homework-10-solutions - ) ⊤ d . But then if the new dual,...

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MthSc810 – Mathematical Programming Fall 2011, solutions to homework #10. Note: the solutions of Exercises 1 and 2 are the same (for Exercise 1, it is actually a subset) as the solutions of the problems in midterm II. Exercise 3. Solve Problem 4.11 from the textbook. Solution. The problem in standard form and its dual (with dual variables u and v ) are: min c x s.t. A x = b d x = f x 0 . max u b + vf s.t. u A + v d c . If the primal is infeasible, then the dual is either unbounded or infeasible. By relaxing d x = f , we obtain the following primal-dual pair: min c x s.t. A x = b x 0 . max u b s.t. u A c . The dual, like the primal, admits an optimal solution x , u with ±nite cost c x = ( u
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Unformatted text preview: ) ⊤ d . But then if the new dual, obtained from the original one by eliminating a variable, is feasible, the original one cannot be infeasible: the feasi-ble set F ′ of the new dual is the intersection (not the projection!) of the feasible set F of the original dual with { ( u ∈ R m , v ∈ R : v = 0 } . To be more precise, after this intersection the variable v is projected out since F ′ is m-dimensional, while F has m +1 dimensions. Since it is feasible the original dual must be feasi-ble too. Hence the original dual cannot be infeasible, and it must be unbounded....
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This note was uploaded on 03/14/2012 for the course MTHSC 810 taught by Professor Staff during the Fall '08 term at Clemson.

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