On the tightness of SDP relaxations of QCQPs

Abstract

Quadratically constrained quadratic programs (QCQPs) are a fundamental class of optimization problems well-known to be NP-hard in general. In this paper we study conditions under which the standard semidefinite program (SDP) relaxation of a QCQP is tight. We begin by outlining a general framework for proving such sufficient conditions. Then using this framework, we show that the SDP relaxation is tight whenever the quadratic eigenvalue multiplicity, a parameter capturing the amount of symmetry present in a given problem, is large enough. We present similar sufficient conditions under which the projected epigraph of the SDP gives the convex hull of the epigraph in the original QCQP. Our results also imply new sufficient conditions for the tightness (as well as convex hull exactness) of a second order cone program relaxation of simultaneously diagonalizable QCQPs.

Appendices

Proof of Proposition 1

Proposition 1

For any SD-QCQP, we have

$$\begin{aligned} \mathcal D_\text{ SOCP } = \mathcal D_\text{ SDP } \quad \text {and}\quad {{\,\mathrm{Opt}\,}}_\text{ SOCP } = {{\,\mathrm{Opt}\,}}_\text{ SDP }. \end{aligned}$$

Proof

The second identity follows immediately from the first identity, thus it suffices to prove only the former.

Let $(x, t)\in \mathcal D_\text{ SDP }$. By definition, there exists $X\in {{\mathbb {S}}}^N$ such that the following system is satisfied

$$\begin{aligned} {\left\{ \begin{array}{ll} Y:=\begin{pmatrix} 1 &{} x^\top \\ x &{} X \end{pmatrix}\\ \left\langle Q_0, Y \right\rangle \le 2t\\ \left\langle Q_i, Y \right\rangle \le 0 ,\,\forall i\in \llbracket m_I \rrbracket \\ \left\langle Q_i,Y \right\rangle = 0 ,\,\forall i\in \llbracket m_I+1,m \rrbracket \\ Y\succeq 0. \end{array}\right. } \end{aligned}$$

Taking a Schur complement of 1 in the matrix Y, we see that $X\succeq xx^\top$. In particular, we have that $X_{j,j} \ge x_j^2$ for all $j\in \llbracket N \rrbracket$. Define the vector y by $y_j = X_{j,j}\ge x_j^2$. Then, noting that $\left\langle {{\,\mathrm{Diag}\,}}(a_i), X \right\rangle = \left\langle a_i, y \right\rangle$ for all $i\in \llbracket 0,m \rrbracket$, we conclude that $(x,t)\in \mathcal D_\text{ SOCP }$.

Let $(x, t)\in \mathcal D_\text{ SOCP }$. By definition, there exists $y\in {{\mathbb {R}}}^N$ such that the following system is satisfied

$$\begin{aligned} {\left\{ \begin{array}{ll} \left\langle a_0, y \right\rangle + 2\left\langle b_0, x \right\rangle + c_0 \le 2t\\ \left\langle a_i, y \right\rangle + 2\left\langle b_i, x \right\rangle + c_i \le 0 ,\,\forall i\in \llbracket m_I \rrbracket \\ \left\langle a_i, y \right\rangle + 2\left\langle b_i, x \right\rangle + c_i = 0 ,\,\forall i\in \llbracket m_I+1,m \rrbracket \\ y_j\ge x_j^2,\,\forall j\in \llbracket N \rrbracket . \end{array}\right. } \end{aligned}$$

Define $X\in {{\mathbb {S}}}^N$ such that $X_{j,j} = y_j$ for all $j\in \llbracket N \rrbracket$ and $X_{j,k} = x_jx_k$ for $j\ne k$. From the definition of $\mathcal D_\text{ SOCP }$, the relation $y_j \ge x_j^2$ holds for all $j\in \llbracket N \rrbracket$, therefore

$$\begin{aligned} \begin{pmatrix} 1 &{}\quad x^\top \\ x &{}\quad X \end{pmatrix}\succeq \begin{pmatrix} 1 &{}\quad x^\top \\ x &{}\quad xx^\top \end{pmatrix}\succeq 0. \end{aligned}$$

Finally, noting that $\left\langle {{\,\mathrm{Diag}\,}}(a_i), X \right\rangle = \left\langle a_i, y \right\rangle$ for all $i\in \llbracket 0,m \rrbracket$, we conclude that $(x,t)\in \mathcal D_\text{ SDP }$. $\square$

Proof of Theorem 8

Theorem 8

Suppose Assumption 1 holds. Define the hyperplane $H = \left\{ (x,t)\in {{\mathbb {R}}}^{N+1}:\, 2t = {{\,\mathrm{Opt}\,}}_\text{ SDP }\right\}$. If the quadratic eigenvalue multiplicity k satisfies $k \ge m+1$, then ${{\,\mathrm{conv}\,}}({\mathcal {D}}\cap H) = {\mathcal {D}}_\text{ SDP } \cap H$. In particular, ${{\,\mathrm{Opt}\,}}= {{\,\mathrm{Opt}\,}}_\text{ SDP }$.

Proof

Suppose $({{\hat{x}}},{{\hat{t}}})\in \mathcal D_\text{ SDP }\cap H$. Then by Lemma 1 and optimality of ${{\hat{t}}}$, we have that $2{{\hat{t}}} = \sup _{\gamma \in \varGamma }q(\gamma ,{{\hat{x}}})$, i.e.,

$$\begin{aligned} 2{{\hat{t}}}&= \sup _{\gamma \in {{\mathbb {R}}}^m} \left\{ q(\gamma ,{{\hat{x}}}) :\, \begin{array}{l} A(\gamma )\succeq 0\\ \gamma _i \ge 0 ,\,\forall i\in \llbracket m_I \rrbracket \end{array}\right\} \\&= \sup _{\gamma \in {{\mathbb {R}}}^m} \left\{ q(\gamma ,{{\hat{x}}}) :\, \begin{array}{l} \mathbb {A}(\gamma )\succeq 0\\ \gamma _i \ge 0 ,\,\forall i\in \llbracket m_I \rrbracket \end{array}\right\} . \end{aligned}$$

The second line follows as $A(\gamma )\succeq 0$ if and only if $\mathbb {A}(\gamma )\succeq 0$. Note that Assumption 1 allows us to apply strong conic duality to the program on the second line. Furthermore, this dual SDP achieves its optimal value, i.e., there exists $Z\in {\mathbb {S}}^n$ such that $({{\hat{x}}},{{\hat{t}}}, Z)$ satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} q_0({{\hat{x}}}) + \left\langle \mathbb {A}_0, Z \right\rangle = 2{{\hat{t}}}\\ q_i({{\hat{x}}}) + \left\langle \mathbb {A}_i, Z \right\rangle \le 0,\,\forall i\in \llbracket m_I \rrbracket \\ q_i({{\hat{x}}}) + \left\langle \mathbb {A}_i, Z \right\rangle =0,\,\forall i\in \llbracket m_I+1,m \rrbracket \\ Z\succeq 0. \end{array}\right. } \end{aligned}$$

(19)

We will show by induction on ${{\,\mathrm{rank}\,}}(Z)$ that for any $({{\hat{x}}},{{\hat{t}}}, Z)$ satisfying (19), we have $({{\hat{x}}},{{\hat{t}}})\in {{\,\mathrm{conv}\,}}({\mathcal {D}}\cap H)$. The claim clearly holds when ${{\,\mathrm{rank}\,}}(Z) = 0$.

Now suppose $r:={{\,\mathrm{rank}\,}}(Z)\ge 1$. Let $({{\hat{x}}},{{\hat{t}}},Z)$ satisfy (19). Write $Z = \sum _{i=1}^{r}z_iz_i^\top$ where each $z_i$ is nonzero. Fix $z:=z_1$.

We claim that the following system in $y\in {{\mathbb {R}}}^k$ is feasible:

$$\begin{aligned} {\left\{ \begin{array}{ll} \left\langle A_i{{\hat{x}}}+b_i, y\otimes z \right\rangle = 0 ,\,\forall i\in \llbracket m \rrbracket \\ y\in \mathbf{S}^{k-1}. \end{array}\right. } \end{aligned}$$

(20)

Indeed, the linear constraints impose at most m homogeneous linear equalities in $k\ge m+1$ variables. In particular, there exists a nonzero solution y to the linear constraints. This y may then be scaled to satisfy $y\in \mathbf{S}^{k-1}$.

Note then that for all $i \in \llbracket 1,m \rrbracket$,

$$\begin{aligned} q_i({{\hat{x}}} \pm y\otimes z) + \left\langle \mathbb {A}_i, Z - zz^\top \right\rangle&= ({{\hat{x}}} \pm y\otimes z)^\top A_i ({{\hat{x}}} \pm y\otimes z) + 2 b_i^\top ({{\hat{x}}} \pm y\otimes z) \\&\quad + c_i + \left\langle \mathbb {A}_i, Z - zz^\top \right\rangle \\&= q_i({{\hat{x}}}) \pm 2\left\langle A_i {{\hat{x}}} + b_i, y\otimes z \right\rangle + \left\langle \mathbb {A}_i, Z \right\rangle \\&= q_i({{\hat{x}}}) + \left\langle \mathbb {A}_i, Z \right\rangle . \end{aligned}$$

Consequently, $({{\hat{x}}} \pm y \otimes z, {{\hat{t}}}, Z - zz^\top )$ satisfies all of the constraints in (19) except possibly the first. We now verify that the first constraint is also satisfied: From

$$\begin{aligned} q_0({{\hat{x}}} \pm y \otimes z) + \left\langle \mathbb {A}_0, Z - zz^\top \right\rangle&= q_0({{\hat{x}}}) \pm 2\left\langle A_0 {{\hat{x}}} + b_0, y\otimes z \right\rangle + \left\langle \mathbb {A}_0, zz^\top \right\rangle \\&\quad + \left\langle \mathbb {A}_0, Z - zz^\top \right\rangle \\&= q_0({{\hat{x}}}) + \left\langle \mathbb {A}_0, Z \right\rangle \pm 2\left\langle A_0 {{\hat{x}}} + b_0, y\otimes z \right\rangle \\&= 2{{\hat{t}}} \pm 2\left\langle A_0 {{\hat{x}}} + b_0, y\otimes z \right\rangle , \end{aligned}$$

we deduce that $({{\hat{x}}}\pm y \otimes z, 2{{\hat{t}}} \pm 2\left\langle A_0 {{\hat{x}}} + b_0, y\otimes z \right\rangle )\in {\mathcal {D}}_\text{ SDP }$. Then, by minimality of ${{\hat{t}}}$ in ${\mathcal {D}}_\text{ SDP }$, we infer that $\left\langle A_0{{\hat{x}}} + b_0, y\otimes z \right\rangle = 0$.

We deduce that $({{\hat{x}}}\pm y\otimes z, {{\hat{t}}}, Z - zz^\top )$ satisfies (19). Furthermore, we have ${{\,\mathrm{rank}\,}}(Z-zz^\top )= r-1$. By induction, $({{\hat{x}}}\pm y\otimes z,{{\hat{t}}})\in {{\,\mathrm{conv}\,}}({\mathcal {D}}\cap H)$. We conclude that $({{\hat{x}}},{{\hat{t}}})\in {{\,\mathrm{conv}\,}}({\mathcal {D}}\cap H)$. $\square$