IPRSOCP: A Primal-Dual Interior-Point Relaxation Algorithm for Second-Order Cone Programming

Abstract

Inspired by the smoothing barrier augmented Lagrangian function in Liu et al. (Math Methods Oper Res 96(3):351–382, 2022), we propose a primal-dual interior-point relaxation algorithm for second-order cone programming, called IPRSOCP. Two features of the IPRSOCP algorithm are as follows. One is that the iterative points of the proposed algorithm need not lie inside the interior region, convening the use of warm-start. The other is that an explicit form of the Schur complement matrix is explored such that the low-rank structure of the Schur complement matrix can be used to improve numerical stability and efficiency. Under suitable assumptions, it is shown that the barrier parameter in the IPRSOCP algorithm tends to zero and the generated sequence of iterations is globally convergent to the solution. Numerical results demonstrate that the IPRSOCP algorithm is competitive with the open-source benchmark solvers, SeDuMi, SDPT3, and ECOS, in terms of robustness and efficiency.

Appendix A: Some Proofs

Proof of Lemma 2

Proof

Define the spectral decomposition $\varvec{v}_i = \eta _1 \varvec{d}_1 + \eta _2 \varvec{d}_2$. Then

$$\begin{aligned} \begin{aligned} \varvec{u}_{i}&=\left( \left( \varvec{s}_{i}-\rho \varvec{x}_{i}\right) ^{2}+4 \rho \mu \varvec{e}_{i}\right) ^{\frac{1}{2}} \\&=\left( \eta _{1}^{2} \varvec{d}_{1}+\eta _{2}^{2} \varvec{d}_{2}+4 \rho \mu \varvec{d}_{1}+4 \rho \mu \varvec{d}_{2}\right) ^{\frac{1}{2}} \\&=\sqrt{\eta _{1}^{2}+4 \rho \mu }\varvec{d}_{1}+\sqrt{\eta _{2}^{2}+4 \rho \mu } \varvec{d}_{2}. \end{aligned} \end{aligned}$$

Therefore, we can obtain

$$\begin{aligned} \begin{aligned}&\varvec{z}_{i}=\frac{\varvec{u}_{i}-\varvec{v}_{i}}{2 \rho }=\frac{\sqrt{\eta _{1}^{2}+4 \rho \mu }-\eta _{1}}{2 \rho } \varvec{d}_{1}+\frac{\sqrt{\eta _{2}^{2}+4 \rho \mu }-\eta _{2}}{2 \rho } \varvec{d}_{2}, \\ {}&\varvec{y}_{i}=\frac{\varvec{u}_{i}+\varvec{v}_{i}}{2 \rho }=\frac{\sqrt{\eta _{1}^{2}+4 \rho \mu }+\eta _{1}}{2 \rho } \varvec{d}_{1}+\frac{\sqrt{\eta _{2}^{2}+4 \rho \mu }+\eta _{2}}{2 \rho } \varvec{d}_{2}. \end{aligned} \end{aligned}$$

$\varvec{z}_i$ and $\varvec{y}_i$ share the same Jordan frame $\{\varvec{d}_1,\varvec{d}_2\}$ and are thus operator commutative.

Because the eigenvalues of $\varvec{z}_i$ and $\varvec{y}_i$ are non-negative, we have $\varvec{z}_{i},\, \varvec{y}_{i} \in K_{i}$, and

$$\begin{aligned}&\varvec{z}_{i} \circ \varvec{y}_{i} \\ =&\left( \frac{\sqrt{\eta _{1}^{2}+4 \rho \mu }-\eta _{1}}{2 \rho } \varvec{d}_{1}+ \frac{\sqrt{\eta _{2}^{2}+4 \rho \mu }-\eta _{2}}{2 \rho } \varvec{d}_{2}\right) \circ \left( \frac{\sqrt{\eta _{1}^{2}+4 \rho \mu }+\eta _{1}}{2 \rho } \varvec{d}_{1} +\frac{\sqrt{\eta _{2}^{2}+4 \rho \mu }+\eta _{2}}{2 \rho } \varvec{d}_{2}\right) \\ =&\frac{\mu }{\rho } \varvec{d}_{1}+\frac{\mu }{\rho } \varvec{d}_{2}=\frac{\mu }{\rho } \varvec{e}_{i},\,i = 1,2,\cdots ,n, \end{aligned}$$

which completes the proof.

Proof of Theorem 5

Proof

The directional derivative of the merit function $\phi _{\left( \mu , \rho ^{(k)}\right) }(\varvec{w})$ at the point $\left( \mu ^{(k)}, \varvec{w}^{(k)}\right)$ along the direction $\left( \Delta \mu ^{(k)}, \Delta \varvec{w}^{(k)}\right)$ is

$$\begin{aligned} \phi _{\left( \mu ^{(k)}, \rho ^{(k)}\right) }^{\prime }\left( \varvec{w}^{(k)}; \Delta \mu ^{(k)}, \Delta \varvec{w}^{(k)}\right) =-2 \phi _{\left( \mu ^{(k)}, \rho ^{(k)}\right) }\left( \varvec{w}^{(k)}\right) . \end{aligned}$$

(62)

The Taylor expansion of $\phi _{\left( \mu ^{(k)}+\alpha \Delta \mu ^{(k)}, \rho ^{(k)}\right) }\left( \varvec{w}^{(k)}+\alpha \Delta \varvec{w}^{(k)}\right)$ with respect to $\alpha$ at $\alpha =0$ shows that

$$\begin{aligned} \begin{aligned}&\phi _{\left( \mu ^{(k)}+\alpha \Delta \mu ^{(k)}, \rho ^{(k)}\right) } \left( \varvec{w}^{(k)}+\alpha \Delta \varvec{w}^{(k)} \right) - \phi _{\left( \mu ^{(k)}, \rho ^{(k)}\right) } \left( \varvec{w}^{(k)}\right) \\&\quad =\alpha \phi _{\left( \mu ^{(k)}, \rho ^{(k)}\right) }^{\prime } \left( \varvec{w}^{(k)}; \Delta \mu ^{(k)}, \Delta \varvec{w}^{(k)}\right) +o(\alpha ) \\&\quad =-2 \tau \alpha \phi _{\left( \mu ^{(k)}, \rho ^{(k)}\right) } \left( \varvec{w}^{(k)}\right) -2(1-\tau ) \alpha \phi _{\left( \mu ^{(k)}, \rho ^{(k)}\right) } \left( \varvec{w}^{(k)}\right) +o(\alpha ). \end{aligned} \end{aligned}$$

(63)

Since $\tau < 1$ and $\phi _{\left( \mu ^{(k)}, \rho ^{(k)}\right) } \left( \varvec{w}^{(k)}\right) > 0$, then (45) holds for all sufficiently small $\alpha > 0$.

Proof of Theorem 6

Proof

By Lemma 8, we have $\phi _{(\mu ^{(k)},\rho ^{(k)})}(\varvec{w}^{(k)})\leqslant \phi _{(\mu ^{(0)},\rho ^{(0)})}(\varvec{w}^{(0)})$ for all $k = 1,2,\cdots$ By the definition of the merit function $\phi _{(\mu ^{(k)},\rho ^{(k)})}(\varvec{w}^{(k)})$, we can obtain

$$\begin{aligned} \frac{1}{2}\Vert \varvec{z}^{(k)}-\varvec{x}^{(k)}\Vert ^2\leqslant \phi _{(\mu ^{(0)},\rho ^{(0)})}(\varvec{w}^{(0)}), \end{aligned}$$

which together with Assumption 2 implies that $\{\varvec{z}^{(k)}\}$ is bounded.

Define the spectral decomposition $\varvec{v}_i= \eta _1 \varvec{d}_1 + \eta _2 \varvec{d}_2$. By the definition of $\varvec{y}_i$, we have

$$\begin{aligned} \varvec{y}_i = \dfrac{\sqrt{\eta _1^2+4\rho \mu }+\eta _1}{2\rho } \varvec{d}_1 + \dfrac{\sqrt{\eta _2^2+4\rho \mu }+\eta _2}{2\rho } \varvec{d}_2. \end{aligned}$$

Since $\Vert \varvec{d}_i\Vert =\dfrac{1}{\sqrt{2}},\,i=1,2$, it is sufficient to prove that $\dfrac{\sqrt{\eta _1^2+4\rho \mu }+\eta _1}{2\rho }$ and $\dfrac{\sqrt{\eta _2^2+4\rho \mu }+\eta _2}{2\rho }$ are bounded. Note that

$$\begin{aligned}{} & {} \dfrac{\sqrt{\eta _1^2+4\rho \mu }+\eta _1}{2\rho }\\{} & {} \quad \leqslant \dfrac{|\eta _1|}{\rho } + \sqrt{\dfrac{\mu }{\rho }}\\{} & {} \quad \leqslant \dfrac{|\varvec{s}_{i0}-\rho \varvec{x}_{i0}|+\Vert \bar{\varvec{s}_i}-\rho \bar{\varvec{x}}_i\Vert }{\rho } + \sqrt{\dfrac{\mu }{\rho }}\\{} & {} \quad \leqslant \dfrac{\sqrt{2}\Vert \varvec{s}_i-\rho \varvec{x}_i\Vert }{\rho }+ \sqrt{\dfrac{\mu }{\rho }}\\{} & {} \quad \leqslant \dfrac{\sqrt{2}\Vert \varvec{s}_i\Vert }{\rho }+\sqrt{2}\Vert \varvec{x}_i\Vert + \sqrt{\dfrac{\mu }{\rho }}\\{} & {} \quad \leqslant \sqrt{2}\dfrac{\max \{1,\Vert \varvec{x}_i\Vert \}}{\sigma }+\sqrt{2}\Vert \varvec{x}_i\Vert + \sqrt{\dfrac{\mu }{\rho }}. \end{aligned}$$

The third inequality holds because $|a|+|b|\leqslant \sqrt{2}\sqrt{a^2+b^2}$. Thus, $\dfrac{\sqrt{\eta _1^2+4\rho \mu }+\eta _1}{2\rho }$ is bounded under Assumption 2. The boundedness of $\dfrac{\sqrt{\eta _2^2+4\rho \mu }+\eta _2}{2\rho }$ can be similarly deduced. So $\{\varvec{y}^{(k)}\}$ is bounded.

Due to $\rho ^{(k)}\varvec{y}^{(k)}_{i}\varvec{z}^{(k)}_{i}=\mu ^{(k)}\varvec{e}_i$, we have $\Vert \varvec{y}^{(k)}_{i}\varvec{z}^{(k)}_{i}\Vert =\dfrac{\mu ^{(k)}}{\rho ^{(k)}}$. Combined with the boundedness of $\{\varvec{y}^{(k)}_i\}$ and $\{\varvec{z}^{(k)}_i\}$, one has the desired inequalities.