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