\documentclass{ictlab}
%\documentclass[solutions]{ictlab}

\RCS $Revision: 1.6 $

\usepackage{verbatim,key,alltt,amstext,answer2}
\usepackage[hang,bf,nooneline]{caption2}
\usepackage[pdfpagemode=None,pdfauthor={Nick Urbanik}]{hyperref}

\newcommand*{\labTitle}{Revision Exercises with Perl, routing,
  switching and LDAP}
\renewcommand*{\floatpagefraction}{0.75} % default is .5, to increase
% density.
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\renewcommand*{\textfraction}{0.1} % default is 0.2

\providecommand*{\IP}{\acro{IP}\xspace}
\providecommand*{\PDU}{\acro{PDU}\xspace}
\providecommand*{\IETF}{\acro{IETF}\xspace}
\providecommand*{\MAC}{\acro{MAC}\xspace}

\begin{document}

%% \section{Background}
%% \label{sec:background}


%% \subsection{Perl}
%% \label{sec:perl}

%% Refer to the summary of Perl, from the subject web site.  I will not
%% repeat that here.


%% \subsection{Routing}
%% \label{sec:routing}


%% \begin{itemize*}
%% \item 
%% \end{itemize*}


%% \section{Questions}
%% \label{sec:questions}

\section{Perl}
\label{sec:perl}

\begin{enumerate}
\item Write a regular expression to match two or more ``\texttt{x}''s
  followed by one or more ``\texttt{y}''s followed by any number of
  ``\texttt{z}''s.

\item[\textbf{Solution:}] All three of these are correct:
\begin{verbatim}
/xx+y+z*/
/xxx*yy*z*/
/x{2,}y{1,}z{0,}/
\end{verbatim}

\item Write a regular expression to match any number of backslashes
  ``\bs'' followed by any number of asterisks ``\texttt{*}''.  Note
  that ``any number'' might be zero.

\item[\textbf{Solution:}]
\begin{verbatim}
/\\*\**/
\end{verbatim}

  \begin{figure}[htb]
\begin{verbatim}
MSRP6604R1    Student Progression Rank Details                Page 1 of 4
Academic Year : 2002/03          Course Board Control No.          APS1238

Progression Course : 95551/2    in HW
                          95570/2    in LW
                          95573/2    in LW
                          95731/2    in LW
Code        Student No.    Name                 Attained  Loc Course/YearClass
95551/2     987654321      WAN Siu Ming (970)       41.0   TY      41230/1G
95551/2     976543210      FUNG Siu Ming (618)      42.1   TY      41456/2B
95551/2     965432109      YUET Siu Ming (639)      43.2   TY      41789/3B
95551/2     954321098      WONG Siu Ming, Joe (571) 44.3   TY      41012/1G
95551/2     943210987      KLIE Siu Ming (486)      45.4   TY      41345/2G
95551/2     932109876      WANG Siu Hing (370)      46.5   TY      41678/3G
95551/2     921098765      WONG Siu Ming (436)      47.6   TY      41901/1G
95551/2     910987654      FONG Siu Ming (359)             TY      41234/3G
95570/2     909876543      HO Siu Ming (133)        49.8   TY      41567/2B
\end{verbatim}
    \caption{A text file containing data}
    \label{fig:data}
  \end{figure}
\item Figure~\vref{fig:data} shows a text file containing data.  Write
  a Perl program to print all student numbers.

\item[\textbf{Solution:}]
\begin{verbatim}
while ( <> ) {
    print "$1\n" if /(\d{9})/;
}
\end{verbatim}%$

\item Write a perl program to print the course, year and class, each
  in a separate column, separated by a tab character.

\item[\textbf{Solution:}]
\begin{verbatim}
while ( <> ) {
    print "$1\t$2\t$3\n" if /\s(\d{5})\/(\d)([a-zA-Z])/;
}
\end{verbatim}%$

\item Write a perl program to generate a password for each student
  according to the following algorithm:
  \begin{itemize}
  \item Take the student's name (but without the number in brackets);
  \item Remove all spaces and non-alphabetical characters from the
    name;
  \item Use the first eight characters left, append them to the last
    four digits of the student number.
  \end{itemize}
  Your program should print:
  \begin{itemize}
  \item the student number;
  \item the full name of the student;
  \item the student's generated password
  \end{itemize}
  all separated by a single tab character.  There should be one line
  of output per student.
  
\item[\textbf{Solution:}] Here is one of many possible solutions.  Here
  I have tried to be clear rather than very concise.
\begin{verbatim}
#! /usr/bin/perl -w

use strict;

while ( <> ) {
    if ( /(\d{9})\s+(\S.*)\s\(/ )
    {
        my ( $student_id, $name ) = ( $1, $2 );
        $student_id =~ /(\d{4})$/;
        my $pw_first = $1;
        my $pw_second = $name;
        $pw_second =~ s/[^a-zA-Z]//g;
        $pw_second =~ /^(.{8})/;
        $pw_second = $1;
        my $passwd = "$pw_first$pw_second";
        print "$student_id\t$name\t$passwd\n";
    }
}
\end{verbatim}%\)
\end{enumerate}

Now some people were confused by the pattern match binding operator,
\texttt{=}$\sim$, and by the substitute operator, \texttt{s/../../}.
Here is a little program that gives some examples of both:
\begin{verbatim}
#! /usr/bin/perl -w

use strict;

# Normally a match is made against $_.
# If you want to match against another string, use the binding
# operator, =~
# Here are some examples:

my $string = 'abc';
$string =~ /.(.)./;
my $what_is_this = $1;
# Notice that no change happened in $string:
print "\$string=$string, \$what_is_this=$what_is_this\n";
# see what the substitution operator does:
$string =~ s/b/B/;
print "\$string=$string\n";
# with no g modifier:
$string =~ s/./Z/;
print "\$string=$string\n";
# Now see what adding the g modifier does:
$string =~ s/./X/g;
print "\$string=$string\n";
# $ ./show-match 
# $string=abc, $what_is_this=b
# $string=aBc
# $string=ZBc
# $string=XXX
\end{verbatim}

\section{Routing and Switching}
\label{sec:routing-and-switching}

\begin{enumerate}
\item Before a set of candidate network routes can be aggregated, they
  must be in the \underline{\mbox{\hspace*{30mm}}}, and the routes in
  binary format must have the \underline{\mbox{\hspace*{30mm}}}.
  
\item[\textbf{Solution:}] Before a set of candidate network routes can
  be aggregated, they must be in the \underline{same network}, and the
  routes in binary format must have the \underline{same prefix}.

  \begin{figure}[htb]
    \centering%
    \includegraphics[width=0.8\linewidth]{routing-question-1}
    \caption{A network with five routers and ten subnets.}
    \label{fig:routing-question-1}
  \end{figure}
\item Figure~\vref{fig:routing-question-1} shows a network with 5
  routers and 10 subnets.  You may select \IP addresses from the block
  of addresses 172.12.0.0/19.  You must leave at least one quarter of
  these addresses available for other purposes.  The requirements are
  that each of subnets 1, 2,\ldots, 8 must support up to 130
  computers, while subnets 9 and 10 must each support up to 600
  computers.
  \begin{enumerate}
  \item Allocate a suitable block of addresses to each of the ten
    subnets that will allow maximum route aggregation.

  \item[\textbf{Solution:}] General strategy: determine the lower and
  upper limits on each subnet.

  The minimum size of each of the first 8 subnets is 256, i.e., a /24
  subnet, as $2^8$ is the lowest power of 2 that contains 130.

  Let us allocate the lowest 8 /24 blocks:

  \begin{tabular}[t]{@{}ll@{}}
    \toprule%
    \emph{subnet} & \emph{network} \\
    \midrule%
    subnet 1 & 172.12.0.0/24 \\
    subnet 2 & 172.12.1.0/24 \\
    subnet 3 & 172.12.2.0/24 \\
    subnet 4 & 172.12.3.0/24 \\
    subnet 5 & 172.12.4.0/24 \\
    subnet 6 & 172.12.5.0/24 \\
    subnet 7 & 172.12.6.0/24 \\
    subnet 8 & 172.12.7.0/24 \\
    \bottomrule
  \end{tabular}

  The minimum size of subnets 9 and 10 is $2^{10} = 1024$, since
  $2^{\lceil \log_2 600 \rceil} = 2^{10}$, giving a subnet size of /22.

  Let us allocate the next lowest 2 /22 blocks:

  \begin{tabular}[t]{@{}ll@{}}
    \toprule%
    \emph{subnet} & \emph{network} \\
    \midrule%
    subnet 9 & 172.12.8.0/22 \\
    subnet 10 & 172.12.12.0/22 \\
    \bottomrule
  \end{tabular}

  This would use only half of the available addresses.

  \item \label{que:no-summarisation}Given your selection in the
    previous part, with \emph{no} route summarisation active on the
    routers, list the routes that would be advertised by router A at
    X, by router B at W, by router C at V, by router D at Y, and by
    router E at Z\@.
    
  \item[\textbf{Solution:}] \emph{Without} route summarisation, router
    A will advertise one route for each of its five subnets:

  \begin{tabular}[t]{@{}l@{}}
    172.12.0.0/24 \\
    172.12.1.0/24 \\
    172.12.2.0/24 \\
    172.12.3.0/24 \\
    172.12.4.0/24
  \end{tabular}
  
  and router B will advertise one route for each of \emph{its} three
  subnets:

   \begin{tabular}[t]{@{}l@{}}
    172.12.5.0/24 \\
    172.12.6.0/24 \\
    172.12.7.0/24
  \end{tabular}

  Router C will advertise one route for each of its two subnets:

   \begin{tabular}[t]{@{}l@{}}
    172.12.8.0/22 \\
    172.12.12.0/22
  \end{tabular}

  Router D will advertise one route for each of the eight subnets behind
  routers A and B:

  \begin{tabular}[t]{@{}l@{}}
    172.12.0.0/24 \\
    172.12.1.0/24 \\
    172.12.2.0/24 \\
    172.12.3.0/24 \\
    172.12.4.0/24 \\
    172.12.5.0/24 \\
    172.12.6.0/24 \\
    172.12.7.0/24
  \end{tabular}

  Finally router E will advertise one route for each of the ten
  subnets behind routers A, B and C:

  \begin{tabular}[t]{@{}l@{}}
    172.12.0.0/24 \\
    172.12.1.0/24 \\
    172.12.2.0/24 \\
    172.12.3.0/24 \\
    172.12.4.0/24 \\
    172.12.5.0/24 \\
    172.12.6.0/24 \\
    172.12.7.0/24 \\
    172.12.8.0/22 \\
    172.12.12.0/22
 \end{tabular}

  
  \item What would be a necessary requirement for the routers to
    support route aggregation?

  \item[\textbf{Solution:}] The routers should be running a classless
  routing protocol, such as \OSPF or \RIP{}2.

  \item Repeat part~\ref{que:no-summarisation}, but for the case where
    the routers all support route summarisation.

  \item[\textbf{Solution:}] For router~A, the first four routes will
    be aggregated (summarised), so there will be two routes
    advertised:

    \begin{tabular}[t]{@{}l@{}}
      172.12.0.0/22 \\
      172.12.4.0/24
    \end{tabular}
    
    Router B will summarise the last two routes:

    \begin{tabular}[t]{@{}l@{}}
      172.12.5.0/24 \\
      172.12.6.0/23
    \end{tabular}
    
    Router C will summarise both routes into one:

    \begin{tabular}[t]{@{}l@{}}
      172.12.8.0/21
    \end{tabular}
    
    Router D will summarise the two routes from router A and the two
    routes from router B into one route:

    \begin{tabular}[t]{@{}l@{}}
      172.12.0.0/21
    \end{tabular}
    
    Router E will aggregate the route from router C with the route
    from router D into one single route:
     \begin{tabular}[t]{@{}l@{}}
       172.12.0.0/20
     \end{tabular}
     
     Note that here I have ignored the routing to any \IP addresses
     used on the serial links on the routers.
  \end{enumerate}
  
%% \item Distinguish between a \emph{broadcast domain} and a
%%   \emph{collision domain}.
  
%% \item For each of the following, list at least two network devices
%%   that can separate a single network into:
%%   \begin{enumerate}
%%   \item two collision domains
%%   \item two broadcast domains
%%   \end{enumerate}

\item When using portable notebook computers that should remain on one
  subnet, what type of \VLAN should be employed?  Are there any
  difficulties in supporting such a \VLAN arrangement?

\item[\textbf{Solution:}] They should use a dynamic \VLAN
  configuration.  No matter what data port the notebook is plugged
  into, it will belong to one particular \VLAN, and hence, in one
  subnet.

  The main drawback of this system is maintaining the database of \MAC
  address --- \VLAN mappings that must be stored in the switches that
  support this dynamic \VLAN.
  
%% \item List two advantages of using switches in a large network, and
%%   one important limitation.

  \begin{figure}[htb]
    \centering%
    \includegraphics[width=0.3\linewidth]{switches-q1}
    \caption{Two switches, each with five ports.}
    \label{fig:switches-q1}
  \end{figure}
\item Two switches each have five ports, as shown in
  figure~\vref{fig:switches-q1}.  Both switches are to support three
  \VLAN{}s, \VLAN{}1, \VLAN{}2 and \VLAN{}3.  Computers will be
  connected to each of these three \VLAN{}s on each of the two
  switches.
  \begin{enumerate}
  \item Each \VLAN is to support a separate subnet.  How would you
  connect the switches so that computers on each \VLAN can communicate
  with each other?

  Support your answer with a diagram showing the connection of the
  switches.  Label the \VLAN{}s.  Indicate whether each port is a
  trunk port or an access port.

\item[\textbf{Solution:}] See figure~\vref{fig:switches-q1-answer}.
  \begin{figure}[htb]
    \centering%
    \includegraphics[width=0.5\linewidth]{switches-q1-answer}
    \caption{Two switches, each with five ports connected with a
      router.  The trunk ports on the right are connected together to
      connect the \VLAN{}s on the two switches together.  Each \VLAN
      is a separate subnet.  The router connects these subnets
      together.}
    \label{fig:switches-q1-answer}
  \end{figure}
  
\item One additional port is added to each switch.  Now two users with
  portable computers, one belonging to \VLAN{}2 and the other
  belonging to \VLAN{}3 want to be able to plug into the newly added
  port on either switch and remain in their respective \VLAN without
  any manual configuration.  How should you configure the new ports in
  each switch to facilitate this?

\item[\textbf{Solution:}] We would configure them as dynamic \VLAN{}s,
  and add one notebook's \MAC address to dynamic \VLAN{}2, the other
  to dynamic \VLAN{}3.

  If notebook-1 has \MAC address \MAC-1, and notebook-2 has \MAC
  address \MAC-2, and we want notebook-1 on \VLAN{}2, and notebook-2
  on \VLAN{}3, then we must enter the following data into the switch's
  dynamic \VLAN configuration:

  \begin{tabular}[t]{@{}cc@{}}
    \toprule%
    \MAC address & Which \VLAN \\
    \midrule%
    \MAC-1 & \VLAN{}2 \\
    \MAC-2 & \VLAN{}3 \\
    \bottomrule
  \end{tabular}
  
  \end{enumerate}
\end{enumerate}


\section{Directories and LDAP}
\label{sec:ldap}

Many of you have not completed the tutorial sheet ``\LDAP Filters and
Searching \LDAP directories.''  I suggest that you complete it.

\end{document}