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bs1tuaz

bs1tuaz

Answered question

2022-05-21

Suppose that the polytope P = { x A x b } is totally dual integral (TDI), and that the inequality a ~ T x b ~ is satisfied at every point of P.
Does this imply that the system { A x b ,     a ~ T x b ~ } is TDI?

Answer & Explanation

zwichtsu

zwichtsu

Beginner2022-05-22Added 9 answers

Total dual integrality is a property of a system and not of a polyhedron. For every rational polyhedron P there exists a TDI system A x b such that P = { x R n   :   A x b }. You need b to be integral in order to be able to deduce that P is integral.
Adding a redundant inequality to a TDI system results to a TDI system. Consider the primal problem with an integral c
max .   c x s . t .   A x b
Since the Primal system before adding the redundant inequality is TDI there exists an integral optimal solution u for the corresponding Dual Problem.
Now consider the dual problem
max . u b + v b ¯ s . t . u A = c ,   v a ¯ = c , u 0 ,   v 0
To prove that the Primal system is TDI we have to construct an optimal integer solution u∗ for the Dual problem. Notice that adding a redudant inequality does not change the optimal value of the Primal problem. Then the solution u = ( u , 0 ) is a feasible optimal integral solution for the Dual.
Nevertheless adding redundant inequalities to systems which are not TDI may result to a TDI system. The following system is clearly non TDI.
( 1 1 1 1 ) ( x 1 x 2 ) ( 0 0 )
Still we can produce a TDI system for the same polyhedron by adding a redudant inequality to obtain the TDI system
( 1 1 1 1 1 0 ) ( x 1 x 2 ) ( 0 0 0 )

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