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\documentclass[a4paper,9pt]{article}
\author{Camil Staps\\\small{s4498062}}
\title{Networking\\\large{Assignment 7}}
\date{May 22, 2016}
\usepackage{polyglossia}
\setmainlanguage{english}
\usepackage{geometry}
\usepackage[hidelinks]{hyperref}
\usepackage{amsmath}
\usepackage{enumitem}
\usepackage{subcaption}
\usepackage{verbatimbox}
\begin{document}
\maketitle
\section{Cyclic Redundancy Checks}
\begin{enumerate}[label=\alph*)]
\item It computes the remainder of $(D\cdot2^r\oplus R) \div G$, and checks
that it equals $0$.
\item See \autoref{fig:longdiv:b}. The answer is \texttt{111}.
\item See \autoref{fig:longdiv:c}. The answer is \texttt{010}.
\item See \autoref{fig:longdiv:d}. The answer is \texttt{111}.
\end{enumerate}
\begin{figure}[h]
\small
\begin{subfigure}[b]{.32\linewidth}
\centering
\begin{verbbox}
1001010101 000
1001
0000010101 000
1001
0000000111 000
100 1
0000000011 100
10 01
0000000001 110
1 001
0000000000 111
\end{verbbox}
\theverbbox%
\caption{\label{fig:longdiv:b}}
\end{subfigure}
\begin{subfigure}[b]{.32\linewidth}
\centering
\begin{verbbox}
0101101010 000
1001
0001001010 000
1001
0000000010 000
10 01
0000000000 010
\end{verbbox}
\theverbbox%
\caption{\label{fig:longdiv:c}}
\end{subfigure}
\begin{subfigure}[b]{.32\linewidth}
\centering
\begin{verbbox}
1010100000 000
1001
0011100000 000
1001
0001110000 000
1001
0000111000 000
1001
0000011100 000
1001
0000001110 000
1001
0000000111 000
100 1
0000000011 100
10 01
0000000001 110
1 001
0000000000 111
\end{verbbox}
\theverbbox%
\caption{\label{fig:longdiv:d}}
\end{subfigure}
\caption{Computing CRC using long division\label{fig:longdiv}}
\end{figure}
\section{Address Resolution Protocol}
\begin{enumerate}
\item We call the left router R and the right router S. Let $R_i$ be the
interface of router $R$ for subnet $i$. Then:
A --- 192.168.1.100; B --- 192.168.1.200;\\
C --- 192.168.2.100; D --- 192.168.2.200;\\
E --- 192.168.3.100; F --- 192.168.3.200;\\
R$_1$ --- 192.168.1.1; R$_2$ --- 192.168.2.1;\\
S$_2$ --- 192.168.2.2; S$_3$ --- 192.168.3.2;\\
\item
A --- 00.00.00.00.00.A0; B --- 00.00.00.00.00.B0;\\
C --- 00.00.00.00.00.C0; D --- 00.00.00.00.00.D0;\\
E --- 00.00.00.00.00.E0; F --- 00.00.00.00.00.F0;\\
R$_1$ --- 88.88.88.00.00.00; R$_2$ --- 88.88.88.00.88.00;\\
S$_2$ --- 88.88.88.88.00.00; S$_3$ --- 88.88.88.88.88.00.
\item Host E looks up the MAC address of the first hop router (S$_3$), and
sends the datagram there. Router S receives the datagram, figures out that
the next hop is router R, looks up the MAC address of R$_2$ and sends the
datagram there. Router R then forwards the datagram in a frame to host B on
interface R$_1$. Concretely, the following frames are sent:
\begin{table}[h]
\centering
\begin{tabular}{l l l l}
Src. IP & Src. MAC & Dest. IP & Dest. MAC \\\hline
192.168.3.100 & 00.00.00.00.00.E0 & 192.168.3.2 & 88.88.88.88.88.00\\
192.168.2.2 & 88.88.88.88.88.00 & 192.168.2.1 & 88.88.88.00.88.00\\
192.168.1.1 & 88.88.88.00.88.00 & 192.168.1.200 & 00.00.00.00.00.B0\\
\end{tabular}
\end{table}
\item The same happens, but before E can send the frame to S$_3$, it has to
get to know its address. It sends an ARP request to the broadcast address
with the request for 192.168.3.2's (that is, S$_3$'s) MAC address. The
router replies with an ARP packet. After this, E has all the addresses it
needs (and all other nodes as well), so the same as above happens.
\begin{table}[h]
\centering
\begin{tabular}{l l l l l}
Src. IP & Src. MAC & Dest. IP & Dest. MAC & Protocol\\\hline
192.168.3.100 & 00.00.00.00.00.E0 & 192.168.3.2 & FF.FF.FF.FF.FF.FF & ARP\\
192.168.3.2 & 88.88.88.88.88.00 & 192.168.3.100 & 00.00.00.00.00.E0 & ARP\\
192.168.3.100 & 00.00.00.00.00.E0 & 192.168.3.2 & 88.88.88.88.88.00 & IP\\
192.168.2.2 & 88.88.88.88.88.00 & 192.168.2.1 & 88.88.88.00.88.00 & IP\\
192.168.1.1 & 88.88.88.00.88.00 & 192.168.1.200 & 00.00.00.00.00.B0 & IP\\
\end{tabular}
\end{table}
\end{enumerate}
\section{Self-learning}
\begin{enumerate}[label=(\roman*)]
\item The switch learns the MAC address of host B. Since it doesn't know
where host E is, it forwards the frame to all links (except the one of B).
\item The switch learns the MAC address of host E. Since it knows where host
B is, it forwards the frame only to that host.
\item The switch learns the MAC address of host A. Since it knows where host
B is, it forwards the frame only to that host.
\item The switch knows where host A is, it forwards the frame only to that
host.
\end{enumerate}
\section{All things learned put together}
%todo
\end{document}
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