# stationary point of the solution of recursive linear programming

The problem I am working on is:

Given an $n$ dimensional vector $r \in \mathcal{R}^n$, and a convex set $G=\{\mu \in \mathcal{R}^n | \mu_i \ge 0, ~ \mu^T \mathbf{1}=1, ~ A\mu =0 \}$ where $\mathbf{1}= [1, 1, ..., 1]^T$. ($A \in \mathcal{R}^{m \times n}$ and $m < n$) We assume that the set $G$ is not an empty set.

Suppose $\bar{\mu} = \arg \max_{\mu \in G} r^T\mu$ given $r$. Then can we say $\tilde{\mu} = \arg \max_{\mu \in G} \bar{\mu}^T\mu$ is identical to $\bar{\mu}$?

In other words, when we fix the feasible set of linear programming, we can treat LP as a mapping from a vector (in the cost function) to another vector (solution). Then, what is the stationary points to this mapping when the feasible set is a subset of a simplex and plane intersecting the origin.

• Recommend better place to ask Optimization (especially, Linear/integer programming) questions. OR-Exchange.org – S. Phil Kim Sep 24 '16 at 2:11

Not true in general. Consider $n=3$ and $A=[2, -1, 6]$. Then, $G$ is the line segment with end points $(1/3, 2/3, 0)$ and $(0, 6/7, 1/7)$. Consider $r$ such that $\bar {\mu} = (1/3, 2/3, 0)$. (For example, $r=(1, -1, 0)$.). Then, $\tilde {\mu}$ is unique and it is $(0, 6/7, 1/7)$.