ORIE 3300/5300
Fall 2011
Example of the Simplex Method
Consider the following linear program in standard equality form:
maximize
x
1
+ 3
x
2
subject to
x
1
+
x
2
+
x
3
= 3

x
1
+
x
2
+
x
4
= 1
2
x
1

x
2
+
x
5
= 3
x
1
,
x
2
,
x
3
,
x
4
,
x
5
≥
0
Let us see how this linear program can be solved using the Simplex Method.
Introducing the new variable
z
=
x
1
+ 3
x
3
to keep track of the objective
value, we get the initial system of equations
z

x
1

3
x
2
= 0
x
1
+
x
2
+
x
3
= 3

x
1
+
x
2
+
x
4
= 1
2
x
1

x
2
+
x
5
= 3
This system is already a tableau. The basis is
{
3
,
4
,
5
}
, the corresponding
basic feasible solution is
x
= [0 0 3 1 3]
T
, and the objective value is
z
= 0.
(
Make sure you understand and agree with these statements!)
In particular,
Step 0 of the Simplex Method (initialization) is already complete. Let us
carry out the other steps.
Iteration 1
Step 1
(Check optimality): The reduced costs are ¯
c
1
= 1, ¯
c
2
= 3, so
do not
stop
.
Step 2
(Choose entering index): Choose
k
∈
N
with ¯
c
k
>
0. Currently
N
=
{
1
,
2
}
and the reduced costs are both positive, so either index would
work. But we have to pick one, so let us choose the smallest index (“smallest
subscript rule”),
k
= 1.
1
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View Full DocumentStep 3
(Check for unboundedness): Since
k
= 1, we have
¯
A
k
=
¯
a
3
k
¯
a
4
k
¯
a
5
k
=
1

1
2
,
which is not
≤
0. So
do not stop
. (Note: the rows of the tableau are indexed
by the basic indices
i
∈
B
=
{
3
,
4
,
5
}
, and so are the elements of
¯
A
k
. This is
why we write ¯
a
3
k
, ¯
a
4
k
, ¯
a
5
k
instead of, for instance, ¯
a
1
k
, ¯
a
2
k
, ¯
a
3
k
.)
Step 4
(Choose leaving index): Compute the ratios
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 '08
 TODD
 Optimization, basic feasible solution, basic index

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