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5.4 Statement of Algorithm

5.4 Statement of Algorithm. We shall discuss the details of the Simplex Algorithm from two related points of view, namely Process Reasoning Consult Table 5.2 for the details. 1. Process: Add Slack Variables Reasoning: ??????? 2. Process: Optimality Test

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5.4 Statement of Algorithm

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  1. 5.4 Statement of Algorithm • We shall discuss the details of the Simplex Algorithm from two related points of view, namely • Process • Reasoning • Consult Table 5.2 for the details.

  2. 1. Process: Add Slack Variables • Reasoning: ??????? • 2. Process: Optimality Test • Reasoning: ??????? • 3. Process: New Basic Variable • Reasoning: ??????? • 4. Process: New NonBasic Variable • Reasoning: ???????

  3. 5. Process: New Basic Feasible Solution • Reasoning: ?????????

  4. Construct a feasible extreme point (NILN) Add slack variables Sign of reduced costs Is this point optimal ? Yes Stop Greedy Rule and Ratio Test Iterate No Pivot Move along an edge to a better extreme point

  5. 5.4.1 Example

  6. Step 1: We add slack variables and construct the canonical form. This yields the first basic feasible solution.

  7. We rewrite this formulation in a table form - so called Simplex Tableau.

  8. Make sure that you fully understand the “logic” of the layout.

  9. Step 2: • There are nonnegative reduced costs, so we continue. • Step 3: • We select the nonbasic variable with the most negative reduced cost (x1) as the new basic variable.

  10. Step 4: • We conduct the ratio test on the column of the new basic variable. Row 3 yields the minimum ratio so we take out of the basic the basic variable associated with row 3, namely x6. • Step 5: • We perform the pivot operation on (i=3,j=1).

  11. Old tableau new Tableau

  12. Step 2: • There are negative reduced costs, so we continue. • Step 3: • We select the nonbasic variable with the most negative reduced cost (x3) as the new basic variable.

  13. Step 4: • We conduct the ratio test on the column of the new basic variable. The minimum ratio is attained at row 2, thus we take out of the basis the basic variable associated with row 2, namely x5. • Step 5: • We conduct the pivot operation on (i=2,j=3).

  14. Old Tableau New Tableau

  15. Step 2: • All the reduced costs are nonnegative. Thus, we Stop! The current solution is optimal. • Report: The optimal solution is: • x*=(1,0,12,28,0,0,5) • and the optimal value of the objective function is equal to z*=300. • Don’t forget ot check the results!!!!!!!

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