Smoothing Methods for Nonlinear Complementarity Problems

Abstract

In this paper, we present some new smoothing techniques to solve general nonlinear complementarity problems. Under a weaker condition than monotonicity as on the original problems, we prove convergence of our methods. We also present an error estimate under a general monotonicity condition. Some numerical tests confirm the efficiency of the proposed methods.

Appendix

We give in this appendix a brief description of each test example and report some numerics obtained by using the following projection method; see [15, Sect. 12.1]:

$$x^{k+1}= \max \bigl(0, x^k -D^{-1}F \bigl(x^k \bigr) \bigr), \quad k=0,1,\ldots. $$

We choose D=λI, where λ>0 is a constant and I is the n×n identity matrix. Table 2 presents the best obtained results when varying the value of λ (λ=0.1,1,10,20,50,100).

• The two first examples P1 and P2 [9] correspond to strongly monotone function F(x)=(F ₁(x),…,F _n (x))^T with $F_{i}(x)=-x_{i+1}+2x_{i}-x_{i-1}+ \frac{1}{3}x_{i}^{3}-b_{i},\quad i=1,\ldots ,n$, (x ₀=x _n+1=0) and b _i =(−1)ⁱ (resp. $b_{i}=\frac{(-1)^{i}}{\sqrt{i}}$), i=1,…,n for P1 (resp. P2).

• P3 is another strongly monotone test problem from [10] where F(x)=(F ₁(x),…,F _n (x))^T with $F_{i}(x)=-x_{i+1}+2x_{i}-x_{i-1}+ \arctan(x_{i}) +(i-\frac{n}{2})$, i=1,…,n, (x ₀=x _n+1=0).

• P4 and P5 are known as the degenerate and the non-degenerate examples of Kojima-Shindo [12]. P4 and P5 are, respectively, defined by

  $$\begin{aligned} &F_4(x)=\left ( \begin{array}{c} 3x_1^2+2x_1x_2+2x_2^2+x_3+3x_4-6\\ 2x_1^2+x_1+x_2^2+10x_3+2x_4-2\\ 3x_1^2+x_1x_2+2x_2^2+2x_3+9x_4-9\\ x_1^2+3x_2^2+2x_3+3x_4-3 \end{array} \right ), \\ &F_5(x)=\left ( \begin{array}{c} 3x_1^2+2x_1x_2+2x_2^2+x_3+3x_4-6\\ 2x_1^2+x_1+x_2^2+10x_3+2x_4-2\\ 3x_1^2+x_1x_2+2x_2^2+2x_3+3x_4-1\\ x_1^2+3x_2^2+2x_3+3x_4-3 \end{array} \right ). \end{aligned}$$

• P5 has a unique solution $x^{*}=(\frac{\sqrt{6}}{2},0,0,\frac {1}{2})$ with $F(x^{*})=(0,2+\frac{\sqrt{6}}{2},3,0)$ while P4 has two optimal solutions $x^{*}=(\frac{\sqrt{6}}{2},0,0,\frac{1}{2})$ with $F(x^{*})=(0,2+\frac{\sqrt {6}}{2},0,0)$ and x ^∗∗=(1,0,3,0) with F(x ^∗∗)=(0,31,0,4). The first optimal solution of P4 is degenerate since $x^{*}_{3}=F_{3}(x^{*})=0$.

• A complete description of P6 and P7 can be found in [13, 14]. These two examples correspond to the Nash–Cournot test problem with N=5 and N=10.

  Let $x \in \mathbb {R}^{N}$, Q=∑x _i and define the functions C _i (x _i ) and p(Q) as follows:

  $$p(Q) = 5000^{\frac{1}{\gamma}}Q^{\frac{-1}{\gamma}}, \qquad C_i(x_i) = c_ix_i + \frac{b_i}{1 + b_i}L_i^{\frac{1}{b_i}}x_i^{\frac{b_i+1}{b_i}}. $$

  The NCP-function is given by F _i (x)=C _i ′(x _i )−p(Q)−x _i p′(Q), i=1,…,N, or in a vectorial form $F(x) = [ c + L^{\frac{1}{b}}x^{\frac{1}{b}}-p(Q)(e -\frac{x}{\gamma Q}) ]$ with c _i , L _i , b _i , γ>0 and γ≥1.

  For our numerics, we used - N=5, c=[10,8,6,4,2]^T, b=[1.2,1.10,1,0.9,0.8]^T, L=[5,5,5,5,5]^T, e=[1,1,1,1,1]^T and γ=1.1.

• N=10, c=[5,3,8,5,1,3,7,4,6,3]^T, b=[1.2,1,0;9,0.6,1.5,1,0.7,1.1,0.95,0.75]^T, L=[10,10,10,10,10,10,10,10,10,10]^T, e=[1,1,1,1,1,1,1,1,1,1]^T and γ=1.2.

• P8, P9 and P10 are also described in [13, 14]. They correspond, respectively, to the HpHard test problem with n=20, n=30 and n=100.

The corresponding function F(x) is of the form F(x)=(AA ^T+B+D)x+q; here the matrices A,B, and D are randomly generated as follows: any entry of the square n×n matrix A and of the n×n skew-symmetric matrix B is uniformly generated from ]−5,5[, and any entry of the diagonal matrix D is uniformly generated from ]0,3[. The vector q is uniformly generated from ]−500;0[.

Table 2 Results for the projection method

The matrix AA ^T+B+D is positive definite and the function F is strongly monotone. We used the M-files proposed in [13] to generate A,B,D and q. We implemented the projection method for solving the previous test problems in the same conditions and using the same material as for our methods. The following table gives the best obtained results when varying the value of λ (λ=0.1,1,10,20,50,100). In each computation we used a vector of ones as starting point. The column Iter corresponds to the number of iterations of the projection method and cannot be compared to Initer or Outiter in Table 1. The other columns correspond to the same things in Table 1 and can be used for comparison.