# Find the final tableau knowing the optimal solution

Consider the following linear program $$\displaystyle \max z=5x_1+2x_2+3x_3\\ s.t. x_1+5x_2+2x_3\le b_1\\ x_1-5x_2-6x_3\le b_2 \\ x_1,x_1,x_3\ge0$$

If the optimal solution is reached at $$x_1=30,x_5=10,$$ write directly the complete optimal tableau, whitout executing the simplex method. And then find the values of $$b_1$$ and $$b_2$$.

Attempt

The optimal tableau should have the following structure

$$\begin{array}{r|rrrrr|r} & x_1 & x_2 & x_3 & x_4 & x_5 & \\ \hline z & 0 & & & & 0 & 150 \\ \hline x_1 & 1 & & ? & & 0 & 30 \\ x_5 & 0 & & & & 1 & 10 \\ \end{array}$$

I don't know how to calculate what is inside the tableau. How will I calculate $$B^{-1}$$? Please help me.

I know the last simplex tableau in matrix form is this

$$\begin{array}{|r r|r} c_BB^{-1}N-c & c_BB^{-1} & c_BB^{-1}b \\ \hline B^{-1}N & B^{-1} & B^{-1}b \end{array}$$

• Write the tableau first, then apply dual simplex. You may find a MathJax template for simplex tableau on Mathematics Meta. Commented Nov 11, 2018 at 19:57
• @GNUSupporter8964民主女神地下教會 thanks for the hint! I'll work on it. Commented Nov 14, 2018 at 0:31
• @GNUSupporter8964民主女神地下教會 but in the exercise its write directly the complete optimal tableau. With this I understand that no need to apply dual simplex is needed. am I wrong? Commented Jan 2, 2019 at 3:25
• Thx for your update to the question. When I posted my comment, the optimal tableau wasn't there. Commented Jan 14, 2019 at 12:58
• @GNUSupporter8964民主女神地下教會 :) nop It wasn't but the statement was already there. Commented Jan 17, 2019 at 4:34

From the given optimal solution $$z(x_1,x_2,x_3,x_4,x_5)=z(30,0,0,0,10)=150$$, you can find: $$x_1+5x_2+2x_3+x_4=b_1=\color{red}{30},\\ x_1-5x_2-6x_3+x_5=b_2=\color{red}{40},$$ where $$x_4$$ and $$x_5$$ are the slack variables.
Since in the optimal table only the slack variable $$x_4$$ is replaced with $$x_1$$, then the intersection of $$x_4$$ row and $$x_1$$ column was the pivot element, which corresponds to the original value $$1$$. Hence the row remained from the initial table: $$\begin{array}{r|rrrrr|r} & x_1 & x_2 & x_3 & x_4 & x_5 & \\ \hline z & 0 & & & & 0 & 150 \\ \hline x_1 & \boxed 1 & \color{red}5 & \color{red}2 & \color{red}1 & 0 & 30 \\ x_5 & 0 & & & & 1 & 10 \\ \end{array}$$ In order to get $$0$$ as the first element in the row $$z=5x_1+2x_2+3x_3$$, where the first element was $$-5$$ initially, pivot row must have been multiplied by $$5$$ and added to it: $$\begin{array}{r|rrrrr|r} & x_1 & x_2 & x_3 & x_4 & x_5 & \\ \hline z & 0 & \color{red}{23} & \color{red}7 & \color{red}5 & 0 & 150 \\ \hline x_1 & \boxed 1 & \color{red}5 & \color{red}2 & \color{red}1 & 0 & 30 \\ x_5 & 0 & & & & 1 & 10 \\ \end{array}$$ In order to get $$0$$ as the first element in the row $$x_5$$, where the first element was $$1$$ initially, pivot row must have been multiplied by $$-1$$ and added to it: $$\begin{array}{r|rrrrr|r} & x_1 & x_2 & x_3 & x_4 & x_5 & \\ \hline z & 0 & \color{red}{23} & \color{red}7 & \color{red}5 & 0 & 150 \\ \hline x_1 & \boxed 1 & \color{red}5 & \color{red}2 & \color{red}1 & 0 & 30 \\ x_5 & 0 & \color{red}{-10} & \color{red}{-8} & \color{red}{-1} & 1 & 10 \\ \end{array}$$ Note: I am not sure if this is ejecuting (executing). But it is reverse engineering (backwards working).