This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
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....
View Full Document
- Fall '08