A second-order cone program (SOCP) is an optimization problem of the form $$ \begin{array}{ll} \mbox{minimize} & f^Tx\\ \mbox{subject to} & \|A_ix + b_i\|_2 \leq c_i^Tx + d_i, \quad i=1,\ldots,m \\ & Fx = g, \end{array} $$ where $x \in \mathcal{R}^{n}$ is the optimization variable and $f \in \mathcal{R}^n$, $A_i \in \mathcal{R}^{n_i \times n}$, $b_i \in \mathcal{R}^{n_i}$, $c_i \in \mathcal{R}^n$, $d_i \in \mathcal{R}$, $F \in \mathcal{R}^{p \times n}$, and $g \in \mathcal{R}^p$ are problem data.
An example of an SOCP is the robust linear program $$ \begin{array}{ll} \mbox{minimize} & c^Tx\\ \mbox{subject to} & (a_i + u_i)^Tx \leq b_i \textrm{ for all } \|u_i\|_2 \leq 1, \quad i=1,\ldots,m, \end{array} $$ where the problem data $a_i$ are known within an $\ell_2$-norm ball of radius one. The robust linear program can be rewritten as the SOCP $$ \begin{array}{ll} \mbox{minimize} & c^Tx\\ \mbox{subject to} & a_i^Tx + \|x\|_2 \leq b_i, \quad i=1,\ldots,m, \end{array} $$
When we solve a SOCP, in addition to a solution $x^\star$, we obtain a dual solution $\lambda_i^\star$ corresponding to each second-order cone constraint. A non-zero $\lambda_i^\star$ indicates that the constraint $ \|A_ix + b_i\|_2 \leq c_i^Tx + d_i$ holds with equality for $x^\star$ and suggests that changing $d_i$ would change the optimal value.
In the following code, we solve an SOCP with CVXPY.
# Import packages.
import cvxpy as cp
import numpy as np
# Generate a random feasible SOCP.
m = 3
n = 10
p = 5
n_i = 5
np.random.seed(2)
f = np.random.randn(n)
A = []
b = []
c = []
d = []
x0 = np.random.randn(n)
for i in range(m):
A.append(np.random.randn(n_i, n))
b.append(np.random.randn(n_i))
c.append(np.random.randn(n))
d.append(np.linalg.norm(A[i]@x0 + b, 2) - c[i].T@x0)
F = np.random.randn(p, n)
g = F@x0
# Define and solve the CVXPY problem.
x = cp.Variable(n)
# We use cp.SOC(t, x) to create the SOC constraint ||x||_2 <= t.
soc_constraints = [
cp.SOC(c[i].T@x + d[i], A[i]@x + b[i]) for i in range(m)
]
prob = cp.Problem(cp.Minimize(f.T@x),
soc_constraints + [F@x == g])
prob.solve()
# Print result.
print("The optimal value is", prob.value)
print("A solution x is")
print(x.value)
for i in range(m):
print("SOC constraint %i dual variable solution" % i)
print(soc_constraints[i].dual_value)
The optimal value is -9.582695716265503 A solution x is [ 1.40303325 2.4194569 1.69146656 -0.26922215 1.30825472 -0.70834842 0.19313706 1.64153496 0.47698583 0.66581033] SOC constraint 0 dual variable solution [ 0.61662526 0.35370661 -0.02327185 0.04253095 0.06243588 0.49886837] SOC constraint 1 dual variable solution [ 0.35283078 -0.14301082 0.16539699 -0.22027817 0.15440264 0.06571645] SOC constraint 2 dual variable solution [ 0.86510445 -0.114638 -0.449291 0.37810251 -0.6144058 -0.11377797]