= Solution
Reading the arrows directly from the original PDF gives, in particular, $c\to a$ and $e\to d$. Starting from zero <flow>, use the following <augmenting paths> and bottlenecks, in order:
* $s\to a\to d\to t$: $6$.
* $s\to b\to e\to t$: $5$.
* $s\to c\to e\to t$: $5$.
* $s\to b\to c\to d\to t$: $2$.
* $s\to b\to c\to e\to t$: $1$.
Each path is available in the current <residual network>. The resulting nonzero edge flows are
$$
\begin{aligned}
f_{sa}&=6,&f_{sb}&=8,&f_{sc}&=5,\\
f_{ad}&=6,&f_{bc}&=3,&f_{be}&=5,\\
f_{cd}&=2,&f_{ce}&=6,&f_{dt}&=8,&f_{et}&=11.
\end{aligned}
$$
The flows on $c\to a$ and $e\to d$ are zero. Capacity constraints and <flow conservation> hold at each interior vertex; net source outflow is $6+8+5=19$.
Take $S=\{s,a,b,c\}$. Its outgoing edges are $a\to d$, $c\to d$, $b\to e$, $c\to e$, with total capacity $6+2+5+6=19$. Thus the matching <flow> and <cut of a flow network> certificates give
$$
\boxed{\text{maximum flow value}=19,\qquad\text{minimum cut }(\{s,a,b,c\},\{d,e,t\})}.
$$
These are also exactly the source-reachable vertices in the final <residual network>. The original drawing below labels each edge by flow/capacity; the highlighted edges realise the minimum <cut of a flow network>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ib/paper-4-flow.png]
{title=A maximum flow and its edge capacities}
{description=Verified maximum flow, with each arrow labelled flow/capacity and the minimum cut marked in orange.}
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