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\usepackage{amsmath} %erweiterte Mathe-Zeichen
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\usepackage[utf8]{inputenc} %Umlaute & Co
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\usepackage{amssymb, amsthm} %noch stärker erweiterte Mathe-Zeichen
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{\bf #4}\\
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{\Large\bf Sheet #1}

{(Hand in #3)}
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%counts the exercisenumber
\newcounter{n}

%Kommando für Aufgaben
%\Aufgabe{AufgTitel}{Punktezahl}
\newcommand{\Aufgabe}[2]{\stepcounter{n}
    \textbf{Exercise \arabic{n}: #1} (#2 Points)}

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\newcommand{\ok}{\begin{corr}
            $\checkmark$
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\begin{document}
%\header{BlattNr}{Tutor}{Abgabedatum}{Vorlesungsname}{Namen}{Semester}{Anzahl Aufgaben}
\header{7}{}{2015-12-02}{Kommunikationsnetze}{\textit{Jonas Jaszkowic, 3592719}\\\textit{Jan-Peter Hohloch, 3908712}}{WS 15/16}{4}
\vspace{1cm}
\Aufgabe{Banyan Switches}{20}\\
We assume $N=2^k,\ k\in\mathds{N}_0$, otherwise there are cases where the first stage is blocking.\\[0.2cm]
We have $2N$ port values, $N$ are $< N$, $N$ are $\geq N$.\\
Assuming there might be an $a_i,b_{N-i+1}$ where both have the same most significant bit (here we use the assumption from above) we distinguish two cases:
\begin{enumerate}[c1]
    \item $a_i\geq N \wedge b_{N-i+1}\geq N$\\
        To assign numbers to all ports $a_k$ we need an additional amount of $N-i$ numbers $\geq N$ ($k>i$). For the ports $b_k$ we need $i$ ($k\geq N-i+1$). In sum that are $1+N-i+i=N+1>N\ \lightning$
    \item $a_i< N \wedge b_{N-i+1}< N$\\
        To assign numbers to all ports $a_k$ we need $i$ numbers $<N$ ($k\leq i$). For the ports $b_k$ we need $N-i+1$ ($k<N-i+1$). In sum that are $N-i+1+i=N+1>N\ \lightning$
\end{enumerate}
So all possibilities where there is blocking in first stage aren't possible which means the first stage is non-blocking.\ok\qed
\\[0.5cm]
\Aufgabe{Flow Control and ARQ Protocols}{22}
\begin{enumerate}
    \item Flow Control means the controlling of the transmission of data. This is necessary for example if the receiver is slower or has higher traffic than the sender and can't keep up. For flow control ACK message can be used to acknowledge received messages (used in the examples from he lecture).
    \begin{corr}
        adjustment of sending rate
    \end{corr}
    \item The specific package may be received anywhere in the window, i.e. newer packages may be received before older ones. However this does not directly lead to retransmission but only if the window for receiving is left. To implement this window a buffer is necessary.
    \begin{corr}
        \begin{itemize}
            \item packages may be lost
            \item transmission errors
            \item reordering
            \item[$\rightarrow$] received packages have to be saved
        \end{itemize}
    \end{corr}
    \item $n,m\in\mathds{N}$\\
        \begin{tabular}{c||c|c|c}
            &SAW&GBN&SR\\\hline
            sender window & 1 & n & n\\
            receiver window & 1 & 1 & m\\
            repetition & if necessary & all since last ACK & if necessary
        \end{tabular}\\[0.5cm]
        As we see in the table the main difference is the size of sending and receiving window. By this differences all the other differences can be explained (e.g. accepting only the matching frame or all in the window).\\
        Common for all protocols is that they use ACK messages to acknowledge receiving (implement flow control by this) and that they only have one sender and one receiver.
        \begin{corr}
            \begin{itemize}
                \item common:
                \begin{itemize}
                    \item implementation of flow control
                    \item use send/receive buffer
                    \item retransmission after timeout
                    \item ACK for each package
                \end{itemize}
                \item differences:
                \begin{itemize}
                    \item windows (see table)
                    \item GBN drop packages out of order
                    \item $n,m > 1$
                    \item SR pass packages to higher layer as soon as they are ordered
                \end{itemize}
            \end{itemize}
        \end{corr}
\end{enumerate}
\Aufgabe{Framing}{22}
\begin{enumerate}
    \item Byte-oriented protocols determine the size of flags to $8k$ bit for $k\in\mathds{N}_0$. For bit-oriented protocols they may have any size. Both protocols use the according stuffing technique to prevent a control pattern from occurring in the data. In opposite to character-oriented protocols both of them don't use a special encoding and are transparent by that.
    \item Bit stuffing means the preventing of the occurrence of control bit sequences (flags) in the data by adding 0-bits after sequences of 5 1s after a 0 (0111111 is the control sequence), by this it is broken and won't be interpreted. This added 0s have to be deleted in unstuffing process.\\
    Byte stuffing for the same purpose has an ESC-byte. The byte following this one won't be interpreted as control byte. Also these escape-bytes have to be removed in the unstuffing.
    \item PPP is byte-oriented and so uses byte stuffing.
\end{enumerate}
\Aufgabe{ALOHA and Slotted ALOHA}{36}
\begin{enumerate}
    \item $\frac{64kbit/s}{256bit/s}=25\textcorr{0}$ messages can be transmitted according to the bandwidth. The maximum of thoughput we reach at $G=0.5$, so the maximum number of terminals is $x^*=\left\lfloor\frac{25\textcorr{0}}{2}\right\rfloor=12\textcorr{5}$.
    \item For Slotted ALOHA the maximum is at $G=1$, so the best number of terminals is $x'^*=25\textcorr{0}$
    \item The formula is $x^*=\left\lfloor\frac{N}{2tm}\right\rfloor$, where $m$ is the message size, $t$ the messages per second. This leads to:
    \begin{enumerate}
        \item doubling $t$ halves $x^*$
        \item doubling $m$ halves $x^*$
        \item doubling $N$ doubles $x^*$
    \end{enumerate}
    All those with the limitation of the floor-function so only approximately.
\end{enumerate}
    \begin{corr}
        Musterlösung mit Übertragungszeit statt Anzahl der Messages
    \end{corr}
\end{document}