A Trust Region Algorithm Using Curve-Linear
Searching Direction for Unconstrained
Optimization
Shu-ping Yang
School of Mathematical Science and Computing Technology, Central South University, Changsha, 410083 [email protected]
Xiu-gui Yuan, Zai-ming Liu
School of Mathematical Science and Computing Technology, Central South University, Changsha, 410083 [email protected]
Abstract—In the paper, aimed at the shortcoming of trust
region method, we proposed a algorithm using negative curvature direction as its searching direction. The convergence of the algorithm was given. Furthermore, combing trust region method and curve-linear searching techniques, a trust region algorithm, using general curve-linear searching direction, was proposed. We proved its efficiency and feasibility. The algorithm has adjustability and can select or update its searching direction according to the iteration. This allows the algorithm that has the properties of curve-linear searching method and the global convergence of trust region method. Finally, we indicate that some searching directions of common methods can be as a special searching direction of the general method.
Index Terms—nonlinear programming, unconstrained
optimization, trust region method, curve-linear searching method, searching direction, quadratic model, directions of negative curvature
I. UNCONSTRAINED OPTIMIZATION PROBLEMS[1] In many problems of unconstrained optimization such as :
(P1)
min ( )
nx R
f x
∈Its solution has been appealed to many peoples to do it. People created many algorithms aiming at question (P1) and presented some methods as the trust region method; Newton method; DFP method and BFGS method and so on. At these methods, people using straight line as its searching direction to searching line in general. with these methods , they have some defect as “saw tooth phenomenon” and “ local convergence”, some people presented the trust region method and curve-linear searching techniques to quality the global convergence of iterative method Curve-linear searching method is a way of searching direction down to a curve, which will avoid certain defects compared to rectilinear direction searching[2]. In trust region method, we gain new iteration step length based on a partial model of minimized objective function on a constrained ellipsoid domain centered at the current iteration point, and the
diameter of the ellipsoid is determined by the pattern of anticipation objective function of the model. Researches have been done by M. J. D. Powell [3], J. E. Dennis jr. and H. H. W. Mei[4], J. J. More [5], D. C. Sorensen[6, 7], D .C. Sorensen and J. J. More [8], G. A. Shulty. etc[9] and Yuan Y. etc[10-13]. It is needed to calculating
one of the partial models of
f x
( )
at the point ofx
k:1
( )
2
T T
k k k k
Q w
=
f
+
g w
+
w G w
It is solved by solving its equivalence problems: It equals to:
(P2)
min
{
Q
k( )
w
;
w
≤ ∆
k}
(P3)
λ
≥
0
s.t.G
k+
λ
I
is a positive semi-definite matrix,and
(
)
( )
k
k
G
I w
g
w
λ
λ
+
= −
⎧⎪
⎨
≤ ∆
⎪⎩
In many studies, it is generally assumed
that
G
k+
λ
I
is a positive definite matrix, so the problemoccurs when
G
k is not a positive semi-definite matrixand is possibly a local minimum point. If
g
k=
0
, we cangain zero solution to question(P3)exclusively, and the
iterative procedure discontinues. When
G
k is not apositive semi-definite matrix, More & Sorensen[8], which means situations hard to cope with appears,
indicating that
G
k has one negative eigenvalueλ
1 , whereg
k is orthogonal to the null space of(
G
k−
λ
1I
)
,
and
(
k 1)
k kG
−
λ
I
+g
< ∆
largest negative eigen value. The solution is that the iterative step length be selected as:
(
1)
k k k k k
p
= −
G
−
λ
I
+g
+
ξ ν
, whereν
k is eigenvector ofg
k relative toλ
1 .Selectξ
k s.tp
k= ∆
kThese methods above only involve single directions, so that if problems occur, the algorithm discontinues. Aiming at these problems, we present corresponding improved methods, using a curve-linear searching direction consisting of two descent directions.
II. IMPROVEMENT OF TRUST REGION ALGORITHMS
Algorithm 1
(1) Where
0
< ≤ <
µ η
1, 0
<
γ
1< <
1
γ
2,
x
0,
∆
0 (2) Given thatx
k,
∆
k, calculateg
k,
G
k(3) Decompose symmetric matrix
G
k by Bunch-parlettdecomposition method [13]
T k k k
k
L
D
L
G
=
Where
L
k is a triangular identity matrix(
( ) ( ))
1
,...
k n k
k
diag
d
d
D
=
(4) (a) If
g
k=
0
andd
(
i
n
)
k
i
0
1
,
2
...
)
(
≥
=
,
let
x
∗=
x
k, then stop. If not, go to theStep(b).
(b) If
G
kis not a positive semi-definite matrix, andg
k isorthogonal to the null space of
( ) 1
k k
G
−
λ
I
, where ) ( 1
k
λ
the largest negative eigenvalue, then go to the Step(c). If not, switch to Step (5).(c) Use Fletcher-Freeman[14]method to determine
negative curvature direction
d
k。(d) Let
x
k+1=
x
( )
α
k=
x
k+
α
kd
k,α
k is determinedby the conditions below:
(
)
(
)
[
k]
T k k k T k k
T k
k
d
d
d
G
d
g
d
x
f
+
≥
+
∇
α
η
α
(
)
( )
1
22
T T
k k k k k k k k
f x
+
α
d
≤
f x
+
µα
g d
+
µα
d G d
k
:
= +
k
1;
switch to Step (2) (5)i) Establish a secondary model:
( )
w
f
g
w
w
G
w
Q
k k kT T k2
1
+
+
=
ii) Find the solution of equation below:
( )
{
}
arg min
;
k k k
p
=
Q
w
w
≤ ∆
iii) Calculate
k k k
pred
ared
=
ρ
.
If
ρ
k<
µ
, then∆ = ∆
k:
γ
1 kand switch to Step ( );if
ρ
k≥
µ
, thenx
k+1=
x
k+
p
k,
k
:
=
k
+
1
;if
ρ
k>
η
, then∆
k+1:
=
γ
2∆
k, or else∆
k+1:
=
∆
kandswitch to Step (4). Theorem 1
If
f
R
R
n
→
:
, where
n
R
D
x
0∈
⊂
and
L x
( )
0=
{
x
∈
D f x
;
( )
≤
f x
( )
0}
is a compact subset ofD
.There is( )
0;
k k
x
∈
L x
G
≤
M
,
{ }
x
k is a iteration pointsequence based on Algorithm 1. Then
lim
T0 ,
lim
T0
k k k k k
k→∞
g d
=
k→∞d G d
=
(1)Prove:
If
{
}
( )
1
;
0,
Null(
1)
k
k k k k
D
=
x g
=
g
⊥
G
−
λ
I
and ( )
1
Null(
G
k−
λ
kI
)
is the null space of
( ) 1
k k
G
−
λ
I
.
Let
D
2=
{ }
x
k−
D
1, and{ }
x
k is an infinite point set.That if
{ }
x
k is finite, then based on Algorithm 1,(
k k)
kk
k
w
G
x
w
w
ared
=
−
+
θ
2
1
,
1
,
2
T T
k k k k k k
pred
=
p w
−
w G w
0,
0
k k
g
=
d
=
Then conflict arises. At least one of
D
1andD
2 is infinite.If
D
1is infinite,L x
( )
0 is a compact set, then{ }
x
khas at least one accumulation point. Let
x
k→
x
∗, basedon algorithm 1:,
⎥⎦
⎤
⎢⎣
⎡
+
−
≥
−
+ T k kk k k
T k k k
k
f
g
d
d
G
d
f
1 22
1
α
α
µ
and
d
kTG
kd
k≤
0
,
d
kTg
k≤
0
( )
x
f
∵
is continuous function, then(
→
∞
)
→
−
f
+k
f
k k 10
,
so that2
lim
k kT k k0, lim
k kT k0
k→∞
α
d G d
=
k→∞α
g d
=
Let
lim
k0
k→∞
α
=
, based on the continuity of(
x
kd
k)
f
+
α
∇
and algorithm 1, step (d):(
)
(
)
1
lim
1
T
k k k k
T T
k
k k k k k k
f x
d
d
g d
d G d
α
η
α
→∞
∇
+
=
≤ <
Then conflict arises. So
α
k≥
ε
,
(
k
≥
k
1)
andlim
kT k k0, lim
Tk k0
k→∞
d G d
=
k→∞g d
=
.If
D
2 is infinite, based on Algorithm 2:x
k∈
D
2 is drawn by trust region algorithm, and based on the convergence of the algorithm, obviouslylim
kT k k0
k→∞
d G d
=
is supported. Then (1) is proved.Based on the conclusion of Theorem 1 the theorem below can be proved:
Theorem 2 Under conditions of Theorem 1, and
{ }
d
k is limited, then∃ >
c
0
, s.t.(
→
∗
)
→
≥
c
g
x
x
∗k
d
k k,
k,
Then( )
=
0
∇
f
x
∗ andG
( )
x
∗ is positive semi-definite.III. GENERAL SEARCHING METHODS
To question (P1),
let
x
( )
α
=
x
+
φ
1( )
α
s
+
φ
2( )
α
d
, where( )
α
φ
1 andφ
2( )
α
are continuous functions under condition thatα β
∈
(whereβ
is a compact set inR
).Given thatφ
1( )
0
=
φ
2( )
0
=
0
(equals tox
( )
α
=
x
);s,
d
are descent directions off
( )
x
at the point ofx
.Create a secondary model off
( )
x
at the point ofx
:( )
p
f
( )
x
g
p
p
Gp
Q
T T2
1
+
+
=
(2)Where
p
( ) ( )
α
=
x
α
−
x
=
φ
1( )
α
s
+
φ
2( )
α
d
(3) To guarantee thatQ
( )
p
≤
f
( )
x
, then( )
α
α
β
φ
2≥
0
,
∈
φ
1( )
α
≥
0
(4) s.t. Algorithm 1 is modified. To make it further, we gaina general trust region algorithm using curve-linear searching direction for unconstrained optimization: Algorithm 2
Step1-3 are the same as (1)-(3) in Algorithm 1. Step4 is same as (4) (a)-(c) in Algorithm 1. (d) Let
x
k+1=
x
( )
α
k=
x
k+
p
( )
α
k .k
α
is determined by the conditions below:( )
(
)
(
)
TT T
k k k k k k k k
f x
p
α
d
η α
⎡
d G d
g d
⎤
∇
+
≥
⎣
+
⎦
( )
(
)
( )
1
22
k k
T T
k k k k k k
f x
p
f x
g d
d G d
α
µα
µα
+
≤
+
+
,:
1
k
= +
k
; switch to Step2.Definition 1
f
0( )
x
is defined as generalized derivativeof
f
( )
x
at the point ofx
,if 0
( )
(
)
( )
0
lim sup
t
f x t
f x
f
x
t
→ ++ −
=
.Definition 2
ζ
is defined as generalized successive derivative off
( )
x
at the point ofx
, if( )
x
≥
ζ
f
0 .Note generalized successive derivative off
( )
x
at the point ofx
asα
βf
( )
x
.That is( )
{
ζ
( )
ζ
}
α
βf
x
=
;
f
0x
≥
.Create the secondary model as below:
( )
( )
(
( )
( )
)
( )
( )
(
( )
( )
(
)
( )
( )
(
)
)
1 2
2
1 2
2
1 2
1 2
0
0
1
0
0
2
1
0
0
2
0
0
T T
T T
T
F
f x
g s
g d
g s
g d
s
d
G
s
d
α
µ
µ
α
α
ν
ν
α
µ
µ
µ
µ
=
+
+
⎡
⎤
+
⎣
+
⎦
⎡
⎤
+
⎣
+
⎦
⎡
⎤
×
⎣
+
⎦
(5)
Obvious (5) is a generalized pattern of (2). To satisfy that
F
( )
α
≤
f
( )
x
, then( )
0
0,
( )
0
0,
(
1, 2
)
i i
i
µ
≥
ν
≥
=
Let
( )
0( ) ( )
0( ) (
)
,
,
1, 2
i i i i
i
µ α
=
φ α ν α
=
ν α
=
, then:( )
1( )
1
0
0
lim
t
t
t
φ
µ
→ +
=
, 2 2( )
0
lim
t
t
t
φ
µ
→ +
=
(
i
=
1, 2 ,
)
ζ φ α
∈
( )
Theorem 3Assume that
f
( )
x
is a second order continuous differentiable function onR
n ,φ
1( ) ( )
α
,
φ
2α
are positive continuous functions onβ
(a compact subset onR
)→
R
,g
( )
α
∈
α
βf x
(
( )
α
)
is continuous from the right of pointx
.If:( ) ( )
0
,
1
,
0
≥
0
,
( )
0
≥
0
,
(
=
1
,
2
)
∈
µ
iν
iI
µ
,if
µ
1( )
0
=
µ
2( )
0
=
0
, and at least one of( ) ( )
0
,
20
1
ν
ν
is not zero, andd
,
s
are limited. And:( )
( )
( )
( )
1 2
1 2
2 2
0 0
lim
0 , lim
0
α α
φ α
φ α
µν
µν
α
α
→ +
≥
→ +≥
.Then
∃ ∈
α β
to satisfy the condition that:( )
(
)
( )
( )
( )
,
(
0,
)
f x
α
≤
f x
+
µ
⎡
⎣
F
α
−
f x
⎤
⎦
α
∈
α
(6) Prove:
Let
µ
1( ) ( )
0 ,
µ
20
≠
0
, ∵g
( )
α
∈
α
βf
(
x
( )
α
)
Based on the definition:( )
x
f
(
x
( )
α
)
g
( )
α
(
x
x
( )
α
)
( )
(
x
) ( )
f
x
( ) ( )
g
s
( )
g
d
f
α
−
≤
φ
1α
Tα
+
φ
20
T[ ]
(
( )
) ( )
( ) ( )
( )
( )
[
( ) ( )
]
[
F
f
x
g
s
g
d
F
f
x
]
x
f
x
f
Q
TT
+
−
−
≤
−
−
−
=
α
µ
α
φ
α
φ
α
µ
α
α
2 1 Then( )
( )
( )
( )
( )
(
) ( )
( )
1 0 0 2 0 1 2lim
lim
...
lim
lim
1
0
0
0
T
T
T T
Q
g S
F
f x
g d
g s
g d
α α α
α
φ α
α
α
φ α
α
µ
α
α
µ µ
µ
→ + → + → +≤
+
−
−
⎡
⎤
= −
⎣
+
⎦
<
If
µ
1( )
0
=
µ
2( )
0
=
0
, and at least one of( ) ( )
0
,
20
1
ν
ν
is not zero.
∵
g
( )
α
is continuous from the right of pointx
, that is( )
α
→
g
(
α
→
0
+
)
g
, andd
,
s
are limited( )
T T
g
α
s
g s
∴
→
,g
T( )
α
d
→
g
Td
(
α
→
0
+
)
∴If
ε
>
0
,ζ
>
0
;if
α ζ
<
, theng
T( )
α
s
≤
g
Ts
+
ε
,( )
α
d
≤
g
d
+
ε
g
T T( )
( ) ( )
( ) ( )
( )
( )
(
)
( )
( )
1 2 1 2 T TQ
g
d
g
s
F
f x
α
ϕ α
α
ϕ α
α
ϕ α
ϕ α ε µ
α
∴
=
+
+
+
−
⎡
⎣
−
⎤
⎦
And
g s
T≤
0
,
g d
T≤
0
,φ
1( )
α
≥
0
,
φ
2( )
α
≥
0
(
α
∈
β
)
( )
( )
( )
( )
( )
1 2 2 0 0 21 0 2 1
lim
lim
0
lim
0
0
T T
Q
g S
g d
α α α
α
φ α
α
α
φ α
µν
µν
α
→ + → + → +∴
≤
⎛
⎛
⎞
⎞
−
+
−
⎜
⎜
⎟
⎟
⎜
⎝
⎠
⎟
⎝
⎠
<
Above all,
∃ ∈
α β
to letα
∈
(
0,
α
]
and satisfy :( )
(
x
) ( )
f
x
[
F
( ) ( )
f
x
]
f
α
−
≤
µ
α
−
.Equation (6) is proved.
Based on Theorem 3, doing linear search along the curve
p
( )
α
can efficiently let the value of the function decrease, and in the meantime, its flexibility allow us to build various models according to the situations during calculating, so that it can reduce the occurrence of the phenomena so-called saw tooth .IV. THE SELECTION OF
s
kANDd
kThere are two methods of selecting
s
kandd
k: 1) Lets
k= −
g
k,s d
kT k≥
0
(that isg d
kT k≤
0
) 2) Lets
k= −
g
k,s
kTβ
kd
k≥
0
(that isg
kTβ
kd
k≥
0
)∵
2 2 2 2 2( )
(
)
(
) (
)
(
)(
)
kk k k k k
T T
k k k k k k k k k
k
k k k k
ared
f k
f x
s
d
g
s
d
s
d
x
s
d
α
α
α
α
α
α
α
α
=
−
+
+
=
+
+
+
∇
+
Where
x
k=
x
k+
θ α
(
k2s
k+
α
kd
k)
(0
≤ ≤
θ
1
)∴
2 2 2 2
( ) ( ) ( )( )
1 1
2( ) 3 4
2 2 aredk r k predk k T T
gk k sk k kd k sk k kd f x k sk k kd
T T T T T
g d g s d d s d s s
k k k k k k k k k k k k k k k k k
α α α α α α α α β α β α β = = + + + ∇ + + + + + Then:
i)where
g d
kT k≺
0
, 0lim
1
k kr
α →=
ii)where
g d
kT k=
0,
g
k≠
0,
s
k= −
g
k,lim 0
1
2 ( 2 ) 2 ( )( 2 ) 2
lim
1 1
2 3 4
0 ( )
2 2 1 r k k k T
g s d f x s d
k k k k k k k k k k
T T k T T
g s d d s d s s
k k k k k k k k k k k k k k
α α α α α α α α β α β α β = → + + ∇ + → + + + = iii)where
0
k ks
= −
g
=
,2 2 2
1
(
)
2
1
2
k Tk k k
k
T
k k k k
d
f x d
r
d
d
α
α
β
∇
=
0lim
1
k kr
α →∴
=
So the method is trusted. Equations relative to
α
:Let
w
=
α
2s
k+
α
d
k 2 3 41
( )
(
)
2
1
2
T T T
k k k k k k k
T T
k k k k k k
g d
g s
d
d
s
d
s
s
α
α
α
β
α
β
α
β
Φ
=
+
+
+
+
Suppose
{
2}
( )
α
min
( );
α α
s
kα
d
k kΦ
=
Φ
+
≤ ∆
(P4)Question (P4) equals to that:
∃ ≥
γ
0
, let:2 2
[ ( )( )] (2 ) 0, (7.1)
, (7.2)
T
k k k k k k
k k k
g I d s s d
s d
β
γ
α
α
α
α
α
⎧ + + + + = ⎪ ⎨ + = ∆ ⎪⎩In question (P4), the constrained condition 2
k k k
s
d
α
+
α
≤ ∆
can be further strengthened to:① Assume that
α
≥
0
∵
2 2 2
k k k k k k
s
d
s
d
s
d
α
−
α
≤
α
+
α
≤
α
+
α
∴
2k k k
s
d
Find the solutions of equation (8): 2
4
0
2
k k k k
k
d
d
s
s
α
−
+
+ ∆
≤ ≤
Therange of
α
is:2
4
0
2
k k k k
k
d
d
s
s
α
−
+
+ ∆
≤ ≤
(9)② Assume that
α
≤
0
2
k k k
s
d
α
−
α
≤ ∆
(8’) Find the solutions of equation (8’):2
2
4
2
4
2
k k k k
k
k k k k
k
d
d
s
s
d
d
s
s
α
−
+ ∆
≤
+
+ ∆
≤
The range of
α
is:2
4
0
2
k k k k
k
d
d
s
s
α
−
+ ∆
≤ ≤
(9’)According to the above, the range of
α
is: 24
0
2
k k k k
k
d
d
s
s
α
−
+
+ ∆
≤
≤
(7.1) can be transformed into
2 3
[2
]
3
[
] 2
[
]
0
T T T T
k k k k k k k k k
T T T T
k k k k k k k k k k
g d
g s
d d
d
d
s
d
s d
s
s
s s
α
γ
β
α
β
γ
α
β
γ
+
+
+
+
+
+
+
=
(10) Based on (10)
2
2[
] 3 3[
]
[2
]
0
T T T T
k k k k k k k k k k
T T T T
k k k k k k k k k
s
s
s s
s
d
s d
g s
d
d
d d
g d
β
γ
α
β
γ
α
β
γ
α
+
+
+
+
+
+
+
=
Let
,
2
,
3[
],
2[
]
T
k k
T T T
k k k k k k k
T T
k k k k k
T T
k k k k k
a
g d
b
g s
d
d
d d
c
s
d
s d
d
s
s
s s
β
γ
β
γ
β
γ
=
=
+
+
=
+
=
+
Then According to formula giving roots of cubic equations:
3 3 2
2 3
3
2 2
3 3 2
2 3
3 3
2 2
1 2 1 2 1
[ ( )] [ ( )] [ ( )]
2 3 27 2 3 27 3 3
1 2 1 2 1
[ ( )] [ ( )] [ ( )]
2 3 27 2 3 27 3 3
( )
bc c bc c c
a a b
d d d d d d d d
bc c bc c c
a a b
d d d d d d d d
α
ϕ γ
= − + + − + + −
+ − − + + − + + −
=
s.t.
Φ
( )
γ
=
α
2s
k+
α
d
kBased on Newton iteration method, we gain the root
γ
kofΦ
( )
γ
− ∆ =
k0
.Then
α
k=
ϕ γ
2(
k)
s
k+
ϕ γ
(
k)
d
kIf we Select
s d
k,
k s.t.,
0
T
k k k k
s
= −
g g d
=
. And then (7) can be transformed into
2
[2
]
3[
]
2[
]
0
T T T
k k k k k k k
T T T T
k k k k k k k k k k
g s
d d
d
d
s
d
s d
s
s
s s
γ
β
α
β
γ
α
β
γ
+
+
+
+
+
+
=
⇒
Whenα
≠
0
,2
2
3[
]
2[
]
0
T T T T T
k k k k k k k k k k k k
T T
k k k k k
g s
d d
d
d
s
d
s d
s
s
s s
γ
β
β
γ
α
β
γ
α
+
+
+
+
+
+
=
According to the formula of extraction of root :
3
4[ ]
2
9( ) 8( )(2 )
4[ ]
3
4[ ]
2
9( ) 8( )( 2 )
4[ ]
3
T gk k kd
T T
sk k ks sk sk
T T T T T T
Sk k kd sk k ks sk sk gk sk dk dk dk k kd
T T
sk k ks sk sk T
gk k kd
T T
gk k kg gk gk
T T T T T T
gk k kd gk k kg gk gk gk gk dk dk dk k kd
T T
gk k kg gk gk T
gk
β α
β γ
β β γ γ β
β γ β
β γ
β β γ γ β
β γ −
= ±
+
− + + +
+
= ±
+
− + − + +
+
= 9( )2 8 ( ) ( 2 ( )
4 [ ]
T T T T
d g d g I g g g d I d
k k k k k k k k k k k k k
T
gk k I gk
β β β γ β γ
β γ
± − + − + +
+
Among this formula, (
±
) is determined by (9) and (9’) . And then (7.2) can be transformed into
α
2d d
kT k+
α
4g g
kT k= ∆
kSo we can select
s d
k,
k conveniently to determinea
k,and to optimization searching calculate with algorithm 1or 2
ACKNOWLEDGMENT
This work was supported by Project 10971230 of the National Science Foundation.
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