How can I solve a linear program with parameters? I have the following problem:
$$\begin{array}{ll} \text{maximize} & x_1 + x_2 + x_3\\ \text{subject to} & a x_1 + b x_2 + c x_3 \leq 1\\ & a^2 x_1 + b^2 x_2 + c^2 x_3 \leq 1\\ & x_1,x_2,x_3 \geq 0\end{array}$$
where $a < b < c$ more than $1$ real parameters. What is the point? How to solve this?
 A: Considering $1<a<b<c$ then we can see that $$a x_1 + b x_2 + c x_3 \leq  a^2 x_1 + b^2 x_2 + c^2 x_3$$
Provided that $x_i \geq 0.$ This proves that the first constraint is redundant. So problem is equivalent to the following linear programin
$$\begin{array}{ll} \text{maximize} & x_1 + x_2 + x_3\\ \text{subject to} & a^2 x_1 + b^2 x_2 + c^2 x_3 \leq 1\\ & x_1,x_2,x_3 \geq 0\end{array}$$
A simple analysis reveals that maximum is attained where $x_1$ be as large as possible. Which is $x_1 = \frac{1}{a^2}$. So the optimal solution is $(\frac{1}{a^2}, 0 , 0).$
Other way of thinking about it is looking at its graph, or there are only three extreme points.
A: It seems that $a,b,c>1$ and $a,b,c \in \mathbb R$. Now apply the simplex alogrithm. To start I give you the initial tableau 
$$\begin{array}{|c|c|c|c|c|c|c|} \hline x_1 & x_2 & x_3 & s_1 & s_2 &RHS \\ \hline -1 & -1 & -1 & 0  & 0 &0 \\ \hline a & b & c & 1& 0 & 1  \\ \hline \color{red}{a^2}& b^2& c^2 &0 &1 &1   \\ \hline\end{array}$$
The coefficients of the objective function are all equal. The simplex algorithm gives us free choice for the pivot column. I choose the first one ($x_1$). And  the pivot row is the  last one since $\min \left(\frac1a; \frac1{a^2}\right)=\frac1{a^2}$
