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7.4 THE DUAL THEOREMS

7.4 THE DUAL THEOREMS. Primal Problem. Dual Problem. b is not assumed to be non-negative. 7.4.1 Weak Duality Theorem. If x is a feasible solution to the primal problem from (7.13) and y is a feasible solution to the dual problem, then cx £ by. Proof :

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7.4 THE DUAL THEOREMS

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  1. 7.4 THE DUAL THEOREMS Primal Problem Dual Problem b is not assumed to be non-negative

  2. 7.4.1 Weak Duality Theorem • If x is a feasible solution to the primal problem from (7.13) and y is a feasible solution to the dual problem, then cx £ by. • Proof: If x is a feasible solution to the Primal Problem, then we must have: Ax £ b So, if y is a feasible solution to the Dual Problem, we must have (why?) yAx £ yb

  3. Similarly, if y is feasible for the Dual, then yA ≥ c and therefore (why?) for any feasible solution, x, of the Primal Problem we have yAx ≥ cx It follows then that cx £ by.

  4. This result implies that the optimal objective function values for the primal and dual systems are bounded above and below respectively by the other problem. • What are the implications for the optimal solutions? • We shall deal with this question shortly.

  5. 7.4.2 Lemma • If a linear programming problem has an unbounded objective, then its dual problem has no feasible solution. Observation: • It is possible that both the primal and the dual problem are infeasible.

  6. 7.4.3 Lemma • Let x* be a feasible solution to the primal problem and y* a feasible solution to the dual. Then cx* = by* implies that x* is an optimal solution for the primal and y* is optimal for the dual. Proof: • From the Weak Duality theorem we have cx £ by hence we also have cx £ by*

  7. for any feasible solution x of the primal and any optimal solution y* of the dual. Thus, for any feasible solution x* of the primal for which cx*=by* we have (why?) cx £ cx* for any feasible solution, x, of the primal. This implis that x* is an optimal solution for the primal.

  8. 7.4.4 Strong Duality Theorem • If an optimal solution exists for either the primal or its dual, then an optimal solution exists for both and the corresponding optimal objective function values are equal, namely Z* = w*. • Proof: Bottom Line: Let B-1* be the “optimal” value of B-1 (final tableau) and define: y*:=cBB-1*

  9. Then show that y* is feasible for the dual and y*b = z*. • In view of the Weak Duality Theorem, this will imply that y* is optimal for the dual. • Alternatively, ....... have a look at the tableau itself ....

  10. y = cBB-1 for any simplex tableau not necessarily the final one This is implied by our beloved recipe: r = cBB-1D - c

  11. Important Observations • IN ANY TABLEAU, THE DUAL VARIABLES CORRESPONDING TO THE PRIMAL BASIC FEASIBLE SOLUTION ARE THE REDUCED COSTS FOR THE PRIMAL SLACK VARIABLES. • IN APPLYING THE SIMPLEX METHOD TO THE PRIMAL PROBLEM, WE ALSO SOLVE ITS DUAL.

  12. 7.4.6 Example

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