the linear constraint ${\textbf{u}}- {\textbf{s}}= {\textbf{0}}$ in the new augmented Lagrangian, the new subproblem with respect to $({\textbf{u}},{\textbf{s}})$ is $$\begin{gathered}
\label{eq:ksvm_admm_ussub}
    \min_{{\textbf{u}},{\textbf{s}}\geq {\textbf{0}}} \quad \frac{\rho_1}{2}\|B^T{\textbf{u}}+ {\textbf{w}}^{k+1}\|_2^2 + \gamma_3^k{\textbf{d}}^T{\textbf{u}}\\
    + \frac{\mu_3}{2}\|{\textbf{d}}^T{\textbf{u}}- b^{k+1} + 1 - q^k\|_2^2 + \gamma_5^T({\textbf{s}}-{\textbf{u}}) + \frac{\mu_5}{2}\|{\textbf{u}}-{\textbf{s}}\|_2^2.\end{gathered}$$ Given an ${\textbf{s}}^k \geq {\textbf{0}}$, we can compute ${\textbf{u}}^{k+1}$ by solving a $k_1 \times k_1$ linear system $$\label{eq:ksvm_admm_linsys_u}
    (\rho_1 BB^T + \mu_3 {\textbf{d}}{\textbf{d}}^T + \mu_5 I){\textbf{u}}^{k+1} = \textbf{r}_{\textbf{u}},$$ where $\textbf{r}_{\textbf{u}}= -\rho_1B{\textbf{w}}^{k+1} + \mu_3 {\textbf{d}}b^{k+1} + \mu_3{\textbf{d}}(q^k-1) - {\textbf{d}}\gamma_3^k + \gamma_5 + \mu_5{\textbf{s}}^k$. We assume that $B$ has full row-rank. This is a reasonable assumption since otherwise there is at least one redundant domain knowledge constraint and we can simply remove it. The number of domain knowledge constraints ($k_1$ and $k_2$) are usually small, so the system (\[eq:ksvm\_admm\_linsys\_u\]) can be solved exactly and efficiently by Cholesky factorization.

Solving for ${\textbf{s}}^{k+1}$ corresponding to ${\textbf{u}}^{k+1}$ is easy, observing that problem (\[eq:ksvm\_admm\_ussub\]) is separable in the elements of ${\textbf{s}}$. For each element $s_i$, the optimal solution to the one-dimensional quadratic problem with a non-negative constraint on $s_i$ is given by $\max(0,u_i - \frac{(\gamma_5)_i}{\mu_5})$. Writing in the vector form, ${\textbf{s}}^{k+1} = \max({\textbf{0}}, {\textbf{u}}^{k+1} - \frac{\gamma_5^k}{\mu_5})$. Similarly, we solve for ${\textbf{v}}^{k+1}$ by introducing a non-negative slack variable 