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# Graphical Method

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### Graphical Method

1. 1. GRAPHICAL METHOD
2. 2. Limitation of Graphical Method <ul><li>Graphical solution is limited to linear programming models containing only two decision variables. </li></ul>
3. 3. Procedure <ul><li>Step I: Convert each inequality as equation </li></ul><ul><li>Step II: Plot each equation on the graph </li></ul><ul><li>Step III: Shade the ‘ Feasible Region ’. Highlight the common Feasible region. </li></ul><ul><ul><li>Feasible Region : Set of all possible solutions. </li></ul></ul><ul><li>Step IV: Compute the coordinates of the corner points (of the feasible region). These corner points will represent the ‘ Feasible Solution ’. </li></ul><ul><ul><li>Feasible Solution : If it satisfies all the constraints and non negativity restrictions. </li></ul></ul>
4. 4. Procedure (Cont…) <ul><li>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 ’. </li></ul><ul><ul><li>Optimal Solution : If it optimizes (maximizes or minimizes) the objective function. </li></ul></ul><ul><ul><li>Unbounded Solution : If the value of the objective function can be increased or decreased indefinitely, Such solutions are called Unbounded solution. </li></ul></ul><ul><ul><li>Inconsistent Solution : It means the solution of problem does not exist. This is possible when there is no common feasible region. </li></ul></ul>
5. 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. 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. 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. 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. 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. 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. 11. <ul><li>Objective Function : Max Z= 5x 1 +7x 2 </li></ul><ul><li>Corner Points Value of Z </li></ul><ul><li>A – (0,0) 0 </li></ul><ul><li>B – (6,0) 30 </li></ul><ul><li>C – (6,2) 44 </li></ul><ul><li>D – (5,3) 46 </li></ul><ul><li>E – (0,6.33) 44.33 </li></ul><ul><li>Optimal Point : (5,3) </li></ul><ul><li>Optimal Value : 46 </li></ul>Example (Cont…)
12. 12. Example <ul><ul><li>Max Z=3 P1 + 5 P2 </li></ul></ul><ul><ul><li>s.t. P1 < 4 </li></ul></ul><ul><ul><li>P2 < 6 </li></ul></ul><ul><ul><li>3 P1 + 2 P2 < 18 </li></ul></ul><ul><ul><li>P1, P2 > 0 </li></ul></ul>
13. 13. Example P1 P2 0 Every point is in this nonnegative quadrant <ul><ul><li>Max Z=3 P1 + 5 P2 </li></ul></ul><ul><ul><li>s.t. P1 < 4 </li></ul></ul><ul><ul><li>P2 < 6 </li></ul></ul><ul><ul><li>3 P1 + 2 P2 < 18 </li></ul></ul><ul><ul><li>P1, P2 > 0 </li></ul></ul>
14. 14. Example (Cont…) P1 P2 (0,0) <ul><ul><li>Max Z= 3 P1 + 5 P2 </li></ul></ul><ul><ul><li>s.t. P1 < 4 </li></ul></ul><ul><ul><li>P2 < 6 </li></ul></ul><ul><ul><li>3 P1 + 2 P2 < 18 </li></ul></ul><ul><ul><li>P1, P2 > 0 </li></ul></ul>(4,0)
15. 15. Example (Cont…) P1 P2 (0,0) <ul><ul><li>Max Z= 3 P1 + 5 P2 </li></ul></ul><ul><ul><li>s.t. P1 < 4 </li></ul></ul><ul><ul><li>P2 < 6 </li></ul></ul><ul><ul><li>3 P1 + 2 P2 < 18 </li></ul></ul><ul><ul><li>P1, P2 > 0 </li></ul></ul>(4,0) (0,6)
16. 16. Example (Cont…) P1 P2 (0,0) <ul><ul><li>Max Z= 3 P1 + 5 P2 </li></ul></ul><ul><ul><li>s.t. P1 < 4 </li></ul></ul><ul><ul><li>P2 < 6 </li></ul></ul><ul><ul><li>3 P1 + 2 P2 < 18 </li></ul></ul><ul><ul><li>P1, P2 > 0 </li></ul></ul>(4,0) (0,6) (2,6) (4,3) ( 6,0)
17. 17. Example (Cont…) P1 P2 (0,0) (4,0) (0,6) (2,6) (4,3) ( 6,0) A B C D E
18. 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. 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. 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. 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. 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. 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. 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. 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. 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.