14
The revised simplex method
Consider once again the standard equalityform linear program
maximize
c
T
x
=
z
subject to
Ax
=
b
x
≥
0
,
Corresponding to any basis
B
is a tableau that is unique except for the
order in which we write the equations. Theorem 12.4 gave a formula for that
tableau, that we can rewrite slightly as follows:
z

k
∈
N
(
c
k

y
T
A
k
)
x
k
=
y
T
b
x
B
+
A

1
B
k
∈
N
x
k
A
k
=
x
*
B
,
where the vector
y
solves the equation
A
T
B
y
=
c
B
, and the vector
x
*
B
(whose
components are the current values of the basic variables) is
A

1
B
b
. Recall that
N
is a list of the nonbasic indices, the vector
A
k
is the
k
th column of the
matrix
A
; as usual,
A
B
is the basis matrix, and analogously, the vector
c
B
has entries
c
i
as the index
i
runs through the basis
B
.
Using this notation, consider how an iteration of the simplex method
proceeds. We begin by choosing an entering variable
x
k
(where
k
∈
N
) with
positive reduced cost
c
k

y
T
A
k
. To do this, we must first calculate the vector
y
.
Step 1:
Solve the equation
A
T
B
y
=
c
B
for the vector
y
.
Step 2:
Choose an entering index
k
∈
N
such that
c
k
> y
T
A
k
.
The next step is the ratio test, for which we need the column vector whose
components are the coefficients of the entering variable
x
k
in the body of the
tableau. We can see from the formula for the tableau that this vector, which
we call
d
, is just
A

1
B
A
k
.
Step 3:
Solve the equation
A
B
d
=
A
k
for the vector
d
.
Step 4:
Calculate
t
←
min
x
*
i
d
i
:
i
∈
B, d
i
>
0
,
73
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and choose a leaving index
r
from among those indices
i
attaining this
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 Fall '07
 BLAND
 Linear Algebra, Linear Programming, Optimization, Simplex algorithm

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