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Complexity of Proximal Augmented Lagrangian for Nonconvex Optimization with Nonlinear Equality Constraints

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We analyze worst-case complexity of a Proximal augmented Lagrangian (Proximal AL) framework for nonconvex optimization with nonlinear equality constraints. When an approximate first-order (second-order) optimal point is obtained in the subproblem, an \(\epsilon \) first-order (second-order) optimal point for the original problem can be guaranteed within \({\mathcal {O}}(1/ \epsilon ^{2 - \eta })\) outer iterations (where \(\eta \) is a user-defined parameter with \(\eta \in [0,2]\) for the first-order result and \(\eta \in [1,2]\) for the second-order result) when the proximal term coefficient \(\beta \) and penalty parameter \(\rho \) satisfy \(\beta = {\mathcal {O}}(\epsilon ^\eta )\) and \(\rho = \varOmega (1/\epsilon ^\eta )\), respectively. We also investigate the total iteration complexity and operation complexity when a Newton-conjugate-gradient algorithm is used to solve the subproblems. Finally, we discuss an adaptive scheme for determining a value of the parameter \(\rho \) that satisfies the requirements of the analysis.

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  1. Circumstances under which the penalty parameter sequence of ALGENCAN is bounded are discussed in [1, Section 5].


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Research supported by Award N660011824020 from the DARPA Lagrange Program, NSF Awards 1628384, 1634597, and 1740707; and Subcontract 8F-30039 from Argonne National Laboratory.

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Correspondence to Yue Xie.

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Appendix: Proofs of Elementary Results

Appendix: Proofs of Elementary Results

Proof of Theorem 1

Since \(x^*\) is a local minimizer of (1), it is the unique global solution of

$$\begin{aligned} \min \, f(x) + \frac{1}{4} \Vert x - x^* \Vert ^4 \quad \hbox {subject to}\;\; c(x) = 0, \;\; \Vert x - x^* \Vert \le \delta , \end{aligned}$$

for \(\delta > 0\) sufficiently small. For the same \(\delta \), we define \(x_k\) to be the global solution of

$$\begin{aligned} \min \, \quad f(x) + \frac{\rho _k}{2} \Vert c(x) \Vert ^2 + \frac{1}{4} \Vert x - x^* \Vert ^4 \quad \hbox {subject to}\;\; \Vert x - x^* \Vert \le \delta , \end{aligned}$$

for a given \(\rho _k\), where \(\{ \rho _k \}_{k \ge 1}\) is a positive sequence such that \(\rho _k \rightarrow +\infty \). Note that \(x_k\) is well defined because the feasible region is compact and the objective is continuous. Suppose that z is any accumulation point of \(\{ x_k \}_{k \ge 1}\), that is, \(x_k \rightarrow z\) for \(k \in {\mathcal {K}}\), for some subsequence \({\mathcal {K}}\). Such a z exists because \(\{ x_k \}_{k \ge 1}\) lies in a compact set, and moreover, \(\Vert z - x^* \Vert \le \delta \). We want to show that \(z = x^*\). By the definition of \(x_k\), we have for any \(k \ge 1\) that

$$\begin{aligned} f(x^*)&= f(x^*) + \frac{\rho _k}{2}\Vert c(x^*) \Vert ^2 + \frac{1}{4} \Vert x^* - x^* \Vert ^4 \nonumber \\&\ge f(x_k) + \frac{\rho _k}{2}\Vert c(x_k) \Vert ^2 + \frac{1}{4} \Vert x_k - x^* \Vert ^4 \ge f(x_k) + \frac{1}{4} \Vert x_k - x^* \Vert ^4. \end{aligned}$$

By taking the limit over \({\mathcal {K}}\), we have \(f(x^*) \ge f(z) + \frac{1}{4} \Vert z - x^* \Vert ^4\). From (70), we have

$$\begin{aligned} \frac{\rho _k}{2}\Vert c(x_k) \Vert ^2 \le f(x^*) - f(x_k) \le f(x^*) - \inf _{k \ge 1} f(x_k) < +\infty . \end{aligned}$$

By taking limits over \({\mathcal {K}}\), we have that \(c(z) = 0\). Therefore, z is the global solution of (68), so that \(z = x^*\).

Without loss of generality, suppose that \(x_k \rightarrow x^*\) and \(\Vert x_k - x^* \Vert < \delta \). By first and second-order optimality conditions for (69), we have

$$\begin{aligned}&\nabla f(x_k) + \rho _k \nabla c(x_k) c(x_k) + \Vert x_k - x^* \Vert ^2 (x_k - x^*) = 0, \nonumber \\&\quad \nabla ^2 f(x_k) + \rho _k \sum _{i=1}^m c_i(x_k) \nabla ^2 c_i(x_k) + \rho _k \nabla c(x_k) [ \nabla c(x_k) ]^T \end{aligned}$$
$$\begin{aligned}&+ 2(x_k-x^*)(x_k-x^*)^T + \Vert x_k-x^* \Vert ^2 I \succeq 0. \end{aligned}$$

Define \(\lambda _k \triangleq \rho _k c(x_k)\) and \(\epsilon _k \triangleq \max \{ \Vert x_k - x^* \Vert ^3, 3 \Vert x_k - x^* \Vert ^2, \sqrt{ 2( f(x^*) - \inf _{k \ge 1} f(x_k) )/\rho _k} \}\). Then by (71), (72), (73) and Definition 2, \(x_k\) is \(\epsilon _k\)-2o. Note that \(x_k \rightarrow x^*\) and \(\rho _k \rightarrow +\infty \), so \(\epsilon _k \rightarrow 0^+\). \(\square \)

Proof of Lemma 1

We prove by contradiction. Otherwise for any \(\alpha \) we could select sequence \(\{ x_k \}_{k \ge 1} \subseteq S_{\alpha }^0\) such that \(f(x_k) + \frac{\rho _0}{2} \Vert c(x_k) \Vert ^2 < - k\). Let \(x^*\) be an accumulation point of \(\{ x_k \}_{k \ge 1}\) (which exists by compactness of \(S_\alpha ^0\)). Then there exists index K such that \(f(x^*) + \frac{\rho _0}{2} \Vert c(x^*) \Vert ^2 \ge -K + 1 > f(x_k) + \frac{\rho _0}{2} \Vert c(x_k) \Vert ^2 + 1\) for all \(k \ge K\), which contradicts the continuity of \(f(x) + \frac{\rho _0}{2} \Vert c(x) \Vert ^2\). \(\square \)

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Xie, Y., Wright, S.J. Complexity of Proximal Augmented Lagrangian for Nonconvex Optimization with Nonlinear Equality Constraints. J Sci Comput 86, 38 (2021).

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