TITLE: Computationally proving a linear programming solution is unique?
QUESTION [7 upvotes]: I have a simple linear programming problem min $c^{T}x$ subject to $Ax\leq b$. That gives me the solution I am looking for when solving in maple. My only problem is that I do not know how to check, with maple, whether or not this solution is unique. I have read MANY threads about this topic and no one seems to be able to provide a simple way to check with a computational program such as maple whether or not the solution is unique. 
Any sort of for all $q\in \mathbb{R}^{n}$ check that there exists an $\epsilon$ such that the solution maximizes the perturbed problem is useless for me as I need a definite answer and obviously can't check for all $q\in \mathbb{R}^{n}$. 
I have read this paper www.jstor.org/stable/822972.
But, I cannot understand how to implement this algorithm based on the way it is presented, if anyone could simply phrase this algorithm I think it would be beneficial for future reference to others.
It is simple enough to provide my exact problem here:
$$
\min_{x\in \mathbb{R}^{8}} 36x_{1}+30x_{2}+50x_{3}+51x_{4}+36x_{5}+55x_{6}+33x_{7}+30x_{8}
$$
s.t
\begin{cases}
x_{1}+x_{5}\leq 2x_{6}\\
2x_{8}\leq x_{3}+x_{7}\\
2x_{1}+x_{7}\leq3x_{5}\\
3x_{6}+3x_{7}\leq 2x_{2}+4x_{8}\\
2x_{2}\leq x_{1}+x_{3}\\
x_{5}+x_{7}\leq 2x_{6}\\
40x_{1}+80x_{2}+60x_{5}+60x_{7}+80x_{8}\leq 40\\
\sum_{i=1}^{8}x_{i}=1,\;\;\;\; x_{i}\geq 0\;\; i=1,...,8\\
\end{cases}
The solution is $x^{*}=(0,\frac{1}{4},\frac{1}{2},0,0,0,0,\frac{1}{4})$ which is exactly what my counter example requires, but I need uniqueness.

REPLY [2 votes]: There is a very simple (while not that efficient) way to check this.
Note that your solution yields the function value $c^T x^*=\frac{30}{4}+\frac{50}{2}+\frac{30}{4}=40$.
Now consider the linear program
$$\max_{x\in \mathbb{R}^{8}} 1x_{1}$$
s.t.
$$\begin{cases}
36x_{1}+30x_{2}+50x_{3}+51x_{4}+36x_{5}+55x_{6}+33x_{7}+30x_{8}=40\\
x_{1}+x_{5}\leq 2x_{6}\\
2x_{8}\leq x_{3}+x_{7}\\
2x_{1}+x_{7}\leq3x_{5}\\
3x_{6}+3x_{7}\leq 2x_{2}+4x_{8}\\
2x_{2}\leq x_{1}+x_{3}\\
x_{5}+x_{7}\leq 2x_{6}\\
40x_{1}+80x_{2}+60x_{5}+60x_{7}+80x_{8}\leq 40\\
\sum_{i=1}^{8}x_{i}=1,\;\;\;\; x_{i}\geq 0\;\; i=1,...,8\\
\end{cases}$$
The first constraint will say that you are searching a different optimal solution (same function value). But the new function you maximize, namely $x_1$ means that you are looking for an optimal solution with maximal $x_1$ value.
You can do that twice for each coordinate to see if your solution is unique. In your case your solution is not optimal, in fact there is a solution with $x_1\approx0.1363636       $.