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\title{ Challenges in Computer Science \\
\large{Assignment 2 - multi-objective question}}
\author{Rick van der Zwet\\
  \texttt{<hvdzwet@liacs.nl>}\\
  \\
  LIACS\\
  Leiden Universiteit\\
  Niels Bohrweg 1\\
  2333 CA Leiden\\
  Nederland}
\date{\today}
\begin{document}
\maketitle
\tableofcontents
\chapter{Introduction}
\label{foo}
The assignment -given during the college of 23 April 2007- defined 3
multi-objective questions. One of them needs further investigation:
\begin{quote}
Your task is to design a fish tank. The width, height, and length of
the box-shaped tank are your decision variables. The volume has to be
maximized, while the surface area has to be minimized {\small{because it is
directly proportional to the cost of the tank.}} Moreover, for aesthetical
reasons, the ratio between the height and length should be 2/3.  Another
constraint: the volume is not to exceed 300 liters and has to be more or
equal to 60 liters.
\end{quote}

\chapter{Assumptions}
\begin{itemize}
\item The box is 'closed', the surface area of a fish tank will be the sum of
all (6) areas of the box.
\item No surface area  will be used to attach the sides to each other.
\item Both decision variables equal, a weight factor of $1$.
\end{itemize}
The surface area of a fish tank will be
\begin{equation}
a = 2 * w * h + 2 * w * l + 2 * h * l
\end{equation}
or more simplified
\begin{equation}
a = 2 * ( w * h + w * l + h * l)
\label{eq:sa}
\end{equation}
The volume will be
\begin{equation}
v = w * h * l
\label{eq:v}
\end{equation}


\chapter{Strategy}
We could very easy reduce the problem from 3 unknown factors to 2
unknown factors.  The ratio between the height and length should be
$2/3$ which makes the height linear dependant of the width. Substitution
in Eq. \ref{eq:sa} leads to
\begin{equation}
a = 2 * ( w * ( 2/3 * l) + w * l + (2/3 * l ) * l)
\end{equation}
After substitution in Eq. \ref{eq:v} the volume will become $v = w * l *
(2/3 * l)$.  With 2 unknown factors it will be perfect to use a Pareto
front. We will map $(w,l) => (v,a)$, with respect of the limitation $60
\leq v < 300$.
\chapter{Model}

\begin{figure}[ht]
\begin{center}
\resizebox{\columnwidth}{!}{\includegraphics{pareto}}
\end{center}
\caption{
    Pareto-front of fish-box
}
\label{fig:pareto}
\end{figure}

Fig. \ref{fig:pareto} has be generated by calculating many values -with
respect to the limitations- and plot them into the figure.
As we are optimising for the smallest surface area and the largest
volume the vectors will originate from the left-top bottom of the graph
($(400,0)$). Both decision variables are as important, so the
pareto-lines will be circles. The shorter the length of the vector the
better the solution. The blue dashed line will be the first line which
intersect a few points, so these point are the optimal solutions.

\appendix
\fvset{numbers=left}
\chapter{pareto.pl}
\VerbatimInput{pareto.pl}
\chapter{pareto.gp}
\VerbatimInput{pareto.gp}
\end{document}