Request a call back

There is a factory located at each of the two places  P and Q. From these location, a certain commodity is delivered to each of these 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 :   To From                                     Cost (in Rs) A B C P 16 10 15 Q 10 12 10   How many units should be transported from each factory to each depot in order that the transportation cost in minimum. Formulate the above LPP mathematically and the solve it.
Asked by Topperlearning User | 31 Jul, 2016, 05:47: PM

Let the factory at P transports x units of commodity to depot at A and y units to depot at B. Then, the mathematical model to the LPP is as follows :

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 = 16x + 10y +1 0(5-x) + 12(5-y) + 10(x+y-4) + 15(8-x-y)

= x - 7y + 190

Minimize Z = x - 7y + 190

Subject to     x + y < 8

x + y < 4

x < 5

y < 5

and              x > 0, y > 0

To solve this LPP graphically, we first convert the in equations into equations and draw the corresponding lines. The feasible region of the LPP is shaded in Fig

The coordinates of the corner points of the feasible region A2 A3 PQ B3 B2 are A2 (4, 0), A3 (5, 0), P(5, 3), Q (3, 5), B3 (0, 5) and B2 (0, 4). These points have been obtained by solving the corresponding intersecting line simultaneously.

The values of the objective function at these points are given in the following table :

 Point (x, y)                                  Value of the objective function                                                                               Z= x - 7y + 190 A2 (4, 0)                                      Z = 4 - 7 x 0 + 190 = 194                    A3 (5, 0)                                      Z = 5 - 7 x 0 + 190 = 195                    P (5, 3)                                       Z = 5 - 7 x 3 + 190 = 174                    Q (3, 5)                                       Z = 3 - 7 x 5 + 190 = 158                    B3 (0, 5)                                      Z = 0 - 7 x 5 + 190 = 155                    B2 (0, 4)                                      Z = 0 - 7 x 4 + 190 = 162

Clearly, Z is minimum at x = 0, y = 5. The minimum value of Z is 155.

Thus, the optimal transportation strategy will be to deliver 0, 5 and 3 units from the factory at P and 5, 0  and 1 unit from the factory at Q to the depots at A, B and C respectively. The minimum transportation cost in this case in Rs 155.

Answered by | 31 Jul, 2016, 07:47: PM

## Concept Videos

CBSE 12-science - Maths
Asked by gargisha381 | 26 May, 2022, 12:24: PM