Symmetric SOR Method for Absolute Complementarity Problems

Abstract

We study symmetric successive overrelaxation (SSOR) method for absolute complementarity problems. Solving this problem is equivalent to solving the absolute value equations. Some examples are given to show the implementation and efficiency of the method.

1. Introduction

Absolute complementarity problem seeks real vectors 𝑥≥0 and 𝐴𝑥−|𝑥|−𝑏≥0, such that ⟨𝑥,𝐴𝑥−|𝑥|−𝑏⟩=0, (1) where 𝐴∈𝑅^𝑛×𝑛 and 𝑏∈𝑅^𝑛. The complementarity theory was introduced and studied by Lemke [1] and Cottle and Dantzig [2]. The complementarity problems have been generalized and extended to study a wide class of problems, which arise in pure and applied sciences; see [1–9] and the references therein. Equally important is the variational inequality problem, which was introduced and studied in the early sixties.

In this paper, we suggest and analyze SSOR [5] method for absolute complementarity problem which was introduced by Noor et al. [10]. The convergence analysis of the proposed method is considered under some suitable conditions. We show that the absolute complementarity problems are equivalent to variational inequalities. Results are very encouraging. The ideas and the technique of this paper may stimulate further research in these areas.

Let 𝑅^𝑛 be the finite dimension Euclidean space, whose the 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∈𝐾^∗, such that 𝑥∈𝐾^∗,𝐴𝑥−|𝑥|−𝑏∈𝐾^∗,⟨𝐴𝑥−|𝑥|−𝑏,𝑥⟩=0, (2) where 𝐾^∗={𝑥∈𝑅^𝑛∶⟨𝑥,𝑦⟩≥0,𝑦∈𝐾} is the polar cone of a closed convex cone 𝐾 in 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 (2) can be viewed as an extension of the complementarity problem considered by Karamardian [6].

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 𝐴𝑥−|𝑥|−𝑏,𝑦−𝑥≥0,∀𝑦∈𝐾. (3) The problem (3) is called the absolute value variational inequality, which is a special form of the mildly nonlinear variational inequalities [11]. If 𝐾=𝑅^𝑛, then the problem (3) is equivalent to find 𝑥∈𝑅^𝑛 such that 𝐴𝑥−|𝑥|−𝑏=0. (4) To propose and analyze algorithms for absolute complementarity problems, we need the following definitions.

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

Definition 2. If 𝐴∈𝑅^𝑛×𝑛 is positive definite, then(i)there exists a constant 𝛾>0, such that ⟨𝐴𝑥,𝑥⟩≥𝛾‖𝑥‖²,∀𝑥∈𝑅^𝑛; (5)(ii)there exists a constant 𝛽>0 such that ‖𝐴𝑥‖≤𝛽‖𝑥‖,∀𝑥∈𝑅^𝑛. (6)

2. Absolute Complementarity Problems

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

Lemma 3 (see [12]). 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, (7) if and only if 𝑢=𝑃_𝐾𝑧, (8) where 𝑃_𝐾 is the projection of 𝑅^𝑛 onto the closed convex set 𝐾.

Lemma 4 (see [10]). If 𝐾 is the positive cone in 𝑅^𝑛, then 𝑥∈𝐾 is a solution of absolute variational inequality (3) if and only if 𝑥∈𝐾 is the solution of the complementarity problem (2).

The next result proves the equivalence between variational inequality (3) and the fixed point.

Lemma 5 (see [10]). If 𝐾 is closed convex set in 𝑅^𝑛, then 𝜌>0; 𝑥∈𝐾 satisfies (3) if and only if 𝑥∈𝐾 satisfies the relation 𝑥=𝑃_𝐾𝑥−𝜌[𝐴𝑥−|𝑥|−𝑏], (9) where 𝑃_𝐾 is the projection of 𝑅^𝑛 onto the closed convex set 𝐾.

Now using Lemmas 4 and 5, the absolute complementarity problem (2) can be transformed to fixed-point problem as 𝑥=𝑃_𝐾𝑥−𝜌[𝐴𝑥−|𝑥|−𝑏]. (10)

Theorem 6 (see [10]). 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)/(𝛽²−1), 𝛽>1, 𝛾>1, then there exists a unique solution 𝑥∈𝐾, such that 𝐴𝑥−|𝑥|−𝑏,𝑦−𝑥≥0∀𝑦∈𝐾, (11) where 𝐾 is a closed convex set in 𝑅^𝑛.

To define the projection operator 𝑃_𝐾, we consider the special case when 𝐾=[0,𝑐] is a closed convex set in 𝑅^𝑛, as follows.

Definition 7 (see [3]). Let 𝐾=[0,𝑐] is a closed convex set in 𝑅^𝑛. Then, the projection operator 𝑃_𝐾𝑥 is defined as 𝑃_𝐾𝑥_𝑖=minmax0,𝑥_𝑖,𝑐_𝑖,𝑖=1,2,…,𝑛. (12)

Lemma 8 (see [3]). For any 𝑥 and 𝑦 in 𝑅^𝑛, the following facts hold:(i)𝑃_𝐾(𝑥+𝑦)≤𝑃_𝐾𝑥+𝑃_𝐾𝑦;(ii)𝑃_𝐾𝑥−𝑃_𝐾𝑦≤𝑃_𝐾(𝑥−𝑦);(iii)𝑥≤𝑦⇒𝑃_𝐾𝑥≤𝑃_𝐾𝑦;(iv)𝑃_𝐾𝑥+𝑃_𝐾(−𝑥)≤|𝑥|, with equality, if and only if −𝑐≤𝑥≤𝑐.
Now one splits the matrix 𝐴 as 𝐴=𝐷−𝐿−𝑈, (13) where 𝐷 is the diagonal matrix and 𝐿 and 𝑈 are strictly lower and strictly upper triangular matrices, respectively. Let 0<𝜔<2; using (13), one suggests the SSOR method for solving (3) as follows.

Algorithm 9. Consider the following.

Step  1. Choose an initial vector 𝑥₀∈𝑅^𝑛 and a parameter 𝜔∈𝑅₊. Set 𝑘=0.

Step  2. Calculate 𝑥_𝑘+1=𝑃_𝐾𝑥_𝑘−𝐷⁻¹×−𝜔𝐿𝑥_𝑘+1+(𝜔(2−𝜔)𝐴+𝜔𝐿)𝑥_𝑘−𝜔(2−𝜔)𝑥_𝑘+𝑏. (14)

Step  3. If 𝑥_𝑘+1=𝑥_𝑘, then stop; else, set𝑘=𝑘+1 and go to step  2.

Algorithm 10. Consider the following.

 Step  1. Choose an initial vector 𝑥₀∈𝑅^𝑛 and a parameter 𝜔∈𝑅₊. Set 𝑘=0.

Step  2. Calculate 𝑥_𝑘+1=𝑃_𝐾𝑥_𝑘−𝐷⁻¹×−𝜔𝑈𝑥_𝑘+1+(𝜔(2−𝜔)𝐴+𝜔𝑈)𝑥_𝑘−𝜔(2−𝜔)𝑥_𝑘+𝑏. (15)

Step  3. If 𝑥_𝑘+1=𝑥_𝑘, then stop; else, set 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_𝐾𝑥−𝐷⁻¹×(𝜔())𝑥(𝑥|.−𝜔𝐿𝜉+2−𝜔𝐴+𝜔𝐿−𝜔2−𝜔)(|+𝑏) (16) We also assume that the set 𝜑=𝑥∈𝑅^𝑛∶𝑥≥0,𝐴𝑥−|𝑥|−𝑏≥0 (17) of the absolute complementarity problem is nonempty. To prove the convergence of Algorithm 9, we need the following result.

Theorem 11. Consider the operator 𝑔∶𝑅^𝑛→𝑅^𝑛 as defined in (16). Assume that 𝐴∈𝑅^𝑛×𝑛 is an 𝐿-matrix. Also assume that 0<𝜔≤1. Then for any 𝑥∈𝜑, it holds that.(i)𝑔(𝑥)≤𝑥;(ii)𝑥≤𝑦⇒𝑔(𝑥)≤𝑔(𝑦);(iii)𝜉=𝑔(𝑥)∈𝜑.

Proof. To prove (i), we need to prove that 𝜉_𝑖≤𝑥_𝑖,𝑖=1,2,…,𝑛 (18) with 𝜉_𝑖 satisfying 𝜉_𝑖=𝑃_𝐾𝑥_𝑖−𝑎⁻¹_𝑖𝑖×−𝜔^𝑖−1_𝑗=1𝐿_𝑖𝑗𝜉_𝑗−𝑥_𝑗()×(|𝑥|)+𝜔2−𝜔𝐴𝑥−−𝑏_𝑖. (19) To prove the required result, we use mathematical induction. For this, let 𝑖=1: 𝜉₁=𝑃_𝐾𝑥₁−𝑎⁻¹₁₁𝜔(2−𝜔)(𝐴𝑥−|𝑥|−𝑏)₁. (20)
Since 𝐴𝑥−|𝑥|−𝑏≥0, 0<𝜔≤1; therefore, 𝜉₁≤𝑥₁.
For 𝑖=2, we have 𝜉₂=𝑃_𝐾𝑥₂−𝑎⁻¹₂₂×−𝜔𝐿₂₁𝜉₁−𝑥₁()×(|𝑥|)+𝜔2−𝜔𝐴𝑥−−𝑏₂. (21) Here, 𝐴𝑥−|𝑥|−𝑏≥0, 0<𝜔≤2, 𝐿₂₁≥0 and 𝜉₁−𝑥₁≤0. This implies that 𝜉₂≤𝑥₂.
Suppose that 𝜉_𝑖≤𝑥_𝑖for𝑖=1,2,…,𝑘−1, (22) we have to prove that the statement is true for 𝑖=𝑘; that is, 𝜉_𝑘≤𝑥_𝑘. (23) Consider 𝜉_𝑘=𝑃_𝐾𝑥_𝑘−𝑎⁻¹_𝑘𝑘×−𝜔^𝑘−1_𝑗=1𝐿_𝑘𝑗𝜉_𝑗−𝑥_𝑗()×(|𝑥|)+𝜔2−𝜔𝐴𝑥−−𝑏_𝑘=𝑃_𝐾𝑥_𝑘−𝑎⁻¹_𝑘𝑘×−𝜔𝐿_𝑘1𝜉₁−𝑥₁+𝐿_𝑘2𝜉₂−𝑥₂+⋯+𝐿_𝑘𝑘−1𝜉_𝑘−1−𝑥_𝑘−1+𝜔(2−𝜔)(𝐴𝑥−|𝑥|−𝑏)_𝑘. (24) Since 𝐴𝑥−|𝑥|−𝑏≥0, 0<𝜔_𝑘≤2, 𝐿_𝑘1,𝐿_𝑘2,…,𝐿_𝑘𝑘−1≥0 and 𝜉_𝑖≤𝑥_𝑖 for 𝑖=1,2,…,𝑘−1; from (24), we can write 𝜉_𝑘≤𝑥_𝑘. (25) Hence, (i) is proved.
Now we prove (ii), for this let us suppose that 𝜉=𝑔(𝑥) and 𝜙=𝑔(𝑦). We will prove that 𝑥≤𝑦⟹𝜉≤𝜙. (26)
As 𝜉=𝑃_𝐾𝑥−𝐷⁻¹×(𝜔())𝑥(𝑥|,−𝜔𝐿𝜉+2−𝜔𝐴+𝜔𝐿−𝜔2−𝜔)(|+𝑏) (27) so 𝜉_𝑖 can be written as 𝜉_𝑖=𝑃_𝐾−𝑎⁻¹_𝑖𝑖×−𝜔^𝑖−1_𝑗=1𝐿_𝑖𝑗𝜉_𝑗+𝜔𝑎_𝑖𝑖𝑥_𝑖+((×𝜔−𝜔2−𝜔))_{𝑖−1𝑗=1}𝐿_𝑖𝑗𝑥_𝑗()×−𝜔2−𝜔_{𝑛𝑗=1𝑗≠𝑖}𝑈_𝑖𝑗𝑥_𝑗−𝜔(2−𝜔)𝑥_𝑖−𝜔(2−𝜔)𝑏_𝑖=𝑃_𝐾(1−𝜔)𝑥_𝑖−𝑎⁻¹_𝑖𝑖×−𝜔^𝑖−1_𝑗=1𝐿_𝑖𝑗𝜉_𝑗+((×𝜔−𝜔2−𝜔))_{𝑖−1𝑗=1}𝐿_𝑖𝑗𝑥_𝑗()×−𝜔2−𝜔_{𝑛𝑗=1𝑗≠𝑖}𝑈_𝑖𝑗𝑥_𝑗−𝜔(2−𝜔)𝑥_𝑖−𝜔(2−𝜔)𝑏_𝑖. (28) Similarly, for 𝜙_𝑖 we have 𝜙_𝑖=𝑃_𝐾(1−𝜔)𝑦_𝑖−𝑎⁻¹_𝑖𝑖×−𝜔^𝑖−1_𝑗=1𝐿_𝑖𝑗𝜙_𝑗+((×𝜔−𝜔2−𝜔))_{𝑖−1𝑗=1}𝐿_𝑖𝑗𝑦_𝑗()×−𝜔2−𝜔_{𝑛𝑗=1𝑗≠𝑖}𝑈_𝑖𝑗𝑦_𝑗−𝜔(2−𝜔)𝑦_𝑖−𝜔(2−𝜔)𝑏_𝑖. (29) For 𝑖=1, we have 𝜙₁=𝑃_𝐾(1−𝜔)𝑦₁−𝑎⁻¹₁₁𝜔()×−2−𝜔_{𝑛𝑗=1𝑗≠𝑖}𝑈_1𝑗𝑦_𝑗−𝑦₁−𝑏₁≥𝑃_𝐾(1−𝜔)𝑥₁−𝑎⁻¹₁₁𝜔()×−2−𝜔_{𝑛𝑗=1𝑗≠𝑖}𝑈_1𝑗𝑥_𝑗−𝑥₁−𝑏₁=𝜉₁. (30) Since 𝑦₁≥𝑥₁, therefore −|𝑦₁|≤−|𝑥₁|. 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)𝑦_𝑘−𝑎⁻¹_𝑘𝑘×^𝑘−1_𝑗=1𝐿_𝑘𝑗𝜙_𝑗+((×𝜔−𝜔2−𝜔))_{𝑘−1𝑗=1}𝐿_𝑘𝑗𝑦_𝑗()×−𝜔2−𝜔_{𝑛𝑗=1𝑗≠𝑖}𝑈_𝑘𝑗𝑦_𝑗−𝜔(2−𝜔)𝑦_𝑘−𝜔(2−𝜔)𝑏_𝑘≥𝑃_𝐾(1−𝜔)𝑥_𝑘−𝑎⁻¹_𝑘𝑘×^𝑘−1_𝑗=1𝐿_𝑘𝑗𝜉_𝑗+((×𝜔−𝜔2−𝜔))_{𝑘−1𝑗=1}𝐿_𝑘𝑗𝑥_𝑗()×−𝜔2−𝜔_{𝑛𝑗=1𝑗≠𝑖}𝑈_𝑘𝑗𝑥_𝑗−𝜔(2−𝜔)𝑥_𝑘−𝜔(2−𝜔)𝑏_𝑘=𝜉_𝑘. (31) 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. (32)
Let 𝜆=𝑔(𝜉)=𝑃_𝐾(𝜉−𝐷⁻¹[𝜔𝐿(𝜆−𝜉)+𝜔(2−𝜔)(𝐴𝜉−|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_𝐾𝜉_𝑖−𝑎⁻¹_𝑖𝑖×−𝜔^𝑖−1_𝑗=1𝐿_𝑖𝑗𝜆_𝑗−𝜉_𝑗()×𝜉+𝜔2−𝜔𝐴𝜉−−𝑏_𝑖 (33) For 𝑖=1,𝜉₁≥0 by definition of 𝑔. Suppose that (𝐴𝜉−|𝜉|−𝑏)_𝑖<0, so 𝜆₁=𝑃_𝐾𝜉₁−𝑎⁻¹₁₁𝜔(2−𝜔)𝐴𝜉−𝜉−𝑏₁>𝑃_𝐾𝜉₁=𝜉₁, (34) which contradicts the fact that 𝜆≤𝜉. Therefore, (𝐴𝜉−|𝜉|−𝑏)_𝑖≥0.
Now we prove it for any 𝑘 in 𝑖=1,2,…,𝑛. Suppose the contrary (𝐴𝜉−|𝜉|−𝑏)_𝑖<0; then 𝜆_𝑘=𝑃_𝐾𝜉_𝑘−𝑎⁻¹_𝑘𝑘×−𝜔^𝑘−1_𝑗=1𝐿_𝑘𝑗𝜆_𝑗−𝜉_𝑗()×𝜉+𝜔2−𝜔𝐴𝜉−−𝑏_𝑘. (35) As it is true for all 𝛼∈[0,1], it should be true for 𝛼=0. That is, 𝜆_𝑘=𝑃_𝐾𝜉_𝑘−𝑎⁻¹_𝑘𝑘𝜔(2−𝜔)𝐴𝜉−𝜉−𝑏_𝑘>𝑃_𝐾𝜉_𝑘=𝜉_𝑘, (36) which contradicts the fact that 𝜆≤𝜉. So, (𝐴𝜉−|𝜉|−𝑏)_𝑘≥0, for any 𝑘 in 𝑖=1,2,…,𝑛.
Hence, 𝜉=𝑓(𝑥)∈𝜑.

Now we prove the convergence criteria of Algorithm 9 when the matrix 𝐴 is an 𝐿-matrix as stated in the next result.

Theorem 12. Assume that 𝐴∈𝑅^𝑛×𝑛 is an 𝐿-matrix and 0<𝜔≤1. Then for any initial vector 𝑥₀∈𝜑, the sequence {𝑥_𝑘},𝑘=0,1,2,…, defined by Algorithm 9 has the following properties:(i)0≤𝑥_𝑘+1≤𝑥_𝑘≤𝑥₀; 𝑘=0,1,2,…;(ii)lim_𝑘→∞𝑥_𝑘=𝑥^∗ is the unique solution of the absolute complementarity problem.

Proof. Since 𝑥₀∈𝜑, by (i) of Theorem 11, we have 𝑥₁≤𝑥₀ and 𝑥₁∈𝜑. Recursively using Theorem 11, we obtain 0≤𝑥_𝑘+1≤𝑥_𝑘≤𝑥₀;𝑘=0,1,2,….. (37) From (i), we observe that the sequence {𝑥_𝑘} is monotone bounded; therefore, it converges to some 𝑥^∗∈𝑅^𝑛₊ satisfying 𝑥^∗=𝑃_𝐾𝑥^∗−𝐷⁻¹×−𝜔𝐿𝑥^∗+𝜔(2−𝜔)(𝐴+𝜔𝐿)𝑥^∗−𝜔(2−𝜔)𝑥^∗+𝑏=𝑃_𝐾𝑥^∗−𝐷⁻¹𝜔()×2−𝜔𝐴𝑥^∗−𝑥^∗−𝑏. (38) Hence, 𝑥^∗ is the solution of the absolute complementarity problem (2).

Note. The convergence of Algorithm 10 has the same steps as given in Theorems 11 and 12.

3. Numerical Results

In this section, we consider several examples to show the efficiency of the proposed methods. The convergence of SSOR method is guaranteed for 𝐿-matrices only, but it is also possible to solve different type of systems. All the experiments are performed with Intel(R) Core 2 × 2.1 GHz, 1 GB RAM, and the codes are written in MATLAB 7.

Example 13 (see [10]). Consider the ordinary differential equation 𝑑²𝑥𝑑𝑡²−|𝑥|=1−𝑡²𝑥(0)(1),0≤𝑥≤1,=0𝑥=1. (39) The exact solution is 𝑥(𝑡)=0.7378827425sin(𝑡)−3cos(𝑡)+3−𝑡²𝑥<0,−0.7310585786𝑒^−𝑡−0.2689414214𝑒^𝑡+1+𝑡²𝑥>0. (40)
We take 𝑛=10; the matrix 𝐴 is given by 𝑎_𝑖,𝑗=−242,for𝑗=𝑖121for𝑗=𝑖+1,𝑖=1,2,…,𝑛−1𝑗=𝑖−1,𝑖=2,3,…,𝑛0,otherwise. (41)
The constant vector 𝑏 is given by 𝑏=120,121117,121112,121105,12196,12185,12172,12157,12140,121−14620121^𝑇. (42) Here, 𝐴 is not an 𝐿-matrix. The comparison between the exact solution and the approximate solutions is given in Figure 1.
In Figure 1, we see that the SSOR method converges rapidly to the approximate solution of absolute complementarity problem (2) as compared to GAOR method.
In the next example, we compare SSOR method with iterative method by Noor et al. [13].

Figure 1 

Comparison between GAOR method and SSOR method.

Example 14 (see [13]). Let the matrix 𝐴 be given by 𝑎_𝑖,𝑗=8,for𝑗=𝑖−1for𝑗=𝑖+1,𝑖=1,2,…,𝑛−1𝑗=𝑖−1,𝑖=2,3,…,𝑛0,otherwise. (43) Let 𝑏=(6,5,5,…,5,6)^𝑇, the problem size 𝑛, ranging from 4 to 1024. The stopping criteria are ‖𝐴𝑥−|𝑥|−𝑏‖<10⁻⁶. We choose initial guess 𝑥₀ as 𝑥₀=(0,0,…,0)^𝑇. The computational results are shown in Table 1.
In Table 1, TOC denotes the total time taken by CPU. The rate of convergence of SSOR method is better than that of iterative method [13].

Table 1 

4. Conclusion

In this paper, we have discussed symmetric SOR method for solving absolute complementarity problem. The comparison with other methods showed the efficiency of the method. The results and ideas of this paper may be used to solve the variational inequalities and related optimization problems.