A fresh CP look at mixed-binary QPs: new formulations and relaxations

Immanuel M. Bomze; Jianqiang Cheng; Peter J. C. Dickinson; Abdel Lisser

doi:10.1007/s10107-017-1109-8

Mathematical Programming

November 2017, Volume 166, Issue 1–2, pp 159–184 | Cite as

A fresh CP look at mixed-binary QPs: new formulations and relaxations

Authors
Authors and affiliations

Immanuel M. Bomze
Jianqiang Cheng
Peter J. C. Dickinson
Abdel LisserEmail author

Full Length Paper Series A

First Online: 21 January 2017

Received: 15 May 2016

Accepted: 03 January 2017

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Abstract

Triggered by Burer’s seminal characterization from 2009, many copositive reformulations of mixed-binary QPs have been discussed by now. Most of them can be used as proper relaxations, if the intractable co(mpletely)positive cones are replaced by tractable approximations. While the widely used approximation hierarchies have the disadvantage to use positive-semidefinite (psd) matrices of orders which rapidly increase with the level of approximation, alternatives focus on the problem of keeping psd matrix orders small, with the aim to avoid memory problems in the interior point algorithms. This work continues this approach, proposing new reformulations and relaxations. Moreover, we provide a thorough comparison of the respective duals and establish a monotonicity relation among their duality gaps. We also identify sufficient conditions for strong duality/zero duality gap in some of these formulations and generalize some of our observations to general conic problems.

Keywords

Copositivity Completely positive Conic optimization Quadratic optimization Reformulations Nonlinear optimization Nonconvex optimization

Mathematics Subject Classification

90C20 90C26 90C30

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Notes

Acknowledgements

The authors would like to thank the anonymous handling Associate Editor and two anonymous referees for their very useful comments and suggestions which helped to improve our paper significantly. This research benefited from the support of the “FMJH Program Gaspard Monge in Optimization and Operations Research”, and from the support to this program by EDF.

Appendix

Proof of Theorem 6

The characterizations of ${{\mathrm{int}}}\left( \mathcal {L}^\perp +\mathcal {K}^*\right) $ in Theorem 6 follow trivially from the definition of the interior and from $\mathcal {K}^*$ being a convex cone with nonempty interior (see e.g. [2, Theorem 2.3]).

If Slater’s condition, 1, holds then we can approach the optimal value to (17) from the interior, and thus the closure is not necessary (i.e., for all $y\in {{\mathrm{feas}}}$ (17) and all $\varepsilon \in ]0,1]$ we have $\mathsf {q}-(y+\varepsilon (\widehat{y}-y))\mathsf {d}_0\in {{\mathrm{int}}}\left( \mathcal {L}^\perp +\mathcal {K}^*\right) $, and thus $y+\varepsilon (\widehat{y}-y)\in {{\mathrm{feas}}}$ (18)). This implies that ${{\mathrm{opt}}}$ (18) $={{\mathrm{opt}}}$ (17) $={{\mathrm{opt}}}$ (15), with the last equality following from Theorem 4.

We are thus left to prove the equivalence relations. In doing this we will repeatedly use the result that for a closed convex pointed cone $\mathcal {C}\subseteq \mathbb {R}^n$ we have ${{\mathrm{int}}}\mathcal {C}^*= \{ \mathsf {z}\in \mathbb {R}^n : \langle \mathsf {x},\mathsf {z}\rangle >0 \text { for all } \mathsf {x}\in \mathcal {C}{\setminus }\{\mathsf {o}\} \}$ (see e.g. [2, Equation (2.1)]). We split the proof of the equivalence relations into four parts:

1 $\Leftrightarrow $ 2:: This follows directly from the observation
$$\begin{aligned}&\left\{ \mathsf {q}\in \mathbb {R}^n : \langle \mathsf {q}, \mathsf {u}\rangle >0 \text { for all } \mathsf {u}\in \mathcal {L}\cap \mathcal {K}{\setminus }\{\mathsf {o}\} \;\mathrm{s.t.}\; \langle \mathsf {d}_0, \mathsf {u}\rangle = 0 \right\} \\&\quad = {{\mathrm{int}}}(\{\mathsf {u}\in \mathcal {L}\cap \mathcal {K}:\langle \mathsf {d}_0,\mathsf {u}\rangle =0\}^*)\\&\quad = {{\mathrm{int}}}({{\mathrm{cl}}}({{\mathrm{span}}}\{\mathsf {d}_0\}+\mathcal {L}^\perp +\mathcal {K}^*))\\&\quad = {{\mathrm{int}}}({{\mathrm{span}}}\{\mathsf {d}_0\}+(\mathcal {L}^\perp +\mathcal {K}^*))\\&\quad = {{\mathrm{span}}}\{\mathsf {d}_0\}+{{\mathrm{int}}}(\mathcal {L}^\perp +\mathcal {K}^*)\\&\quad = \left\{ \mathsf {q}\in \mathbb {R}^n : \exists y\in \mathbb {R}\;\mathrm{s.t.}\; \mathsf {q}-y\mathsf {d}_0\in {{\mathrm{int}}}(\mathcal {L}^\perp +\mathcal {K}^*) \right\} . \end{aligned}$$
2 $\Rightarrow $ 4:: Suppose for the sake of contradiction that statement 2 is true but there exists $\mu \in \mathbb {R}$ such that the set $\left\{ \mathsf {x}\in {{\mathrm{feas}}}(15) : \langle \mathsf {q},\mathsf {x}\rangle \le \mu \right\} $ is not compact. This set is the intersection of closed sets and thus is itself closed. Therefore it must be unbounded, i.e. for all $i\in \mathbb {N}$ there exists $\mathsf {x}_i\in {{\mathrm{feas}}}$ (15) such that $\Vert \mathsf {x}_i\Vert _2>i$ and $\langle \mathsf {q},\mathsf {x}_i \rangle \le \mu $. For all i, letting $\mathsf {u}_i=\frac{1}{\Vert \mathsf {x}_i\Vert _2}\mathsf {x}_i$ we have $\Vert \mathsf {u}_i\Vert _2=1$ and thus the sequence $\{\mathsf {u}_i:i\in \mathbb {N}\}$ is limited to a compact set. Therefore there is a limiting subsequence tending towards a limiting point $\mathsf {u}$. For all i we have $\mathsf {u}_i\in \mathcal {L}\cap \mathcal {K}$ and $\Vert \mathsf {u}_i\Vert _2=1$, and as $\mathcal {L}\cap \mathcal {K}$ is closed this implies that $\mathsf {u}\in \mathcal {L}\cap \mathcal {K}{\setminus }\{\mathsf {o}\}$. We also have $\langle \mathsf {d}_0, \mathsf {u}\rangle = \lim _{i\rightarrow \infty } \tfrac{1}{\Vert \mathsf {x}_i\Vert _2}\langle \mathsf {d}_0,\mathsf {x}_i\rangle = 0$ and $\langle \mathsf {q},\mathsf {u}\rangle \le \lim _{i\rightarrow \infty } \tfrac{\mu }{\Vert \mathsf {x}_i\Vert _2} = 0$. This gives the contradiction that statement 2 does not hold.
4 $\wedge {{\mathrm{feas}}}(15) \ne \emptyset \Rightarrow $ 3:: This follows from considering an arbitrary $\mathsf {x}\in {{\mathrm{feas}}}$ (15) and letting $\mu =\langle \mathsf {q},\mathsf {x}\rangle $.
3 $\Rightarrow $ 2:: Suppose there exists $\mu \in \mathbb {R}\cup \{\infty \}$ such that the set $\left\{ \mathsf {x}\in {{\mathrm{feas}}}(15) : \langle \mathsf {q},\mathsf {x}\rangle \le \mu \right\} $ is nonempty and compact, and additionally suppose for the sake of contradiction that there exists a $\mathsf {u}\in \mathcal {L}\cap \mathcal {K}{\setminus }\{\mathsf {o}\}$ such that $\langle \mathsf {q},\mathsf {u}\rangle \le 0 = \langle \mathsf {d}_0,\mathsf {u}\rangle $. Then there exists $\mathsf {v}\in {{\mathrm{feas}}}$ (15) such that $\langle \mathsf {q},\mathsf {v}\rangle \le \mu $, and for all $\lambda \ge 0$ we have $\mathsf {v}+\lambda \mathsf {u}\in {{\mathrm{feas}}}$ (15) and $\langle \mathsf {q}, \mathsf {v}+\lambda \mathsf {u}\rangle \le \langle \mathsf {q}, \mathsf {v}\rangle \le \mu $, which is a contradiction. $\square $

Proof of Lemma 3

To establish Lemma 3, first note that $\mathcal {K}$ is a closed convex pointed cone. For $\varepsilon \in ]0,1]$, letting ${\mathsf Y}=\varepsilon {\mathsf I}+ (1-\varepsilon )\mathsf {e}_0\mathsf {e}_0^\top $, we have ${\mathsf Y}\in {{\mathrm{int}}}\mathcal {S}_+^{n+1-m}$ and ${{\mathsf R}}{\mathsf Y}{{\mathsf R}}^\top = \varepsilon {{\mathsf R}}^\top {{\mathsf R}}+ (1-\varepsilon ) \begin{pmatrix}1\\ \mathsf {x}_0\end{pmatrix}\begin{pmatrix}1\\ \mathsf {x}_0\end{pmatrix}^\top $ which is strictly positive for $\varepsilon >0$ small enough. Therefore ${\mathsf Y}\in {{\mathrm{int}}}\mathcal {K}$ for $\varepsilon >0$ small enough, completing the proof that $\mathcal {K}$ is a proper cone. The remainder of the proof then follows from the following equalities:

$$\begin{aligned}&\left( \mathcal {S}_+^{n+1-m}\right) ^*= \mathcal {S}_+^{n+1-m}, \end{aligned}$$

(34)

$$\begin{aligned}&\left( {{\mathsf R}}^\top \mathcal {N}^{n+1}{{\mathsf R}}\right) ^*= \left\{ {\mathsf Y}\in \mathcal {S}^{n+1-m} : \langle {\mathsf Y}, {{\mathsf R}}^\top {\mathsf X}{{\mathsf R}}\rangle \ge 0 \text { for all } {\mathsf X}\in \mathcal {N}^{n+1} \right\} \nonumber \\&\quad = \left\{ {\mathsf Y}\in \mathcal {S}^{n+1-m} : \langle {{\mathsf R}}{\mathsf Y}{{\mathsf R}}^\top , {\mathsf X}\rangle \ge 0 \text { for all } {\mathsf X}\in \mathcal {N}^{n+1} \right\} \nonumber \\&\quad = \left\{ {\mathsf Y}\in \mathcal {S}^{n+1-m} : {{\mathsf R}}{\mathsf Y}{{\mathsf R}}^\top \in \mathcal {N}^{n+1} \right\} , \end{aligned}$$

(35)

$$\begin{aligned}&\left\{ {\mathsf Y}\in \mathcal {S}^{n+1-m} : {{\mathsf R}}{\mathsf Y}{{\mathsf R}}^\top \in \mathcal {N}^{n+1} \right\} ^*\nonumber \\&\quad = \left( {{\mathsf R}}^\top \mathcal {N}^{n+1}{{\mathsf R}}\right) ^{**} = {{\mathrm{cl}}}\left( {{\mathsf R}}^\top \mathcal {N}^{n+1}{{\mathsf R}}\right) ={{\mathsf R}}^\top \mathcal {N}^{n+1}{{\mathsf R}}, \end{aligned}$$

(36)

$$\begin{aligned}&\mathcal {K}^*= {{\mathrm{cl}}}\left( \left( \mathcal {S}_+^{n+1-m}\right) ^*+ \left\{ {\mathsf Y}\in \mathcal {S}^{n+1-m} : {{\mathsf R}}{\mathsf Y}{{\mathsf R}}^\top \in \mathcal {N}^{n+1} \right\} ^*\right) \nonumber \\&\quad = {{\mathrm{cl}}}\left( \mathcal {S}_+^{n+1-m}+{{\mathsf R}}^\top \mathcal {N}^{n+1}{{\mathsf R}}\right) \end{aligned}$$

(37)

$$\begin{aligned}&\quad = \mathcal {S}_+^{n+1-m}+{{\mathsf R}}^\top \mathcal {N}^{n+1}{{\mathsf R}}. \end{aligned}$$

(38)

The equalities (34) and (35) follow from the well known result that both $\mathcal {S}_+^{n+1-m}$ and $\mathcal {N}^{n+1}$ are self-dual (see e.g. [14, Theorem 1.32]). The equalities in (36) follow from (35), [2, Corollary 2.1] and $\mathcal {N}^{n+1}$ being a polyhedron implying that ${{\mathsf R}}^\top \mathcal {N}^{n+1}{{\mathsf R}}$ is a polyhedron. Finally, the equality (37) follows from [2, Corollary 2.2] and equality (38) follows from [14, Corollary 1.12]. $\square $

Proof of Theorem 9

First we consider problem (25) for arbitrary $k{ \in \! [{1}\! : \! {2}]}$, $i{ \in \! [{1}\! : \! {4}]}$. From Theorem 6 we have that Slater’s condition holds if and only if

$$\begin{aligned}&\left\langle \begin{pmatrix} 0 &{} \mathsf {c}^\top \\ \mathsf {c}&{} {\mathsf Q}\end{pmatrix}, \begin{pmatrix} x_0 &{} \mathsf {x}^\top \\ \mathsf {x}&{} {\mathsf X}\end{pmatrix} \right\rangle >0 \quad \text {for all } \begin{pmatrix} x_0 &{} \mathsf {x}^\top \\ \mathsf {x}&{} {\mathsf X}\end{pmatrix}\\&\quad \in \mathcal {B}_k\cap \mathcal {L}_i\cap \mathcal {CP}^{n+1}{\setminus }\{{\mathsf O}\} \text { such that } x_0 = 0. \end{aligned}$$

or equivalently

$$\begin{aligned} \langle {\mathsf Q}, {\mathsf X}\rangle >0 \quad \text {for all } {\mathsf X}\in \mathcal {S}^{n}{\setminus }\{{\mathsf O}\} \text { such that } \begin{pmatrix} 0 &{} \mathsf {o}^\top \\ \mathsf {o}&{} {\mathsf X}\end{pmatrix}\in \mathcal {B}_k\cap \mathcal {L}_i\cap \mathcal {CP}^{n+1}. \end{aligned}$$

From Proposition 1, this condition is independent of k and l, and from Lemma 1 it can be shown that $\begin{pmatrix} 0 &{} \mathsf {o}^\top \\ \mathsf {o}&{} {\mathsf X}\end{pmatrix}\in \mathcal {B}_k\cap \mathcal {L}_i\cap \mathcal {CP}^{n+1}$ if and only if

$$\begin{aligned} X_{jk}&= 0 \quad \text { for all } j{ \in \! [{1}\! : \! {n}]}{\setminus } J,\; k{ \in \! [{1}\! : \! {n}]},\\ \widetilde{{\mathsf X}}&\in {{\mathrm{conv}}}\{ \mathsf {v}\mathsf {v}^\top : \mathsf {v}\in \mathbb {R}^{|J|}_+ ,\; \widetilde{\mathsf {a}}_i^\top \mathsf {v}=0 \quad \text { for all } i{ \in \! [{1}\! : \! {m}]} \}, \end{aligned}$$

where $\widetilde{{\mathsf X}}$ is the principal submatrix of ${\mathsf X}$ corresponding to the indices J. This completes the proof for problem (25).

We next consider problem (26) for arbitrary $k{ \in \! [{1}\! : \! {2}]}$, $i{ \in \! [{1}\! : \! {4}]}$. Similarly to before, from Theorem 6 we have that Slater’s condition holds if and only if

$$\begin{aligned} \langle {\mathsf Q}, {\mathsf X}\rangle >0 \quad \text {for all } {\mathsf X}\in \mathcal {S}^{n}{\setminus }\{{\mathsf O}\} \text { such that } \begin{pmatrix} 0 &{} \mathsf {o}^\top \\ \mathsf {o}&{} {\mathsf X}\end{pmatrix}\in \mathcal {B}_k\cap \mathcal {L}_i\cap \mathcal {S}_+^{n+1}\cap \mathcal {N}^{n+1}. \end{aligned}$$

From Proposition 1, this condition is independent of k and i, and from Lemma 1 it can be shown that $\begin{pmatrix} 0 &{} \mathsf {o}^\top \\ \mathsf {o}&{} {\mathsf X}\end{pmatrix}\in \mathcal {L}_2\cap \mathcal {N}^{n+1}$ if and only if

$$\begin{aligned} X_{jk}&= 0 \quad \text { for all } j{ \in \! [{1}\! : \! {n}]}{\setminus } J,\; k{ \in \! [{1}\! : \! {n}]},\\ \widetilde{{\mathsf X}}&\in \mathcal {N}^{|J|}\\ \widetilde{{\mathsf X}}\widetilde{\mathsf {a}}_j&= 0 \quad \text { for all } j{ \in \! [{1}\! : \! {m}]}, \end{aligned}$$

and we then additionally have $\begin{pmatrix} 0 &{} \mathsf {o}^\top \\ \mathsf {o}&{} {\mathsf X}\end{pmatrix}\in \mathcal {B}_k$. Therefore $\begin{pmatrix} 0 &{} \mathsf {o}^\top \\ \mathsf {o}&{} {\mathsf X}\end{pmatrix}\in \mathcal {B}_k\cap \mathcal {L}_i\cap \mathcal {S}_+^{n+1}\cap \mathcal {N}^{n+1}$ if and only if

$$\begin{aligned} X_{jk}&= 0 \quad \text { for all } j{ \in \! [{1}\! : \! {n}]}{\setminus } J,\; k{ \in \! [{1}\! : \! {n}]},\\ \langle \widetilde{\mathsf {a}}_j\widetilde{\mathsf {a}}_j^\top ,\widetilde{{\mathsf X}}\rangle&= 0 \quad \text { for all } j{ \in \! [{1}\! : \! {m}]},\\ \widetilde{{\mathsf X}}&\in \mathcal {S}_+^{|J|}\cap \mathcal {N}^{|J|}, \end{aligned}$$

which completes the proof for problem (26).

Finally we consider problem (27) for all $k{ \in \! [{1}\! : \! {2}]}$. From Theorem 6 we have that Slater’s condition holds if and only if

$$\begin{aligned}&\left\langle {{\mathsf R}}^\top \begin{pmatrix} 0 &{} \mathsf {c}^\top \\ \mathsf {c}&{} {\mathsf Q}\end{pmatrix} {{\mathsf R}}, {\mathsf Y}\right\rangle >0 \text { for all } {\mathsf Y}\in \mathcal {S}_+^{n+1-m}{\setminus }\{{\mathsf O}\}\, \mathrm{s.t.}\, {{\mathsf R}}{\mathsf Y}{{\mathsf R}}^\top \in \mathcal {B}_k,\; {{\mathsf R}}{\mathsf Y}{{\mathsf R}}^\top \in \mathcal {N}^{n+1},\\&\quad \left\langle {{\mathsf R}}^\top \begin{pmatrix} 1 &{} \mathsf {o}^\top \\ \mathsf {o}&{} {\mathsf O}\end{pmatrix} {{\mathsf R}}, {\mathsf Y}\right\rangle = 0, \end{aligned}$$

or equivalently

$$\begin{aligned}&\left\langle \begin{pmatrix} 0 &{} \mathsf {c}^\top \\ \mathsf {c}&{} {\mathsf Q}\end{pmatrix} , \begin{pmatrix} x_0 &{} \mathsf {x}^\top \\ \mathsf {x}&{} {\mathsf X}\end{pmatrix} \right\rangle >0 \text { for all } \begin{pmatrix} x_0 &{} \mathsf {x}^\top \\ \mathsf {x}&{} {\mathsf X}\end{pmatrix}\in \mathcal {B}_k\cap \left( {{\mathsf R}}\mathcal {S}_+^{n+1-m}{{\mathsf R}}^\top \right) \\&\quad \cap \mathcal {N}^{n+1}{\setminus }\{{\mathsf O}\} \;\mathrm{s.t.}\; x_0 = 0. \end{aligned}$$

From Theorem 1, this is equivalent to the previous case for $i=1$. $\square $

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Authors and Affiliations

Immanuel M. Bomze
- 1
Jianqiang Cheng
- 2
Peter J. C. Dickinson
- 3
Abdel Lisser
- 4
Email authorView author's OrcID profile

1.Institut Für Statistik Und Operations Research (ISOR)Universität WienViennaAustria
2.Department of Systems and Industrial EngineeringUniversity of ArizonaTucsonUSA
3.Department of Applied MathematicsUniversity of TwenteEnschedeThe Netherlands
4.Laboratoire de Recherche En Informatique (LRI)Université Paris Sud - XIOrsay CedexFrance

A fresh CP look at mixed-binary QPs: new formulations and relaxations

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Mathematics Subject Classification

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