Graphical Method of Solving Linear Programming Problem

GRAPHICAL METHOD OF SOLVING LINEAR PROGRAMMING PROBLEM

Presentedby: Guruvayur Maharana Rahul Rah ul Sin Singha ghania nia Soum So umy ya Ra Ranj njan an Da Dass Abha Ab hay y Gu Gupt pta a

Problem A company produces two types of hats. Every Every hat A require require twice as much labour labour time as the second hat B. If the company 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 graphically to get the optimal solution.

Solution Here let x1 and x2 be the no. of units of type A and type B hats resp especti ectiv vely ely. The objec jective function is: Max Z = 8x1 + 5x2 1 Subj Subjec ectt to: 2x1 + x2 <=500 x1 < = 150 x2 < = 250 x1,x2 >=0

2 3 4

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 the object ective function Corner Points

Value of Z = 8x1 + 5x2

O(0,0)

0

A(150,0)

1200

B(150,200)

2200

C(125,250)

2250 (Max Z)

D(0,250)

12 50

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>= 60 3x1 + 7x2<= 84 x1 + 2x2>= 18 x1,x2>= 0

Solution Minimize

Z= 6x1 + 14x2..1

Subject to

5x1 + 4x2>= 602 3x1 + 7x2<= 843 x1 + 2x2>= 184 x1,x2>= 0

First First rewrite the inequalities of the constraints into an equation and plot the lines in the graph. 5x1 5x1 + 4x2 4x2 = 60 60

pass passes es thr through ough(0 (0,1 ,15) 5)&( &(12 12,0 ,0))

3x1 3x1 + 7x2 7x2 = 84 84

pass passes es thr through ough(2 (28, 8,0) 0)&( &(0, 0,12 12))

x1 + 2x2 = 18

passes 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 substitute it in the the objective function. Corner Points


A(18,0)

108

B(28,0)

168

C(84/23,240/23)

168

D(8,5)

118

(Min Z)

Therefore the minimum value of Z is attained at A(18,0). Thus the optimal solution is x1=18 and x2=0.

Problem A retired retired person person wants wants to invest invest upto upto an amount of Rs. 30,000 in fixed 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 most Rs. Rs. 12,000 12,000 in Bond B and atleas atleastt Rs. 6,000 in Bond A. He also wants the amount invest invested ed in Bond A to be be atleast atleast equal to the the amount invested invested in Bond B. What should the broker recommend if the investor wants to maximize his return on investment? Solve graphically.

Solution Let x1 & x2 be the amt. invested invested in Bond A & Bond B respectively respectivel y. Thus, we have h ave Max Z = 0.07x1 + 0.10x21 Subject to

x1 + x2<= 30,000..2 x1 >= 6,000.3 x2<= 12,000.4 x1  x2>=0 x2>=0  .5 .5 x1, x2>=0

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

x1 + x2 = 30,000 x1 = 6,000 x2 = 12,000 x1  x2 = 0

passes through(30000,0)(0,30 ,30000) passes through(6000,0) passes through(0,12000) through(0,12000)

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)


Value of Z = 0.07x1 + 0.1x2

A(6000,0)

420

B(6000,6000)

1020

C(12000,12000)

2040

D(18000,12000)

2460

E(30000,0)

2100

(Max Z)

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..2 x1 + x2>= 61..3 x1 + 3x2>= 94 x1,x2>= 0

First 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


x1 + 3x2 = 9


Thus, ABCD is the corner points.

8 7

A 0,7

6 B 1,5

5 4 3 2

C4515

1 D 9,0

1

2

3

4

5

6

7

8

9 +

2x1 + x2>=7

x1 + x2>=6

>=

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)



A(0,7)

35

B(1,5)

28

C(4.5,1.5)

21

D(9,0)

27

(Max Z)

The value of the objective function at the corner points A,B,C & D are 35, 28, 21 & 27. But there exists an infinite 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.

Graphical Method of Solving Linear Programming Problem

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