OCR D2 — Question 4 11 marks

Exam BoardOCR
ModuleD2 (Decision Mathematics 2)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNetwork Flows
TypeCalculate cut capacity
DifficultyStandard +0.3 This is a standard network flows question testing routine application of cut calculations and the labelling procedure algorithm. Part (a)-(b) require simple arithmetic of arc capacities across given cuts, while (c)-(d) involve mechanical application of the flow-augmentation algorithm—all standard D2 textbook exercises with no novel problem-solving required. Slightly easier than average A-level due to being algorithmic rather than requiring mathematical insight.
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]{89e3545e-fa4b-47dd-8651-7c8f998df9e7-3_725_1303_274_340} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} Figure 2 above shows a capacitated, directed network. The number on each arc indicates the capacity of that arc.
  1. Calculate the values of cuts \(C _ { 1 }\) and \(C _ { 2 }\).
  2. Find the minimum cut and state its value. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{89e3545e-fa4b-47dd-8651-7c8f998df9e7-3_645_1316_1430_338} \captionsetup{labelformat=empty} \caption{Fig. 3}
    \end{figure} Figure 3 shows a feasible flow through the same network.
  3. State the values of \(x , y\) and \(z\).
  4. Using this as your initial flow pattern, use the labelling procedure to find a maximal flow. You should list each flow-augmenting route you use together with its flow. State how you know that you have found a maximal flow.

Question 4:
Part (a)
AnswerMarks
\(C_1 = 80\); \(C_2 = 94\)B2
Part (b)
AnswerMarks
Minimum cut: \(\{S, A, B, C, D, F\} \mid \{E, T\} = 57\)M1 A1
Part (c)
AnswerMarks
\(x = 15,\ y = 10,\ z = 36\)B2
Part (d)
AnswerMarks Guidance
Augment \(SCET\) by 2 and \(SCAET\) by 1 giving maximum flow diagram shownM2 A2
Max flow \(= 57\)
This is maximum flow as it is equal to the minimum cutB1 (11 marks total)
# Question 4:

## Part (a)
| $C_1 = 80$; $C_2 = 94$ | B2 | |

## Part (b)
| Minimum cut: $\{S, A, B, C, D, F\} \mid \{E, T\} = 57$ | M1 A1 | |

## Part (c)
| $x = 15,\ y = 10,\ z = 36$ | B2 | |

## Part (d)
| Augment $SCET$ by 2 and $SCAET$ by 1 giving maximum flow diagram shown | M2 A2 | |
| Max flow $= 57$ | | |
| This is maximum flow as it is equal to the minimum cut | B1 | **(11 marks total)** |

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\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]{89e3545e-fa4b-47dd-8651-7c8f998df9e7-3_725_1303_274_340}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

Figure 2 above shows a capacitated, directed network. The number on each arc indicates the capacity of that arc.\\
(a) Calculate the values of cuts $C _ { 1 }$ and $C _ { 2 }$.\\
(b) Find the minimum cut and state its value.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{89e3545e-fa4b-47dd-8651-7c8f998df9e7-3_645_1316_1430_338}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}

Figure 3 shows a feasible flow through the same network.\\
(c) State the values of $x , y$ and $z$.\\
(d) Using this as your initial flow pattern, use the labelling procedure to find a maximal flow. You should list each flow-augmenting route you use together with its flow.

State how you know that you have found a maximal flow.\\

\hfill \mbox{\textit{OCR D2  Q4 [11]}}