move placement of example box

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hannes.kuchelmeister
2020-05-09 10:56:15 +02:00
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@@ -149,45 +149,6 @@ They select the current state at the beginning of the process. Then users repeat
The case study used in this thesis is a simplified version from forestry \todo[]{hier evtl ergänzen: wo kommt der Use Case her / aus welchem Forschungsprojekt / warum ist er interessant?}. The case study used in this thesis is a simplified version from forestry \todo[]{hier evtl ergänzen: wo kommt der Use Case her / aus welchem Forschungsprojekt / warum ist er interessant?}.
The used characteristics and attributes are shown in \autoref{fig:Concept:ForestExample}. Additionally, as examples preferences, a configuration state and a finished configuration are given. The used characteristics and attributes are shown in \autoref{fig:Concept:ForestExample}. Additionally, as examples preferences, a configuration state and a finished configuration are given.
\begin{figure}
\begin{mdframed}[
nobreak=true,
frametitle={Example for Forestry Use Case},
linecolor=black,
frametitlerulecolor=black,
frametitlebackgroundcolor=gray!5
]
In this example there is a small group of users. The use case is a piece of forest and variables are for example harvesting activity, which trees to grow and accessibility for people.
\begin{align}
\begin{split}
V = \{ & \textit{indigenous}, \textit{resilient}, \textit{usable}, \textit{effort}, \textit{quantity}, \textit{price}, \textit{accessibility} \},
\end{split} \notag \\
\mathfrak{D}(\textit{indigenous}) = \{ & \text{low}, \text{moderate}, \text{high}\}, \notag \\
\mathfrak{D}(\textit{resilient}) = \{ & \text{low}, \text{moderate}, \text{high}\}, \notag \\
\mathfrak{D}(\textit{usable}) = \{ & \text{low}, \text{moderate}, \text{high}\}, \notag \\
\mathfrak{D}(\textit{effort}) = \{ & \text{manual}, \text{harvester}, \text{autonomous}\}, \notag \\
\mathfrak{D}(\textit{quantity}) = \{ & \text{low}, \text{moderate}, \text{high}\}, \notag \\
\mathfrak{D}(\textit{price}) = \{ & \text{low}, \text{moderate}, \text{high}\}, \notag\\
\mathfrak{D}(\textit{accessibility}) = \{ & \text{low}, \text{moderate}, \text{high}\},\notag \\
U = \{ & 1,2\} \notag\\
P = \{ & P_1, P_2\} \notag\\
\begin{split}
P_1 = \{ & (\text{manual}, 0.8), (\text{harvester}, 0.3) \} \\
& \cup \{ (d,0.5)\ |\ d \in \mathfrak{D}(i),\ i \in V,\ i \notin \{ \text{manual}, \text{harvester}\} \ \} \
\end{split} \notag \\
P_2 = \{ & (d,0.5)\ |\ d \in \mathfrak{D}(i),\ i \in V \} \notag \\
S = \{ & (\textit{indigenous}, \text{low}), (\textit{quantity}, \text{moderate}) \} \notag \\
\begin{split}
S_F = \{ & (\textit{indigenous}, \text{low}), (\textit{resilient}, \text{low}), (\textit{usable},\text{low}), (\textit{effort}, \text{manual}), \\
& (\textit{quantity}, \text{low}), (\textit{price},\text{high}),(\textit{accessibility}, \text{low}) \}
\end{split} \notag
\end{align}
\end{mdframed}
\caption[Forestry Use Case]{An example of use case in forestry that includes two people.}
\label{fig:Concept:ForestExample}
\end{figure}
\section{Recommendation Generation} \section{Recommendation Generation}
\label{sec:Concept:SolutionGeneration} \label{sec:Concept:SolutionGeneration}
@@ -234,6 +195,45 @@ where $aggr$ is the aggregation function and $score_{user}(P_i, s)$ is the confi
score_{user}(P_i, s) = average(\{x \ | \ (characteristic, x) \in P_i \land characteristic \in s \}) \notag . score_{user}(P_i, s) = average(\{x \ | \ (characteristic, x) \in P_i \land characteristic \in s \}) \notag .
\end{equation} \end{equation}
\begin{figure}[htb]
\begin{mdframed}[
nobreak=true,
frametitle={Example for Forestry Use Case},
linecolor=black,
frametitlerulecolor=black,
frametitlebackgroundcolor=gray!5
]
In this example there is a small group of users. The use case is a piece of forest and variables are for example harvesting activity, which trees to grow and accessibility for people.
\begin{align}
\begin{split}
V = \{ & \textit{indigenous}, \textit{resilient}, \textit{usable}, \textit{effort}, \textit{quantity}, \textit{price}, \textit{accessibility} \},
\end{split} \notag \\
\mathfrak{D}(\textit{indigenous}) = \{ & \text{low}, \text{moderate}, \text{high}\}, \notag \\
\mathfrak{D}(\textit{resilient}) = \{ & \text{low}, \text{moderate}, \text{high}\}, \notag \\
\mathfrak{D}(\textit{usable}) = \{ & \text{low}, \text{moderate}, \text{high}\}, \notag \\
\mathfrak{D}(\textit{effort}) = \{ & \text{manual}, \text{harvester}, \text{autonomous}\}, \notag \\
\mathfrak{D}(\textit{quantity}) = \{ & \text{low}, \text{moderate}, \text{high}\}, \notag \\
\mathfrak{D}(\textit{price}) = \{ & \text{low}, \text{moderate}, \text{high}\}, \notag\\
\mathfrak{D}(\textit{accessibility}) = \{ & \text{low}, \text{moderate}, \text{high}\},\notag \\
U = \{ & 1,2\} \notag\\
P = \{ & P_1, P_2\} \notag\\
\begin{split}
P_1 = \{ & (\text{manual}, 0.8), (\text{harvester}, 0.3) \} \\
& \cup \{ (d,0.5)\ |\ d \in \mathfrak{D}(i),\ i \in V,\ i \notin \{ \text{manual}, \text{harvester}\} \ \} \
\end{split} \notag \\
P_2 = \{ & (d,0.5)\ |\ d \in \mathfrak{D}(i),\ i \in V \} \notag \\
S = \{ & (\textit{indigenous}, \text{low}), (\textit{quantity}, \text{moderate}) \} \notag \\
\begin{split}
S_F = \{ & (\textit{indigenous}, \text{low}), (\textit{resilient}, \text{low}), (\textit{usable},\text{low}), (\textit{effort}, \text{manual}), \\
& (\textit{quantity}, \text{low}), (\textit{price},\text{high}),(\textit{accessibility}, \text{low}) \}
\end{split} \notag
\end{align}
\end{mdframed}
\caption[Forestry Use Case]{An example of use case in forestry that includes two people.}
\label{fig:Concept:ForestExample}
\end{figure}
The example in \autoref{fig:Concept:ForestExample} contains two users. The first user has preferences for the characteristic \emph{manual} of the feature with $0.8$ and the characteristic \emph{harvester} of the same feature with $0.3$. All other characteristics have a preference of $0.5$. The second user's preferences are $0.5$ for all characteristics. The finished configuration that is supposed to be rated in this example contains the characteristics \emph{low} for each feature except for \emph{effort} and \emph{quantity} which are set to \emph{manual} and \emph{high}. The score fore the finished configuration $S_F$ of user one is $0.54$. This score is the average of all seven features. User one rates all characteristics of all features as $0.5$ except two characteristics for \emph{effort}. Thus, all feature scores for this user are $0.5$ except the score for \emph{effort} is $0.8$ because of the user's preference of $0.8$ for the characteristic \emph{manual}. The resulting average score per feature of $0.54$ is the user's score for this configuration. User two rates all characteristics with $0.5$ therefore the resulting average is $0.5$. The example in \autoref{fig:Concept:ForestExample} contains two users. The first user has preferences for the characteristic \emph{manual} of the feature with $0.8$ and the characteristic \emph{harvester} of the same feature with $0.3$. All other characteristics have a preference of $0.5$. The second user's preferences are $0.5$ for all characteristics. The finished configuration that is supposed to be rated in this example contains the characteristics \emph{low} for each feature except for \emph{effort} and \emph{quantity} which are set to \emph{manual} and \emph{high}. The score fore the finished configuration $S_F$ of user one is $0.54$. This score is the average of all seven features. User one rates all characteristics of all features as $0.5$ except two characteristics for \emph{effort}. Thus, all feature scores for this user are $0.5$ except the score for \emph{effort} is $0.8$ because of the user's preference of $0.8$ for the characteristic \emph{manual}. The resulting average score per feature of $0.54$ is the user's score for this configuration. User two rates all characteristics with $0.5$ therefore the resulting average is $0.5$.
The group configuration score is dependent on the used aggregation strategy. Multiplication results in a score of $0.54 \cdot 0.5 = 0.27$. The score for average is $\frac{1}{2}(0.54 + 0.5) = 0.52$ and for least misery $\min \{0.54, 0.5\} = 0.5$. The group configuration score is dependent on the used aggregation strategy. Multiplication results in a score of $0.54 \cdot 0.5 = 0.27$. The score for average is $\frac{1}{2}(0.54 + 0.5) = 0.52$ and for least misery $\min \{0.54, 0.5\} = 0.5$.