OCR D2 — Question 5 22 marks

Exam BoardOCR
ModuleD2 (Decision Mathematics 2)
Marks22
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNetwork Flows
TypeFind minimum cut
DifficultyStandard +0.3 This is a standard max-flow min-cut problem using the labelling procedure (Ford-Fulkerson algorithm), which is a core D2 topic. The network is small with only 6 nodes and straightforward structure. Part (a) is routine transcription, parts (b-c) apply textbook algorithms mechanically, and part (d) requires stating the max-flow min-cut theorem. While multi-step, it requires no novel insight—just careful execution of a learned procedure, making it slightly easier than average.
Spec7.02p Networks: weighted graphs, modelling connections7.04a Shortest path: Dijkstra's algorithm

The following matrix gives the capacities of the pipes in a system. \begin{array}{c|c|c|c|c|c|c|c} To & & S & T & A & B & C & D
From & & & & & & &
\hline S & & -- & -- & 16 & 26 & -- & --
T & & -- & -- & -- & -- & -- & --
A & & -- & -- & -- & -- & 13 & 5
B & & -- & 16 & -- & -- & -- & 11
C & & -- & 11 & -- & -- & -- & --
D & & -- & 11 & -- & -- & -- & --
\end{array}
  1. Represent this information as a digraph. [3 marks]
  2. Find the minimum cut, expressing it in the form \(\{\) \(\}|\{\) \(\}\), and state its value. [2 marks]
  3. Starting from having no flow in the system, use the labelling procedure to find a maximal flow through the system. You should list each flow-augmenting route you use, together with its flow. [5 marks]
  4. Explain how you know that this flow is maximal. [1 mark]
[11 marks]

AnswerMarks Guidance
(a) Network diagram with vertices S, A, B, C, D, T and edges: SA = 16, AC = 13, CT = 11, AD = 5, DT = 11, SB = 26, BD = 11, BT = 16M1 A2
(b) Minimum cut = {S, A, B, C, D} \{T} = 38 M1 A1
(c) Augment SACT by 11, SBT by 16, SBDT by 10 and SADT by 1, giving maximum flow diagram with flow values marked on edgesM2 A3
(d) Maximum as = to minimum cutB1 (11)
**(a)** Network diagram with vertices S, A, B, C, D, T and edges: SA = 16, AC = 13, CT = 11, AD = 5, DT = 11, SB = 26, BD = 11, BT = 16 | M1 A2 | |

**(b)** Minimum cut = {S, A, B, C, D} \| {T} = 38 | M1 A1 | |

**(c)** Augment SACT by 11, SBT by 16, SBDT by 10 and SADT by 1, giving maximum flow diagram with flow values marked on edges | M2 A3 | |

**(d)** Maximum as = to minimum cut | B1 | (11) |
The following matrix gives the capacities of the pipes in a system.

\begin{array}{c|c|c|c|c|c|c|c}
To & & S & T & A & B & C & D \\
From & & & & & & & \\
\hline
S & & -- & -- & 16 & 26 & -- & -- \\
T & & -- & -- & -- & -- & -- & -- \\
A & & -- & -- & -- & -- & 13 & 5 \\
B & & -- & 16 & -- & -- & -- & 11 \\
C & & -- & 11 & -- & -- & -- & -- \\
D & & -- & 11 & -- & -- & -- & -- \\
\end{array}

\begin{enumerate}[label=(\alph*)]
\item Represent this information as a digraph. [3 marks]
\item Find the minimum cut, expressing it in the form $\{$ $\}|\{$ $\}$, and state its value. [2 marks]
\item Starting from having no flow in the system, use the labelling procedure to find a maximal flow through the system. You should list each flow-augmenting route you use, together with its flow. [5 marks]
\item Explain how you know that this flow is maximal. [1 mark]
\end{enumerate}

[11 marks]

\hfill \mbox{\textit{OCR D2  Q5 [22]}}