OCR D2 — Question 4 9 marks

Exam BoardOCR
ModuleD2 (Decision Mathematics 2)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNetwork Flows
TypeFind missing flow values
DifficultyStandard +0.3 This is a standard network flows question requiring application of the labelling procedure (Ford-Fulkerson algorithm) and max-flow min-cut theorem. While it involves multiple parts and systematic working, these are routine algorithmic procedures taught directly in D2 with no novel problem-solving required. The initial part (a) simply applies conservation of flow at nodes, making it slightly easier than average overall.
Spec7.04f Network problems: choosing appropriate algorithm

  1. A sheet is provided for use in answering this question.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{b8eb80d5-5af5-4a8b-8335-6fae95f3aa73-3_881_1310_319_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.

AnswerMarks Guidance
ContentMarks Guidance
(a) \(x = 2, y = 14\)B2
(b) (i), (ii) Augment SCT by 2 and SBECADT by 3 giving network diagram with: edges A-D (capacity 13), S-A (capacity 18), S-B (capacity 19), A-C (capacity 5), B-C (capacity 2), C-D (capacity 2), D-T (capacity 15), B-E (capacity 1), C-E (capacity 17), E-T (capacity 18), D-E (capacity 20), with all intermediate capacities labeledM2 A3
maximum flow = 53M2 A3
(c) (i) minimum cut = 53, passing through DT, CT and ETB1
(ii) max flow = min cut; it is not possible to get any more flow across this cutB1 (9)
| Content | Marks | Guidance |
|---------|-------|----------|
| (a) $x = 2, y = 14$ | B2 | |
| (b) (i), (ii) Augment SCT by 2 and SBECADT by 3 giving network diagram with: edges A-D (capacity 13), S-A (capacity 18), S-B (capacity 19), A-C (capacity 5), B-C (capacity 2), C-D (capacity 2), D-T (capacity 15), B-E (capacity 1), C-E (capacity 17), E-T (capacity 18), D-E (capacity 20), with all intermediate capacities labeled | M2 A3 | |
| maximum flow = 53 | M2 A3 | |
| (c) (i) minimum cut = 53, passing through DT, CT and ET | B1 | |
| (ii) max flow = min cut; it is not possible to get any more flow across this cut | B1 | (9) |
\begin{enumerate}
  \item A sheet is provided for use in answering this question.
\end{enumerate}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{b8eb80d5-5af5-4a8b-8335-6fae95f3aa73-3_881_1310_319_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.\\
(a) Find the values of $x$ and $y$.\\
(b) (i) Use the labelling procedure to find the maximum flow through this network, listing each flow-augmenting route you use together with its flow.\\
(ii) Show your maximum flow pattern and state its value.\\
(c) (i) Find a minimum cut, listing the arcs through which it passes.\\
(ii) Explain why this proves that the flow found in part (b) is a maximum.\\

\hfill \mbox{\textit{OCR D2  Q4 [9]}}