PROBLEM SUMMARY

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. *13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. *28. *29. *30. 31. 32. 33. 34. Simplex short answer Simplex discussion short answer Simplex short answer Simplex short answer Simplex short answer Simplex short answer 4 tableaus 2 tableaus 3 tableaus 3 tableaus 2 tableaus 5 tableaus 5 tableaus 5 tableaus 6 tableaus 4 tableaus 3 tableaus 3 tableaus 3 tableaus Simplex short answer 3 tableaus 4 tableaus graphical analysis 2 tableaus 2 tableaus 6 tableaus 5 tableaus graphical analysis Mixed constraint model transformation Mixed constraint model transformation 5 tableaus 3 tableaus 3 tableaus 4 tableaus 50. 51. 52. 53. 54. *49. *47. *48. 46. 44. 45. 43. 35. 36. 37. 38. 39. 40. 41. *42. 3 tableaus, multiple optimal 4 tableaus 3 tableaus, multiple optimal 2 tableaus, infeasible 2 tableaus, unbounded 4 tableaus, pivot row and column ties, multiple optimal Infeasible problem Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Dual formation and interpretation, sensitivity analysis Sensitivity analysis, cj and qi Sensitivity analysis with duality Sensitivity analysis with duality Sensitivity analysis with duality

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PROBLEM SOLUTIONS

1. a) x1 = 10, x2 = 40, s3 = 30, z = 420 b) Yes; all cj – zj row values are zero or negative. c) x3 = 0; s2 = 0 d) Maximize Z = 10x1 + 2x2 + 6x3 e) 3Á f) Since there are three decision variables, a three-dimensional graph is required. a) minimization; because zj – cj is being calculated on the bottom row and not cj – zj b) x1 = 20, x3 = 10, s1 = 10, z = 240 c) Yes; all zj – cj values are negative or zero. d) Minimize Z = 6x1 + 20x2 + 12x3 e) 3 constraints f) Yes, one; because there are 3 constraints but only 2 surplus variables. Since both ≤ and ≥ constraints have slack or surplus variables, and an equality does not, then one of the three constraints was an =. g) x2 = 0 a) maximization; because of “cj-zj” b) x1 = 12, x2 = 0, x3 = 0, x4 = 15 c) s1 = 20, s2 = 0, s3 = 0, s4 = 45 d) If x2 is selected as the entering basic variable, Z will increase by 20 for every unit of x2 entered into the solution. e) This solution is not optimal because there are positive values in the cj - zj row.

2.

3.

254

cj 0 50 60 45

Basic Variables s1 x4 x1 x3 zj cjzj

Quantity 20 15 12 45/8 13,785/8

60 x1 0 0 1 0 60 0

50 x2 1 0 1/2 0 30 20

45 x3 0 0 0 1 45 0

50 x4 0 1 0 6 50 0

0 s1 1 0 0 0 0 0

0 s2 0 1 0 -3/4 65/4 -65/4

0 s3 0 0 1/10 0 6 -6

0 s4 0 0 0 1/8 45/8 -45/8

cj 50 50 60 45

Basic Variables x2 x4 x1 x3 zj cj-zj

Quantity 20 15 2 45/8 2,123.125

60 x1 0 0 1 0 60 0

50 x2 1 0 0 0 30 0

45 x3 0 0 0 1 45 0

50 x4 0 1 0 6 50 0

0 s1 1 0 -1/2 0 20 20

0 s2 0 1 0 -3/4 65/4 -65/4

0 s3 0 0 1/10 0 6 -6

0 s4 0 0 0 1/8 45/8 -45/8

Optimal 4. a) minimization; because “zj – cj ” and a positive “M ” in the objective function. b) x3 = 0 c) “M/6 - 5/3” has no real meaning since it includes “M ”; it simply represents a large net decrease in cost if s2 is entered into the solution. d) two; since there are two artificial variables remaining in the tableau, it will take at least two more tableaus to eliminate them, and, they must be eliminated to insure a feasible solution. e) no; because there are positive values in the zj – cj row.

cj 4 10 M

Basic Variables x3 x2 A3 zj zj-cj

Quantity 30 10 70 7M+220

8 x1 2/3 1/3 -2/3 -2M/3 + 18/3 -2M/3-2

10 x2 0 1 0 10 0

4 x3 0 0 1 4 0

0 s1 -1 0 1 M-4 M-4

0 s2 1/6 -1/6 -1/6 -M/6-1 -M/6-1

0 s2 0 0 -1 -M -M

M A3 0...