Learning Curve…

MB0048 : A furniture manufacturer makes two products: chairs and tables. Processing of these products is done on two machines A and B. A chair requires 2 hours on machine A and 6 hours on machine B. A table requires 5 hours on machine A and no time on machine B. There are 16 hours per day available on machine A and 30 hours on machine B. Profit gained by the manufacturer from a chair and a table is Rs 2 and Rs 10, respectively. What should be the daily production of each of the two products?

Posted on: October 20, 2011

MB0048 : A furniture manufacturer makes two products: chairs and tables. Processing of these products is done on two machines A and B. A chair requires 2 hours on machine A and 6 hours on machine B. A table requires 5 hours on machine A and no time on machine B. There are 16 hours per day available on machine A and 30 hours on machine B. Profit gained by the manufacturer from a chair and a table is Rs 2 and Rs 10, respectively. What should be the daily production of each of the two products?

Solution

Let x1 = no of hours on machine A & x2 =no of hours on machine B

Then the problem can be formulated as a P model as follows:-

Objective function, Maximise  Z = 2x1 + 10x2

Constraint equations: –

2x1 + 6x2 < 16

5x1                < 30                              

x1, x2 > 0

Step I

Treating all the constraints as equality, the first constraint is

2x1 + 6x2 < 16

Put x1 = 0 ⇒ = x2 = 2.7 ∴ The point is (0, 2.7)

Put x2 = 0 ⇒ = x1 = 8 ∴ The point is (8, 0)                                 2.7

Step II

Treating all the constraints as equality, the first constraint is

5x1                < 30                                                                                                                             0                                                      6A     8c

X1 = 6 ∴ The point is (6, 0)

Step III

The intersection of the above graphic denotes (A, B, C) the feasible region for the given problem.

Step IV

At corner points (O, A, B, C), find the profit value from the objective function. That point which maximize the profit is the optimal point.

Corner points

W-Ordinates

Objective Function

Z = 2x1 + 10x2

Value

A (6, 0) Z = 2(6) +0 12
B (6, 2/3) Z = 2(6) + 10 (2/3) 19
C (8, 0) Z = 2(8) + 0 16

Result

The optimal solution is:

No of hours on machine A = 6

No of hours on machine B =2/3

Total profit, = 19 which is the maximum.

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