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Matrix-free algorithm for the optimization of multidisciplinary systems

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Abstract

Multidisciplinary engineering systems are usually modeled by coupling software components that were developed for each discipline independently. The use of disparate solvers complicates the optimization of multidisciplinary systems and has been a long-standing motivation for optimization architectures that support modularity. The individual discipline feasible (IDF) formulation is particularly attractive in this respect. IDF achieves modularity by introducing optimization variables and constraints that effectively decouple the disciplinary solvers during each optimization iteration. Unfortunately, the number of variables and constraints can be significant, and the IDF constraint Jacobian required by most conventional optimization algorithms is prohibitively expensive to compute. Furthermore, limited-memory quasi-Newton approximations, commonly used for large-scale problems, exhibit linear convergence rates that can struggle with the large number of design variables introduced by the IDF formulation. In this work, we show that these challenges can be overcome using a reduced-space inexact-Newton-Krylov algorithm. The proposed algorithm avoids the need for the explicit constraint Jacobian and Hessian by using a Krylov iterative method to solve the Newton steps. The Krylov method requires matrix-vector products, which can be evaluated in a matrix-free manner using second-order adjoints. The Krylov method also needs to be preconditioned, and a key contribution of this work is a novel and effective preconditioner that is based on approximating a monolithic solution of the (linearized) multidisciplinary system. We demonstrate the efficacy of the algorithm by comparing it with the popular multidisciplinary feasible formulation on two test problems.

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Acknowledgements

This material is based upon work supported by the National Science Foundation under Grant No. 1332819. The authors gratefully acknowledge this support.

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Correspondence to Alp Dener.

Appendices

Appendix

Proof of Theorem 1

Consider an MDA with d disciplines. Let the equation for the i th discipline be given by

$${\boldsymbol{\mathcal{R}}}_{i}\left( \boldsymbol{w}_{i}, \{ {\boldsymbol{\mathcal{E}}}_{ij}(\boldsymbol{w}_{j}) | j\,=\,1,\ldots,d,\! j\!\neq\! i \} \right), \qquad \forall i \,=\, 1,\ldots,d. $$

where the set notation in the second argument indicates that the residual depends on the states of the other d − 1 disciplines via the nonlinear mappings \({\boldsymbol {\mathcal {E}}}_{ij}\). Next, introduce the coupling variables, \({{\bar {\boldsymbol {w}}}}_{ij}\), which are defined by

$$\begin{array}{@{}rcl@{}} {\boldsymbol{\mathcal{C}}}_{ij}(\boldsymbol{w}_{j},{{\bar{\boldsymbol{w}}}}_{ij}) &\equiv& {\boldsymbol{\mathcal{E}}}_{ij}(\boldsymbol{w}_{j}) - {{\bar{\boldsymbol{w}}}}_{ij} = 0,\\ \forall i,j &=& 1,\dots,d, \; j\neq i. \end{array} $$

Using the above conditions, the residual of the i th discipline can be expressed as a function of the coupling variables:

$${\boldsymbol{\mathcal{R}}}_{i}\left( \boldsymbol{w}_{i}, \{{{\bar{\boldsymbol{w}}}}_{ij} | j=1,\ldots,d, j\neq i \}\right), \qquad \forall i = 1,\ldots,d. $$

Let \(\boldsymbol {w}^{T} = [\boldsymbol {w}_{1}^{T},\ldots ,\boldsymbol {w}_{d}^{T}]\) and \({{\bar {\boldsymbol {w}}}}^{T} = [{{\bar {\boldsymbol {w}}}}_{12}^{T},\ldots ,{{\bar {\boldsymbol {w}}}}_{d,d-1}^{T}]\) denote the compound vectors containing all the states and coupling variables, respectively, and consider the compound residual given by

$$\left( \begin{array}{c} {\boldsymbol{\mathcal{R}}}(\boldsymbol{w},{{\bar{\boldsymbol{w}}}}) \\ {\boldsymbol{\mathcal{C}}}(\boldsymbol{w},{{\bar{\boldsymbol{w}}}}) \end{array}\right) =\left( \begin{array}{c} {\boldsymbol{\mathcal{R}}}_{1}(\boldsymbol{w}_{1}, {{\bar{\boldsymbol{w}}}}_{12}, \ldots, {{\bar{\boldsymbol{w}}}}_{1d}) \\ {\vdots} \\ {\boldsymbol{\mathcal{R}}}_{d}(\boldsymbol{w}_{d}, {{\bar{\boldsymbol{w}}}}_{d1}, \ldots, {{\bar{\boldsymbol{w}}}}_{d,d-1}) \\ {\boldsymbol{\mathcal{C}}}_{12}(\boldsymbol{w}_{2},{{\bar{\boldsymbol{w}}}}_{12}) \\ {\vdots} \\ {\boldsymbol{\mathcal{C}}}_{d,d-1}(\boldsymbol{w}_{d-1},{{\bar{\boldsymbol{w}}}}_{d,d-1}) \end{array}\right). $$

The Jacobian of this compound residual is the matrix

$${\textsf{D}} \equiv \left[\begin{array}{cc} \frac{\partial {\boldsymbol{\mathcal{R}}}}{\partial \boldsymbol{w}} & \frac{\partial {\boldsymbol{\mathcal{R}}}}{\partial {{\bar{\boldsymbol{w}}}}} \\ \frac{\partial {\boldsymbol{\mathcal{C}}}}{\partial \boldsymbol{w}} & -{\textsf{I}} \end{array}\right]. $$

The (2,2) block of D is the negative identity, which is clearly invertible, and the (1,1) block of D is the block diagonal matrix d i a g(A 1,…, A d ), which is also invertible by assumption that the A i are invertible. Therefore, we can form the Schur complement of D with respect to either block.

The Schur complement with respect to the (2,2) block of D is the matrix

$${\textsf{S}}_{(2,2)} \equiv \frac{\partial {\boldsymbol{\mathcal{R}}}}{\partial \boldsymbol{w}} + \frac{\partial {\boldsymbol{\mathcal{R}}}}{\partial {{\bar{\boldsymbol{w}}}}} \frac{\partial {\boldsymbol{\mathcal{C}}}}{\partial \boldsymbol{w}}. $$

Now, the (i, j) block of second term above is, for ij,

$$\left( \frac{\partial {\boldsymbol{\mathcal{R}}}}{\partial {{\bar{\boldsymbol{w}}}}} \frac{\partial {\boldsymbol{\mathcal{C}}}}{\partial \boldsymbol{w}} \right)_{i,j} = \frac{\partial {\boldsymbol{\mathcal{R}}}_{i}}{\partial {{\bar{\boldsymbol{w}}}}_{ij}} \frac{\partial {\boldsymbol{\mathcal{E}}}_{ij}}{\partial \boldsymbol{w}_{j}} = \frac{\partial {\boldsymbol{\mathcal{R}}}_{i}}{\partial \boldsymbol{w}_{j}}, $$

where the last equality follows from the chain rule. There is no contribution to the (i, i) block from the second term, since there is no coupling variable between a discipline and itself. Conversely, recall that the first term in the Schur complement is \({\mathsf {A}}_{i} = \partial {\boldsymbol {\mathcal {R}}}_{i}/\partial \boldsymbol {w}_{i}\), which has no off-diagonal blocks. Summarizing, the Schur complement with respect to the (2,2) block of D is the monolithic MDA Jacobian

$${\textsf{S}}_{(2,2)} = \left( \frac{\partial {\boldsymbol{\mathcal{R}}}}{\partial \boldsymbol{w}} + \frac{\partial {\boldsymbol{\mathcal{R}}}}{\partial {{\bar{\boldsymbol{w}}}}} \frac{\partial {\boldsymbol{\mathcal{C}}}}{\partial \boldsymbol{w}} \right)_{i,j} = \frac{\partial {\boldsymbol{\mathcal{R}}}_{i}}{\partial \boldsymbol{w}_{j}}. $$

Next, we consider the Schur complement of D with respect to the (1,1) block:

$${\textsf{S}}_{(1,1)} \equiv -{\textsf{I}} - \frac{\partial {\boldsymbol{\mathcal{C}}}}{\partial \boldsymbol{w}} \left[\frac{\partial {\boldsymbol{\mathcal{R}}}}{\partial \boldsymbol{w}}\right]^{-1} \frac{\partial {\boldsymbol{\mathcal{R}}}}{\partial {{\bar{\boldsymbol{w}}}}}. $$

Now, the direct sensitivities of w i with respect to \({{\bar {\boldsymbol {w}}}}_{ij}\) can be found by taking the total derivative of \({\boldsymbol {\mathcal {R}}}_{i} = \mathbf {0}\):

$$\begin{array}{@{}rcl@{}} &&\frac{d {\boldsymbol{\mathcal{R}}}_{i}}{d {{\bar{\boldsymbol{w}}}}_{ij}} = \frac{\partial {\boldsymbol{\mathcal{R}}}_{i}}{\partial {{\bar{\boldsymbol{w}}}}_{ij}} + \frac{\partial {\boldsymbol{\mathcal{R}}}_{i}}{\partial \boldsymbol{w}_{i}} \frac{d \boldsymbol{w}_{i}}{d {{\bar{\boldsymbol{w}}}}_{ij}} = 0 \\ &&\Rightarrow\qquad \frac{d \boldsymbol{w}_{i}}{d {{\bar{\boldsymbol{w}}}}_{ij}} = - \left[\frac{\partial {\boldsymbol{\mathcal{R}}}_{i}}{\partial \boldsymbol{w}_{i}}\right]^{-1} \frac{\partial {\boldsymbol{\mathcal{R}}}_{i}}{\partial {{\bar{\boldsymbol{w}}}}_{ij}}. \end{array} $$

Consequently, substituting this expression, the Schur complement with respect to the (1,1) block of D simplifies to

$${\textsf{S}}_{(1,1)} = -{\textsf{I}} + \frac{\partial {\boldsymbol{\mathcal{C}}}}{\partial \boldsymbol{w}} \frac{d \boldsymbol{w}}{d {{\bar{\boldsymbol{w}}}}} = \frac{d {\boldsymbol{\mathcal{C}}}}{d {{\bar{\boldsymbol{w}}}}}, $$

which is the total derivative of the coupling constraints with respect to the coupling variables, i.e. the IDF Jacobian.

The desired result now follows. If the MDA Jacobian is invertible, i.e. the Schur complement S (2,2) is invertible, then D is invertible. If D is invertible, then the IDF Jacobian must be invertible, since it is the Schur complement S (1,1). The reverse implication is analogous. This completes the proof.

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Dener, A., Hicken, J.E. Matrix-free algorithm for the optimization of multidisciplinary systems. Struct Multidisc Optim 56, 1429–1446 (2017). https://doi.org/10.1007/s00158-017-1734-0

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