1. GRAPHICAL METHOD

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

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.