Let $x\rightarrow x^
$\begingroup$ Not necessarily. Imagine a program with no valid $x$. Such a program never has an optimal solution (for any objective function). $\endgroup$
Commented Oct 8, 2014 at 18:21$\begingroup$ can you tell me more about that? I'm kinda new to this topic. Thank you. Maybe you can post an answer. $\endgroup$
Commented Oct 8, 2014 at 18:23$\begingroup$ I have now tried to find concrete programs for the various cases. Feel free to ask for clarification. $\endgroup$
Commented Oct 8, 2014 at 18:46If we have an unconstrained problem with objective $x^Tc$ we have no optimum because we can get "as good as we want": Chose $x_n = -nc$ to obtain a sequence of values $-n\|c\| \to-\infty$. If we are constrained, the admissible region is a polytope of the form $Ax = b$: $$\min_ x^T c \tag P$$ If the admissible region $\$ is bounded and non-empty, we always have a solution, no matter what $c$ we chose (esp. $-c$ also gives a solution). If it's empty, we never have a solution and if it's unbounded we can give examples such that (P) is unbounded for both $c, -c$, bounded for exactly one of $c,-c$ or bounded for both $c$ and $-c$, so we can't really say anything.
Examples:
$A = \pmatrix, b = 0, c = \pmatrix$: unbounded for $-c$, unique for $c$ $A = \pmatrix, b = 1, c = \pmatrix$: solutions for both, $c, -c$
$A = \pmatrix, b = \pmatrix, c=\pmatrix$: unbounded for both $c, -c$
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