Chapter 4.12, Problem 1P

Explanation of Solution

Solving the LP using Big M method:

Given,

$\min Z=4x_1+4x_2+x_3$

Subject to,

$\begin{array}{l}{\text{x}}_{\text{1}}{\text{+x}}_{\text{2}}{\text{+x}}_{\text{3}}\le \text{2}\\ {\text{2x}}_{\text{1}}{\text{+x}}_{\text{2}}\le \text{3}\\ {\text{2x}}_{\text{1}}{\text{+x}}_{\text{2}}{\text{+3x}}_{\text{3}}\le \text{3}\\ {\text{x}}_{\text{1}}{\text{,x}}_{\text{2}}{\text{,x}}_{\text{3}}\ge \text{0}\\ Z_j\\ {\text{R}}_{\text{1}}{\text{(new)=R}}_{\text{1}}\text{(old)÷2}\\ {\text{R}}_{\text{2}}{\text{(new)=R}}_{\text{2}}{\text{(old)+R}}_{\text{1}}\text{(new)}\end{array}$

Introduce slack variables S₁, S₂ of cost coefficients zero each, surplus variable S₃ and artificial variable A₁ of cost coefficient Big M

After introducing slack variables, the LP problem becomes,

${\text{Minimize Z=4x}}_{\text{1}}{\text{+4x}}_{\text{2}}{\text{+x}}_{\text{3}}{\text{+(0)S}}_{\text{1}}{\text{+(0)S}}_{\text{2}}{\text{+(-M)A}}_{\text{1}}$

Subject to

$\begin{array}{l}{x}_1+x_2+x_3+S_1=2\\ 2x_1+x_2+S_2=3\\ 2x_1+x_2+3x_3-S_3+A_1=3\\ x_1,x_2,x_3,A_1,S_1,S_2,S_3\ge 0\end{array}$

Table 1:

Cj	B. V	b	4 x1	4 x2	1 x3	0 S1	0 S2	0 S3	-M A1	$\theta =\frac{b}{x_3}$
0 0 -M	$S_1$ $S_2$ $A_1$	$\begin{array}{l}2\\ 3\\ 3\end{array}$	$\begin{array}{l}1\\ 2\\ 2\end{array}$	$\begin{array}{l}1\\ 1\\ 1\end{array}$	$\begin{array}{l}1\\ 0\\ 3\end{array}$	$\begin{array}{l}1\\ 0\\ 0\end{array}$	$\begin{array}{l}0\\ 1\\ 0\end{array}$	$\begin{array}{l}0\\ 0\\ -1\end{array}$	$\begin{array}{l}0\\ 0\\ 1\end{array}$	$\begin{array}{l}2\\ \infty \\ 1\end{array}$
	$Z_j-C_j$		$-2M-4$	$-M-4$	$-3M-1$	$0$	$0$	$M>0$	$0$

Table 2:

Cj	B. V	b	4 x1	4 x2	1 x3	0 S1	0 S2	0 S3	-M A1	$\begin{array}{l}\theta =\frac{b}{x_2}\\ \end{array}$
$\begin{array}{l}0\\ 0\\ 1\end{array}$	$S_1$ $S_2$ $x_1$	$\begin{array}{l}1\\ 3\\ 1\end{array}$	$\begin{array}{l}1/3\\ 2\\ 2/3\end{array}$	$\begin{array}{l}2/3\\ 1\\ 1/3\end{array}$	$\begin{array}{l}1\\ 0\\ 3\end{array}$	$\begin{array}{l}1\\ 0\\ 0\end{array}$	$\begin{array}{l}0\\ 1\\ 0\end{array}$	$\begin{array}{l}0\\ 0\\ -1\end{array}$	$\begin{array}{l}0\\ 0\\ 1\end{array}$	$\begin{array}{l}2\\ \infty \\ 1\end{array}$
	$Z_j-C_j$		$-\frac{10}{3}$	$-\frac{11}{3}$	$0$	$0$	$0$	$-1/3$	$\frac{1}{3}+M$

Table 3:

Cj

B