The given information can be exhibited diagrammatically as follows:
Let the factory at P transports x units of commodity to depot at A and yunits to depot at B. Since the factory at P has the capacity of 8 units of the commodity. Therefore, the left out (8 – x – y) units will be transported to depot at C.
Since the requirements are always non-negative quantities. Therefore,
x ≥ 0, y ≥ 0 and 8 – x – y ≥ 0 ⇒ x ≥ 0, y ≥ 0 and x + y ≤ 8
Since the weekly requirement of the depot at A is 5 units of the commodity and x units are transported from the factory at P. Therefore, the remaining (5 – x) units are to be transported from the factory at Q. Similarly, 5 – yunits of the commodity will be transported from the factory at Q to the depot at B. But the factory at Q has the capacity of 6 units only, therefore the remaining 6 – (5 – x + 5 – y) = x + y – 4 units will be transported to the depot at C. As the requirements at the depots at A, B and C are always non-negative.
∴ 5 – x ≥ 0, 5 – y ≥ 0 and x + y – 4 ≥ 0
⇒ x ≤ 5, y ≤ 5 and x + y ≥ 4.
The transportation cost from the factory at P to the depots at A, B and Care respectively Rs 16x, 10y and 15(8 – x – y). Similarly, the transportation cost from the factory at Q to the depots at A, B and C are respectively Rs 10(5 – x), 12(5 – y) and 10(x + y – 4). Therefore, the total transportation cost Z is given by
Z = 16x + 10y + 15(8 – x – y) + 10(5 – x) + 12(5 – y) + 10(x + y – 4)
Hence, the above LPP can be stated mathematically as follows:
Find x and y which
Minimize Z = x – 7y + 190
Subject to
x + y ≤ 8
x + y ≥ 4
x ≤ 5
y ≤ 5
and, x ≥ 0, y ≥ 0
