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# s5 - C O350 L INEAR P ROGRAMMING S OLUTIONS TO ASSIGNMENT 5...

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CO350 L INEAR P ROGRAMMING - S OLUTIONS 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 0 } 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 0 := B \ { j } ∪ { k } be the new basis. Then the reduced cost for x j for the basis B 0 is ˆ c j = - ¯ c k ¯ a jk . As k is the entering variable, ¯ c k > 0 . Moreover x j is the leaving variable, so ¯ a jk > 0 . So ˆ c j = - ¯ c k ¯ a jk < 0 . 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 0 } 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 0 ) 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 > 0 , so x 1 enters. Solve A B d = A 1 : 1 1 1 0 ! d = 2 0 !

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