A differentiable path-following algorithm for computing perfect stationary points

Abstract

This paper is concerned with the computation of perfect stationary point, which is a strict refinement of stationary point. A differentiable homotopy method is developed for finding perfect stationary points of continuous functions on convex polytopes. We constitute an artificial problem by introducing a continuously differentiable function of an extra variable. With the optimality conditions of this problem and a fixed point argument, a differentiable homotopy mapping is constructed. As the extra variable becomes close to zero, the homotopy path naturally provides a sequence of closely approximate stationary points on perturbed polytopes, and converges to a perfect stationary point on the original polytope. Numerical experiments are implemented to further illustrate the effectiveness of our method.

Appendix

Theorem 3

(Mas-Colell’s fixed point Theorem, [19]) Let$\Phi$be a non-empty, compact and convex set and$H: \Phi \times [0,1]\rightarrow \Phi$an upper semi-continuous mapping. Then the set$H^{-1}=\{(z,t)\in \Phi \times [0,1]\, | \,z\in H(z,t)\}$contains a connected set$H^c$such that$(\Phi \times \{1\}) \cap {H^c} \ne \emptyset$and$(\Phi \times \{0\}) \cap {H^c} \ne \emptyset$.

Theorem 4

(Transversality Theorem, [20]) Let$L:S\times {\mathbb {R}}^l\rightarrow {\mathbb {R}}^s$be$C^r$, where$S\subset {\mathbb {R}}^n$is an open set and$r\ge 1+\max \{0,n-s\}$. If zero is a regular value ofL, then zero is a regular value of$L(\cdot ,{\hat{w}}):S\rightarrow {\mathbb {R}}^s$for almost all fixed${\hat{w}}\in {\mathbb {R}}^l$.

The following proves zero is a regular value of $H(x,\gamma ,1)$. This property is used in Theorem 1.

Proof

Notice that f is a (at least) twice differentiable mapping from the polytope ${{{\mathcal {P}}}}=\{x\in {\mathbb {R}}^n |Ax \le b\}$ to ${\mathbb {R}}^n$, $a_i^{\top }=[a_{i1},\ldots ,a_{in}]\in {\mathbb {R}}^n$ being the ith row of matrix A for $i\in M$, and $H(x,\gamma ,t)$ is the left side of (9). As $t=1$, we have $H(x,\gamma ,1)=[l(x,\gamma ,1)^\top ,q(x,\gamma ,1)^\top ]^\top$, where $l(x,\gamma ,1)=-\sum \nolimits _{i=1}^{m} z_i(\gamma ,1)a_i$, and $q(x,\gamma ,1)=[q_i(x,\gamma ,1)]_{i\in M}$ with $q_i(x,\gamma ,1)=a_i^{\top }x+s_i(\gamma ,1)+\eta _0-b_i$. The Jacobian matrix is given by

$$\begin{aligned} JH(x,\gamma ,1)=\left[ \begin{array}{lc} 0&{}\quad -A^\top B \\ A&{}\quad C\\ \end{array}\right] \in {\mathbb {R}}^{(n+m)\times (n+m)}, \end{aligned}$$

where B and C are $m\times m$ diagonal matrices with the ith entry being $B_{ii}=\frac{\partial z_i(\gamma ,1)}{\partial \gamma _i}$ and $C_{ii}=\frac{\partial s_i(\gamma ,1)}{\partial \gamma _i}$, respectively. One can easily verify that B and C do not have full rank if $\gamma _i=0$ for some $i\in M$. As a result, the Jacobian matrix may not have full rank. However, as we will show, this is not a generic situation.

For $(x,\gamma ,u,\eta ,1)\in {{{\mathcal {P}}}}\times {\mathbb {R}}^m\times {\mathbb {R}}^{m}\times {\mathbb {R}}^m$, replace $H(x,\gamma ,1)$ with $H(x,\gamma ,u,\eta ,1)$. The Jacobian matrix of $H(x,\gamma ,u,\eta ,1)$ is given by

$$\begin{aligned} JH(x,\gamma ,u,\eta ,1)=\left[ \begin{array}{lccc} 0&{}-A^{\top } B&{}-A^{\top }Z&{}0 \\ A&{}C&{}0&{}I_m\\ \end{array}\right] , \end{aligned}$$

where B and C are defined as above, and $Z=\text {diag}({z_1(\gamma ,1),\ldots ,z_m(\gamma ,1)})\in {\mathbb {R}}^{m\times m}$. Since $\theta (1)=1$, we have $z(\gamma ,1)>0$, and $A^{\top }Z$ has full row rank with $\text {rank}(A^{\top }Z)=n$. Therefore, $JH(x,\gamma ,u,\eta ,1)$ is of full row rank and zero is a regular value of $H(x,\gamma ,u,\eta ,1)$. By a direct application of transversality theorem, for generic $u\in {\mathbb {R}}^{m}$ and $\eta \in {\mathbb {R}}^{m}$, zero is a regular value of $H(x,\gamma ,1)$. $\square$