AQA D2 2011 June — Question 5 14 marks

Exam BoardAQA
ModuleD2 (Decision Mathematics 2)
Year2011
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNetwork Flows
TypeState maximum flow along specific routes
DifficultyModerate -0.5 This is a standard network flows question testing routine application of max-flow min-cut theorem and Ford-Fulkerson labelling procedure. While it requires multiple steps (finding a cut, initial flows on specific routes, applying labelling procedure, handling a capacity constraint), these are all textbook techniques from D2 with no novel problem-solving required. The structured parts guide students through the algorithm systematically, making it easier than average A-level maths questions overall.
Spec7.04a Shortest path: Dijkstra's algorithm7.04b Minimum spanning tree: Prim's and Kruskal's algorithms7.04c Travelling salesman upper bound: nearest neighbour method7.04d Travelling salesman lower bound: using minimum spanning tree7.04e Route inspection: Chinese postman, pairing odd nodes7.04f Network problems: choosing appropriate algorithm

5 The network shows the evacuation routes along corridors in a college, from two teaching areas to the exit, in case of a fire alarm sounding. \includegraphics[max width=\textwidth, alt={}, center]{1aca4e91-d1b3-4a78-8880-e42a4fbf3ddb-14_729_1013_434_497} The two teaching areas are at \(A\) and \(G\) and the exit is at \(X\). The number on each edge represents the maximum number of people that can travel along a particular corridor in one minute.
  1. Find the value of the cut shown on the diagram.
  2. Find the maximum flow along each of the routes \(A B D X , G F B X\) and \(G H E X\) and enter their values in the table on Figure 3 opposite.
    1. Taking your answers to part (b) as the initial flow, use the labelling procedure on Figure 3 to find the maximum flow through the network. You should indicate any flow augmenting routes in the table and modify the potential increases and decreases of the flow on the network.
    2. State the value of the maximum flow, and, on Figure 4, illustrate a possible flow along each edge corresponding to this maximum flow.
  3. During one particular fire drill, there is an obstruction allowing no more than 45 people per minute to pass through vertex \(B\). State the maximum number of people that can move through the network per minute during this fire drill. \begin{figure}[h]
    \captionsetup{labelformat=empty} \caption{Figure 3} \includegraphics[alt={},max width=\textwidth]{1aca4e91-d1b3-4a78-8880-e42a4fbf3ddb-15_905_1559_331_292}
    \end{figure} Figure 4 \includegraphics[max width=\textwidth, alt={}, center]{1aca4e91-d1b3-4a78-8880-e42a4fbf3ddb-15_689_851_1370_598} Maximum flow is \(\_\_\_\_\) people per minute.

Question 5:
Part (a):
AnswerMarks Guidance
Cut value \(= 27 + 20 + 6 + 24 = 77\) (edges crossing cut in correct direction)B1 \(\mathbf{77}\)
Part (b):
AnswerMarks Guidance
Route \(ABDX\): flow \(= 40\) (limited by \(AB=16\)... minimum of edges)B1 \(ABDX = 16\)
Route \(GFBX\): flow limited by minimum capacity along routeB1 \(GFBX = 19\)
Route \(GHEX\): flow limited by minimum capacity along routeB1 \(GHEX = 20\)
Part (c)(i):
AnswerMarks
Correct application of labelling procedure showing augmenting pathM1
Correct potential increases/decreases shownA1
At least one valid augmenting route foundA1
All updates correctA1 A1
Part (c)(ii):
AnswerMarks Guidance
Maximum flow \(= 77\) people per minuteB1 Correct value
Correct flow values on Figure 4 consistent with maximumB1
Part (d):
AnswerMarks Guidance
With \(B\) restricted to 45, maximum flow through \(B = 45\); total maximum \(= 45 + 20 = 65\)M1 A1 65 people per minute
# Question 5:

## Part (a):
| Cut value $= 27 + 20 + 6 + 24 = 77$ (edges crossing cut in correct direction) | B1 | $\mathbf{77}$ |

## Part (b):
| Route $ABDX$: flow $= 40$ (limited by $AB=16$... minimum of edges) | B1 | $ABDX = 16$ |
| Route $GFBX$: flow limited by minimum capacity along route | B1 | $GFBX = 19$ |
| Route $GHEX$: flow limited by minimum capacity along route | B1 | $GHEX = 20$ |

## Part (c)(i):
| Correct application of labelling procedure showing augmenting path | M1 | |
| Correct potential increases/decreases shown | A1 | |
| At least one valid augmenting route found | A1 | |
| All updates correct | A1 A1 | |

## Part (c)(ii):
| Maximum flow $= 77$ people per minute | B1 | Correct value |
| Correct flow values on Figure 4 consistent with maximum | B1 | |

## Part (d):
| With $B$ restricted to 45, maximum flow through $B = 45$; total maximum $= 45 + 20 = 65$ | M1 A1 | **65 people per minute** |

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5 The network shows the evacuation routes along corridors in a college, from two teaching areas to the exit, in case of a fire alarm sounding.\\
\includegraphics[max width=\textwidth, alt={}, center]{1aca4e91-d1b3-4a78-8880-e42a4fbf3ddb-14_729_1013_434_497}

The two teaching areas are at $A$ and $G$ and the exit is at $X$.

The number on each edge represents the maximum number of people that can travel along a particular corridor in one minute.
\begin{enumerate}[label=(\alph*)]
\item Find the value of the cut shown on the diagram.
\item Find the maximum flow along each of the routes $A B D X , G F B X$ and $G H E X$ and enter their values in the table on Figure 3 opposite.
\item \begin{enumerate}[label=(\roman*)]
\item Taking your answers to part (b) as the initial flow, use the labelling procedure on Figure 3 to find the maximum flow through the network. You should indicate any flow augmenting routes in the table and modify the potential increases and decreases of the flow on the network.
\item State the value of the maximum flow, and, on Figure 4, illustrate a possible flow along each edge corresponding to this maximum flow.
\end{enumerate}\item During one particular fire drill, there is an obstruction allowing no more than 45 people per minute to pass through vertex $B$. State the maximum number of people that can move through the network per minute during this fire drill.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
  \includegraphics[alt={},max width=\textwidth]{1aca4e91-d1b3-4a78-8880-e42a4fbf3ddb-15_905_1559_331_292}
\end{center}
\end{figure}

Figure 4\\
\includegraphics[max width=\textwidth, alt={}, center]{1aca4e91-d1b3-4a78-8880-e42a4fbf3ddb-15_689_851_1370_598}

Maximum flow is $\_\_\_\_$ people per minute.
\end{enumerate}

\hfill \mbox{\textit{AQA D2 2011 Q5 [14]}}