# The Generalized Trust-Region Sub-Problem with Additional Linear Inequality Constraints—Two Convex Quadratic Relaxations and Strong Duality

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## Abstract

**:**

## 1. Introduction

- (i)
- Under a regularity condition, we present two convex quadratic relaxations (CQRs) under two different conditions for problem (1) that minimize a linear objective function subject to two convex quadratic constraints with a fixed number of additional linear inequality constraints. Our CQRs are inspired by the one proposed for GTRS in Reference [29]. Then we derive sufficient conditions under which problem (1) is equivalent to exactly one of the CQRs and the optimal solution of (1) can be recovered from an optimal solution of the CQRs. These sufficient conditions are easy to verify and involve only one (any) optimal solution of CQRs. We also show that under these conditions the attractive features of GTRS such as strong Lagrangian duality and exact SDO-relaxation hold for (1). It should be noted that in the case of GTRS, the CQRs reduce to the ones proposed in Reference [29]. The CQRs are always exact for GTRS but in the presence of linear constraints, they are not exact in general.
- (ii)
- Exploiting the results in (i), we also derive sufficient conditions that are expressed in terms of the data of the model problem (1) for exactness of the CQRs, strong Lagrangian duality and consequently for tightness of the SDO-relaxation. In the case of eTRS, these sufficient conditions reduce to the one presented in Reference [17] that is the existing best results in the literature. As a consequence, we present necessary and sufficient conditions for global optimality of problem (P) under the new condition together with the Slater condition. We also obtain a form of S-lemma for the system of two quadratic and a fixed number of linear inequalities.
- (iii)
- The sufficient conditions in (i) and (ii) ensure the exactness of the CQRs and the SDO-relaxation of problem (1). It is worth noting that solving large-scale semidefinite problems is still an intractable task. In contrast, the CQRs are significantly more tractable than SDOs, and advanced commercial software is available to solve them [30].

**Notation**

**1.**

## 2. Convex Quadratic Relaxation, Global Minimization and Strong Duality

**Assumption**

**1.**

**Assumption**

**2.**

**Lemma**

**1.**

**Proof.**

**Condition**

**1.**

**Condition**

**2.**

**Lemma**

**2.**

**Proof.**

**Example**

**1.**

**Theorem**

**1.**

- (1)
- ${h}_{1}\left({x}^{*}\right)={h}_{2}\left({x}^{*}\right)={t}^{*}$.
- (2)
- ${h}_{1}\left({x}^{*}\right)<{t}^{*}$ and there exists nonzero $z\in \mathit{Null}(A+{\lambda}_{2}B)$ such that ${(a+{\lambda}_{2}b)}^{T}z=0$ and ${c}_{i}^{T}z\le 0$ for $i=1,\dots ,m$.
- (3)
- ${h}_{2}\left({x}^{*}\right)<{t}^{*}$ and ${\widehat{\lambda}}_{1}=0$.
- (4)
- ${h}_{2}\left({x}^{*}\right)<{t}^{*}$, ${\widehat{\lambda}}_{1}>0$ and there exists nonzero $z\in \mathit{Null}(A+{\lambda}_{1}B)$ such that ${(a+{\lambda}_{1}b)}^{T}z=0$ and ${c}_{i}^{T}z\le 0$ for $i=1,\dots ,m$.

**Proof.**

**Theorem**

**2.**

- (1)
- ${q}_{2}\left({x}^{*}\right)=0$.
- (2)
- ${q}_{2}\left({x}^{*}\right)<0$ and there exists nonzero $z\in \mathit{Null}(A+{\widehat{\lambda}}_{1}B)$ such that ${(a+{\widehat{\lambda}}_{1}b)}^{T}z=0$ and ${c}_{i}^{T}z\le 0$ for $i=1,\dots ,m$.

**Proof.**

## 3. New Conditions for Strong Duality and Exact SDO-Relaxation

**Consider**

**3.**

- 1.
- Condition 1 holds, ${\widehat{\lambda}}_{1}=0$ and there exists nonzero $z\in \mathit{Null}(A+{\lambda}_{2}B)$ such that ${(a+{\lambda}_{2}b)}^{T}z\le 0$ and ${c}_{i}^{T}z\le 0$ for $i=1,\dots ,m$.
- 2.
- Condition 1 holds, ${\widehat{\lambda}}_{1}>0$, there exist nonzero ${z}_{1}\in \mathit{Null}(A+{\lambda}_{1}B)$ and ${z}_{2}\in \mathit{Null}(A+{\lambda}_{2}B)$, such that ${(a+{\lambda}_{1}b)}^{T}{z}_{1}\le 0$, ${(a+{\lambda}_{2}b)}^{T}{z}_{2}\le 0$, ${c}_{i}^{T}{z}_{1}\le 0$ and ${c}_{i}^{T}{z}_{2}\le 0$ for $i=1,\dots ,m$.
- 3.
- Condition 2 holds and there exists nonzero $z\in \mathit{Null}(A+{\widehat{\lambda}}_{1}B)$ such that ${(a+{\widehat{\lambda}}_{1}b)}^{T}z\le 0$ and ${c}_{i}^{T}z\le 0$ for $i=1,\dots ,m$.

**Lemma**

**3.**

**Proof.**

**Remark**

**1.**

**Remark**

**2.**

**Example**

**2.**

**Remark**

**3.**

**Corollary**

**1.**

**Proof.**

**Lemma**

**4.**

- (1)
- ${x}^{T}Bx+2{b}^{T}x+\beta \le 0,{c}_{i}^{T}x\le {d}_{i},i=1,\dots ,m\phantom{\rule{1.em}{0ex}}\Rightarrow \phantom{\rule{1.em}{0ex}}{x}^{T}Ax+2{a}^{T}x+\gamma \ge 0.$
- (2)
- There exist ${\lambda}_{i}\ge 0,i=0,\dots ,m$ such that$$({x}^{T}Ax+2{a}^{T}x+\gamma )+{\lambda}_{0}({x}^{T}Bx+2{b}^{T}x+\beta )+\sum _{i=1}^{m}{\lambda}_{i}({c}_{i}^{T}x-{d}_{i})\ge 0,\phantom{\rule{1.em}{0ex}}\forall x\in {\mathbb{R}}^{n}.$$

**Proof.**

## 4. Conclusions

## Author Contributions

## Funding

## Conflicts of Interest

## Appendix A

#### Appendix A.1. Proof of Lemma 1

#### Appendix A.2. Proof of Lemma 2

#### Appendix A.3. Proof of Theorem 1

- (1)
- ${h}_{1}\left({x}^{*}\right)={h}_{2}\left({x}^{*}\right)={t}^{*}$ implies that $({\widehat{\lambda}}_{1}-{\lambda}_{2})({x}^{{*}^{T}}B{x}^{*}+2{b}^{T}{x}^{*}+\beta )=0.$ Furthermore, since ${\widehat{\lambda}}_{1}\ne {\lambda}_{2}$, we obtain$$\begin{array}{c}\hfill {x}^{{*}^{T}}B{x}^{*}+2{b}^{T}{x}^{*}+\beta =0.\end{array}$$It follows from (A12) and (A13) that $({x}^{*},{t}^{*})$ is also feasible for (8) and since (6) is a relaxation of (8), then $({x}^{*},{t}^{*})$ solves (8), ${p}^{*}={p}_{1}^{*}$ and thus ${x}^{*}$ solves (1). To prove strong duality, set ${\lambda}^{*}={\mu}_{1}^{*}{\widehat{\lambda}}_{1}+{\mu}_{2}^{*}{\lambda}_{2}$. Since ${\mu}_{1}^{*}\ge 0$, ${\mu}_{2}^{*}\ge 0$ and ${\mu}_{1}^{*}+{\mu}_{2}^{*}=1$, then ${\lambda}^{*}\in [{\widehat{\lambda}}_{1},{\lambda}_{2}]$ and thus$$\begin{array}{c}\hfill A+{\lambda}^{*}B\u2ab00.\end{array}$$Also, we have$$\begin{array}{cc}\hfill {p}^{*}\ge {d}^{*}:& =\underset{{\gamma}_{i}\ge 0,i=0,\dots ,m}{max}\underset{x}{min}\left\{{q}_{1}\left(x\right)+{\gamma}_{0}{q}_{2}\left(x\right)+\sum _{i=1}^{m}{\gamma}_{i}({c}_{i}^{T}x-{d}_{i})\right\}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \ge \underset{x}{min}\left\{{q}_{1}\left(x\right)+{\lambda}^{*}{q}_{2}\left(x\right)+\sum _{i=1}^{m}{s}_{i}^{*}({c}_{i}^{T}x-{d}_{i})\right\}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& ={q}_{1}\left({x}^{*}\right)+{\lambda}^{*}{q}_{2}\left({x}^{*}\right)+\sum _{i=1}^{m}{s}_{i}^{*}({c}_{i}^{T}{x}^{*}-{d}_{i})\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& ={q}_{1}\left({x}^{*}\right)\ge {p}^{*},\hfill \end{array}$$
- (2)
- In this case, ${\mu}_{2}^{*}=1$ and hence, ${h}_{2}\left({x}^{*}\right)={t}^{*}$. Then ${h}_{1}\left({x}^{*}\right)<{t}^{*}$ and ${h}_{2}\left({x}^{*}\right)={t}^{*}$ imply that $({\widehat{\lambda}}_{1}-{\lambda}_{2})({x}^{{*}^{T}}B{x}^{*}+2{b}^{T}{x}^{*}+\beta )<0.$ Since ${\widehat{\lambda}}_{1}<{\lambda}_{2}$, we obtain$$\begin{array}{c}\hfill {x}^{{*}^{T}}B{x}^{*}+2{b}^{T}{x}^{*}+\beta >0.\end{array}$$By the assumption, there exists nonzero $z\in \mathrm{Null}(A+{\lambda}_{2}B)$ such that ${(a+{\lambda}_{2}b)}^{T}z=0$ and ${c}_{i}^{T}z\le 0$ for $i=1,\dots ,m$. Consider the following quadratic equation of variable $\alpha $:$$\begin{array}{c}\hfill {q}_{2}({x}^{*}+\alpha z)={\alpha}^{2}{z}^{T}Bz+2\alpha ({z}^{T}B{x}^{*}+{b}^{T}z)+{x}^{{*}^{T}}B{x}^{*}+2{b}^{T}{x}^{*}+\beta =0.\end{array}$$The fact that ${z}^{T}Bz<0$ (see Lemma 3.4 of Reference [22]) with (A16) imply that the above equation has a positive root ${\alpha}^{*}$. Set ${\overline{x}}^{*}={x}^{*}+{\alpha}^{*}z$. We have ${q}_{2}\left({\overline{x}}^{*}\right)=0$ and since ${c}_{i}^{T}z\le 0,i=1,\dots ,m$, we also have ${c}_{i}^{T}{\overline{x}}^{*}\le {d}_{i}$, $i=1,\dots ,m$. Furthermore, since $z\in \mathrm{Null}(A+{\lambda}_{2}B)$ and ${(a+{\lambda}_{2}b)}^{T}z=0$, we have$$\begin{array}{c}\hfill {h}_{2}\left({\overline{x}}^{*}\right)={\overline{x}}^{{*}^{T}}(A+{\lambda}_{2}B){\overline{x}}^{*}+2{(a+{\lambda}_{2}b)}^{T}{\overline{x}}^{*}+{\lambda}_{2}\beta ={h}_{2}\left({x}^{*}\right)={t}^{*}.\end{array}$$It follows from (A18) and ${q}_{2}\left({\overline{x}}^{*}\right)=0$ that ${q}_{1}\left({\overline{x}}^{*}\right)={t}^{*}$ and consequently ${h}_{1}\left({\overline{x}}^{*}\right)={t}^{*}$. These indicate that $({\overline{x}}^{*},{t}^{*})$ is an optimal solution of (6) which is also feasible for (8). Since (6) is a relaxation of (8), $({\overline{x}}^{*},{t}^{*})$ solves (8), ${p}^{*}={p}_{1}^{*}$ and thus ${\overline{x}}^{*}$ solves (1). The same approach as in part (1) can be applied to show that strong duality holds for (1) and the Lagrangian dual problem is solvable.
- (3)
- In this case, ${\mu}_{1}^{*}=1$ and hence, ${h}_{1}\left({x}^{*}\right)={t}^{*}$. Also, ${h}_{2}\left({x}^{*}\right)<{t}^{*}$ and ${h}_{1}\left({x}^{*}\right)={t}^{*}$ imply that $({\lambda}_{2}-{\widehat{\lambda}}_{1})({x}^{{*}^{T}}B{x}^{*}+2{b}^{T}{x}^{*}+\beta )<0.$ Then ${\lambda}_{2}>{\widehat{\lambda}}_{1}$ results in$$\begin{array}{c}\hfill {x}^{{*}^{T}}B{x}^{*}+2{b}^{T}{x}^{*}+\beta <0.\end{array}$$It follows from (A19) and (A12) that $({x}^{*},{t}^{*})$ is also feasible for (8) and since (6) is a relaxation of (8), ${x}^{*}$ solves (1) and ${p}^{*}={p}_{1}^{*}$. Then by setting ${\lambda}^{*}=0$, the same approach as in part (1) can be applied to show that strong duality holds for (1) and the Lagrangian dual problem is solvable.
- (4)
- By the assumption, there exists nonzero $z\in \mathrm{Null}(A+{\lambda}_{1}B)$ such that ${(a+{\lambda}_{1}b)}^{T}z=0$ and ${c}_{i}^{T}z\le 0$ for $i=1,\dots ,m$. Consider the following quadratic equation of variable $\alpha $:$$\begin{array}{c}\hfill {q}_{2}({x}^{*}+\alpha z)={\alpha}^{2}{z}^{T}Bz+2\alpha ({z}^{T}B{x}^{*}+{b}^{T}z)+{x}^{{*}^{T}}B{x}^{*}+2{b}^{T}{x}^{*}+\beta =0.\end{array}$$The fact that ${z}^{T}Bz>0$ (see Lemma 3.4 of Reference [22]) with (A19) imply that the above equation has a positive root ${\alpha}^{*}$. Then following the same discussion as in part (2) where ${\lambda}_{2}$ is replaced by ${\lambda}_{1}$ and ${h}_{2}\left({\overline{x}}^{*}\right)$ in (A18) is replaced by ${h}_{1}\left({\overline{x}}^{*}\right)$, it can be shown that ${\overline{x}}^{*}:={x}^{*}+\alpha z$ solves problem (1), ${p}^{*}={p}_{1}^{*}$, strong duality holds for problem (1) and the Lagrangian dual problem is solvable.

#### Appendix A.4. Proof of Theorem 2

- (1)
- In this case, since ${q}_{2}\left({x}^{*}\right)=0$ and (7) is a relaxation of (1), then ${q}_{1}\left({x}^{*}\right)={p}_{2}^{*}\le {p}^{*}\le {q}_{1}\left({x}^{*}\right)$ and consequently ${p}^{*}={p}_{2}^{*}={q}_{1}\left({x}^{*}\right)$. Next by setting ${\lambda}^{*}={\mu}_{1}^{*}+{\widehat{\lambda}}_{1}$, the same approach as in part (1) of Theorem 1 can be applied to show that strong duality holds for problem (1) and the Lagrangian dual problem is solvable.
- (2)
- By the assumption, there exists nonzero $z\in \mathrm{Null}(A+{\widehat{\lambda}}_{1}B)$ such that ${(a+{\widehat{\lambda}}_{1}b)}^{T}z=0$ and ${c}_{i}^{T}z\le 0$ for $i=1,\dots ,m$. Consider the following quadratic equation of variable $\alpha $:$${q}_{2}({x}^{*}+\alpha z)={\alpha}^{2}{z}^{T}Bz+2\alpha ({z}^{T}B{x}^{*}+{b}^{T}z)+{x}^{{*}^{T}}B{x}^{*}+2{b}^{T}{x}^{*}+\beta =0.$$The fact that ${z}^{T}Bz>0$ with ${q}_{2}\left({x}^{*}\right)<0$ implies that the above equation has a positive root ${\alpha}^{*}$. Set ${\overline{x}}^{*}={x}^{*}+{\alpha}^{*}z$. We have ${q}_{2}\left({\overline{x}}^{*}\right)=0$, ${c}_{i}^{T}{\overline{x}}^{*}\le 0$ for $i=1,\dots ,m$ and$$\begin{array}{c}\hfill {h}_{1}\left({\overline{x}}^{*}\right)={\overline{x}}^{{*}^{T}}(A+{\widehat{\lambda}}_{1}B){\overline{x}}^{*}+2{(a+{\widehat{\lambda}}_{1}b)}^{T}{\overline{x}}^{*}+{\widehat{\lambda}}_{1}\beta ={h}_{1}\left({x}^{*}\right),\end{array}$$

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**MDPI and ACS Style**

Almaadeed, T.A.; Taati, A.; Salahi, M.; Hamdi, A.
The Generalized Trust-Region Sub-Problem with Additional Linear Inequality Constraints—Two Convex Quadratic Relaxations and Strong Duality. *Symmetry* **2020**, *12*, 1369.
https://doi.org/10.3390/sym12081369

**AMA Style**

Almaadeed TA, Taati A, Salahi M, Hamdi A.
The Generalized Trust-Region Sub-Problem with Additional Linear Inequality Constraints—Two Convex Quadratic Relaxations and Strong Duality. *Symmetry*. 2020; 12(8):1369.
https://doi.org/10.3390/sym12081369

**Chicago/Turabian Style**

Almaadeed, Temadher A., Akram Taati, Maziar Salahi, and Abdelouahed Hamdi.
2020. "The Generalized Trust-Region Sub-Problem with Additional Linear Inequality Constraints—Two Convex Quadratic Relaxations and Strong Duality" *Symmetry* 12, no. 8: 1369.
https://doi.org/10.3390/sym12081369