diff --git a/ad/AD-Gruppe_8_Koehler_Krabbe_Martens_Blatt5.tex b/ad/AD-Gruppe_8_Koehler_Krabbe_Martens_Blatt5.tex
index 5ed1c08..1e2a697 100644
--- a/ad/AD-Gruppe_8_Koehler_Krabbe_Martens_Blatt5.tex
+++ b/ad/AD-Gruppe_8_Koehler_Krabbe_Martens_Blatt5.tex
@@ -99,4 +99,145 @@ Jim Martens (6420323)}
 	\subsection{} %a
 	\subsection{} %b
 \section{} %6
+		\begin{tikzpicture}[shorten >=1pt,node distance=1.1cm]
+			\node[state] (s) {s};		
+			\node[state] (a) [above right=1.5 and 1.0 of s] {A};
+			\node[state] (b) [below right=1.5 and 1.0 of s] {B};
+			\node[state] (c) [above right=0.5 and 1.0 of b] {C};
+			\node[state] (d) [below right=0.5 and 1.0 of b] {D};
+			\node[state] (e) [above right=0.5 and 1.0 of a] {E};
+			\node[state] (f) [below right=0.5 and 1.0 of a] {F};
+			\node[state] (g) [above right=0.5 and 2.0 of c] {G};
+			
+			\path[every node/.style={font=\scriptsize},->] 	
+				(s) edge node [above] {30} (a)
+				(s) edge node [below] {18} (b)
+				(a) edge node [above] {20} (e)
+				(a) edge node [below] {22} (f)
+				(b) edge node [above] {21} (c)
+				(b) edge node [below] {38} (d)
+				(f) edge node [above] {9} (g)
+				(c) edge node [below] {9} (g);
+		\end{tikzpicture}
+		
+		Da dies ein gerichteter Graph ohne Zyklen ist und auch keine negativen Kantengewichte vorkommen, kann der Dijkstra-Algorithmus angewendet werden. Da von 9 bis 17 Uhr Fahrer bezahlt werden müssen, kommen nur die Pfade in Betrachtung, die zu einer Senke führen. Senken sind D, E und G. Derjenige dieser Knoten welcher als erster vom Algorithmus gefunden wird (über den eindeutig bestimmbaren Weg), ist auch der kürzeste Pfad vom Startknoten aus, der die Bedingungen erfüllt.
+		
+		Es ergibt sich folgende Abfolge:
+		\begin{verbatim}
+		S = {s}
+		d(s) = 0
+		while 1.
+		    U = {A, B}
+		    for all u in U -> u = A
+		        for all pre(u) in S that are predecessors of u -> pre(u) = s
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 0 + 30
+		    for all u in U -> u = B
+		        for all pre(u) in S that are predecessors of u -> pre(u) = s
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 0 + 18
+		    u* = B
+		    d(u*) = 18
+		    S = {s, B}
+		while 2.
+		    U = {A, C, D}
+		    for all u in U -> u = A
+		        for all pre(u) in S that are predecessors of u -> pre(u) = s
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 0 + 30
+		    for all u in U -> u = C
+		        for all pre(u) in S that are predecessors of u -> pre(u) = B
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 18 + 21
+		    for all u in U -> u = D
+		        for all pre(u) in S that are predecessors of u -> pre(u) = B
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 18 + 38
+		    u* = A
+		    d(u*) = 30
+		    S = {S, B, A}
+		while 3.
+		    U = {C, D, E, F}
+		    for all u in U -> u = C
+		        for all pre(u) in S that are predecessors of u -> pre(u) = B
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 18 + 21
+		    for all u in U -> u = D
+		        for all pre(u) in S that are predecessors of u -> pre(u) = B
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 18 + 38
+		    for all u in U -> u = E
+		        for all pre(u) in S that are predecessors of u -> pre(u) = A
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 30 + 20
+		    for all u in U -> u = F
+		        for all pre(u) in S that are predecessors of u -> pre(u) = A
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 30 + 22
+		    u* = C
+		    d(u*) = 39
+		    S = {s, B, A, C}
+		while 4.
+		    U = {D, E, F, G}
+		    for all u in U -> u = D
+		        for all pre(u) in S that are predecessors of u -> pre(u) = B
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 18 + 38
+		    for all u in U -> u = E
+		        for all pre(u) in S that are predecessors of u -> pre(u) = A
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 30 + 20
+		    for all u in U -> u = F
+		        for all pre(u) in S that are predecessors of u -> pre(u) = A
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 30 + 22
+		    for all u in U -> u = G
+		        for all pre(u) in S that are predecessors of u -> pre(u) = C
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 39 + 9
+		    u* = G
+		    d(u*) = 48
+		    S = {s, B, A, C, G}
+		while 5.
+		    U = {D, E, F}
+		    for all u in U -> u = D
+		        for all pre(u) in S that are predecessors of u -> pre(u) = B
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 18 + 38
+		    for all u in U -> u = E
+		        for all pre(u) in S that are predecessors of u -> pre(u) = A
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 30 + 20
+		    for all u in U -> u = F
+		        for all pre(u) in S that are predecessors of u -> pre(u) = A
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 30 + 22
+		    u* = E
+		    d(u*) = 50
+		    S = {s, B, A, C, G, E}
+		while 6.
+		    U = {D, F}
+		    for all u in U -> u = D
+		        for all pre(u) in S that are predecessors of u -> pre(u) = B
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 18 + 38
+		    for all u in U -> u = F
+		        for all pre(u) in S that are predecessors of u -> pre(u) = A
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 30 + 22
+		    u* = F
+		    d(u*) = 52
+		    S = {s, B, A, C, G, E, F}
+		while 7.
+		    U = {D}
+		    for all u in U -> u = D
+		        for all pre(u) in S that are predecessors of u -> pre(u) = B
+		            d'(u, pre(u)) = d(pre(u)) + w(pre(u), u)
+		            d'(u, pre(u)) = 18 + 38
+		    u* = D
+		    d(u*) = 56
+		    S = {s, B, A, C, G, E, F, D}
+		\end{verbatim}
+		
+		Wie zu sehen ist wird die Senke G zuerst erreicht. Folgt man dem Weg zu G, so ergibt sich, dass der kürzeste Pfad von s über B und C nach G führt. Weniger Kosten als 48 sind daher nicht möglich.
 \end{document}