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Lecture on “Linear Programming”

Lecture on “Linear Programming”. Difficulties in Simplex Procedure. 1. Tie in selecting key column/key row. 2. Inequality of greater than or equal to kind. 3. Equality constraints. 4. Negative RHS. 5. Unrestricted variables. 6. Restricted variables.

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Lecture on “Linear Programming”

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  1. Lecture • on • “Linear Programming”

  2. Difficulties in Simplex Procedure 1. Tie in selecting key column/key row. 2. Inequality of greater than or equal to kind. 3. Equality constraints. 4. Negative RHS. 5. Unrestricted variables. 6. Restricted variables. 7. Minimization Problem.

  3. (1) Tie in selecting key column/Key row (A) Tie in selecting key column • Decision variable and slack or surplus variable Select Decision variable column • Decision variable and Decision variable Select any one column • Slack/Surplus variable and Slack/Surplus variable Select any one column

  4. (B) Tie in selecting key row • aik are elements in key column. • Compare the ratio of aij to aik for tied rows first in identity from L to R and then in body. • Key row Min. Positive ratio.

  5. Simplex Method -A Case for Tie in selecting key row Standard Form: Max Z = 2x1+1x2+0w1+0w2+0w3 4x1+3x2+w1+0w2+0w3 = 12 4x1+x2+0w1+w2+0w3 = 8 4x1-x2+0w1+0w2+w3 = 8

  6. Tableau-I cj 2 1 0 0 0 ci xi 0 w1 0 w2 0 w3 bi x1 x2 w1 w2 w3 Ratio 3 2 2 12 4 3 1 0 0 8 4 1 0 1 0 8 4 -1 0 0 1 Tie IjZ = 0 0 0 0 -2 -1 • Ij = (Zj-cj) = (aij.ci) - cj

  7. Tableau-I cj 2 1 0 0 0 ci xi 0 w1 0 w2 0 w3 bi x1 x2 w1 w2 w3 Ratio 3 2 2 12 4 3 1 0 0 8 4 1 0 1 0 8 4 -1 0 0 1 Tie IjZ = 0 0 0 0 -2 -1

  8. Tableau-I : Rewritten in required form cj 0 0 0 2 1 ci xi 0 w1 0 w2 0 w3 bi w1 w2 w3 x1 x2 Ratio 3 2 2 12 1 0 0 4 3 8 0 1 0 4 1 8 0 0 1 4 -1 Tie IjZ = 0 0 0 0 -2 -1 Ratio : 0/4 - R2 0/4 - R3 1/4 = 0.25 0/4 = 0 Key Row

  9. (2)Constraints with inequality “greater than or equalto” kind

  10. (3) Equality Constraints. OR

  11. (4)Negative R.H.S. (5)Unrestricted Variables. If xis unrestricted then x is changed to (x’ - x”), where x’ & x” are positive.

  12. (6)Restricted Variables.

  13. (7) Minimization Problem

  14. Methods to Solve Minimization Problem • A. Maximization Method • B. Minimization Method • C. Two Phase Method • D. Dual Simplex Method

  15. (A) Maximization Method : Convert Minimization Problem into Maximization Problem & solve

  16. (B) Minimization Method: Same methodology as Maximization except criterion of selecting ‘key column’ & Coeff. of ‘A’ changed to : Key column  Max. + ve Index Number. Co-eff. of ‘A’ in Obj. eq. = +M

  17. (C) Two Phase Method: In Phase- I : For a Maximization Problem Consider Cj = -1 for only A, else Cj = 0 and Get Final Usual Tableau In Phase- II : Consider All Cj = Original values in Final Tableau of Phase I and Get Optimal Solution

  18. (D) Dual Simplex Method This method is applicable to any Standard Maximizaion Problem with - Ve RHS First “Key Row” is selected w.r.t. Max. – Ve RHS Key Column is selected w.r.t. Max. (I/ aij) for -ve aij only

  19. (A)Maximization Method Convert this into maximization problem.

  20. Convert this into Standard form.

  21. Standard Form : Max (-Z) = -3x1-5x2+0s1+(- M)A1+0w1+0w2 3x1+2x2-s1+A1+0w1+0w2 = 18 1x1+0x2-0s1+0A1+1w1+0w2 = 4 0x1+1x2-0s1+0A1+0w1+1w2 = 6

  22. Tableau - I cj -3 -5 0 -M 0 0 ci xibi x1 x2 s1 A1 w1 w2 -M A1 0 w1 0 w2 18 3 2 -1 1 0 0 4 1 0 0 0 1 0 6 0 1 0 0 0 1 0 0 0 -3M -2M M +3 +5 IjZ= -18M • Ij = (Zj-cj) = (aij.ci) - cj

  23. Tableau - II cj -3 -5 0 -M 0 0 ci xibi x1 x2 s1 A1 w1 w2 2 -1 1 0 0 -3 -M A1 -3 x1 0 w2 6 4 6 0 1 0 0 0 1 1 0 0 0 0 1 0 IjZ= -6M -12 -2M M 3M +5 -3 0 0

  24. Tableau - III cj -3 -5 0 -M 0 0 ci xibi x1 x2 s1 A1 w1 w2 -5 x2 -3 x1 0 w2 3 4 3 0 IjZ= -27 0 5/2 M -5/2 9/2 0 Hence, Optimal Solution is x1=4, x2=3 giving Zmax = -27 , Zmin=27.

  25. (B)Minimization Method Convert this into Standard form.

  26. Standard Form: Min Z = 3x1+5x2+0s1+MA1+0w1+0w2 3x1+2x2-s1+A1+0w1+0w2 = 18 1x1+0x2-0s1+0A1+1w1+0w2 = 4 0x1+1x2-0s1+0A1+0w1+1w2 = 6

  27. Tableau - I cj 3 5 0 M 0 0 ci xibi x1 x2 s1 A1 w1 w2 Ratio 6 4  M A1 0 w1 0 w2 18 4 6 3 2 -1 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 3M 2M -M -3 -5 IjZ= 18M • Key Column Max Ij

  28. Tableau - II cj 3 5 0 M 0 0 ci xibi x1 x2 s1 A1 w1 w2 Ratio 3  6 M A1 6 0 2 -1 1 -0 0 3 x1 4 1 0 0 0 1 0 0 w2 6 0 1 0 0 0 1 IjZ= 6M 0 2M -M 0 -3M 0 +12 -5 +3

  29. Tableau - III cj 3 5 0 M 0 0 ci xibi x1 x2 s1 A1 w1 w2 5 x2 3 x1 0 w2 3 4 3 IjZ = 27 0 0 -1 -M -3M 0 +1 +3 Hence, Optimal Solution is x1=4, x2=3 giving Z = 27.

  30. (C) Two Phase Method Convert this into maximization problem.

  31. Convert this into Standard form.

  32. Standard Form: (Phase- I) cj = -1 for only A, else cj = 0 Max (-Z) = 0x1+0x2+0s1-1A1+0w1+0w2 3x1+2x2-s1+A1+0w1+0w2 = 18 1x1+0x2 -0s1+0A1 +w1+0w2 = 4 0x1+1x2- 0s1+0A1+0w1+w2 = 6

  33. Phase I : Tableau - I cj 0 0 0 -1 0 0 ci xibi x1 x2 s1 A1 w1 w2 -1 A1 0 w1 0 w2 18 3 2 -1 1 0 0 4 1 0 0 0 1 0 6 0 1 0 0 0 1 0 0 0 -3 -2 1 IjZ= -18 • Ij = (Zj-cj) = (aij.ci)-cj

  34. Phase I : Tableau - II cj 0 0 0 -1 0 0 ci xi bi x1 x2 s1 A w1 w2 2 -1 1 0 0 -3 0 0 1 -1 A1 0 x1 0 w2 6 4 6 0 1 0 0 1 0 0 0 1 IjZ= -6 -2 1 3 0 0 0

  35. Phase I : Tableau - III cj 0 0 0 -1 0 0 ci xibi x1 x2 s1 A1 w1 w2 0 0 1 0 x2 0 x1 0 w2 3 4 3 0 1 0 1 0 0 -1/2 1/2 -3/2 0 0 1 3/2 1/2 -1/2 IjZ= 0 0 0 0 1 0 0

  36. Phase II : Tableau - I cj -3 -5 0 -M 0 0 ci xibi x1 x2 s1 A1 w1 w2 0 0 1 -5 x2 -3 x1 0 w2 3 4 3 0 1 0 1 0 0 -1/2 1/2 -3/2 0 0 1 3/2 1/2 -1/2 -5/2 +M IjZ= -27 0 0 5/2 9/2 0 Hence, Optimal Solution is x1=4, x2=3 giving Zmax = -27 , Zmin=27

  37. (D) Dual Simplex Method Convert this into maximization problem.

  38. Convert this into Standard format.

  39. Convert this into Standard form.

  40. Standard Form: Max (-Z) = -3x1-5x2+0w1+0w2+0w3 -3x1-2x2+w1+0w2+0w3 = -18 x1+0x2+0w1+w2+0w3= 4 0x1+1x2+0w1+0w2+w3 = 6

  41. To prepare initial Tableau: Tableau - I cj -3 -5 0 0 0 ci xi 0 w1 0 w2 0 w3 bi x1 x2 w1 w2 w3 -18 -3 -2 1 0 0 4 1 0 0 1 0 6 0 1 0 0 1 3 5 IjZ = 0 0 0 0 • Key Column Max Ratio Ratio: 3/-3 5/-2 =-1 =-2.5

  42. Tableau - II cj -3 -5 0 0 0 ci xi -3 x1 0 w2 5 w3 bi x1 x2 w1 w2 w3 6 0 1 0 0 0 1 2/3 -1/3 1 0 0 -2 1/3 -2/3 6 1 0 3 1 0 0 0 IjZ =-18 Ratio: 3/(-2/3) =-9/2

  43. Tableau - III cj -3 -5 0 0 0 ci xi -3 x1 -5 x2 0 w3 bi x1 x2 w1 w2 w3 4 3 3 0 0 0 3/2 1/2 IjZ = -27 Hence, Optimal Solution is x1=4, x2=3 giving Zmax= -27 and Zmin = 27.

  44. Cases for • ‘Alternative Optimal Solution’ • ‘Unbounded Solution’, • ‘Infeasible Solution’ and • ‘Unrestricted Variable’ through Simplex.

  45. Case for Alternative Optimal Solution Standard Form: Max Z = 6x1+4x2+0w1+0w2+0s1+(-M)A1 2x1+3x2+w1+0w2+0s1+0A1 = 30 3x1+2x2+0w1+w2+0s1+0A1 = 24 x1+x2+0w1+0w2- s1+A1 = 3

  46. cj 6 4 0 0 0 -M ci xibi x1 x2 w1 w2 s1 A1 Ratio 8.4 - 12 0 w1 14 0 5/3 1 -2/3 0 0 0 s1 5 0 -1/3 0 1/3 1 -1 6 x1 8 1 2/3 0 1/3 0 0 IjZ= 48 0 0 0 2 0 M Optimal Solutionis x1=8, x2=0giving Z = 48. Final Tableau

  47. cj 6 4 0 0 0 -M ci xibi x1 x2 w1 w2 s1 A1 4 x2 42/5 0 s1 39/5 6 x1 12/5 IjZ= 48 0 0 0 2 0 M AlternativeOptimal Solutionis x1=12/5 , x2= 42/5giving Z =48.

  48. Case for Unbounded Solution Standard Form: Max Z = 3x1+5x2+0w1+0w2+0s1+(-M)A1 x1-2x2+w1+0w2+0s1+0A1 = 6 x1+0x2+0w1+w2+0s1+0A1 = 10 0x1+x2+0w1+0w2- s1+A1 = 1

  49. cj 3 5 0 0 0 -M ci xibi x1 x2 w1 w2 s1 A1 Ratio -  1 0 w1 6 1 -2 1 0 0 0 0 w2 10 1 0 0 1 0 0 -M A1 1 0 1 0 0 -1 1 IjZ= -M -3 -M 0 0 M 0 -5 Tableau-I

  50. cj 6 4 0 0 0 -M ci xi 0 w1 0 w2 5 x2 bi x1 x2 w1 w2 s1 A1 8 10 1 1 0 1 0 -2 -2 1 0 0 1 0 0 0 1 0 0 -1 1 IjZ= 5 -3 0 0 0 -5 2M +5 All aij<= 0 w.r.t. key column. Hence, Solution is unbounded.

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