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5.3 Optimality Test

5.3 Optimality Test. We continue with the process until there are no non-basic variables with negative reduced costs, namely until the Greedy Rule fails. Hence:

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5.3 Optimality Test

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  1. 5.3 Optimality Test • We continue with the process until there are no non-basic variables with negative reduced costs, namely until the Greedy Rule fails. Hence: An optimal solution has been found when all the nonbasic variables have nonnegative coefficients in the z-row of the simplex tableau. More formally:

  2. 5.3.1 Theorem If the value of the objective function at an extreme point of the feasible region is equal to or better than the values that it attains at all the adjacent extreme points in the feasible region, then the said extreme points is an optimal solution.

  3. Conclusion: If all the reduced costs are nonnegative, the current solution is optimal.

  4. Stopping Rule Stop if all the reduced costs are nonnegative

  5. Example (Continued) • Optimality Test: • There is at least one negative reduced cost, hence we have to continue with the algorithm.

  6. Iteration: • We have three steps: • “Variable in”: Greedy Rule • “Variable Out”: Ratio Test • “Transformation”: Pivot Operation. • Variable in: • There is only one variable with negative reduced costs, namely x2. So we set j=2 and put x2 in the basis.

  7. Variable out: • We conduct the ratio test using the right-hand side column (RHS) and the column of the new basic variable (x2).

  8. x2 1 1 0 Ratio 10 15 -- RHS 10 15 15 • Thus, the minimum ratio is attained at the first row. We therefore take out of the basis the basic variable associated with row 1, namely x3. We set i=1. • Transformation: • We have to conduct the pivot operation on (i=1, j=2).

  9. Old Tableau New Tableau

  10. The current basic feasible solution is x=(15,10,0,5,0). The value of the objective function at this point is equal to z=90. • Optimality Test: There is at least one nonbasic variable with negative reduced cost so we continue.

  11. Iteration: • Variable in: • The variable with the most negative reduced cost is x5, so we place x5 in the basis and set j=5. • Variable out: • We conduct the ratio test on the column of x5.

  12. x5 RHS Ratio -2 1 1 10 5 15 -- 5 15 • The minimum ratio is attained at i=2, thus we take out of the basis the basic variable associated with row 2, namely x4. • Transformation: • We conduct the pivot operation on (i=2,j=5).

  13. Old Tableau • The new basic variable is x=(10,20,0,0,5) and the value of the objective function at x is equal to z=100. New Tableau

  14. Optimality Test: • All the reduced costs are nonnegative, hence we stop. The current solution is optimal. • An Important Note: (NILN) When you stop, you have to do two things: • 1. Write down the optimal solution and optimal value of the objective function, namely x* and z*.

  15. 2. Check these values. • Constraints are satisfied? • the value of z* is consistent with the value of x* ?

  16. Example (Continued) • 1. From the tableau x*=(10,20,0,0,5) and z*=100. • 2. z* =? c1x1 +c2x2 4x10 + 3x20 = 40 + 60 = 100 =z* (OK). Constraints: 2x1* + x2* <=? 40 (2x10 + 1x20 = 40, OK) x1* + x2* <=? 30 (1x10 + 1x20 = 30, OK)

  17. Warning (NILN) You are expected to check the results you obtain

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