# An improved error bound for linear complementarity problems for B-matrices

## Abstract

A new error bound for the linear complementarity problem when the matrix involved is a B-matrix is presented, which improves the corresponding result in (Li et al. in Electron. J. Linear Algebra 31(1):476-484, 2016). In addition some sufficient conditions such that the new bound is sharper than that in (García-Esnaola and Peña in Appl. Math. Lett. 22(7):1071-1075, 2009) are provided.

## Introduction

Given an $$n\times n$$ real matrix M and $$q\in R^{n}$$, the linear complementarity problem (LCP) is to find a vector $$x\in R^{n}$$ satisfying

$$x\geqslant0,\qquad Mx+q\geqslant0, \qquad(Mx+q)^{T}x=0$$
(1)

or to show that no such vector x exists. We denote this problem (1) by $$\operatorname{LCP}(M, q)$$. The $$\operatorname{LCP}(M, q)$$ arises in many applications such as finding Nash equilibrium point of a bimatrix game, the network equilibrium problem, the contact problem and the free boundary problem for journal bearing etc.; for details, see .

It is well known that the $$\operatorname{LCP}(M, q)$$ has a unique solution for any vector $$q\in R^{n}$$ if and only if M is a P-matrix . Here a matrix M is called a P-matrix if all its principal minors are positive. For the $$\operatorname{LCP}(M, q)$$, one of the interesting problems is to estimate

$$\max_{d\in[0,1]^{n}}\big\| (I -D+DM)^{-1}\big\| _{\infty},$$
(2)

which can be used to bound the error $$\|x-x^{*}\|_{\infty}$$ , that is,

$$\big\| x-x^{*}\big\| _{\infty} \leqslant\max_{d\in[0,1]^{n}}\big\| (I -D+DM)^{-1}\big\| _{\infty} \big\| r(x)\big\| _{\infty},$$

where $$x^{*}$$ is the solution of the $$\operatorname{LCP}(M, q)$$, $$r(x)=\min\{ x,Mx+q\}$$, $$D=\operatorname{diag}(d_{i})$$ with $$0\leqslant d_{i} \leqslant1$$ for each $$i\in N$$, $$d=[d_{1},d_{2},\ldots,d_{n}]^{T}\in[0,1]^{n}$$, and the min operator $$r(x)$$ denotes the componentwise minimum of two vectors.

When the matrix M for the $$\operatorname{LCP}(M, q)$$ belongs to P-matrices or some subclass of P-matrices, various bounds for (2) were proposed; e.g., see [2, 615] and the references therein. Recently, García-Esnaola and Peña in  provided an upper bound for (2) when M is a B-matrix as a subclass of P-matrices. Here, a matrix $$M=[m_{ij}]\in R^{n, n}$$ is called a B-matrix  if for each $$i\in N=\{1,2,\ldots,n\}$$,

$$\sum_{k\in N}m_{ik}>0, \quad\text{and}\quad \frac{1}{n} \biggl(\sum_{k\in N}m_{ik} \biggr)>m_{ij}\quad \text{for any }j\in N\text{ and }j\neq i.$$

### Theorem 1

, Theorem 2.2

Let $$M=[m_{ij}]\in R^{n, n}$$ be a B-matrix with the form

$$M=B^{+}+C,$$
(3)

where

$$B^{+} =[b_{ij}]= \left [ \textstyle\begin{array}{c@{\quad}c@{\quad}c} m_{11}-r_{1}^{+} &\cdots &m_{1n}-r_{1}^{+} \\ \vdots & &\vdots \\ m_{n1}-r_{n}^{+} &\cdots &m_{nn}-r_{n}^{+} \end{array}\displaystyle \right ],\qquad C=\left [ \textstyle\begin{array}{c@{\quad}c@{\quad}c} r_{1}^{+} &\cdots &r_{1}^{+} \\ \vdots & &\vdots \\ r_{n}^{+} &\cdots &r_{n}^{+} \end{array}\displaystyle \right ],$$
(4)

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

$$\max_{d\in [0,1]^{n}}\big\| (I-D+DM)^{-1}\big\| _{\infty} \leqslant\frac{n-1}{\min\{\beta,1\}},$$
(5)

where $$\beta=\min_{i\in N}\{\beta_{i}\}$$ and $$\beta_{i}=b_{ii}-\sum_{j\neq i}|b_{ij}|$$.

It is not difficult to see that the bound (5) will be inaccurate when the matrix M has very small value of $$\min_{i\in N}\{b_{ii}-\sum_{j\neq i}|b_{ij}|\}$$; for details, see [17, 18]. To conquer this problem, Li et al., in  gave the following bound for (2) when M is a B-matrix, which improves those provided by Li and Li in [17, 18].

### Theorem 2

, Theorem 2.4

Let $$M=[m_{ij}]\in R^{n, n}$$ be a B-matrix with the form $$M=B^{+}+C$$, where $$B^{+}=[b_{ij}]$$ is the matrix of (4). Then

$$\max_{d\in [0,1]^{n}}\big\| (I-D+DM)^{-1}\big\| _{\infty} \leqslant \sum_{i=1}^{n}\frac{n-1}{\min\{\bar{\beta}_{i},1\}} \prod_{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}},$$
(6)

where $$\bar{\beta}_{i}=b_{ii}-\sum_{j=i+1}^{n}|b_{ij}|l_{i}(B^{+})$$ with $$l_{k}(B^{+})=\max_{k\leq i\leq n} \{\frac{1}{|b_{ii}|}\sum_{j=k,\atop j\neq i}^{n}|b_{ij}| \}$$, and $$\prod_{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}}=1$$ if $$i=1$$.

In this paper, we further improve error bounds on the $$\operatorname{LCP}(M,q)$$ when M belongs to B-matrices. The rest of this paper is organized as follows: In Section 2 we present a new error bound for (2), and then prove that this bound is better than those in Theorems 1 and 2. In Section 3, some numerical examples are given to illustrate our theoretical results obtained.

## Main result

In this section, an upper bound for (2) is provided when M is a B-matrix. Firstly, some definitions, notation and lemmas which will be used later are given as follows.

A matrix $$A=[a_{ij}]\in C^{n,n}$$ is called a strictly diagonally dominant (SDD) matrix if $$|a_{ii}|>\sum_{j\neq i}^{n}|a_{ij}|$$ for all $$i=1,2,\ldots,n$$. A matrix $$A=[a_{ij}]\in R^{n,n}$$ is called a nonsingular M-matrix if its inverse is nonnegative and all its off-diagonal entries are nonpositive . In  it was proved that a B-matrix has positive diagonal elements, and a real matrix A is a B-matrix if and only if it can be written in the form (3) with $$B^{+}$$ being a SDD matrix. Given a matrix $$A=[a_{ij}]\in C^{n,n}$$, let

$$\begin{gathered} w_{ij}(A)=\frac{|a_{ij}|}{|a_{ii}|-\sum_{k=j+1,\atop k\neq i}^{n}|a_{ik}|},\quad i\neq j, \\ w_{i}(A)=\max_{j\neq i}\bigl\{ w_{ij}(A) \bigr\} , \\ m_{ij}(A)=\frac{|a_{ij}|+\sum_{ k=j+1,\atop k\neq i}^{n}|a_{ik}|w_{k}(A)}{|a_{ii}|},\quad i\neq j. \end{gathered}$$
(7)

### Lemma 1

, Theorem 14

Let $$A=[a_{ij}]$$ be an $$n\times n$$ row strictly diagonally dominant M-matrix. Then

$$\big\| A^{-1}\big\| _{\infty}\leqslant\sum_{i=1}^{n} \Biggl(\frac{1}{a_{ii}-\sum_{k=i+1}^{n}|a_{ik}|m_{ki}(A)}\prod_{j=1}^{i-1} \frac{1}{1-u_{j}(A)l_{j}(A)} \Biggr),$$

where $$u_{i}(A)=\frac{1}{|a_{ii}|}\sum_{j=i+1}^{n}|a_{ij}|$$, $$l_{k}(A)=\max_{k\leq i\leq n} \{\frac{1}{|a_{ii}|}\sum_{j=k,\atop j\neq i}^{n}|a_{ij}| \}$$, $$\prod_{j=1}^{i-1}\frac{1}{1-u_{j}(A)l_{j}(A)}=1$$ if $$i=1$$, and $$m_{ki}(A)$$ is defined as in (7).

### Lemma 2

, Lemma 3

Let $$\gamma> 0$$ and $$\eta\geqslant0$$. Then, for any $$x\in [0,1]$$,

$$\frac{1}{1-x+\gamma x} \leqslant \frac{1}{\min\{\gamma,1\}}$$

and

$$\frac{\eta x}{1-x+\gamma x} \leqslant \frac{\eta}{\gamma}.$$

### Lemma 3

, Lemma 5

Let $$A=[a_{ij}]$$ with $$a_{ii}>\sum_{j=i+1}^{n}|a_{ij}|$$ for each $$i\in N$$. Then, for any $$x_{i}\in[0,1]$$,

$$\frac{1-x_{i}+a_{ii}x_{i}}{1-x_{i}+a_{ii}x_{i}-\sum_{j=i+1}^{n}|a_{ij}|x_{i}} \leqslant \frac{a_{ii}}{a_{ii}-\sum_{j=i+1}^{n}|a_{ij}|}.$$

Lemmas 2 and 3 will be used in the proofs of the following lemma and Theorem 3.

### Lemma 4

Let $$M=[m_{ij}]\in R^{n, n}$$ be a B-matrix with the form $$M=B^{+}+C$$, where $$B^{+}=[b_{ij}]$$ is the matrix of (4). And let $$B_{D}^{+}=I-D+DB^{+}=[\tilde{b}_{ij}]$$ where $$D=\operatorname{diag}(d_{i})$$ with $$0\leqslant d_{i} \leqslant1$$. Then

$$w_{i}\bigl(B_{D}^{+}\bigr)\leqslant \max _{j\neq i} \biggl\{ \frac{|b_{ij}|}{b_{ii}-\sum_{k=j+1,\atop k\neq i}^{n}|b_{ik}|} \biggr\}$$

and

$$m_{ij}\bigl(B_{D}^{+}\bigr)\leqslant v_{ij}\bigl(B^{+}\bigr)< 1,$$

where $$w_{i}(B_{D}^{+})$$, $$m_{ij}(B_{D}^{+})$$ are defined as in (7), and

$$v_{ij}\bigl(B^{+}\bigr)=\frac{1}{b_{ii}} \Biggl(|b_{ij}|+\sum_{k=j+1,\atop k\neq i}^{n} \biggl(|b_{ik}|\cdot\max_{h\neq k} \biggl\{ \frac{|b_{kh}|}{b_{kk}-\sum_{l=h+1,\atop l\neq k}^{n}|b_{kl}|} \biggr\} \biggr) \Biggr).$$

### Proof

Note that

$$\bigl[B_{D}^{+}\bigr]_{ij}=\tilde{b}_{ij}= \left \{ \textstyle\begin{array}{l@{\quad}l} 1-d_{i}+d_{i}b_{ij}, &i=j,\\ d_{i}b_{ij}, &i\neq j. \end{array}\displaystyle \right .$$

Since $$B^{+}$$ is SDD, $$b_{ii}-\sum_{k=j+1,\atop k\neq i}^{n}|b_{ik}|> |b_{ij}|$$ for each $$i\neq j$$. Hence, by Lemma 2 and (7), it follows that

\begin{aligned}[b] w_{i}\bigl(B_{D}^{+}\bigr)&=\max _{j\neq i} \bigl\{ w_{ij}\bigl(B_{D}^{+} \bigr) \bigr\} =\max_{j\neq i} \biggl\{ \frac{|b_{ij}|d_{i}}{1-d_{i}+b_{ii}d_{i}-\sum_{k=j+1,\atop k\neq i}^{n}|b_{ik}|d_{i}} \biggr\} \\ &\leqslant\max_{j\neq i} \biggl\{ \frac{|b_{ij}|}{b_{ii}-\sum_{k=j+1,\atop k\neq i}^{n}|b_{ik}|} \biggr\} < 1. \end{aligned}
(8)

Furthermore, it follows from (7), (8) and Lemma 2 that for each $$i\neq j$$ ($$j< i\leqslant n$$)

\begin{aligned} m_{ij}\bigl(B_{D}^{+}\bigr)&=\frac{|b_{ij}|\cdot d_{i}+\sum_{k=j+1,\atop k\neq i}^{n}|b_{ik}|\cdot d_{i}\cdot w_{k}(B_{D}^{+})}{1-d_{i}+b_{ii}\cdot d_{i}} \\ &\leqslant \frac{1}{b_{ii}}\cdot \Biggl(|b_{ij}|+\sum _{k=j+1,\atop k\neq i}^{n}|b_{ik}|\cdot w_{k} \bigl(B_{D}^{+}\bigr) \Biggr) \\ &\leqslant\frac{1}{b_{ii}} \Biggl(|b_{ij}|+\sum _{k=j+1,\atop k\neq i}^{n} \biggl(|b_{ik}|\cdot\max _{h\neq k} \biggl\{ \frac{|b_{kh}|}{b_{kk}-\sum_{l=h+1,\atop l\neq k}^{n}|b_{kl}|} \biggr\} \biggr) \Biggr) \\ &=v_{ij}\bigl(B^{+}\bigr) \\ &< \frac{1}{b_{ii}} \Biggl(|b_{ij}|+\sum _{k=j+1,\atop k\neq i}^{n}|b_{ik}| \Biggr)< 1. \end{aligned}

The proof is completed. □

By Lemmas 1, 2, 3 and 4, we give the following bound for (2) when M is a B-matrix.

### Theorem 3

Let $$M=[m_{ij}]\in R^{n, n}$$ be a B-matrix with the form $$M=B^{+}+C$$, where $$B^{+}=[b_{ij}]$$ is the matrix of (4). Then

$$\max_{d\in [0,1]^{n}}\big\| (I-D+DM)^{-1}\big\| _{\infty} \leqslant \sum_{i=1}^{n}\frac{n-1}{\min\{\widehat{\beta}_{i},1\}} \prod_{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}},$$
(9)

where $$\widehat{\beta}_{i}=b_{ii}-\sum_{k=i+1}^{n}|b_{ik}|\cdot v_{ki}(B^{+})$$ with $$v_{ki}(B^{+})$$ is defined in Lemma 4, $$\bar{\beta}_{i}$$ is defined in Theorem 2, and $$\prod_{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}}=1$$ if $$i=1$$.

### Proof

Let $$M_{D}=I-D+DM$$. Then

$$M_{D}=I-D+DM=I-D+D\bigl(B^{+}+C\bigr)=B_{D}^{+}+C_{D},$$

where $$B_{D}^{+}=I-D+DB^{+}=[\tilde{b}_{ij}]$$ and $$C_{D}=DC$$. Similarly to the proof of Theorem 2.2 in , we find that $$B_{D}^{+}$$ is an SDD M-matrix with positive diagonal elements and that

$$\big\| M_{D}^{-1}\big\| _{\infty}\leqslant\big\| \bigl(I + \bigl(B^{+}_{D}\bigr)^{-1}C_{D} \bigr)^{-1} \big\| _{\infty}\big\| \bigl(B^{+}_{D} \bigr)^{-1} \big\| _{\infty}\leqslant(n-1) \big\| \bigl(B^{+}_{D} \bigr)^{-1} \big\| _{\infty}.$$
(10)

Next, we give an upper bound for $$\|(B^{+}_{D} )^{-1} \|_{\infty}$$. By Lemma 1, we have

$$\big\| \bigl(B^{+}_{D} \bigr)^{-1} \big\| _{\infty}\leqslant\sum_{i=1}^{n} \Biggl( \frac{1}{1-d_{i}+b_{ii}d_{i}-\sum_{k=i+1}^{n}|b_{ik}|\cdot d_{i}\cdot m_{ki}(B^{+}_{D})}\prod_{j=1}^{i-1} \frac{1}{1-u_{j}(B^{+}_{D})l_{j}(B^{+}_{D})} \Biggr),$$
(11)

where

$$u_{j}\bigl(B^{+}_{D}\bigr)=\frac{\sum_{k=j+1}^{n}|b_{jk}|d_{j}}{1-d_{j}+b_{jj}d_{j}},\qquad l_{k}\bigl(B^{+}_{D}\bigr)=\max_{k\leq i\leq n} \biggl\{ \frac{\sum_{j=k,\atop j\neq i}^{n}|b_{ij}|d_{i}}{1-d_{i}+b_{ii}d_{i}} \biggr\} ,$$

and

$$m_{ki}\bigl(B^{+}_{D}\bigr)=\frac{|b_{ki}|\cdot d_{k}+\sum_{l=i+1,\atop l\neq k}^{n}|b_{kl}|\cdot d_{k}\cdot w_{l}(B_{D}^{+})}{1-d_{k}+b_{kk}\cdot d_{k}}$$

with $$w_{l}(B_{D}^{+})=\max_{h\neq l} \{\frac{|b_{lh}|d_{l}}{1-d_{l}+b_{ll}d_{l}-\sum_{s=h+1,\atop s\neq l}^{n}|b_{ls}|d_{l}} \}$$.

By Lemmas 2 and 4, we can easily see that, for each $$i\in N$$,

\begin{aligned}[b]\frac{1}{1-d_{i}+b_{ii}d_{i}-\sum_{k=i+1}^{n}|b_{ik}|\cdot d_{i}\cdot m_{ki}(B^{+}_{D})}&\leqslant \frac{1}{\min \{b_{ii}-\sum_{k=i+1}^{n}|b_{ik}|\cdot m_{ki}(B^{+}_{D}),1 \}} \\ &\leqslant \frac{1}{\min \{b_{ii}-\sum_{k=i+1}^{n}|b_{ik}|\cdot v_{ki}(B^{+}),1 \}} \\ &=\frac{1}{\min \{\widehat{\beta }_{i},1 \}}, \end{aligned}
(12)

and that, for each $$k\in N$$,

$$l_{k}\bigl(B^{+}_{D}\bigr)=\max _{k\leq i\leq n} \biggl\{ \frac{\sum_{j=k,\atop j\neq i}^{n}|b_{ij}|d_{i}}{1-d_{i}+b_{ii}d_{i}} \biggr\} \leqslant \max _{k\leq i\leq n} \Biggl\{ \frac{1}{b_{ii}}\sum _{j=k,\atop j\neq i}^{n}|b_{ij}| \Biggr\} =l_{k} \bigl(B^{+}\bigr)< 1.$$
(13)

Furthermore, according to Lemma 3 and (13), it follows that, for each $$j\in N$$,

$$\frac{1}{1-u_{j}(B^{+}_{D})l_{j}(B^{+}_{D})}=\frac {1-d_{j}+b_{jj}d_{j}}{1-d_{j}+b_{jj}d_{j}-\sum_{k=j+1}^{n}|b_{jk}|\cdot d_{j}\cdot l_{j}(B^{+}_{D})}\leqslant\frac{b_{jj}}{\bar{\beta}_{j}}.$$
(14)

By (11), (12) and (14), we have

$$\big\| \bigl(B^{+}_{D} \bigr)^{-1} \big\| _{\infty}\leqslant \frac{1}{\min \{\widehat{\beta}_{1},1 \}}+\sum_{i=2}^{n} \Biggl(\frac{1}{\min \{\widehat{\beta}_{i},1 \}}\prod_{j=1}^{i-1} \frac {b_{jj}}{\bar{\beta}_{j}} \Biggr).$$
(15)

The conclusion follows from (10) and (15). □

The comparisons of the bounds in Theorems 2 and 3 are established as follows.

### Theorem 4

Let $$M=[m_{ij}]\in R^{n, n}$$ be a B-matrix with the form $$M=B^{+}+C$$, where $$B^{+}=[b_{ij}]$$ is the matrix of (4). Let $$\bar{\beta}_{i}$$ and $$\widehat{\beta}_{i}$$ be defined in Theorems 2 and 3, respectively. Then

$$\sum_{i=1}^{n}\frac{n-1}{\min\{\widehat{\beta}_{i},1\} }\prod _{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}} \leqslant \sum _{i=1}^{n}\frac{n-1}{\min\{\bar{\beta}_{i},1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}}.$$

### Proof

Note that

$$\bar{\beta}_{i}=b_{ii}-\sum_{j=i+1}^{n}|b_{ij}|l_{i} \bigl(B^{+}\bigr),\qquad \widehat{\beta }_{i}=b_{ii}- \sum_{k=i+1}^{n}|b_{ik}|v_{ki} \bigl(B^{+}\bigr),$$

and $$B^{+}$$ is a SDD matrix, it follows that for each $$i\neq j$$ ($$j< i\leqslant n$$)

\begin{aligned} v_{ij}\bigl(B^{+}\bigr)&=\frac{1}{b_{ii}} \Biggl(|b_{ij}|+\sum_{k=j+1,\atop k\neq i}^{n} \biggl(|b_{ik}|\cdot\max_{h\neq k} \biggl\{ \frac{|b_{kh}|}{b_{kk}-\sum_{l=h+1,\atop l\neq k}^{n}|b_{kl}|} \biggr\} \biggr) \Biggr) \\ &< \frac{1}{b_{ii}}\sum_{k=j,\atop k\neq i}^{n}|b_{ik}| \\ &\leqslant \max_{j\leqslant i\leqslant n} \Biggl\{ \frac{1}{b_{ii}}\sum _{k=j,\atop k\neq i}^{n}|b_{ik}| \Biggr\} =l_{j} \bigl(B^{+}\bigr). \end{aligned}

Hence, for each $$i\in N$$

$$\widehat{\beta}_{i}=b_{ii}-\sum _{k=i+1}^{n}|b_{ik}|v_{ki} \bigl(B^{+}\bigr)>b_{ii}-\sum_{k=i+1}^{n}|b_{ik}|l_{i} \bigl(B^{+}\bigr)=\bar{\beta}_{i},$$

which implies that

$$\frac{1}{\min\{\widehat{\beta}_{i},1\}}\leqslant \frac{1}{\min\{\bar{\beta}_{i},1\}}.$$

This completes the proof. □

Remark here that, when $$\bar{\beta}_{i}<1$$ for all $$i\in N$$, then

\begin{aligned} \frac{1}{\min\{\widehat{\beta}_{i},1\}}< \frac{1}{\min\{\bar{\beta}_{i},1\}}, \end{aligned}

which yields

$$\sum_{i=1}^{n}\frac{n-1}{\min\{\widehat{\beta}_{i},1\} }\prod _{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}} < \sum _{i=1}^{n}\frac{n-1}{\min\{\bar{\beta}_{i},1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}}.$$

Next it is proved that the bound (9) given in Theorem 3 can improve the bound (5) in Theorem 1 (Theorem 2.2 in ) in some cases.

### Theorem 5

Let $$M=[m_{ij}]\in R^{n, n}$$ be a B-matrix with the form $$M=B^{+}+C$$, where $$B^{+}=[b_{ij}]$$ is the matrix of (4). Let β, $$\bar{\beta}_{i}$$ and $$\widehat{\beta}_{i}$$ be defined in Theorems 1, 2 and 3, respectively, and let $$\alpha=1+\sum_{i=2}^{n}\prod_{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}}$$ and $$\widehat{\beta}=\min_{i\in N}\{\widehat{\beta}_{i}\}$$. If one of the following conditions holds:

1. (i)

$$\widehat{\beta}>1$$ and $$\alpha<\frac{1}{\beta}$$;

2. (ii)

$$\widehat{\beta}<1$$ and $$\alpha\beta<\widehat{\beta}$$,

then

$$\sum_{i=1}^{n}\frac{n-1}{\min\{\widehat{\beta}_{i},1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}}< \frac{n-1}{\min\{\beta,1\}}.$$

### Proof

When $$\widehat{\beta}>1$$ and $$\alpha<\frac{1}{\beta}$$, we can easily get

$$\sum_{i=1}^{n}\frac{n-1}{\min\{\widehat{\beta}_{i},1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}}< \frac{n-1}{\min\{\widehat{\beta},1\}}\sum_{i=1}^{n}\prod _{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}}= (n-1)\alpha< \frac{n-1}{\beta}\leqslant\frac{n-1}{\min\{\beta,1\}}.$$

Similarly, for $$\widehat{\beta}<1$$ and $$\alpha\beta<\widehat{\beta}$$, the conclusion can be proved directly. □

## Numerical examples

Two examples are given to show that the bound in Theorem 3 is sharper than those in Theorems 1 and 2.

### Example 1

Consider the family of B-matrices in :

$$M_{k} = \left [ \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} 1.5 &0.5 &0.4 &0.5 \\ -0.1 &1.7 &0.7 &0.6 \\ 0.8 &-0.1\frac{k}{k+1} &1.8 &0.7 \\ 0 & 0.7 &0.8 & 1.8 \end{array}\displaystyle \right ],$$

where $$k\geqslant1$$. Then $$M_{k}=B_{k}^{+}+C_{k}$$, where

$$B_{k}^{+} = \left [ \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} 1 &0 &-0.1 &0 \\ -0.8 &1 &0 &-0.1 \\ 0 &-0.1\frac{k}{k+1}-0.8 &1 &-0.1\\ -0.8 & -0.1 &0 & 1 \end{array}\displaystyle \right ].$$

By computations, we have $$\beta=\frac{1}{10(k+1)}$$, $$\bar{\beta}_{1}=\bar{\beta}_{2}=\frac {90k+91}{100k+100}$$, $$\bar{\beta}_{3}=0.99$$, $$\bar{\beta}_{4}=1$$, $$\hat{\beta}_{1}=\frac{820k+828}{900k+900}$$, $$\hat{\beta}_{2}=0.99$$, $$\hat{\beta}_{3}=1$$ and $$\hat{\beta}_{4}=1$$. Then it is easy to verify that $$M_{k}$$ satisfies the condition (ii) of Theorem 5. Hence, by Theorem 1 (Theorem 2.2 in ), we have

$$\max_{d\in [0,1]^{4}}\big\| (I-D+DM_{k})^{-1}\big\| _{\infty} \leqslant \frac{4-1}{\min\{\beta,1\}}=30(k+1).$$

It is obvious that

$$30(k+1)\longrightarrow+\infty,\quad \text{when }k\longrightarrow+\infty.$$

By Theorem 2, we find that, for any $$k\geqslant1$$,

$$\begin{gathered} \max_{d\in [0,1]^{4}}\big\| (I-D+DM_{k})^{-1}\big\| _{\infty} \\ \quad\leqslant 3 \biggl(\frac{1}{\bar{\beta}_{1}}+\frac{1}{\bar{\beta}_{2}}\cdot \frac{1}{\bar {\beta}_{1}}+\frac{1}{\bar{\beta}_{3}}\cdot\frac{1}{\bar{\beta}_{1}\bar{\beta}_{2}} +\frac{1}{\bar{\beta}_{1}\bar{\beta}_{2}\bar{\beta}_{3}} \biggr) \\ \quad=3 \biggl(\frac{100k+100}{90k+91}+\frac{(100k+100)^{2}}{(90k+91)^{2}} +\frac{2(100k+100)^{2}}{0.99(90k+91)^{2}} \biggr)< 14.5193. \end{gathered}$$

By Theorem 3, we find that, for any $$k\geqslant1$$,

$$\begin{gathered} \max_{d\in [0,1]^{4}}\big\| (I-D+DM_{k})^{-1}\big\| _{\infty} \\ \quad\leqslant 3 \biggl(\frac{1}{\hat{\beta}_{1}}+\frac{1}{\hat{\beta}_{2}}\cdot \frac{1}{\bar {\beta}_{1}}+\frac{1}{\bar{\beta}_{1}\bar{\beta}_{2}} +\frac{1}{\bar{\beta}_{1}\bar{\beta}_{2}\bar{\beta}_{3}} \biggr) \\ \quad=3 \biggl(\frac{900k+900}{820k+828}+\frac{(100k+100)}{0.99(90k+91)} +\frac{1.99(100k+100)^{2}}{0.99(90k+91)^{2}} \biggr) \\ \quad< 3 \biggl(\frac {100k+100}{90k+91}+\frac{(100k+100)^{2}}{(90k+91)^{2}} +\frac{2(100k+100)^{2}}{0.99(90k+91)^{2}} \biggr). \end{gathered}$$

In particular, when $$k=1$$,

$$\begin{gathered} 3 \biggl(\frac{900k+900}{820k+828}+\frac{(100k+100)}{0.99(90k+91)} +\frac{1.99(100k+100)^{2}}{0.99(90k+91)^{2}} \biggr) \approx13.9878, \\3 \biggl(\frac{100k+100}{90k+91}+\frac{(100k+100)^{2}}{(90k+91)^{2}} +\frac{2(100k+100)^{2}}{0.99(90k+91)^{2}} \biggr) \approx14.3775, \end{gathered}$$

and the bound (5) in Theorem 1 is

$$\frac{4-1}{\min\{\beta,1\}}=30(k+1)=60.$$

When $$k=2$$,

$$\begin{gathered} 3 \biggl(\frac{900k+900}{820k+828}+\frac{(100k+100)}{0.99(90k+91)} +\frac{1.99(100k+100)^{2}}{0.99(90k+91)^{2}} \biggr) \approx14.0265, \\3 \biggl(\frac{100k+100}{90k+91}+\frac{(100k+100)^{2}}{(90k+91)^{2}} +\frac{2(100k+100)^{2}}{0.99(90k+91)^{2}} \biggr) \approx14.4246, \end{gathered}$$

and the bound (5) in Theorem 1 is

$$\frac{4-1}{\min\{\beta,1\}}=30(k+1)=90.$$

### Example 2

Consider the following family of B-matrices:

$$M_{k} = \left [ \textstyle\begin{array}{c@{\quad}c} \frac{1}{k} &\frac{-a}{k} \\ 0 &\frac{1}{k} \end{array}\displaystyle \right ],$$

where $$\frac{\sqrt{5}-1}{2}< a<1$$ and $$\frac{2-a^{2}}{1+a}< k<1$$. Then $$M_{k}=B_{k}^{+}+C$$ with C is the null matrix.

By simple computations, we can get

$$\beta=\frac{1-a}{k},\qquad\bar{\beta}_{1}=\frac{1-a^{2}}{k},\qquad\bar{ \beta}_{2}=\frac {1}{k},\qquad\hat{\beta}_{1}= \frac{1}{k}\quad \text{and} \quad\hat{\beta}_{2}=\frac{1}{k}.$$

It is not difficult to verify that $$M_{k}$$ satisfies the condition (i) of Theorem 5. Thus, the bound (6) of Theorem 2 (Theorem 2.4 in ) is

$$\sum_{i=1}^{2}\frac{2-1}{\min\{\bar{\beta}_{i},1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}}= \frac{k+1}{1-a^{2}},$$

which is larger than the bound

$$\frac{1}{\min\{\beta,1\}}=\frac{k}{1-a}$$

given by (5) in Theorem 1 (Theorem 2.2 in ). However, by Theorem 3 we can get

$$\max_{d\in[0,1]^{2}}\big\| (I-D+DM_{k})^{-1}\big\| _{\infty} \leqslant \frac{2-a^{2}}{1-a^{2}},$$

which is smaller than the bound (5) in Theorem 1, i.e.,

$$\frac{2-a^{2}}{1-a^{2}}< \frac{k}{1-a}.$$

In particular, when $$a=\frac{4}{5}$$ and $$k=\frac{8}{9}$$, the bounds in Theorems 1 and 2 are, respectively,

$$\frac{1}{\min\{\beta,1\}}=\frac{k}{1-a}=\frac{360}{81}$$

and

$$\sum_{i=1}^{2}\frac{2-1}{\min\{\bar{\beta}_{i},1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}}= \frac{k+1}{1-a^{2}}=\frac{425}{81},$$

while the bound (9) in Theorem 3 is

$$\sum_{i=1}^{2}\frac{2-1}{\min\{\hat{\beta}_{i},1\}}\prod _{j=1}^{i-1}\frac{b_{jj}}{\bar{\beta}_{j}}= \frac{2-a^{2}}{1-a^{2}}=\frac{306}{81}.$$

These two examples show that the bound in Theorem 3 is sharper than those in Theorems 1 and 2.

## Conclusions

In this paper, we give a new error bound for the linear complementarity problem when the matrix involved is a B-matrix, which improves those bounds obtained in  and . Numerical examples are given to illustrate the corresponding results.

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## Acknowledgements

This work is partly supported by National Natural Science Foundations of China (11601473, 31600299), Young Talent fund of University Association for Science and Technology in Shaanxi, China (20160234), the Research Foundation of Baoji University of Arts and Sciences (ZK2017021), and CAS ‘Light of West China’ Program.

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Correspondence to Chaoqian Li.

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