A primal-dual interior point method for large-scale free material optimization

Abstract

Free Material Optimization (FMO) is a branch of structural optimization in which the design variable is the elastic material tensor that is allowed to vary over the design domain. The requirements are that the material tensor is symmetric positive semidefinite with bounded trace. The resulting optimization problem is a nonlinear semidefinite program with many small matrix inequalities for which a special-purpose optimization method should be developed. The objective of this article is to propose an efficient primal-dual interior point method for FMO that can robustly and accurately solve large-scale problems. Several equivalent formulations of FMO problems are discussed and recommendations on the best choice based on the results from our numerical experiments are presented. Furthermore, the choice of search direction is also investigated numerically and a recommendation is given. The number of iterations the interior point method requires is modest and increases only marginally with problem size. The computed optimal solutions obtain a higher precision than other available special-purpose methods for FMO. The efficiency and robustness of the method is demonstrated by numerical experiments on a set of large-scale FMO problems.

Appendices

Appendix 1

In this Appendix we derive the dual formulation (8) of the minimum weighted compliance problem (3). Similar result for minimax problems can be found in [7]. Analogous all-quadratic formulations of minimum compliance truss topology optimization problems are described, for example, in [1]. The linear elasticity static structural analysis problem can be written as

$$\begin{aligned} \begin{aligned} \underset{u_\ell }{\sup }&\left\{ 2f_\ell ^Tu_\ell -u_\ell ^TA(E)u_\ell \right\} \end{aligned} \end{aligned}$$

which is a quadratic problem with negative definite Hessian and hence a concave maximization problem. The optimality condition is $A(E)u_\ell =f_\ell$ and the optimal value is $f_\ell ^Tu_\ell$ if $f_\ell \in \mathcal {R}(A(E))$ and $-\infty$ otherwise. Due to the stated assumptions we replace the $\sup$ with $\max$ in the following. Therefore, the minimum compliance problem (3) is equivalent to

(42)

The Lagrangian $\mathcal {L}$ associated with (42) is

$$\begin{aligned} \begin{aligned} \mathcal {L}(E,u_\ell ,\alpha ,\bar{\beta },\underline{\beta })&= \sum \limits _{\ell \in L}w_\ell \underset{u_\ell }{\max } \left\{ 2f_\ell ^Tu_\ell -u_\ell ^TA(E)u_\ell \right\} +\sum \limits _{i=1}^m \underline{\beta }_i(-Tr(E_i)+\underline{\rho })\\&\quad +\sum \limits _{i=1}^m \bar{\beta }_i(Tr(E_i)-\bar{\rho })+\alpha \left( \sum \limits _{i=1}^mTr(E_i)-\bar{V}\right) \\&= \underset{u_1,\ldots ,u_{n_L}}{\max }\sum \limits _{\ell \in L}w_\ell (2f_\ell ^Tu_\ell -u_\ell ^TA(E)u_\ell )+\sum \limits _{i=1}^m \underline{\beta }_i(-Tr(E_i)+\underline{\rho })\\&\quad +\sum \limits _{i=1}^m \bar{\beta }_i(Tr(E_i)-\bar{\rho })+\alpha \left( \sum \limits _{i=1}^mTr(E_i)-\bar{V}\right) \\&= \underset{u_1,\ldots ,u_{n_L}}{\max }\left( \sum \limits _{\ell \in L}w_\ell (2f_\ell ^Tu_\ell \!-\!u_\ell ^TA(E)u_\ell )\!+\!\sum \limits _{i=1}^m \underline{\beta }_i(-Tr(E_i)+\underline{\rho })\right. \\&\quad \left. +\sum \limits _{i=1}^m \bar{\beta }_i(Tr(E_i)-\bar{\rho })+\alpha \left( \sum \limits _{i=1}^mTr(E_i)-\bar{V}\right) \right) \\&=\underset{u_1,\ldots ,u_{n_L}}{\max }\Big (\sum \limits _{\ell \in L}2w_\ell f_\ell ^Tu_\ell -\alpha \bar{V}+\underline{\rho }\sum \limits _{i=1}^m \underline{\beta }_i-\bar{\rho }\sum \limits _{i=1}^m\bar{\beta }_i\\&\quad +\sum _{i=1}^m \Big \langle E_i,(\alpha -\underline{\beta }_i+ \bar{\beta }_i)I-\sum _{\ell \in L}\sum _{k=1}^{n_G}w_\ell B_{i,k}^Tu_\ell u_\ell ^T B_{i,k}\Big \rangle \Big ). \end{aligned} \end{aligned}$$

The corresponding dual function is

$$\begin{aligned}&g(u_1,\ldots ,u_{n_L},\bar{\beta },\underline{\beta },\alpha )\\&\quad =\underset{ E_1,\ldots ,E_m\succeq 0}{\text {minimize}}\;\underset{u_1,\ldots ,u_{n_L}}{\max }\Big (\sum _{\ell \in L}2w_\ell f_\ell ^Tu_\ell -\alpha \bar{V}+\underline{\rho }\sum _{i=1}^m \underline{\beta }_i-\bar{\rho }\sum _{i=1}^m\bar{\beta }_i\\&\quad \quad +\sum _{i=1}^m \Big \langle E_i,(\alpha -\underline{\beta }_i+ \bar{\beta }_i)I-\sum _{\ell \in L}\sum _{k=1}^{n_G}w_\ell B_{i,k}^Tu_\ell u_\ell ^T B_{i,k}\Big \rangle \Big )\\&\quad = \left\{ \begin{array}{l l} \underset{u_1,\ldots ,u_{n_L}}{\max }(\sum _{\ell \in L}2w_\ell f_\ell ^Tu_\ell -\alpha \bar{V}+\underline{\rho }\sum _{i=1}^m \underline{\beta }_i-\bar{\rho }\sum _{i=1}^m\bar{\beta }_i) &{} \quad \text {if}\;(43)\;\text {holds}\\ -\infty &{} \quad \text {otherwise}.\\ \end{array} \right. \end{aligned}$$

Below is the condition that the dual function $g$ attains its minimum value.

$$\begin{aligned} \begin{aligned} \sum _{\ell \in L}\sum _{k=1}^{n_G}w_\ell B_{i,k}^Tu_\ell u_\ell ^T B_{i,k}\preceq (\alpha -\underline{\beta }_i+ \bar{\beta }_i)I,\quad i=1,\ldots ,m. \end{aligned} \end{aligned}$$

(43)

The dual formulation of the minimum compliance problem (3) becomes

$$\begin{aligned} \begin{array}{lll} &{} \underset{u_1,\ldots ,u_{n_L},\alpha \ge 0,\bar{\beta }\ge 0,\underline{\beta }\ge 0}{\text {sup}} &{} -\alpha \bar{V}+2\sum _{\ell \in L}w_\ell f_\ell ^Tu_\ell +\underline{\rho }\sum _{i=1}^m \underline{\beta }_i-\bar{\rho }\sum _{i=1}^m\bar{\beta }_i\\ &{} \text {subject to} &{} \sum _{\ell \in L}\sum _{k=1}^{n_G}w_\ell B_{i,k}^Tu_\ell u_\ell ^T B_{i,k}-(\alpha \!-\!\underline{\beta }_i\!+\! \bar{\beta }_i)I\preceq 0,\\ &{}&{} i\!=\!1,\ldots ,m. \end{array} \end{aligned}$$

Appendix 2

The following products are in tensor notation.

• 1. $(\nabla _{X_rX_s}^2\mathcal {L}_\mu (X,u,s,\lambda )\varDelta X_s)_{ij}=(\nabla _{X_rX_s}^2\mathcal {L}_\mu (X,u,s,\lambda ))_{ijpq}(\varDelta X_s)_{pq}$, for $r,s=1,\ldots ,m$, for $i,j=1,\ldots ,d_r$, and for $p,q=1,\ldots ,d_s$.

• 2. $(\nabla _{X_ru}^2\mathcal {L}_\mu (X,u,s,\lambda )\varDelta X_r)_{i}\!=\!(\nabla _{uX_r}^2\mathcal {L}_\mu (X,u,s,\lambda ))_{ipq}(\varDelta X_r)_{pq}$, for $r\!=\!1,\ldots ,m$, for $p,q=1,\ldots ,d_r$, and for $i=1,\ldots ,n$.

• 3. $(\nabla _{X_ru}^2\mathcal {L}_\mu (X,u,s,\lambda )^T\varDelta u)_{ij}=(\nabla _{X_ru}^2\mathcal {L}_\mu (X,u,s,\lambda ))_{ijp}(\varDelta u)_{p}$, for $r=1,\ldots ,m$, for $i,j=1,\ldots ,d_r$, and for $p=1,\ldots ,n$.

• 4. $(\nabla _{X_r}g(X,u)^T\varDelta \lambda )_{ij}=(\nabla _{X_r}(g(X,u)^T\varDelta \lambda ))_{ij}$, for $r=1,\ldots ,m$, and for $i,j=1,\ldots ,d_r$.

• 5. $(\nabla _{X_r}g(X,u)\varDelta X_r)_i=(\nabla _{X_r}g_i(X,u))_{pq}(\varDelta X_r)_{pq}$, for $r=1,\ldots ,m$, for $p,q=1,\ldots ,d_r$, and for $i=1,\ldots ,k$.