An Interval Matrix Based Generalized Newton Method for Linear ...

Open Journal of Applied Sciences, 2015, 5, 443-449 Published Online August 2015 in SciRes. http://www.scirp.org/journal/ojapps http://dx.doi.org/10.4236/ojapps.2015.58044

An Interval Matrix Based Generalized Newton Method for Linear Complementarity Problems Hai-Shan Han, Lan-Ying College of Mathematics, Inner Mongolia University for the Nationalities, Tongliao, China Email: [email protected] Received 6 July 2015; accepted 9 August 2015; published 12 August 2015 Copyright © 2015 by authors and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

Abstract The penalty equation of LCP is transformed into the absolute value equation, and then the existence of solutions for the penalty equation is proved by the regularity of the interval matrix. We propose a generalized Newton method for solving the linear complementarity problem with the regular interval matrix based on the nonlinear penalized equation. Further, we prove that this method is convergent. Numerical experiments are presented to show that the generalized Newton method is effective.

Keywords Linear Complementarity Problem, Nonlinear Penalized Equation, Interval Matrix, Generalized Newton Method

1. Introduction The linear complementarity problem, denoted by LCP ( M , q ) , is to find a vector x ∈ R n such that x ≥ 0, Mx − q ≥ 0, x T ( Mx − q ) = 0

(1.1)

where M ∈ R n×n is a given matrix and q ∈ R n is a given vector. This problem serves as a unified formulation of linear and quadratic programming problems as well as of two-person (noncooperative) matrix-games, and has several important applications in economics and engineering sciences; see Cottle, Pang, and Stone [1] and its references.

How to cite this paper: Han, H.-S. and L.-Y. (2015) An Interval Matrix Based Generalized Newton Method for Linear Complementarity Problems. Open Journal of Applied Sciences, 5, 443-449. http://dx.doi.org/10.4236/ojapps.2015.58044

H.-S. Han, L.-Y.

There exist several methods for solving LCP ( M , q ) , such as projection method, multi splitting method, inte-

rior point method, and the nonsmooth Newton method, smoothing Newton method, homotopy method etc. See [1]-[6] and its references. In [7], it given a nonlinear penalized Equation (1.2) corresponding to linear complementarity problem (1.1). Find xλ ∈ R n such that 1

Mxλ + λ [ xλ ]+k = q

(1.2)

(

where λ > 1 is the penalized parameter, k > 0 , [u ]+ = max {u , 0} and yα = y1α , y2α , , ynα = y

( y1 , y2 , , yn )

T

)

T

for any

∈R . n

The nonlinear penalized problems (1.2) corresponding to the linear complementarity problem (1.1), which its research has achieved good results. In 1984, Glowinski [1] studied nonlinear penalized Equation (1.2) in R n , and proved the convergence of penalized equation that matrix A was symmetric positive definite. In 2006, Wang et al. [8] presented a power penalty function approach to the linear complementarity problem arising from pricing American options. It is shown that the solution to the penalized equation converges to that of the linear complementarity problem with matrix is positive definite. In 2008, Yang [7] proved that solution to this penalized Equations (1.2) converged to that of the LCP at an exponential rate for a positive definite matrix case where the diagonal entries were positive and off-diagonal entries were not greater than zero. The same year, Wang and Huang [9] presented a penalty method for solving a complementarity problem involving a secondorder nonlinear parabolic differential operator, and defined a nonlinear parabolic partial differential equation (PDE) approximating the variational inequality using a power penalty term with a penalty constant λ > 1 , a power parameter k >0 and a smoothing parameter ε . And prove that the solution to the penalized PDE converges to that of the variational inequality in an appropriate norm at an arbitrary exponential rate of the form O([λ − k + ε (1 + λε 1/ k )]1/2 ) . Under some assumptions, Li [10] [11] proved that the solution to this equation converges to that of the linear complementarity problem with A is a strict row diagonally dominant upper triangular P-matrix when the penalty parameter approaches to infinity and the convergence rate was also exponential. It is worth mentioning that the penalty technique has been widely used solving nonlinear programming, but it seems that there is a limited study for the LCP. Although the studies solving for the linear complementarity problem based on the nonlinear penalized equation have good results. But there is no method that is given for solving the nonlinear penalized equation. Throughout the paper, we propose a generalized Newton method for solving the nonlinear penalized equation with under the suppose [ M , M + λ I ] is regular. So the method can better to solve linear complementarity problem. We will show that the proposed method is convergent. Numerical experiments are also given to show the effectiveness of the proposed method.

2. Preliminaries Some words about our notation: I refers to the identity matrix, and all vectors are column vectors, x T refers to the transpose of the x ,

⋅

represents the 2-norm. For x ∈ R n ,

[ x ]+ = max { x, 0} , that generalized Jaco-

D ( x ) , where D ( x ) denotes diagonal matrix, On the diagonal elements with component 1.0 or bian ∂ [ x ]+ =

σ ∈ [ 0,1] corresponding to the component of x which is positive , negative or zero, respectively.

The definition of interval matrix arises from the linear interval equations [12], given two matrices A = ( Aij )

I and A = ( Aij ) , an interval matrix A=  A, A= 

i, j . A

I

is called regular if each A ∈ A

I

Lemma 1: Assume that interval matrix

{ A : A ≤ A ≤ A} , where

A ≤ A refers to Aij ≤ Aij for each

is nonsingular.

[M , M + λ I ]

is regular. Then for diagonal matrix D ( xλ ) and real

number λ > 1 , M + λ D ( xλ ) is nonsingular. Proof: By definition of D = diag ( x ) , for every x, we have D ∈ [ 0, I ] then

444

M + λ D ∈[M , M + λ I ]

H.-S. Han, L.-Y.

By the assumptions, we have M + λ D ( xλ ) is nonsingular.

□

Lemma 2: Let x, y ∈ R . Then n

[ x ]+ − [ y ]+

≤ x− y

Definition 1 [1]: A matrix M ∈ R n×n is said to be a P-matrix if all its principal minors are positive. Lemma 3 [1]: A matrix M ∈ R n×n is a P-matrix if and only if the LCP ( q, M ) has a unique solution for all vectors q ∈ R n .

3. Generalized Newton Method In this section, we will propose that a new generalized Newton method based on the nonlinear penalized equation for solving the linear complementarity problem. Because when k > 1 , penalty term of the nonlinear penalized equation (1.2) is none Lipschitz continuously, hence we only discusses a case that k = 1 . So from nonlinear penalized equation (1.2), we have that

Mxλ + λ [ xλ ]+ = q

(3.1)

These case penalty problems for the continuous Variational Inequality and the linear complementarity problems are discussed in [2] [13]. Let us note

f ( xλ ) = Mxλ + λ [ xλ ]+ − q, λ > 1

(3.2)

Thus, nonlinear penalized equation (3.1) is equivalent to the equation f ( xλ ) = 0 . A generalized Jacobian ∂f ( xλ ) of f ( xλ ) is given by

∂f ( xλ ) = M + λ D ( xλ ) where D ( xλ ) = ∂ [ xλ ]+ is a diagonal matrix whose diagonal entries are equal 1,0 or a real number σ ∈ [ 0,1] depending on whether the corresponding component of xλ is positive, negative, or zero. The generalized

Newton method for finding a solution of the equation f ( xλ ) = 0 consists of the following iteration:

( )

( )( x

f xλi + ∂f xλi

Then

i +1

λ

( )

)

− xλi = 0

( )

(3.3)

( )

∂f xλi xλi +1 = ∂f xλi xλi − f xλi

( )

( )

 M + λk D xλi  xλi +1 = M + λk D xλi  xλi −  Mxλi + λk  xλi  − q   +      

( )

= λk D xλ xλi − λk  xλi  + q + i

( )

Since D xλi xλi =  xλi  , consequently +

( )

 M + λk D xλi  xλi +1 = q   q equivalent to 0 ≤ z ⊥ Az + b ≥ 0 , where Proposition 1 Mxλ + λ [ xλ ]+ = z =Mxλ − q, A =( M + λ I ) M −1 , b =λ M −1q

2 xλ , xλ ≥ 0 1 Proof: Since xλ + xλ = , then [= xλ ]+ ( xλ + xλ ) , we have  < 0, x 0 2 λ 

445

(3.4)

H.-S. Han, L.-Y.

Mxλ + λ [ xλ ]+ − q= Mxλ + λ

1 1  1  xλ + xλ ) − q=  M + λ I  xλ + λ xλ − q= 0 ( 2 2  2 

( 2M + λ I ) xλ + λ

xλ − 2q = 0

By [14], its equivalent to

0 ≤ ( 2 M + λ I ) xλ − λ xλ − 2q ⊥ ( 2 M + λ I ) xλ + λ xλ − 2q ≥ 0 0 ≤ Mxλ − q ⊥ ( M + λ I ) xλ − q ≥ 0 z Mxλ − q then Let=

(3.5)

( M + λ I ) xλ − q= ( M + λ I ) M −1 ( z + q ) − q= ( M + λ I ) M −1 z + λ M −1q .

q has a unique solution if and only if the interval matrix Proposition 2 Mxλ + λ [ xλ ]+ = regular. Proof: Since M , M + λ I ∈ [ M , M + λ I ] , by theorem 1.2 of [12], if

(M + λI ) M

−1

[M , M + λ I ]

is P-matrix, which implies that the LCP has a unique solution for any q ∈ R

□

[M , M + λ I ]

is

is regular, then n

[1], from the re-

lation between the (3.1) and the LCP (3.5), we can easily deduce that the (3.1) is uniquely solvable for any q ∈ Rn . □ Algorithm 3 Step 1: Choose an arbitrary initial point xλ00 ∈ R n , ε > 0 and given λ0 > 1 , µ > 1 , σ ∈ [ 0,1] , let k := 0 ; Step 2: for the λk , computer xλi +k 1 by solving  M + λk D xλi  xλi +1 = b k  k 

( )

Step 3: If Step4: If

(3.6)

xλi +k 1 − xλi k < ε , let xk = xλi k go to step 4. Otherwise, i = i + 1 go to step 2.

xk +1 − xk < ε

Mxk − q < ε , terminate, x = xk is solution of LCP. Otherwise let

and

λk +1 = µλk , xλk +1 = xk let k =: k + 1 , go to 2. 0

4. The Convergence of the Algorithm We will show that the sequence

{x }

∞ i λk i =1

generated by the generalized Newton iteration (3.6) converges to an

accumulation point xk associated with λk . First, we establish boundness of the sequence

{x } i λk

for any

λk > 0 generated by the Newton iterates (3.6) and hence the existence of accumulation point at each generalized Newton iteration. Theorem 3: Suppose that the interval matrix

[ M , M + λk I ]

is regular. Then, the sequence

{x } i λk

generated

by Algorithm 3 is bounded. Consequently, there exits an accumulation points xk such that b.  M + λk D ( xk )  xk =

Proof: Suppose that sequence

{x } ⊂ {x } ij

λk

i

λk

{x } i λk

is unbounded, Thus, there exists an infinite nonzero subsequence

such that

D(x ) { x } → ∞,= ij λk

ij λk

D and D ∈ [ 0, I ]

where D is main diagonal element of diagonal matrix which is 1, 0, σ ∈ [ 0,1] .  xi j   λ  We know subsequence  i k  is bounded. Hence, exists convergence subsequence and assume that conxj   λk   vergence point is x , and satisfy

446

H.-S. Han, L.-Y.

( Letting j → ∞ yields

A + λk D

i

xλjk

)

b = i xλjk

ij

xλk

( M + λ D ) x =0,

x =1.

k

[ M , M + λk I ]

Since M + λk D ∈ [ M , M + λk I ] and the interval matrix

is regular, we know that

( M + λ D ) is exists and hence x = 0 , contradicting to the fact that x = 1 . Consequently, the sequence {x } is bounded and there exists an accumulation point x of {x } such that −1

k

i λk

i λk

k

( M + λ D ( x ))

= xk

k

−1

□

b

k

Under a somewhat restrictive assumption we can establish finite termination of the generalized Newton iteration at a penalized equation solution as follows. −1 1 Theorem 4: Suppose that the interval matrix [ M , M + λk I ] is regular and M + λk D xλi k holds < 2λk

(

( ))

for all sufficiently large λk , then the generalized Newton iteration (3.6) linearly converges from any starting point xλ0k to a solution xk of the nonlinear penalized equation (3.1).

( )

Proof: Suppose that xk is a solution of nonlinear penalized equation. By the lemma 1, M + λk D xλk

( )

nonsingular. To simply notation, let D = D ( xk ) , D i = D xλi k

( M + λ D ( x )) x

and since

k

k

k

is

= b,

( )

 M + λk D xλi  xλi +1 = b have Dxk = [ xk ]+ , D i xλi k =  xλi k  , hence k  k  +

(

)

(

M xλi +k 1 − xk =−λk D i xλi +k 1 + b + λk Dxk − b =−λk D i xλi +k 1 − [ xk ]+

( ( = −λ (  x  = −λ (  x 

)

= −λk D i xλi +k 1 − xλi k + xλi k − [ xk ]+

( ) ) −[x ] ) − λ D ( x − x ) − λ D ( x − x ) −λ (  x  − [ x ] ) − λ D ( x − x ) (M + λ D )( x − x ) = ( x − x ) = ( M + λ D ) ( −λ (  x  − [ x ] ) + λ D ( x − x ) ) ( M + λ D ( x )) < 21λ and by the lemma 2, we have i

k

λk

k

λk

+

− [ xk ]+ − λk D i xλi +k 1 − xk + xk − xλi k

i

i

k

i +1 λk

i

k

i

k +

+

k

i λk

k

i −1

i +1

λk

since

)

)

k

i

k

+

k +

i

k

k +

+

k

i λk

k

i

λk

k

λk

k

i

k

i +1

i

λk

k

i

λk

k

k

−1

λk

k

(x

i +1

λk

− xk

) ≤ (M + λ D ) i

k

−1

( 2λ ( x

i

− xk

))

) (x

i

− xk

) < (x

(

= 2λk ⋅ M + λk D i

k

−1

λk

λk

i

λk

− xk

)

(4.2)

Letting i → ∞ and taking limits in both sides of the last inequality above, we have

(x (x

i +1

− xk

i

− xk

λk

λk

) )