diff --git a/ex10/kn10.pdf b/ex10/kn10.pdf
index 826b70f..8838fca 100644
--- a/ex10/kn10.pdf
+++ b/ex10/kn10.pdf
Binary files differ
diff --git a/ex10/kn10.tex b/ex10/kn10.tex
index 810210a..475bf9a 100644
--- a/ex10/kn10.tex
+++ b/ex10/kn10.tex
@@ -108,16 +108,12 @@
       \end{array}
      $
      \end{center}
-    \item Packet p6 is the first which is out-of-profile. This packet is delayed and appended to the waiting queue. Each time a token is added to the bucket, the system tries to send this delayed packet. Simultaneously incoming packets which are not in the queue are preferred. At time $t=3$ the buckets content is $C=1$ and packet p7 is arriving, consuming the last token. Therefore, the packets in the queue have to wait again. At $t=3.5$ the buckets content is $C=0$, the arriving packet p8 is delayed and appended to the waiting queue. At time $t=4$, the bucket contains again $C=1$, this last token is consumed by incoming packet p9. Thus, the queue does not change.% At time $t=4.5$ no new packet is incoming, since no token is added, $C=0$ stays the same.
-    At time $t=5$ the bucket contains is $C=1$. No new packet is incoming, therefore the delayed packet p6 can finally be sent consuming the token.
-    % At time $t=5.5$, to token is added and no packet is incoming, $C=0$ stays the same.
-    At time $t=6$ the bucket contains is $C=1$, since no packet is incoming, the delayed packet p8 can finally be sent consuming the last token.\\
-    %TODO: aren't packages in queue sent first?
-    By this we get the following times for transmission:\\
+    \item Packet p6 is the first which is out-of-profile. This packet is delayed and appended to the waiting queue. Each time a token is added to the bucket, the system tries to send this delayed packet.\\
+    Each arriving package is enqueued. By this we get the following times for transmission:\\
     \begin{tabular}
         {c|ccccccccc}
         p & p1 & p2 & p3 &p4 & p5 & p6 & p7 & p8 & p9\\\hline
-        t & 0  &0.5 &  1 &1.5&2   & 5 & 3 & 6 & 4
+        t & 0  &0.5 &  1 &1.5&2   & 3 & 4 & 5 & 6
     \end{tabular}
 \end{enumerate}\pagebreak
 \Aufgabe{Wi-Fi - Problems}{5+10+5}
@@ -131,14 +127,16 @@
     \begin{center}
       \includegraphics[width=\textwidth]{exposedstation.png}
     \end{center}
+    %TODO: Quellen
 \end{enumerate}
 \Aufgabe{Wi-Fi - CSMA/CA, RTS/CTS}{10+10+10}
+%TODO: labled figure
 \begin{enumerate}
     \item Time to transmit one packet and receiving the acknowledgment:
     \begin{align*}
       T_1 &= \frac{l_d}{N_d} + \frac{l_a}{N_a} \\
-         &= \frac{12000~bit}{11 \cdot 10^{6} ~Mbit/s} + \frac{112~bit}{1 \cdot 10^{6} ~Mbit/s} \\
-         &= \frac{12}{11} \cdot 10^{-3} + 0.112 \cdot 10^{-3} \\
+         &= \frac{12000~bit}{11 \cdot 10^{6} ~bit/s} + \frac{112~bit}{1 \cdot 10^{6} ~bit/s} \\
+         &= \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: