4
Basic feasible solutions
To summarize the main idea from the last section, a basis for an
m
by
n
matrix
A
is a list of numbers chosen from
{
1
,
2
, , .
. . , n
}
such that the matrix
A
B
with columns indexed by this list is invertible. The corresponding basic
solution of a system
Ax
=
b
is the unique solution of this system satisfying
x
j
= 0 for all indices
j
6∈
B
.
Consider now the constraints of a linear programming problem in stan
dard equality form:
Ax
=
b
x
≥
0
.
A
basic feasible solution
of this system is a basic solution that is also feasi
ble. Such solutions play a special role in linear programming. To illustrate,
consider the simple example (2.1):
(4.1)
maximize
x
1
+
x
2
subject to
x
1
≤
2
x
1
+ 2
x
2
≤
4
x
1
,
x
2
≥
0
.
As before, we can sketch the feasible region:
If we transform the constraints of this linear programming problem to stan
dard equality form, by introducing slack variables as we described in Section
2, we arrive at the following system:
x
1
+
x
3
= 2
x
1
+ 2
x
2
+
x
4
= 4
x
1
,
x
2
,
x
3
,
x
4
≥
0
.
25
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View Full DocumentFeasible solutions of this system correspond onetoone with feasible solutions
of the linear programming problem (4.1), in an obvious way. If we think of
this new system as
Ax
=
b
, the matrix
A
has ﬁve possible bases, not including
permutations: [1
,
2]
,
[1
,
3]
,
[1
,
4]
,
[2
,
3]
,
[3
,
4]. The ﬁve corresponding basic
solutions are
x
1
x
2
x
3
x
4
=
2
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 '08
 TODD
 Linear Programming, Vector Space, Optimization, Line segment, Constraint, extreme point

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