The Problems Solved

The primal and dual pair of conic optimization problems over a self-dual cone are defined as

$$\begin{array}{lrccrcr} & \min & \langle c, x \rangle & \hsp... ...= b & \hspace{.75cm} & & \mathcal{A}^* y + z = c & \end{array}$$

where

• $K$ is a closed, convex cone in a euclidean space $X$.
• $\mathcal{A}: X \rightarrow \mathbb{R}^{m}$ is a linear operator, and $\mathcal{A}^*$ is its adjoint.
• $b \in \mathbb{R}^{m}$, and $c \in X$.

In the case of a semidefinite-quadratic-linear program these are defined as follows:

• The space $X$: $x \in X \Leftrightarrow x = ( x^s_1, \dots, x^s_{n_s}, x^q_1, \dots, x^q_{n_q}, x^\ell)$, where
  • $x^s_1, \dots, x^s_{n_s}$ are symmetric matrices (possibly of various sizes).
  • $x^q_1, \dots, x^q_{n_q}$ are vectors (again, possibly of various sizes).
  • $x^\ell$ is a vector.
• The cone $K$: $x \in K \Leftrightarrow x^s_j \succeq 0 \; \forall j, \; x^q_k \geq_q 0 \; \forall k, \; x^\ell \geq 0$, where
  • $u \succeq 0$ means that the symmetric matrix $u$ is positive semidefinite.
  • $v \geq_q 0$ means that the vector $v$ is in a quadratic cone (also known as the second-order cone, Lorentz cone or ice cream cone) of appropriate size. That is, if $v \in \mathbb{R}^{k}$, then $v \geq_q 0 \Leftrightarrow v_1 \geq \vert\!\vert v_{2:k} \vert\!\vert _2$.
  • $w \geq 0$ means that the vector $w$ is componentwise nonnegative.
• The inner product $\langle . , . \rangle$: For $x, z \in K$
  $$\langle x, z \rangle = \sum_{j=1}^{n_s} x^s_j \bullet z^s_j + \sum_{k=1}^{n_q} x^q_k * z^q_k + x^\ell * z^\ell,$$
  where
  • For matrices $a$ and $b$, $a \bullet b = {\rm trace }~a^{T}b = \sum_{i,j} a_{ij} b_{ij}$.
  • For vectors $a$ and $b$, $a * b = a^T b = \sum_{i} a_{i} b_{i}$.

Thus (P) and (D) become

$$\begin{array}{rlllllllllllll} \min & \sum_{j=1}^{n_s} c^s_j \... ...^q_k \geq_q 0 \; \forall k & \quad x^\ell \geq 0 \\ \end{array} SQLP-P$$

and

$$\begin{array}{rllllllllll} \max & \sum_{i=1}^{m} b_i y_i \\ {... ...ceq 0 & \quad z^q_k \geq_q 0&\quad z^\ell \geq 0 \\ \end{array} SQLP-D$$

Thus the feasible set is a product of semidefinite, quadratic and nonnegative orthant cones, intersected with an affine subspace. It is possible that one or more of the three parts of the problem is absent, i.e., any of $n_s$, $n_q$, or the length of $x^\ell$ may be zero.

The rotated quadratic cone is defined as

$$\{ \, v \in \mathbb{R}^{k} \, \vert \, v_1 v_2 \geq \vert\!\vert v_{3:k} \vert\!\vert \, \}.$$

It is simply a rotation of the usual quadratic cone, but for the purpose of modeling quadratic inequalities, it is more convenient to use, thus several participating codes support this cone.