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'Corner Point Theorem'
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| Title of object: |
Corner Point Theorem |
| Canonical Name: |
CornerPointTheorem |
| Type: |
Theorem |
| Created on: |
2006-02-02 17:01:52 |
| Modified on: |
2006-02-03 12:25:13 |
| Classification: |
msc:90C05 |
| Defines: |
corner point |
| Synonyms: |
Corner Point Theorem=fundamental theorem of linear programming |
Preamble:
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Content:
Let $P$ be a primal linear programming problem with $f$ the objective function and polyhedron $R$ its feasible region. A \emph{corner point}, or extreme point of $P$ is a vertex of $R$.
\textbf{Corner Point Theorem.} If $P$ has an optimal solution $a<\infty$, then there is a corner point $p$ of $P$ such that $f(p)=a$. If another corner point $q$ also satisfies $f(x)=a$, then $f(\overline{pq})=\lbrace a\rbrace$. If $r$ is a third corner point such that $f(r)=a$, then $f(\triangle pqr)=\lbrace a\rbrace$.
Note that the line segment and triangle mentioned above are necessarily subsets of $R$.
A cousin of the above theorem is the following:
\textbf{Theorem.} If $P$ has an optimal solution $a<\infty$ occuring at an interior point of an edge $E$ (or a face $F$) on the boundary of the feasible region $R$, then $f(E)=\lbrace a\rbrace$ (or $f(F)=\lbrace a\rbrace$). In fact, if the solution occurs at an interior point of $R$, then the solution is satisfied by all points of $R$: $f(R)=\lbrace a\rbrace$. In other words, the objective function $f$ is constant on $R$.
On way to visualize the above theorems is to simplify them into the case when the objective function is a line $f(x)=mx+b$ on the ``$x-y$ plane'' and the feasible region is a line segment $I=[x_0,x_1]$ on the $x$-axis. It is easy to see now that the maximum (or minimum) of $f$ occurs at either $x_0$ or $x_1$. If the optimal value occurs either at both end points, or at an interior point $x_2\in(x_0,x_1)$, then $f$ is a horizontal line segment on $I$.
An application of the above theorems can be demonstrated by the following example: If the feasible region $R$ is a unit square and if corner points $(0,0),(1,1)$ satisfy the optimal solution of $P$, then all points on $\lbrace (x,y)\mid x=y\rbrace \cap R$ satisfy the solution. In particular, $(\frac{1}{2},\frac{1}{2})$, an interior point of $R$, satisfies the solution. As a result, all points of $R$ satisfy the solution. |
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