1
$\begingroup$

If I have a two variable linear problem, and I convert it to standard form (adding slack and surplus variables, etc), then any solution to the constraints equation system is a vertex of the convex set of feasible solutions.

And the same holds true for a linear problem of three variables. Converting it to standard form, any solution to the equation system is a polyhedron vertex.

Is this true?

(I'm only asking for a true or false, not more articles about maths where I'd actually have to read and take the time to understand, which would take more 10000x more time than reading a "true" or a "false", thanks. I'm not a mathematician)

$\endgroup$
1
$\begingroup$

This is precisely the Fundamental Theorem of Linear Programming.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Strictly speaking it is false.

What is true is the following: if the linear program has a bounded solution (i.e., if its optimal value is not infinity), then the optimal value can be achieved by a vertex. If the optimal value is achieved at exactly one point, then that point is indeed a vertex.

The optimal value could potentially be attained at more than one vertices, which in turn would mean that it is also attained at all points in their convex hull (facets of the feasible polyhedron). Therefore, an optimal point for an LP may not always be a vertex. But you do know that there exists a vertex that also achieves the same optimal value.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.