Limitation of Graphical Method

Graphical solution is limited to linear programming models containing only two decision variables.

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.

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.

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

xx22

xx11

Example Max Max zz = 5 = 5xx11 + 7 + 7xx22

s.t. s.t. xx11 << 6 6

22xx11 + 3 + 3xx22 << 19 19

xx11 + + xx22 << 8 8

xx11 , , xx22 >> 0 0

Every point is in this nonnegative quadrant

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

xx22

xx11

xx11 << 6 6

(6, 0)(6, 0)

Example (Cont…)Max Max zz = 5 = 5xx11 + 7 + 7xx22

s.t. s.t. xx11 << 6 6

22xx11 + 3 + 3xx22 << 19 19

xx11 + + xx22 << 8 8

xx11 , , xx22 >> 0 0

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

22xx11 + 3 + 3xx22 << 19 19

xx22

xx11

(0, 6.33)(0, 6.33)

(9.5 , 0)(9.5 , 0)

Max Max zz = 5 = 5xx11 + 7 + 7xx22

s.t. s.t. xx11 << 6 6

22xx11 + 3 + 3xx22 << 19 19

xx11 + + xx22 << 8 8

xx11 , , xx22 >> 0 0

Example (Cont…)

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

xx22

xx11

xx11 + + xx22 << 8 8

(0, 8)(0, 8)

(8, 0)(8, 0)

Max Max zz = 5 = 5xx11 + 7 + 7xx22

s.t. s.t. xx11 << 6 6

22xx11 + 3 + 3xx22 << 19 19

xx11 + + xx22 << 8 8

xx11 , , xx22 >> 0 0

Example (Cont…)

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

22xx11 + 3 + 3xx22 << 19 19

xx22

xx11

xx11 + + xx22 << 8 8

xx11 << 6 6

Max Max zz = 5 = 5xx11 + 7 + 7xx22

s.t. s.t. xx11 << 6 6

22xx11 + 3 + 3xx22 << 19 19

xx11 + + xx22 << 8 8

xx11 , , xx22 >> 0 0

Example (Cont…)

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

xx22

xx11

Example (Cont…)

A (0,0) (6,0)B

(6,2)C

(5,3)D

(0,6.33)E

Objective Function : Max Z= 5x1+7x2

Corner Points Value of ZA – (0,0) 0B – (6,0) 30C – (6,2) 44D – (5,3) 46E – (0,6.33) 44.33

Optimal Point : (5,3)Optimal Value : 46

Example (Cont…)

Example

Max Z=3 P1 + 5 P2s.t. P1 < 4 P2 < 6 3 P1 + 2 P2 < 18 P1, P2 > 0

Example

P1

P2

0

Every point is in this nonnegative quadrant

Max Z=3 P1 + 5 P2s.t. P1 < 4 P2 < 6 3 P1 + 2 P2 < 18 P1, P2 > 0

Example (Cont…)

P1

P2

(0,0)

Max Z= 3 P1 + 5 P2s.t. P1 < 4 P2 < 6 3 P1 + 2 P2 < 18 P1, P2 > 0

(4,0)

Example (Cont…)

P1

P2

(0,0)

Max Z= 3 P1 + 5 P2s.t. P1 < 4 P2 < 6 3 P1 + 2 P2 < 18 P1, P2 > 0

(4,0)

(0,6)

Example (Cont…)

P1

P2

(0,0)

Max Z= 3 P1 + 5 P2s.t. P1 < 4 P2 < 6 3 P1 + 2 P2 < 18 P1, P2 > 0

(4,0)

(0,6) (2,6)

(4,3)

(6,0)

Example (Cont…)

P1

P2

(0,0)

(4,0)

(0,6) (2,6)

(4,3)

(6,0)A B

C

DE

Example (Cont…)Objective Function : Max Z= 3 P1 + 5 P2

Corner Points Value of ZA – (0,0) 0B – (4,0) 12C – (4,3) 27D – (2,6) 36E – (0,6) 30

Optimal Point : (2,6)Optimal Value : 36

Example

Min Min z z = 5 = 5xx11 + 2 + 2xx22

s.t. 2s.t. 2xx11 + 5 + 5xx22 >> 10 10

44xx11 - - xx22 >> 12 12

xx11 + + xx22 >> 4 4

xx11, , xx22 >> 0 0

55

44

33

22

11

55

44

33

22

11

1 2 3 4 5 61 2 3 4 5 6 1 2 3 4 5 61 2 3 4 5 6

xx22xx22

44xx11 - - xx22 >> 12 1244xx11 - - xx22 >> 12 12

22xx11 + 5 + 5xx22 >> 10 1022xx11 + 5 + 5xx22 >> 10 10

xx11xx11

ExampleExample Min Min z z = 5 = 5xx11 + 2 + 2xx22

s.t. 2s.t. 2xx11 + 5 + 5xx22 >> 10 10

44xx11 - - xx22 >> 12 12

xx11 + + xx22 >> 4 4

xx11, , xx22 >> 0 0

xx1 + 1 + xx2 2 >> 4 4

55

44

33

22

11

55

44

33

22

11

1 2 3 4 5 61 2 3 4 5 6 1 2 3 4 5 61 2 3 4 5 6

xx22xx22

xx11xx11

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 = 5xx11 + 2 + 2xx22

s.t. 2s.t. 2xx11 + 5 + 5xx22 >> 10 10

44xx11 - - xx22 >> 12 12

xx11 + + xx22 >> 4 4

xx11, , xx22 >> 0 0

A (35/11 , 8/11)A (35/11 , 8/11)

B (5,0)B (5,0)

Example (Cont…)

Objective Function : Max Z= 5x1+2x2

Corner Points Value of ZA – (35/11, 8/11) 191/11 (17.364)B – (5,0) 25

Optimal Point : (35/11,8/11)Optimal Value : 191/11=17.364

Example

Max Max z z = 3 = 3xx11 + 4 + 4xx22

s.t. s.t. xx11 + + xx22 >> 5 5

33xx11 + + xx22 >> 8 8

xx11, , xx22 >> 0 0

ExampleExample

xx11

33xx11 + + xx22 >> 8 8

xx11 + + xx22 >> 5 5

55

55

88

2.672.67

Max Max z z = 3 = 3xx11 + 4 + 4xx22

s.t. s.t. xx11 + + xx22 >> 5 5

33xx11 + + xx22 >> 8 8

xx11, , xx22 >> 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..

ExampleExample

Max Max zz = 2 = 2xx11 + 6 + 6xx22

s.t. 4s.t. 4xx11 + 3 + 3xx22 << 12 12

22xx11 + + xx22 >> 8 8

xx11, , xx22 >> 0 0

ExampleExampleMax Max zz = 2 = 2xx11 + 6 + 6xx22

s.t. 4s.t. 4xx11 + 3 + 3xx22 << 12 12

22xx11 + + xx22 >> 8 8

xx11, , xx22 >> 0 0xx22

xx11

44xx11 + 3 + 3xx22 << 12 12

22xx11 + + xx22 >> 8 8

33 44

44

88In 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.