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Two-Phase Method for Solving Infeasible Linear Programming Problems, Exams of Mathematics

The two-phase method, an approach to solve infeasible linear programming problems using the simplex algorithm. The method introduces an auxiliary problem to find a feasible initial point for the original problem. The steps to write the auxiliary problem, create tableaus, and apply the simplex algorithm.

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

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Download Two-Phase Method for Solving Infeasible Linear Programming Problems and more Exams Mathematics in PDF only on Docsity! 8 The two-phase method. 8.1 The two-phase method. The simplex algorithm assumes that the initial point is feasible in the primal problem. If b is greater than or equal to zero, then the origin is feasible. If the origin is not feasible, then it is necessary to determine some other initial point that is feasible. It is possible to introduce an auxiliary primal problem specifically designed to help in this task. The first phase of the two- phase method applies the simplex algorithm to the auxiliary problem. The second phase uses the feasible point generated in the first phase as initial point in the original problem. There is typically a need for elementary row operations to bring the tableau into the form required by the simplex algorithm. 8.2 The auxiliary problem. Consider the primal problem 1 2 1 2 1 2 1 2 2 2 max 9 8x when 3 3 , 0 x x x x x x x − − ≤ −− − − − ≤ − ≥ . The feasible set does not include the origin. The augmented system has the two equations 1 2 1 1 2 2 2 2 3 3 x x y x x y − − + = − − − + = − . Write the system in the form 1 2 1 1 1 2 2 2 2 2 3 3 x x y z x x y z + − + = + − + = , using the auxiliary non-negative variables 1 2,z z . The auxiliary problem seeks to minimize 1 2z z+ without violating the previous system. 8.3 Tableaux. If the minimizer is given by 1 2 0z z= = , then the previous system yields a solution that is feasible in the original problem. The tableau associated with the problem to maximize 1 2z z− − is given by 1 2 1 0 1 0 2 3 1 0 1 0 1 3 0 0 0 0 1 1 0 − − − − . Observe how there are no basic variables. This is readily fixed by adding the first and the second row to the last row. It is convenient to extend the tableau with one more row for this particular
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