Chapter 4.14, Problem 2P

Explanation of Solution

Solving the LP using simplex algorithm:

Given,

Maximize $z=2x_1+x_2$

Subject to,

$\begin{array}{l}3x_1+x_2\le 6\\ x_1+x_2\le 4\\ x_1\ge 0,x_{2}urs\end{array}$

Converting to standard form:

As, $x_2$ is URS, introduce the non-negative variables $x_2\text{'}$ and $x_2\text{'}\text{'}$. So that,

$x_2=x_2\text{'}-x_2\text{'}\text{'}$ where, $x_2\text{'},x_2\text{'}\text{'}\ge 0$

The standard form of LP problem becomes,

Maximize, $z=2x_1+x_2\text{'}-x_2\text{'}\text{'}$

Subject to,

$\begin{array}{l}3x_1+x_2\text{'}-x_2\text{'}\text{'}\le 0\\ x_1+x_2\text{'}-x_2\text{'}\text{'}\le 4\end{array}$

and $x_1,x_2\text{'},x_2\text{'}\text{'}\ge 0$

Converting LP to canonical form:

The LP is converted to canonical form by adding slack, surplus and artificial variables as appropriate. This is done as given below,

As the constraint “$3x_1+x_2\text{'}-x_2\text{'}\text{'}\le 0$” is “$\le$” type, the user should add S₁.

For the constraint “$x_1+x_2\text{'}-x_2\text{'}\text{'}\le 4$” add S₂.

Thus the LP becomes,

Maximize $z=2x_1+x_2\text{'}-x_2\text{'}\text{'}+0S_1+0S_2$

Subject to,

$\begin{array}{l}3x_1+x_2\text{'}-x_2\text{'}\text{'}+S_1=6\\ x_1+x_2\text{'}-x_2\text{'}\text{'}+S_2=4\\ x_1,x_2\text{'},x_2\text{'}\text{'},S_1,S_2\ge 0\end{array}$

Iteration 1:

		$C_j$	2	1	-1	0	0	Min Ratio
$\begin{array}{l}\\ B\end{array}$	$\begin{array}{l}\\ C_B\end{array}$	$\begin{array}{l}\\ X_B\end{array}$	$\begin{array}{l}\\ x_1\end{array}$	$\begin{array}{l}\\ x_2\text{'}\end{array}$	$\begin{array}{l}\\ x_2\text{'}\text{'}\end{array}$	$\begin{array}{l}\\ S_1\end{array}$	$\begin{array}{l}\\ S_2\end{array}$	$\frac{X_B}{x_1}$
$S_1$	0	6	(3)	1	-1	1	0	$\frac{6}{3}=2\to$
$S_2$	0	4	1	1	-1	0	1	$\frac{4}{1}=4$
z=0		$Z_j$	0	0	0	0	0
		$Z_j-C_j$	$-2$	-1	1	0	0

Negative min $Z_j-C_j=-2$ and its column index 1

$x_1$ is entering variable.

Minimum ratio is 2 and its row index is 1,

$S_1$ is leaving basis variable