# Please suggest vectors to construct an infeasible solution.

 -1 Consider the standard primal linear programming problem: $$\text{Max} \ z = cx$$ $$\text{s.t.} \ Ax \le b, x \ge 0$$ where $A = \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$ The standard dual linear programming problem is $$\text{Min} \ Z = b'y = y'b$$ $$\text{s.t.} \ A'y \ge c', y \ge 0$$ Find vectors $$b = [b_1, b_2], c = [c_1, c_2]$$ such that both the primal and dual are infeasible. Note: $$c$$ is a row vector. I think the rest are column vectors. $$x \ge 0$$ means $$x_i \ge 0$$ I tried using $$b = [-1 -1], c = -b$$ but this seems too simple. Is it right? Wrong? Why? I happened noticed that we need to have $$-b_2 \le x_1 - x_2 \le b_1$$ and $$-c_2 \le y_1 - y_2 \le c_1$$ Must it be then that $$-b_2 \le b_1, -c_2 \le c_1$$ ? In that case, it doesn't seem like $b = [-1, -1], c= -b$ satisfy those inequalities Please suggest what I can use instead, or please explain why I am correct. asked 11 May '16, 14:10 BCLC 8●3●13 accept rate: 0%
Be the first one to answer this question!
 toggle preview community wiki

By Email:

Markdown Basics

• *italic* or _italic_
• **bold** or __bold__
• image?![alt text](/path/img.jpg "Title")
• numbered list: 1. Foo 2. Bar
• to add a line break simply add two spaces to where you would like the new line to be.
• basic HTML tags are also supported

Tags: