diff --git a/ex09/kn09.pdf b/ex09/kn09.pdf
index 79dd795..6817cab 100644
--- a/ex09/kn09.pdf
+++ b/ex09/kn09.pdf
Binary files differ
diff --git a/ex09/kn09.tex b/ex09/kn09.tex
index 87e6f26..6dcc40a 100644
--- a/ex09/kn09.tex
+++ b/ex09/kn09.tex
@@ -104,7 +104,7 @@
             (C) edge (D)
             (S) edge node[above right]{\tiny{3}}(E)
             (E) edge (F);
-    \end{tikzpicture}\vspace*{-2cm}
+    \end{tikzpicture}\vspace*{-2cm}\textcorr{evtl. LANs zusammenfassen}
     \item $B_4$: \begin{tabular}{c|c}
     adr  & port \\\hline
     A & 1\\
@@ -114,7 +114,7 @@
     E&3\\
     F&3
     \end{tabular}
-    \item Brides and LANs are involved as nodes. The bridge-IDs have to be configured, which port to use and which to block is learned. The bridges broadcast their ID each bridge determines if it is the root bridge and which ports have to be used and which are blocked. The port nearest to the root and the designated ports (shortest connection for segment to root) are used, others blocked.
+    \item \textcorr{tree is spanned...} Brides and LANs\textcorr{ bzw. only bridges} are involved as nodes. The bridge-IDs \textcorr{may} be configured, which port to use and which to block is learned. The bridges broadcast their ID each bridge determines if it is the root bridge and which ports have to be used and which are blocked. The port nearest to the root and the designated ports (shortest connection for segment to root) are used, others blocked.
     \item Network:\\
     \begin{tikzpicture}[auto,node distance=2.0cm]
         \node[state] (A) {A};
@@ -164,7 +164,7 @@
             (B2) edge node[above right]{\tiny{f:2}} (F)
             (E) edge[dotted] node[above left]{\tiny{b:2}} (B3)
             (B3) edge node[below]{\tiny{f:1}} (A);
-    \end{tikzpicture}
+    \end{tikzpicture}\\\textcorr{maybe give root-port}
     \item $B_1$: \begin{tabular}{c|c}
     adr  & port \\\hline
     A & 1\\
@@ -210,7 +210,7 @@
             (B2) edge node[above right]{\tiny{f:2}} (F)
             (E) edge node[above left]{\tiny{f:2}} (B3)
             (B3) edge node[below]{\tiny{f:1}} (A);
-    \end{tikzpicture}
+    \end{tikzpicture}\\\textcorr{$B_4$ should use port 2 since $B_2$ is root}
     \item $B_2$: \begin{tabular}{c|c}
     adr  & port \\\hline
     A & 2\\
@@ -239,9 +239,9 @@
 	\item
 	\begin{enumerate}
 		\item \textbf{Photonic layer}: physical specification of optical fiber channel. Encoding with unipolar NRZ, 1 is presence of light, 0 is absence of light.
-		\item \textbf{Section layer}: handles framing, scrambling and error control.
-		\item \textbf{Line layer}: provides access to STS-1 signals for adding or dropping individual multiplexed signals on a line.
-		\item \textbf{Path layer}: combines n STS-1 signals to an STS-n signal. Combines lower bit rate signals into synchronous payload envelopes.
+		\item \textbf{Section layer}: handles framing, scrambling and error control. \textcorr{Terminated at regenerator}
+		\item \textbf{Line layer}: provides access to STS-1 signals for adding or dropping individual multiplexed signals on a line. \textcorr{Terminated at ADM}
+		\item \textbf{Path layer}: combines n STS-1 signals to an STS-n signal. Combines lower bit rate signals into synchronous payload envelopes. \textcorr{Reaches form STS multiplexer to STS multiplexer}
 	\end{enumerate}
 	\item A SONET regenerator is something like a repeater for optical signals. Whereas a repeater is level one, a regenerator acts on level two because it is also responsible for framing, scrambling and error control. The frame is regenerated, management information is added. A repeater would only recover every single bit.
 	\item SONET is called a synchronous network because it uses STS (\textbf{synchronous} transport signal) multiplexing and demultiplexing. That means the receiver has to be synchronized with the sender to interpret the signal correctly.
@@ -249,9 +249,9 @@
 \Aufgabe{ATM}{10+5}
 \begin{enumerate}
     \item One cell has 53byte. So at most we have to wait for he transmission of 53byte at each switch. With a transmission rate of 43 Mbit/s we get:\\
-    $\frac{5\cdot2\cdot(53\cdot8)}{43 Mbit/s}=9.86\cdot 10^{-5}s$\\
+    $\frac{5\cdot2\cdot(53\cdot8)}{43 Mbit/s}=9.86\cdot 10^{-5}s$\ok\\
     Minimal is half of the time (no waiting but only transmission at each switch):\\
-    $\frac{5\cdot(53\cdot8)}{43 Mbit/s}=4.93\cdot 10^{-5}s$
-    \item $20byte+20byte+8byte=48byte$ which is exactly the payload of an ATM cell, so we need one cell.
+    $\frac{5\cdot(53\cdot8)}{43 Mbit/s}=4.93\cdot 10^{-5}s$\ok
+    \item $20byte+20byte+8byte=48byte$ which is exactly the payload of an ATM cell, so we need one cell.\ok
 \end{enumerate}
 \end{document}
diff --git a/korr.txt b/korr.txt
new file mode 100644
index 0000000..83c3492
--- /dev/null
+++ b/korr.txt
@@ -0,0 +1,2 @@
+Exercise 1 93,00 1.1: Layer 4 unklar, 1.2 Erklärung nicht ausreichend
+Exercise 2 95,00  gar nicht gefordert, die Stichworte hätten gereicht 1.2: 1.3: 1.4: 1.5: 1.6:Inner und outer header vertauscht 1,7:Inner und outer header vertauscht