From 7d44f9b2ad08d8fd53282add899fa0ebaa3bc941 Mon Sep 17 00:00:00 2001
From: Camil Staps
Date: Sun, 22 May 2016 17:48:05 +0200
Subject: Start assignment 7

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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
+			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.
+	\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 --- 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 %todo
+	\item %todo
+\end{enumerate}
+
+\section{Self-learning}
+\begin{enumerate}[label=(\roman*)]
+	\item %todo
+	\item %todo
+	\item %todo
+	\item %todo
+\end{enumerate}
+
+\section{All things learned put together}
+%todo
+
+\end{document}
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