# Simplex algorithm

The simplex algorithm operates on linear programs in the canonical form

A linear program in standard form can be represented as a *tableau* of the form

Let a linear program be given by a canonical tableau. The simplex algorithm proceeds by performing successive pivot operations each of which give an improved basic feasible solution; the choice of pivot element at each step is largely determined by the requirement that this pivot improves the solution.

Since the entering variable will, in general, increase from 0 to a positive number, the value of the objective function will decrease if the derivative of the objective function with respect to this variable is negative. Equivalently, the value of the objective function is increased if the pivot column is selected so that the corresponding entry in the objective row of the tableau is positive.

If all the entries in the objective row are less than or equal to 0 then no choice of entering variable can be made and the solution is in fact optimal. It is easily seen to be optimal since the objective row now corresponds to an equation of the form

By changing the entering variable choice rule so that it selects a column where the entry in the objective row is negative, the algorithm is changed so that it finds the minimum of the objective function rather than the maximum.

Once the pivot column has been selected, the choice of pivot row is largely determined by the requirement that the resulting solution be feasible. First, only positive entries in the pivot column are considered since this guarantees that the value of the entering variable will be nonnegative. If there are no positive entries in the pivot column then the entering variable can take any non-negative value with the solution remaining feasible. In this case the objective function is unbounded below and there is no minimum.

Columns 2, 3, and 4 can be selected as pivot columns, for this example column 4 is selected. The values of *z* resulting from the choice of rows 2 and 3 as pivot rows are 10/1 = 10 and 15/3 = 5 respectively. Of these the minimum is 5, so row 3 must be the pivot row. Performing the pivot produces

For the next step, there are no positive entries in the objective row and in fact

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

Select column 5 as a pivot column, so the pivot row must be row 4, and the updated tableau is

Now select column 3 as a pivot column, for which row 3 must be the pivot row, to get

The artificial variables are now 0 and they may be dropped giving a canonical tableau equivalent to the original problem:

This is, fortuitously, already optimal and the optimum value for the original linear program is −130/7.