The phase I procedure uses the true right-hand side column and the artificial objective row. Any negative entry in the right-hand side column may be chosen as the leaving variable. Once picked, a ratio test must be done, either by eyeballing or on scratch paper, to determine the entering variable using the artificial objective row. Once the leaving and entering variables have been determined, click on the corresponding button to actually do the pivot. If the entering variable was chosen correctly, the artificial objective row will still have all negative entries after the pivot. If it was chosen incorrectly, there will be positive entries. These will be highlighted in yellow. If a yellow highlight appears in the artificial objective row you might want to click on the Undo button to undo the last pivot. Continue with the phase I procedure until all of the entries in the right-hand-side column are nonnegative; that is, until there are no more pink highlights in this column. At this point, you are ready for phase II.

Phase II is the ordinary simplex method applied to a primal-feasible
dictionary. For this phase, we ignore both the
artificial column and the artificial row.
Possible entering variables are highlighted in pink in the objective row.
Choose one of these as the entering variable and then do a ratio test to
determine the corresponding leaving variable. Click on the associated pivot
button to do the pivot. If the leaving variable was selected correctly, then
the new dictionary will still be primal feasible. If it is infeasible, you
will see at least one pink cell in the right-hand-side column. If that
happens, you might want to hit the Undo button to back up one iteration.
Continue with these phase II pivots until there are no more pink highlights
remaining. At this point, the dictionary is optimal. To tell the applet that
you recognize that the current dictionary is optimal, click on the
*Optimal* button. The applet will confirm if you are correct.

*Infeasibility*. If during phase I a ratio test reveals that no
variable can enter the basis, then the problem is infeasible. If this happens,
click on the *Infeasible* button to tell the applet that you recognize
this situation.
*Unboundedness*. If during phase II a ratio test reveals that no
variable can leave the basis, then the problem is unbounded. If this happens,
click on the *Unbounded* button to tell the applet that you recognize
this situation.

The applet shows in the lower-left corner
the interval of *mu* values for which the current
dictionary is optimal.
This interval is determined by the requirements that each coefficient of
a variable in the objective function be nonpositive and each right-hand side be
nonnegative. These requirements translate into a number of inequalities on
*mu* and the interval consists of those *mu* values that satisfy
every inequality.

The first step is to identify which row or column is responsible for the
lower bound on *mu*. If it is a column, then the variable associated
with that column will be the entering variable and the leaving variable must be
determined using a usual ratio test. On the other hand, if it is a row, then
the basic variable associated with that row will be the leaving variable and
the entering variable must be determined using a ratio test for the dual
simplex method. Once the entering and leaving variable have been identified,
do the pivot by clicking on the correct pivot button. As always,
the dictionary will automatically update. In addition, a new interval of
*mu* values will appear. If the correct pivot button was selected
(generally there will only be one correct choice), then the new interval will
be an interval adjacent and to the left of the old interval. In most other
cases, the lower limit of the interval will be larger than the upper limit.
That is, the true interval will be empty. In this case, the interval will be
highlighted in yellow to indicate that something is wrong. You probably will
want to hit the *Undo* when this happens.

Once the *mu* interval includes zero, the current dictionary is optimal
for the original problem.