application of linear programming

8/7/2019 Application of Linear Programming

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Application of Linear Programming:

Linear Programming is used in many managerial decisions that are part of the scheduled

decision making of the firm. Here are some of the examples where Linear Programming can

be used:

� Product mix

� Make-buy

� Media selection

� Marketing research

� Portfolio selection

� Shipping & transportation, etc.

Now let us have a look at how Linear Programming works in a real life scenario. Given below

is a data of a company which is facing a dilemma of selecting the right product mix for its

production capacity so that they can maximize the profit by producing optimal quantity two

products.

Dilemma:

The firm manufactures two products A and B, each of which must be processed through

Department-1 and then Department-2. Product A takes 3 hours per unit of Department-1

and 4 hours of Department-2, while Product-B consumes 2 hours per unit of Department-1

and 6 hours per unit of Department-2.

120 labor hours at the maximum are available at Department-1, while 260 labor hours at

Department-2 per week. Profit margin for Product A is $5 per unit and for Product B is $6

per unit.

We have to find the optimal quantities of Product -A and Product-B at which the profit of thecompany is Maximum.

Product-A Product-B Available Labor Hours

Department-1 3 hours/unit 2 hours/unit 120 hours

Department-2 4 hours/unit 6 hours/unit 260 hours

Profit Margins $5/unit $6/unit

To solve this dilemma, we first have to convert this data into mathematical expression by

taking some basic assumptions.

1. Let X and Y represent Product-A and Product-B

2. Let our objective function be: Z = 5X + 6Y (maximization)

3. Let our constraint function be: (3X + 2Y <= 120) and (4X + 6Y <=260)

4. X and Y can be >= 0



Now lets put this mathematical expression to work.

We now have to graph this optimization function so that we can determine the optimal

point. To do this we will start with plotting our constraints.

We will label constraint (3X + 2Y = 120) as Line-1, and constraint (4X + 6Y = 260) as Line-2

To plot Line-1 and Line-2, we require coordinates.

For Line-1:

3X + 2Y = 120

For X intercept, put Y = 0

3X + 2(0) = 120

3X = 120

X = 40

Therefore, (X, 0) = (40, 0)

For Y intercept, put X = 0

3(0) + 2Y = 120

2Y = 120

Y = 60

Therefore, (0, Y) = (0, 60)

For Line-2:

4X + 6Y = 260

For X intercept, put Y = 0

4X + 6(0) = 2604X = 260

X = 65

Therefore, (X, 0) = (65, 0)

For Y intercept, put X = 0

4(0) + 6Y = 260

6Y = 260

Y = 43.34

Therefore, (0, Y) = (0, 43.34)



Identification of Corner Points for Feasible Region:

y A (0, 0)

y B (40, 0)

y C (20, 30)

y D (0, 43.34)

Calculation for Point-C:

C is the point of intersection of Line -1 and Line-2

To find C, we will solve the equation of Line -1 and Line-2 simultaneously

3X + 2Y = 120 - Line-1

4X + 6Y = 260 - Line-2

Multiply Equation of Line-1 by 3

3X + 2Y = 120 x 34X + 6Y = 260

+9X + 6Y = +360

-4X 6Y = -260

5X 0 = 100

5X = 100

X = 20

Put the value of X in Equation of Line -1

3X + 2Y = 120

3(20) + 2(Y) = 120

60 + 2Y = 120

2Y = 60

Y = 30

Therefore, C = (X, Y) = (20, 30)

The objective function of this optimization problem is (Z = 5X + 6Y)

Points 5 (X) + 6 (Y) = Z

A (0, 0) 0 + 0 = 0

B (40, 0) 200 + 0 = 200

C (20 , 30) 100 + 180 = 280*

D (0, 43.34) 0 + 260 = 260

*The objective point C (20, 30) is producing optimal value of objective function i.e.,

maximum profit; therefore point C (20, 30) is the optimal point, meaning if the company

produce 20 units of Product -A and 30 units of Product-B it will yield the maximum profit.

application of linear programming

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