Consider the standard primal linear programming problem:

$$\text{Max} \ z = cx$$

$$\text{s.t.} \ Ax \le b, x \ge 0$$


alt text

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.


  1. \(c\) is a row vector. I think the rest are column vectors.

  2. \(x \ge 0\) 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

$$-b_2 \le x_1 - x_2 \le b_1$$


$$-c_2 \le y_1 - y_2 \le c_1$$

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's gravatar image

accept rate: 0%

edited 11 May '16, 14:31

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Asked: 11 May '16, 14:10

Seen: 310 times

Last updated: 11 May '16, 14:31

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