The document provides a summary of a presentation on solving linear programming problems (LPP) using the graphical method. It defines LPP and the graphical method. It then walks through the steps to solve an example LPP problem graphically, including formulating the problem, framing the graph, plotting the constraints, finding the optimal solution point, and determining the maximum value. The summary concludes that the optimal solution for the example problem is 5 male workers and 6 female workers, with a maximum total return of Rs. 1,01,000.

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