The document describes the graphical method for solving linear programming problems with two decision variables. It provides the step-by-step procedure which involves plotting the constraints on a graph to identify the feasible region, determining the corner points of this region which represent the feasible solutions, substituting these points into the objective function to find the optimal value, and identifying the optimal solution. It then provides examples demonstrating this process and different types of solutions that can arise such as unbounded, infeasible, and optimal.

1. 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. 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x 2 x 1 Example Max z = 5 x 1 + 7 x 2 s.t. x 1 < 6 2 x 1 + 3 x 2 < 19 x 1 + x 2 < 8 x 1 , x 2 > 0 Every point is in this nonnegative quadrant

6. 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x 2 x 1 x 1 < 6 (6, 0) Example (Cont…) Max z = 5 x 1 + 7 x 2 s.t. x 1 < 6 2 x 1 + 3 x 2 < 19 x 1 + x 2 < 8 x 1 , x 2 > 0

7. 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 2 x 1 + 3 x 2 < 19 x 2 x 1 (0, 6.33) (9.5 , 0) Max z = 5 x 1 + 7 x 2 s.t. x 1 < 6 2 x 1 + 3 x 2 < 19 x 1 + x 2 < 8 x 1 , x 2 > 0 Example (Cont…)

8. 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x 2 x 1 x 1 + x 2 < 8 (0, 8) (8, 0) Max z = 5 x 1 + 7 x 2 s.t. x 1 < 6 2 x 1 + 3 x 2 < 19 x 1 + x 2 < 8 x 1 , x 2 > 0 Example (Cont…)

9. 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 2 x 1 + 3 x 2 < 19 x 2 x 1 x 1 + x 2 < 8 x 1 < 6 Max z = 5 x 1 + 7 x 2 s.t. x 1 < 6 2 x 1 + 3 x 2 < 19 x 1 + x 2 < 8 x 1 , x 2 > 0 Example (Cont…)

10. 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x 2 x 1 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 D E

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 z = 5 x 1 + 2 x 2 s.t. 2 x 1 + 5 x 2 > 10 4 x 1 - x 2 > 12 x 1 + x 2 > 4 x 1 , x 2 > 0

20. 5 4 3 2 1 1 2 3 4 5 6 x 2 4 x 1 - x 2 > 12 2 x 1 + 5 x 2 > 10 x 1 Example Min z = 5 x 1 + 2 x 2 s.t. 2 x 1 + 5 x 2 > 10 4 x 1 - x 2 > 12 x 1 + x 2 > 4 x 1 , x 2 > 0 x 1 + x 2 > 4

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

24. Example x 1 3 x 1 + x 2 > 8 x 1 + x 2 > 5 5 5 8 2.67 Max z = 3 x 1 + 4 x 2 s.t. x 1 + x 2 > 5 3 x 1 + x 2 > 8 x 1 , x 2 > 0 Feasible Region This feasible region is unbounded, hence z can be increased infinitely. So this problem is having a Unbounded Solution .

25. Example Max z = 2 x 1 + 6 x 2 s.t. 4 x 1 + 3 x 2 < 12 2 x 1 + x 2 > 8 x 1 , x 2 > 0

26. Example Max z = 2 x 1 + 6 x 2 s.t. 4 x 1 + 3 x 2 < 12 2 x 1 + x 2 > 8 x 1 , x 2 > 0 x 2 x 1 4 x 1 + 3 x 2 < 12 2 x 1 + x 2 > 8 3 4 4 8 In this example, common feasible region does not exist and hence the problem is not having a optimal solution. This is the case of infeasible solution.