L1 merit globalization with finite penalty and backtracking budgets

At $x\in U$, choose a positive-definite SQP model matrix and solve the exact quadratic subproblem by Exact exhaustive active-set solve for the SQP quadratic subproblem. If it reports infeasibility or failure, return that status. On a solution return, retain its step $p$, full QP multiplier vectors $({\nu }^{QP},{\lambda }^{QP})$, retained equality basis $E$, and selected inequality support $J$. The stacked rows $D{h}_E(x)$ and $D{g}_J(x)$ are independent, and the selected rows satisfy $h_E(x)+D{h}_E(x)p=0$ and $g_J(x)+D{g}_J(x)p=0$. Given parameters ${\rho }_0>0$, $\gamma >1$, $\beta \in (0,1)$, $\sigma \in (0,1)$ and finite integer caps $R,J_{\max }\ge 1$, keep the resulting step $p$ fixed, initialize flags $P_{\mathrm{p}\mathrm{o}\mathrm{s}}=\mathrm{f}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{e}$ and $P_{\mathrm{d}\mathrm{o}\mathrm{m}}=\mathrm{f}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{e}$, and perform attempts $r=0,\dots ,R-1$ with ${\rho }_r={\rho }_0{\gamma }^r$:

1. Compute $D_r=D_{{\rho }_r}(x,p)$ from L1 exact-penalty merit and predicted decrease for SQP. If $D_r\le 0$, go to the next penalty. Otherwise set $P_{\mathrm{p}\mathrm{o}\mathrm{s}}=\mathrm{t}\mathrm{r}\mathrm{u}\mathrm{e}$.
2. If $x+p\in U$, apply Minimum-norm second-order correction for identified active constraints to the auxiliary program with equality map $h_E$ and the same $f,g,U,B$, selecting inequalities $J$; thus its complete active map is $c_A=(h_E,g_J)$. The original QP step $p$ also solves this auxiliary program's QP, since the consistent equality-row reduction preserves the linearized feasible set. This call makes no residual-order claim for omitted nonlinear equality rows; every merit evaluation below still uses the full original $h,g$. If it returns a corrected displacement $d$ and $x+d\in U$, set $P_{\mathrm{d}\mathrm{o}\mathrm{m}}=\mathrm{t}\mathrm{r}\mathrm{u}\mathrm{e}$ and test ${\psi }_{{\rho }_r}(x+d)\le {\psi }_{{\rho }_r}(x)-\sigma D_r$. If it passes, return accepted-soc with $x+d$, $p$, $({\nu }^{QP},{\lambda }^{QP})$, $E$, and $J$. If $x+p\mathrm{\notin }U$, skip the correction call.
3. For $j=0,\dots ,J_{\max }-1$, set $t_j={\beta }^j$. Whenever $x+t_{j}p\in U$, set $P_{\mathrm{d}\mathrm{o}\mathrm{m}}=\mathrm{t}\mathrm{r}\mathrm{u}\mathrm{e}$ and test ${\psi }_{{\rho }_r}(x+t_{j}p)\le {\psi }_{{\rho }_r}(x)-\sigma t_j{D}_r\mathrm{.}$ On the first passing test, return accepted with $x+t_{j}p$, $t_j$, $p$, $({\nu }^{QP},{\lambda }^{QP})$, $E$, and $J$.

After all attempts, if $P_{\mathrm{p}\mathrm{o}\mathrm{s}}$ is false, return nonpositive-predicted-decrease-budget. If $P_{\mathrm{p}\mathrm{o}\mathrm{s}}$ is true but $P_{\mathrm{d}\mathrm{o}\mathrm{m}}$ is false, return domain-exit: no SOC or ordinary trial could be evaluated in $U$. Otherwise return line-search-budget-exhausted. These finite tests do not guarantee acceptance or convergence.