ORiON (ORSSA style sheet)

\documentclass[11pt]{article}
\usepackage{orioneng,epsfig}

\newcommand{\be}{\begin{equation}}
\newcommand{\ee}{\end{equation}}
\newcommand{\bea}{\begin{eqnarray}}
\newcommand{\eea}{\end{eqnarray}}
\newcommand{\ul}{\underline}

\title{Your paper} 

\author{SE Visagie\thanks{Department of Logistics, University of Stellenbosch, Private bag X1, Matieland, 7602, %%@
Republic of South Africa, fax: 021 8082409, email: {\tt [email protected]}} \and ADR Outeur\thanks{Department of %%@
Department, University of University}}

\shorttitle{ORiON: Example}
\shortauthor{SE Visagie \& ADR Outeur}

\begin{document}
\oriontitle

\begin{Abstract}
The format for papers published in ORiON is provided. Examples of how environments, such as tables, should be %%@
presented are also provided. The editorial format and standards of ORiON are maintained throughout this paper. The %%@
purpose is not to provide a mathematically correct or coherent content to this document, but rather to create a paper %%@
using specific examples of standard \LaTeX\ elements.
\end{Abstract}

\begin{keywords} 
Style sheet, examples, \LaTeX
\end{keywords}

\section{Introduction}
This paper contains examples of several functions used in ORiON's \LaTeX\ package. The purpose of this paper is not to %%@
provide readers with sensible content or proper logic. The theorems and definitions used in this paper may therefore %%@
appear not to make sense. The purpose is, however, not to provide the correct theorems and definitions, but rather to %%@
provide examples of these entities. The examples demonstrate how the standard environments and elements are created %%@
within the ORiON package. This paper also provides examples of the preferred format in which equations, references and %%@
other standard elements are handled in ORION.

The examples provided in this paper are not exhaustive. Should an author wish to include specific or exceptional %%@
situations in a paper, the business manager of ORiON can be contacted for guidance. Frequently asked questions will be %%@
addressed as examples in further updates of this paper.

There are various manuals available for \LaTeX\ and \TeX \cite{doob, lamp}. Should these not provide sufficient %%@
guidance, the internet can also be used as a source of information. Text is entered into the source document in the %%@
same manner as for standard \LaTeX\ documents. The environments in Table~\ref{oron:t1} have already been defined.
\begin{table}[htb]\begin{center}\begin{tabular}{lll}
\verb=algorithm= & \verb=axiom= & \verb=corollary=  \\ 
\verb=definition= & \verb=example= & \verb=heuristic=\\ 
\verb=lemma= & \verb=proposition= & \verb=theorem= \\
\end{tabular}\end{center}
\caption{Pre-defined environments in {\tt ORiONeng.sty}}
\label{oron:t1}
\end{table}

\section{Text}
To solve LPs with the help of the simplex algorithm, knowledge of linear algebra is required. Amongst other things, %%@
the concept of linearly dependent vectors is important.
\begin{definition}
If $\ul{a}_j$ with $j = 1, 2, 3, \ldots, r$ is a set of vectors, then the equation
$$
k_1 \ul{a}_1 + k_2 \ul{a}_2 + \dots + k_r \ul{a}_r = \ul{0} 
$$
will have at least one solution, which is $k_1=k_2 = \ldots = k_r = 0$. If this is the only solution, the set, %%@
$\ul{a}_j$, is called linearly independent. If other solutions exist, the set is called linearly dependent.
\end{definition}
A set of column vectors, $\ul{a}_j$, with $j = 1, 2, 3, \ldots, k$, is linear independent or linearly dependent. %%@
Assume the set of vectors is linearly dependent. Under these circumstances it is possible to reduce the number of %%@
positive variables step by step until the columns which correspond with the remaining positive variables are linearly %%@
independent. If $\ul{a}_j$, with $j=1, 2, 3, \ldots, n$, is linearly dependent, real numbers $\alpha_j$ (not all equal %%@
to zero) exist, so that
\be
\sum_{j=1}^{k} \alpha_j \ul{a}_j = \ul{0}. \label{simv21}
\ee
Equation~(\ref{simv21}) can be used to reduce an $x_j$, say $x_r$, in 
\be
\sum_{j=1}^{k} x_j \ul{a}_j = \ul{b}, \ \ \ \ x_j > 0 \label{simv22}
\ee
to zero. Suppose any vector $\ul{a}_r$ of the $k$ vectors in equation~(\ref{simv21}), for which $\alpha_r \not= 0$, %%@
can be expressed in terms of the remaining $k-1$ vectors. Thus 
\be
\ul{a}_r = - \sum_{j\not= r} \frac{\alpha_j}{\alpha_r}\ul{a}_j. \label{simv23}
\ee
Substitute equation~(\ref{simv23}) into equation~(\ref{simv22}), so that 
\be
\sum_{j=1}^{p} \left(x_j - x_r\frac{\alpha_j}{\alpha_r} \right)\ul{a}_j  = \ul{b}. \label{simv24}
\ee
This solution does not have more than $k-1$ non-zero variables. If $\ul{a}_r$ is selected arbitrarily, there would be %%@
no assurance that some of the variables may not perhaps be negative. If $\ul{a}_r$ is, however, selected carefully, %%@
the remaining $k-1$ variables will still be non-negative.  Select $\ul{a}_r$ so that
\be
x_j - x_r\frac{\alpha_j}{\alpha_r} \geq 0  \ \ \ j = 1, \ldots, p. \label{simv25}
\ee


For any $j$, where $\alpha_j = 0$, equation~(\ref{simv25}) will automatically hold. If $\alpha_j \not= 0$, then %%@
equation~(\ref{simv25}), divided by $\alpha_j$, will give one of the following inequalities:
\bea
\frac{x_j}{\alpha_j} - \frac{x_r}{\alpha_r} & \geq & 0 \ \ \ \mbox{ if } \alpha_j > 0,  \mbox{ or}\label{simv26a}\\
\frac{x_j}{\alpha_j} - \frac{x_r}{\alpha_r} & \leq & 0 \ \ \ \mbox{ if } \alpha_j < 0. \label{simv26b}
\eea 
From (\ref{simv26a}) and (\ref{simv26b}) follows that $\frac{x_r}{\alpha_r}$ can not be greater than any %%@
$\frac{x_j}{\alpha_j} > 0$ and also not less than any $\frac{x_j}{\alpha_j} < 0$. If we choose the vector $\ul{a}_r$ %%@
so that
\be
\frac{x_r}{\alpha_r} = \min_j \left\{ \frac{x_j}{\alpha_j}, \ \ \ \alpha_j > 0 \right\},\label{simv27}
\ee
then all the variables in equation~(\ref{simv24}) will be non-negative. A feasible solution with no more then $k-1$ %%@
non-zero variables has been found. If this process of reduction is continued, the remaining $x_j$'s will eventually be %%@
linearly independent and the solution will be a basic feasible solution. 

The theorem which summarises the theory of the solution of LPs with the simplex algorithm, is the fundamental theorem %%@
of LP.
\begin{theorem}[Fundamental theorem of linear programming (LP)] \label{simst1} 
This theorem consists of two parts:
\begin{enumerate}
\item If a feasible solution exists for an LP, a basic feasible solution exists.
\item If an optimal solution exists for an LP, an optimal basic feasible solution exists.
\end{enumerate}
\end{theorem}
Only a partial proof for Theorem~\ref{simst1} is presented.

{\bf Proof:} \ \ Assume that a feasible solution exists with $k<n$ variables having, a positive value. Number these %%@
variables in such a way that the first $k$ variables have a positive value. The solution can then be formulated as
$$
\sum_{j=1}^{k} x_j \ul{a}_j = \ul{b}
$$ 
with $x_j > 0$ where $j = 1, 2, \ldots, k$ and $x_j = 0$ for $j = k+1, \ldots, n$. The vectors $\ul{a}_j$ which are %%@
associated with the positive values can either be linearly independent or linearly dependent \cite{nich}.

For a more detailed proof, see \cite{hadl}.

It is obviously much easier to solve LPs using software packages. A few of these programs are listed in %%@
Table~\ref{orionvb:t2}.
\begin{table}[htb]
\begin{center}\begin{tabular}{ll} \hline
Software & Advantages\\ \hline \hline
LINDO \& LINGO & Programmable \\
	& Built-in mathematical functions\\
QM for Windows	& User-friendly interface\\
	& Interface with Excel\\ \hline 
\end{tabular}\end{center}
\caption{LP software with their respective advantages.}
\label{orionvb:t2}
\end{table}

\section{Algorithms}
The binary search algorithm starts its search in the centre of an alphabetical list of words. The algorithm compares %%@
the word in the middle of the list with the word that must be found (WRD) to test whether the middle word is, in fact, %%@
WRD. If this is not the case, it means that WRD is either after or before the word in the middle of the alphabetical %%@
list. If WRD comes alphabetically before the middle word, the word halfway between the first word and the middle word %%@
is tested to determine whether that word is WRD, while the other half (after the middle word) of the list is %%@
discarded. If WRD comes after the middle word, the word halfway between the middle and the last word is tested, while %%@
the first half of the list is discarded. This process of dividing the remaining list in half is repeated until the %%@
word is located or the list has been exhausted. This implies that only one half of the list has to be tested after the %%@
first step, one quarter of the list after the second step, one eighth of the list after the third step, etc. (The %%@
middle word is the word in the $\left\lfloor \frac{1+n}{2} \right\rfloor$-th position in the list.)
\begin{algorithm}[Binary search algorithm] \label{algalg3} 
Find a word in an alphabetical list of words by comparing the middle word with the word that must be found and by %%@
dividing the list, if required, into sublists.
\begin{enumerate}
\item Let $\alpha$ and $\beta$ be the first and last positions of the sublist. Initially $\alpha=1$ and $\beta=n$. %%@
Thus $\alpha \leftarrow 1$ and $\beta \leftarrow n$. WRD $\leftarrow$ ``the word to be located".
\item  Repeat this step until $\alpha \leq \beta$, else go to \ref{algb}.
\begin{enumerate}
\item \label{alga} Determine $k$, the middle position (between $\alpha$ and $\beta$). 
$$k \leftarrow \left\lfloor \frac{\alpha+\beta}{2} \right\rfloor$$
\item If $W(k)$ = WRD, stop. Output: $k$.
\item If WRD $>$ $W(k)$, then $\beta \leftarrow k-1$. Go to \ref{alga}. 
\item If WRD $<$ $W(k)$, then $\alpha \leftarrow k+1$. Go to \ref{alga}.  
\end{enumerate} 
\item \label{algb}Stop. Output: ``Not in the list".
\end{enumerate}
\end{algorithm}

Let us examine the (worst case) complexity of the binary search algorithm. Note that step~1 and step~3 are both %%@
performed only once. In both of these steps, a constant number of calculations are performed. This means that the %%@
complexity of these steps is $O(1)$. Let $B(n)$ be the maximum number of times that step~2 must be repeated. After %%@
step~2 has been performed for the first time, the remaining list will have a maximum of $\lfloor\frac{n}{2}\rfloor$ %%@
elements to consider. This means that
\be
B(n) = 1 + B\left(\left\lfloor\frac{n}{2}\right\rfloor\right). \label{algv2}
\ee
We use induction to prove that
\be
B(n) \leq 1 + \log_2n \label{algv1}
\ee
for each positive integer $n$. Equation~(\ref{algv1}) applies if $n = 1$. Accept that this result holds for an %%@
integer, $k$. We must then show that this inequality holds for $k + 1$. We therefore accept that
\be
B(k) \leq 1 + \log_2k. \label{algv3}
\ee
From equation~(\ref{algv2}) we know that
$$
B(k+1) = 1 + B\left(\left\lfloor\frac{k+1}{2}\right\rfloor\right),
$$
but from equation~(\ref{algv3}) it follows that
$$
B\left(\left\lfloor\frac{k+1}{2}\right\rfloor\right) \leq 1 + %%@
\log_2\left(\left\lfloor\frac{k+1}{2}\right\rfloor\right).
$$
We also know that
$$
\log_2\left(\left\lfloor\frac{k+1}{2}\right\rfloor\right) \leq \log_2\left(\frac{k+1}{2}\right). 
$$
This means that
\begin{eqnarray*}
B(k+1) &\leq& 2 +  \log_2\left(\frac{k+1}{2}\right)\\
	&\leq& 2 + \log_2(k+1) - \log_2(2)\\
	&\leq& 1 + \log_2(k+1),
\end{eqnarray*}
which gives the desired result that $B(n) = O(\log_2n)$.

\section{Matrices}
The variable, $x_3$, enters the base with a value of $\frac{-20}{-1} = 20$, which is smaller than or equal to the %%@
upper bound. Thus a simplex iteration can be performed. The column below $x_3$ is given by
\[B_{1}^{-1}p_{3} =\left[\begin{array}{rrr}
-1&0&1\\
0&1&0\\
0&0&-1
\end{array}\right]
\left[\begin{array}{r} 1\\-1\\0 \end{array}\right]  
=\left[\begin{array}{r} -1\\-1\\0 \end{array}\right].
\]
The elementary matrix is given by
\[
E_{1}=\left[\begin{array}{rrr}
\left(\frac{1}{-1}\right)&0&0\\
-\left(\frac{-1}{-1}\right)&1&0\\
-\left(\frac{0}{-1}\right)&0&1
\end{array}\right].
\]
We can then calculate
\[ B_{2}^{-1}=E_{1}B_{1}^{-1}=\left[\begin{array}{rrr}
-1&0&0\\
-1&1&0\\
0&0&1
\end{array}\right]\left[\begin{array}{rrr}
-1&0&1\\
0&1&0\\
0&0&-1
\end{array}\right]=\left[\begin{array}{rrr}
1&0&-1\\
1&1&-1\\
0&0&-1
\end{array}\right].\]
After the iteration, the respective values are given by
\[ \begin{array}{c}
A'=\left[\begin{array}{rrrrrr}
-1&-1&1&-1&0&0\\
1&0&-1&0&1&0\\
-1&-1&0&0&0&1
\end{array}\right] \ \ \
B_{2}^{-1}=\left[\begin{array}{rrr}
1&0&-1\\
1&1&-1\\
0&0&-1
\end{array}\right] \ \ \ 
\ul{b}=\left[\begin{array}{r}
-60 \\
 30 \\
-80
\end{array}\right] \\[0.9cm]
\begin{array}{rclcrcl}
\ul{c}_{bv}(2)&=&\left[-1,0,-1 \right]& \ \ \ \ \ \ \ \ &\ul{c}_{nbv}(2)&=&\left[-2,0,0\right] \\[.3cm]
\ul{x}_{bv}(2)&=&\left[x_{3}, s_{2}, \widetilde{x}_{2} \right]& & \ul{x}_{nbv}(2)& = & %%@
\left[\widetilde{x}_{1},s_{3},\widetilde{s}_{1} \right].
\end{array} 
\end{array} \]
Go to step~5.

Step~5: \\
The current right hand sides are given by 
\[ \ul{b}(2) = B_{2}^{-1}\ul{b}=\left[\begin{array}{rrr}
1&0&-1\\
1&1&-1\\
0&0&-1
\end{array}\right] \left[\begin{array}{r}
-60 \\
 30 \\
-80
\end{array}\right] = \left[\begin{array}{c}
20 \\
50 \\
80
\end{array}\right].\]
The basic variables $x_{3}$, $s_{2}$ and $\widetilde{x}_{2}$ have the values 20, 50 and 80 respectively. None of these %%@
variables occur at a value above their upper bounds.

Step~6:\\
All the right hand sides are positive, which means that an optimal solution has been found. 

The optimal solution is given by
\bea
x_{1}&=&80 - \widetilde{x}_{1} = 80-0=80 \nonumber\\
x_{2}&=&80 - \widetilde{x}_{2} = 80-80=0 \nonumber\\
x_{3}&=&20 \nonumber\\
s_{1}&=&0 - \widetilde{s}_{1} = 0-0=0 \nonumber\\
s_{2}&=&50 \nonumber\\
s_{3}&=&0. \nonumber
\eea
The optimal objection function value is
\bea
\mbox{Objection function} & = & c_{1}x_{1}+c_{2}x_{2}+c_{3}x_{3}
\nonumber \\
&=& (-2)(80) + (-1)(0) + (1)(20) \nonumber\\
&=& -140. \nonumber
\eea 

\begin{thebibliography}{xx}
\bibitem{char}{{\sc Chartrand G \& Oellermann OR}, 1993, {\em Applied and Algorithmic Graph Theory}, McGraw Hill, %%@
Singapore, pp 43--46}
\bibitem{doob}{{\sc Doob M}, 1996, {\em \TeX \ starting from 1}, Springer-Verlag, New York.}
\bibitem{gassharr}{{\sc Gass SI \& Harris CM}, 1996, {\em Encyclopedia of Operations Research and Management Science}, %%@
Kluwer Academic Publishers, Boston.}
\bibitem{hadl}{\sc Hadley G}, 1962, {\em Linear programming}, Addison-Wesley Publishing Company Inc., Reading, %%@
Massachusetts.
\bibitem{lamp}{\sc Lamport L}, 1994, {\em \LaTeX \ A document preparation system: User's guide and reference manual}, %%@
Addison-Wesley Longman Inc., Reading, Massachusetts.
\bibitem{nich}{\sc Nickolson WK}, 1995, {\em Linear algebra with applications}, third edition, PWS Publishing Company, %%@
Boston.
\end{thebibliography}
\end{document}

\typeout{Stylblad vir voorlegging van artikels in ORiON:  01/01/2004}

%Definieer omgewings
\newtheorem{theorem}{Theorem}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{heuristic}[theorem]{Heuristic}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

%Stel die bladsygrootte
\oddsidemargin 0mm
\evensidemargin 10mm
\topmargin -7mm
\headsep 0.4cm
\footskip 0.6cm
\textwidth 150mm
%\advance\textwidth -2in
%\advance\textwidth -\oddsidemargin
%\advance\textwidth -\evensidemargin
\textheight 220mm
%\advance\textheight -2in
%\advance\textheight -\footskip
\parskip 2mm plus 1pt


%Maak die artikel tweekantig
\@twosidetrue \@mparswitchtrue

%Maak artikel een kolom
\onecolumn

%Geen indent
\parindent 0mm

%Inligting op die eerste bladsy
%Datum ontvang, hersien en aanvaar.
\def\volume#1{\gdef\@volume{#1}}
\def\vnumber#1{\gdef\@vnumber{#1}}
\def\pages#1{\gdef\@pages{#1}}
\def\received#1{\gdef\@received{#1}}
\def\revised#1{\gdef\@revised{#1}}
\def\accepted#1{\gdef\@accepted{#1}}
\def\taal#1{\gdef\@taal{#1}}

\newcommand{\datesRRA}{\vspace{-1cm}\begin{center} {\footnotesize \em Received: \@received ; Revised: \@revised ; %%@
Accepted: \@accepted} \end{center}}
\newcommand{\datesRA}{\vspace{-1cm}\begin{center} {\footnotesize \em Received: \@received ; Accepted: \@accepted} %%@
\end{center}}

%Definieer 'n nuwe omgewing vir opsomming
\newenvironment{ops}[1]{
 \footnotesize
 \begin{center}%
  {\bfseries #1 \vspace{-.5em}\vspace{\z@}}%
 \end{center}%
 \begin{center}\begin{minipage}{13cm} \hspace{-1.5mm}}
 {\end{minipage} \end{center}}

%Maak abstract
\newcommand\abst{Abstract}
\newenvironment{Abstract}{\begin{ops}{\abst}\noindent}{\end{ops}}

%Maak opsomming
\newcommand\opsom{Opsomming}
\newenvironment{opsomming}{\begin{ops}{\opsom}\noindent}{\end{ops}}

%Maak Sleutelwoorde
\newcommand\sleut{Sleutelwoorde}
\newenvironment{sleutelwoorde}{\footnotesize {\bf \sleut: \ \ }}{\vspace*{-0.2cm}}

%Maak Key words
\newcommand\keywordsname{Key words}
\newenvironment{keywords}{\footnotesize {\bf \keywordsname: \ \ }}{\vspace*{-0.2cm}}

%Definieer 'n nuwe omgewing vir bewyse
\newenvironment{proof}{\vspace*{2mm plus 1pt} {\bf Proof:}\ \ }{\vspace*{2mm plus 1pt}}
\newenvironment{bewys}{\vspace*{2mm plus 1pt} {\bf Bewys:}\ \ }{\vspace*{2mm plus 1pt}}

%Maak running heads
\def\shortauthor#1{\gdef\@shortauthor{#1}}
\def\shorttitle#1{\gdef\@shorttitle{#1}}
\renewcommand{\ps@plain}{\let\@mkboth\@gobbletwo%
  \renewcommand{\@oddhead}{}%
  \renewcommand{\@evenhead}{}%
  \renewcommand{\@evenfoot}{\hfil\footnotesize\textrm{\thepage}\hfil}%
  \renewcommand{\@oddfoot}{\@evenfoot}}

    \def\ps@headings{%
      \let\@oddfoot\@empty\let\@evenfoot\@empty
      \def\@evenhead{\rlap{\thepage}\hfil\upshape\small \em Authors' identities suppressed: Blind refereeing copy
       \hfil}%
      \def\@oddhead{\hbox{}\hfil{\upshape\small \em\@shorttitle}\hfil\llap{\thepage}}% %!
      \let\@mkboth\markboth
      \def\sectionmark##1{%
        \markboth {\uppercase{\ifnum \c@secnumdepth >\z@
            \thesection.\hskip 1em\relax\fi
            ##1}}{}}%
      \def\subsectionmark##1{%
        \markright {\ifnum \c@secnumdepth >\@ne
            \thesubsection\hskip 1em\relax \fi
            ##1}}}

\def\ps@myheadings{%
    \let\@oddfoot\@empty\let\@evenfoot\@empty
    \def\@evenhead{\rlap{\thepage}\hfil\upshape\small\em Authors' identities suppressed: Blind refereeing copy
    \hfil}%!
    \def\@oddhead{\hbox{}\hfil{\upshape\small\em\@shorttitle}\hfil\llap{\thepage}}% %!
    \let\@mkboth\@gobbletwo
    \let\sectionmark\@gobble
    \let\subsectionmark\@gobble
    }

%Maak oriontitel
\newcommand\oriontitle{\par
  \begingroup
    \renewcommand\thefootnote{\@fnsymbol\c@footnote}%
    \def\@makefnmark{\rlap{\@textsuperscript{\normalfont\@thefnmark}}}%
    \long\def\@makefntext##1{\parindent 1em\noindent
            \hb@[email protected]{%
                \hss\@textsuperscript{\normalfont\@thefnmark}}##1}%
    \if@twocolumn
      \ifnum \col@number=\@ne
        \@maketitle
      \else
        \twocolumn[\@maketitle]%
      \fi
    \else
      \newpage
      \global\@topnum\z@   % Voorkom dat figure tot voor die titel geskuif word.
      \@maketitle
    \fi
    \thispagestyle{plain}\@thanks
  \endgroup
  \setcounter{footnote}{0}%
  \global\let\thanks\relax
  \global\let\maketitle\relax
  \global\let\@maketitle\relax
  \global\let\@thanks\@empty
  \global\let\@author\@empty
  \global\let\@date\@empty
  \global\let\@title\@empty
  \global\let\title\relax
  \global\let\author\relax
  \global\let\date\relax
  \global\let\and\relax
}
\def\@maketitle{%
  %\cleardoublepage
  \thispagestyle{plain}
  %\null
  %\vskip 2em%
  \raisebox{1.2cm}{\begin{minipage}{2cm}
  \footnotesize \begin{tabular}{l} \hline \hline
   Operations\\ Research\\ Society of\\
  South Africa \rule{1mm}{0mm}\\  \hline \hline
  \end{tabular} \end{minipage}}
  \hfill
  \raisebox{1.2cm}{\begin{tabular}{c}
  {\footnotesize Submitted}\\
  {\footnotesize for publication in} \\
  {\footnotesize ORiON}
  \end{tabular}}
  \hfill
  \raisebox{1.2cm}{\begin{minipage}{2cm}
  \footnotesize  \begin{tabular}{r} \hline \hline
   Operasionele\\ Navorsings-\\ vereniging van\\ Suid-Afrika\\ \hline \hline
  \end{tabular} \end{minipage}} \hspace{1mm} \\
  %\vspace*{-0.2cm}
  \begin{center}%
    \let \footnote \thanks
    {\huge \@title \par}%
    \vskip 1.5em%
    {\large
      \lineskip .5em%
      \begin{tabular}[t]{c}%
        Authors' identities suppressed: Blind refereeing copy
      \end{tabular}\par}%
    %\vskip 1em%
    %{\large }%
  \end{center}%
  \par
  \vskip 1.5em}

%Maak bold in Wiskunde
\def\bold#1{\mbox{\boldmath$#1$\unboldmath}}
\def\Bold#1{\mbox{\boldmath$\displaystyle #1$\unboldmath}}


%\renewcommand\section{\@startsection{section}{1}{\z@}%
%                       {-18\p@ \@plus -4\p@ \@minus -4\p@}%
%                       {12\p@ \@plus 4\p@ \@minus 4\p@}%
%                       {\normalfont\large\bfseries\boldmath
%                        \rightskip=\z@ \@plus 8em\pretolerance=10000 }}

\renewcommand\section{\@startsection {section}{1}{\z@}%
                                   {-3.5ex \@plus -1ex \@minus -.2ex}%
                                   {1.5ex \@plus .1ex}%
                                   {\normalfont\Large\bfseries}}
\renewcommand\subsection{\@startsection{subsection}{2}{\z@}%
                                     {-3.25ex\@plus -1ex \@minus -.2ex}%
                                     {.1ex \@plus .1ex}%
                                     {\normalfont\large\bfseries}}
\renewcommand\subsubsection{\@startsection{subsubsection}{3}{\z@}%
                                     {-3.25ex\@plus -1ex \@minus -.2ex}%
                                     {.1ex \@plus .2ex}%
                                     {\normalfont\normalsize\bfseries}}
\renewcommand\paragraph{\@startsection{paragraph}{4}{\z@}%
                                    {3.25ex \@plus1ex \@minus.2ex}%
                                    {-1em}%
                                    {\normalfont\normalsize\bfseries}}
\renewcommand\subparagraph{\@startsection{subparagraph}{5}{\parindent}%
                                       {3.25ex \@plus1ex \@minus .2ex}%
                                       {-1em}%
                                      {\normalfont\normalsize\bfseries}}

%Kies die regte blasyopskrifte
\pagestyle{myheadings}

%Figuur se opskrif
\setlength\abovecaptionskip{-5\p@}
\long\def\@makecaption#1#2{%
  \vskip\abovecaptionskip
  \sbox\@tempboxa{{\bf #1:} {\small \sl #2}}%
  \ifdim \wd\@tempboxa >\hsize
    {\bf #1:} {\small \sl #2}\par
  \else
    \global \@minipagefalse
    \hb@xt@\hsize{\hfil\box\@tempboxa\hfil}%
  \fi
  \vskip\belowcaptionskip}

%Maak onderkante gelyk
\flushbottom

%Maak die opskrifte skoon:
\received{??Geen datum??}
\revised{??Geen datum??}
\accepted{??Geen datum??}
\shorttitle{??Geen verkorte titel??}
\shortauthor{??Geen verkorte outeurs??}
\volume{??Geen volume??}
\vnumber{??Geen nommer??}
\pages{??Geen bladsye??}

\renewcommand\appendix{\par
  \setcounter{section}{0}%
  \setcounter{subsection}{0}%
  \gdef\thesection{Appendix~\@Alph\c@section:}}