graphical method
TRANSCRIPT
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
KRATIKA DHOOT
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