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diff --git a/ex10/kn10.pdf b/ex10/kn10.pdf
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diff --git a/ex10/kn10.tex b/ex10/kn10.tex
index 0a57ca3..2dd7b48 100644
--- a/ex10/kn10.tex
+++ b/ex10/kn10.tex
@@ -87,7 +87,7 @@
 
 \begin{document}
 %\header{BlattNr}{Tutor}{Abgabedatum}{Vorlesungsname}{Namen}{Semester}{Anzahl Aufgaben}
-\header{10}{}{2016-01-13}{Kommunikationsnetze}{\textit{Jonas Jaszkowic, 3592719}\\\textit{Jan-Peter Hohloch, 3908712}}{WS 15/16}{4}
+\header{10}{}{2016-01-13}{Kommunikationsnetze}{\textit{Jan-Peter Hohloch, 3908712}\\\textit{Jonas Jaszkowic, 3592719}}{WS 15/16}{4}
 \vspace{1cm}
 \Aufgabe{Token Bucket}{10+10}
 \begin{enumerate}
@@ -134,7 +134,6 @@
     %TODO: Quellen
 \end{enumerate}\pagebreak
 \Aufgabe{Wi-Fi - CSMA/CA, RTS/CTS}{10+10+10}
-%TODO: labeled figure
 \begin{enumerate}
     \item Time to transmit one packet and receiving the acknowledgement:
     \begin{align*}
@@ -143,7 +142,8 @@
          &= \frac{12}{11} \cdot 10^{-3}s + 0.112 \cdot 10^{-3}s \\
          &= 1.203 ms
     \end{align*}
-    The maximum throuput $D_1$ in $bit/s$ is therefore:
+    \includegraphics[width=.5\textwidth]{31.png}\\
+    The maximum throughput $D_1$ in $bit/s$ is therefore:
     \begin{align*}
       D_1 = (1 / 1.203 ms) * (12000~bit + 112~bit) = 10.07 \cdot 10^{6} bit/s
     \end{align*}
@@ -154,6 +154,7 @@
           &= 1573 \mu s \\
           &= 1.573 ms
     \end{align*}
+    \includegraphics[width=.5\textwidth]{32.png}\\
     The average throughput $D_2$ in $bit/s$ is therefore:
     \begin{align*}
       D_2 = (1 / 1.573 ms) * (12000 + 112 bit) = 7.70 \cdot 10^{6} bit/s
@@ -165,6 +166,7 @@
           &= 292 \mu s + 1573 \mu s \\
           &= 1.865 ms
     \end{align*}
+    \includegraphics[width=.5\textwidth]{33.png}\\
      The maximum throughput $D_3$ in $bit/s$ is therefore:
      \begin{align*}
        D_3 = (1 / 1.865 ms) * (12000 + 112 + 160 + 112) = 6.640 \cdot 10^{6} bit/s