An implicit filtering algorithm for derivative-free multiobjective optimization with box constraints

Abstract

This paper is concerned with the definition of new derivative-free methods for box constrained multiobjective optimization. The method that we propose is a non-trivial extension of the well-known implicit filtering algorithm to the multiobjective case. Global convergence results are stated under smooth assumptions on the objective functions. We also show how the proposed method can be used as a tool to enhance the performance of the Direct MultiSearch (DMS) algorithm. Numerical results on a set of test problems show the efficiency of the implicit filtering algorithm when used to find a single Pareto solution of the problem. Furthermore, we also show through numerical experience that the proposed algorithm improves the performance of DMS alone when used to reconstruct the entire Pareto front.

Appendix: Technical results

In the appendix we prove two technical results that are used for the convergence analysis.

Proposition 2

Let $f:\mathbb {R}^n \rightarrow \mathbb {R}$ be continuously differentiable and let $x\in \mathcal {F}$. Let $\{z_k\}\subset \mathcal {F}$ and $\{h_k\}\subset \mathbb {R}^+$ be sequences such that

$$\begin{aligned} \lim _{k\rightarrow \infty }z_k=x\quad \quad \lim _{k\rightarrow \infty }h_k=0. \end{aligned}$$

(28)

Assume that, for $i=1,\ldots ,n$, at least one of the following condition holds

$$\begin{aligned}&z_k+h_ke_i\in \mathcal {F},\\&z_k-h_ke_i\in \mathcal {F}. \end{aligned}$$

Then we have

$$\begin{aligned} \lim _{k\rightarrow \infty }\nabla _{h_k}f(z_k)=\nabla f(x). \end{aligned}$$

Proof

Let $i\in \{1,\ldots ,n\}$ and define the following subsets

$$\begin{aligned} K_1= & {} \left\{ k:z_k+h_ke_i\in \mathcal {F}, \ z_k-h_ke_i\notin \mathcal {F}\right\} ,\\ K_2= & {} \left\{ k: z_k\pm h_ke_i\in \mathcal {F}\right\} ,\\ K_3= & {} \left\{ k:z_k-h_ke_i\in \mathcal {F}, \ z_k+h_ke_i\notin \mathcal {F}\right\} . \end{aligned}$$

By definition of approximated gradient we have

$$\begin{aligned} \frac{\partial _h f(z_k)}{\partial x_i}= \left\{ \begin{array}{ll} \frac{f(z_k+h_ke_i)-f(z_k)}{h_k}&{} k\in K_1\\ \frac{f(z_k+h_ke_i)-f(z_k-h_ke_i)}{2h_k}&{}k\in K_2\\ \frac{f(z_k)-f(z_k-h_ke_i)}{h_k}&{}k\in K_3 \end{array} \right. \end{aligned}$$

Suppose that $K_1$ is an infinite subset. For all $k\in K_1$, by the Mean Value Theorem, we can write

$$\begin{aligned} \frac{\partial _h f(z_k)}{\partial x_i}=\frac{\partial f(\xi _k)}{\partial x_i}, \end{aligned}$$

where $\xi _k=z_k+\theta _k h_ke_i$, with $\theta _k\in (0,1)$. Taking the limits for $k\in K_1$ and $k\rightarrow \infty$, recalling (28) and the continuity of the gradient, we obtain

$$\begin{aligned} \lim _{k\in K_1,k\rightarrow \infty } \frac{\partial _h f(z_k)}{\partial x_i}=\frac{\partial f(x)}{\partial x_i}. \end{aligned}$$

By repeating the same reasonings using the sets $K_2$ and $K_3$, we have

$$\begin{aligned} \lim _{k\rightarrow \infty } \frac{\partial _h f(z_k)}{\partial x_i}=\frac{\partial f (x)}{\partial x_i}, \end{aligned}$$

and the thesis is proved. $\square$

Proposition 3

Consider Problem (1), let $F:\mathbb {R}^n \rightarrow \mathbb {R}^m$ be continuously differentiable, $x\in \mathcal {F}$, and let $\theta :\mathcal {F}\times R^+ \rightarrow \mathbb {R}$ be defined as in (9). Then:

• (i) $\theta (x,h)\le 0$ for all $x \in \mathcal {F}$ and $h>0$;

• (ii) let $\{z_k\}\subset \mathcal {F}$ and $\{h_k\}\subset \mathbb {R}^+$ be sequences satisfying the assumptions of Proposition 2; we have

  $$\begin{aligned} \lim _{k\rightarrow \infty }\theta (z_k,h_k)=\theta (x). \end{aligned}$$

Proof

(i) Given $x,y\in \mathcal {F}$ and $h>0$, we consider the function g defined as follows:

$$\begin{aligned} g(y,h,x) = \max _{i=1,\ldots ,m}\nabla _h f_{i}(x)^\top (y-x), \end{aligned}$$

and note that

$$\begin{aligned} \theta (x,h) = \min _{y \in \mathcal {F}} g(y,h,x). \end{aligned}$$

Then $\theta (x,h)\le 0$ follows easily from $g(x,h,x) =0$.

(ii) We preliminary observe that

$$\begin{aligned} | \max _{i} a_i - \max _{ i} b_i |\le \Vert a-b\Vert , \quad \mathrm{for ~any~} \quad a,b\in \mathbb {R}^m. \end{aligned}$$

Let us define

$$\begin{aligned}&y(x)\in \arg \min _{y\in \mathcal {F}} ~ \max _{i=1,\ldots ,m} \nabla f_i(x)^\top (y-x),\\&y_k\in \arg \min _{y\in \mathcal {F}} ~ \max _{i=1,\ldots ,m} \nabla _{h_k} f_i(z_k)^\top (y-z_k), \end{aligned}$$

so that

$$\begin{aligned}&\max _{i=1,\ldots ,m} \nabla f_i(x)^\top (y(x)-x)\le \max _{i=1,\ldots ,m} \nabla f_i(x)^\top (y_k-x)\\&\max _{i=1,\ldots ,m} \nabla _{h_k} f_i(z_k)^\top (y_k-z_k)\le \max _{i=1,\ldots ,m} \nabla _{h_k} f_i(z_k)^\top (y(x)-z_k). \end{aligned}$$

Denote by $J_{h_k}(z_k)$ the approximated Jacobian $J_{h_k}(z_k)=[ \nabla _{h_k}f_1(z_k),\ldots , \nabla _{h_k} f_m(z_k) ]^\top$. We can write

$$\begin{aligned} \theta (z_k,h_k)-\theta (x)= & {} \max _{i} \nabla _{h_k} f_i(z_k)^\top (y_k-z_k)- \max _{i} \nabla f_i(x)^\top (y(x)-x)\\\le & {} \max _{i} \nabla _{h_k} f_i(z_k)^\top (y(x)-z_k)- \max _{i} \nabla f_i(x)^\top (y(x)-x)\\\le & {} \Vert J_{h_k}(z_k)^\top (y(x)-z_k) - J(x)^\top (y(x)-x)\Vert \\\le & {} \Vert (J_{h_k}(z_k) - J(x))^\top y(x)\Vert \\&+ \Vert J(x)^\top x - J_{h_k}(z_k)^\top z_k + J(x)^\top z_k - J(x)^\top z_k\Vert \\\le & {} \Vert (J_{h_k}(z_k) - J(x))^\top y(x)\Vert + \Vert J(x)^\top (z_k-x)\Vert \\&+ \, \Vert (J_{h_k}(z_k)-J(x))^\top z_k \Vert . \end{aligned}$$

A quite similar bound, with $y_k$ in place of y(x), can be obtained for $\theta (x)-\theta (z_k,h_k)$. Then, as $z_k$ and $y_k$ belong to the compact set $\mathcal {F}$, by Proposition 2, $|\theta (z_k,h_k)-\theta (x)|\rightarrow 0$ for $k\rightarrow \infty$. $\square$