Graphical Method

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  • 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.