7 The simplex algorithm by example.
7.1 A second pivoting rule.
Consider the primal problem
12
12 12
12
39
max 2 3 when 2 8
,0
xx
xx xx
xx
+≤
++≤
≥
.
The corresponding initial tableau is given by
13109
21018
23000
.
Since both 1
x and 2
x are non-basic variables, it is possible to pivot using either variable. It is
reasonable to examine the current objective and choose the variable with the largest coefficient.
In the example 2
x has the larger coefficient. In terms of the tableau, the second pivoting rule
selects the column with the largest value in the bottom row to the left of the vertical line. This
choice leads to the ‘fastest’ increase in the objective.
7.2 The next tableau.
Combine the two pivoting rules and pivot to get
1/3 1 1/3 0 3
5/3 0 1/3 1 5
10 10 9
−
−−
.
The new solution is given by 12
0, 3
xx==
. There is a positive coefficient in the ‘new’ objective
11
9xy+−. There is a need for a second pivot.
7.3 The final tableau.
Apply the original pivoting rule and get the tableau
0 1 2/5 1/5 2
1 0 1/5 3/5 3
0 0 4/5 3/5 12
−
−
−− −
.
The ‘new’ objective is given by 43
55
12
12 yy−−
with no hope of improvement. This must be the
final tableau and the proposed solution is given by the feasible 12
ˆˆ
3, 2xx==. To verify the
solution observe that the dual problem is given by
