I want to maximize $z = 12x + 20y$ with the following constraints: $$\eqalign{ & 2x + 8y \le 180 \cr & 4x + 4y \le 120 \cr & x,y \ge 0 \cr} $$
I need the simplex tableau to find optimal solution but also I need to determine the dual problem and find its optimal.
Here is a detailed solution => https://i.hizliresim.com/Xqm406.png
s1 and s2 are slack variables. Cj is coefficient of the variables. In every iteration find column contains the maximum Cj-Zj and find the row of minimum ratio. Then you have a pivot at the intersection cell
The dual problem is :
minimize $ z = 180x + 120y$ due to : $$2x + 4y \ge 12$$
$$8x + 4y \ge 20$$
Edit : To calculate sensitivity of first constraint find the delegate column. The delegate of first constraint is $s1$. So get the values in column below $s1$ to get the lower bound. $$\begin{bmatrix}1/6\\-1/6\end{bmatrix}*\begin{bmatrix}a&b\end{bmatrix}=\begin{bmatrix}20\\10\end{bmatrix}$$
Here $a$ and $b$ must be positive and we must select the minimum of the positive values $min(a,b)$. So $a = 120$ and $b = -60$. We get $a$. Then lower bound is $180 - 120 = 60$
For the upper bound :
$$\begin{bmatrix}1/6\\-1/6\end{bmatrix}*\begin{bmatrix}-a&-b\end{bmatrix}=\begin{bmatrix}20\\10\end{bmatrix}$$
Here $a$ and $b$ must be positive too and we must select the minimum of the positive values $min(a,b)$. So $a = -120$ and $b = 60$. We get $b$. Then the upper bound is $180 + 60 = 240$
Finally we have $$ 60 \le C1 \le 240$$
You can start from the following tableau: \begin{array}{ccccc|c} x&y&s_1&s_2&z&RHS\\ 1&4&1&0&0&90\\ 1&1&0&1&0&30\\ -12&-20&0&0&1&0 \end{array}
