Maximize $\vec c*\vec x$ subject to $A\vec x\le \vec b$ and $\vec x \ge \vec 0$
$\vec c$=$\begin{bmatrix}1 \\2\\1\end{bmatrix}$, $\vec x$=$\begin{bmatrix}x_1 \\x_2\\x_2\end{bmatrix}$
A=$\begin{bmatrix} 1 & 1 & 1 \\\ -2 & -2 & 1 \\\ 0 & -1 & -1 \end{bmatrix}$,
$\vec b$=$\begin{bmatrix}4 \\ -1 \\ 2\end{bmatrix}$
The final dictionary is
$x_2=4-x_1-x_3-x_4$
$x_5=7-3x_3-2x_4$
$x_6=2-x_1-x_4$
$z=8-x_1-x_3-2x_4$
Find the largest $\epsilon$ for any $0\le t\le \epsilon$ s.t. $\vec x^*$ is still optimal to the given LP by changing $\vec c$ to $\vec c$ + $\begin{bmatrix}t \\ 0 \\ 0\end{bmatrix}$
This one I have no idea how to do
Let $\vec y^*$ be the optimal sol to the dual problem. Find the largest $\epsilon$ s.t. for any $t$ that $-\epsilon \le t \le \epsilon$, the vector $\vec y^*$ is still optimal to the new dual problem obtained by changing $\vec b$ to $\vec b$ + $\begin{bmatrix}t \\ 0 \\ 0\end{bmatrix}$
By strong duality you can get the solution to the dual problem by getting the negative coefficients of the corresponding slack variables in the final objective solution so optimal sol to the dual is {2,0,0}. So for this to still be optimal $\vec b=\begin{bmatrix}4+t \\ -1 \\ 2\end{bmatrix}$ times [2,0,0] has to still be the smallest value you can get that is within the constraints. I'm still a little stuck here.
