UNIT 2
Major steps in the solution of L.P.P by Graphical Major steps in the solution of L.P.P by Graphical Method.
Step:1 draw x1 and x2 axes (which are mutually perpendicular) on a graph sheet.
Step:2 draw a line and identity the region connected with its corresponding to each
constraints.
Step:3 identity the solutions space which is the region that is common to all constraints
including the non negative restrictions.
Step:4 find the value of Z at each vertex of the solution space.
Step:5 identity the optimum solution.
Example:1
Solve by graphical method.
Max Z = 3x1 + 4x2
Subject to constraints
4x1 + 2x2 ≤ 80
2x1 + 5x2 ≤ 180 and x1, x2 ≥ 0
Solution:
Consider the first constraint
4x1 + 2x2 ≤ 80
Treating as a strict equation, we have
4x1 + 2x2 = 80
Put x1 = 0 0 + 2x2 = 80
x2 =
80
2
= 40
put x2 = 0 4x1 + 0 = 80
x1 =
80
4
= 20
x1 0 20
x2 40 0
Consider the second constraint
2x1 + 5x2 ≤ 180
Treating as a strict equation, we have
2x1 + 5x2 = 180
Put x1 = 0 0 + 5x2 = 180.Step:1 draw x1 and x2 axes (which are mutually perpendicular) on a graph sheet.
Step:2 draw a line and identity the region connected with its corresponding to each
constraints.
Step:3 identity the solutions space which is the region that is common to all constraints
including the non negative restrictions.
Step:4 find the value of Z at each vertex of the solution space.
Step:5 identity the optimum solution.
Example:1
Solve by graphical method.
Max Z = 3x1 + 4x2
Subject to constraints
4x1 + 2x2 ≤ 80
2x1 + 5x2 ≤ 180 and x1, x2 ≥ 0
Solution:
Consider the first constraint
4x1 + 2x2 ≤ 80
Treating as a strict equation, we have
4x1 + 2x2 = 80
Put x1 = 0 0 + 2x2 = 80
x2 =
80
2
= 40
put x2 = 0 4x1 + 0 = 80
x1 =
80
4
= 20
x1 0 20
x2 40 0
Consider the second constraint
2x1 + 5x2 ≤ 180
Treating as a strict equation, we have
2x1 + 5x2 = 180
Put x1 = 0 0 + 5x2 = 180x2 =
180
5
= 36
put x2 = 0 2x1 + 0 = 180
O(0,0) 20 40 60 80 100 X1
4x1 + 2x2 = 80
The values of the objective function at the vertex are
Corner Point Coordinates Max Z = 3x1+ 4x2
O (0, 0) Z = 3(0)+ 4(0)= 0
A (20, 0) Z = 3(20)+ 4(0)= 60
C (2.5, 35) Z = 3(2.5)+ 4(35)= 147.5
E (0, 36) Z = 3(0)+ 4(36) =144
Hence the maximum value of Z=147.5 and the solution of the given L.P.P
x1 = 2.5 and x2= 35
A.P.Mukhil
23UCM026
11-02-2024