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1. A toy company manufactures two types of dolls, a basic version doll-A and a deluxe version doll-B. Each doll of type B takes twice as long to produce as one of type A, and the company would have time to make maximum of 1000 per day. The supply of plastic is sufficient to produce 1000 dolls per day(both A & B combined). The deluxe version requires a facny dress of which there are only 500 per day available. If the company makes a profit of Rs 3.00 and Rs 5.. per doll, respectively on doll A and B, then how many of each doll should be produced per day in order to maximise the total profit. Formulate this problem.

Ans.

Let x1 and x2 be the number of dolls produced per day of type A and B respectively.

Let the doll A require “t” hrs to make.
So the doll B would require “2t” hours to make.
So the total time to manufacture x1 & x2 dolls should not exceed 1000 per day. Therefore,
x1 + 2x2 ≤ 1000 (time constrain).
Other constrains are:
x1 + x2 ≤ 1000 (Plastic constrain).
x2 ≤ 500 (Dress constrain).
And the non-negative constrains are
x1 ≥0, x2 ≥0

Maximize the profit function:
p = 3x1 + 5x2

2. What are the advantages of Linear programming techniques?

Ans. Advantages—

1. The linear programming technique helps to make the best possible use of available productive resources (such as time, labour, machines etc.)

2. It improves the quality of decisions. The individual who makes use of linear programming methods becomes more objective than subjective.

3. It also helps in providing better tools for adjustment to meet changing conditions.

4. It highlights the bottlenecks in the production processes.

5. As the problem is linear in nature the computational power required is less compared to non-linear methods. Therefore, large problems like optimization of an entire refinery can be performed using LP technique using desk top computers.

3. Solve the following Assignment Problem

|Operations | | | | | | |M1 |M2 |M3 |M4 | |O1 |10 |15 |12 |11 | |O2 |9 |10 |9 |12 | |O3 |15 |16 |16 |17 |

Ans:

1) Introducing the dummy row
|Operations | | | | | | |M1 |M2 |M3 |M4 | |O1 |10 |15 |12 |11 | |O2 |9 |10 |9 |12 | |O3 |15 |16 |16 |17 | |O4 |0 |0 |0 |0 |

2) Row reduced matrix

|Operations |  |  |  |  | |  |M1 |M2 |M3 |M4 | |O1 |0 |5 |2 |1 | |O2 |0 |1 |0 |3 | |O3 |0 |1 |1 |2 | |O4 |0 |0 |0 |0 |

3) Column reduced matrix

Table same as step 2.

4) Draw the minimum number of straight lines needed to cover all zeros. Since three lines suffice, the solution is not optimal. (There are four operators, i.e. O1, O2, O3 and O4 to be assigned so need minimum four line)

|Operations |  |  |  |  | |  |M1 |M2 |M3 |M4 | |O1 |0 |5 |2 |1 | |O2 |0 |1 |0 |3 | |O3 |0 |1...
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