A Full-Newton Step Infeasible Interior-Point Method for the Special Weighted Linear Complementarity Problem

Abstract

As an extension of the complementarity problem (CP), the weighted complementarity problem (wCP) is a large class of equilibrium problems with wide applications in science, economics, and engineering. If the weight vector is zero, the wCP reduces to a CP. In this paper, we present a full-Newton step infeasible interior-point method (IIPM) for the special weighted linear complementarity problem over the nonnegative orthant. One iteration of the algorithm consists of one feasibility step followed by a few centering steps. All of them are full-Newton steps, and hence, no calculation of the step size is necessary. The iteration bound of the algorithm is as good as the best-known polynomial complexity of IIPMs for linear optimization.

Proof of Lemma 4.3

By (15), (19) and (20), we have

$$\begin{aligned} (v^{f})^{2}=\frac{w(t)(e+d_{x}d_{s})}{w(t_{+})}=\frac{w(t)(e+d_{x}d_{s})}{(1-\theta )w(t)+\theta w}. \end{aligned}$$

Since $\omega (v)<\dfrac{1}{\sqrt{2}},$ we obtain from (18) that $\Vert d_{x}d_{s}\Vert _{\infty }\le 2\omega (v)^{2}<1,$ and hence $e+d_{x}d_{s}>0.$ By letting $u:=\sqrt{e+d_{x}d_{s}},$ we have $v^{f}=\sqrt{\dfrac{w(t)}{(1-\theta )w(t)+\theta w}}u.$ Then, from (20),

$$\begin{aligned} 4\delta (v^{f})^{2}= & {} \left\| \sqrt{\dfrac{w(t)}{(1-\theta )w(t)+\theta w}}u-\sqrt{\dfrac{(1-\theta )w(t)+\theta w}{w(t)}}u^{-1}\right\| ^{2}\\= & {} \left\| \dfrac{\theta (w(t)-w)}{\sqrt{w(t)[(1-\theta )w(t)+\theta w]}}u+\sqrt{\dfrac{(1-\theta )w(t)+\theta w}{w(t)}}(u-u^{-1})\right\| ^{2}\\= & {} \theta ^{2}e^{T}\left[ \dfrac{(w(t)-w)^{2}}{w(t)((1-\theta )w(t)+\theta w)}u^{2}\right] +e^{T}\left[ \dfrac{((1-\theta )w(t)+\theta w)}{w(t)}(u-u^{-1})^{2}\right] \\&+2\theta e^{T}\left[ \dfrac{(w(t)-w)}{w(t)}u(u-u^{-1})\right] . \end{aligned}$$

Taking into account the fact that $u^{2}=e+d_{x}d_{s},$ the three terms in the last relation can be written, respectively, as follows:

$$\begin{aligned}&\theta ^{2}e^{T}\left[ \dfrac{(w(t)-w)^{2}}{w(t)((1-\theta )w(t)+\theta w)}u^{2}\right] =\theta ^{2}e^{T}\left[ \dfrac{(w(t)-w)^{2}}{w(t)((1-\theta )w(t)+\theta w)}(e+d_{x}d_{s})\right] \\&\quad =\theta ^{2}e^{T}\left[ \dfrac{(w(t)-w)^{2}}{w(t)((1-\theta )w(t)+\theta w)}\right] +\theta ^{2}e^{T}\left[ \dfrac{(w(t)-w)^{2}}{w(t)((1-\theta )w(t)+\theta w)}d_{x}d_{s}\right] ,\\&e^{T}\left[ \dfrac{((1-\theta )w(t)+\theta w)}{w(t)}(u-u^{-1})^{2}\right] \\&\quad =e^{T}\left[ \dfrac{((1-\theta )w(t)+\theta w)}{w(t)}(u^{2}+u^{-2}-2e)\right] \\&\quad =e^{T}\left[ \dfrac{((1-\theta )w(t)+\theta w)}{w(t)}(d_{x}d_{s}+\dfrac{e}{e+d_{x}d_{s}}-e)\right] \\&\quad =e^{T}\left[ \dfrac{((1-\theta )w(t)+\theta w)}{w(t)}d_{x}d_{s}\right] +e^{T}\left[ \dfrac{((1-\theta )w(t)+\theta w)}{w(t)}\cdot \dfrac{e}{e+d_{x}d_{s}}\right] \\&\qquad -e^{T}\left[ \dfrac{(1-\theta )w(t)+\theta w}{w(t)}\right] , \end{aligned}$$

and

$$\begin{aligned}&2\theta e^{T}\left[ \dfrac{(w(t)-w)}{w(t)}u(u-u^{-1})\right] \\&\quad =2\theta e^{T}\left[ \dfrac{(w(t)-w)}{w(t)}u^{2}\right] -2\theta e^{T}\left[ \dfrac{w(t)-w}{w(t)}\right] \\&\quad =2\theta e^{T}(eu^{2})-2\theta e^{T}\left[ \dfrac{w}{w(t)}u^{2}\right] -2\theta e^{T}e+2\theta e^{T}\left[ \dfrac{w}{w(t)}\right] \\&\quad =2\theta e^{T}(e+d_{x}d_{s})-2\theta e^{T}\left[ \dfrac{w}{w(t)}(e+d_{x}d_{s})\right] -2\theta n+2\theta e^{T}\left[ \dfrac{w}{w(t)}\right] \\&\quad =2\theta n+2\theta d_{x}^{T}d_{s}-2\theta e^{T}\left[ \dfrac{w}{w(t)}\right] -2\theta e^{T}\left[ \dfrac{w}{w(t)}d_{x}d_{s}\right] -2\theta n+2\theta e^{T}\left[ \dfrac{w}{w(t)}\right] \\&\quad =2\theta e^{T}\left[ \dfrac{(w(t)-w)}{w(t)}d_{x}d_{s}\right] . \end{aligned}$$

Substitution yields

$$\begin{aligned} 4\delta (v^{f})^{2}= & {} \theta ^{2}e^{T}\left[ \dfrac{(w(t)-w)^{2}}{w(t)((1-\theta )w(t)+\theta w)}\right] +\theta ^{2}e^{T}\left[ \dfrac{(w(t)-w)^{2}}{w(t)((1-\theta )w(t)+\theta w)}d_{x}d_{s}\right] \\&+e^{T}\left[ \dfrac{((1-\theta )w(t)+\theta w)}{w(t)}d_{x}d_{s}\right] +e^{T}\left[ \dfrac{((1-\theta )w(t)+\theta w)}{w(t)}\cdot \dfrac{e}{e+d_{x}d_{s}}\right] \\&-e^{T}\left[ \dfrac{((1-\theta )w(t)+\theta w)}{w(t)}\right] +2\theta e^{T}\left[ \dfrac{(w(t)-w)}{w(t)}d_{x}d_{s}\right] \\= & {} e^{T}\left\{ \left[ \dfrac{\theta ^{2}(w(t)-w)^{2}}{w(t)((1-\theta )w(t)+\theta w)} +\dfrac{(1+\theta )w(t)-\theta w}{w(t)}\right] d_{x}d_{s}\right\} \\&+e^{T}\left[ \dfrac{\theta ^{2}(w(t)-w)^{2}}{w(t)((1-\theta )w(t)+\theta w)} -\dfrac{(1-\theta )w(t)+\theta w}{w(t)}\right] \\&+e^{T}\left\{ \left[ (1-\theta )e+\theta \dfrac{w}{w(t)}\right] \dfrac{e}{e+d_{x}d_{s}}\right\} . \end{aligned}$$

Since

$$\begin{aligned}&\dfrac{\theta ^{2}(w(t)-w)^{2}}{w(t)((1-\theta )w(t)+\theta w)} +\dfrac{(1+\theta )w(t)-\theta w}{w(t)}\\&\quad =\dfrac{\theta ^{2}(w(t)^{2}+w^{2}-2w(t)w)+((1+\theta )w(t)-\theta w)((1-\theta )w(t)+\theta w)}{w(t)((1-\theta )w(t)+\theta w)}\\&\quad =\dfrac{\theta ^{2}w(t)^{2}+\theta ^{2}w^{2}-2\theta ^{2}w(t)w+(1-\theta ^{2})w(t)^{2}-\theta ^{2}w^{2}+2\theta ^{2}w(t)w}{w(t)((1-\theta )w(t)+\theta w)}\\&\quad =\dfrac{w(t)}{(1-\theta )w(t)+\theta w} \end{aligned}$$

and

$$\begin{aligned}&\dfrac{\theta ^{2}(w(t)-w)^{2}}{w(t)((1-\theta )w(t)+\theta w)} -\dfrac{(1-\theta )w(t)+\theta w}{w(t)}\\&\quad =\dfrac{\theta ^{2}(w(t)-w)^{2}-((1-\theta )w(t)+\theta w)^{2}}{w(t)((1-\theta )w(t)+\theta w)}\\&\quad =\dfrac{[\theta (w(t)-w)+(1-\theta )w(t)+\theta w][\theta (w(t)-w)-(1-\theta )w(t)-\theta w]}{w(t)((1-\theta )w(t)+\theta w)}\\&\quad =\dfrac{\theta (w(t)-w)-(1-\theta )w(t)-\theta w}{(1-\theta )w(t)+\theta w}\\&\quad =\dfrac{\theta (w(t)-w)}{(1-\theta )w(t)+\theta w}-e, \end{aligned}$$

we have

$$\begin{aligned} 4\delta (v^{f})^{2}= & {} e^{T}\left[ \dfrac{w(t)}{(1-\theta )w(t)+\theta w}d_{x}d_{s}\right] +e^{T}\left[ \dfrac{\theta (w(t)-w)}{(1-\theta )w(t)+\theta w}-e\right] \\&+e^{T}\left\{ \left[ (1-\theta )e+\theta \dfrac{ w}{w(t)}\right] \dfrac{e}{e+d_{x}d_{s}}\right\} \\= & {} e^{T}\left[ \dfrac{w(t)}{(1-\theta )w(t)+\theta w}d_{x}d_{s}\right] +e^{T}\left[ \dfrac{\theta (w(t)-w)}{(1-\theta )w(t)+\theta w}-e\right] \\&-e^{T}\left\{ \left[ (1-\theta )e+\theta \dfrac{ w}{w(t)}\right] \dfrac{d_{x}d_{s}}{e+d_{x}d_{s}}\right\} +e^{T}\left[ (1-\theta )e+\theta \dfrac{ w}{w(t)}\right] \\= & {} e^{T}\left[ \dfrac{w(t)}{(1-\theta )w(t)+\theta w}d_{x}d_{s}\right] +e^{T}\left[ \frac{\theta (w(t)-w)}{(1-\theta )w(t)+\theta w}+\theta \frac{w}{w(t)}-\theta e\right] \\&-(1-\theta )e^{T}\left( \dfrac{d_{x}d_{s}}{e+d_{x}d_{s}}\right) -\theta e^{T}\left[ \dfrac{w}{w(t)}\cdot \dfrac{d_{x}d_{s}}{e+d_{x}d_{s}}\right] \\= & {} e^{T}\left[ \dfrac{w(t)}{(1-\theta )w(t)+\theta w}d_{x}d_{s}\right] +\theta ^{2} e^{T}\left[ \dfrac{(w(t)-w)^{2}}{w(t)((1-\theta )w(t)+\theta w)}\right] \\&-(1-\theta )e^{T}\left( \dfrac{d_{x}d_{s}}{e+d_{x}d_{s}}\right) -\theta e^{T}\left[ \dfrac{w}{w(t)}\cdot \dfrac{d_{x}d_{s}}{e+d_{x}d_{s}}\right] . \end{aligned}$$

Hence, it follows from (4), (17) and (18) that

$$\begin{aligned} 4\delta (v^{f})^{2}\le & {} \dfrac{1}{1-\theta }\sum \limits _{i=1}^{n}|d_{x_{i}}d_{s_{i}}|+(1-\theta )\dfrac{\sum \limits _{i=1}^{n}|d_{x_{i}}d_{s_{i}}|}{1-2\omega (v)^{2}} +\theta \sum \limits _{i=1}^{n}\dfrac{w_{i}}{w_{i}(t)}\cdot \dfrac{|d_{x_{i}}d_{s_{i}}|}{1-2\omega (v)^{2}}\\&+\theta ^{2}\sum \limits _{i=1}^{n}\left( \dfrac{t}{t_{0}}\right) ^{2}\cdot \dfrac{(w^{0}_{i}-w_{i})^{2}}{w_{i}(t)((1-\theta )w_{i}(t)+\theta w_{i})}\\\le & {} \dfrac{2\omega (v)^{2}}{1-\theta }+(1-\theta )\dfrac{2\omega (v)^{2}}{1-2\omega (v)^{2}} +\theta \max \limits _{1\le i\le n}\left\{ 1,\dfrac{w_{i}}{w^{0}_{i}}\right\} \dfrac{2\omega (v)^{2}}{1-2\omega (v)^{2}}\\&+\theta ^{2}n\max \limits _{1\le i\le n}\left\{ \left( \dfrac{w^{0}_{i}}{w_{i}}-1\right) ^{2},\left( 1-\dfrac{w_{i}}{w^{0}_{i}}\right) ^{2},\dfrac{1}{1-\theta }\right\} . \end{aligned}$$

Moreover, in the case of $w_{i}=0$ for some i,  we have

$$\begin{aligned} \left( \dfrac{t}{t_{0}}\right) ^{2}\cdot \dfrac{(w^{0}_{i}-w_{i})^{2}}{w_{i}(t)((1-\theta )w_{i}(t)+\theta w_{i})} =\left( \dfrac{t}{t_{0}}\right) ^{2}\cdot \dfrac{(w^{0}_{i})^{2}}{(1-\theta )\cdot \left( \dfrac{t}{t_{0}}w^{0}_{i}\right) ^{2}}=\dfrac{1}{1-\theta }, \end{aligned}$$

which implies that relation (21) holds for $w\ge 0.$ $\square$