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1498 JOURNAL OF COMPUTERS, VOL. 8, NO. 6, JUNE 2013
Step 3
Computation of the feasible direction with descent
Where
δ
ek
k
d :
k k T −1
d = d0 − δ
kAk( Ak Ak)
ek
(9)
T | Jk
|
= (1, ,1) ∈R
, and
|| d || ( d ) H d
k k T k
0 0 k 0 k T −1
T k
k
= , μ =−( Ak Ak) Ak∇f( x )
kT k
2 | μ ek
||| d0
|| + 1
Step 4. Computation of the high-order revised direction
k
d :
where
d A A A d e g x d (10)
k T −1
k k k
=− ( ) (||
0
|| τ
k k k k
+
J
( + )),
k
k k k k k k T k
g
( x + d ) = g ( x + d ) −g ( x ) −∇g ( x ) d .
J J J
k k k J k
Step 5. Line search:
Compute t k
, the first number t in the sequence
1 1 1
{1, , , ,...}
satisfying
2 4 8
2
f ( x k + td k + t d k ) ≤ f( x k ) + αt∇f( x k ) T d
k , (11)
g x td t d j I
(12)
k k 2 k
j
( + + ) ≤0, ∈ .
Step 6. Update:
Obtain H
k +
by updating the positive definite matrix
1
H k
using some quasi-Newton formulas. Set
k 1 k k 2 k
x + = x + t d + t d , and k = k+ 1 . Go back to step 1.
k
Throughout this paper, following basic assumptions
are assumed.
H2.1 The feasible set X ≠Φ, and functions f ( x ),
g
j
( x),
j∈ I are twice continuously differentiable.
H2.2 ∀ x ∈ X , the vectors { ∇g ( x), j∈I( x)}
are
linearly independent.
Lemma 2.1 Suppose that H2.1and H2.2 hold, then
1) For any iteration, there is no infinite cycle in step 1.
2) If a sequence { x k } of points has an accumulation
point, then there exists a constant _ ε > 0
ε
kik ,
_
> ε for k large enough.
j
such that
Proof.
1) Suppose that the desired conclusion is false, that is
to say, there exists some k, such that there is an infinite
cycle in Step 1, then we obtain, ∀ i = 1, 2, , that Aki
,
is
not of full rank, i.e., it holds that
det( A A ) = 0, i = 1,2, , (13)
T
ki , ki ,
And by (6), we can know that Jki
, + 1
⊆ Jki
,
. Since there
are only finitely many choices for J
ki ,
⊆ I , it is sure that
~
J ≡ J L for i large enough. From (6) and (13),
ki , + 1 ki , k
with i →∞, we obtain
~
L k
k = I ( x ), det( A A ) = 0.
T
k k
I( x ) I( x )
This is a contradiction to H 2.2, which shows that the
statement is true.
2) Suppose K is an infinite index set such that
*
{ x } → x . We suppose that the conclusion is false,
k k∈K
i.e., there exists
for
~
L
k
Let
~
'
'
K ⊆ K K
(| | =∞ ) , such that
ε
ki
→ k∈K k →∞
'
,
0, , .
k
Lk = Jk, i k −1. From the definition of ε
ki , k
, it holds,
k∈
K ' ,
k large enough, that
~
T T k
~ ~ ε
ki , k j k
Lk
Lk
det( A A ) = 0, −2 ≤ g ( x ) ≤0, j∈ L . (14)
Since there are only finitely many choices for sets
⊆ I , it is sure that there exists
such that
~ ~
''
k
, ( )
'' ' ''
K ⊆ K (| K | =∞ ) ,
L ≡ L k∈ K , for k large enough.
Denote ~ *
A = { ∇g ( x )| j∈ L
~
} , then, let
from (14), it holds that
j
~ T ~ ~
* *
=
j
= ∈ ⊆
k∈K '' , k →∞ ,
det( A A) 0, g ( x ) 0, j L I( x ).
This is a contradiction to H 2.2, too, which shows that
the statement is true.
Lemma 2.2 For the QP sub-problem (8) at x
k , if
k
d
0
= 0 , then x k
k
is a KKT point of (1). If d0 ≠ 0 , then
k
x computed in step 4 is a feasible direction with descent
k
of (1) at x .
Proof.
By the KKT conditions of QP sub-problem (8), we
have
k k k
∇ f( x ) + Hkd0
+ Akb
= 0,
k k T k
p +∇ g ( x ) d = 0, j∈J
,
j j 0
k
k
If d
0
= 0 , we obtain
k k k
∇ f ( x ) + A b = 0, p = 0, j∈
J ,
k j k
k
Thereby, from (7) and x ∈ X,
∀ k implies that
k
k
g ( x ) = 0, v ≥0, j∈
J .
j j k
b k B k k
k
f x v
In addition, we have =− ∇ ( ) = , in a word,
we obtain
k k
∇ f ( x ) + A b
k k
= 0, g ( x ) = 0, b ≥0, j∈
J ,
k j j k
k
Let bj
= 0, j∈ I \ Jk
, which shows that x k is a KKT
point of (1).
k
If d0 ≠ 0 , we have
k T k T k k
g ( x ) d = A d = −p − δ e ,
So,
Jk
k k k
k T k k T k kT k
∇ f ( x ) d = − ( d ) H d + b p ,
0 0 k 0
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