| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Discrete (Further AS Paper 2 Discrete) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Network Flows |
| Type | Find missing flow values |
| Difficulty | Standard +0.3 This is a standard max-flow/min-cut network question requiring basic flow conservation principles and the max-flow min-cut theorem. Part (a) applies flow conservation at source/sink (routine), parts (b)(i-ii) use flow conservation at nodes (trivial once you understand the setup), and part (c) requires identifying a cut—all standard textbook techniques for Further Maths discrete. While it's a Further Maths topic, the question itself involves straightforward application of well-rehearsed algorithms rather than problem-solving insight. |
| Spec | 7.02p Networks: weighted graphs, modelling connections |
| Answer | Marks |
|---|---|
| 8(a) | Adds together current flows on |
| Answer | Marks | Guidance |
|---|---|---|
| flow through the network is 25 | 3.1b | M1 |
Total flow from the source
| Answer | Marks | Guidance |
|---|---|---|
| Obtains flows of 25 and 18 + x | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| the sink to show that x = 7 | 2.1 | R1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 8(b)(i) | States 5 | 1.1b |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 8(b)(ii) | States 4 | 1.1b |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 8(c) | Identifies the cut through arcs BT, | |
| CT and ET | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone missing units | 2.1 | R1 |
| Subtotal | 2 | |
| Question total | 7 | |
| Q | Marking instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Adds together current flows on
BT, CT and ET or flows on SA,
SC and SD or states the current
flow through the network is 25 | 3.1b | M1 | Total flow into sink
= BT + CT + ET
= 12 + 3 + 10
= 25
Total flow from the source
= SA + SC + SD
= 10 + 8 + x
= 18 + x
The total flow from source must be
equal to the total flow into sink, so
25 = 18 + x, hence x = 7
Obtains flows of 25 and 18 + x | 1.1b | A1
Completes reasoned argument
referring to the total flow from the
source equalling the total flow into
the sink to show that x = 7 | 2.1 | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 8(b)(i) ---
8(b)(i) | States 5 | 1.1b | B1 | y = 5
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 8(b)(ii) ---
8(b)(ii) | States 4 | 1.1b | B1 | z = 4
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 8(c) ---
8(c) | Identifies the cut through arcs BT,
CT and ET | 3.1b | M1 | A cut through the arcs BT, CT and
ET has a value of 27 m3 s–1
As the maximum flow through a
network is less than or equal to the
value of any cut in the network, by
the maximum flow-minimum cut
theorem, the maximum flow is at
most 27 m3 s–1
Completes reasoned argument
using the maximum flow-minimum
cut theorem to establish that the
maximum flow through the
network is at most 27 m3 s–1
Condone missing units | 2.1 | R1
Subtotal | 2
Question total | 7
Q | Marking instructions | AO | Marks | Typical solutionThe diagram below shows a network of pipes.
\includegraphics{figure_8}
The uncircled numbers on each arc represent the capacity of each pipe in m³ s⁻¹
The circled numbers on each arc represent an initial feasible flow, in m³ s⁻¹, through the network.
The initial flow through pipe $SD$ is $x$ m³ s⁻¹
The initial flow through pipe $DC$ is $y$ m³ s⁻¹
The initial flow through pipe $CB$ is $z$ m³ s⁻¹
\begin{enumerate}[label=8 (\alph*)]
\item By considering the flows at the source and the sink, explain why $x = 7$
[3 marks]
\item \begin{enumerate}[label=(\roman*)]
\item State the value of $y$
[1 mark]
\item State the value of $z$
[1 mark]
\end{enumerate}
\item Prove that the maximum flow through the network is at most 27 m³ s⁻¹
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Discrete 2024 Q8 [7]}}