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Unformatted text preview: CO350—Linear Optimization Assignment 6 Solutions 1. (a) Show that if there is a tie for the choice of leaving variable in an iteration of the simplex method, then the new basis is degenerate. Deduce that there is never a tie for the choice of leaving variable in the perturbation method. Solution: Let x k be the entering variable and x r be the leaving variable for such an iteration of the simplex method. Let x j be a variable (other than x r ) which is tied for the choice of leaving variable. Then, t = min i b i a ik : a ik > = b r a rk = b j a jk . For the new basic solution, we have x j = b j a jk t = 0, so the new basis is degenerate. Since the perturbed LP used in the perturbation method has no degenerate basis, this situation cannot happen in the perturbation method. (b) Show that if there is never a tie for the choice of leaving variable in the course of solving a problem with the simplex method, then the algorithm terminates. Solution: If there is never a tie for the choice of leaving variable, then the simplex method is the same as the perturbation method. Since the perturbation always terminates, the simplex method also terminates in this case. 2. (a) Let B and B = B ∪{ k }\{ r } be two feasible bases for the LP max { c T x : Ax = b,x ≥ } . Show that it is possible to have an iteration of the simplex method which starts with basis B and ends with basis B if and only if there is a vector d such that Ad = 0, c T d > 0, d k > 0, d r < 0, and d i = 0 for i / ∈ B ∪ { k } ....
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 Winter '07
 S.Furino,B.Guenin
 Linear Programming, Optimization, DI, Duality

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