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Unformatted text preview: CO350 LINEAR PROGRAMMING  SOLUTIONS TO ASSIGNMENT 5 Exercise 1. Prove that if x j is the leaving variable during one iteration of the simplex method then x j cannot be the entering variable at the next iteration. Solution: Consider the tableau max { z = c T N x N + v : x B + A N x N = b,x } for a basis B . Suppose j B and x j is the leaving variable for an iteration of the simplex method starting from the basis B . Let x k be the entering variable and B := B \{ j }{ k } be the new basis. Then the reduced cost for x j for the basis B is c j = c k a jk . As k is the entering variable, c k > . Moreover x j is the leaving variable, so a jk > . So c j = c k a jk < . Hence x j cannot be chosen as the entering variable for the next iteration. Exercise 2. Consider the linear program max { c T x : Ax = b,x } where A = " 2 1 1 0 0 1 0 1 # , b = " 6 2 # , and c = [3 , 1 , 1 , 1] T . Solve this linear program using the revised simplex method, starting with the feasible basic solution x * = (0 , 2 , 4 , 0) T . If the problem has an optimal solution, provide one. If the problem is unbounded, provide a parametric set of feasible solutions x ( t ) (for t ) such that c T x ( t ) + as t + . Show all of your work. Solution: Iteration 1 : Basis B = { 2 , 3 } . Solve A T B y = c B : 1 1 1 0 ! y = 1 1 ! So y = [1 , 0] T . The reduced cost for x 1 is c 1 = c 1 y T A 1 = 1 > , so x 1 enters....
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 Winter '07
 S.Furino,B.Guenin

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