Editing Simplex algorithm (section)

===Example===
Consider the linear program
:Minimize
::<math>Z = -2 x - 3 y - 4 z\,</math>

:Subject to
::<math>\begin{align}
 3 x + 2 y + z &= 10\\
 2 x + 5 y + 3 z &= 15\\
 x,\, y,\, z &\ge 0
\end{align}</math>

It differs from the previous example by having equality instead of inequality constraints. The previous solution <math>x=y=0\, , z=5</math> violates the first constraint. 
This new problem is represented by the (non-canonical) tableau
:<math>
  \begin{bmatrix}
    1 & 2 & 3 & 4 &  0 \\   
    0 & 3 & 2 & 1 & 10 \\
    0 & 2 & 5 & 3 & 15
  \end{bmatrix}
</math>

Introduce artificial variables ''u'' and ''v'' and objective function ''W''&nbsp;=&nbsp;''u''&nbsp;+&nbsp;''v'', giving a new tableau
:<math>
  \begin{bmatrix}
    1 & 0 & 0 & 0 & 0 & -1 & -1 &  0 \\  
    0 & 1 & 2 & 3 & 4 &  0 &  0 &  0 \\   
    0 & 0 & 3 & 2 & 1 &  1 &  0 & 10 \\
    0 & 0 & 2 & 5 & 3 &  0 &  1 & 15
  \end{bmatrix}
</math>

The equation defining the original objective function is retained in anticipation of Phase II.

By construction, ''u'' and ''v'' are both basic variables since they are part of the initial identity matrix. However, the objective function ''W'' currently assumes that ''u'' and ''v'' are both&nbsp;0. In order to adjust the objective function to be the correct value where ''u''&nbsp;=&nbsp;10 and ''v''&nbsp;=&nbsp;15, add the third and fourth rows to the first row giving
:<math>
  \begin{bmatrix}
    1 & 0 & 5 & 7 & 4 & 0 & 0 & 25 \\  
    0 & 1 & 2 & 3 & 4 & 0 & 0 &  0 \\   
    0 & 0 & 3 & 2 & 1 & 1 & 0 & 10 \\
    0 & 0 & 2 & 5 & 3 & 0 & 1 & 15
  \end{bmatrix}
</math>

Select column 5 as a pivot column, so the pivot row must be row 4, and the updated tableau is 
:<math>
  \begin{bmatrix}
    3 & 0 &  7 &   1 & 0 & 0 & -4 &   15 \\  
    0 & 3 & -2 & -11 & 0 & 0 & -4 & -60 \\   
    0 & 0 &  7 &   1 & 0 & 3 & -1 &   15 \\
    0 & 0 &  2 &   5 & 3 & 0 &  1 &   15
  \end{bmatrix}
</math>

Now select column 3 as a pivot column, for which row 3 must be the pivot row, to get
:<math>
  \begin{bmatrix}
    7 & 0 & 0 & 0 & 0 & -7 & -7 & 0 \\  
    0 & 7 & 0 & -25 & 0 &  2 & -10 & -130 \\   
    0 & 0 & 7 &   1 & 0 &  3 &  -1 &   15 \\
    0 & 0 & 0 & 11 & 7 & -2 &   3 &   25
  \end{bmatrix}
</math>

The artificial variables are now 0 and they may be dropped giving a canonical tableau equivalent to the original problem:
:<math>
  \begin{bmatrix}
    7 & 0 & -25 & 0 &  -130 \\   
    0 & 7 &   1 & 0 &    15 \\
    0 & 0 &  11 & 7 &    25 
  \end{bmatrix}
</math>

This is, fortuitously, already optimal and the optimum value for the original linear program is&nbsp;−130/7. This value is "worse" than -20 which is to be expected for a problem which is more constrained.