APRESENTATION

ONSOLVING LPP

BYGRAPHICAL METHOD

Submitted By:Kratika DhootMBA- 2nd sem

What is LPP ???

• Optimization technique• To find optimal value of objective function, i.e.

maximum or minimum• “LINEAR” means all mathematical functions

are required to be linear…• “PROGRAMMING” refers to Planning, not

computer programming…

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What is graphical method ???• One of the LPP method• Used to solve 2 variable problems of LPP…

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Steps for graphical method…FORMULATE THE

PROBLEM( for objective &

constraints functions)

FRAME THE GRAPH( one variable on

horizontal & other at vertical axes)

PLOT THE CONSTRAINTS(inequality to be as equality;

give arbitrary value to variables & plot the point on graph )

PLOT THE GRAPH( one variable on

horizontal & other at vertical axes)

OUTLINE THE SOLUTION AREA

( area which satisfies the constraints)

CIRCLE POTENTIAL SOLUTION POINTS( the intersection

points of all constraints)

SUBSTITUTE & FIND OPTIMIZED SOLUTION

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LET US TAKE AN EXAMPLE!!!

SMALL SCALE ELECTRICAL

REGULATORS INDUSTRY

ACCOMPLISHED BY SKILLED MEN &

WOMEN WORKERS

BUT NUMBER OF WORKERS CAN’T

EXCEED 11

MALE WORKERS ARE PAID Rs.6,000pm &

FEMALE WORKERS ARE PAID Rs.5,000pm

SALARY BILL NOT MORE THAN Rs.

60,000 pm

DATA COLLECTED FOR THE

PERFORMANCE

DATA INDICATED MALE MEMBERS CONTRIBUTES Rs.10,000pm &

FEMALE MEMBERS CONTRIBUTES Rs.8,500pm

DETERMINE No. OF MALES & FEMALES TO BE EMPLOYED IN ORDER TO MAXIMIZE TOTAL

RETURN

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STEP 1-FORMULATE THE PROBLEM

Objective Function :- Let no. of males be x & no. of females be y

Maximize Z = Contribution of Male members + contribution of Female members

Subjected To Constraints :-Max Z = 10,000x + 8,500y

x + y ≤ 11 ………..(1)6,000x + 5,000y ≤ 60,000 ………..(2)

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STEP 2- FRAME THE GRAPH

• Let no. of Male Workers(x) be on horizontal axis & no. of Female Workers (y) be vertical axis..

No. of females

No. of males

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STEP 3- PLOT THE CONSTRAINTS• To plot the constraints, we will opt an arbitrary

value to the variables as:-x + y ≤ 11:- converting as x + y = 11

6,000x+5,000y≤60,000:- converting as 6x + 5y= 60

x 0 11

y 11 0

x 0 10

y 12 0

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No. of females

No. of males

0

6

4

10

8

1412

2

2

4 6 10 128

x + y ≤ 11 6x + 5y ≤60

●

●

●

●

( 0 , 11 )

( 11 , 0 )

( 0 , 12 )

( 10 , 0 )

STEP 4- PLOT THE GRAPHNo. of females

No. of males

0

6

4

10

8

1412

2

2

4 6 10 128

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No. of females

No. of males0

6

4

10

8

12

2

2

4 6 10 128

●

●

●

●

●

OPTIMAL SOLUTION POINT

( 5, 6 )

STEP 5- FIND THE OPTIMAL SOLUTION

FEASIBLE REGION

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STEP 6- CIRCLE POTENTIAL OPTIMAL POINTS

No. of females

No. of males0

6

4

10

8

12

2

2

4 6 10 128

( 5, 6 )

( 10 , 0 )

( 0 , 11 )

( 0 , 0 )

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STEP 7- SUBSTITUE & OPTIMIZE

Max Z = 10,000x + 8,500y

POTENTIAL OPTIMAL PTS.

Z = 10,000x + 8,500y MAXIMUM Z

(0,0) 10,000(0) + 8,500(0) 0

(0,11) 10,000(0)+8,500(11) 93,500

(5,6) 10,000(5)+8,500(6) 1,01,000

(10,0) 10,000(10)+8,500(0) 1,00,000

1,01,000

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CONCLUSION• Thus, maximum total return is about

Rs.1,01,000 by adopting 5 male workers & 6 female workers.

• Hence, optimal solution for LPP is :-No. of male workers = 5No. of female workers = 6Max. Z = Rs. 1,01,000

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Let us take other example!!!

• Find the maximum value of objective functionZ= 4x + 2y

s.t. x + 2y ≥ 43x + y ≥ 7-x + 2y ≤ 7& x ≥ 0 & y ≥ 0

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PLOT THE CONSTRAINTS

x + 2y = 4

3x + y = 7

-x + 2y = 7

x 0 4

y 2 0

x 0 7/3

y 7 0

x 0 -7

y 7/2 0

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PLOTTING CONSTRAINTS TO

GRAPH

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x + 2y ≥ 4x 0 4

y 2 0

740

3

2

5

4

76

1

1

2 3 5 6

Y

X

( 0 , 2 )

( 4 , 0 )

• •

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3x + y ≥ 7 x 0 7/3

y 7 0

•

• 4

0

3

2

5

4

76

1

1

2 3 5 6

Y

X7

( 0 , 7 )

( 7/3 , 0 )

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-x + 2y ≤ 7x 0 -7

y 7/2 0

-4 -1

Y

X

3

2

5

4

76

-7

1

-6 -5 -3 -20

( 0 , 7/2 )

( -7 , 0 )

•

•

KRATIKA DHOOT

There is a common portion or common points which intersects by all 3 regions of lines

40

3

5

4

76

1 2 3 5 6

21

Y

X7

-1-2-3-4-5-6-7

x + 2y ≥ 4

3x + y ≥ 7

-x + 2y ≤ 7

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( 4 ,0 )

-5 -24

0

3

54

76

1 2 3 5 6

21

Y

X7

-1-3-4-6-7

( 1 , 4 )

( 2 , 1 )

CIRCLE THE POTENTIAL POINTS!!!

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STEP 7- SUBSTITUE & OPTIMIZE

Max Z = 4x + 2y

POTENTIAL OPTIMAL PTS.

Z = 4x + 2y MAXIMUM Z

(1,4) 4(1) + 2(4) 12

(2,1) 4(2)+2(1) 10

(4,0) 4 (4)+ 2(0) 1616

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CONCLUSION

Hence, the optimal solution is:X = 4Y = 0

Max z = 16

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PRACTICE QUESTIONS … (1) Maximize f(x) = x1 + 2x2

subject to: x1 + 2x2 ≤ 3x1 + x2 ≤ 2x1 ≤ 1

& x1 , x2 ≥ 0

(2) Maximize z = 4x+2ysubject to: 4x+6y≥12

2x+4y≤4& x≥0 ; y≥0

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THANK YOU !!!

KRATIKA DHOOT