Add proportional greedy rule

This commit is contained in:
Tobias Eidelpes 2020-05-16 15:26:09 +02:00
parent f23706b93b
commit f4a59983fc

View File

@ -295,6 +295,36 @@ voters that are represented by the subset.
voters $v_1$ and $v_5$ and project $p_4$ voters $v_2$, $v_3$ and $v_4$. voters $v_1$ and $v_5$ and project $p_4$ voters $v_2$, $v_3$ and $v_4$.
\end{example} \end{example}
The third rule, which places a heavy emphasis on cost versus benefit, is similar
to the greedy rule but instead of disregarding the satisfaction per cost that a
project provides, it seeks to maximize the sum of satisfaction divided by cost
for a project $p\in P$:
\begin{equation}
\frac{\sum_{v\in V}sat(P_v,A\cup\{p\}) - \sum_{v\in V}sat(P_v,A)}{c(p)}
\end{equation}
\textcite{talmonFrameworkApprovalBasedBudgeting2019} call this type of
aggregation rule \emph{proportional greedy rule}. Example~\ref{ex:prop greedy}
shows how the outcome of a budgeting scenario might look like compared to using
a simple greedy rule or a max rule. Since the proportional greedy rule is a
variation of the simple greedy rule, it is therefore also solvable in polynomial
time. The variation of computing the satisfaction per unit of cost does not
change the complexity since it only adds an additional step which can be done in
constant time.
\begin{example}\label{ex:prop greedy}
We again have the same set of projects $P = \{ p_2,p_3,p_4,p_5,p_6 \}$, the
same budget limit of $B = 10$ and the five voters: $v_1 = \{ p_2,p_5,p_6
\}$, $v_2 = \{ p_2, p_3,p_4,p_5 \}$, $v_3 = \{ p_3,p_4,p_5 \}$, $v_4 = \{
p_4,p_5 \}$ and $v_5 = \{ p_6 \}$. If we combine the satisfaction function
$sat_\#$ from equation~\ref{eq:3} with the proportional greedy rule, we get
the same result as with the simple greedy rule of $\{ p_4,p_5 \}$. While the
simple greedy rule selects first $p_5$ and then $p_4$, the proportional
greedy rule first selects $p_4$ and then $p_5$. The rule
$\mathcal{R}_{sat_\$}^p$ yields the same result as $\mathcal{R}_{sat_\$}^g$
and $\mathcal{R}_{sat_\$}^m$ of $\{ p_4,p_5 \}$. $\mathcal{R}_{sat_{0/1}}^p$
however, gives $\{ p_2,p_3,p_4 \}$.
\end{example}
\printbibliography \printbibliography
\end{document} \end{document}