| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Discrete (Further AS Paper 2 Discrete) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Network Flows |
| Type | Calculate cut capacity |
| Difficulty | Moderate -0.5 This is a straightforward application of cut capacity calculation in network flows. Part (a) requires identifying edges crossing a given cut and summing capacities (routine procedure), part (b) asks for a cut with specific value (pattern recognition), and part (c) applies the max-flow min-cut theorem (standard recall). While it's Further Maths content, the question requires only direct application of learned algorithms with no problem-solving insight or novel reasoning. |
| Spec | 7.04f Network problems: choosing appropriate algorithm |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(110 + 120 + 45 + 55 + 70 = 400 \text{ m}^3\text{s}^{-1}\) | B1 | Determines correct value of the cut; condone missing units; AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\{A, B, C, D, E, G, H, I\}\ \{F\}\) | B1 | Writes down the correct cut; AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| As \(300 < 400\), the maximum flow through the network is less than or equal to \(300 \text{ m}^3\text{s}^{-1}\) by the maximum flow–minimum cut theorem | B1F | Deduces maximum flow cannot exceed minimum of their answer to (a) and \(300\); condone strict inequality but not equality; AO2.2a |
| Explains answer with reference to the maximum flow–minimum cut theorem | E1F | Must be weak inequality; AO2.4 |
## Question 2:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $110 + 120 + 45 + 55 + 70 = 400 \text{ m}^3\text{s}^{-1}$ | B1 | Determines correct value of the cut; condone missing units; AO1.1b |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\{A, B, C, D, E, G, H, I\}\ \{F\}$ | B1 | Writes down the correct cut; AO1.1b |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| As $300 < 400$, the maximum flow through the network is less than or equal to $300 \text{ m}^3\text{s}^{-1}$ by the maximum flow–minimum cut theorem | B1F | Deduces maximum flow cannot exceed minimum of their answer to (a) and $300$; condone strict inequality but not equality; AO2.2a |
| Explains answer with reference to the maximum flow–minimum cut theorem | E1F | Must be weak inequality; AO2.4 |2 The diagram shows a network of pipes.
Each pipe is labelled with its upper capacity in $\mathrm { m } ^ { 3 } \mathrm {~s} ^ { - 1 }$\\
\includegraphics[max width=\textwidth, alt={}, center]{ecbeedf5-148e-40ad-b8a2-a7aa3db4a115-03_424_1262_445_388}
2
\begin{enumerate}[label=(\alph*)]
\item Find the value of the cut $\{ A , C , D , G , H \} \{ B , E , F , I \}$
2
\item Write down a cut with a value of $300 \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }$
2
\item Using the values from part (a) and part (b), state what can be deduced about the maximum flow through the network.
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Discrete 2022 Q2 [4]}}