After introducing slack variables s and t, the initial basic feasible solution is represented by the table below (basic variables are s = 6 and t = 6, and the objective function value is 0).
|
|
–4
|
–6
|
0
|
0
|
0
|
|
s
|
3
|
2
|
1
|
0
|
6
|
|
t
|
2
|
3
|
0
|
1
|
6
|
|
|
x
|
y
|
s
|
t
|
RHS
|
After some simplex iterations, the following table is obtained
|
|
0
|
0
|
0
|
2
|
12
|
|
s
|
5/3
|
0
|
1
|
–2/3
|
2
|
|
y
|
2/3
|
1
|
0
|
1/3
|
2
|
|
|
x
|
y
|
s
|
t
|
RHS
|
From this, one can conclude that
(a) The LP has a unique optimal solution
(b) The LP has an optimal solution that is not unique
(c) The LP is infeasible
(d) The LP is unbounded