# Duality

Conic duality is the starting point for MOI's duality conventions. When all functions are affine (or coordinate projections), and all constraint sets are closed convex cones, the model may be called a conic optimization problem.

For a minimization problem in geometric conic form, the primal is:

\begin{align} & \min_{x \in \mathbb{R}^n} & a_0^T x + b_0 \\ & \;\;\text{s.t.} & A_i x + b_i & \in \mathcal{C}_i & i = 1 \ldots m \end{align}

and the dual is a maximization problem in standard conic form:

\begin{align} & \max_{y_1, \ldots, y_m} & -\sum_{i=1}^m b_i^T y_i + b_0 \\ & \;\;\text{s.t.} & a_0 - \sum_{i=1}^m A_i^T y_i & = 0 \\ & & y_i & \in \mathcal{C}_i^* & i = 1 \ldots m \end{align}

where each $\mathcal{C}_i$ is a closed convex cone and $\mathcal{C}_i^*$ is its dual cone.

For a maximization problem in geometric conic form, the primal is:

\begin{align} & \max_{x \in \mathbb{R}^n} & a_0^T x + b_0 \\ & \;\;\text{s.t.} & A_i x + b_i & \in \mathcal{C}_i & i = 1 \ldots m \end{align}

and the dual is a minimization problem in standard conic form:

\begin{align} & \min_{y_1, \ldots, y_m} & \sum_{i=1}^m b_i^T y_i + b_0 \\ & \;\;\text{s.t.} & a_0 + \sum_{i=1}^m A_i^T y_i & = 0 \\ & & y_i & \in \mathcal{C}_i^* & i = 1 \ldots m \end{align}

A linear inequality constraint $a^T x + b \ge c$ is equivalent to $a^T x + b - c \in \mathbb{R}_+$, and $a^T x + b \le c$ is equivalent to $a^T x + b - c \in \mathbb{R}_-$. Variable-wise constraints are affine constraints with the appropriate identity mapping in place of $A_i$.

For the special case of minimization LPs, the MOI primal form can be stated as:

\begin{align} & \min_{x \in \mathbb{R}^n} & a_0^T x &+ b_0 \\ & \;\;\text{s.t.} &A_1 x & \ge b_1\\ && A_2 x & \le b_2\\ && A_3 x & = b_3 \end{align}

By applying the stated transformations to conic form, taking the dual, and transforming back into linear inequality form, one obtains the following dual:

\begin{align} & \max_{y_1,y_2,y_3} & b_1^Ty_1 + b_2^Ty_2 + b_3^Ty_3 &+ b_0 \\ & \;\;\text{s.t.} &A_1^Ty_1 + A_2^Ty_2 + A_3^Ty_3 & = a_0\\ && y_1 &\ge 0\\ && y_2 &\le 0 \end{align}

For maximization LPs, the MOI primal form can be stated as:

\begin{align} & \max_{x \in \mathbb{R}^n} & a_0^T x &+ b_0 \\ & \;\;\text{s.t.} &A_1 x & \ge b_1\\ && A_2 x & \le b_2\\ && A_3 x & = b_3 \end{align}

and similarly, the dual is:

\begin{align} & \min_{y_1,y_2,y_3} & -b_1^Ty_1 - b_2^Ty_2 - b_3^Ty_3 &+ b_0 \\ & \;\;\text{s.t.} &A_1^Ty_1 + A_2^Ty_2 + A_3^Ty_3 & = -a_0\\ && y_1 &\ge 0\\ && y_2 &\le 0 \end{align}
Warning

For the LP case, the signs of the feasible dual variables depend only on the sense of the corresponding primal inequality and not on the objective sense.

## Duality and scalar product

The scalar product is different from the canonical one for the sets PositiveSemidefiniteConeTriangle, LogDetConeTriangle, RootDetConeTriangle.

If the set $C_i$ of the section Duality is one of these three cones, then the rows of the matrix $A_i$ corresponding to off-diagonal entries are twice the value of the coefficients field in the VectorAffineFunction for the corresponding rows. See PositiveSemidefiniteConeTriangle for details.

## Dual for problems with quadratic functions

For quadratic programs with only affine conic constraints,

\begin{align*} & \min_{x \in \mathbb{R}^n} & \frac{1}{2}x^TQ_0x + a_0^T x + b_0 \\ & \;\;\text{s.t.} & A_i x + b_i & \in \mathcal{C}_i & i = 1 \ldots m. \end{align*}

with cones $\mathcal{C}_i \subseteq \mathbb{R}^{m_i}$ for $i = 1, \ldots, m$, consider the Lagrangian function

$$$L(x, y) = \frac{1}{2}x^TQ_0x + a_0^T x + b_0 - \sum_{i = 1}^m y_i^T (A_i x + b_i).$$$

Let $z(y)$ denote $\sum_{i = 1}^m A_i^T y_i - a_0$, the Lagrangian can be rewritten as

$$$L(x, y) = \frac{1}{2}{x}^TQ_0x - z(y)^T x + b_0 - \sum_{i = 1}^m y_i^T b_i.$$$

The condition $\nabla_x L(x, y) = 0$ gives

$$$0 = \nabla_x L(x, y) = Q_0x + a_0 - \sum_{i = 1}^m y_i^T b_i$$$

which gives $Q_0x = z(y)$. This allows to obtain that

$$$\min_{x \in \mathbb{R}^n} L(x, y) = -\frac{1}{2}x^TQ_0x + b_0 - \sum_{i = 1}^m y_i^T b_i$$$

so the dual problem is

$$$\max_{y_i \in \mathcal{C}_i^*} \min_{x \in \mathbb{R}^n} -\frac{1}{2}x^TQ_0x + b_0 - \sum_{i = 1}^m y_i^T b_i.$$$

If $Q_0$ is invertible, we have $x = Q_0^{-1}z(y)$ hence

$$$\min_{x \in \mathbb{R}^n} L(x, y) = -\frac{1}{2}z(y)^TQ_0^{-1}z(y) + b_0 - \sum_{i = 1}^m y_i^T b_i$$$

so the dual problem is

$$$\max_{y_i \in \mathcal{C}_i^*} -\frac{1}{2}z(y)^TQ_0^{-1}z(y) + b_0 - \sum_{i = 1}^m y_i^T b_i.$$$

\begin{align*} & \min_{x \in \mathbb{R}^n} & \frac{1}{2}x^TQ_0x + a_0^T x + b_0 \\ & \;\;\text{s.t.} & \frac{1}{2}x^TQ_ix + a_i^T x + b_i & \in \mathcal{C}_i & i = 1 \ldots m. \end{align*}

with cones $\mathcal{C}_i \subseteq \mathbb{R}$ for $i = 1 \ldots m$, consider the Lagrangian function

$$$L(x, y) = \frac{1}{2}x^TQ_0x + a_0^T x + b_0 - \sum_{i = 1}^m y_i (\frac{1}{2}x^TQ_ix + a_i^T x + b_i)$$$

A pair of primal-dual variables $(x^\star, y^\star)$ is optimal if

• $x^\star$ is a minimizer of$$$\min_{x \in \mathbb{R}^n} L(x, y^\star).$$$That is,$$$0 = \nabla_x L(x, y^\star) = Q_0x + a_0 - \sum_{i = 1}^m y_i^\star (Q_ix + a_i).$$$
• and $y^\star$ is a maximizer of$$$\max_{y_i \in \mathcal{C}_i^*} L(x^\star, y).$$$That is, for all $i = 1, \ldots, m$, $\frac{1}{2}x^TQ_ix + a_i^T x + b_i$ is either zero or in the normal cone of $\mathcal{C}_i^*$ at $y^\star$. For instance, if $\mathcal{C}_i$ is $\{ z \in \mathbb{R} : z \le 0 \}$, this means that if $\frac{1}{2}x^TQ_ix + a_i^T x + b_i$ is nonzero at $x^\star$ then $y_i^\star = 0$. This is the classical complementary slackness condition.

If $\mathcal{C}_i$ is a vector set, the discussion remains valid with $y_i(\frac{1}{2}x^TQ_ix + a_i^T x + b_i)$ replaced with the scalar product between $y_i$ and the vector of scalar-valued quadratic functions.

## Dual for square semidefinite matrices

The set PositiveSemidefiniteConeTriangle is a self-dual. That is, querying ConstraintDual of a PositiveSemidefiniteConeTriangle constraint returns a vector that is itself a member of PositiveSemidefiniteConeTriangle.

However, the dual of PositiveSemidefiniteConeSquare is not so straight forward. This section explains the duality convention we use, and how it is derived.

tl;dr

If you have a PositiveSemidefiniteConeSquare constraint, the result matrix $A$ from ConstraintDual is not positive semidefinite. However, $A + A^\top$ is positive semidefinite.

Let $\mathcal{S}_+$ be the cone of symmetric semidefinite matrices in the $\frac{n(n+1)}{2}$ dimensional space of symmetric $\mathbb{R}^{n \times n}$ matrices. That is, $\mathcal{S}_+$ is the set PositiveSemidefiniteConeTriangle. It is well known that $\mathcal{S}_+$ is a self-dual proper cone.

Let $\mathcal{P}_+$ be the cone of symmetric semidefinite matrices in the $n^2$ dimensional space of $\mathbb{R}^{n \times n}$ matrices. That is $\mathcal{P}_+$ is the set PositiveSemidefiniteConeSquare.

In addition, let $\mathcal{D}_+$ be the cone of matrices $A$ such that $A+A^\top \in \mathcal{P}_+$.

$\mathcal{P}_+$ is not proper because it is not solid (it is not $n^2$ dimensional), so it is not necessarily true that $\mathcal{P}_+^{**} = \mathcal{P}_+$.

However, this is the case, because we will show that $\mathcal{P}_+^{*} = \mathcal{D}_+$ and $\mathcal{D}_+^{*} = \mathcal{P}_+$.

First, let us see why $\mathcal{P}_+^{*} = \mathcal{D}_+$.

If $B$ is symmetric, then

$$$\langle A,B \rangle = \langle A^\top, B^\top \rangle = \langle A^\top, B\rangle$$$

so

$$$2\langle A, B \rangle = \langle A, B \rangle + \langle A^\top, B \rangle = \langle A + A^\top , B \rangle.$$$

Therefore, $\langle A,B\rangle \ge 0$ for all $B \in \mathcal{P}_+$ if and only if $\langle A+A^\top,B\rangle \ge 0$ for all $B \in \mathcal{P}_+$. Since $A+A^\top$ is symmetric, and we know that $\mathcal{S}_+$ is self-dual, we have shown that $\mathcal{P}_+^{*}$ is the set of matrices $A$ such that $A+A^\top \in \mathcal{P}_+$.

Second, let us see why $\mathcal{D}_+^{*} = \mathcal{P}_+$.

Since $A \in \mathcal{D}_+$ implies that $A^\top \in \mathcal{D}_+$, $B \in \mathcal{D}_+^{*}$ means that $\langle A+A^\top,B\rangle \ge 0$ for all $A \in \mathcal{D}_+$, and hence $B \in \\mathcal{P}_+$.

To see why it should be symmetric, simply notice that if $B_{i,j} < B_{j,i}$, then $\langle A,B\rangle$ can be made arbitrarily small by setting $A_{i,j} = A_{i,j} + s$ and $A_{j,i} = A_{j,i} - s$, with $s$ arbitrarily large, and $A$ stays in $\mathcal{D}_+$ because $A+A^\top$ does not change.

Typically, the primal/dual pair for semidefinite programs is presented as:

\begin{align} \min & \langle C, X \rangle \\ \text{s.t.} \;\; & \langle A_k, X\rangle = b_k \forall k \\ & X \in \mathcal{S}_+ \end{align}

with the dual

\begin{align} \max & \sum_k b_k y_k \\ \text{s.t.} \;\; & C - \sum A_k y_k \in \mathcal{S}_+ \end{align}

If we allow $A_k$ to be non-symmetric, we should instead use:

\begin{align} \min & \langle C, X \rangle \\ \text{s.t.} \;\; & \langle A_k, X\rangle = b_k \forall k \\ & X \in \mathcal{D}_+ \end{align}

with the dual

\begin{align} \max & \sum b_k y_k \\ \text{s.t.} \;\; & C - \sum A_k y_k \in \mathcal{P}_+ \end{align}

This is implemented as:

\begin{align} \min & \langle C, Z \rangle + \langle C - C^\top, S \rangle \\ \text{s.t.} \;\; & \langle A_k, Z \rangle + \langle A_k - A_k^\top, S \rangle = b_k \forall k \\ & Z \in \mathcal{S}_+ \end{align}

with the dual

\begin{align} \max & \sum b_k y_k \\ \text{s.t.} \;\; & C+C^\top - \sum (A_k+A_k^\top) y_k \in \mathcal{S}_+ \\ & C-C^\top - \sum(A_k-A_k^\top) y_k = 0 \end{align}

and we recover $Z = X + X^\top$.