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The dual simplex method
We have seen how, when a linear programming problem in standard equality
form has a nondegenerate basic optimal solution, the optimal solution of the
dual problem provides sensitivity information with respect to small changes
to the righthand sides. We begin this section by revisiting the question of
exactly what we mean by “small” in this context.
We begin with the example from Section 18, which was a simple modifi
cation of the example in Section 4, once again:
maximize
x
1
+
x
2
subject to
x
1
≤
2
x
1
+
3
x
2
≤
4
x
1
,
x
2
≥
0
.
We can picture the optimal solution graphically as shown below.
If we introduce slack variables as usual, the problem becomes
maximize
x
1
+
x
2
=
z
subject to
x
1
+
x
3
=
2
x
1
+
3
x
2
+
x
4
=
4
x
1
,
x
2
,
x
3
,
x
4
≥
0
.
After solving the problem by the simplex method, we arrive at the following
tableau:
z
+
2
3
x
3
+
1
3
x
4
=
8
3
x
1
+
x
3
=
2
x
2

1
3
x
3
+
1
3
x
4
=
2
3
.
113
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The corresponding basic feasible solution is nondegenerate and optimal. We
can easily check, using the revised simplex method calculations, or comple
mentary slackness, that the optimal solution of the dual problem is
y
*
1
=
2
3
, y
*
2
=
1
3
. Hence, if, for example, the righthand side 4 changes to 4 +
for
any small number
, then by Theorem 18.1 the optimal value increases by
y
*
2
=
3
.
Reviewing the proof of Theorem 18.1, we see the explanation: the new
tableau corresponding to the old optimal basis
B
= [1
,
2] is unchanged except
for the righthand sides:
z
+
2
3
x
3
+
1
3
x
4
=
8
3
+
3
x
1
+
x
3
=
(
A

1
B
b
)
1
x
2

1
3
x
3
+
1
3
x
4
=
(
A

1
B
b
)
2
where the vector
b
is the new righthand side in the linear programming
problem:
b
= [2
,
4 +
]
T
. After calculating the inverse of the basis matrix
A

1
B
, we find
A

1
B
b
=
1
0

1
3
1
3
2
4 +
=
2
2
3
+
3
,
so the new tableau is
z
+
2
3
x
3
+
1
3
x
4
=
8
3
+
3
x
1
+
x
3
=
2
x
2

1
3
x
3
+
1
3
x
4
=
2
3
+
3
.
As predicted by Theorem 18.1, this tableau remains feasible, and hence op
timal, providing
is small.
In fact, we can be more precise:
the tableau
remains optimal providing
≥ 
2.
We can interpret this observation geometrically. If the righthand side 4
changes slightly to 4+ , then the diagonal line (corresponding to
x
1
+3
x
2
= 4)
in the picture of the feasible region above moves slightly, but the optimal
solution remains at the intersection of this line with the vertical line
x
1
= 2.
But what if
is not small enough to guarantee feasibility in the tableau?
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 '08
 TODD
 Linear Programming, Optimization, Algebraically, dual simplex method

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