app7-regularization.tex

%-------------------------------------------------------------------------------
\chapter{Solving a hierarchy with regularization}
\label{app.regularization}
\acresetall
%-------------------------------------------------------------------------------

Consider a simple hierarchy
%
\begin{hierarchy}[hr.reg]
    \level $\ubarV{\objb}_1 \le \objA_1 \x \le \barV{\objb}_1$
    \level $\ubarV{\objb}_2 \le \objA_2 \x \le \barV{\objb}_2$
    \level $\objA_3 \x = \V{\objb}_3$
\end{hierarchy}
%
which must be solved with regularization of the objective on the second level
using the equality objective on the last level. We can solve this hierarchy in
two steps using a solver, which does not support regularization.


The first step is to solve the following hierarchy consisting of two levels
%
\begin{hierarchy}[hr.reg_step1]
    \level $\ubarV{\objb}_1 \le \objA_1 \x \le \barV{\objb}_1$
    \level $\ubarV{\objb}_2 \le \objA_2 \x \le \barV{\objb}_2$\\
           $\gamma_r \objA_3 \x = \gamma_r \V{\objb}_3$
\end{hierarchy}
%
where $\gamma_r$ is regularization factor. Solution of \cref{hr.reg_step1}
yields the vector of violations $\violation_2^{\star}$ of the second objective:
%
\begin{equation}
    \ubarV{\objb}_2 \le \objA_2 \x - \violation_2^{\star} \le \barV{\objb}_2.
\end{equation}
%
We decompose $\violation_2^{\star}$ into three components
%
\begin{equation}
    \violation_2^{\star} = \ubar{\violation}_2^{\star} + \tilde{\violation}_2^{\star} + \bar{\violation}_2^{\star}
\end{equation}
%
Vector $\ubar{\violation}_2^{\star}$ contains violations of the lower bounds
and zeros for the constraints, where lower bounds are satisfied. Vectors
$\tilde{\violation}_2^{\star}$ and $\bar{\violation}_2^{\star}$ are formed in
the same way for equality and inequality constraints.


The second step is to solve a hierarchy with relaxed bounds on the second
level:
%
\begin{hierarchy}[hr.reg_step2]
    \level $\ubarV{\objb}_1 \le \objA_1 \x \le \barV{\objb}_1$
    \level $\ubarV{\objb}_2 + \ubar{\violation}_2^{\star} + \tilde{\violation}_2^{\star}
            \le
            \objA_2 \x
            \le
            \barV{\objb}_2 + \bar{\violation}_2^{\star} + \tilde{\violation}_2^{\star}$
    \level $\objA_3 \x = \V{\objb}_3$
\end{hierarchy}
%


The procedure is easily generalized to a larger number of levels with and
without regularization. The number of hierarchies, which must be solved, is
equal to the number of regularized levels plus one.