intruduced tkitz and removed postscripts should compile with normal latxmk now

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Sheppy 2015-10-19 17:08:39 +02:00
parent 3565b1c745
commit 7faa0241c9

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@ -2,6 +2,9 @@
\documentclass{article} \documentclass{article}
\usepackage{amsmath} \usepackage{amsmath}
%\usepackage{epsf} %\usepackage{epsf}
%tkitz stuff
\usepackage{tikz}
\usepackage{listings}
\DeclareMathSizes{10}{10}{10}{10} \DeclareMathSizes{10}{10}{10}{10}
\setlength{\parindent}{0pt} \setlength{\parindent}{0pt}
\title{Klausur M\"arz 2015} \title{Klausur M\"arz 2015}
@ -37,10 +40,178 @@
\textbf{a)}\\ \textbf{a)}\\
- selbsttaktend: Empfänger kann Sendertakt aus Signal gewinnen \\ - selbsttaktend: Empfänger kann Sendertakt aus Signal gewinnen \\
- gleichstromfrei: kein Gleichanteil im elektrischen Signal \\ - gleichstromfrei: kein Gleichanteil im elektrischen Signal \\
\textbf{b)}\\ \textbf{b)}\\\\
%Postscript include \begin{tikzpicture}[xscale=2]
%\epsfxsize=10pt \draw (0,0) -- (0.25,0) -- (0.25,0.5) -- (0.75,0.5) -- (0.75,0) -- (1,0) --
%\includegraphics[scale=1]{/postscripts/test.eps} (1,0.5) -- (1.25,0.5) -- (1.25,0) -- (1.5,0) -- (1.5,0.5) -- (1.75,0.5) --
(1.75,0) -- (2.25,0) -- (2.25,0.5) -- (2.5,0.5) -- (2.5,0) -- (2.75,0) --
(2.75,0.5) -- (3.25,0.5) -- (3.25,0) -- (3.5,0) -- (3.5,0.5) -- (3.75,0.5) --
\end{document} (3.75,0) -- (4,0);
\end{tikzpicture}\\
-\hspace{3mm}0\hspace{8mm}1\hspace{8.5mm}1\hspace{8mm}1\hspace{9mm}0\hspace{8mm}
0\hspace{8mm}1\hspace{8mm}1 \;\;- \\
\section{TCP}
\subsection{Round-Trip-Time}
\[ RTT = 2*d_{prop} + 10ms = 2*45ms+10ms = 100ms \]
\subsection{TCP Verlauf}
$A->B: SYN, Seq=1, Ack=XX \\
A<-B: SYN, ACK, Seq=XX+1, Ack=2 \\
A->B: ACK, Seq=2, Ack=XX+2 \\
\textbf{\textit{Verbindung erfolgreich aufgebaut}} \\
A<-B: Data, Seq=XX+2, Ack=3 \hspace*{1cm}|\;\;CW = 1 \\
A->B: ACK, Seq=3, Ack=XX+3 \\
\textbf{\textit{BANG!}} \\
A<-B: Data, Seq=XX+3, Ack=4 \\
A<-B: Data, Seq=XX+4, Ack=5 \hspace*{1cm}|\;\; CW = 2 \\
\textbf{\textit{Demnach weiterer Verlauf nach tau:}} \\
A<-B: Data, Seq=XX+3, Ack=4 \hspace*{1cm}|\;\; CW = 1 \\
A->B: ACK, Seq=4, Ack=XX+4 \\
A<-B: Data, Seq=XX+4, Ack=5 \hspace*{1cm}|\;\; CW = 2 \\
A<-B: Data, Seq=XX+5, Ack=6 \\
A->B: ACK, Seq=5, Ack=XX+5 \\
A->B: ACK, Seq=6, Ack=XX+6 \\
$\\
Nach dem 4. Paket wurden 5000 Bytes uebertragen.
\subsection{Leistungsanalyse}
$d_{prop} = 45ms$\\\\
$d_{trans} = \frac{L}{R} = 1ms\;\;denn\;10\frac{MBit}{s} =
10^6 \frac{Bit}{s} = 1250\frac{Byte}{s}$\\\\
$\frac{O}{L} = \frac{5000}{1250} = 4$\;\;\;\;$t = 2RTT + 4*1ms + 4*45ms = 384 (mit\;RTT\;als\;100ms)$
\newpage
\section{Programieraufgabe TCP-FIN}
\begin{lstlisting}[language=java]
public class TCPFIN extends base {
final int FIN_WAIT_1 = 1;
final int FIN_WAIT_2 = 2;
final int CLOSING = 3;
final int TIME_WAIT = 4;
final int CLOSED = 5;
int state;
void close() throws IOException {
state = FIN_WAIT_1;
send(FIN);
}
void receive(int flag) throws IOException {
if (flag == FIN) {
if (state == FIN_WAIT_1) {
state = CLOSING;
} else if (state == FIN_WAIT_2) {
state = TIME_WAIT;
startTimer();
}
send(ACK);
} else if (flag == ACK) {
if (state == FIN_WAIT_1) {
state = FIN_WAIT_2;
} else if (state == CLOSING) {
state = TIME_WAIT;
startTimer();
}
}
}
void timeout() throws WrongStateException {
if (state != TIME_WAIT) {
throw new WrongStateException();
}
state = CLOSED;
}
}
\end{lstlisting}
\newpage
%unformated codeblock end
\section{Routingverfahren}
\subsection{Routingverfahren}
\subsubsection{Darstellung}
\begin{tikzpicture}
%Knoten
\draw(0,2) circle [radius=0.3];
\draw(3,2) circle [radius=0.3];
\draw(6,2) circle [radius=0.3];
\draw(0,0) circle [radius=0.3];
\draw(3,0) circle [radius=0.3];
%Beschriftung Knoten
\node[purple] at (0,2) {P};
\node[purple] at (3,2) {A};
\node[purple] at (6,2) {S};
\node[purple] at (0,0) {C};
\node[purple] at (3,0) {B};
%Linien
\draw (0.3,0) -- (2.7,0);
\draw (0,0.3) -- (0,1.7);
\draw (0.3,2) -- (2.7,2);
\draw (3,0.3) -- (3,1.7);
\draw (3.3,2) -- (5.7,2);
\draw (3.3,0) -- (6,1.7);
\draw (3,0.3) -- (0,1.7);
%Beschriftungen Linien
\node [left,red] at (0,1) {1};
\node [above,red] at (1.5,2) {8};
\node [above,red] at (1.5,1) {4};
\node [below,red] at (1.5,0) {2};
\node [left,red] at (3,1) {2};
\node [red] at (4.5,1) {4};
\node [above,red] at (4.5,2) {1};
\end{tikzpicture}
\subsubsection{Minimal aufspannender Baum}
\begin{tabular}{| l | c | c | c | c | c |}
\hline
S & N' & D(A),p(A) & D(B),p(B) & D(C),p(C) & D(P),p(P) \\ \hline
test&test &test &test &test &test \\ \hline
\end{tabular}
\end{document}