A parallelizable augmented Lagrangian method applied to large-scale non-convex-constrained optimization problems

Abstract

We contribute improvements to a Lagrangian dual solution approach applied to large-scale optimization problems whose objective functions are convex, continuously differentiable and possibly nonlinear, while the non-relaxed constraint set is compact but not necessarily convex. Such problems arise, for example, in the split-variable deterministic reformulation of stochastic mixed-integer optimization problems. We adapt the augmented Lagrangian method framework to address the presence of nonconvexity in the non-relaxed constraint set and to enable efficient parallelization. The development of our approach is most naturally compared with the development of proximal bundle methods and especially with their use of serious step conditions. However, deviations from these developments allow for an improvement in efficiency with which parallelization can be utilized. Pivotal in our modification to the augmented Lagrangian method is an integration of the simplicial decomposition method and the nonlinear block Gauss–Seidel method. An adaptation of a serious step condition associated with proximal bundle methods allows for the approximation tolerance to be automatically adjusted. Under mild conditions optimal dual convergence is proven, and we report computational results on test instances from the stochastic optimization literature. We demonstrate improvement in parallel speedup over a baseline parallel approach.

A Technical lemmas for establishing optimal convergence of SDM-GS

Given initial $(x^0,z^0) \in X \times Z \subset \mathbb {R}^n \times \mathbb {R}^q$, we consider the generation of the sequence $\left\{ (x^k,z^k) \right\}$ with iterations computed using Algorithm 4, whose target problem is given by

$$\begin{aligned} \min _{x,z} \left\{ F(x,z) : x \in X, z \in Z \right\} , \end{aligned}$$

(32)

where $(x,z) \mapsto F(x,z)$ is convex and continuously differentiable over $X \times Z$, and sets X and Z are closed and convex, with X bounded and $z \mapsto F(x,z)$ is inf-compact for each $x \in X$.

We define the directional derivative with respect to x as

$$\begin{aligned} F_x'(x,z;d) := \lim _{\alpha \downarrow 0} \frac{F(x+\alpha d,z) - F(x,z)}{\alpha }. \end{aligned}$$

Of interest is the satisfaction of the following local stationarity condition at $x \in X$:

$$\begin{aligned}&F_x'({x},{z};d) \ge 0 \quad \text {for all}\; d \in X - \left\{ {x} \right\} \end{aligned}$$

(33)

for any limit point $(x,z)=(\bar{x},\bar{z})$ of some sequence $\left\{ (x^k,z^k) \right\}$ of feasible solutions to problem (32). For the sake of nontriviality, we shall assume that the x-stationarity condition (33) never holds at $(x,z)=(x^k,z^k)$ for any $k \ge 0$. Thus, for each $x^k$, $k \ge 0$, there always exists a $d^k \in X-\left\{ x^k \right\}$ for which $F_x'(x^k,z^k;d^k) < 0$.

Direction Assumptions (DAs) For each iteration $k \ge 0$, given $x^k \in X$ and $z^k \in Z$, we have $d^k$ chosen so that (1) $x^k + d^k \in X$; and (2) $F_x'(x^k,z^k;d^k) < 0$.

Gradient Related Assumption (GRA) Given a sequence $\left\{ (x^k,z^k) \right\}$ with $\lim _{k \rightarrow \infty } (x^k,z^k) = (\overline{x},\overline{z})$, and a bounded sequence $\left\{ d^k \right\}$ of directions, then the existence of a direction $\overline{d} \in X - \left\{ \overline{x} \right\}$ such that $F_{x}'(\overline{x},\overline{z};\overline{d}) < 0$ implies that

$$\begin{aligned} \limsup _{k \rightarrow \infty } F_{x}'\left( x^k,z^k;d^k\right) < 0. \end{aligned}$$

(34)

In this case, we say that $\left\{ d^k \right\}$ is gradient related to $\left\{ x^k \right\}$. This gradient related condition is similar to the one defined in [3]. The sequence of directions ${d^k}$ is typically gradient related to $\left\{ x^k \right\}$ by construction. (See Lemma 7.)

To state the last assumption, we require the notion of an Armijo rule step length $\alpha ^k \in (0,1]$ given $(x^k,z^k,d^k)$ and parameters $\beta ,\sigma \in (0,1)$.

Remark 6

Under mild assumptions on F such as continuity that guarantee the existence of finite $F_x'(x,z;d)$ for all $(x,z,d) \in \left\{ (x,z,d) : x \in X, d \in X-\left\{ x \right\} , z \in Z \right\}$, we may assume that the while loop of Lines 3–5 terminates after a finite number of iterations. Thus, we have $\alpha ^k \in (0,1]$ for each $k \ge 1$.

The last significant assumption is stated as follows.

Sufficient Decrease Assumption (SDA) For sequences $\left\{ (x^k,z^k,d^k) \right\}$ and step lengths $\left\{ \alpha ^k \right\}$ computed according to Algorithm 4, we assume for each $k \ge 0$, that $(x^{k+1},z^{k+1})$ satisfies

$$\begin{aligned} F\left( x^{k+1},z^{k+1}\right) \le F\left( x^k + \alpha ^k d^k,z^k\right) . \end{aligned}$$

Lemma 6

For problem (32), let $F : \mathbb {R}^{n_x} \times \mathbb {R}^{n_z} \mapsto \mathbb {R}$ be convex and continuously differentiable, $X \subset \mathbb {R}^{n_x}$ convex and compact, and $Z \subseteq \mathbb {R}^{n_z}$ closed and convex. Furthermore, assume for each $x \in X$ that $z \mapsto F(x,z)$ is inf-compact. If a sequence $\left\{ (x^k,z^k,d^k) \right\}$ satisfies the DA, the GRA, and the SDA for some fixed $\beta ,\sigma \in (0,1)$, then the sequence ${(x^k,z^k)}$ has limit points $(\overline{x},\overline{z})$, each of which satisfies the stationarity condition (33).

Proof

The existence of limit points $(\overline{x},\overline{z})$ follows from the compactness of X, the inf-compactness of $z \mapsto F(x,z)$ for each $x \in X$, and the SDA. In generating $\left\{ \alpha ^k \right\}$ according to the Armijo rule as implemented in Lines 2–5 of Algorithm 4, we have

$$\begin{aligned} \frac{F\left( x^k + \alpha ^k d^k,z^k\right) - F\left( x^k,z^k\right) }{\alpha ^k} \le \sigma F_x'\left( x^k,z^k ; d^k\right) . \end{aligned}$$

(35)

By the DA, $F_x'(x^k,z^k ; d^k) < 0$ and since $\alpha ^k > 0$ for each $k \ge 1$ by Remark 6, we infer from (35) that $F(x^k + \alpha ^k d^k,z^k) < F(x^k,z^k)$. By construction, we have $F(x^{k+1},z^{k+1}) \le F(x^k + \alpha ^k d^k,z^k) < F(x^k,z^k).$. By the monotonicity $F(x^{k+1},z^{k+1}) < F(x^k,z^k)$ and F being bounded from below on $X \times Z$, we have $\lim _{k \rightarrow \infty } F(x^k,z^k) = \bar{F} > -\infty$. Therefore,

$$\begin{aligned} \lim _{k \rightarrow \infty } F\left( x^{k+1},z^{k+1}\right) - F\left( x^k,z^k\right) = 0, \end{aligned}$$

which implies

$$\begin{aligned} \lim _{k \rightarrow \infty } F\left( x^k + \alpha ^k d^k,z^k\right) - F\left( x^k,z^k\right) = 0. \end{aligned}$$

(36)

We assume for sake of contradiction that $\lim _{k \rightarrow \infty } (x^k,z^k) = (\overline{x},\overline{y})$ does not satisfy the stationarity condition (33). By GRA, we have that $\left\{ d^k \right\}$ is gradient related to $\left\{ x^k \right\}$; that is,

$$\begin{aligned} \limsup _{k \rightarrow \infty } F_x'\left( x^k,z^k ; d^k\right) < 0. \end{aligned}$$

(37)

Thus, it follows from (35)–(37) that $\lim _{k \rightarrow \infty } \alpha ^k = 0$.

Consequently, after a certain iteration $k \ge \bar{k}$, we can define $\left\{ \bar{\alpha }^k \right\}$, $\bar{\alpha }^k = \alpha ^k / \beta$, where $\bar{\alpha }^k \le 1$ for $k \ge \bar{k}$, and so we have

$$\begin{aligned} \sigma F_x'\left( x^k,z^k ; d^k\right) < \frac{F\left( x^k + \bar{\alpha }^k d^k,z^k\right) - F\left( x^k,z^k\right) }{\bar{\alpha }^k}. \end{aligned}$$

(38)

Since F is continuously differentiable, the mean value theorem may be applied to the right-hand side of (38) to get

$$\begin{aligned} \sigma F_x'\left( x^k,z^k ; d^k\right) < F_x'\left( x^k +\widetilde{\alpha }^k d^k, z^k ; d^k\right) , \end{aligned}$$

(39)

for some $\widetilde{\alpha }^k \in [0,\overline{\alpha }^k]$.

Again, using the assumption $\limsup _{k \rightarrow \infty } F_x'(x^k,z^k ; d^k) < 0$, and also the compactness of $X - X$, we take a limit point $\overline{d}$ of $\left\{ d^k \right\}$, with its associated subsequence index set denoted by $\mathcal {K}$, such that $F_x'(\overline{x},\overline{z},\overline{d}) < 0$. Taking the limits over the subsequence indexed by $\mathcal {K}$, we have $\lim _{k \rightarrow \infty , k \in \mathcal {K}} F_x'(x^k,z^k ; d^k) = F_x'(\overline{x},\overline{z} ; \overline{d})$ and $\lim _{k \rightarrow \infty , k \in \mathcal {K}} F_x'(x^k +\widetilde{\alpha }^k d^k, z^k ; d^k) = F_x'(\overline{x},\overline{z} ; \overline{d})$. These two limits holds since (1) $(x,z) \mapsto F_x'(x,z;d)$ for each $d \in X - X$ is continuous and (2) $d \mapsto F_x'(x,z;d)$ is locally Lipschitz continuous for each $(x,z) \in X \times Z$ (e.g., Proposition 2.1.1 of [60]); these two facts together imply that $(x,z;d) \mapsto F_x'(x,z;d)$ is continuous. Then, inequality (39) becomes in the limit as $k \rightarrow \infty$, $k \in \mathcal {K}$,

$$\begin{aligned} \sigma F_x'(\overline{x},\overline{z} ; \overline{d}) \le F_x'(\overline{x},\overline{z} ; \overline{d}) \quad \Longrightarrow \quad 0 \le (1-\sigma ) F_x'(\overline{x},\overline{z} ; \overline{d}). \end{aligned}$$

Since $(1-\sigma ) > 0$ and $F_x'(\overline{x},\overline{z} ; \overline{d})<0$, we have a contradiction. Thus, $\overline{x}$ must satisfy the stationary condition (33). $\square$

Remark 7

Noting that $F_x'(x^k,z^k;d^k) = \nabla _x F(x^k,z^k) ^\top d^k$ under the assumption of continuous differentiability of F, one means of constructing $\left\{ d^k \right\}$ is as follows:

$$\begin{aligned} d^k \leftarrow {{\mathrm{argmin}}}_d \left\{ \nabla _x F(x^k,z^k)^\top d : d \in X-\left\{ x^k \right\} \right\} . \end{aligned}$$

(40)

Lemma 7

Given sequence $\left\{ (x^k,z^k) \right\}$ with $\lim _{k \rightarrow \infty } (x^k,z^k) = (\overline{x},\overline{z})$, let each $d^k$, $k \ge 1$, be generated as in (40). Then $\left\{ d^k \right\}$ is gradient related to $\left\{ x^k \right\}$.

Proof

By the construction of $d^k$, $k \ge 1$, we have

$$\begin{aligned} F_x'(x^k,z^k ; d^k) \le F_x'(x^k,z^k ; d) \quad \forall \; d \in X-\left\{ x^k \right\} . \end{aligned}$$

Taking the limit, we have

$$\begin{aligned} \limsup _{k \rightarrow \infty } F_x'(x^k, z^k ; d^k) \le \limsup _{k \rightarrow \infty } F_x'(x^k, z^k ; d) \le F_x'(\overline{x}, \overline{z}; d) \quad \forall \; d \in X-\left\{ \overline{x} \right\} , \end{aligned}$$

where the last inequality follows from the upper semicontinuity of the function $(x,z,d) \mapsto F_x'(x,z;d)$, which holds in our setting due, primarily, to Proposition 2.1.1 (b) of  [60] given that F is assumed to be convex and continuous on $\mathbb {R}^n$. Taking

$$\begin{aligned} \overline{d} \in {{\mathrm{argmin}}}_d \left\{ F_x'(\overline{x},\overline{z} ; d) : d \in X - \left\{ \overline{x} \right\} \right\} , \end{aligned}$$

we have by the assumed nonstationarity that $F_x'(\overline{x},\overline{z};\overline{d}) < 0$. Thus, $\limsup _{k \rightarrow \infty } F_x'(x^k, z^k ; d^k) < 0$, and so GRA holds. $\square$