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In this section, we prove the existence
of the step length using the method described in
[18]. See also [11] for an analysis of the semidefinite case.
Here is the algorithm which we analyze in this section.
Algorithm 2.1
Given
,
,
for
Set
Solve (
16) for
Find the largest
, such that
Update
end.
Denote the maximal and minimal eigenvalues of by
and
,
respectively. From [9], we have that
Since is
the diagonal matrix consisting of
, it follows
that
and
are
obtained from
We now start to find upper bounds for vectors. We consider
Algorithm 2.1, but we use (16) with the
NT direction instead of (15).
First, we introduce a useful quantity defined by

(21) 
where is the step length and is the
centering parameter in the th iteration. According to Lemma
3.13 in [4] and the definitions of
in
Algorithm 2.1, we have
We choose the initial point by the following:

(22) 
where
The
following lemma gives the bound on
Lemma 2.2
Let the initial point be given by (
22). For all
, there exists
, such that
Proof.
For simplicity, we ignore the
iteration index
on the vectors. Let
be any primaldual
solution and define
It is easy to verify that
satisfies the following three equations:

(23) 
Hence,

(24) 
From (
24), we have

(25) 
Since
is selfdual and
is a solution, we
have that
After inserting the above equations into (
25), we obtain
the following inequality,
Next, we define the constant
by
where
are the first two entries of
By (
22), we have that
One can also
derive the following inequality based on the definition of the
initial point,
Therefore,
Since
,
Furthermore, because
it
follows that
Since we specifically discuss the system (16) with
the NT search direction, we define the following notation for the
scaled vectors at the th iteration:
The next lemma
gives bounds on the scaled vectors
and
Lemma 2.3
Applying the NT scaling scheme to (
16), i.e., using
, there is a positive constant
, such that
Proof.
Let us define
It is easy to verify that
satisfies (
23)
and (
24). In order to obtain the fourth equation in
(
16), we use the following expression
Since we consider the NT direction, i.e.,
, we
multiply by
from the left and obtain
Similarly,
We now consider the 2norm of the vector,
On the other hand,
Combining the above two observations, we obtain an upper bound,
We now need to show that each term in the right hand side of the
above inequality is
First, we consider
From the property of
and the
definition of
, we derive
Hence,
Thus,
For the remaining terms in the inequality, we have

(26) 
Since
max
, we find upper bounds for
and
So,
since
Thus,
Now (
26) is bounded by
max
Hence,
If the initial point satisfies (22), then the next
lemma follows immediately.
Lemma 2.4
Under the same conditions as in the previous lemma and if the
initial point satisfies the (
22), then
and
Proof.
From(
22), we have that
where
Then, we can derive the
following two estimates,
It is easy to check that
The next lemma shows the existence of the step length
Lemma 2.5
There exists
such that
for all
where
is defined as
Proof.
For simplicity, we consider the quadratic cone case
only. By Lemma
2.4, we have
From
the last two equations of the system (
16), we
obtain
In order to consider
we estimate
by
We now consider the conditions of
. We note that
If
holds, then
Therefore,
min 
(27) 
A similar result in [9] is based on feasible initial
points, however, the condition (27) is good for
infeasible initial points.
Next: Application Problems
Up: Interior Point Methods for
Previous: Introduction and Definitions
Hans D. Mittelmann
20030910