A $\mathcal {O}(1/k^{3/2})$ hybrid proximal extragradient primal–dual interior point method for nonlinear monotone mixed complementarity problems

Abstract

We present a Newton-type hybrid proximal extragradient primal–dual interior point method for solving smooth monotone mixed complementarity problems. Dual variables for the nonnegativity constraints are introduced. The ergodic complexity of the method is $\mathcal {O}(1/k^{3/2})$. The method performs two types of iterations: underelaxed hybrid proximal extragradient iterations and short-steps primal–dual interior point iterations.

Appendix

In this section we prove Lemma 4 and a technical result.

Proof of Lemma 4.

Proof

Since $y, s>0$, $\mu , \nu >0$ and A is positive semi-definite, pre-multiplying the last ${m}$ equations of (20) by $S^{-1}$ we obtain an equivalent linear system of equations, which has is a positive definite matrix of coefficients and, therefore, a unique solution. Thus, the first claim of the lemma holds.

• (a) Left-multiply both sides of (20) by $(d^x,d^y,0)^T$ and use the assumption of A being positive semi-definite and Cauchy–Schwartz inequality to conclude that

  $$\begin{aligned} \nu \Vert (d^x,d^y)\Vert ^2-\mu \langle {d^y},{d^s}\rangle \le \langle {(d^x,d^y)},{(b^x,b^y)}\rangle \le \dfrac{\nu \Vert (d^x,d^y)\Vert ^2}{2} +\dfrac{\Vert (b^x,b^y)\Vert ^2}{2\nu }. \end{aligned}$$

  We have that

  $$\begin{aligned} \mu \Vert D^yd^s\Vert +\mu \langle {d^y},{d^s}\rangle&\le \frac{\mu }{2}\sum _{i=1}^n \frac{s_i}{y_i}(d^y_i)^2+\frac{y_i}{s_i}(d^s_i)^2+\mu \langle {d^y},{d^s}\rangle \\&=\frac{1}{2}\left\| (\mu YS)^{-1/2}\left( \mu Sd^y+\mu Yd^s\right) \right\| ^2\\&\le \frac{\Vert \mu \left( Sd^y+\mu Yd^s\right) \Vert ^2}{2(1-\Vert \mu Ys-e\Vert )} =\frac{\Vert b^s\Vert ^2}{2(1-\Vert \mu Ys-e\Vert )} \end{aligned}$$

  where the first inequality follows from the fact that the 2-norm is bounded by the 1-norm and relation $2ab\le a^2+b^2$ for any $a,b\in \mathbb {R}$, the second inequality follows from relations $0<1-\Vert \mu Ys-e\Vert \le \mu y_i s_i$ $i=1,2,\ldots ,n$, and the last equality follows from the definition of the Newton step. To end the proof, combine the above inequalities.

• (b) Define $(y_\alpha ,s_\alpha )=(y,s)+\alpha (d^y,d^s)$ for $\alpha \ge 0$ and observe that, since $\mu Sd^y+\mu Yd^s=b^s=-(\mu Ys-e)$, it holds

  $$\begin{aligned} \mu Y_\alpha s_\alpha -e =(1-\alpha ) (\mu Ys-e) + \alpha ^2\mu D^yd^s. \end{aligned}$$

  In view of the assumptions and item (a), we have that $\mu \Vert D^yd_s\Vert <1$. Therefore,

  $$\begin{aligned} \Vert \mu Y_\alpha s_\alpha -e\Vert \le (1-\alpha ) \Vert \mu Ys-e\Vert + \alpha ^2\mu \Vert D^yd^s\Vert<1-\alpha +\alpha ^2<1\ \ \forall \,\alpha \in [0,1]. \end{aligned}$$

  Thus, the components of $(y_\alpha ,s_\alpha )$, which are continuous functions of $\alpha$, do not vanish for any $\alpha \in [0,1]$. Since $(y_0,s_0)=(y,s)>0$, we conclude that $(y_{\alpha =1},s_{\alpha =1})=(y+d^y,s+d^s)>0$.

Next we prove a technical lemma.

Lemma 8

If $\alpha _1,\dots ,\alpha _k>0$ and $\sum _{i=1}^k\alpha _i^2\le a,$ then

$$\begin{aligned} \sum _{i=1}^k\frac{1}{\alpha _i}\ge \sqrt{ \frac{k^3}{a} }. \end{aligned}$$

Proof

Define $f:\mathbb {R}^k_{++}\rightarrow \mathbb {R}$ and $\alpha ^*\in \mathbb {R}^k_{++}$,

$$\begin{aligned} f(\alpha )=\sum _{i=1}^k\frac{1}{\alpha _i}+\frac{1}{2} \left( \frac{k}{a}\right) ^{3/2}\left[ \sum _{i=1}^k\alpha _i^2-a \right] ,\ \alpha _1^*=\alpha _2^*=\cdots =\alpha _k^*=\sqrt{\frac{\textstyle a}{\textstyle k}}. \end{aligned}$$

Observe that f is strictly convex in $\mathbb {R}^k_{++}$, $\nabla f(\alpha ^*)=0$ and $\sum _{i=1}^k(\alpha _i^*)^2=a$; therefore, if $\alpha _1,\dots ,\alpha _k>0$ and $\sum _{i=1}^k\alpha _i^2\le a$, then it holds

$$\begin{aligned} \sum _{i=1}^k\frac{1}{\alpha _i}&\ge f(\alpha _1,\dots ,\alpha _k) \ge f(\alpha ^*) =\sum _{i=1}^k\frac{1}{\alpha _i^*}=\sqrt{\frac{\textstyle k^3}{\textstyle a}}, \end{aligned}$$