Chapter 6.2, Problem 2P

Explanation of Solution

Given:

Let $x_1$ and $x_2$ be the basic variables and the analog of the LP is obtained by adding the slack variables $s_1$ and $s_2$.

$\begin{array}{l}\text{max}\text{z=}-{\text{x}}_{\text{1}}{\text{+x}}_{\text{2}}\\ {\text{subject to 2x}}_1+x_2+s_1=4\\ x_1+x_2+s_2=2\\ x_1,x_2\ge 0\end{array}$

Writing the matrix of basic variables and vector b,

$\begin{array}{l}\text{BV=}\left\{{\text{x}}_{\text{1,}}{\text{x}}_{\text{2}}\right\}\\ \text{b=}\left[\begin{array}{l}4\\ 2\end{array}\right]\end{array}$

Writing the B matrix,

$B=\left[\begin{array}{l}11\\ 10\end{array}\right]$

Compute the $B^{-1}$,

${\text{B}}^{\text{-1}}\text{=}\left[\begin{array}{l}0\text{1}\\ \text{1}-1\end{array}\right]$

Writing $c_{Bv}$ matrix as,

$c_{Bv}{B}^{-1}=[01]$

Find the coefficient of ${\text{s}}_{\text{1}}{\text{ and s}}_{\text{2}}$ in the optimal tableau.

$\begin{array}{l}{\text{Coefficient of x}}_{\text{1}}\text{in row 0=First element of the above matrix}\\ =2\end{array}$

$\begin{array}{l}{\text{Coefficient of s}}_2\text{ in row 0=First element of the above matrix}\\ \end{array}$