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APRESENTATION
ONSOLVING LPP
BYGRAPHICAL METHOD
Submitted By:Kratika DhootMBA- 2nd sem
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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…
KRATIKA DHOOT
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What is graphical method ???• One of the LPP method• Used to solve 2 variable problems of LPP…
KRATIKA DHOOT
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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
KRATIKA DHOOT
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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
KRATIKA DHOOT
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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)
KRATIKA DHOOT
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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
KRATIKA DHOOT
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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
KRATIKA DHOOT
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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 )
KRATIKA DHOOT
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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
KRATIKA DHOOT
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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
KRATIKA DHOOT
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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 )
• •
KRATIKA DHOOT
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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 )
KRATIKA DHOOT
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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
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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
KRATIKA DHOOT
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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!!!
KRATIKA DHOOT
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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
KRATIKA DHOOT
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CONCLUSION
Hence, the optimal solution is:X = 4Y = 0
Max z = 16
KRATIKA DHOOT
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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
KRATIKA DHOOT
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THANK YOU !!!
KRATIKA DHOOT