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Unformatted text preview: 1 15.093J/2.098J Optimization Methods Assignment 3 Solutions Exercise 3.1 BT, Exercise 4.27. Prove (a) (b) Consider the problem min x 1 s.t. Ax 0 x 0 For every feasible solution of this problem, x 1 = 0. Thus, the problem has a finite optimum and the optimal objective is 0. The dual problem max p 0 s.t. p A 1 1 p A p 0 also has a zero finite optimum. It means that the dual problem is feasible. Thus there exists a vector p 0 such that p A and p A 1 1 or we can say p A 1 < 0. Prove (b) (a) Let p 0 be a vector that satisfies all conditions in (b). Consider the problem min ( p A 1 ) x 1 s.t. Ax 0 x 0 Its dual problem is then max p 0 s.t. p A 1 p A 1 p A p 0 The dual problem is feasible, p = p 0 is a feasible solution. It has the finite optimum of zero. Thus the primal problem is also feasible and has the optimal solution of zero. So, for all feasible solution x , Ax , x , we have: ( p A 1 ) x 1 0. p A 1 < 0 and x , thus we need to have x 1 = 0. So we have (a) (b). Exercise 3.2 BT, Exercise 4.29. Consider the problem min 0 x s.t. Ax b The dual problem is then min b p s.t. p A = p 0 This problem is feasibe, p = 0 is a feasible solution. The primal is infeasible, the the dual must be unbounded. Assume that all constraints are linearly independent; otherwise, we can remove the dependent constraints and work with a system with less constraints. The dual polyhedron is in the fist orthant of R m , thus it does not contain a line, which means the polyhedron has at least an extreme point. Staring the simplex method with this extreme point, the simplex method will stop at a basis B at which we can find a non-basic index j such that the corresponding feasible direction calculated from the simplex method allows the objective to go to . This direction can be simply calculated with the basis B and the column A j . Consider the modified dual problem with only these n + 1 columns, we have: the problem is still unbounded....
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- Spring '08