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Having read the link:

Why maximum/minimum of linear programming occurs at a vertex?

I understand why the optimal solution of any linear programming problem must be on the corner or lies on a face of a convex polygon. But my question is about a proof given below:

http://en.wikipedia.org/wiki/Fundamental_theorem_of_linear_programming

I do not know why the next argument was

$$x^\ast - \frac{\epsilon}{2} \frac{c}{||c||} \in P$$

Sorry to ask this question, but I can't imagine where the term

$$\frac{\epsilon}{2} \frac{c}{||c||}$$

is coming from. Anyone, please enlighten me. Thanks.

Isn't it the contradiction will still be true even if we have, $$x^\ast - \frac{c}{||c||}$$.

The problem may be this construction may not be in $P$. Now, I really want to visualize the expressions $\frac{\epsilon}{2} \frac{c}{||c||} $ and $\frac{c}{||c||}$. Thanks.

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Keneth Adrian
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1 Answers1

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In the proof, the assumption that $x^\ast$ is in the interior of $P$ is shown to lead to a contradiction.

Let

$$z = x^\ast - \frac{\epsilon}{2} \frac{c}{||c||}$$

Then

$$||z - x^\ast|| = \frac{\epsilon}{2} \frac{||c||}{||c||}= \frac{\epsilon}{2} < \epsilon,$$

and

$$z \in B_{\epsilon}(x^\ast) \subset P.$$

This was chosen to obtain a feasible point $z \in P$, such that $c^Tz < c^Tx^\ast$

Since $x^\ast$ is optimal we have a contradiction, implying that $x^\ast$ cannot lie in the interior of $P$.

RRL
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  • thanks for the explanation, but can I clarify what is the difference between c and ||c|| ? is that the modulus? and $z$ must be in the interior of the ball centered at x* with radius $\epsilon$? ( I think I somehow get it) – Keneth Adrian Nov 02 '14 at 07:19
  • @KenethAdrian: Yes, $||c||$ is the norm (magnitude) of the vector c. With that choice we get a point in the ball where the objective function is smaller than the minimum at $x^{\ast}$, a contradiction. – RRL Nov 02 '14 at 07:31
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    Yes, $|c|$ is the norm (or length) of the vector $c$. So $\frac{c}{|c|}$ as just a fancy way of saying "a unit vector in the direction of $c$". – bubba Nov 02 '14 at 07:39
  • so really instead of $\frac{\epsilon}{2}$, they could chosen $\alpha \epsilon$ as long as $\alpha < 1$. – Kiran K. Feb 07 '16 at 16:47
  • @Kiran K: Yes you just need to get a contradiction by producing a point that is both in the interior of $P$ and such that $c^Tz < c^Tx^*$. – RRL Feb 07 '16 at 23:01
  • Hello! What is the function "B-epsilon"? Thanks! – stats_noob Apr 02 '22 at 01:12
  • @stats_noob: $B_\epsilon(x^)$ is the open ball with center $x^$ and radius $\epsilon$. In general for $c \in \mathbb{R}^n$ we have $B_r(c) := {x \in \mathbb{R}^n,:, |x-c| < r}$ – RRL Apr 02 '22 at 17:59