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

2. 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

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 OUTLINE THE
PROBLEM SOLUTION AREA
( for objective & ( area which satisfies
constraints functions) the constraints)
CIRCLE POTENTIAL
FRAME THE GRAPH PLOT THE GRAPH SOLUTION POINTS
( one variable on ( one variable on ( the intersection
horizontal & other at horizontal & other points of all
vertical axes) at vertical axes) constraints)
PLOT THE CONSTRAINTS
(inequality to be as equality; SUBSTITUTE & FIND
give arbitrary value to variables OPTIMIZED SOLUTION
& plot the point on graph )
KRATIKA DHOOT

5. LET US TAKE AN EXAMPLE!!!
SMALL SCALE
ACCOMPLISHED BY BUT NUMBER OF
ELECTRICAL
SKILLED MEN & WORKERS CAN’T
REGULATORS
WOMEN WORKERS EXCEED 11
INDUSTRY
SALARY BILL NOT MALE WORKERS ARE DATA COLLECTED
MORE THAN Rs. PAID Rs.6,000pm & FOR THE
60,000 pm FEMALE WORKERS ARE PERFORMANCE
PAID Rs.5,000pm
DATA INDICATED MALE MEMBERS DETERMINE No. OF MALES &
CONTRIBUTES Rs.10,000pm & FEMALES TO BE EMPLOYED IN
FEMALE MEMBERS CONTRIBUTES ORDER TO MAXIMIZE TOTAL
Rs.8,500pm 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
Max Z = 10,000x + 8,500y
Subjected To Constraints :-
x + y ≤ 11 ………..(1)
6,000x + 5,000y ≤ 60,000 ………..(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
• To plot the constraints, we will opt an arbitrary
value to the variables as:-
x + y ≤ 11:- converting as x + y = 11
x 0 11
y 11 0
6,000x+5,000y≤60,000:- converting as 6x + 5y= 60
x 0 10
y 12 0
KRATIKA DHOOT

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

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

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

12. STEP 7- SUBSTITUE & OPTIMIZE
Max Z = 10,000x + 8,500y
POTENTIAL Z = 10,000x + 8,500y MAXIMUM Z
OPTIMAL PTS.
(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
1,01,000
(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
Z= 4x + 2y
s.t. x + 2y ≥ 4
3x + y ≥ 7
-x + 2y ≤ 7
&x≥0&y≥0
KRATIKA DHOOT

15. PLOT THE CONSTRAINTS
x 0 4
x + 2y = 4
y 2 0
x 0 7/3
3x + y = 7 y 7 0
x 0 -7
-x + 2y = 7 y 7/2 0
KRATIKA DHOOT

16. PLOTTING
CONSTRAINTS TO
GRAPH
KRATIKA DHOOT

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

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

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

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

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

22. STEP 7- SUBSTITUE & OPTIMIZE
Max Z = 4x + 2y
POTENTIAL Z = 4x + 2y MAXIMUM Z
OPTIMAL PTS.
(1,4) 4(1) + 2(4) 12
(2,1) 4(2)+2(1) 10
(4,0) 4 (4)+ 2(0) 16
16
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