\section{The Ford-Fulkerson Algorithm}
We now have all the tools we need to construct an algorithm that finds a maximal flow. \\
It works as follows:
\begin{enumerate}
	\item[\texttt{00}] Take a weighted directed graph $G$.
	\item[\texttt{01}] Find any flow $F$ in $G$
	\item[\texttt{02}] Calculate $R$, the residual of $F$.
	\item[\texttt{03}] ~~~~If $S$ and $T$ are not connected in $R$, $F$ is a maximal flow. \texttt{HALT}.
	\item[\texttt{04}] Otherwise, find another flow $F_0$ in $R$.
	\item[\texttt{05}] Add $F_0$ to $F$
	\item[\texttt{06}] \texttt{GOTO 02}
\end{enumerate}

\problem{}
Run the Ford-Fulkerson algorithm on the following graph. \\
There is extra space on the next page.

\begin{center}
	\begin{tikzpicture}
		% Nodes
		\begin{scope}[layer = nodes]
			\node[main] (S) at (-5mm, 0mm) {$S$};
			\node[main] (A) at (20mm, 20mm) {$A$};
			\node[main] (B) at (20mm, 0mm) {$B$};
			\node[main] (C) at (20mm, -20mm) {$C$};
			\node[main] (D) at (50mm, 20mm) {$D$};
			\node[main] (E) at (50mm, 0mm) {$E$};
			\node[main] (F) at (50mm, -20mm) {$F$};
			\node[main] (T) at (75mm, 0mm) {$T$};
		\end{scope}

		% Edges
		\draw[->]
			(S) edge node[label] {$8$} (A)
			(S) edge node[label] {$7$} (B)
			(S) edge node[label] {$4$} (C)

			(A) edge node[label] {$2$} (B)
			(B) edge node[label] {$5$} (C)

			(A) edge node[label] {$3$} (D)
			(A) edge node[label] {$9$} (E)
			(B) edge node[label] {$6$} (E)
			(C) edge node[label] {$7$} (E)
			(C) edge node[label] {$2$} (F)

			(E) edge node[label] {$3$} (D)
			(E) edge node[label] {$4$} (F)

			(D) edge node[label] {$9$} (T)
			(E) edge node[label] {$5$} (T)
			(F) edge node[label] {$8$} (T)
		;

	\end{tikzpicture}
\end{center}

\begin{solution}
	The maximum flow is $17$.
\end{solution}

\vspace{5mm}

\pagebreak

\begin{center}
	\begin{tikzpicture}
		% Nodes
		\begin{scope}[layer = nodes]
			\node[main] (S) at (-5mm, 0mm) {$S$};
			\node[main] (A) at (20mm, 20mm) {$A$};
			\node[main] (B) at (20mm, 0mm) {$B$};
			\node[main] (C) at (20mm, -20mm) {$C$};
			\node[main] (D) at (50mm, 20mm) {$D$};
			\node[main] (E) at (50mm, 0mm) {$E$};
			\node[main] (F) at (50mm, -20mm) {$F$};
			\node[main] (T) at (75mm, 0mm) {$T$};
		\end{scope}

		% Edges
		\draw[->]
			(S) edge node[label] {$8$} (A)
			(S) edge node[label] {$7$} (B)
			(S) edge node[label] {$4$} (C)

			(A) edge node[label] {$2$} (B)
			(B) edge node[label] {$5$} (C)

			(A) edge node[label] {$3$} (D)
			(A) edge node[label] {$9$} (E)
			(B) edge node[label] {$6$} (E)
			(C) edge node[label] {$7$} (E)
			(C) edge node[label] {$2$} (F)

			(E) edge node[label] {$3$} (D)
			(E) edge node[label] {$4$} (F)

			(D) edge node[label] {$9$} (T)
			(E) edge node[label] {$5$} (T)
			(F) edge node[label] {$8$} (T)
		;

	\end{tikzpicture}
\end{center}

\vfill
\pagebreak

\problem{}
You are given a large network. How would you quickly find an upper bound for the number of iterations the Ford-Fulkerson algorithm will need to find a maximum flow?

\begin{solution}
	Each iteration adds at least one unit of flow. So, we will find a maximum flow in at most $\min(\text{flow out of } S,~\text{flow into } T)$ iterations.

	\vspace{2ex}

	A simpler answer could only count the flow on $S$.

\end{solution}

\vfill
\pagebreak