# A Penalized-Equation-Based Generalized Newton Method for Solving Absolute-Value Linear Complementarity Problems

Authors: Yuan Li; Hai-Shan Han; Dan-Dan Yang
Journal: Journal of Mathematics
 (2014)

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## Abstract

We consider a class of absolute-value linear complementarity problems. We propose a new approximation reformulation of absolute value linear complementarity problems by using a nonlinear penalized equation. Based on this approximation reformulation, a penalized-equation-based generalized Newton method is proposed for solving the absolute value linear complementary problem. We show that the proposed method is globally and superlinearly convergent when the matrix of complementarity problems is positive definite and its singular values exceed 1. Numerical results show that our proposed method is very effective and efficient.


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## Body

## 1. Introduction

LetF:Rn→Rn be a given function. The complementarity problems, CP(F) for short, is to find a solution of the system
(1)x≤0,F(x)≤0,F(x)Tx=0.The CP(F) is called the linear complementarity problems (for short LCP) if F is an affine mapping of the form
(2)F(x)=Ax-b,
where A∈Rn×n and b∈Rn. Otherwise, the CP(F) is called the nonlinear complementarity problems (NCP(F)).The systematic study of the finite-dimensional CP(F) began in the mid-1960s; in a span of five decades, the subject has developed into a very fruitful discipline in the field of mathematical programming. The developments include a rich mathematical theory, a host of effective solution algorithms, a multitude of interesting connections to numerous disciplines, and a wide range of important applications in engineering and economics (see, e.g., [1–4] and the references therein).Generalized Newton method (semismooth Newton method) is one of efficacious algorithms for solving CP(F). The main idea of semismooth Newton method is based on an equivalent reformulation of the complementarity problems consisting of a nonsmooth equation and then solving the nonsmooth equation by Newton type method (see, e.g., [5, 6]). Most reformulations of the CP(F) are based on the Fischer-Burmeister function [7] (see, e.g., [8–10] and the references therein). Chen et al. [11] introduced a penalized Fischer-Burmeister function and proposed a new semismooth Newton method based on this new NCP function. Kanzow and Kleinmichel [12] proposed a new, one-parametric class of NCP functions based on Fischer-Burmeister function and gave a semismooth Newton method via these NCP functions. Kanzow [13] researched an inexact semismooth Newton method based on Fischer-Burmeister function and penalized Fischer-Burmeister function. Ito and Kunisch [14] studied a semismooth Newton method based on the max-type NCP function.All generalized Newton methods mentioned above involve the continuously differentiable assumption onF in CP(F). The existed generalized Newton methods proposed are based on the equivalent reformulation via NCP functions. To the best of our knowledge, until now, there exist very few literature resources to study the complementary problems when the involved function F is not differentiable. However, as in many practical problems, F is not differentiable; for instance, F(x)=Ax-|x|-b; see Noor et al. [15]. This is our focus in this paper.In this paper, we consider the following absolute-value linear complementarity problems (for short AVLCP (A,b)): find x∈Rn such that
(3)x≤0,Ax-|x|-b≤0,(Ax-|x|-b)Tx=0,
where A∈Rn×n and b∈Rn.The AVLCP(A,b) is a special case when F is a piecewise-linear function F=Ax-|x|-b in CP(F) and can be viewed as an extension of the LCP. This complementarity problem was first introduced and studied by Noor et al. [15] in 2012. Noor et al. proposed a generalized AOR method via establishing the equivalence between AVLCP(A,b) and the fixed point problem using the projection operator, but the convergence rate is linear. Moreover, the study on AVLCP(A,b) is in its infancy and, to the authors knowledge, there has been no work except for the above-mentioned results of Noor et al. [15]. These observations motivated us to further improve the theory and numerical system for solving AVLCP(A,b).We use the penalty technique to show that the AVLCP(A,b) in Rn are approximately equivalent to a nonlinear penalized equation, which was first introduced for solving AVLCP(A,b). It is worth mentioning that the penalty technique has been widely used in solving nonlinear programming, but it seems that there is a limited study for complementarity problems (see [16–18]). We show that the solution to this penalized equation converges to that of the AVLCP(A,b) at an exponential rate when the penalized parameter tends to infinity. We again use the generalized Jacobian based on subgradient to analyze a generalized Newton method for solving the nonlinear penalized equation under some mild assumptions. The algorithm will be shown to be superlinearly convergent and can start from an arbitrary point. Preliminary numerical experiments are also given to show the effectiveness and efficiency of the proposed method.The rest of this paper is organized as follows. In Section2, we present some notations and the well-known results. In Section 3, we provide a penalized equation for approximating the AVLCP(A,b) and its properties. In Section 4, a generalized Newton method is introduced for solving the penalized equation. In Section 5, we introduce the numerical results of our methods.


## 2. Preliminaries

For convenience, we will now briefly explain some of the terminologies that will be used in the next section.Rn denotes the n-dimensional Euclidean space. All vectors in Rn are column vectors. Let A=(aij)∈Rn×n be an n×n real matrix. The scalar product of two vectors x and y is denoted by xTy. For p>1, the p-norm of x=(x1,x2,…,xn)T∈Rn is defined as ∥x∥p=(∑i=1n‍|xi|p)1/p. When p=2, the p-norm becomes the 2-norm ∥x∥=(xTx)1/2. For any x=(x1,x2,…,xn)∈Rn and α∈R, xα=(x1α,x2α,…,xnα). |x| stands for the vector in Rn of absolute values of components of x. sign⁡(x) will denote the vector with components equal to 1, 0, or −1 depending on whether the corresponding component of x is positive, zero, or negative. diag⁡(sign⁡(x)) will denote a diagonal matrix corresponding to sign⁡(x). The plus function [x]+, which replaces the negative components of x by zeros, is a projection operator that projects x onto the nonnegative orthant; namely, [x]+=max⁡{x,0}. For a solvable matrix equation By=d, we will use the MATLAB backslash B∖d to denote a solution y. The generalized Jacobians ∂|x| of |x| and ∂[x]+1/l of [x]+1/l based on a subgradient [20] of their components are given by the diagonal matrices D(x) and D′(x), respectively, where D(x)=diag⁡(sign⁡(x)) and D′(x) is the the diagonal matrix whose diagonal entries are equal to (1/l)xi(1/l)-1, 0, or a real number σ∈[0,1] depending on whether the corresponding component of x is positive, negative, or zero.Definition 1 (see [21]).
LetA=∈Rn×n be a matrix; the matrix A is called (1)
positive definite if there exists a constantγ>0 such that xTAx≥γ∥x∥2 for any x∈Rn,(2)
bounded if there exists a constantβ>0 such that ∥Ax∥≤β∥x∥ for any x∈Rn.Lemma 2 (Hölder’s inequality).
Letx,y∈Rn. Then
(4)|xTy|≤∥x∥p∥y∥q=(∑i=1n‍|xi|p)1/p(∑i=1n‍|yi|q)1/q,
where p>1 and q>1 are real numbers such that (1/p)+(1/q)=1.Lemma 3 (see [15]).
LetK be a closed and convex set in Rn. A vector x solves the AVLCP (A,b) if and only if x solves the following absolute-value variational inequalities:
(5)〈Ax-|x|-b,y-x〉≥0,∀y∈K.Lemma 4.
LetA∈Rm×n, x∈Rn. Then
(6)∥Ax∥≤∥A∥F∥x∥,
where ∥A∥F denotes the Frobenius norm ∥A∥F2=tr⁡(ATA)=∑i=1m‍∑j=1n‍|aij|2.Lemma 5 (see [22]).
The singular values of the matrixA∈Rn×n exceed 1 if and only if the minimum eigenvalue of ATA exceeds 1.Lemma 6 ((Banach perturbation lemma) [21]).
LetA,C∈Rn×n and assume that A is invertible with ∥A-1∥≤α. If ∥A-C∥≤β and αβ<1, then C is also invertible and
(7)∥C-1∥≤α1-αβ.Lemma 7 (see [23]).
Letx,y∈Rn. Then
(8)∥|x|-|y|∥≤2∥x-y∥.Lemma 8.
Letx,y∈Rn. Then
(9)∥[x]+-[y]+∥≤∥x-y∥.


## 3. A Penalized-Equation Approximation Reformulation of AVLCP(A,b)

In this section, we construct a nonlinear penalized equation corresponding to absolute-value linear complementarity problem (3).Findxλ∈Rn such that
(10)Axλ-|xλ|+λ[xλ]+1/l=b,
where λ>1 is the penalized parameter and l≥1.We will prove that the solution to the penalized equation (10) converges to that of the AVLCP(A,b). Thus, we make the following assumptions on the system matrix A:(A1)
A is positive definite and γ>1 in Definition 1;(A2)
the entries ofA satisfy aii>0, aij≤0 for all i,j=1,2,…,n with i≠j.Under assumption (A1), the solution of AVLCP(A,b) is unique [15]. Our main results in this section are as follows. First, we start our discussion with the following lemma.Lemma 9.
Letxλ be the solution to nonlinear penalized equation (10). Then there exists a positive constant C0, independent of n, xλ, and λ such that
(11)∥[xλ]+∥≤C0λl/2,
where λ and l are parameters used in (10).Proof.
Left-multiplying both sides of (10) by [xλ]+T gives
(12)[xλ]+TAxλ-[xλ]+T|xλ|+λ[xλ]+T[xλ]+1/l=[xλ]+Tb.
Without loss of generality, we assume that [xλ]+=(u1T,0)T, where u1∈Rm, 0≤m≤n, u1≥0. Other cases can be transformed into this by reordering the system.
Whenm=0, then [xλ]+=0; thus (11) is trivially satisfied. We only consider the case when m≥1. In this case xλ can be decomposed into xλ=(u1T,u2T)T, where u2∈Rn-m, u2≤0, We now decompose A into
(13)A=(A11A12A21A22),
where A11∈Rm×m, A12∈Rm×(n-m), A21∈R(n-m)×m, and A22∈R(n-m)×(n-m); then (12) becomes
(14)u1TA11u1+u1TA12u2-[xλ]+T|xλ|+λ∥[xλ]+∥pp=[xλ]+Tb,
where p=1+(1/l).
From assumption (A1) andu1≥0, we have u1TA11u1≥0, and
(15)u1TA11u1=[xλ]+TA[xλ]+≥γ∥[xλ]+∥2≥∥[xλ]+∥2
holds, where the last inequality follows from γ>1. From assumption (A2) and u2≤0, we also have u1TA12u2≥0. Note that [xλ]+T|xλ|=∥[xλ]+∥2; one has that
(16)u1TA11u1-[xλ]+T|xλ|≥0.
It follows from (14), (16), and u1TA12u2≥0 that we have the following inequality:
(17)λ∥[xλ]+∥pp≤[xλ]+Tb.
By using Lemma2 in (17), we get
(18)λ∥[xλ]+∥pp≤[xλ]+Tb≤∥[xλ]+∥p∥b∥q,
where q=l+1 satisfying (1/p)+(1/q)=1. Since p-1=1/l, we thus have the following inequality:
(19)∥[xλ]+∥p≤1λl∥b∥ql.
Since all norms onRn are equivalent for a fixed positive integer n, then it follows that there exists a positive constant a0 such that
(20)∥[xλ]+∥≤a0∥[xλ]+∥p,
and thus the left hand in (16) can be written as
(21)u1TA11u1+u1TA12u2-[xλ]+T|xλ|+λ∥[xλ]+∥pp≥u1TA11u1+u1TA12u2-[xλ]+T|xλ|+λa0p∥[xλ]+∥p.
Combining with (14), we obtain
(22)u1TA11u1+u1TA12u2-[xλ]+T|xλ|+λa0p∥[xλ]+∥p≤[xλ]+Tb,
together with (16); dropping the first three terms in (22) and using Lemma 2, we have
(23)λa0p∥[xλ]+∥p≤∥[xλ]+∥p∥b∥q.
Since p=1+(1/l)∈(1,2], we thus have from the above inequality
(24)∥[xλ]+∥≤1λ(l+1)/p(a0p∥b∥ql+1)1/p≤1λl/2(a0p∥b∥ql+1)1/p=C0λl/2,
where C0=(a0p∥b∥ql+1)1/p. Thus, the proof of this lemma is completed.Using Lemma9, we can establish the relationship between solutions of penalized equation (10) and solutions of the AVLCP(A,b).Theorem 10.
Letx* and xλ be the solution to AVLCP(A,b) and nonlinear penalized equation (10), respectively. Then there exists a positive constant C, independent of n, x*, xλ, and λ such that
(25)∥x*-xλ∥≤Cλl/2,
where λ and l are parameters used in (10).Proof.
Let[xλ]-=-min⁡{xλ,0}; then
(26)xλ=[xλ]+-[xλ]-,
and the vector x*-xλ can be decomposed as
(27)x*-xλ=x*+[xλ]--[xλ]+=rλ-[xλ]+,
where rλ=x*+[xλ]-.
Letω=x*-rλ; one has that
(28)ω=x*-rλ=-[xλ]-≤0,
and therefore ω∈K={y∈Rn∣y≤0}.
Replacingy in (5) by ω gives(29)-rλT(Ax*-|x*|)≥-rλTb;
left-multiplying (10) by rλT we have
(30)rλTAxλ-rλT|xλ|+λrλT[xλ]+1/l=rλTb;
adding up to (29) and (30), we get
(31)rλTA(xλ-x*)+rλT(|x*|-|xλ|)+λrλT[xλ]+1/l≥0;
further we get from (31) that
(32)rλTA(xλ-x*)+rλT(|x*|-|xλ|)≥0,
where
(33)rλT[xλ]+1/l=(x*+[xλ]-)T[xλ]+1/l=(x*)T[xλ]+1/l≤0.
One further has that(34)rλTA(xλ-x*)+rλT(|x*|-|xλ|)=rλTA(xλ-x*)+rλT(-x*-|xλ|+xλ-xλ)=rλTA(xλ-x*)+rλT(xλ-x*)-rλT(xλ+|xλ|)=rλTA(xλ-x*)+rλT(xλ-x*)-2rλT[xλ]+=rλTA([xλ]+-rλ)+rλT([xλ]+-rλ)-2rλT[xλ]+=rλTA[xλ]+-rλTArλ-rλTrλ-rλT[xλ]+;
thus
(35)rλTA[xλ]+-rλT[xλ]+≥rλTArλ+rλTrλ≥(γ+1)∥rλ∥2;
the last inequality is true by the positive definiteness of A.
Hence(36)rλTA[xλ]+-rλT[xλ]+≥0.
On the other hand, using the Cauchy-Schwarz inequality, Lemma4, and (11) to (36), we get
(37)rλTA[xλ]+-rλT[xλ]+≤|rλTA[xλ]+|+|rλT[xλ]+|≤∥rλ∥·∥A[xλ]+∥+∥rλ∥·∥[xλ]+∥≤∥rλ∥·∥A∥F·∥[xλ]+∥+∥rλ∥·∥[xλ]+∥=∥rλ∥·∥[xλ]+∥·(∥A∥F+1)≤∥rλ∥·C0λl/2·(∥A∥F+1);
this implies that
(38)∥rλ∥≤C0(∥A∥F+1)(γ+1)λl/2=C′λl/2,
where C′=C0(∥A∥F+1)/(γ+1). From (11), (27), (38), and the triangle inequality, we obtain
(39)∥x*-xλ∥≤∥rλ∥+∥[xλ]+∥≤Cλl/2.
We complete this theorem.


## 4. A Penalized-Equation-Based Generalized Newton Method and Its Convergence

In this section, we present a generalized Newton method for solving nonlinear penalized equation (10). We begin by defining the vector function specified by the nonlinear penalized equation (10) as follows:
(40)g(xλ)=Axλ-|xλ|+λ[xλ]+1/l-b,λ>1.Letxλ=(u1,u2,…,un)T; a generalized Jacobian ∂g(xλ) of g(xλ) is given by
(41)∂g(xλ)=A-D(xλ)+λD′(xλ),
where D(xλ)=∂|xλ|=diag⁡(sign⁡(xλ)) and D′(xλ)=∂[xλ]+1/l is a diagonal matrix whose diagonal entries are equal to (1/l)ui(1/l)-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 g(xλ)=0 consists of the following iteration:
(42)g(xλi)+∂g(xλi)(xλi+1-xλi)=0.Replacingg(xλi) by its definition (40) and setting ∂g(xλi) by (41) give
(43)[A-D(xλi)+λD′(xλi)]xλi+1=b+λ(1l-1)[xλi]+1/l.
Thus, solving for xλi+1 gives
(44)xλi+1=[A-D(xλi)+λD′(xλi)]∖(b+λ((1l)-1)[xλi]+1/l),
which is our final generalized Newton iteration for solving the nonlinear penalized equation (10). In the following, we can establish the penalized-equation-based generalized Newton method for solving AVLCP(A, b).Algorithm 11 (penalized-equation-based generalized Newton algorithm).
We have the following.Step 1.
Given constantsɛ>0, λ0>1, μ>1, l≥1, and σ∈[0,1] and a starting point xλ00∈Rn, set k:=0.Step 2.
Calculatexλki+1 from the generalized Newton equation
(45)[A-D(xλki)+λkD′(xλki)]xλki+1=b+λk(1l-1)[xλki]+1/l,
starting from xλk0 associated with λk.Step 3.
IfD(xλki+1)=D(xλki), D′(xλki+1)=D′(xλki) and ∥xλki+1-xλki∥≤ɛ, stop; otherwise, set i:=i+1 and go to Step 2. Denote the accumulation point of {xλki} by x¯k.Step 4.
If∥[x¯k]+∥≤ɛ, then stop; otherwise, let λk+1=μλk, choose new starting point xλk+10=x¯k, set k:=k+1, and go to Step 2.


### 4.1. Existence of Accumulation Point at Each Generalized Newton Iteration

We will show that the sequence{xλki}i=1+∞ generated by generalized Newton iteration (45) converges to an accumulation point x¯k associated with λk. We firstly give the following sufficient conditions that the generalized Newton iteration (45) is well defined.Lemma 12.
If the singular values ofA∈Rn×n exceed 1. Then (A-D(xλki)+λkD′(xλki))-1 exists for any diagonal matrices D(xλki) and D′(xλki), where λk>1, D(xλki) is the diagonal matrix with diagonal elements equal to ±1 or 0 and D′(xλki) is the diagonal matrix with diagonal elements equal to (1/l)ui(1/l)-1, 0, or a real number σ∈[0,1].Proof.
If(A-D(xλki)+λkD′(xλki)) is singular, then x≠0 exists such that
(46)(A-D(xλki)+λkD′(xλki))x=0.
Thus, we have the following contradiction
(47)xTx<xTATAx=(Ax)TAx=xT(D(xλki)-λkD′(xλki))T×(D-λkD′(xλki))x≤xTx
for l=1 and λk∈(1,2), where the first inequality follows from Lemma 5. Hence, (A-D(xλki)+λkD′(xλki)) is nonsingular.We now establish boundness of the generalized Newton iteration (45) and thus existence of an accumulation.Theorem 13.
Let the singular values ofA∈Rn×n exceed 1. Then the iteration xλki+1=[A-D(xλki)+λkD′(xλki)]∖(b+λk((1/l)-1)[xλki]+1/l) of the Algorithm 11 is well defined and bounded. Consequently, there exists an accumulation point x¯k such that (A-D(x¯k)+λkD′(x¯k))x¯k=b+λk((1/l)-1)[x¯k]+1/l.Proof.
By Lemma12, (A-D(xλki)+λkD′(xλki))-1 exists; hence, the generalized Newton iteration xλki+1=[A-D(xλki)+λkD′(xλki)]∖(b+λk((1/l)-1)[xλi]+1/l) is well defined.
We now prove the boundedness of the iterative sequence{xλki}. Suppose that {xλki} is unbounded; then there exists a subsequence {xλkij+1} with ∥xλkij+1∥→∞ and nonzero such that D(xλkij)=D~, D′(xλkij)=D′~, where D~ and D′~ are fixed diagonal matrices extended from the finite number of possible configurations for D(xλkij) in the sequence {D(xλkij)} and D′(xλkij) in the sequence {D′(xλkij)}, respectively, such that the bounded subsequence {xλkij+1/∥xλkij+1∥} converges to x~. Hence
(48)(A-D~+λkD′~)xλkij+1∥xλkij+1∥=b+λk((1/l)-1)[xλi]+1/l∥xλkij+1∥.
By lettingj→∞, we obtain
(49)(A-D~+λkD′~)x~=0,∥x~∥=1,
since ∥xλkij+1∥→∞. This contradicts the nonsingularity (A-D~+λkD′~) which follows from Lemma 12. Therefore, the iterative sequence {xλki} is bounded and there exists an accumulation point x¯k such that (A-D(x¯k)+λkD′(x¯k))x¯k=b+λk((1/l)-1)[x¯k]+1/l.We then establish the finite termination of generalized Newton iteration (45).Theorem 14.
Let the singular values ofA∈Rn×n exceed 1. If D(xλki+1)=D(xλki), D′(xλki+1)=D′(xλki), and ∥xλki+1-xλki∥≤ɛ for some i for the well defined iteration (45) in Algorithm 11, then xλki+1 solves the nonlinear penalized equation (10).Proof.
The generalized Newton iteration(50)xλki+1=[A-D(xλki)+λkD′(xλki)]∖(b+λk(1l-1)ggggggggggggggggggggggggggg×[xλki]+1/l1l)
is well defined by Lemma 12, and if D(xλki+1)=D(xλki), D′(xλki+1)=D′(xλki) and ∥xλki+1-xλki∥≤ɛ, then
(51)0=[A-D(xλki)+λkD′(xλki)]xλki+1-b-λk(1l-1)[xλki]+1/l=Axλki+1-D(xλki)xλki+1+λkD′(xλki)xλki+1-b-λk(1l-1)[xλki]+1/l=Axλki+1-D(xλki+1)xλki+1+λkD′(xλki+1)xλki+1-b-λk(1l-1)[xλki]+1/l=Axλki+1-|xλki+1|+λk[xλki]+1/l-b+λk(D′(xλki+1)xλki+1-D′(xλki)xλki)=Axλki+1-|xλki+1|+λk[xλki]+1/l-b+λkD′(xλki+1)(xλki+1-xλki)=Axλki+1-|xλki+1|+λk[xλki+1]+1/l-b+O(ɛ),
giving the result.Furthermore, we have the following results.Theorem 15.
Suppose that the singular values ofA∈Rn×n exceed 1 and x¯k is the unique solution of the nonlinear penalized equation (10). Then, for any xλk such that
(52)(x¯k)i>0⟹(xλk)i>0,(x¯k)i<0⟹(xλk)i<0,hhhhhhhhhhhhhhhgghhhhhh∀i∈{1,2,…,n},
the generalized Newton iteration (45) reaches x¯k in one iteration.Proof.
The theorem can be proved in a similar way to the one in [24, Lemma 2.1]. We omit it here.Remark 16.
Not that the “sign match” property (52) holds if xλk is sufficiently near x¯k. Hence, by Theorem 15, global convergence is then obvious for generalized Newton iteration (45).Now, we discuss the globally linear convergence of{xλki}i=1+∞ generated by generalized Newton iteration (45).Theorem 17.
If∥(A+λkD′(xλki))-1∥<1/(4+λk(1+∥D′(xλki)∥)) holds for all sufficiently large λk and ∥D(xλki)∥≠0, then the generalized Newton iteration (45) converges linearly from any starting point xλk0 to a solution x¯k of the nonlinear penalized equation (10).Proof.
Since∥(A+λkD′(xλki))-1∥·∥D(xλki)∥<1 according to the assumption and the definition of D(xλki), then (A+λkD′(xλki)-D(xλki))-1 exists for any xλki by Lemma 6. We also have by the same lemma that
(53)∥[(A+λkD′(xλki))-D(xλki)]-1∥≤∥(A+λkD′(xλki))-1∥1-∥(A+λkD′(xλki))-1∥·∥D(xλki)∥<1/(4+λk(1+∥D′(xλki)∥))1-(1/1/(4+λk(1+∥D′(xλki)∥)))·1=13+λk(1+∥D′(xλki)∥).
Letx¯k be a solution of the nonlinear penalized equation (10). To simplify notation, let D¯=D(x¯k), Di=D(xλki), D′¯=D′(x¯k), and D′i=D′(xλki). Noting that |x¯k|=D¯x¯k, |xλki|=Dixλki, (1/l)[x¯k]+1/l=D′¯x¯k, and (1/l)[xλki]+1/l=D′ixλki, we have
(54)A(xλki+1-x¯k)=Dixλki+1-λkD′ixλki+1+b+λk(1l-1)[xλki]+1/l-(D¯x¯k-λkD′¯x¯k+b+λk(1l-1)[x¯k]+1/l)=Dixλki+1-λkD′ixλki+1-D¯x¯k+λkD′¯x¯k+λk(1l-1)([xλki]+1/l-[x¯k]+1/l)=(Dixλki+1-D¯x¯k)-λk(D′ixλki+1-D′¯x¯k)+λk(1l-1)([xλki]+1/l-[x¯k]+1/l)=(Dixλki+1-|x¯k|)-λk(D′ixλki+1-1l[x¯k]+1/l)+λk(1l-1)([xλki]+1/l-[x¯k]+1/l)=Di(xλki+1-xλki+xλki)-|x¯k|-λk[D′i(xλki+1-xλki+xλki)-1l[x¯k]+1/l]+λk(1l-1)([xλki]+1/l-[x¯k]+1/l)=|xλki|-|x¯k|+Di(xλki+1-x¯k+x¯k-xλki)-1lλk([xλki]+1/l-[x¯k]+1/l)-λkD′i(xλki+1-x¯k+x¯k-xλki)+λk(1l-1)([xλki]+1/l-[x¯k]+1/l),
where the first inequality follows from (A-D¯+λkD′¯)x¯k=b+λk((1/l)-1)[x¯k]+1/l and (A-Di+λkD′i)xλki+1=b+λk((1/l)-1)[xλki]+1/l.
Hence, one has that(55)(A-Di+λkD′i)(xλki+1-x¯k)=|xλki|-|x¯k|-Di(xλki-x¯k)-1lλk([xλki]+-[x¯k]+)+λkD′i(xλki-x¯k)+λk(1l-1)([xλki]+1/l-[x¯k]+1/l).
Thus
(56)xλki+1-x¯k=(A-Di+λkD′i)-1×[[xλki]+1/l|xλki|-|x¯k|-Di(xλki-x¯k)fff-1lλk([xλki]+-[x¯k]+)fff+λkD′i(xλki-x¯k)gfg+λk(1l-1)([xλki]+1/l-[x¯k]+1/l)].
Applying Lemmas7 and 8 and ∥[xλki]+-[x¯k]+∥≤ɛ when i is sufficiently large, we get
(57)∥xλki+1-x¯k∥≤∥(A-Di+λkD′i)-1∥hhhh·((3+λkl)∥xλki-x¯k∥hffffffffffffh+λk∥D′i(xλki-x¯k)∥hffffffffffffh+λk(1-1l)∥[xλki]+-[x¯k]+∥+O(ɛ)λkl)≤(3+λk+λk∥D′i∥+O(ɛ))·∥(A-Di+λkD′i)-1∥·∥xλki-x¯k∥.
Lettingi→∞,ɛ→0 and taking limits in both sides of the last inequality above, we have
(58)∥xλki+1-x¯k∥∥xλki-x¯k∥≤(3+λk(1+∥D′(xλki)∥))·∥(A-Di(xλki)+λkD′i(xλki))-1∥<1,
where the last inequality in (58) follows from ∥(A-D(xλki)+λkD′(xλki))-1∥<1/(3+λk(1+∥D′i∥)).
Consequently, the sequence{xλki} converges linearly to a solution x¯k.In the above proof, the choice ofx¯k is arbitrary; hence we have the following result.Corollary 18.
Assume that∥(A+λkD′(xλki))-1∥<1/(1+λki(1+∥D′(xλki)∥)) and ∥D(xλki)∥≠0 for any sufficiently large λk. Then the nonlinear penalized equation (10) has a unique solution for any b∈Rn.Finally, we give the globally superlinear convergence of{xλki}i=1+∞ generated by generalized Newton iteration (45).Theorem 19.
Assume that∥(A+λkD′(xλki))-1∥<1/(1+λki(1+∥D′(xλki)∥)) and ∥D(xλki)∥≠0 for any sufficiently large λk. Then the generalized Newton iteration (45) converges superlinearly from any starting point xλk0 to a solution x¯k of the nonlinear penalized equation (10).Proof.
According to Lemma6, one has that
(59)∥[(A+λkD′(xλki))-D(xλki)]-1∥≤∥(A+λkD′(xλki))-1∥1-∥(A+λkD′(xλki))-1∥·∥D(xλki)∥≤∥(A+λkD′(xλki))-1∥1-∥(A+λkD′(xλki))-1∥<1λki(1+∥D′(xλki)∥),
where the last inequality follows from the sequence ∥(A+λkD′(xλki))-1∥≤1/(1+λki(1+∥D′(xλki)∥)).
Combining this with (58) in Theorem 15, we have
(60)∥xλki+1-x¯k∥∥xλki-x¯k∥<(3+λk(1+∥D′(xλki)∥))·∥(A-D(xλki)+λkD′(xλki))-1∥(61)<3+λk(1+∥D′(xλki)∥)λki(1+∥D′(xλki)∥).
Letting i→∞ and taking limits in (61), we can see that the sequence {xλki}i=1+∞ generated by generalized Newton iteration (45) converges superlinearly to a solution x¯k.


### 4.2. Convergence of Penalized-Equation-Based Generalized Newton Method

In this subsection, we will focus on the convergence of Algorithm11. We first present the global convergence.Theorem 20.
Let the singular values ofA∈Rn×n exceed 1. Then the sequence {x¯k} generated by Algorithm 11 is bounded. Consequently, there exists an accumulation point x¯* of the nonlinear penalized equation (10).Proof.
Sincex¯k is an accumulation point of {xλki}i=1+∞, it follows from Theorem 13 that the sequence {xλki}i=1+∞ is bounded, we can thus obtain the boundness of {x¯k}. Hence there exists an accumulation point x¯* of {x¯k} such that the nonlinear penalized equation (10) holds, giving the results.We then establish the linear convergence of Algorithm11.Theorem 21.
Under the assumption that∥(A+λkD′(x¯k))-1∥<1/(1+λk(1+∥D′(x¯k)∥)), ∥D(x¯k)∥≠0, Algorithm 11 converges linearly from any starting point xλ00 to a solution x¯* of the nonlinear penalized equation (10).Proof.
Taking into account “match property” (52), the theorem can be proved in a similar way to that of Theorem 15.Finally, we establish the superlinear convergence of Algorithm11.Theorem 22.
Under the assumption that∥(A+λkD′(x¯k))-1∥<1/(1+λkm(1+∥D′(x¯k)∥))(m>1), ∥D(x¯k)∥≠0, Algorithm 11 converges superlinearly from any starting point xλ00 to a solution x¯* of the nonlinear penalized equation (10).Proof.
Taking into account “match property” (52), the theorem can be proved in a similar way to that of Theorems 15 and 17.


## 4.1. Existence of Accumulation Point at Each Generalized Newton Iteration

We will show that the sequence{xλki}i=1+∞ generated by generalized Newton iteration (45) converges to an accumulation point x¯k associated with λk. We firstly give the following sufficient conditions that the generalized Newton iteration (45) is well defined.Lemma 12.
If the singular values ofA∈Rn×n exceed 1. Then (A-D(xλki)+λkD′(xλki))-1 exists for any diagonal matrices D(xλki) and D′(xλki), where λk>1, D(xλki) is the diagonal matrix with diagonal elements equal to ±1 or 0 and D′(xλki) is the diagonal matrix with diagonal elements equal to (1/l)ui(1/l)-1, 0, or a real number σ∈[0,1].Proof.
If(A-D(xλki)+λkD′(xλki)) is singular, then x≠0 exists such that
(46)(A-D(xλki)+λkD′(xλki))x=0.
Thus, we have the following contradiction
(47)xTx<xTATAx=(Ax)TAx=xT(D(xλki)-λkD′(xλki))T×(D-λkD′(xλki))x≤xTx
for l=1 and λk∈(1,2), where the first inequality follows from Lemma 5. Hence, (A-D(xλki)+λkD′(xλki)) is nonsingular.We now establish boundness of the generalized Newton iteration (45) and thus existence of an accumulation.Theorem 13.
Let the singular values ofA∈Rn×n exceed 1. Then the iteration xλki+1=[A-D(xλki)+λkD′(xλki)]∖(b+λk((1/l)-1)[xλki]+1/l) of the Algorithm 11 is well defined and bounded. Consequently, there exists an accumulation point x¯k such that (A-D(x¯k)+λkD′(x¯k))x¯k=b+λk((1/l)-1)[x¯k]+1/l.Proof.
By Lemma12, (A-D(xλki)+λkD′(xλki))-1 exists; hence, the generalized Newton iteration xλki+1=[A-D(xλki)+λkD′(xλki)]∖(b+λk((1/l)-1)[xλi]+1/l) is well defined.
We now prove the boundedness of the iterative sequence{xλki}. Suppose that {xλki} is unbounded; then there exists a subsequence {xλkij+1} with ∥xλkij+1∥→∞ and nonzero such that D(xλkij)=D~, D′(xλkij)=D′~, where D~ and D′~ are fixed diagonal matrices extended from the finite number of possible configurations for D(xλkij) in the sequence {D(xλkij)} and D′(xλkij) in the sequence {D′(xλkij)}, respectively, such that the bounded subsequence {xλkij+1/∥xλkij+1∥} converges to x~. Hence
(48)(A-D~+λkD′~)xλkij+1∥xλkij+1∥=b+λk((1/l)-1)[xλi]+1/l∥xλkij+1∥.
By lettingj→∞, we obtain
(49)(A-D~+λkD′~)x~=0,∥x~∥=1,
since ∥xλkij+1∥→∞. This contradicts the nonsingularity (A-D~+λkD′~) which follows from Lemma 12. Therefore, the iterative sequence {xλki} is bounded and there exists an accumulation point x¯k such that (A-D(x¯k)+λkD′(x¯k))x¯k=b+λk((1/l)-1)[x¯k]+1/l.We then establish the finite termination of generalized Newton iteration (45).Theorem 14.
Let the singular values ofA∈Rn×n exceed 1. If D(xλki+1)=D(xλki), D′(xλki+1)=D′(xλki), and ∥xλki+1-xλki∥≤ɛ for some i for the well defined iteration (45) in Algorithm 11, then xλki+1 solves the nonlinear penalized equation (10).Proof.
The generalized Newton iteration(50)xλki+1=[A-D(xλki)+λkD′(xλki)]∖(b+λk(1l-1)ggggggggggggggggggggggggggg×[xλki]+1/l1l)
is well defined by Lemma 12, and if D(xλki+1)=D(xλki), D′(xλki+1)=D′(xλki) and ∥xλki+1-xλki∥≤ɛ, then
(51)0=[A-D(xλki)+λkD′(xλki)]xλki+1-b-λk(1l-1)[xλki]+1/l=Axλki+1-D(xλki)xλki+1+λkD′(xλki)xλki+1-b-λk(1l-1)[xλki]+1/l=Axλki+1-D(xλki+1)xλki+1+λkD′(xλki+1)xλki+1-b-λk(1l-1)[xλki]+1/l=Axλki+1-|xλki+1|+λk[xλki]+1/l-b+λk(D′(xλki+1)xλki+1-D′(xλki)xλki)=Axλki+1-|xλki+1|+λk[xλki]+1/l-b+λkD′(xλki+1)(xλki+1-xλki)=Axλki+1-|xλki+1|+λk[xλki+1]+1/l-b+O(ɛ),
giving the result.Furthermore, we have the following results.Theorem 15.
Suppose that the singular values ofA∈Rn×n exceed 1 and x¯k is the unique solution of the nonlinear penalized equation (10). Then, for any xλk such that
(52)(x¯k)i>0⟹(xλk)i>0,(x¯k)i<0⟹(xλk)i<0,hhhhhhhhhhhhhhhgghhhhhh∀i∈{1,2,…,n},
the generalized Newton iteration (45) reaches x¯k in one iteration.Proof.
The theorem can be proved in a similar way to the one in [24, Lemma 2.1]. We omit it here.Remark 16.
Not that the “sign match” property (52) holds if xλk is sufficiently near x¯k. Hence, by Theorem 15, global convergence is then obvious for generalized Newton iteration (45).Now, we discuss the globally linear convergence of{xλki}i=1+∞ generated by generalized Newton iteration (45).Theorem 17.
If∥(A+λkD′(xλki))-1∥<1/(4+λk(1+∥D′(xλki)∥)) holds for all sufficiently large λk and ∥D(xλki)∥≠0, then the generalized Newton iteration (45) converges linearly from any starting point xλk0 to a solution x¯k of the nonlinear penalized equation (10).Proof.
Since∥(A+λkD′(xλki))-1∥·∥D(xλki)∥<1 according to the assumption and the definition of D(xλki), then (A+λkD′(xλki)-D(xλki))-1 exists for any xλki by Lemma 6. We also have by the same lemma that
(53)∥[(A+λkD′(xλki))-D(xλki)]-1∥≤∥(A+λkD′(xλki))-1∥1-∥(A+λkD′(xλki))-1∥·∥D(xλki)∥<1/(4+λk(1+∥D′(xλki)∥))1-(1/1/(4+λk(1+∥D′(xλki)∥)))·1=13+λk(1+∥D′(xλki)∥).
Letx¯k be a solution of the nonlinear penalized equation (10). To simplify notation, let D¯=D(x¯k), Di=D(xλki), D′¯=D′(x¯k), and D′i=D′(xλki). Noting that |x¯k|=D¯x¯k, |xλki|=Dixλki, (1/l)[x¯k]+1/l=D′¯x¯k, and (1/l)[xλki]+1/l=D′ixλki, we have
(54)A(xλki+1-x¯k)=Dixλki+1-λkD′ixλki+1+b+λk(1l-1)[xλki]+1/l-(D¯x¯k-λkD′¯x¯k+b+λk(1l-1)[x¯k]+1/l)=Dixλki+1-λkD′ixλki+1-D¯x¯k+λkD′¯x¯k+λk(1l-1)([xλki]+1/l-[x¯k]+1/l)=(Dixλki+1-D¯x¯k)-λk(D′ixλki+1-D′¯x¯k)+λk(1l-1)([xλki]+1/l-[x¯k]+1/l)=(Dixλki+1-|x¯k|)-λk(D′ixλki+1-1l[x¯k]+1/l)+λk(1l-1)([xλki]+1/l-[x¯k]+1/l)=Di(xλki+1-xλki+xλki)-|x¯k|-λk[D′i(xλki+1-xλki+xλki)-1l[x¯k]+1/l]+λk(1l-1)([xλki]+1/l-[x¯k]+1/l)=|xλki|-|x¯k|+Di(xλki+1-x¯k+x¯k-xλki)-1lλk([xλki]+1/l-[x¯k]+1/l)-λkD′i(xλki+1-x¯k+x¯k-xλki)+λk(1l-1)([xλki]+1/l-[x¯k]+1/l),
where the first inequality follows from (A-D¯+λkD′¯)x¯k=b+λk((1/l)-1)[x¯k]+1/l and (A-Di+λkD′i)xλki+1=b+λk((1/l)-1)[xλki]+1/l.
Hence, one has that(55)(A-Di+λkD′i)(xλki+1-x¯k)=|xλki|-|x¯k|-Di(xλki-x¯k)-1lλk([xλki]+-[x¯k]+)+λkD′i(xλki-x¯k)+λk(1l-1)([xλki]+1/l-[x¯k]+1/l).
Thus
(56)xλki+1-x¯k=(A-Di+λkD′i)-1×[[xλki]+1/l|xλki|-|x¯k|-Di(xλki-x¯k)fff-1lλk([xλki]+-[x¯k]+)fff+λkD′i(xλki-x¯k)gfg+λk(1l-1)([xλki]+1/l-[x¯k]+1/l)].
Applying Lemmas7 and 8 and ∥[xλki]+-[x¯k]+∥≤ɛ when i is sufficiently large, we get
(57)∥xλki+1-x¯k∥≤∥(A-Di+λkD′i)-1∥hhhh·((3+λkl)∥xλki-x¯k∥hffffffffffffh+λk∥D′i(xλki-x¯k)∥hffffffffffffh+λk(1-1l)∥[xλki]+-[x¯k]+∥+O(ɛ)λkl)≤(3+λk+λk∥D′i∥+O(ɛ))·∥(A-Di+λkD′i)-1∥·∥xλki-x¯k∥.
Lettingi→∞,ɛ→0 and taking limits in both sides of the last inequality above, we have
(58)∥xλki+1-x¯k∥∥xλki-x¯k∥≤(3+λk(1+∥D′(xλki)∥))·∥(A-Di(xλki)+λkD′i(xλki))-1∥<1,
where the last inequality in (58) follows from ∥(A-D(xλki)+λkD′(xλki))-1∥<1/(3+λk(1+∥D′i∥)).
Consequently, the sequence{xλki} converges linearly to a solution x¯k.In the above proof, the choice ofx¯k is arbitrary; hence we have the following result.Corollary 18.
Assume that∥(A+λkD′(xλki))-1∥<1/(1+λki(1+∥D′(xλki)∥)) and ∥D(xλki)∥≠0 for any sufficiently large λk. Then the nonlinear penalized equation (10) has a unique solution for any b∈Rn.Finally, we give the globally superlinear convergence of{xλki}i=1+∞ generated by generalized Newton iteration (45).Theorem 19.
Assume that∥(A+λkD′(xλki))-1∥<1/(1+λki(1+∥D′(xλki)∥)) and ∥D(xλki)∥≠0 for any sufficiently large λk. Then the generalized Newton iteration (45) converges superlinearly from any starting point xλk0 to a solution x¯k of the nonlinear penalized equation (10).Proof.
According to Lemma6, one has that
(59)∥[(A+λkD′(xλki))-D(xλki)]-1∥≤∥(A+λkD′(xλki))-1∥1-∥(A+λkD′(xλki))-1∥·∥D(xλki)∥≤∥(A+λkD′(xλki))-1∥1-∥(A+λkD′(xλki))-1∥<1λki(1+∥D′(xλki)∥),
where the last inequality follows from the sequence ∥(A+λkD′(xλki))-1∥≤1/(1+λki(1+∥D′(xλki)∥)).
Combining this with (58) in Theorem 15, we have
(60)∥xλki+1-x¯k∥∥xλki-x¯k∥<(3+λk(1+∥D′(xλki)∥))·∥(A-D(xλki)+λkD′(xλki))-1∥(61)<3+λk(1+∥D′(xλki)∥)λki(1+∥D′(xλki)∥).
Letting i→∞ and taking limits in (61), we can see that the sequence {xλki}i=1+∞ generated by generalized Newton iteration (45) converges superlinearly to a solution x¯k.


## 4.2. Convergence of Penalized-Equation-Based Generalized Newton Method

In this subsection, we will focus on the convergence of Algorithm11. We first present the global convergence.Theorem 20.
Let the singular values ofA∈Rn×n exceed 1. Then the sequence {x¯k} generated by Algorithm 11 is bounded. Consequently, there exists an accumulation point x¯* of the nonlinear penalized equation (10).Proof.
Sincex¯k is an accumulation point of {xλki}i=1+∞, it follows from Theorem 13 that the sequence {xλki}i=1+∞ is bounded, we can thus obtain the boundness of {x¯k}. Hence there exists an accumulation point x¯* of {x¯k} such that the nonlinear penalized equation (10) holds, giving the results.We then establish the linear convergence of Algorithm11.Theorem 21.
Under the assumption that∥(A+λkD′(x¯k))-1∥<1/(1+λk(1+∥D′(x¯k)∥)), ∥D(x¯k)∥≠0, Algorithm 11 converges linearly from any starting point xλ00 to a solution x¯* of the nonlinear penalized equation (10).Proof.
Taking into account “match property” (52), the theorem can be proved in a similar way to that of Theorem 15.Finally, we establish the superlinear convergence of Algorithm11.Theorem 22.
Under the assumption that∥(A+λkD′(x¯k))-1∥<1/(1+λkm(1+∥D′(x¯k)∥))(m>1), ∥D(x¯k)∥≠0, Algorithm 11 converges superlinearly from any starting point xλ00 to a solution x¯* of the nonlinear penalized equation (10).Proof.
Taking into account “match property” (52), the theorem can be proved in a similar way to that of Theorems 15 and 17.


## 5. Numerical Results

In this section, we consider several examples to show the efficiency of the proposed method by running in MATLAB 7.5 with Intel(R) Core (TM) of 2× 2.70 GHz and RAM of 2.0 GB. Throughout these computational experiments, the parameters used in the algorithm are set as ɛ=10-6, λ0=10, μ=2, and l=1. The accumulation point of Algorithm 11 is written as x¯*.Example 1.
Let the matrixA of AVLCP(A,b) be given by
(62)A=(2-1-13),
and b=(-1,0)T. The solution of AVLCP(A,b) is x*=(-4/11,-1/11)T. The computational results are shown in Table 1.Table 1
The numerical results of Example1.
Initial pointxλ0s
Numbers of Newton iterations
x
-
*
(
0,0
)
T
2
(
-
4
1
1
l
,
-
1
11
)
T
(
0,1
)
T
2
(
-
4
1
1
l
,
-
1
11
)
T
(
-
1
,
-
1
)
T
1
(
-
4
1
1
l
,
-
1
11
)
T
(
2
,
-
1
)
T
2
(
-
4
1
1
l
,
-
1
11
)
TExample 2.
Let the matrixA of AVLCP(A,b) be given by
(63)A=(4-100-14-100-14-100-14),
and b=(-1,-1,0,0)T. The solution of AVLCP(A,b) is x*=(-139/551,-144/551,-30/551,-6/551)T. The computational results are shown in Table 2.Table 2
The numerical results of Example2.
Initial pointxλ0s
Numbers of Newton iterations
x
-
*
(
-
1
,
-
1
,
-
1
,
-
1
)
T
1
(
-
139
551
,
-
144
551
,
-
30
551
,
-
6
55
1
l
)
T
(
0,0
,
0,0
)
T
2
(
-
139
551
,
-
144
551
,
-
30
551
,
-
6
55
1
l
)
T
(
1
,
-
1,1
,
-
1
)
T
2
(
-
139
551
,
-
144
551
,
-
30
551
,
-
6
55
1
l
)
T
(
1,2
,
-
1,0
)
T
2
(
-
139
551
,
-
144
551
,
-
30
551
,
-
6
55
1
l
)
TExample 3.
Let the matrixA of AVLCP(A,b) be given by
(64)A=(4000-1-10400-1-1004-1-1000-1400-1-1-1040-1-10004),
and b=(-4,4,-4,-4,4,-4)T. The solution of AVLCP(A,b) is x*=(-1,0,-1,-1,0,-1)T. The computational results are shown in Table 3.Table 3
The numerical results of Example3.
Initial pointxλ0s
Numbers of Newton iterations
x
-
*
=
(
u
-
1
*
,
u
-
2
*
,
u
-
3
*
,
u
-
4
*
,
u
-
5
*
,
u
-
6
*
)
T
(
0,0
,
0,0
,
0,0
)
T
21
u
-
1
*
=
-
1
u
-
2
*
=
5.72204
e
-
7
u
-
3
*
=
-
1
u
-
4
*
=
-
1
u
-
5
*
=
3.8147
e
-
7
u
-
6
*
=
-
1
(
-
1
,
-
1
,
-
1
,
-
1
,
-
1
,
-
1
)
T
21
u
-
1
*
=
-
1
u
-
2
*
=
5.72204
e
-
7
u
-
3
*
=
-
1
u
-
4
*
=
-
1
u
-
5
*
=
3.8147
e
-
7
u
-
6
*
=
-
1
(
-
2,0
,
-
2
,
-
2,0
,
-
2
)
T
20
u
-
1
*
=
-
1
u
-
2
*
=
5.72204
e
-
7
u
-
3
*
=
-
1
u
-
4
*
=
-
1
u
-
5
*
=
3.8147
e
-
7
u
-
6
*
=
-
1
(
1,1
,
1,1
,
1,1
)
T
21
u
-
1
*
=
-
1
u
-
2
*
=
5.72204
e
-
7
u
-
3
*
=
-
1
u
-
4
*
=
-
1
u
-
5
*
=
3.8147
e
-
7
u
-
6
*
=
-
1Example 4.
Let the matrixA of AVLCP(A,b) be given by
(65)aij={8,forj=i,-1,for{j=i+1,i=1,2,…,n-1,j=i-1,i=2,3,…,n,0,otherwise.
Let b=(6,5,5,…,5,6)T. Then choose initial point xλ0s as xλ0s=(0,0,…,0)T. We compared our algorithm with the existing methods in [15, 19]. The computational results are shown in Table 4.Table 4
Computational results of Example4.
Order
Numbers of iterations
Iterative method [19]
GAOR method [15]
Algorithm11
4
10
10
8
8
11
11
9
16
11
11
9
32
12
11
9
64
12
11
9
128
12
11
9
256
12
11
9
512
12
11
9
1024
13
11
9From Table1 to Table 4, we can see that our method has some nice convergence which coincides with our results.


## 6. Conclusion

In this paper, we propose a new approximation to absolute-value linear complementarity problems (3) by using the nonlinear penalized equation (10), based on which a generalized Newton method is proposed for solving this penalized equation. Under suitable assumptions, the algorithm is shown to be both globally and superlinearly convergent. The numerical results presented showed that the generalized Newton method proposed by us is efficient. The results and ideas of this paper may be used to solve the absolute variational inequalities and related optimization problems.



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*Source: 560578-2014-09-09.xml*