Edexcel D1 — Question 3 11 marks

Exam BoardEdexcel
ModuleD1 (Decision Mathematics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNetwork Flows
TypeFind missing flow values
DifficultyStandard +0.3 This is a standard D1 network flows question requiring application of the labelling procedure and max-flow min-cut theorem. While it involves multiple parts and systematic working, these are routine algorithmic procedures taught directly in the specification with no novel problem-solving required. The conceptual demand is low compared to typical A-level pure maths questions.
Spec7.04f Network problems: choosing appropriate algorithm

3. This question should be answered on the sheet provided. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{acc09687-11a3-4392-af17-3d4d331d5ab4-04_883_1317_317_315} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} Figure 2 shows a capacitated, directed network.
The numbers in bold denote the capacities of each arc.
The numbers in circles show a feasible flow of 48 through the network.
  1. Find the values of \(x\) and \(y\).
    1. Use the labelling procedure to find the maximum flow through this network, listing each flow-augmenting route you use together with its flow.
    2. Show your maximum flow pattern and state its value.
    1. Find a minimum cut, listing the arcs through which it passes.
    2. Explain why this proves that the flow found in part (b) is a maximum.
      (2 marks)

AnswerMarks Guidance
(a)\(x = 2, y = 14\) M2 A1
(b) (i)e.g. augment \(SC7\) by 2 and \(SBECADT\) by 3
(b) (ii)Diagram showing augmentations with maximum flow = 53 M3 A3, A1
(c) (i)minimum cut = 53, passing through \(DT\), \(CT\) and \(ET\) B1
(c) (ii)max flow = min cut; it is not possible to get any more flow across this cut B1
**(a)** | $x = 2, y = 14$ | M2 A1 | |

**(b) (i)** | e.g. augment $SC7$ by 2 and $SBECADT$ by 3 | | |

**(b) (ii)** | Diagram showing augmentations with maximum flow = 53 | M3 A3, A1 | |

**(c) (i)** | minimum cut = 53, passing through $DT$, $CT$ and $ET$ | B1 | |

**(c) (ii)** | max flow = min cut; it is not possible to get any more flow across this cut | B1 | (11) |
3. This question should be answered on the sheet provided.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{acc09687-11a3-4392-af17-3d4d331d5ab4-04_883_1317_317_315}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

Figure 2 shows a capacitated, directed network.\\
The numbers in bold denote the capacities of each arc.\\
The numbers in circles show a feasible flow of 48 through the network.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $x$ and $y$.
\item \begin{enumerate}[label=(\roman*)]
\item Use the labelling procedure to find the maximum flow through this network, listing each flow-augmenting route you use together with its flow.
\item Show your maximum flow pattern and state its value.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find a minimum cut, listing the arcs through which it passes.
\item Explain why this proves that the flow found in part (b) is a maximum.\\
(2 marks)
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{Edexcel D1  Q3 [11]}}