GRAPHICAL METHOD OF SOLVING LINEAR PROGRAMMING PROBLEM

Presentedby:Guruvayur Maharana

Rahul SinghaniaSoumya Ranjan Das

Abhay Gupta

ProblemA company produces two types of hats. Every

hat A require twice as much labour time as the second hat B. If the company procures only hat B then it can produce a total of 500 hats a day. The market limits daily sales of the hat A and B of 150 and 250 hats. The profits on hat A and B are Rs. 8 and Rs. 5 respectively. Solve graphically to get the optimal solution.

SolutionHere let x1 and x2 be the no. of units of type A

and type B hats respectively.The objective function is:

Max Z = 8x1 + 5x2 ……………1

Subject to:2x1 + x2 <=500 ……………………2

x1 <=150 ……………………3 x2 <=250 ……………………4

x1,x2 >=0

First rewrite the inequalities of the constraints into an equation and plot the lines in the graph.

2x1 + x2 = 500 passes through (0,500)(250,0) x1 = 150 passes through (150,0) x2 = 250 passes through (0,250)

Thus, OABCD is the feasible region.

The point B can be found out by solving eq.2 & eq.3, so we get point B=(150,200)

The point C can be found out by solving eq.2 & eq.4, so we get point C=(125,250)

Now as we have all the corner points we substitute it in the objective function

Corner Points Value of Z = 8x1 + 5x2O(0,0) 0A(150,0) 1200B(150,200) 2200C(125,250) 2250 (Max Z)D(0,250) 1250

Therefore the maximum value of Z is attained at C(125,250).

Thus the optimal solution is x1=125 and x2=250.

Thus, the company has to produce 125 hats of type A and 250 hats of type B

Problem

Solve Graphically:Minimize Z= 6x1 + 14x2

Subject to 5x1 + 4x2>= 603x1 + 7x2<= 84 x1 + 2x2>= 18 x1,x2>= 0

Solution

Minimize Z= 6x1 + 14x2…………..1

Subject to 5x1 + 4x2>= 60…………23x1 + 7x2<= 84…………3 x1 + 2x2>= 18…………4 x1,x2>= 0

First rewrite the inequalities of the constraints into an equation and plot the lines in the graph.

5x1 + 4x2 = 60 passes through(0,15)&(12,0)3x1 + 7x2 = 84 passes through(28,0)&(0,12)x1 + 2x2 = 18passes through(18,0)&(0,9)

Thus, ABCD is the feasible region.

The point C can be found out by solving eq.2 & eq.3, so we get point C=(84/23,240/23)

The point D can be found out by solving eq.2 & eq.4, so we get point D=(18,5)

Now as we have all the corner points we substitute it in the objective function.

Corner Points Value of Z = 6x1 + 14x2A(18,0) 108 (Min Z)B(28,0) 168C(84/23,240/23) 168D(8,5) 118

Therefore the minimum value of Z is attained at A(18,0).

Thus the optimal solution is x1=18 and x2=0.

ProblemA retired person wants to invest upto an amount

of Rs. 30,000 in fixed income securities. His broker recommends investing in two bonds: Bond A yielding 7% and Bond B yielding 10%. After some consideration he decides to invest at most Rs. 12,000 in Bond B and atleast Rs. 6,000 in Bond A. He also wants the amount invested in Bond A to be atleast equal to the amount invested in Bond B. What should the broker recommend if the investor wants to maximize his return on investment? Solve graphically.

SolutionLet x1 & x2 be the amt. invested in Bond A & Bond B

respectively.

Thus, we haveMax Z = 0.07x1 + 0.10x2………………………1

Subject to x1 + x2<= 30,000………………..2x1 >= 6,000……………….3 x2<= 12,000……………….4x1 – x2>=0………………………….5 x1, x2>=0

First rewrite the inequalities of the constraints into an equation and plot the lines in the graph.

x1 + x2 = 30,000 passes through(30000,0)(0,30000)x1 = 6,000 passes through(6000,0)

x2 = 12,000 passes through(0,12000)x1 – x2 = 0

Thus, the feasible region is ABCDE

The point B can be found out by solving eq.5 & eq.3, so we get point B=(6000,6000)

The point C can be found out by solving eq.5 & eq.4, so we get point C=(12000,12000)

The point D can be found out by solving eq.2 & eq.4, so we get point D=(18000,12000)

Now as we have all the corner points we substitute it in the objective function.

Corner Points Value of Z = 0.07x1 + 0.1x2A(6000,0) 420B(6000,6000) 1020C(12000,12000) 2040D(18000,12000) 2460 (Max Z)E(30000,0) 2100

Therefore the minimum value of Z is attained at D(18000,12000)

Thus the optimal solution is x1=18,000 and x2=12,000.

Unbounded Problem

Solve Graphically:Maximize Z= 3x1 + 5x2………………….1

Subject to 2x1 + x2>= 7…………………..2x1 + x2>= 61…………………..3 x1 + 3x2>= 9…………………4 x1,x2>= 0

First rewrite the inequalities of the constraints into an equation and plot the lines in the graph.

2x1 + x2 = 7 passes through(0,7)&(3.5,0)x1 + x2 = 6 passes through(0,6)&(6,0)x1 + 3x2 = 9 passes through(9,0)&(0,3)

Thus, ABCD is the corner points.

1 2 3 4 5 6 7 8 9

8

7

6

5

4

3

2

1

A(0,7)

B(1,5)

C(4.5,1.5)

D(9,0)

2x1 + x2>=7 x1 + x2>=6

x1 + 3x2>=9

The point B can be found out by solving eq.3 & eq.4, so we get point B=(1,5)

The point C can be found out by solving eq.5 & eq.4, so we get point C=(4.5,1.5)

Now as we have all the corner points we substitute it in the objective function.

Corner Points Value of Z = 3x1 + 5x2A(0,7) 35 (Max Z)B(1,5) 28C(4.5,1.5) 21D(9,0) 27

The value of the objective function at the corner points A,B,C & D are 35, 28, 21 & 27. But there exists an infinite number of points in the feasible region where the values of the objective function is more than the values at these four points.

Thus, it follows that the maximum values of the objective function occurs at the point at infinity and hence the problem has an unbound solution.