# Different forms of primal-dual second-order cone program optimization problems

I'm trying to understand the difference between the following two definitions of a SOCP (second-order cone program). The first way I've seen a primal-dual SOCP define is as follows:

The primal problem is, \begin{aligned} &\min_x && c^T x \\ &\text{s.t.} && Ax=b \\ &&& x \in K \end{aligned} \tag{P1}

and dual problem, \begin{aligned} &\max_y && b^T y \\ &\text{s.t.} && A^Ty + s = c \\ &&& s \in K^* \end{aligned} \tag{D1}

where $K$ and $K^{*}$ are second-order cones. The second way I've seen an SOCP defined is as follows (as given under the "Linear Cone Programs" section of the following webpage: http://cvxopt.org/userguide/coneprog.html):

The primal problem is, \begin{aligned} &\min_x && c^T x \\ &\text{s.t.} && Ax=b \\ &&& Gx + s = h \\ &&& s \in K \end{aligned} \tag{P2}

and the dual problem is, \begin{aligned} &\max_{z,y} &&-h^{T}z-b^{T}y \\ &\text{s.t.} && G^{T}z+A^{T}y+c=0 \\ &&& z \in K^{*} \end{aligned} \tag{D2}

How exactly can the above two primal-dual problems define the same thing? I'm rather confused, and any help would be much appreciated.

• There's something wrong with your third problem above---it's missing an objective function. – Michael Grant Feb 16 '18 at 19:41
• Why do we only require $z \in K^*$ in D2? I would have thought there'd be three dual variables, $z, y, w$ where $w$ is the dual variable corresponding to $s \in K$, so that (if I'm not mistaken) the Lagrangian would be $\langle x, G^T z + A^T y + c, x \rangle + \langle s, z + w \rangle - z^T h - y^T b$? – jjjjjj Feb 17 '18 at 16:46

Just looking at P1 and P2, consider taking $G = -I$ and $h=0$. Thus, the the second constraint in P2 gives $$-Ix + s = 0 \iff x = s$$ so now we can formulate P2 as \begin{align} & \underset{x}{\text{minimize}} && c^T x \\ & \text{subject to} && Ax = b \\ &&& x \succeq_K 0 \end{align}