm{and}8-x-y\ge 0 \\ \Rightarrow x\ge 0,y\ge 0\mathrm{and}x+y\le 8 \\ 5-x\ge 0,5-y\ge 0\mathrm{and}x+y-4\ge 0 \\ \Rightarrow x\mathit{\le }5,y\le 5\mathrm{and}x+y\ge 4 \end{gathered}$$

The mathematical formulation of the given problem is
$$\begin{gathered} \mathrm{Minimize}\mathrm{Z}=160x+100y+150\left(8-x-y\right)+100\left(5-x\right)+120\left(5-y\right)+100\left(x+y-4\right) \\ =10x-70y+1900 \end{gathered}$$

subject to

$$\begin{gathered} x+y\le 8 \\ x\le 5 \\ y\le 5 \\ x+y\ge 4 \\ x,y\ge 0 \end{gathered}$$

First we will convert inequations into equations as follows:
x + y = 8, x = 5, y = 5, x + y = 4, x = 0 and y = 0

Region represented by x + y ≤ 8:
The line x + y = 8 meets the coordinate axes at A₁(8, 0) and B₁(0, 8) respectively. By joining these points we obtain the line x + y = 8. Clearly (0,0) satisfies the x + y = 8. So, the region which contains the origin represents the solution set of the inequation x + y ≤ 8.

Region represented by x ≤ 5:
x = 5 is the line that passes C₁(5, 0) and is parallel to the Y axis.The region to the left of the line x = 5 will satisfy the inequation x ≤ 5.

Region represented by y ≤ 5:
y = 5 is the line that passes D₁(0, 5) and is parallel to the X axis.The region below the line y = 5 will satisfy the inequation y ≤ 5.

Region represented by x + y ≥ 4:
The line x + y = 4 meets the coordinate axes at E₁(4, 0) and F₁(0, 4) respectively. By joining these points we obtain the line x + y = 4. Clearly (0,0) does not satisfies the inequation x + y ≥ 4. So,the region which does not contain the origin represents the solution set of the inequation x + y ≥ 4.

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 of the LPP is the shaded portion in the graph



The corner points of the feasible region are E₁(4, 0), C₁(5, 0), G₁(5, 3), H₁(3, 5), D₁(0, 5) and F₁(0, 4).

The values of the objective function at the corner points are

Corner points	Z = $10x-70y+1900$
E1(4, 0)	1940
C1(5, 0)	1950
G1(5, 3)	1740
H1(3, 5)	1580
D1(0, 5)	1550
F1(0, 4)	1620

Thus, the minimum value of Z is 1550 which is attained at D₁(0, 5).

Therefore, from P to A: 0 units, P to B : 5 units , P to C: 3 units
from Q to A : 5 units, Q to B : 0 units, Q to C: 1 unit