Error bounds for linear complementarity problems of weakly chained diagonally dominant B-matrices

Abstract

In this paper, new error bounds for the linear complementarity problem are obtained when the involved matrix is a weakly chained diagonally dominant B-matrix. The proposed error bounds are better than some existing results. The advantages of the results obtained are illustrated by numerical examples.

Introduction

A linear complementarity problem (LCP) is to find a vector $x\in {\mathbb{R}}^{n\times 1}$ such that

$${(Mx+q)}^{T}x=0,Mx+q\ge 0,x\ge 0,$$

where $M=[m_{ij}]\in {\mathbb{R}}^{n\times n}$ and $q\in {\mathbb{R}}^{n\times 1}$. The LCP has various applications in the free boundary problems for journal bearing, the contact problem, and the Nash equilibrium point of a bimatrix game [1–3].

The LCP has a unique solution for any $q\in {\mathbb{R}}^{n\times 1}$ if and only if M is a P-matrix [4]. In [5], Chen et al. gave the following error bound for the LCP when M is a P-matrix:

$${\parallel x-x^{\ast }\parallel }_{\mathrm{\infty }}\le \underset{d\in {[0,1]}^n}{max}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}{\parallel r(x)\parallel }_{\mathrm{\infty }},$$

where $x^{\ast }$ is the solution of the LCP, $r(x)=min\{x,Mx+q\}$, $D=diag(d_i)$ with $0\le d_i\le 1$, and the min operator $r(x)$ denotes the componentwise minimum of two vectors. If M satisfies special structures, then some bounds of ${max}_{d\in {[0,1]}^n}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}$ can be derived [6–11].

Definition 1

[4]

A matrix $M=[m_{ij}]\in {\mathbb{R}}^{n\times n}$ is called a B-matrix if for any $i,j\in \mathbb{N}=\{1,2,\dots ,n\}$,

$$\sum _{k\in N}{m}_{ik}>0,\frac{1}{n}\left(\sum _{k\in N}{m}_{ik}\right)>m_{ij},j\ne i.$$

Definition 2

[12]

A matrix $A=[a_{ij}]\in {\mathbb{R}}^{n\times n}$ is called a weakly chained diagonally dominant (wcdd) matrix if A is diagonally dominant, i.e.,

$$|a_{ii}|\ge r_i(A)=\sum _{j=1,\ne i}^n|a_{ij}|,\mathrm{\forall }i\in \mathbb{N},$$

and for each $i\notin J(A)=\{i\in \mathbb{N}:|a_{ii}|>r_i(A)\}\ne \mathrm{\varnothing }$, there is a sequence of nonzero elements of A of the form $a_{i{i}_1},a_{i_1{i}_2},\dots ,a_{i_{r}j}$ with $j\in J(A)$.

Definition 3

[13]

A matrix $M=[m_{ij}]\in {\mathbb{R}}^{n\times n}$ is called a weakly chained diagonally dominant (wcdd) B-matrix if it can be written in the form $M=B^{+}+C$ with $B^{+}$ a wcdd matrix whose diagonal entries are all positive.

García-Esnaola et al. [8] gave the upper bound for ${max}_{d\in {[0,1]}^n}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}$ when M is a B-matrix: Let $M=[m_{ij}]\in {\mathbb{R}}^{n\times n}$ be a B-matrix with the form

$$M=B^{+}+C,$$

where

$$B^{+}=[b_{ij}]=\left[\begin{array}{ccc}{m}_{11}-r_1^{+}& \cdots & m_{1n}-r_1^{+}\\ ⋮& & ⋮\\ m_{n1}-r_n^{+}& \cdots & m_{nn}-r_n^{+}\end{array}\right],$$ 1

and $r_i^{+}=max\{0,m_{ij}|j\ne i\}$. Then

$$\underset{d\in {[0,1]}^n}{max}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}\le \frac{n-1}{min\{\beta ,1\}},$$ 2

where $\beta ={min}_{i\in \mathbb{N}}\{{\beta }_i\}$ and ${\beta }_i=b_{ii}-{\sum }_{j\ne i}|b_{ij}|$.

To improve the bound in (2), Li et al. [14] presented the following result: Let $M=[m_{ij}]\in {\mathbb{R}}^{n\times n}$ be a B-matrix with the form $M=B^{+}+C$, where $B^{+}=[b_{ij}]$ is defined as (1). Then

$$\underset{d\in {[0,1]}^n}{max}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}\le \sum _{i=1}^n\frac{n-1}{min\{{\overline{\beta }}_i,1\}}\prod _{j=1}^{i-1}(1+\frac{1}{{\overline{\beta }}_j}\sum _{k=j+1}^n|b_{jk}|),$$ 3

where ${\overline{\beta }}_i=b_{ii}-{\sum }_{j=i+1}^n|b_{ij}|l_i(B^{+})$, $l_k(B^{+})={max}_{k\le i\le n}\{\frac{1}{|b_{ii}|}{\sum }_{j=k,\ne i}^n|b_{ij}|\}$ and

$$\prod _{j=1}^{i-1}(1+\frac{1}{{\overline{\beta }}_j}\sum _{k=j+1}^n|b_{jk}|)=1,\text{if }i=1.$$

Recently, when M is a weakly chained diagonally dominant (wcdd) B-matrix, Li et al. [13] gave a bound for ${max}_{d\in {[0,1]}^n}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}$: Let $M=[m_{ij}]\in {\mathbb{R}}^{n\times n}$ be a wcdd B-matrix with the form $M=B^{+}+C$, where $B^{+}=[b_{ij}]$ is defined as (1). Then

$$\underset{d\in {[0,1]}^n}{max}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}\le \sum _{i=1}^n\left(\frac{n-1}{min\{{\tilde{\beta }}_i,1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{{\tilde{\beta }}_j}\right),$$ 4

where ${\tilde{\beta }}_i=b_{ii}-{\sum }_{j=i+1}^n|b_{ij}|>0$ and ${\prod }_{j=1}^{i-1}\frac{b_{jj}}{{\tilde{\beta }}_j}=1$ if $i=1$.

This bound in (4) holds when M is a B-matrix since a B-matrix is a weakly chained diagonally dominant B-matrix [13].

Now, some notation is given, which will be used in the sequel. Let $A=[a_{ij}]\in {\mathbb{R}}^{n\times n}$. For $i,j,k\in \mathbb{N}$, denote

$$\begin{array}{rl}& u_i(A)=\frac{1}{|a_{ii}|}\sum _{j=i+1}^n|a_{ij}|,u_n(A)=0,\\ & b_k(A)=\underset{k+1\le i\le n}{max}\left\{\frac{{\sum }_{j=k,\ne i}^n|a_{ij}|}{|a_{ii}|}\right\},b_n(A)=1,\\ & p_k(A)=\underset{k+1\le i\le n}{max}\left\{\frac{|a_{ik}|+{\sum }_{j=k+1,\ne i}^n|a_{ij}|b_k(A)}{|a_{ii}|}\right\},p_n(A)=1.\end{array}$$

The rest of this paper is organized as follows: In Section 2, we present some new bounds for ${max}_{d\in {[0,1]}^n}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}$ when M is a wcdd B-matrix. Numerical examples are given to verify the corresponding results in Section 3.

Main results

In this section, some new upper bounds for ${max}_{d\in {[0,1]}^n}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}$ are provided when M is a wcdd B-matrix. Firstly, several lemmas, which will be used later, are given.

Lemma 1

[13]

Let $M=[m_{ij}]\in {\mathbb{R}}^{n\times n}$ be a wcdd B-matrix with the form $M=B^{+}+C$, where $B^{+}$ is defined as (1). Then

$${\parallel {(I+{\left(B_D^{+}\right)}^{-1}{C}_D)}^{-1}\parallel }_{\mathrm{\infty }}\le n-1,$$

where $B_D^{+}=I-D+D{B}^{+}$ and $C_D=DC$.

Lemma 2

[15]

Let $A=[a_{ij}]\in {\mathbb{R}}^{n\times n}$ be a wcdd M-matrix with $u_k(A)p_k(A)<1$ ($\mathrm{\forall }k\in \mathbb{N}$). Then

$$\begin{array}{rl}{\parallel A^{-1}\parallel }_{\mathrm{\infty }}& \le max\{\sum _{i=1}^n\left(\frac{1}{a_{ii}(1-u_i(A)p_i(A))}\prod _{j=1}^{i-1}\frac{u_j(A)}{1-u_j(A)p_j(A)}\right),\\ & \sum _{i=1}^n\left(\frac{p_i(A)}{a_{ii}(1-u_i(A)p_i(A))}\prod _{j=1}^{i-1}\frac{1}{1-u_j(A)p_j(A)}\right)\},\end{array}$$

where

$$\prod _{j=1}^{i-1}\frac{u_j(A)}{1-u_j(A)p_j(A)}=1,\prod _{j=1}^{i-1}\frac{1}{1-u_j(A)p_j(A)}=1,\mathit{\text{if }}i=1.$$

Lemma 3

[14]

Let $\gamma >0$ and $\eta \ge 0$. Then, for any $x\in [0,1]$,

$$\frac{1}{1-x+\gamma x}\le \frac{1}{min\{\gamma ,1\}},\frac{\eta x}{1-x+\gamma x}\le \frac{\eta }{\gamma }.$$

Theorem 1

Let $M=[m_{ij}]\in {\mathbb{R}}^{n\times n}$ be a wcdd B-matrix with the form $M=B^{+}+C$, where $B^{+}=[b_{ij}]$ is defined as (1). If, for each $i\in \mathbb{N}$,

$${\hat{\beta }}_i=b_{ii}-\sum _{j=i+1}^n|b_{ij}|p_i\left(B^{+}\right)>0,$$

then

$$\begin{array}{rl}& \underset{d\in {[0,1]}^n}{max}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}\\ & \le max\{\sum _{i=1}^n\frac{n-1}{min\{{\hat{\beta }}_i,1\}}\prod _{j=1}^{i-1}(\frac{1}{{\hat{\beta }}_j}\sum _{k=j+1}^n|b_{jk}|),\sum _{i=1}^n\frac{(n-1)p_i(B^{+})}{min\{{\hat{\beta }}_i,1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{{\hat{\beta }}_j}\},\end{array}$$ 5

where

$$\prod _{j=1}^{i-1}(\frac{1}{{\hat{\beta }}_j}\sum _{k=j+1}^n|b_{jk}|)=1,\prod _{j=1}^{i-1}\frac{b_{jj}}{{\hat{\beta }}_j}=1,\mathit{\text{if }}i=1.$$

Proof

Let $M_D=I-D+DM$. Then

$$M_D=I-D+DM=I-D+D(B^{+}+C)=B_D^{+}+C_D,$$

where $B_D^{+}=I-D+D{B}^{+}$. Similar to the proof of Theorem 2 in [13], we see that $B_D^{+}$ is a wcdd M-matrix with positive diagonal elements and $C_D=DC$, and, by Lemma 1,

$${\parallel M_D^{-1}\parallel }_{\mathrm{\infty }}\le {\parallel {(I+{\left(B_D^{+}\right)}^{-1}{C}_D)}^{-1}\parallel }_{\mathrm{\infty }}{\parallel {\left(B_D^{+}\right)}^{-1}\parallel }_{\mathrm{\infty }}\le (n-1){\parallel {\left(B_D^{+}\right)}^{-1}\parallel }_{\mathrm{\infty }}.$$ 6

By Lemma 2, we have

$$\begin{array}{rl}{\parallel {\left(B_D^{+}\right)}^{-1}\parallel }_{\mathrm{\infty }}& \le max\{\sum _{i=1}^n\frac{1}{(1-d_i+b_{ii}{d}_i)(1-u_i(B_D^{+})p_i(B_D^{+}))}\prod _{j=1}^{i-1}\frac{u_j((B_D^{+}))}{1-u_j((B_D^{+}))p_j(B_D^{+})},\\ & \sum _{i=1}^n\frac{p_i(B_D^{+})}{(1-d_i+b_{ii}{d}_i)(1-u_i((B_D^{+}))p_i(B_D^{+}))}\prod _{j=1}^{i-1}\frac{1}{1-u_j(B_D^{+})p_j(B_D^{+})}\}.\end{array}$$

By Lemma 3, we can easily get the following results: for each $i,j,k\in \mathbb{N}$,

$$\begin{array}{c}{b}_k\left(B_D^{+}\right)=\underset{k+1\le i\le n}{max}\left\{\frac{{\sum }_{j=k,\ne i}^n|b_{ij}|d_i}{1-d_i+b_{ii}{d}_i}\right\}\le \underset{k+1\le i\le n}{max}\left\{\frac{{\sum }_{j=k,\ne i}^n|b_{ij}|}{b_{ii}}\right\}=b_k\left(B^{+}\right),\hfill \\ p_k\left(B_D^{+}\right)=\underset{k+1\le i\le n}{max}\left\{\frac{|b_{ik}|d_i+{\sum }_{j=k+1,\ne i}^n|b_{ij}|d_i{b}_k(B_D^{+})}{1-d_i+b_{ii}{d}_i}\right\}\hfill \\ \phantom{p_k(B_D^{+})}\le \underset{k+1\le i\le n}{max}\left\{\frac{|b_{ik}|+{\sum }_{j=k+1,\ne i}^n|b_{ij}|b_k(B_D^{+})}{b_{ii}}\right\}\hfill \\ \phantom{p_k(B_D^{+})}\le \underset{k+1\le i\le n}{max}\left\{\frac{|b_{ik}|+{\sum }_{j=k+1,\ne i}^n|b_{ij}|b_k(B^{+})}{b_{ii}}\right\}\hfill \\ \phantom{p_k(B_D^{+})}=p_k\left(B^{+}\right),\hfill \end{array}$$

and

$$\begin{array}{rl}\frac{1}{(1-d_i+b_{ii}{d}_i)(1-u_i(B_D^{+})p_i(B_D^{+}))}& =\frac{1}{1-d_i+b_{ii}{d}_i-{\sum }_{j=i+1}^n|b_{ij}|d_i{p}_i(B_D^{+})}\\ & \le \frac{1}{min\{b_{ii}-{\sum }_{j=i+1}^n|b_{ij}|p_i(B^{+}),1\}}\\ & =\frac{1}{min\{{\hat{\beta }}_i,1\}}.\end{array}$$ 7

Furthermore, by Lemma 3, we have

$$\begin{array}{rl}\frac{u_i(B_D^{+})}{1-u_i(B_D^{+})p_i(B_D^{+})}& =\frac{{\sum }_{j=i+1}^n|b_{ij}|d_i}{1-d_i+b_{ii}{d}_i-{\sum }_{j=i+1}^n|b_{ij}|d_i{p}_i(B_D^{+})}\\ & \le \frac{{\sum }_{j=i+1}^n|b_{ij}|}{b_{ii}-{\sum }_{j=i+1}^n|b_{ij}|p_i(B^{+})}\\ & =\frac{1}{{\hat{\beta }}_i}\sum _{j=i+1}^n|b_{ij}|\end{array}$$ 8

and

$$\begin{array}{rl}\frac{1}{1-u_i(B_D^{+})p_i(B_D^{+})}& =\frac{1-d_i+b_{ii}{d}_i}{1-d_i+b_{ii}{d}_i-{\sum }_{j=i+1}^n|b_{ij}|d_i{p}_i(B_D^{+})}\\ & \le \frac{1-d_i+b_{ii}{d}_i}{b_{ii}-{\sum }_{j=i+1}^n|b_{ij}|p_i(B^{+})}\\ & =\frac{b_{ii}}{{\hat{\beta }}_i}.\end{array}$$ 9

By (7), (8), and (9), we obtain

$${\parallel {\left(B_D^{+}\right)}^{-1}\parallel }_{\mathrm{\infty }}\le max\{\sum _{i=1}^n\frac{1}{min\{{\hat{\beta }}_i,1\}}\prod _{j=1}^{i-1}(\frac{1}{{\hat{\beta }}_j}\sum _{k=j+1}^n|b_{jk}|),\sum _{i=1}^n\frac{p_i(B^{+})}{min\{{\hat{\beta }}_i,1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{{\hat{\beta }}_j}\}.$$ 10

Therefore, the result in (5) holds from (6) and (10). □

Since a B-matrix is also a wcdd B-matrix, then by Theorem 1, we find the following result.

Corollary 1

Let $M=[m_{ij}]\in {\mathbb{R}}^{n\times n}$ be a B-matrix with the form $M=B^{+}+C$, where $B^{+}=[b_{ij}]$ is defined as (1). Then

$$\begin{array}{rl}& \underset{d\in {[0,1]}^n}{max}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}\\ & \le max\{\sum _{i=1}^n\frac{n-1}{min\{{\hat{\beta }}_i,1\}}\prod _{j=1}^{i-1}(\frac{1}{{\hat{\beta }}_j}\sum _{k=j+1}^n|b_{jk}|),\sum _{i=1}^n\frac{(n-1)p_i(B^{+})}{min\{{\hat{\beta }}_i,1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{{\hat{\beta }}_j}\},\end{array}$$ 11

where ${\hat{\beta }}_i$ is defined as in Theorem  1.

We next give a comparison of the bounds in (4) and (5) as follows.

Theorem 2

Let $M=[m_{ij}]\in {\mathbb{R}}^{n\times n}$ be a wcdd B-matrix with the form $M=B^{+}+C$, where $B^{+}=[b_{ij}]$ is defined as (1). Let ${\overline{\beta }}_i$, ${\tilde{\beta }}_i$, and ${\hat{\beta }}_i$ be defined as in (3), (4), and (5), respectively. Then

$$\begin{array}{rl}& max\{\sum _{i=1}^n\frac{n-1}{min\{{\hat{\beta }}_i,1\}}\prod _{j=1}^{i-1}(\frac{1}{{\hat{\beta }}_j}\sum _{k=j+1}^n|b_{jk}|),\sum _{i=1}^n\frac{(n-1)p_i(B^{+})}{min\{{\hat{\beta }}_i,1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{{\hat{\beta }}_j}\}\\ & \le \sum _{i=1}^n\left(\frac{n-1}{min\{{\tilde{\beta }}_i,1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{{\tilde{\beta }}_j}\right).\end{array}$$ 12

Proof

Since $B^{+}$ is a wcdd matrix with positive diagonal elements, for any $i\in \mathbb{N}$,

$$0\le p_i\left(B^{+}\right)\le 1,{\tilde{\beta }}_i\le {\hat{\beta }}_i.$$ 13

By (13), for each $i\in \mathbb{N}$,

$$\frac{1}{{\hat{\beta }}_i}\le \frac{1}{{\tilde{\beta }}_i},\frac{1}{min\{{\hat{\beta }}_i,1\}}\le \frac{1}{min\{{\tilde{\beta }}_i,1\}}.$$ 14

The result in (12) follows by (13) and (14). □

Remark 1

• (i)Theorem 2 shows that the bound in (5) is better than that in (4).
• (ii)When n is very large, one needs more computations to obtain these upper bounds by (5) than by (4).

Numerical examples

In this section, we present numerical examples to illustrate the advantages of our derived results.

Example 1

Consider the family of B-matrices in [14]:

$$M_k=\left[\begin{array}{cccc}1.5& 0.5& 0.4& 0.5\\ -0.1& 1.7& 0.7& 0.6\\ 0.8& -0.1\frac{k}{k+1}& 1.8& 0.7\\ 0& 0.7& 0.8& 1.8\end{array}\right],$$

where $k\ge 1$. Then $M_k=B_k^{+}+C_k$, where

$$B_k^{+}=\left[\begin{array}{cccc}1& 0& -0.1& 0\\ -0.8& 1& 0& -0.1\\ 0& -0.1\frac{k}{k+1}-0.8& 1& -0.1\\ -0.8& -0.1& 0& 1\end{array}\right].$$

By (2), we have

$$\underset{d\in {[0,1]}^4}{max}{\parallel {(I-D+D{M}_k)}^{-1}\parallel }_{\mathrm{\infty }}\le \frac{4-1}{min\{\beta ,1\}}=30(k+1).$$

It is obvious that

$$30(k+1)\to +\mathrm{\infty },\text{if }k\to +\mathrm{\infty }.$$

By (3), we get

$$\underset{d\in {[0,1]}^4}{max}{\parallel {(I-D+D{M}_k)}^{-1}\parallel }_{\mathrm{\infty }}\le 15.2675.$$

By Theorem 7 of [11], we have

$$\underset{d\in {[0,1]}^4}{max}{\parallel {(I-D+D{M}_k)}^{-1}\parallel }_{\mathrm{\infty }}\le 13.6777.$$

By Corollary 1 of [13], we have

$$\underset{d\in {[0,1]}^4}{max}{\parallel {(I-D+D{M}_k)}^{-1}\parallel }_{\mathrm{\infty }}\le \sum _{i=1}^4\left(\frac{3}{min\{{\tilde{\beta }}_i,1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{{\tilde{\beta }}_j}\right)\approx 15.2675.$$

By (11), we obtain

$$\underset{d\in {[0,1]}^4}{max}{\parallel {(I-D+D{M}_k)}^{-1}\parallel }_{\mathrm{\infty }}\le 9.9683.$$

In these two cases, the bounds in (2) are equal to 60 ($k=1$) and 90 ($k=2$), respectively.

Example 2

Consider the wcdd B-matrix in [13]:

$$M=\left[\begin{array}{cccc}1.5& 0.2& 0.4& 0.5\\ -0.1& 1.5& 0.5& 0.1\\ 0.5& -0.1& 1.5& 0.1\\ 0.4& 0.4& 0.8& 1.8\end{array}\right].$$

Then $M=B^{+}+C$, where

$$B^{+}=\left[\begin{array}{cccc}1& -0.3& -0.1& 0\\ -0.6& 1& 0& -0.4\\ 0& -0.6& 1& -0.4\\ -0.4& -0.4& 0& 1\end{array}\right].$$

By (4), we get

$$\underset{d\in {[0,1]}^4}{max}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}\le 41.1111.$$

By (5), we have

$$\underset{d\in {[0,1]}^4}{max}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}\le 21.6667.$$

Conclusions

In this paper, we present some new upper bounds for ${max}_{d\in {[0,1]}^n}{\parallel {(I-D+DM)}^{-1}\parallel }_{\mathrm{\infty }}$ when M is a weakly chained diagonally dominant B-matrix, which improve some existing results. A numerical example shows that the given bounds are efficient.