A new approximation hierarchy for polynomial conic optimization

Abstract

In this paper we consider polynomial conic optimization problems, where the feasible set is defined by constraints in the form of given polynomial vectors belonging to given nonempty closed convex cones, and we assume that all the feasible solutions are non-negative. This family of problems captures in particular polynomial optimization problems (POPs), polynomial semi-definite polynomial optimization problems (PSDPs) and polynomial second-order cone-optimization problems (PSOCPs). We propose a new general hierarchy of linear conic optimization relaxations inspired by an extension of Pólya’s Positivstellensatz for homogeneous polynomials being positive over a basic semi-algebraic cone contained in the non-negative orthant, introduced in Dickinson and Povh (J Glob Optim 61(4):615–625, 2015). We prove that based on some classic assumptions, these relaxations converge monotonically to the optimal value of the original problem. Adding a redundant polynomial positive semi-definite constraint to the original problem drastically improves the bounds produced by our method. We provide an extensive list of numerical examples that clearly indicate the advantages and disadvantages of our hierarchy. In particular, in comparison to the classic approach of sum-of-squares, our new method provides reasonable bounds on the optimal value for POPs, and strong bounds for PSDPs and PSOCPs, even outperforming the sum-of-squares approach in these latter two cases.

Appendix: Proofs of Example 4.4

Lemma A.1

We have ${{\,\mathrm{val}\,}}$(18)$=n-1$.

Proof

If we consider $\mathbf {x}=2\mathbf {e}_1$ then we have $G(2\mathbf {e}_1)=\mathrm {Diag}(0,1,\ldots ,1)\in \mathcal {PSD}^{n}$ and thus ${{\,\mathrm{val}\,}}$(18) $\le \mathrm {trace}~G(2\mathbf {e}_1) = n-1$. Now assume for the sake of contradiction that ${{\,\mathrm{val}\,}}$(18)$< n-1$. This is possible only if there exists $\mathbf {x}\in \mathbb {R}_+^n$ such that at least two of the on-diagonal entries of $G(\mathbf {x})$ are strictly less than one (otherwise, we have $n-1$ diagonal entries that are at least 1, and the remaining is nonnegative, hence $\mathrm {trace}~G(\mathbf {x})\ge n-1$). Without loss of generality the first two on-diagonal entries of $G(\mathbf {x})$ are strictly less than one, implying that $x_1,x_2> \frac{4}{3}$. The positive semi-definiteness of G implies that the determinant of the submatrix of G corresponding of the first two rows and columns is non-negative, but

$$\begin{aligned} G_{1,1}(\mathbf {x})\cdot G_{2,2}(\mathbf {x})-G_{1,2}(\mathbf {x})^2< 1 - \tfrac{16}{9}<0, \end{aligned}$$

which is a contradiction. Hence, ${{\,\mathrm{val}\,}}$(18)$=n-1$. $\square$

Considering the approximation hierarchies discussed in this paper for problem (18), we have the following results:

Lemma A.2

The optimal value of our new hierarchy, (15) (without the additional constraint (8) in the original problem), for $r\in \{0,1\}$ is equal to zero.

Proof

Our hierarchy, (15), for $r=0$ and $r=1$ reduces to the following respectively:

$$\begin{aligned} \max _{\lambda ,Y} \quad&\lambda \nonumber \\ \text {s.t.}\quad&\mathrm {trace}\, G(\mathbf {x}) - \lambda \ge _c \langle Y, G(\mathbf {x})\rangle \end{aligned}$$

(21)

$$\begin{aligned}&Y\in \mathcal {PSD}^{n},\nonumber \\ \max _{\lambda ,\{Y\}} \quad&\lambda \nonumber \\ \text {s.t.}\quad&(1+\mathbf {e}^\mathsf {T}\mathbf {x})(\mathrm {trace}\, G(\mathbf {x}) - \lambda ) \ge _c \langle Y_0, G(\mathbf {x})\rangle + \sum _{i=1}^n x_i \langle Y_i, G(\mathbf {x})\rangle \nonumber \\&Y_i\in \mathcal {PSD}^{n} \qquad \text {for all}\quad i=0,\ldots ,n. \end{aligned}$$

(22)

Considering $\lambda =0$ and Y equal to the identity matrix gives us a feasible point of (21), and thus ${{\,\mathrm{val}\,}}$(21)$\ge 0$.

From Proposition 3.3 we have ${{\,\mathrm{val}\,}}$(21)$\le {{\,\mathrm{val}\,}}$(22). We will now complete the proof by showing that ${{\,\mathrm{val}\,}}$(22)$\le 0$.

For an arbitrary feasible point of (22), we consider the following coefficient inequalities explicitly:

This implies that for all i we have $(Y_0)_{ii} \ge \tfrac{4}{3} (Y_i)_{ii} - \tfrac{1}{3} \ge \tfrac{4}{3}-\tfrac{1}{3} = 1,$ which in turn implies that $\lambda \le n - \mathrm {trace}\, Y_0 \le n-n = 0$, and thus ${{\,\mathrm{val}\,}}$(22)$\le 0$. $\square$

Lemma A.3

The optimal value of both of our new hierarchy, (16) (with the additional constraint (8) in the original problem), and the Matrix SOS hierarchy, (14), are equal to zero for $r=0$.

Proof

These problems both reduce to the following problem:

$$\begin{aligned} \max _{\lambda ,Y,v,\mathbf {w},Z} \quad&\lambda \nonumber \\ \text {s.t.}\quad&\mathrm {trace}\, G(\mathbf {x}) - \lambda \ge _c \langle Y, G(\mathbf {x})\rangle + \left\langle \begin{pmatrix} v &{} \mathbf {w}^\mathsf {T}\\ \mathbf {w}&{} Z \end{pmatrix}, \begin{pmatrix} 1 &{} \mathbf {x}^\mathsf {T}\\ \mathbf {x}&{} \mathbf {x}\mathbf {x}^\mathsf {T}\end{pmatrix} \right\rangle \nonumber \\&Y\in \mathcal {PSD}^{n},\quad \begin{pmatrix} v &{} \mathbf {w}^\mathsf {T}\\ \mathbf {w}&{} Z \end{pmatrix}\in \mathcal {PSD}^{n+1}. \end{aligned}$$

(23)

Considering $\lambda =v=0$, $\mathbf {w}=\varvec{0}$, $Z=0$ and Y equal to the identity matrix gives us a feasible point of (23), and thus ${{\,\mathrm{val}\,}}$(23)$\ge 0$.

For an arbitrary feasible point of(23), we consider the following coefficient inequalities explicitly:

Considering the positive semi-definiteness constraints, we also have $0 \le v$ and $0\le z_{ii}$ for all i. Therefore, $y_{ii} \ge 1 + \tfrac{4}{3}z_{ii} \ge 1$ for all i and $\lambda \le n-\mathrm {trace}\, Y - v \le n - n-0 =0$, implying that ${{\,\mathrm{val}\,}}$(23)$\le 0$. $\square$

Lemma A.4

The optimal value of the Matrix SOS hierarchy, (14), is equal to zero for $r=1$.

Proof

This problem reduces to the following problem:

$$\begin{aligned} \max _{\lambda ,Y,\{v_i\},\{\mathbf {w}_i\},\{Z_i\}} \quad&\lambda \\ \text {s.t.}\quad&\mathrm {trace}\, G(\mathbf {x}) - \lambda =_c \langle Y, G(\mathbf {x})\rangle + \left\langle \begin{pmatrix} v_0 &{} \mathbf {w}_0^\mathsf {T}\\ \mathbf {w}_0 &{} Z_0 \end{pmatrix}, \begin{pmatrix} 1 &{} \mathbf {x}^\mathsf {T}\\ \mathbf {x}&{} \mathbf {x}\mathbf {x}^\mathsf {T}\end{pmatrix} \right\rangle +\cdots \\&\cdots + \sum _{i=1}^n x_i \left\langle \begin{pmatrix} v_i &{} \mathbf {w}_i^\mathsf {T}\\ \mathbf {w}_i &{} Z_i \end{pmatrix}, \begin{pmatrix} 1 &{} \mathbf {x}^\mathsf {T}\\ \mathbf {x}&{} \mathbf {x}\mathbf {x}^\mathsf {T}\end{pmatrix} \right\rangle \\&Y\in \mathcal {PSD}^{n},\qquad \begin{pmatrix} v_i &{} \mathbf {w}_i^\mathsf {T}\\ \mathbf {w}_i &{} Z_i \end{pmatrix}\in \mathcal {PSD}^{n+1} \quad \text {for } i=0,\ldots ,n. \end{aligned}$$

For $i,j\in \{1,\ldots ,n\}$ such that $i=j$, considering the coefficient equalities for $x_i^3$ and $x_i^2x_j$ explicitly, for any feasible point to this problem we have $0=(Z_i)_{ii}$ and $0=2(Z_i)_{ij}+(Z_i)_{jj}$. As $Z_i\in \mathcal {PSD}^{n}$, this implies that $Z_i$ is equal to the zero matrix for all i. Therefore, this problem reduces to the problem (23), which we have already shown to have an optimal value equal to zero. $\square$