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# Mathematical formulation of a linear programming problem

The procedure for mathematical formulation of a linear programming problem consists of the following steps.

Mathematical formulation of a linear programming problem:

The procedure for mathematical formulation of a linear programming problem consists of the following steps.

(i) Identify the decision variables.

(ii) Identify the objective function to be maximized or minimized and express it as a linear function of decision variables.

(iii) Identify the set of constraint conditions and express them as linear inequalities or equations in terms of the decision variables.

Example 10.1

A furniture dealer deals only two items viz., tables and chairs. He has to invest Rs.10,000/- and a space to store atmost 60 pieces. A table cost him Rs.500/– and a chair Rs.200/–. He can sell all the items that he buys. He is getting a profit of Rs.50 per table and Rs.15 per chair. Formulate this problem as an LPP, so as to maximize the profit.

Solution:

(i) Variables: Let x1 and x2 denote the number of tables and chairs respectively.

(i) Objective function:

Profit on x1 tables = 50 x1

Profit on x2 chairs = 15 x2

Total profit = 50 x1+15x2

Let Z = 50 x1+15 x2 , which is the objective function.

Since the total profit is to be maximized, we have to maximize Z=50x1+15x2

(iii)  Constraints:

The dealer has a space to store atmost 60 pieces

x1+x2 ≤ 60

The cost of x1 tables = Rs.500 x1

The cost of x2 tables = Rs. 200 x2

Total cost = 500 x1 + 200 x2 ,which cannot be more than 10000

500 x1 + 200 x2 ≤ 10000

5x1+ 2x2 ≤ 100

(iv) Non-negative restrictions: Since the number of tables and chairs cannot be negative, we have x1 ≥0, x2 ≥0

Thus, the mathematical formulation of the LPP is

Maximize Z=50 x1+15x2

Subject to the constrains

x1+x2 ≤60

5x1+2x2≤100

x1, x2≥0

Example 10.2

A company is producing three products P1, P2 and P3, with profit contribution of Rs.20,Rs.25 and Rs.15 per unit respectively. The resource requirements per unit of each of the products and total availability are given below.

Formulate the above as a linear programming model.

Solution:

(i) Variables: Let x1 , x2 and x3 be the number of units of products P1, P2 and P3 to be produced.

(ii) Objective function: Profit on x1 units of the product P1 = 20 x1

Profit on x2 units of the product P2 = 25 x2

Profit on x3 units of the product P3 = 15 x3

Total profit = 20 x1 +25 x2 +15 x3

Since the total profit is to be maximized, we have to maximize Z= 20 x1+25 x2 +15 x3

Constraints:

6x1 +3 x2 +12 x3 ≤ 200

2 x1 +5 x2 +4 x3 ≤ 350

x1 +2 x2 + x3 ≤ 100

Non-negative restrictions: Since the number of units of the products A,B and C cannot be negative, we have x1, x2, x3  0

Thus, we have the following linear programming model.

Maximize Z = 20 x1 +25 x2 +15 x3

Subject to

6 x1 +3 x2 +12 x3≤ 200

2x1+5x2 +4 x3 ≤ 350

x1+2 x2 + x3≤ 100

x1, x2, x3 0

Example 10.3

A dietician wishes to mix two types of food F1 and F2 in such a way that the vitamin contents of the mixture contains atleast 6units of vitamin A and 9 units of vitamin B. Food F1 costs Rs.50 per kg and F2 costs Rs 70 per kg. Food F1 contains 4 units per kg of vitamin A and 6 units per kg of vitamin B while food F2 contains 5 units per kg of vitamin A and 3 units per kg of vitamin B. Formulate the above problem as a linear programming problem to minimize the cost of mixture.

Solution:

(i) Variables: Let the mixture contains x1 kg of food F1 and x2 kg of food F2

(ii) Objective function:

cost of x1 kg of food F1 = 50 x1

cost of x2 kg of food F2 = 70x2

The cost is to be minimized

Therefore    minimize Z= 50 x1+70x2

(iii)  Constraints:

We make the following table from the given data

4x1+5x2 ≥6 (since the mixture contains ‘atleast 6’ units of vitamin A ,we have the inequality of the type ≥ )

6x1+3x2 ≥9(since the mixture contains ‘atleast 9’ units of vitamin B ,we have the inequality of the type ≥ )

(iv)  Non-negative restrictions:

Since the number of kgs of vitamin A and vitamin B are non-negative , we have x1, x2 ≥0

Thus, we have the following linear programming model

Minimize Z = 50 x1 +70x2

subject to 4 x1+5x2 ≥6

6 x1+3x2 ≥9

and x1, x2≥0

Example 10.4

A soft drink company has two bottling plants C1 and C2. Each plant produces three different soft drinks S1,S2 and S3. The production of the two plants in number of bottles per day are:

A market survey indicates that during the month of April there will be a demand for 24000 bottles of S1, 16000 bottles of S2 and 48000 bottles of S3. The operating costs, per day, of running plants C1 and C2 are respectively Rs.600 and Rs.400. How many days should the firm run each plant in April so that the production cost is minimized while still meeting the market demand? Formulate the above as a linear programming model.

Solution:

(i) Variables: Let x1 be the number of days required to run plant C1 and x2 be the number of days required to run plant C2

Objective function: Minimize Z = 600 x1 + 400 x2

(ii) Constraints: 3000 x1 + 1000 x2 ≥ 24000(since there is a demand of 24000 bottles of drink A, production should not be less than 24000)

1000x1 + 1000 x2 ≥ 16000

2000x1+ 6000 x2 ≥ 48000

(iii) Non-negative restrictions: Since be the number of days required of a firm are non-negative, we have x1, x2 ≥0

Thus we have the following LP model.

Minimize Z = 600 x1 + 400 x2

subject to 3000 x1 + 1000 x2 ≥ 24000

1000 x1 + 1000 x2 ≥ 16000

2000 x1 + 6000 x2 ≥ 48000

and x1, x2 ≥0

Tags : Linear programming problem , 11th Business Mathematics and Statistics(EMS) : Chapter 10 : Operations Research
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11th Business Mathematics and Statistics(EMS) : Chapter 10 : Operations Research : Mathematical formulation of a linear programming problem | Linear programming problem