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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 743861, 14 pages
doi:10.1155/2012/743861
Research Article

Generalized AOR Method for Solving Absolute Complementarity Problems

Muhammad Aslam Noor,1 Javed Iqbal,1 Khalida Inayat Noor,1 and E. Al-Said2

1Department of Mathematics, COMSATS Institute of Information Technology, Park Road, Chak Shahzad, Islamabad 44000, Pakistan
2Mathematics Department, College of Science, King Saud University, Riyadh 11451, Saudi Arabia

Received 25 January 2012; Accepted 9 February 2012

Academic Editor: Yonghong Yao

Copyright © 2012 Muhammad Aslam Noor et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We introduce and consider a new class of complementarity problems, which is called the absolute value complementarity problem. We establish the equivalence between the absolute complementarity problems and the fixed point problem using the projection operator. This alternative equivalent formulation is used to discuss the existence of a solution of the absolute value complementarity problem. A generalized AOR method is suggested and analyzed for solving the absolute the complementarity problems. We discuss the convergence of generalized AOR method for the L-matrix. Several examples are given to illustrate the implementation and efficiency of the method. Results are very encouraging and may stimulate further research in this direction.

1. Introduction

Complementarity theory introduced and studied by Lemke [1] and Cottle and Dantzig [2] has enjoyed a vigorous growth for the last fifty years. It is well known that both the linear and nonlinear programs can be characterized by a class of complementarity problems. The complementarity problems have been generalized and extended to study a wide class of problems, which arise in pure and applied sciences; see [1–24] and the references therein. Equally important is the variational inequality problem, which was introduced and studied in the early sixties. The theory of variational inequality has been developed not only to study the fundamental facts on the qualitative behavior of solutions but also to provide highly efficient new numerical methods for solving various nonlinear problems. For the recent applications, formulation, numerical results, and other aspects of the variational inequalities, see [13–22].

Motivated and inspired by the research going on in these areas, we introduced and consider a new class of complmenetarity problems, which is called the absolute value complmenetarity problem. Related to the absolute value complementarity problem, we consider the problem of solving the absolute value variational inequality. We show that if the under lying set is a convex cone, then both these problems are equivalent. If the underlying set is the whole space, then the absolute value problem is equivalent to solving the absolute value equations, which have been studied extensively in recent years.

We use the projection technique to show that the absolute value complementarity problems are equivalent to the fixed point problem. This alternative equivalent form is used to study the existence of a unique solution of the absolute value complementarity problems under some suitable conditions. We again use the fixed point formulation to suggest and analyze a generalized AOR method for solving the absolute value complementarity problems. The convergence analysis of the proposed method is considered under some suitable conditions. Some examples are given to illustrate the efficiency and implementation of the proposed iterative methods. Results are very encouraging. The ideas and the technique of this paper may stimulate further research in these areas.

Let 𝑅 𝑛 be an inner product space, whose inner product and norm are denoted by ⟨ ⋅ , ⋅ ⟩ and ‖ ⋅ ‖ , respectively. For a given matrix 𝐴 ∈ 𝑅 𝑛 × 𝑛 , a vector 𝑏 ∈ 𝑅 𝑛 , we consider the problem of finding 𝑥 ∈ 𝑅 𝑛 such that 𝑥 ∈ 𝐾 , 𝐴 𝑥 − | 𝑥 | − 𝑏 ∈ 𝐾 ∗ , ⟨ 𝐴 𝑥 − | 𝑥 | − 𝑏 , 𝑥 ⟩ = 0 , ( 1 . 1 ) where 𝐾 ∗ = { 𝑥 ∈ 𝑅 𝑛 ∶ ⟨ 𝑥 , 𝑦 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐾 } is the polar cone of a closed and convex cone 𝐾 , 𝐴 ∈ 𝑅 𝑛 × 𝑛 , 𝑏 ∈ 𝑅 𝑛 , and | 𝑥 | will denote the vector in 𝑅 𝑛 with absolute values of components of 𝑥 ∈ 𝑅 𝑛 .

We remark that the absolute value complementarity problem (1.1) can be viewed as an extension of the complementarity problem considered by Lemke [1]. To solve the linear complementarity problems, several methods were proposed. These methods can be divided into two categories, the direct and indirect (iterative) methods. Lemke [1] and Cottle and Dantzig [2] developed the direct methods for solving linear complementarity problems based on the process of pivoting, whereas Mangasarian [10], Noor [14, 15], and Noor et al. [20–22] considered the iterative methods. For recent applications, formulations, numerical methods, and other aspects of the complementarity problems and variational inequalities, see [1–24].

Let 𝐾 be a closed and convex set in the inner product space 𝑅 𝑛 . We consider the problem of finding 𝑥 ∈ 𝐾 such that ⟨ 𝐴 𝑥 − | 𝑥 | − 𝑏 , 𝑦 − 𝑥 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐾 . ( 1 . 2 ) The problem (1.2) is called the absolute value variational inequality, which is a special form of the mildly nonlinear variational inequalities, introduced and studied by Noor [13] in 1975.

If 𝐾 = 𝑅 𝑛 , then the problem (1.2) is equivalent to find 𝑥 ∈ 𝑅 𝑛 such that 𝐴 𝑥 − | 𝑥 | − 𝑏 = 0 , ( 1 . 3 ) which are known as the absolute value equations. These equations have been considered and studied extensively in recent years; see [7–12, 17–20, 23, 24]. We would like to emphasize that Mangasarian [1, 2, 4–11] has shown that the absolute value equations (1.3) are equivalent to the complementarity problems (1.1). Mangasarian [7–11] has used the complementarity approach to solve the absolute value equations (1.3). For other methods, see [7–12, 20–24] and the references therein.

In this paper, we suggest a generalized AOR method for solving absolute complementarity problem, which is easy to implement and gives almost exact solution of (1.3).

We also need the following definitions and concepts.

Definition 1.1. 𝐵 ∈ 𝑅 𝑛 × 𝑛 is called an 𝐿 -matrix if 𝑏 𝑖 𝑖 > 0 for 𝑖 = 1 , 2 , … , 𝑛 , and 𝑏 𝑖 𝑗 ≤ 0 for 𝑖 ≠ 𝑗 , 𝑖 , 𝑗 = 1 , 2 , … , 𝑛 .

Definition 1.2. A matrix 𝐴 ∈ 𝑅 𝑛 × 𝑛 is said to be positive definite matrix, if there exists a constant 𝛾 > 0 , such that ⟨ 𝐴 𝑥 , 𝑥 ⟩ ≥ 𝛾 ‖ 𝑥 ‖ 2 , ∀ 𝑥 ∈ 𝑅 𝑛 , ( 1 . 4 ) and bounded if there exists a constant 𝛽 > 0 such that ‖ 𝐴 𝑥 ‖ ≤ 𝛽 ‖ 𝑥 ‖ , ∀ 𝑥 ∈ 𝑅 𝑛 . ( 1 . 5 )

2. Iterative Methods

To propose and analyze algorithm for absolute complementarity problems, we need the following well-known results.

Lemma 2.1 (see [16]). Let 𝐾 be a nonempty closed convex set in 𝑅 𝑛 . For a given 𝑧 ∈ 𝑅 𝑛 , 𝑢 ∈ 𝐾 satisfies the inequality ⟨ 𝑢 − 𝑧 , 𝑢 − 𝑣 ⟩ ≥ 0 , 𝑣 ∈ 𝐾 , ( 2 . 1 ) if and only if 𝑢 = 𝑃 𝐾 𝑧 , ( 2 . 2 ) where 𝑃 𝐾 is the projection of 𝑅 𝑛 onto the closed convex set 𝐾 .

Lemma 2.2. If 𝐾 is the positive cone in 𝑅 𝑛 , then 𝑥 ∈ 𝐾 is a solution of absolute value variational inequality (1.2) if and only if 𝑥 ∈ 𝐾 solves the absolute value complementarity problem (1.1).

Proof. Let 𝑥 ∈ 𝐾   be the solution of (1.2). Then ⟨ 𝐴 𝑥 − | 𝑥 | − 𝑏 , 𝑦 − 𝑥 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐾 . ( 2 . 3 ) Since 𝐾   is a cone, taking 𝑦 = 0 ∈ 𝐾 and 𝑦 = 2 𝑥 ∈ 𝐾 , we have ⟨ 𝐴 𝑥 − | 𝑥 | − 𝑏 , 𝑥 ⟩ = 0 . ( 2 . 4 ) From inequality (1.2), we have 0 ≤ ⟨ 𝐴 𝑥 − | 𝑥 | − 𝑏 , 𝑦 − 𝑥 ⟩ = ⟨ 𝐴 𝑥 − | 𝑥 | − 𝑏 , 𝑦 ⟩ − ⟨ 𝐴 𝑥 − | 𝑥 | − 𝑏 , 𝑥 ⟩ , = ⟨ 𝐴 𝑥 − | 𝑥 | − 𝑏 , 𝑦 ⟩ , ∀ 𝑦 ∈ 𝐾 , ( 2 . 5 ) from which it follows that 𝐴 𝑥 − | 𝑥 | − 𝑏 ∈ 𝐾 ∗ . Thus we conclude that 𝑥 ∈ 𝐾 is the solution of absolute complementarity problems (1.1).
Conversely, let 𝑥 ∈ 𝐾 be a solution of (1.1). Then 𝑥 ∈ 𝐾 , 𝐴 𝑥 − | 𝑥 | − 𝑏 ∈ 𝐾 ∗ , ⟨ 𝐴 𝑥 − | 𝑥 | − 𝑏 , 𝑥 ⟩ = 0 . ( 2 . 6 ) From (2.5) and (2.6), it follows that ⟨ 𝐴 𝑥 − | 𝑥 | − 𝑏 , 𝑦 − 𝑥 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐾 . ( 2 . 7 ) Hence 𝑥 ∈ 𝐾 is the solution of absolute variational inequality (1.2).

From Lemma 2.1, it follows that both the problems (1.1) and (1.2) are equivalent.

In the next result, we prove the equivalence between the absolute value variational inequality (1.2) and the fixed point.

Lemma 2.3. Let 𝐾 be closed convex set in 𝑅 𝑛 . Then, for a constant 𝜌 > 0 , 𝑥 ∈ 𝐾 satisfies (1.2) if and only if 𝑥 ∈ 𝐾 satisfies the relation 𝑥 = 𝑃 𝐾 [ ] ( 𝑥 − 𝜌 𝐴 𝑥 − | 𝑥 | − 𝑏 ) , ( 2 . 8 ) where 𝑃 𝐾 is the projection of 𝑅 𝑛 onto the closed convex set 𝐾 .

Proof. Let 𝑥 ∈ 𝐾   be the solution of (1.2). Then, for a positive constant 𝜌 > 0 , ⟨ 𝑥 − ( 𝑥 − 𝜌 ( 𝐴 𝑥 − | 𝑥 | − 𝑏 ) ) , 𝑦 − 𝑥 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐾 . ( 2 . 9 ) Using Lemma 2.1, we have 𝑥 = 𝑃 𝐾 [ ] ( 𝑥 − 𝜌 𝐴 𝑥 − | 𝑥 | − 𝑏 ) , ( 2 . 1 0 ) which is the required result.

Now using Lemmas 2.2 and 2.3, we see that the absolute value complementarity problem (1.1) is equivalent to the fixed point problem of the following type: 𝑥 = 𝑃 𝐾 [ ] ( 𝑥 − 𝜌 𝐴 𝑥 − | 𝑥 | − 𝑏 ) . ( 2 . 1 1 ) We use this alternative fixed point formulation to study the existence of a unique solution of the absolute value complementarity problem. Equation (1.1) and this is the main motivation of our next result.

Theorem 2.4. Let 𝐴 ∈ 𝑅 𝑛 × 𝑛 be a positive definite matrix with constant 𝛼 > 0 and continuous with constant 𝛽 > 0 . If 0 < 𝜌 < 2 ( 𝛾 − 1 ) / ( 𝛽 2 − 1 ) , 𝛾 > 1 , then there exists a unique solution 𝑥 ∈ 𝐾 such that ⟨ 𝐴 𝑥 − | 𝑥 | − 𝑏 , 𝑦 − 𝑥 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐾 , ( 2 . 1 2 ) where 𝐾 is a closed convex set in 𝑅 𝑛 .

Proof. Uniqueness: Let 𝑥 1 ≠ 𝑥 2 ∈ 𝐾 be two solutions of (1.2). Then 𝐴 𝑥 1 − | | 𝑥 1 | | − 𝑏 , 𝑦 − 𝑥 1 ≥ 0 , ∀ 𝑦 ∈ 𝐾 , 𝐴 𝑥 2 − | | 𝑥 2 | | − 𝑏 , 𝑦 − 𝑥 2 ≥ 0 , ∀ 𝑦 ∈ 𝐾 . ( 2 . 1 3 ) Taking 𝑦 = 𝑥 2 ∈ 𝐾 in (2.13) and 𝑦 = 𝑥 1 ∈ 𝐾 in (2.6), we have − 𝐴 𝑥 1 + | | 𝑥 1 | | + 𝑏 , 𝑥 1 − 𝑥 2 ≥ 0 , 𝐴 𝑥 2 − | | 𝑥 2 | | − 𝑏 , 𝑥 1 − 𝑥 2 ≥ 0 . ( 2 . 1 4 ) Adding the previous inequalities, we obtain 𝐴 𝑥 1 − 𝑥 2 − | | 𝑥 1 | | + | | 𝑥 2 | | , 𝑥 1 − 𝑥 2 ≤ 0 , ( 2 . 1 5 ) which implies that 𝐴 𝑥 1 − 𝑥 2 , 𝑥 1 − 𝑥 2 − ‖ ‖ 𝑥 1 − 𝑥 2 ‖ ‖ 2 ≤ 0 . ( 2 . 1 6 ) Since 𝐴 is positive definite, from (2.16), we have ‖ ‖ 𝑥 ( 𝛾 − 1 ) 1 − 𝑥 2 ‖ ‖ 2 ≤ 0 . ( 2 . 1 7 ) As 𝛾 > 1 , therefore from (2.17) we have ‖ ‖ 𝑥 1 − 𝑥 2 ‖ ‖ 2 ≤ 0 , ( 2 . 1 8 ) which contradicts the fact that ‖ 𝑥 1 − 𝑥 2 ‖ 2 ≥ 0 ; hence 𝑥 1 = 𝑥 2 .
Existence
Let 𝑥 ∈ 𝐾 be the solution of (1.2). Then ⟨ 𝐴 𝑥 − | 𝑥 | − 𝑏 , 𝑦 − 𝑥 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐾 . ( 2 . 1 9 ) From Lemma 2.3, we have 𝑥 = 𝑃 𝐾 [ ] ( 𝑥 − 𝜌 𝐴 𝑥 − | 𝑥 | − 𝑏 ) . ( 2 . 2 0 ) Define a mapping 𝐹 ( 𝑥 ) = 𝑥 = 𝑃 𝐾 [ ] ( 𝑥 − 𝜌 𝐴 𝑥 − | 𝑥 | − 𝑏 ) . ( 2 . 2 1 ) To show that the mapping 𝐹 ( 𝑥 ) defined by (2.21) has a fixed point, it is enough to prove that 𝐹 ( 𝑥 ) is a contraction mapping. For 𝑥 1 ≠ 𝑥 2 ∈ 𝐾 , consider ‖ ‖ 𝐹 𝑥 1 𝑥 − 𝐹 2 ‖ ‖ = ‖ ‖ 𝑃 𝐾 𝑥 1 − 𝜌 𝐴 𝑥 1 − | | 𝑥 1 | | − 𝑏 − 𝑃 𝐾 𝑥 2 − 𝜌 𝐴 𝑥 2 − | | 𝑥 2 | | ‖ ‖ ≤ ‖ ‖ 𝑥 − 𝑏 1 − 𝜌 𝐴 𝑥 1 − | | 𝑥 1 | | − 𝑥 − 𝑏 2 − 𝜌 𝐴 𝑥 2 − | | 𝑥 2 | | ‖ ‖ ≤ ‖ ‖ 𝑥 − 𝑏 1 − 𝑥 2 − 𝜌 𝐴 𝑥 1 − 𝐴 𝑥 2 ‖ ‖ ‖ ‖ | | 𝑥 + 𝜌 1 | | − | | 𝑥 2 | | ‖ ‖ = ‖ ‖ 𝑥 1 − 𝑥 2 𝑥 − 𝜌 𝐴 1 − 𝑥 2 ‖ ‖ ‖ ‖ 𝑥 + 𝜌 1 − 𝑥 2 ‖ ‖ , ( 2 . 2 2 ) where we have used the fact that 𝑃 𝐾 is nonexpansive. Now using positive definiteness of 𝐴 , we have ‖ ‖ 𝑥 1 − 𝑥 2 𝑥 − 𝜌 𝐴 1 − 𝑥 2 ‖ ‖ 2 = 𝑥 1 − 𝑥 2 𝑥 − 𝜌 𝐴 1 − 𝑥 2 , 𝑥 1 − 𝑥 2 𝑥 − 𝜌 𝐴 1 − 𝑥 2 ≤ ‖ ‖ 𝑥 1 − 𝑥 2 ‖ ‖ 2 𝑥 − 2 𝜌 𝛾 1 − 𝑥 2 𝑥 , 𝐴 1 − 𝑥 2 + 𝜌 2 𝛽 2 ‖ ‖ 𝑥 1 − 𝑥 2 ‖ ‖ 2 = 1 − 2 𝛼 𝜌 + 𝛽 2 𝜌 2 ‖ ‖ 𝑥 1 − 𝑥 2 ‖ ‖ 2 . ( 2 . 2 3 ) From (2.22) and (2.23), we have ‖ ‖ 𝐹 𝑥 1 𝑥 − 𝐹 2 ‖ ‖ ‖ ‖ 𝑥 ≤ 𝜃 1 − 𝑥 2 ‖ ‖ , ( 2 . 2 4 ) where √ 𝜃 = ( 𝜌 + 1 − 2 𝛾 𝜌 + 𝛽 2 𝜌 2 ) . Form 0 < 𝜌 < 2 ( 𝛾 − 1 ) / ( 𝛽 2 − 1 ) and 𝜌 < 1 , we have 𝜃 < 1 .
This shows that 𝐹 ( 𝑥 )   is a contraction mapping and has a fixed point 𝑥 ∈ 𝐾 satisfying the absolute value variational inequality (1.2).

For the sake of simplicity, we consider the special case, when 𝐾 = [ 0 , 𝑐 ] is a closed convex set in 𝑅 𝑛 and we define the projection 𝑃 𝐾 𝑥 as 𝑃 𝐾 𝑥 𝑖 = m i n m a x 0 , 𝑥 𝑖 , 𝑐 𝑖 , 𝑖 = 1 , 2 , … , 𝑛 . ( 2 . 2 5 ) We recall the following well-known result.

Lemma 2.5 (see [3]). For any 𝑥   and 𝑦 in 𝑅 𝑛 , the projection 𝑃 𝐾 𝑥 has the following properties:(i) 𝑃 𝐾 ( 𝑥 + 𝑦 ) ≤ 𝑃 𝐾 𝑥 + 𝑃 𝐾 𝑦 , (ii) 𝑃 𝐾 𝑥 − 𝑃 𝐾 𝑦 ≤ 𝑃 𝐾 ( 𝑥 − 𝑦 ) , (iii) 𝑥 ≤ 𝑦 ⇒ 𝑃 𝐾 𝑥 ≤ 𝑃 𝐾 𝑦 ,(iv) 𝑃 𝐾 𝑥 + 𝑃 𝐾 ( − 𝑥 ) ≤ | 𝑥 | , w i t h e q u a l i t y , i f a n d o n l y i f − 𝑐 ≤ 𝑥 ≤ 𝑐 .

We now suggest the iterative methods for solving the absolute value complmentarity problems (1.1). For this purpose, we decompose the matrix 𝐴   as, 𝐴 = 𝐷 + 𝐿 + 𝑈 , ( 2 . 2 6 ) where 𝐷 is the diagonal matrix, and 𝐿   and 𝑈 are strictly lower and strictly upper triangular matrices, respectively. Let Ω = d i a g ( 𝜔 1 , 𝜔 2 , … , 𝜔 𝑛 ) with 𝜔 𝑖 ∈ 𝑅 + and let 𝛼 be a real number.

Algorithm 2.6. Step 1. Choose an initial vector 𝑥 0 ∈ 𝑅 𝑛 and a parameter Ω ∈ 𝑅 + . S e t 𝑘 = 0 .Step 2. Calculate 𝑥 𝑘 + 1 = 𝑃 𝐾 𝑥 𝑘 − 𝐷 − 1 𝛼 Ω 𝐿 𝑢 𝑘 + 1 + ( Ω 𝐴 − 𝛼 Ω 𝐿 ) 𝑢 𝑘 | | 𝑥 − Ω 𝑘 | | + 𝑏 . ( 2 . 2 7 ) Step 3. If 𝑥 𝑘 + 1 = 𝑥 𝑘 , then stop. Else, s e t 𝑘 = 𝑘 + 1 and go to Step 2.Now we define an operator 𝑔 ∶ 𝑅 𝑛 → 𝑅 𝑛 such that 𝑔 ( 𝑥 ) = 𝜉 , where 𝜉 is the fixed point of the system 𝜉 = 𝑃 𝐾 𝑥 − 𝐷 − 1 [ ] 𝛼 Ω 𝐿 𝜉 + ( Ω 𝐴 − 𝛼 Ω 𝐿 ) 𝑥 − Ω ( | 𝑥 | + 𝑏 ) . ( 2 . 2 8 ) We also assume that the set 𝜑 = { 𝑥 ∈ 𝑅 𝑛 ∶ 𝑥 ≥ 0 , 𝐴 𝑥 − | 𝑥 | − 𝑏 ≥ 0 } ( 2 . 2 9 ) of the absolute value complementarity problem is nonempty.

To prove the convergence of Algorithm 2.6, we need the following result.

Theorem 2.7. Consider the operator 𝑔 ∶ 𝑅 𝑛 → 𝑅 𝑛 as defined in (2.28). Assume that 𝐴 ∈ 𝑅 𝑛 × 𝑛 is an 𝐿 -matrix. Also assume that 0 < 𝜔 𝑖 ≤ 1 , 0 ≤ 𝛼 ≤ 1 . Then, for any 𝑥 ∈ 𝜑 , the following holds:(i) 𝑔 ( 𝑥 ) ≤ 𝑥 , (ii) 𝑥 ≤ 𝑦 ⇒ 𝑔 ( 𝑥 ) ≤ 𝑔 ( 𝑦 ) , (iii) 𝜉 = 𝑔 ( 𝑥 ) ∈ 𝜑 .

Proof. To prove (i), we need to verify that 𝜉 𝑖 ≤ 𝑥 𝑖 , 𝑖 = 1 , 2 , … , 𝑛 hold with 𝜉 𝑖 satisfying 𝜉 𝑖 = 𝑃 𝐾 𝑥 𝑖 − 𝑎 − 1 𝑖 𝑖 𝛼 𝜔 𝑖 𝑖 − 1 𝑗 = 1 𝐿 𝑖 𝑗 𝜉 𝑗 − 𝑥 𝑗 + 𝜔 𝑖 ( 𝐴 𝑥 − | 𝑥 | − 𝑏 ) 𝑖 . ( 2 . 3 0 ) To prove the required result, we use mathematical induction. For this let 𝑖 = 1 , 𝜉 1 = 𝑃 𝐾 𝑥 1 − 𝑎 − 1 1 1 𝜔 1 ( 𝐴 𝑥 − | 𝑥 | − 𝑏 ) 1 . ( 2 . 3 1 ) Since 𝐴 𝑥 − | 𝑥 | − 𝑏 ≥ 0 , 𝜔 𝑖 > 0 , therefore 𝜉 1 ≤ 𝑥 1 . For 𝑖 = 2 , we have 𝜉 2 = 𝑃 𝐾 𝑥 2 − 𝑎 − 1 2 2 𝛼 𝜔 2 𝐿 2 1 𝜉 1 − 𝑥 1 + 𝜔 2 ( 𝐴 𝑥 − | 𝑥 | − 𝑏 ) 2 . ( 2 . 3 2 ) Here 𝐴 𝑥 − | 𝑥 | − 𝑏 ≥ 0 , 𝜔 𝑖 > 0 , 𝐿 2 1 ≤ 0 , and 𝜉 1 − 𝑥 1 ≤ 0 . This implies that 𝜉 2 ≤ 𝑥 2 .
Suppose that 𝜉 𝑖 ≤ 𝑥 𝑖 f o r 𝑖 = 1 , 2 , … , 𝑘 − 1 . ( 2 . 3 3 ) We have to prove that the statement is true, for 𝑖 = 𝑘 , that is, 𝜉 𝑘 ≤ 𝑥 𝑘 . ( 2 . 3 4 ) Consider 𝜉 𝑘 = 𝑃 𝐾 𝑥 𝑘 − 𝑎 − 1 𝑘 𝑘 𝛼 𝜔 𝑘 𝑘 − 1 𝑗 = 1 𝐿 𝑘 𝑗 𝜉 𝑗 − 𝑥 𝑗 + 𝜔 𝑘 ( 𝐴 𝑥 − | 𝑥 | − 𝑏 ) 𝑘 , = 𝑃 𝐾 𝑥 𝑘 − 𝑎 − 1 𝑘 𝑘 𝛼 𝜔 𝑘 𝐿 𝑘 1 𝜉 1 − 𝑥 1 + 𝐿 𝑘 2 𝜉 2 − 𝑥 2 + ⋯ + 𝐿 𝑘 𝑘 − 1 𝜉 𝑘 − 1 − 𝑥 𝑘 − 1 + 𝜔 𝑘 ( 𝐴 𝑥 − | 𝑥 | − 𝑏 ) 𝑘 . ( 2 . 3 5 ) Since 𝐴 𝑥 − | 𝑥 | − 𝑏 ≥ 0 , 𝜔 𝑘 > 0 , 𝐿 𝑘 1 , 𝐿 𝑘 2 , … , 𝐿 𝑘 𝑘 − 1 ≤ 0 , and 𝜉 𝑖 ≤ 𝑥 𝑖 for 𝑖 = 1 , 2 , … , 𝑘 − 1 , from (2.35) we can write 𝜉 𝑘 ≤ 𝑥 𝑘 . Hence (i) is proved.
Now we prove (ii). For this let us suppose that 𝜉 = 𝑔 ( 𝑥 ) and 𝜙 = 𝑔 ( 𝑦 ) . We will prove 𝑥 ≤ 𝑦 ⟹ 𝜉 ≤ 𝜙 . ( 2 . 3 6 ) As 𝜉 = 𝑃 𝐾 𝑥 − 𝐷 − 1 [ ] 𝛼 Ω 𝐿 𝜉 + ( Ω 𝐴 − 𝛼 Ω 𝐿 ) 𝑥 − Ω ( | 𝑥 | + 𝑏 ) , ( 2 . 3 7 ) so 𝜉 𝑖   can be written as 𝜉 𝑖 = 𝑃 𝐾 ⎛ ⎜ ⎜ ⎜ ⎝ 𝑥 𝑖 − 𝑎 − 1 𝑖 𝑖 ⎡ ⎢ ⎢ ⎢ ⎣ 𝛼 𝜔 𝑖 𝑖 − 1 𝑗 = 1 𝐿 𝑖 𝑗 𝜉 𝑗 + 𝜔 𝑖 𝑎 𝑖 𝑖 𝑥 𝑖 + ( 1 − 𝛼 ) 𝜔 𝑖 𝑖 − 1 𝑗 = 1 𝐿 𝑖 𝑗 𝑥 𝑗 + 𝜔 𝑖 𝑛 𝑗 = 1 𝑗 ≠ 𝑖 𝑈 𝑖 𝑗 𝑥 𝑗 − 𝜔 𝑖 | | 𝑥 𝑖 | | − 𝜔 𝑖 𝑏 𝑖 ⎤ ⎥ ⎥ ⎥ ⎦ ⎞ ⎟ ⎟ ⎟ ⎠ = 𝑃 𝐾 ⎛ ⎜ ⎜ ⎜ ⎝ 1 − 𝜔 𝑖 𝑥 𝑖 − 𝑎 − 1 𝑖 𝑖 ⎡ ⎢ ⎢ ⎢ ⎣ 𝛼 𝜔 𝑖 𝑖 − 1 𝑗 = 1 𝐿 𝑖 𝑗 𝜉 𝑗 + ( 1 − 𝛼 ) 𝜔 𝑖 𝑖 − 1 𝑗 = 1 𝐿 𝑖 𝑗 𝑥 𝑗 + 𝜔 𝑖 𝑛 𝑗 = 1 𝑗 ≠ 𝑖 𝑈 𝑖 𝑗 𝑥 𝑗 − 𝜔 𝑖 | | 𝑥 𝑖 | | − 𝜔 𝑖 𝑏 𝑖 ⎤ ⎥ ⎥ ⎥ ⎦ ⎞ ⎟ ⎟ ⎟ ⎠ . ( 2 . 3 8 ) Similarly, for 𝜙 𝑖 , we have 𝜙 𝑖 = 𝑃 𝐾 ⎛ ⎜ ⎜ ⎜ ⎝ 1 − 𝜔 𝑖 𝑦 𝑖 − 𝑎 − 1 𝑖 𝑖 ⎡ ⎢ ⎢ ⎢ ⎣ 𝛼 𝜔 𝑖 𝑖 − 1 𝑗 = 1 𝐿 𝑖 𝑗 𝜙 𝑗 + ( 1 − 𝛼 ) 𝜔 𝑖 𝑖 − 1 𝑗 = 1 𝐿 𝑖 𝑗 𝑦 𝑗 + 𝜔 𝑖 𝑛 𝑗 = 1 𝑗 ≠ 𝑖 𝑈 𝑖 𝑗 𝑦 𝑗 − 𝜔 𝑖 | | 𝑦 𝑖 | | − 𝜔 𝑖 𝑏 𝑖 ⎤ ⎥ ⎥ ⎥ ⎦ ⎞ ⎟ ⎟ ⎟ ⎠ , ( 2 . 3 9 ) and for 𝑖 = 1 , 𝜙 1 = 𝑃 𝐾 ⎛ ⎜ ⎜ ⎜ ⎝ 1 − 𝜔 1 𝑦 1 − 𝑎 − 1 1 1 𝜔 1 ⎡ ⎢ ⎢ ⎢ ⎣ 𝑛 𝑗 = 1 𝑗 ≠ 𝑖 𝑈 1 𝑗 𝑦 𝑗 − | | 𝑦 1 | | − 𝑏 1 ⎤ ⎥ ⎥ ⎥ ⎦ ⎞ ⎟ ⎟ ⎟ ⎠ ≥ 𝑃 𝐾 ⎛ ⎜ ⎜ ⎜ ⎝ 1 − 𝜔 1 𝑥 1 − 𝑎 − 1 1 1 𝜔 1 ⎡ ⎢ ⎢ ⎢ ⎣ 𝑛 𝑗 = 1 𝑗 ≠ 𝑖 𝑈 1 𝑗 𝑥 𝑗 − | | 𝑥 1 | | − 𝑏 1 ⎤ ⎥ ⎥ ⎥ ⎦ ⎞ ⎟ ⎟ ⎟ ⎠ = 𝜉 1 . ( 2 . 4 0 ) Since 𝑦 1 ≥ 𝑥 1 , therefore − | 𝑦 1 | ≤ − | 𝑥 1 | . Hence it is true for 𝑖 = 1 . Suppose it is true for 𝑖 = 1 , 2 , … 𝑘 − 1 ; we will prove it for 𝑖 = 𝑘 ; for this consider 𝜙 𝑘 = 𝑃 𝐾 ⎛ ⎜ ⎜ ⎜ ⎝ 1 − 𝜔 𝑘 𝑦 𝑘 − 𝑎 − 1 𝑘 𝑘 ⎡ ⎢ ⎢ ⎢ ⎣ 𝛼 𝜔 𝑘 𝑘 − 1 𝑗 = 1 𝐿 𝑘 𝑗 𝜙 𝑗 + ( 1 − 𝛼 ) 𝜔 𝑘 𝑘 − 1 𝑗 = 1 𝐿 𝑘 𝑗 𝑦 𝑗 + 𝜔 𝑘 𝑛 𝑗 = 1 𝑗 ≠ 𝑖 𝑈 𝑘 𝑗 𝑦 𝑗 − 𝜔 𝑘 | | 𝑦 𝑘 | | − 𝜔 𝑘 𝑏 𝑘 ⎤ ⎥ ⎥ ⎥ ⎦ ⎞ ⎟ ⎟ ⎟ ⎠ ≥ 𝑃 𝐾 ⎛ ⎜ ⎜ ⎜ ⎝ 1 − 𝜔 𝑘 𝑥 𝑘 − 𝑎 − 1 𝑘 𝑘 ⎡ ⎢ ⎢ ⎢ ⎣ 𝛼 𝜔 𝑘 𝑘 − 1 𝑗 = 1 𝐿 𝑘 𝑗 𝜉 𝑗 + ( 1 − 𝛼 ) 𝜔 𝑘 𝑘 − 1 𝑗 = 1 𝐿 𝑘 𝑗 𝑥 𝑗 + 𝜔 𝑘 𝑛 𝑗 = 1 𝑗 ≠ 𝑖 𝑈 𝑘 𝑗 𝑥 𝑗 − 𝜔 𝑘 | | 𝑥 𝑘 | | − 𝜔 𝑘 𝑏 𝑘 ⎤ ⎥ ⎥ ⎥ ⎦ ⎞ ⎟ ⎟ ⎟ ⎠ = 𝜉 𝑘 . ( 2 . 4 1 ) Since 𝑥 ≤ 𝑦 , and 𝜉 𝑖 ≤ 𝜙 𝑖 for 𝑖 = 1 , 2 , … 𝑘 − 1 , hence it is true for 𝑘 and (ii) is verified.
Next we prove (iii), that is, 𝜉 = 𝑔 ( 𝑥 ) ∈ 𝜑 . ( 2 . 4 2 ) Let 𝜆 = 𝑔 ( 𝜉 ) = 𝑃 𝐾 ( 𝜉 − 𝐷 − 1 Ω [ 𝛼 𝐿 ( 𝜆 − 𝜉 ) + 𝐴 𝜉 − | 𝜉 | − 𝑏 ] ) from (i) 𝑔 ( 𝜉 ) = 𝜆 ≤ 𝜉 . Also by definition of 𝑔 , 𝜉 = 𝑔 ( 𝑥 ) ≥ 0 and 𝜆 = 𝑔 ( 𝜉 ) ≥ 0 .
Now 𝜆 𝑖 = 𝑃 𝐾 𝜉 𝑖 − 𝑎 − 1 𝑖 𝑖 𝛼 𝜔 𝑖 𝑖 − 1 𝑗 = 1 𝐿 𝑖 𝑗 𝜆 𝑗 − 𝜉 𝑗 + 𝜔 𝑖 | | 𝜉 | | 𝐴 𝜉 − − 𝑏 𝑖 . ( 2 . 4 3 ) For 𝑖 = 1 , 𝜉 1 ≥ 0 by definition of 𝑔 . Suppose that ( 𝐴 𝜉 − | 𝜉 | − 𝑏 ) 𝑖 < 0 , so 𝜆 1 = 𝑃 𝐾 𝜉 1 − 𝑎 − 1 1 1 𝜔 1 | | 𝜉 | | 𝐴 𝜉 − − 𝑏 1 > 𝑃 𝐾 𝜉 1 = 𝜉 1 , ( 2 . 4 4 ) which contradicts the fact that 𝜆 ≤ 𝜉 . Therefore, ( 𝐴 𝜉 − | 𝜉 | − 𝑏 ) 𝑖 ≥ 0 .
Now we prove it for any 𝑘 in 𝑖 = 1 , 2 , … , 𝑛 . Suppose the contrary ( 𝐴 𝜉 − | 𝜉 | − 𝑏 ) 𝑖 < 0 ; then 𝜆 𝑘 = 𝑃 𝐾 𝜉 𝑘 − 𝑎 − 1 𝑘 𝑘 𝛼 𝜔 𝑘 𝑘 − 1 𝑗 = 1 𝐿 𝑘 𝑗 𝜆 𝑗 − 𝜉 𝑗 + 𝜔 𝑘 | | 𝜉 | | 𝐴 𝜉 − − 𝑏 𝑘 . ( 2 . 4 5 ) As it is true for all 𝛼 ∈ [ 0 , 1 ] , it should be true for 𝛼 = 0 . That is, 𝜆 𝑘 = 𝑃 𝐾 𝜉 𝑘 − 𝑎 − 1 𝑘 𝑘 𝜔 𝑘 | | 𝜉 | | 𝐴 𝜉 − − 𝑏 𝑘 > 𝑃 𝐾 𝜉 𝑘 = 𝜉 𝑘 , ( 2 . 4 6 ) which contradicts the fact that 𝜆 ≤ 𝜉 . So ( 𝐴 𝜉 − | 𝜉 | − 𝑏 ) 𝑘 ≥ 0 , for any 𝑘 in 𝑖 = 1 , 2 , … , 𝑛 . Hence 𝜉 = 𝑓 ( 𝑥 ) ∈ 𝜑 .

We now consider the convergence criteria of Algorithm 2.6 and this is the main motivation of our next result.

Theorem 2.8. Assume that 𝐴 ∈ 𝑅 𝑛 × 𝑛 is an 𝐿 -matrix. Also assume that 0 < 𝜔 𝑖 ≤ 1 , 0 ≤ 𝛼 ≤ 1 . Then for any initial vector 𝑥 0 ∈ 𝜑 , the sequence { 𝑥 𝑘 } , 𝑘 = 0 , 1 , 2 , … , defined by Algorithm 2.6 has the following properties:(i) 0 ≤ 𝑥 𝑘 + 1 ≤ 𝑥 𝑘 ≤ 𝑥 0 ; 𝑘 = 0 , 1 , 2 , … , (ii) l i m 𝑘 → ∞ 𝑥 𝑘 = 𝑥 ∗ is a unique solution of the absolute value complementarity problem (1.1).

Proof. Since 𝑥 0 ∈ 𝜑 , by (i) of Theorem 2.7, we have 𝑥 1 ≤ 𝑥 0 and 𝑥 1 ∈ 𝜑 . Recursively using Theorem 2.7 we obtain 0 ≤ 𝑥 𝑘 + 1 ≤ 𝑥 𝑘 ≤ 𝑥 0 ; 𝑘 = 0 , 1 , 2 , … . ( 2 . 4 7 ) From (i) we observe that the sequence { 𝑥 𝑘 } is monotone bounded; therefore, it converges to some 𝑥 ∗ ∈ 𝑅 𝑛 + satisfying 𝑥 ∗ = 𝑃 𝐾 𝑥 ∗ − 𝐷 − 1 𝛼 Ω 𝐿 𝑥 ∗ + ( Ω 𝐴 − 𝛼 Ω 𝐿 ) 𝑥 ∗ | | 𝑥 − Ω ∗ | | + 𝑏 = 𝑃 𝐾 𝑥 ∗ − 𝐷 − 1 Ω 𝐴 𝑥 ∗ | | 𝑥 − Ω ∗ | | . − Ω 𝑏 ( 2 . 4 8 ) Hence 𝑥 ∗ is the solution of the absolute value complementarity problem (1.1).

3. Numerical Results

In this section, we consider several examples to show the efficiency of the proposed method. The convergence of the generalized AOR method is guaranteed for L-matrices only but it is also possible to solve different types of systems. The elements of the diagonal matrix   Ω are chosen from the interval [ 𝑎 , 𝑏 ] such that 𝜔 𝑖 = 𝑎 + ( 𝑏 − 𝑎 ) 𝑖 𝑛 , 𝑖 = 1 , 2 , … , 𝑛 , ( 3 . 1 ) where 𝜔 𝑖   is the 𝑖 th diagonal element of Ω . All the experiments are performed with Intel(R) Core(TM) 2 × 2.1 GHz, 1 GB RAM, and the codes are written in Matlab 7.

Example 3.1. We test Algorithm 2.6 on 𝑚 consecutively generated solvable random problems 𝐴 ∈ 𝑅 𝑛 × 𝑛 , and 𝑛 ranging from 10 to 1000. We chose a random matrix 𝐴 from a uniform distribution on [ 0 , 1 ] , such that whose all diagonal elements are equal to 1000 and 𝑥   is chosen randomly from a uniform distribution on [ 0 , 1 ] . The constant vector is computed as 𝑏 = 𝐴 𝑥 − | 𝑥 | . The computational results are shown in Table 1.

tab1
Table 1: Computational results.

Example 3.2. Consider the ordinary differential equation: 𝑑 2 𝑥 𝑑 𝑡 2 − | 𝑥 | = 1 − 𝑥 2 , 0 ≤ 𝑥 ≤ 1 , 𝑥 ( 0 ) = 0 , 𝑥 ( 1 ) = 1 . ( 3 . 2 ) The exact solution is 𝑥 ( 𝑡 ) = 0 . 7 3 7 8 8 2 7 4 2 5 s i n ( 𝑡 ) − 3 c o s ( 𝑡 ) + 3 − 𝑥 2 , 𝑥 < 0 , − 0 . 7 3 1 0 5 8 5 7 8 6 𝑒 − 𝑥 − 0 . 2 6 8 9 4 1 4 2 1 4 𝑒 𝑥 + 1 + 𝑥 2 , 𝑥 > 0 . ( 3 . 3 ) We take 𝑛 = 1 0 ; the matrix 𝐴 is given by 𝑎 𝑖 , 𝑗 = ⎧ ⎪ ⎨ ⎪ ⎩ − 2 4 2 , f o r 𝑗 = 𝑖 , 1 2 1 , f o r 𝑗 = 𝑖 + 1 , 𝑖 = 1 , 2 , … , 𝑛 − 1 , 𝑗 = 𝑖 − 1 , 𝑖 = 2 , 3 , … , 𝑛 , 0 , o t h e r w i s e . ( 3 . 4 ) The constant vector 𝑏 is given by 𝑏 = 1 2 0 , 1 2 1 1 1 7 , 1 2 1 1 1 2 , 1 2 1 1 0 5 , 1 2 1 9 6 , 1 2 1 8 5 , 1 2 1 7 2 , 1 2 1 5 7 , 1 2 1 4 0 , 1 2 1 − 1 4 6 2 0 1 2 1 𝑇 . ( 3 . 5 ) Here 𝐴 is not 𝐿 -matrix. The comparison between the exact solution and the approximate solutions is given in Figure 1.

743861.fig.001
Figure 1

Example 3.3. Let the matrix 𝐴 be given by 𝑎 𝑖 , 𝑗 = ⎧ ⎪ ⎨ ⎪ ⎩ 8 , f o r 𝑗 = 𝑖 , − 1 , f o r 𝑗 = 𝑖 + 1 , 𝑖 = 1 , 2 , … , 𝑛 − 1 , 𝑗 = 𝑖 − 1 , 𝑖 = 2 , 3 , … , 𝑛 , 0 , o t h e r w i s e . ( 3 . 6 ) Let 𝑏 = ( 6 , 5 , 5 , … , 5 , 6 ) 𝑇 , the problem size 𝑛 ranging from 4 to 1024. The stopping criteria are ‖ 𝐴 𝑥 − | 𝑥 | − 𝑏 ‖ < 1 0 − 6 . We choose initial guess 𝑥 0 as 𝑥 0 = ( 0 , 0 , … , 0 ) 𝑇 . The computational results are shown in Table 2.
In Table 2 TOC denotes total time taken by CPU. The rate of convergence of AOR method is better than iterative method [21].

tab2
Table 2: Computational results.

4. Conclusion

In this paper, we have introduced a new class of complementarity problems, known as the absolute value complementarity problems. We have used the projection technique to establish the equivalence between the absolute value variational inequalities, fixed point problems, and the absolute value complementarity problems. This equivalence is used to discuss the existence of a unique solution of the absolute value problems under some suitable conditions. We have also used this alternative equivalent formulation to suggest and analyze an iterative method for solving the absolute value complementarity problems. Some special cases are also discussed. The results and ideas of this paper may be used to solve the variational inequalities and related optimization problems. This is another direction for future research.

Acknowledgments

This paper is supported by the Visiting Professor Program of King Saud University, Riyadh, Saudi Arabia. The authors are also grateful to Dr. S. M. Junaid Zaidi, Rector, COMSATS Institute of Information Technology, Pakistan, for providing the excellent research facilities.

References

  1. C. E. Lemke, “Bimatrix equilibrium points and mathematical programming,” Management Science, vol. 11, pp. 681–689, 1965. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  2. R. W. Cottle and G. B. Dantzig, “Complementary pivot theory of mathematical programming,” Linear Algebra and its Applications, vol. 1, no. 1, pp. 103–125, 1968. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  3. B. H. Ahn, “Iterative methods for linear complementarity problems with upperbounds on primary variables,” Mathematical Programming, vol. 26, no. 3, pp. 295–315, 1983. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  4. R. W. Cottle, J. S. Pang, and R. E. Stone, The Linear Complementarity Problem, Computer Science and Scientific Computing, Academic Press, Boston, Mass, USA, 1992.
  5. S. Karamardian, “Generalized complementarity problem,” Journal of Optimization Theory and Applications, vol. 8, pp. 161–168, 1971. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  6. Y. Li and P. Dai, “Generalized AOR methods for linear complementarity problem,” Applied Mathematics and Computation, vol. 188, no. 1, pp. 7–18, 2007. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  7. O. L. Mangasarian, “Absolute value programming,” Computational Optimization and Applications, vol. 36, no. 1, pp. 43–53, 2007. View at Publisher · View at Google Scholar
  8. O. L. Mangasarian, “Absolute value equation solution via concave minimization,” Optimization Letters, vol. 1, no. 1, pp. 3–8, 2007. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  9. O. L. Mangasarian, “A generalized Newton method for absolute value equations,” Optimization Letters, vol. 3, no. 1, pp. 101–108, 2009. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  10. O. L. Mangasarian, “Solution of symmetric linear complementarity problems by iterative methods,” Journal of Optimization Theory and Applications, vol. 22, no. 4, pp. 465–485, 1977. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  11. O. L. Mangasarian, “The linear complementarity problem as a separable bilinear program,” Journal of Global Optimization, vol. 6, no. 2, pp. 153–161, 1995. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  12. O. L. Mangasarian and R. R. Meyer, “Absolute value equations,” Linear Algebra and its Applications, vol. 419, no. 2-3, pp. 359–367, 2006. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  13. M. A. Noor, On variational inequalities, Ph. D thesis, Brunel university, London, UK, 1975.
  14. M. A. Noor, “Fixed point approach for complementarity problems,” Journal of Mathematical Analysis and Applications, vol. 133, no. 2, pp. 437–448, 1988. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  15. M. A. Noor, “Iterative methods for a class of complementarity problems,” Journal of Mathematical Analysis and Applications, vol. 133, no. 2, pp. 366–382, 1988. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  16. M. A. Noor, “Some developments in general variational inequalities,” Applied Mathematics and Computation, vol. 152, no. 1, pp. 199–277, 2004. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  17. M. A. Noor, “Extended general variational inequalities,” Applied Mathematics Letters, vol. 22, no. 2, pp. 182–186, 2009. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  18. M. A. Noor, K. I. Noor, and E. Al-Said, “Iterative methods for solving nonconvex equilibrium variational inequalities,” Applied Mathematics and Information Sciences, vol. 6, no. 1, pp. 65–69, 2012.
  19. M. A. Noor, K. I. Noor, and Th. M. Rassias, “Some aspects of variational inequalities,” Journal of Computational and Applied Mathematics, vol. 47, no. 3, pp. 285–312, 1993. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  20. M. A. Noor, J. Iqbal, S. Khattri, and E. Al-Said, “A new iterative method for solving absolute value equations,” International Journal of the Physical Sciences, vol. 6, no. 7, pp. 1793–1797, 2011.
  21. M. A. Noor, J. Iqbal, K. I. Noor, and E. Al-Said, “On an iterative method for solving absolute value equations,” submitted to Optimization Letters. View at Publisher · View at Google Scholar
  22. M. A. Noor, J. Iqbal, and E. Al-Said, “Residual iterative method for solving absolute value equations,” Abstract and Applied Analysis. In press.
  23. J. Rohn, “A theorem of the alternatives for the equation Ax+b|x|=b,” Linear and Multilinear Algebra, vol. 52, no. 6, pp. 421–426, 2004. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  24. J. Rohn, “An algorithm for computing all solutions of an absolute valueequation,” Optimization Letters. In press. View at Publisher · View at Google Scholar