Graphical Method

2. Limitation of Graphical Method
Graphical solution is limited to linear
programming models containing only two
decision variables.

3. Procedure
Step I: Convert each inequality as equation
Step II: Plot each equation on the graph
Step III: Shade the ‘Feasible Region’. Highlight the
common Feasible region.
Feasible Region: Set of all possible solutions.
Step IV: Compute the coordinates of the corner points
(of the feasible region). These corner points will
represent the ‘Feasible Solution’.
Feasible Solution: If it satisfies all the constraints
and non negativity restrictions.

4. Procedure (Cont…)
Step V: Substitute the coordinates of the corner points into the
objective function to see which gives the Optimal Value. That
will be the ‘Optimal Solution’.
Optimal Solution: If it optimizes (maximizes or minimizes)
the objective function.
Unbounded Solution: If the value of the objective function
can be increased or decreased indefinitely, Such solutions
are called Unbounded solution.
Inconsistent Solution: It means the solution of problem does
not exist. This is possible when there is no common feasible
region.

5. 88
77
66
55
44
33
22
11
1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10
xx
22
xx
11
Example
Max Max zz = 5 = 5xx
11 + 7 + 7xx
22
s.t. s.t. xx
11 << 6 6
22xx
11 + 3 + 3xx
22 << 19 19
xx
11 + + xx
22 << 8 8
xx
11 , , xx
22 >> 0 0
Every point is in this nonnegative quadrant

6. 88
77
66
55
44
33
22
11
1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10
xx
22
xx
11
xx
11 << 6 6
(6, 0)(6, 0)
Example (Cont…)
Max Max zz = 5 = 5xx
11 + 7 + 7xx
22
s.t. s.t. xx
11 << 6 6
22xx
11 + 3 + 3xx
22 << 19 19
xx
11 + + xx
22 << 8 8
xx
11 , , xx
22 >> 0 0

7. 88
77
66
55
44
33
22
11
1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10
22xx
11 + 3 + 3xx
22 << 19 19
xx
22
xx
11
(0, 6.33)(0, 6.33)
(9.5 , 0)(9.5 , 0)
Max Max zz = 5 = 5xx
11 + 7 + 7xx
22
s.t. s.t. xx
11 << 6 6
22xx
11 + 3 + 3xx
22 << 19 19
xx
11 + + xx
22 << 8 8
xx
11 , , xx
22 >> 0 0
Example (Cont…)

8. 88
77
66
55
44
33
22
11
1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10
xx
22
xx
11
xx
11 + + xx
22 << 8 8
(0, 8)(0, 8)
(8, 0)(8, 0)
Max Max zz = 5 = 5xx
11 + 7 + 7xx
22
s.t. s.t. xx
11 << 6 6
22xx
11 + 3 + 3xx
22 << 19 19
xx
11 + + xx
22 << 8 8
xx
11 , , xx
22 >> 0 0
Example (Cont…)

9. 88
77
66
55
44
33
22
11
1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10
22xx
11 + 3 + 3xx
22 << 19 19
xx
22
xx
11
xx
11 + + xx
22 << 8 8
xx
11 << 6 6
Max Max zz = 5 = 5xx
11 + 7 + 7xx
22
s.t. s.t. xx
11 << 6 6
22xx
11 + 3 + 3xx
22 << 19 19
xx
11 + + xx
22 << 8 8
xx
11 , , xx
22 >> 0 0
Example (Cont…)

10. 88
77
66
55
44
33
22
11
1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10
xx
22
xx
11
Example (Cont…)
A (0,0)
(6,0)B
(6,2)C
(5,3)D
(0,6.33)E

11. Objective Function : Max Z= 5x
1
+7x
2
Corner Points Value of Z
A – (0,0) 0
B – (6,0) 30
C – (6,2) 44
D – (5,3) 46
E – (0,6.33) 44.33
Optimal Point : (5,3)
Optimal Value : 46
Example (Cont…)

12. Example
Max Z=3 P1 + 5 P2
s.t. P1 < 4
P2 < 6
3 P1 + 2 P2 < 18
P1, P2 > 0

13. Example
P1
P2
0
Every point is in this nonnegative quadrant
Max Z=3 P1 + 5 P2
s.t. P1 < 4
P2 < 6
3 P1 + 2 P2 < 18
P1, P2 > 0

14. Example (Cont…)
P1
P2
(0,0)

Max Z= 3 P1 + 5 P2
s.t. P1 < 4
P2 < 6
3 P1 + 2 P2 < 18
P1, P2 > 0
(4,0)

15. Example (Cont…)
P1
P2
(0,0)

Max Z= 3 P1 + 5 P2
s.t. P1 < 4
P2 < 6
3 P1 + 2 P2 < 18
P1, P2 > 0
(4,0)
(0,6)

16. Example (Cont…)
P1
P2
(0,0)

Max Z= 3 P1 + 5 P2
s.t. P1 < 4
P2 < 6
3 P1 + 2 P2 < 18
P1, P2 > 0
(4,0)
(0,6) (2,6)
(4,3)
(6,0)

17. Example (Cont…)
P1
P2
(0,0)

(4,0)
(0,6) (2,6)
(4,3)
(6,0)A
B
C
DE

18. Example (Cont…)
Objective Function : Max Z= 3 P1 + 5 P2
Corner Points Value of Z
A – (0,0) 0
B – (4,0) 12
C – (4,3) 27
D – (2,6) 36
E – (0,6) 30
Optimal Point : (2,6)
Optimal Value : 36

19. Example
Min Min z z = 5 = 5xx
11 + 2 + 2xx
22
s.t. 2s.t. 2xx
11 + 5 + 5xx
22 >> 10 10
44xx
11 - - xx
22 >> 12 12
xx
11 + + xx
22 >> 4 4
xx
11, , xx
22 >> 0 0

20. 55
44
33
22
11
1 2 3 4 5 61 2 3 4 5 6
xx
22
44xx
11 - - xx
22 >> 12 12
22xx
11 + 5 + 5xx
22 >> 10 10
xx
11
ExampleExample
Min Min z z = 5 = 5xx
11 + 2 + 2xx
22
s.t. 2s.t. 2xx
11 + 5 + 5xx
22 >> 10 10
44xx
11 - - xx
22 >> 12 12
xx
11 + + xx
22 >> 4 4
xx
11, , xx
22 >> 0 0
xx1 + 1 + xx2 2 >> 4 4

21. 55
44
33
22
11
1 2 3 4 5 61 2 3 4 5 6
xx
22
xx
11
Feasible RegionFeasible Region
This is the case of This is the case of
‘Unbounded Feasible Region’.‘Unbounded Feasible Region’.
Example (Cont…)Example (Cont…)
Min Min z z = 5 = 5xx
11 + 2 + 2xx
22
s.t. 2s.t. 2xx
11 + 5 + 5xx
22 >> 10 10
44xx
11 - - xx
22 >> 12 12
xx
11 + + xx
22 >> 4 4
xx
11, , xx
22 >> 0 0
A (35/11 , 8/11)A (35/11 , 8/11)
B (5,0)B (5,0)

22. Example (Cont…)
Objective Function : Max Z= 5x
1
+2x
2

Corner Points Value of Z
A – (35/11, 8/11)191/11 (17.364)
B – (5,0) 25
Optimal Point : (35/11,8/11)
Optimal Value : 191/11=17.364

23. Example
Max Max z z = 3 = 3xx
11 + 4 + 4xx
22
s.t. s.t. xx
11 + + xx
22 >> 5 5
33xx
11 + + xx
22 >> 8 8
xx
11, , xx
22 >> 0 0

24. ExampleExample
xx
11
33xx
11 + + xx
22 >> 8 8
xx
11 + + xx
22 >> 5 5
55
55
88
2.672.67
Max Max z z = 3 = 3xx
11 + 4 + 4xx
22
s.t. s.t. xx
11 + + xx
22 >> 5 5
33xx
11 + + xx
22 >> 8 8
xx
11, , xx
22 >> 0 0
Feasible RegionFeasible Region
This feasible region is This feasible region is
unbounded, hence unbounded, hence zz can can
be increased infinitely.be increased infinitely.
So this problem is having a So this problem is having a
Unbounded SolutionUnbounded Solution..

25. ExampleExample
Max Max zz = 2 = 2xx
11 + 6 + 6xx
22
s.t. 4s.t. 4xx
11 + 3 + 3xx
22 << 12 12
22xx
11 + + xx
22 >> 8 8
xx
11, , xx
22 >> 0 0

26. ExampleExample
Max Max zz = 2 = 2xx
11 + 6 + 6xx
22
s.t. 4s.t. 4xx
11 + 3 + 3xx
22 << 12 12
22xx
11 + + xx
22 >> 8 8
xx
11, , xx
22 >> 0 0
xx
22
xx
11
44xx
11 + 3 + 3xx
22 << 12 12
22xx
11 + + xx
22 >> 8 8
3344
44
88
In this example, common In this example, common
feasible region does not exist feasible region does not exist
and hence the problem is not and hence the problem is not
having a optimal solution. having a optimal solution.
This is the case of infeasible This is the case of infeasible
solution.solution.