a presentation on solving lpp by graphical method

1. A
PRESENTATION
ON
SOLVING LPP
BY
GRAPHICAL METHOD
Submitted By:
Kratika Dhoot
MBA- 2nd sem

2. What is LPP ???
• Optimization technique
• Tofind 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

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

4. 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
OPTIMIZEDSOLUTION
KRATIKA DHOOT

5. 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
DATACOLLECTED
FOR THE
PERFORMANCE
DATAINDICATED 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

6. 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 ToConstraints :-
Max Z = 10,000x + 8,500y
x + y ≤ 11
6,000x + 5,000y ≤ 60,000
………..(1)
………..(2)
KRATIKA DHOOT

7. 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
KRATIKA DHOOT

8. STEP 3- PLOT THE CONSTRAINTS
• Toplot 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

9. No. of
females
0
14
12
10
8
6
4
2
x + y ≤ 11 6x + 5y ≤60
● ●
( 0 , 11 )
( 11 , 0 )
● ( 0 , 12 )
( 10 , 0 )
STEP 4- PLOT THE GRAPH
2 4 6 8 10 12 No. of 2 4 6 8 10 12 No. of
males males
0
No. of
females
14
12
●
10
8
6
4
2
KRATIKA DHOOT

10. No. of females
No. of males
0
2
12
10
8
6
4
2
4 6 8
●
●
● ●
10 12
●
OPTIMAL
SOLUTION POINT
( 5, 6 )
STEP 5- FIND THE OPTIMAL SOLUTION
FEASIBLE
REGION
KRATIKA DHOOT

11. STEP 6- CIRCLE POTENTIAL OPTIMAL
POINTS
No. of females
No. of males
12
10
2
8
6
4
2
4 6 10 12
8
( 5, 6 )
( 10 , 0 )
( 0 , 11 )
0
( 0 , 0 )
KRATIKA DHOOT

12. 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
1
,
0
,
0
1
1
,
,
0
0
0
0
0
0
(10,0) 10,000(10)+8,500(0) 1,00,000
KRATIKA DHOOT

13. 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 = 5
No. of female workers = 6
Max. Z = Rs. 1,01,000
KRATIKA DHOOT

14. Let us take other example!!!
• Find the maximum value of objective function
s.t.
Z= 4x + 2y
x + 2y ≥ 4
3x + y ≥ 7
-x + 2y ≤ 7
& x ≥ 0 & y ≥ 0
KRATIKA DHOOT

15. 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
KRATIKA DHOOT

16. PLOTTING
CONSTRAINTS TO
GRAPH
KRATIKA DHOOT

17. x + 2y ≥ 4
x 0 4
y 2 0
4
0
6
5
4
3
2
1
1 2 3 5 6 7
Y 7
X
( 4 , 0 )
•
•( 0 , 2 )
KRATIKA DHOOT

18. 3x + y ≥ 7 x 0 7/3
y 7 0
•
4
0
6
5
4
3
2
1
1 2 3 5 6
Y 7
X
7
( 0 , 7 )
•( 7/3 , 0
)
KRATIKA DHOOT

19. -x + 2y ≤ 7
x 0 -7
y 7/2 0
Y
X
-7
7
6
5
4
3
2
1
-6 -5 -4 -3 -2 -1
0
( -7 , 0 )
•
( 0 , 7/2
•
)
KRATIKA DHOOT

20. There is a common portion or common points which
intersects by all 3 regions of
lines
4
0
1 2 3 5 6
6
5
4
3
2
1
Y 7
X
7
-7 -6 -5 -4 -3 -2 -1
x + 2y ≥ 4
3x + y ≥ 7
-x + 2y ≤ 7
KRATIKA DHOOT

21. 4
0
1 2 3
( 4 ,0 )
5 6
6
5
4
3
2
1
Y 7
X
7
-7 -6 -5 -4 -3 -2 -1
( 1 , 4 )
( 2 , 1 )
CIRCLE THE POTENTIAL POINTS!!!
KRATIKA DHOOT

22. 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) 1
1
6
6
KRATIKA DHOOT

23. CONCLUSION
Hence, the optimal solution is:
X = 4
Y = 0
Max z = 16
KRATIKA DHOOT

24. PRACTICE QUESTIONS …
(1)Maximize f(x) = x1 + 2x2
subject to: x1 + 2x2 ≤ 3
x1 + x2 ≤ 2
x1 ≤1
& x1 , x2 ≥ 0
(2) Maximize z = 4x+2y
subject to: 4x+6y≥12
2x+4y≤4
& x≥0 ; y≥0
KRATIKA DHOOT

25. THANK YOU !!!
KRATIKA DHOOT