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Unformatted text preview: Fall 2010 Optimization I (ORIE 3300/5300) Assignment 10 Solution Problem 1 (a) We wish to show that there does not exist any vector x that satisfies both Ax = b and x 0. To do this, let us suppose to the contrary, that there exists such vector x . Then, we make the following observations: Since Ax = b and b T y < 0, then x T A T y = ( Ax ) T y = b T y < . Since the vector x 0 and the vector A T y 0, then their dot product is also nonneg ative. That is, x T A T y = x T ( A T y ) . However, we see that the two observations above are not consistent. This means that our assumption that there exists a vector x satisfying Ax = b, x 0 must be false. This proves that the linear programming problem has no feasible solution. (b) First, we need to show that the auxiliary problem has a feasible solution. This is easy to do, since x = 0 , v = b is a feasible solution to the auxiliary problem (you can check that x = 0 , v = b satisfies all the constraints). Next, we observe that the objective value corresponding to any feasible solution ( x,v ) of the auxiliary problem has an objective value that is negative or zero: Suppose ( x,v ) is feasible. Then, we know that v 0, which implies that e T v = v 1 v 2 ... v n . Finally, we note that the objective value is exactly zero if and only if v is the zero vector. When v is the zero vector, then the solution ( x,v ) satisfies: Ax + 0 = b,x 0, which means that x is a feasible solution to the original problem. That is, the objective value of the auxiliary problem is zero if and only if the original lin ear programming problem is feasible. This implies that if the original linear programming problem is infeasible, then the optimal objective value of the auxiliary problem is negative....
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 Fall '08
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