Convergence of a Weighted Barrier Algorithm for Stochastic Convex Quadratic Semidefinite Optimization

Abstract

Mehrotra and Özevin (SIAM J Optim 19:1846–1880, 2009) computationally found that a weighted barrier decomposition algorithm for two-stage stochastic conic programs achieves significantly superior performance when compared to standard barrier decomposition algorithms existing in the literature. Inspired by this motivation, Mehrotra and Özevin (SIAM J Optim 20:2474–2486, 2010) theoretically analyzed the iteration complexity for a decomposition algorithm based on the weighted logarithmic barrier function for two-stage stochastic linear optimization with discrete support. In this paper, we extend the aforementioned theoretical paper and its self-concordance analysis from the polyhedral case to the semidefinite case and analyze the iteration complexity for a weighted logarithmic barrier decomposition algorithm for two-stage stochastic convex quadratic SDP with discrete support.

A Complexity Proofs

In this appendix, we give proofs of Theorems 6.1 and 6.2. The following proposition, which is due to [18, Theorem 2.1.1], follows directly from the definition of self-concordance.

Proposition A.1

Let $\mu > 0$, $X \in \mathcal {P}$, and $\delta \triangleq \sqrt{(1/\mu )\nabla ^2\eta (\mu ,X)[\Delta X,\Delta X]}$. Then, for $\delta < 1$, $\tau \in [0,1]$, and $D \in \mathbb {S}^{n_1}$, we have

$$\begin{aligned} (1-\tau \delta )^2\nabla ^2\eta (\mu ,X)[D,D] \le \nabla ^2\eta (\mu ,X+\tau \Delta X)[D,D] \le (1-\tau \delta )^2\nabla ^2\eta (\mu ,X)[D,D]. \end{aligned}$$

(31)

We use two different merit functions to measure the speed of Newton method to estimate the number of Newton steps required for recentering. For the short-step algorithm, we utilize $\delta (\mu ,X)$, and for the long-step algorithm, we use the first-stage objective $\eta (\mu ,X)$. The following lemma is derived from Theorem 2.2.3 in [18] and describes the behavior of the Newton method when applied to $\eta (\mu ,\cdot )$.

Lemma A.1

For any $\mu > 0$ and $X \in \mathcal {P}_\circ$, let $\Delta X$ be the Newton direction calculated by (30), and $\delta$ be defined as $\delta \triangleq \delta (\mu ,X)\triangleq \sqrt{1/(\mu \alpha \min _k \{\pi ^{(k)}\}) \nabla ^2\eta (\mu ,X)[\Delta X,\Delta X]}$. Then, the following statements hold:

• (i) If $\delta < 2-\sqrt{3}$, then $\delta (\mu ,X+\Delta X)\le (\delta /(1-\delta ))^2\le \delta /2$.

• (ii) If $\delta \ge 2-\sqrt{3}$, then $\eta (\mu ,X)-\eta (\mu ,X+\bar{\theta }\Delta X)\ge \mu \alpha \min _k\{ \pi ^{(k)}\} (\delta -\ln (1+\delta ))$, where $\bar{\theta }=1/(1+\delta )$.

1.1 A.1 Complexity Proof of the Short-Step Algorithm

Proving Theorem 6.1 makes use of Lemma A.2.

Lemma A.2

Let $\mu ^+\triangleq \gamma \mu$, where $\gamma = 1 - \sigma \sqrt{\alpha \min _k \left\{ \pi ^{(k)}\right\} /(n_1+\alpha n_2)}$ and $\sigma \le 0.1$. Furthermore, let $\beta = (2 - \sqrt{3})/2$. If $\delta (\mu ,X)\le \beta$, then $\delta (\mu ^+,X)\le 2\beta$.

Proof

Let $\kappa =2\beta = 2 - \sqrt{3}$. It is easy to verify that, for $\sigma < 0.1$, $\mu ^+$ satisfies

$$\begin{aligned}{} & {} \varphi _\kappa \left( \eta ;\mu ,\mu ^+\right) \triangleq \left( \frac{1 + \sqrt{n_2}}{2} +\frac{\sqrt{n_1+\alpha n_2}}{\kappa \sqrt{ \min _k \pi ^{(k)}}}\right) \ln \left( 1 - \sigma \sqrt{\alpha \min _k \left\{ \pi ^{(k)}\right\} (n_1+\alpha n_2)}\right) ^{-1} \\{} & {} \qquad \le \frac{1}{2}\le 1-\frac{\delta (\mu ,X)}{\kappa }. \end{aligned}$$

Applying Theorem 3.1.1 in [18], we get $\delta (\mu ^+,X) < \kappa =2\beta$ as desired. $\square$

From Lemmas A.1 and A.2, we conclude that at each iteration we can reduce the barrier parameter $\mu$ by the factor $\gamma = 1 -\sigma \sqrt{\alpha \min _k\left\{ \pi ^{(k)} \right\} /(n_1+\alpha n_2)}$, with $\sigma \le 0.1$, and that only one Newton step is sufficient to restore proximity to the central path. This proves Theorem 6.1.

1.2 A.2 Complexity Proof of the Long-Step Algorithm

We use $\eta$ as the merit function for the analysis of the long-step algorithm because the iterations that are generated by the less conservative long-step algorithm may violate the condition $\delta <2 - \sqrt{3}$, which is required in item (i) of Lemma A.1.

We assume that we have a point $\mu ^{(j-1)}$ sufficiently close to $X(\mu ^{(j-1)})$. Then, we reduce the barrier parameter from $\mu ^{(j-1)}$ to $\mu ^{(j)} = \gamma \mu ^{(j-1)}$, where $\gamma \in (0,1)$. While searching for a point $X^{(j)}$ that is sufficiently close to $X(\mu ^{(j)})$, our algorithm will generate a finite sequence of points $P_1,P_2,\ldots , P_N \in \mathcal {P}$ and we finally set $X(\mu ^{(j)})= P_N$ . We need to determine an upper bound on N; the number of Newton iterations needed for recentering. We begin by determining an upper bound on the difference $\phi (\mu ^{(j)},X^{(j-1)}) \triangleq \eta (\mu ^{(j)},X^{(j-1)}) -\eta (\mu ^{(j)},X(\mu ^{(j)}))$. Then, by item (ii) of Lemma A.1, we know that at $P_k \in \mathcal {P}$, a Newton step with step size $\bar{\theta } =( 1 + \delta )^{-1}$ decreases $\eta (\mu ^{(j)},P_k)$ at least by a certain amount, which depends on the current value of $\delta$ and $\mu$. A line search might yield even a larger decrease. Consequently, we will get an upper bound on N.

Proving Theorem 6.2 makes use of the following propositions and lemmas. We give Proposition A.2 without proof because the proof is quite similar to that of [29, Lemma 6].

Proposition A.2

For any $\mu > 0$ and $X\in \mathcal {P}_\circ$, we denote $\tilde{\Delta }X \triangleq X-X(\mu )$ and define the parameter $\tilde{\delta }(\mu ,X)$ as $\tilde{\delta }(\mu ,X)\triangleq \sqrt{ 1/(\mu \alpha \min _k \{\pi ^{(k)}\})\nabla ^2\eta (\mu ,X)[\tilde{\Delta }X,\tilde{\Delta } X]}$. If $\tilde{\delta }<1$, then $\phi (\mu ,X)\le \mu ( \tilde{\delta }/(1-\tilde{\delta }) +\ln (1-\tilde{\delta }))$.

Lemma A.3

Let $\tilde{\Delta } X$ and $\tilde{\delta }$ be as defined in Proposition A.2. Let also $\mu > 0$ and $X\in \mathcal {P}_\circ$. If $\tilde{\delta }<1$, then we have $|\phi ^\prime (\mu ,X)|\le -\sqrt{n_1+\alpha n_2}\ln (1-\tilde{\delta })$.

Proof

For any $\mu > 0$, using the chain rule, we can write $\phi ^\prime (\mu ,X)=\eta ^\prime (\mu ,X)-\eta ^\prime (\mu ,X(\mu ))-\nabla \eta ^\prime (\mu ,X(\mu ))[ X^\prime (\mu )]$. According to the optimality conditions (29), we have $\nabla \eta ^\prime (\mu ,X(\mu )) [ X^\prime (\mu )]= 0$. Consequently,

$$\begin{aligned} \phi ^\prime (\mu ,X)=\eta ^\prime (\mu ,X)-\eta ^\prime (\mu ,X(\mu )). \end{aligned}$$

(32)

From (24), we get

$$\begin{aligned} \left\langle \left\{ \nabla \eta (\mu ,X) \right\} ^\prime , \left\{ \nabla ^2\eta (\mu ,X)\right\} ^{-1} \left\{ \nabla \eta (\mu ,X)\right\} ^\prime \right\rangle \le \displaystyle \frac{n_1+\alpha n_2}{\mu }. \end{aligned}$$

(33)

From (32), applying the mean-value theorem, we get

$$\begin{aligned}{} & {} \left| \phi ^\prime (\mu ,X)\right| =\left| \displaystyle \int _0^1 \left\langle \left\{ \nabla \eta (\mu ,X(\mu )+\tau \tilde{\Delta }X)^\textsf{T }\right\} ^\prime ,\tilde{\Delta }X \right\rangle d\tau \right| \nonumber \\{} & {} \quad \le \displaystyle \int _0^1 \sqrt{\left\langle \nabla ^2\eta \left( \mu ,X(\mu )+\tau \tilde{\Delta }X\right) \tilde{\Delta }X,\tilde{\Delta }X\right\rangle }\nonumber \\{} & {} \qquad \times \sqrt{ \left\langle \left\{ \nabla ^2\eta (\mu ,X(\mu )+\tau \tilde{\Delta }X)\right\} ^{-1} \left\{ \nabla \eta (\mu ,X(\mu )+\tau \tilde{\Delta }X)^\textsf{T }\right\} ^\prime , \left\{ \nabla \eta (\mu ,X(\mu )+\tau \tilde{\Delta }X)^\textsf{T }\right\} ^\prime \right\rangle }~~ d\tau .\nonumber \\ \end{aligned}$$

(34)

Using (34), (33), and (31) and noting that $X(\mu )+\tau \tilde{\Delta }X=X-(1-\tau )\tilde{\Delta }X$, we have

$$\begin{aligned}{} & {} \left| \phi ^\prime (\mu ,X)\right| = \displaystyle \int _0^1\frac{\sqrt{ \left\langle \nabla ^2\eta (\mu ,X) \tilde{\Delta } X, \tilde{\Delta }X\right\rangle }}{1-\tilde{\delta }+\tau \tilde{\delta }} \sqrt{\frac{n_1+\alpha n_2}{\mu }}d\tau \\{} & {} \qquad \le \displaystyle \int _0^1\frac{\sqrt{\mu }\tilde{\delta }}{1-\tilde{\delta }+\tau \tilde{\delta }} \sqrt{\frac{n_1+\alpha n_2}{\mu }}d\tau =-\sqrt{n_1+\alpha n_2} \ln \left( 1-\tilde{\delta }\right) . \end{aligned}$$

The proof is complete. $\square$

Lemma A.4

Let $\mu > 0$ and $X\in \mathcal {P}_\circ$ be such that $\tilde{\delta }<1$, where $\tilde{\delta }$ is as defined in Proposition A.2. Let also $\mu ^+=\gamma \mu$ with $\gamma \in (0,1)$. Then, $\eta (\mu ^+,X) -\eta (\mu ^+,X(\mu ^+)) \le \mathcal {O}(n_1+\alpha n_2)\,\mu ^+$.

Proof

Differentiating (32), we get

$$\begin{aligned} \phi ^{\prime \prime }(\mu ,X)=\eta ^{\prime \prime }(\mu ,X)-\eta ^{\prime \prime }(\mu ,X(\mu )) -\left\langle \nabla \eta ^{\prime }(\mu ,X(\mu )), X^\prime (\mu )\right\rangle . \end{aligned}$$

(35)

We take the derivative of the optimality conditions (29) of the first-stage problem (2) to get $X^\prime (\mu )$:

$$\begin{aligned} \begin{array}{rrl} \left\{ \nabla \eta (\mu ,X(\mu ))\right\} ^\prime + \Sigma [X^\prime (\mu )]-\mathcal {A}^\star \lambda ^\prime (\mu )&{}=&{}0,\\ \mathcal {A}X^\prime (\mu )&{}=&{}0. \end{array} \end{aligned}$$

(36)

where $\Sigma =\nabla ^2\eta (\mu ,X(\mu ))$. Solving (36), we have $X^\prime (\mu )=-(\Sigma ^{-1}- \Sigma ^{-1}\mathcal {A}^\textsf{T }( \mathcal {A}\Sigma ^{-1}\mathcal {A}^\textsf{T })\mathcal {A}\Sigma ^{-1}) [\{\nabla \eta (\mu ,X(\mu ))\}^\prime ].$ Next, we have

$$\begin{aligned}{} & {} - \left\langle \left\{ \nabla \eta (\mu ,X)\right\} ^\prime , X^\prime (\mu ) \right\rangle \nonumber \\{} & {} \quad = \left\langle \left( \Sigma ^{-1}- \Sigma ^{-1}\mathcal {A}^\textsf{T }(\mathcal {A}\Sigma ^{-1}\mathcal {A}^\textsf{T })\mathcal {A}\Sigma ^{-1}\right) \left[ \left\{ \nabla \eta (\mu ,X(\mu ))\right\} ^\prime \right] , \left\{ \nabla \eta (\mu ,X(\mu ))\right\} ^\prime \right\rangle \nonumber \\{} & {} \quad \le \left\langle \Sigma ^{-1} \left[ \left\{ \nabla \eta (\mu ,X(\mu ))\right\} ^\prime \right] , \left\{ \nabla \eta (\mu ,X(\mu ))\right\} ^\prime \right\rangle ~\le ~ \displaystyle \frac{1}{\mu }(n_1+\alpha n_2), \end{aligned}$$

(37)

where the last inequality is followed from (24).

Now, we bound the first two terms in the right-hand side of (35). From the definition of $\eta (\mu ,X)$ (see (2)), it is clear that $\eta ^{\prime \prime }(\mu ,X)=\sum _{k=1}^K\rho ^{(k)^{\prime \prime }}(\mu ,X)$.

Now, by differentiating (10) and using (9), we get

$$\begin{aligned} \rho ^{(k)^{\prime }}(\mu ,X)= & {} \frac{1}{\alpha }\text {vec}\left( Y^{(k)}\right) ^\textsf{T }\varvec{\text {H}}^{(k)} \text {vec}\left( Y^{(k)^\prime }\right) + \frac{1}{\alpha }\left\langle F^{(k)}, Y^{(k)^\prime }\right\rangle -\ln \det Y^{(k)} \\{} & {} -\mu \text {vec}\left( I\right) ^\textsf{T }\left( Y^{(k)^{-1}}\otimes I\right) \text {vec}\left( Y^{(k)^\prime }\right) \\= & {} \left( \frac{1}{\alpha } \varvec{\text {H}}^{(k)} \text {vec}\left( Y^{(k)}\right) + \frac{1}{\alpha }\text {vec}\left( F^{(k)}\right) -\mu \text {vec}\left( S^{(k)}\right) \right) ^\textsf{T }\text {vec}\left( Y^{(k)^\prime }\right) \\{} & {} -\ln \det Y^{(k)}. \end{aligned}$$

Therefore,

$$\begin{aligned} \rho ^{(k)^{\prime }}(\mu ,X)= & {} z^{{k}^\textsf{T }} \mathcal {W}^{(k)} \text {vec}\left( Y^{(k)^\prime }\right) -\ln \det Y^{(k)}\\= & {} -\ln \det Y^{(k)} \,\,\left( \text {noting that} \,\, \mathcal {W}^{(k)} \text {vec}\left( Y^{(k)^\prime }\right) = 0 \;\text {using}\;(12)\right) , \end{aligned}$$

and

$$\begin{aligned}{} & {} -\rho ^{(k)^{\prime \prime }}(\mu ,X) =\text {vec}(I)^\textsf{T }\left( Y^{(k)^{-1/2}} \otimes Y^{(k)^{-1/2}}\right) \text {vec}\left( Y^{(k)^\prime }(\mu ,X)\right) \\{} & {} \quad = \text {vec}(I)^\textsf{T }\left( Y^{(k)^{-1/2}} \otimes Y^{(k)^{-1/2}}\right) \mathcal {Q}^{(k)} \left( I-\mathcal {Q}^{(k)} {\mathcal {W}^{(k)^\textsf{T }}}\mathcal {R}^{(k)^{-1}}\mathcal {W}^{(k)} \mathcal {Q}^{(k)} \right) \\{} & {} \qquad \mathcal {Q}^{(k)}\left( Y^{(k)^{-1/2}} \otimes Y^{(k)^{-1/2}}\right) \text {vec}(I)\\{} & {} \quad \le \left\Vert I \right\Vert ^2_2 \left\Vert \mathcal {Q}^{(k)}\left( Y^{(k)^{-1}} \otimes Y^{(k)^{-1}}\right) \mathcal {Q}^{(k)} \right\Vert _2^2\\{} & {} \quad \le \displaystyle \frac{n_2}{\mu } \,\, \left( \text {noting that} \, \, \mathcal {Q}^{(k)}\left( Y^{(k)^{-1}} \otimes Y^{(k)^{-1}}\right) \mathcal {Q}^{(k)}\preceq \mu ^{-1}I\right) . \end{aligned}$$

Thus, for any $X\in \mathcal {P}_\circ$, we have $\eta ^{\prime \prime }(\mu ,X)\ge \alpha n_2/\mu$. Since $\eta (\mu ,X)$ is strictly concave in $\mu$ for any $X\in \mathcal {P}_\circ$, we also have $\eta ^{\prime \prime }(\mu ,X)\le 0$. Now, using the above bounds and the one given in (37), from (35) we obtain $\phi ^{\prime \prime }(\mu ,X)\le (n_1+2\alpha n_2)/\mu$. Then, by Proposition A.2 and Lemma A.3, we get

$$\begin{aligned} \phi (\mu ^+,X)\le & {} \mu \left( \displaystyle \frac{\tilde{\delta }}{1-\tilde{\delta }} \ln (1-\tilde{\delta })\right) -\sqrt{n_1+\alpha n_2}\ln (1-\tilde{\delta })(\mu ^+-\mu )\nonumber \\{} & {} \quad +(n_1+\alpha n_2)\displaystyle \int _\mu ^{\mu ^1}\displaystyle \int _\mu ^\tau \mu ^{-1}(\mu )\;d\mu d\tau \nonumber \\\le & {} \mu \left( \displaystyle \frac{\tilde{\delta }}{1-\tilde{\delta }} \ln (1-\tilde{\delta })\right) -\sqrt{n_1+\alpha n_2}\ln (1-\tilde{\delta })(\mu ^+-\mu ) \nonumber \\{} & {} \quad +(n_1+\alpha n_2)+(n_1+\alpha n_2)\ln \gamma ^{-1}(\mu ^+-\mu ). \end{aligned}$$

(38)

Since $\tilde{\delta }$ and $\gamma$ are arbitrary constants, (38) implies that $\eta (\mu ^+,X)-\eta (\mu ^+,X(\mu ^+))\le \mathcal {O}(n_1+\alpha n_2)\mu ^+$ as desired. $\square$

Note that Proposition A.2 and Lemmas A.3 and A.2 all require $\tilde{\delta }$ to be less than one. However, we cannot evaluate $\tilde{\delta }$ because we do not explicitly know the points $X(\mu )$ forming the central path. Nonetheless, we find that $\delta$ is proportional to $\tilde{\delta }$ as shown in the following proposition, which is due to [14, Lemma 5.5].

Proposition A.3

For any $\mu > 0, X\in \mathcal {P}_\circ$, and $\lambda \in \mathbb {R}^{m_1}$, let $(\Delta X,\Delta \lambda )$ be the Newton direction defined in (30) and $(\tilde{\Delta }X,\tilde{\Delta } \lambda )\triangleq (X-X(\mu ),\lambda -\lambda (\mu ))$. We denote $\delta (\mu ,X)=(\Sigma [\Delta X,\Delta X]/\mu )^{1/2}, \,\, \, \text {and} \,\, \, \tilde{\delta }(\mu ,X)=( \Sigma [\tilde{\Delta } X,\tilde{\Delta } X]/\mu )^{1/2}$, where $\Sigma =\nabla ^2\eta (\mu ,X)$. If ${\delta }\le 1/6$ then $2\delta /3 \le \tilde{\delta }\le 2{\delta }$.

Lemma A.1 implies that each line search should decrease the value of $\eta$ by at least $\mu (\delta -\ln (1-\delta ))$. Therefore, in view of Lemmas A.1 and A.4, it is clear that after reducing $\mu$ by a factor $\gamma \in (0,1)$, at most $\mathcal {O}((n_1+\alpha n_2)/ (\alpha \min _k\{\pi ^{(k)}\}))$ Newton iterations are needed for recentering. In the long-step version of the proposed algorithm, we need to update barrier parameter $\mu$ no more than $\mathcal {O}(\ln \mu ^{(0)}/\epsilon )$ times. Theorem 6.2 now follows from Lemmas A.1(ii) and A.4 and Proposition A.3.