An efficient global optimization algorithm for a class of linear multiplicative problems based on convex relaxation

Abstract

This paper presents an efficient global optimization algorithm to solve a class of linear multiplicative problems (LMP). The algorithm first converts LMP into an equivalent problem (EP) via some variables transformation, and a convex relaxation problem is constructed to derive a lower bound to the optimal value of EP. Consequently, the process of solving LMP is transformed into tackling a series of convex programs. Additionally, a pruning rule is developed to offer a chance to remove the portion of the investigated space which does not contain the optimal solution of EP, and we propose a strategy which provides more choices of the feasible solution to update the upper bound for the optimal value of LMP. We also analyze the convergence of the algorithm and give its complexity result. Finally, our approach has been confirmed to be effective based on numerical results.

Appendix A. The derivation of the relaxations in Cambini et al. (2023)

We first report the two linear relaxations offered by Cambini et al. (2023) for Problem 2. To this end, the following 4p linear programs need to be solved:

$$\begin{aligned} {\underline{c}}_{i}=\min _{x\in D}c_{i}^{\top }x,\ {\overline{c}}_{i}=\max _{x\in D}c_{i}^{\top }x,\ {\underline{d}}_{i}=\min _{x\in D}d_{i}^{\top }x,\ {\overline{d}}_{i}=\max _{x\in D}d_{i}^{\top }x,\ i=1,\ldots ,p. \end{aligned}$$

Then the two linear relaxations of Problem 2 can be expressed as follows:

$$\begin{aligned}{{(\text {LRP1})}:}\left\{ \begin{array}{lll} &{}\min &{}\sum \limits _{i=1}^{p}(-{\underline{c}}_{i}{\underline{d}}_{i}+ ({\underline{c}}_{i} d_{i}+{\underline{d}}_{i}c_{i})^{\top }x) +\sum \limits _{i=1}^{p}({c}_{0i}{d}_{0i}+( {c}_{0i}d_{i}+{d}_{0i}c_{i})^{\top }x)\\ &{}\mathrm {s.t.}&{} x\in D,\ {\underline{c}}_{i}\le c_{i}^{\top }x\le {\overline{c}}_{i},\ {\underline{d}}_{i}\le d_{i}^{\top }x\le {\overline{d}}_{i}, i=1,\ldots , p.\\ \end{array} \right. \end{aligned}$$

$$\begin{aligned}{{(\text {LRP2})}:}\left\{ \begin{array}{lll} &{}\min &{}\sum \limits _{i=1}^{p}(-{\overline{c}}_{i}{\overline{d}}_{i}+ ({\overline{c}}_{i} d_{i}+{\overline{d}}_{i}c_{i})^{\top }x) +\sum \limits _{i=1}^{p}({c}_{0i}{d}_{0i}+( {c}_{0i}d_{i}+{d}_{0i}c_{i})^{\top }x)\\ &{}\mathrm {s.t.}&{} x\in D,\ {\underline{c}}_{i}\le c_{i}^{\top }x\le {\overline{c}}_{i},\ {\underline{d}}_{i}\le d_{i}^{\top }x\le {\overline{d}}_{i}, i=1,\ldots , p.\\ \end{array} \right. \end{aligned}$$

Next, we will introduce the six convex relaxation programs in Cambini et al. (2023), towards this end, we first rewrite $\varphi (x)$ as the following three D.C. functions (D.C. function: the difference between two convex functions):

(i) $\varphi (x)=\sum \nolimits _{i=1}^{p}({c}_{0i}{d}_{0i}+( {c}_{0i}d_{i}+{d}_{0i}c_{i})^{\top }x)+\frac{1}{2}\sum \nolimits _{i=1}^{p} (c_{i}^{\top }x+d_{i}^{\top }x) ^{2}-\frac{1}{2}\sum \nolimits _{i=1}^{p} ((c_{i}^{\top }x)^2+ (d_{i}^{\top }x)^2)$;

(ii) $\varphi (x)=\sum \nolimits _{i=1}^{p}({c}_{0i}{d}_{0i}+( {c}_{0i}d_{i}+{d}_{0i}c_{i})^{\top }x)+\frac{1}{4}\sum \nolimits _{i=1}^{p} (c_{i}^{\top }x+d_{i}^{\top }x) ^{2}-\frac{1}{4}\sum \nolimits _{i=1}^{p} (c_{i}^{\top }x-d_{i}^{\top }x)^2$;

(iii) $\varphi (x)=\sum \nolimits _{i=1}^{p}({c}_{0i}{d}_{0i}+( {c}_{0i}d_{i}+{d}_{0i}c_{i})^{\top }x)+\frac{1}{2}\sum \nolimits _{i=1}^{p} ((c_{i}^{\top }x)^2+ (d_{i}^{\top }x)^2)-\frac{1}{2}\sum \nolimits _{i=1}^{p} (c_{i}^{\top }x-d_{i}^{\top }x)^2$.

Based on (i)–(iii), Problem 2 can be relaxed to the following three convex programs:

$$\begin{aligned}{{(\text {CRP1})}:}\left\{ \begin{array}{lll} &{}&{}\min \sum \limits _{i=1}^{p}({c}_{0i}{d}_{0i}+( {c}_{0i}d_{i}+{d}_{0i}c_{i})^{\top }x)+\frac{1}{2}\sum \limits _{i=1}^{p} (c_{i}^{\top }x+d_{i}^{\top }x) ^{2}\\ &{}&{} +\frac{1}{2}\sum \limits _{i=1}^{p}({\underline{c}}_{i}{\overline{c}}_{i} +{\underline{d}}_{i}{\overline{d}}_{i} -({\underline{c}}_{i}+{\overline{c}}_{i}){c}_{i}^{T}{x}-({\underline{d}}_{i} +{\overline{d}}_{i}){d}_{i}^{T}{x})\\ &{}&{} {\mathrm{s.t.}} x\in D,\ {\underline{c}}_{i}\le c_{i}^{\top }x\le {\overline{c}}_{i},\ {\underline{d}}_{i}\le d_{i}^{\top }x\le {\overline{d}}_{i}, i=1,\ldots , p.\\ \end{array} \right. \end{aligned}$$

$$\begin{aligned}{{(\text {CRP2})}:}\left\{ \begin{array}{lll} &{}\min &{}\sum \limits _{i=1}^{p}({c}_{0i}{d}_{0i}+( {c}_{0i}d_{i}+{d}_{0i}c_{i})^{\top }x)+\frac{1}{4}\sum \limits _{i=1}^{p} (c_{i}^{\top }x+d_{i}^{\top }x) ^{2}\\ &{}&{}+\frac{1}{4}\sum \limits _{i=1}^{p}({\underline{\sigma }}_{i}{\overline{\sigma }}_{i} -({\underline{\sigma }}_{i}+{\overline{\sigma }}_{i})(c_{i}^{\top }x-d_{i}^{\top }x))\\ &{}\mathrm{s.t.}&{} x\in D,\ {\underline{\sigma }}_{i}\le (c_{i}-d_{i})^{\top }x\le {\overline{\sigma }}_{i}, i=1,\ldots , p,\\ \end{array} \right. \end{aligned}$$

where ${\underline{\sigma }}_{i}=\min _{{x}\in D}(c_{i}^{\top }x-d_{i}^{\top }x),$ ${\overline{\sigma }}_{i}=\max _{{x}\in D}(c_{i}^{\top }x-d_{i}^{\top }x), i=1,\ldots , p.$

$$\begin{aligned}{{(\text {CRP3})}:}\left\{ \begin{array}{lll} &{}\min &{}\sum \limits _{i=1}^{p}({c}_{0i}{d}_{0i}+( {c}_{0i}d_{i}+{d}_{0i}c_{i})^{\top }x)+\frac{1}{2}\sum \limits _{i=1}^{p} ((c_{i}^{\top }x)^2+ (d_{i}^{\top }x)^2)\\ &{}&{}+\frac{1}{2}\sum \limits _{i=1}^{p} ({\underline{\sigma }}_{i}{\overline{\sigma }}_{i} -({\underline{\sigma }}_{i}+{\overline{\sigma }}_{i})(c_{i}^{\top }x-d_{i}^{\top }x))\\ &{}\mathrm {s.t.}&{} x\in D,\ {\underline{\sigma }}_{i}\le (c_{i}-d_{i})^{\top }x\le {\overline{\sigma }}_{i}, i=1,\ldots , p.\\ \end{array} \right. \end{aligned}$$

The last three convex relaxation programs are based on the D.C. forms of the objective function $\varphi (x)$ in Problem 2 and the eigenvalue decomposition of the matrix in the quadratic representation of $\varphi (x)$. The details are explained as follows.

Note that the D.C. forms offered by (i)–(ii) can be rewritten as:

(i) $\varphi (x)=\sum \nolimits _{i=1}^{p}({c}_{0i}{d}_{0i}+( {c}_{0i}d_{i}+{d}_{0i}c_{i})^{\top }x)+\frac{1}{2}\sum \nolimits _{i=1}^{p} (c_{i}^{\top }x+d_{i}^{\top }x) ^{2}-x^{\top }Q_{1}x$,

(ii) $\varphi (x)=\sum \nolimits _{i=1}^{p}({c}_{0i}{d}_{0i}+( {c}_{0i}d_{i}+{d}_{0i}c_{i})^{\top }x)+\frac{1}{4}\sum \nolimits _{i=1}^{p} (c_{i}^{\top }x+d_{i}^{\top }x) ^{2}-x^{\top }Q_{2}x$,

where $Q_{1}=\frac{1}{2}\sum \nolimits _{i=1}^{p}(c_{i}c_{i}^{\top }+d_{i}d_{i}^{\top }),\ Q_{2}=\frac{1}{4}\sum \nolimits _{i=1}^{p}(c_{i}-d_{i})(c_{i}-d_{i})^{\top }$ are symmetric positive semidefinite matrices. Therefore, there exists an orthonormal matrix ${\tilde{U}}\in {\mathbb {R}}^{n\times n}$ with its columns ${\tilde{u}}_{1},\ldots ,{\tilde{u}}_{n}\in {\mathbb {R}}^{n}$, and a diagonal matrix ${\tilde{D}}\in {\mathbb {R}}^{n\times n}$ with its diagonal elements ${\tilde{\lambda }}_{1},\ldots ,{\tilde{\lambda }}_{n}\in {\mathbb {R}}$ such that $Q_{1}={\tilde{U}}{\tilde{D}}{\tilde{U}}^{\top }$. Let $\Theta ^{+}=\{i=1,\ldots ,n: {\tilde{\lambda }}_{i}> 0\}$ and ${\tilde{\vartheta }}_{i}=\sqrt{{\tilde{\lambda }}_{i}}\cdot {\tilde{u}}_{i}$ for all $i\in \Theta ^{+},$ then the convex relaxation of Problem 2 can be obtained as follows:

$$\begin{aligned}{{(\text {CRP4})}:}\left\{ \begin{array}{lll} &{}\min &{}\sum \limits _{i=1}^{p}({c}_{0i}{d}_{0i}+( {c}_{0i}d_{i}+{d}_{0i}c_{i})^{\top }x)+\frac{1}{2}\sum \limits _{i=1}^{p} (c_{i}^{\top }x+d_{i}^{\top }x) ^{2}\\ &{}&{}+\sum \limits _{i\in \Theta ^{+}} ({\tilde{\vartheta }}^{L}_{i}{\tilde{\vartheta }}^{U}_{i} -({\tilde{\vartheta }}^{L}_{i}+{\tilde{\vartheta }}^{U}_{i}){\tilde{\vartheta }}_{i}^{\top }x)\\ &{}\mathrm{s.t.}&{} x\in D,\ {\tilde{\vartheta }}^{L}_{i}\le {\tilde{\vartheta }}_{i}^{\top }x\le {\tilde{\vartheta }}^{U}_{i}, i\in \Theta ^{+},\\ \end{array} \right. \end{aligned}$$

where ${\tilde{\vartheta }}^{L}_{i}=\min _{{x}\in D}({\tilde{\vartheta }}_{i}^{T}{x}),$ ${\tilde{\vartheta }}^{U}_{i}=\max _{{x}\in D}({\tilde{\vartheta }}_{i}^{T}{x}), \ i\in \Theta ^{+}.$

Similarly, there exists an orthonormal matrix ${\hat{U}}\in {\mathbb {R}}^{n\times n}$ with its columns ${\hat{u}}_{1},\ldots ,{\hat{u}}_{n}\in {\mathbb {R}}^{n}$, and a diagonal matrix ${\hat{D}}\in {\mathbb {R}}^{n\times n}$ with its diagonal elements ${\hat{\lambda }}_{1},\ldots ,{\hat{\lambda }}_{n}\in {\mathbb {R}}$ such that $Q_{2}={\hat{U}}{\hat{D}}{\hat{U}}^{\top }$. Let $\Gamma ^{+}=\{i=1,\ldots ,n: {\hat{\lambda }}_{i}> 0\}$ and ${\hat{\vartheta }}_{i}=\sqrt{{\hat{\lambda }}_{i}}\cdot {\hat{u}}_{i}$ for all $i\in \Gamma ^{+},$ then we can construct the convex relaxation of Problem 2 as follows:

$$\begin{aligned}{} \mathbf{{(\text {CRP5})}:}\left\{ \begin{array}{lll} &{}\min &{}\sum \limits _{i=1}^{p}({c}_{0i}{d}_{0i}+( {c}_{0i}d_{i}+{d}_{0i}c_{i})^{\top }x)+\frac{1}{4}\sum \limits _{i=1}^{p} (c_{i}^{\top }x+d_{i}^{\top }x) ^{2}\\ &{}&{}+\sum \limits _{i\in \Gamma ^{+}} ({\hat{\vartheta }}^{L}_{i}{\hat{\vartheta }}^{U}_{i} -({\hat{\vartheta }}^{L}_{i}+{\hat{\vartheta }}^{U}_{i}){\hat{\vartheta }}_{i}^{\top }x)\\ &{}\mathrm{s.t.}&{} x\in D,\ {\hat{\vartheta }}^{L}_{i}\le {\hat{\vartheta }}_{i}^{\top }x\le {\hat{\vartheta }}^{U}_{i}, i\in \Gamma ^{+},\\ \end{array} \right. \end{aligned}$$

where ${\hat{\vartheta }}^{L}_{i}=\min _{{x}\in D}({\hat{\vartheta }}_{i}^{T}{x}),$ ${\hat{\vartheta }}^{U}_{i}=\max _{{x}\in D}({\hat{\vartheta }}_{i}^{T}{x}), \ i\in \Gamma ^{+}.$

To report the last convex relaxation of Problem 2, we rewrite $\varphi (x)$ as a quadratic function below:

$$\begin{aligned} \varphi (x)=x^{\top }{\hat{Q}}x+{\hat{a}}^{\top }x+{\hat{a}}_{0}, \end{aligned}$$

where ${\hat{a}}=\sum _{i=1}^{p}(c_{0i}d_{i}+d_{0i}c_{i}),\ {\hat{a}}_{0}=\sum _{i=1}^{p}c_{0i}d_{0i},\ {\hat{Q}}=\frac{1}{2}\sum _{i=1}^{p}(c_{i}d_{i}^{\top }+d_{i}c_{i}^{\top })$ is a symmetric matrix, thus there exists an orthonormal matrix ${\bar{P}}\in {\mathbb {R}}^{n\times n}$ with its columns ${\bar{p}}_{1},\ldots ,{\bar{p}}_{n}\in {\mathbb {R}}^{n}$, and a diagonal matrix ${\bar{D}}\in {\mathbb {R}}^{n\times n}$ with its diagonal elements ${\bar{\lambda }}_{1},\ldots ,{\bar{\lambda }}_{n}\in {\mathbb {R}}$ such that ${\hat{Q}}={\bar{P}}{\bar{D}}{\bar{P}}^{\top }$. Let $\Lambda ^{+}=\{i=1,\ldots ,n: {\bar{\lambda }}_{i}> 0\},\ \Lambda ^{-}=\{i=1,\ldots ,n: {\bar{\lambda }}_{i}< 0\}$ and ${\bar{\vartheta }}_{i}=\sqrt{{\bar{\lambda }}_{i}}\cdot {\bar{p}}_{i},\ i\in \Lambda ^{+}\cup \Lambda ^{-}$, then $x^{\top }{\hat{Q}}x=\sum _{i\in \Lambda ^{+}}({\bar{\vartheta }}_{i}^{\top }x)^{2}-\sum _{i\in \Lambda ^{-}}({\bar{\vartheta }}_{i}^{\top }x)^{2}$, and the convex relaxation of Problem 2 can be obtained as follows:

$$\begin{aligned}{{(\text {CRP6})}:}\left\{ \begin{array}{lll} &{}\min &{}{\hat{a}}^{\top }x+{\hat{a}}_{0}+\sum \limits _{i\in \Lambda ^{+}}({\bar{\vartheta }}_{i}^{\top }x)^{2}+\sum \limits _{i\in \Lambda ^{-}} ({\bar{\vartheta }}^{L}_{i}{\bar{\vartheta }}^{U}_{i} -({\bar{\vartheta }}^{L}_{i}+{\bar{\vartheta }}^{U}_{i}){\bar{\vartheta }}_{i}^{\top }x)\\ &{}\mathrm {s.t.}&{} x\in D,\ {\bar{\vartheta }}^{L}_{i}\le {\bar{\vartheta }}_{i}^{\top }x\le {\bar{\vartheta }}^{U}_{i}, i\in \Lambda ^{-},\\ \end{array} \right. \end{aligned}$$

where ${\bar{\vartheta }}^{L}_{i}=\min _{{x}\in D}({\bar{\vartheta }}_{i}^{T}{x}),$ ${\bar{\vartheta }}^{U}_{i}=\max _{{x}\in D}({\bar{\vartheta }}_{i}^{T}{x}), \ i\in \Lambda ^{-}.$