Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Linear Programming

Question:

A company has two factories located at P and Q and has three depots situated at A, B and C. The weekly requirement of the depots at A, B and C is respectively 5, 5 and 4 units, while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost (in) of transportation per unit is given below.

Cost (in )

To →

From ↓

A

B

C

P

160

100

150

Q

100

120

100

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

Options:

₹1450

₹1550

₹1650

₹1600

Correct Answer:

₹1550

Explanation:

The correct answer is Option (2) → ₹1550

We note that total weekly production (of P and Q) = 8+ 6 = 14 units, and total weekly requirement at depots A, B, C = 5 + 5 + 4 = 14 units, so there is no mismatch between supply and demand.

Let factory P supply x units per week to depot A and y units to depot B so that it supplies $8 - x - y$ units to depot C. Obviously $0 ≤ x ≤ 5, 0 ≤ y ≤ 5, 0 ≤8-x - y ≤ 4$.

The given data can be represented diagrammatically as:

As depot A's requirement is 5 units and it receives x units from factory P, it must receive $(5 - x)$ units from factory Q. Similarly, depot B receives $(5 - y)$ units from factory Q and Depot C receives $4- (8-x − y) = x + y - 4$ units from factory Q.

(As a cross-check, quantity supplied from factory Q to depots $A, B, C =(5x) + (5− y) + (x + y − 4) = 6$ units = capacity of factory Q.)

The total transportation cost (in ₹)

$= 160x + 100y+ 150(8 - x - y) + 100(5-x) + 120(5− y) + 100(x + y − 4)$

$= 10(x - 7y+ 190)$.

Hence, the given problem can be formulated as an L.P.P. as :

Find x and y which minimize

$Z = 10(x-7y+ 190)$ subject to the constraints

$x + y ≥ 4, x + y ≤ 8, x 20, x ≤5, y ≥0, y ≤5$

The feasible region corresponding to these inequalities is shown shaded in the figure given below. It is the bounded (convex) region ABCDEF. We calculate the values of Z (in ₹) at the six corner points: 

Point

  $x$  

  $y$  

$Z=10 (x-7y+190)$

A

4

0

1940

B

5

0

1950

C

5

3

1740

D

3

5

1580

E

0

5

1550 → Minimum

F

0

4

1620

We see that Z is minimum ₹1550 at point E(0, 5). Hence, $x = 0, y = 5$. Thus, for minimum transportation cost, factory P should supply 0, 5, 3 units to depots A, B, C respectively and factory Q should supply 5, 0, 1 units respectively to depots A, B, C.