Let x units and y units of the commodity be transported from the factory at P to the depots at A and B respectively. Then (8 – x – y) units will be transported to depot at C
Hence, we have x ≥ 0, y ≥ 0 and 8 – x – y ≥ 0
i.e. x ≥ 0, y ≥ 0 and x + y ≤ 8
Now, the weekly requirement of the depot at A is 5 units of the commodity. Since x units are transported from the factory at P, the remaining (5 – x) units need to be transported from the factory at Q. Obviously, 5 – x ≥ 0, i.e. x ≤ 5.
Similarly, (5 – y) and 6 – (5 – x + 5 – y) = x + y – 4 units are to be transported from the factory at Q to the depots at B and C respectively.
Thus, 5 – y ≥ 0 , x + y – 4 ≥ 0 i.e. y ≤ 5 , x + y ≥ 4
Total transportation cost Z is given by
Z = 160 x + 100y + 100(5 – x) + 120(5 – y) + 100(x + y – 4) + 150 (8 – x – y)
= 10 (x – 7 y + 190)
Therefore, the problem reduces to
Minimise Z = 10 (x – 7y + 190)
subject to the constraints:
x ≥ 0, y ≥ 0 ... (1)
x + y ≤ 8 ... (2)
x ≤ 5 ... (3)
y ≤ 5 ... (4)
and x + y ≥ 4 ... (5)
The shaded region ABCDEF represented by the constraints (1) to (5) is the feasible region (see below figure). Observe that the feasible region is bounded. The coordinates of the corner points of the feasible region are (0, 4), (0, 5), (3, 5), (5, 3), (5, 0) and (4, 0).
Let us evaluate Z at these points.
From the table, we see that the minimum value of Z is 1550 at the point (0, 5).
Hence, the optimal transportation strategy will be to deliver 0, 5 and 3 units from the factory at P and 5, 0 and 1 units from the factory at Q to the depots at A, B and C respectively.
Corresponding to this strategy, the transportation cost would be minimum, i.e., Rs 1550.
