| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Discrete (Further Paper 3 Discrete) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Network Flows |
| Type | Lower and upper capacity networks |
| Difficulty | Challenging +1.2 This is a Further Maths Decision Mathematics question on network flows with lower/upper capacities. Parts (a) and (b) require understanding of feasibility constraints (3 marks of explanation), while part (c) involves standard flow augmentation technique (5 marks). The topic is specialized but the question follows textbook procedures without requiring novel insight—harder than average A-level due to Further Maths content but routine within that context. |
| Spec | 7.04f Network problems: choosing appropriate algorithm |
| Augmenting Path | Flow |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Max flow into vertex \(A = 7\); min flow out of vertex \(A = 4+3 = 7\); as flow into \(A\) = flow out of \(A\), both \(AD\) and \(AB\) must be at the lower capacity | E1 | Explains correctly using maximum flow into and minimum flow out of \(A\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(AD = 4\) from (a)(i), so \(DE = 1\) and \(DT = 3\). Since flow out of vertex \(E\) is at least 2, and \(DE = 1\), \(BE\) must be at its upper capacity of 1 | E1 | Explains correctly using maximum flow into and minimum flow out of \(E\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Min flow across the cut \(\{S, A, B, C\}/\{D, E, G, T\} = AD + BE + \min(BG) + \min(CG) = 4+1+5+2 = 12 > 11\) | E1 | Explains correctly the statement using a cut in the network |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Finds a potential increase and decrease for each arc with a value on the two arrows for each arc, with \(SA\), \(SC\), \(AB\), \(BC\) and \(CF\) correct | M1 | |
| Determines correctly the potential increase and decrease for each arc with values on all arrows correct (see diagram) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Finds correctly one augmenting path and flow | B1 | e.g. \(SADEFT\): flow 1 |
| Finds correctly two augmenting paths and the flows | B1 | e.g. \(SCBET\): flow 3 |
| Finds correctly three (or four) augmenting paths and the flows, and clearly states the maximum flow in context with units: Maximum flow through the network is 22 litres per second | B1 | e.g. \(SADBET\): flow 1; AO3.2a |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{S, A, B, C, D, E\}/\{F, T\} = 7 + 3 + 9 + 3 = 22\). As the flow (22) is equal to the value of the cut (22), the maximum flow is 22 by the maximum flow-minimum cut theorem. | R1 | Must clearly show cut value calculation AND state equality with flow AND invoke max-flow min-cut theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Flow through network can only increase by 2 litres per second as \(DT\) and \(CF\) are already saturated. Therefore restricted capacity node reduces the maximum flow to \(17 + 2 = 19\) litres per second | R1 (AO2.4) + R1 (AO3.A) | Must argue 7 litres/sec already flowing through node \(E\), only 2 further litres/sec can flow; then conclude maximum flow reduces to 19 litres per second |
## Question 6(a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Max flow into vertex $A = 7$; min flow out of vertex $A = 4+3 = 7$; as flow into $A$ = flow out of $A$, both $AD$ and $AB$ must be at the lower capacity | E1 | Explains correctly using maximum flow into and minimum flow out of $A$ |
---
## Question 6(a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $AD = 4$ from (a)(i), so $DE = 1$ and $DT = 3$. Since flow out of vertex $E$ is at least 2, and $DE = 1$, $BE$ must be at its upper capacity of 1 | E1 | Explains correctly using maximum flow into and minimum flow out of $E$ |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Min flow across the cut $\{S, A, B, C\}/\{D, E, G, T\} = AD + BE + \min(BG) + \min(CG) = 4+1+5+2 = 12 > 11$ | E1 | Explains correctly the statement using a cut in the network |
---
## Question 6(c)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Finds a potential increase and decrease for each arc with a value on the two arrows for each arc, with $SA$, $SC$, $AB$, $BC$ and $CF$ correct | M1 | |
| Determines correctly the potential increase and decrease for each arc with values on all arrows correct (see diagram) | A1 | |
---
## Question 6(c)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Finds correctly one augmenting path and flow | B1 | e.g. $SADEFT$: flow 1 |
| Finds correctly two augmenting paths and the flows | B1 | e.g. $SCBET$: flow 3 |
| Finds correctly three (or four) augmenting paths and the flows, and clearly states the maximum flow in context with units: **Maximum flow through the network is 22 litres per second** | B1 | e.g. $SADBET$: flow 1; AO3.2a |
## Question 6(c)(iii):
$\{S, A, B, C, D, E\}/\{F, T\} = 7 + 3 + 9 + 3 = 22$. As the flow (22) is equal to the value of the cut (22), the maximum flow is 22 by the maximum flow-minimum cut theorem. | R1 | Must clearly show cut value calculation AND state equality with flow AND invoke max-flow min-cut theorem
---
## Question 6(c)(iv):
Flow through network can only increase by 2 litres per second as $DT$ and $CF$ are already saturated. Therefore restricted capacity node reduces the maximum flow to $17 + 2 = 19$ litres per second | R1 (AO2.4) + R1 (AO3.A) | Must argue 7 litres/sec already flowing through node $E$, only 2 further litres/sec can flow; then conclude maximum flow reduces to 19 litres per second
---
6\\
The network shows a system of pipes, where $S$ is the source and $T$ is the sink.\\
The lower and upper capacities, in litres per second, of each pipe are shown on each arc.\\
\includegraphics[max width=\textwidth, alt={}, center]{88669bc0-9d3f-431a-8939-8aef2682412b-09_649_1399_580_424}
6
\begin{enumerate}[label=(\alph*)]
\item There is a feasible flow from $S$ to $T$.
6 (a) (i) Explain why arc $A D$ must be at its lower capacity.\\[0pt]
[1 mark]
6 (a) (ii) Explain why arc $B E$ must be at its upper capacity.\\[0pt]
[1 mark]
6
\item Explain why a flow of 11 litres per second through the network is impossible.\\[0pt]
[1 mark]
6
\item The network in Figure 2 shows a second system of pipes, where $S$ is the source and $T$ is the sink.
The lower and upper capacities, in litres per second, of each pipe are shown on each edge.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{88669bc0-9d3f-431a-8939-8aef2682412b-10_760_1372_680_470}
\end{center}
\end{figure}
Figure 3 shows a feasible flow of 17 litres per second through the system of pipes.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{88669bc0-9d3f-431a-8939-8aef2682412b-10_750_1371_1811_466}
\end{center}
\end{figure}
6 (c) (i) Using Figures 2 and 3, indicate on Figure 4 potential increases and decreases in the flow along each arc.\\[0pt]
[2 marks]
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\includegraphics[alt={},max width=\textwidth]{88669bc0-9d3f-431a-8939-8aef2682412b-11_749_1384_457_426}
\end{center}
\end{figure}
6 (c) (ii) Use flow augmentation on Figure 4 to find the maximum flow from $S$ to $T$.\\
You should indicate any flow augmenting paths clearly in the table below and modify the potential increases and decreases of the flow on Figure 4.\\[0pt]
[3 marks]
\begin{center}
\begin{tabular}{ | l | l | }
\hline
Augmenting Path & Flow \\
\hline
& \\
\hline
& \\
\hline
& \\
\hline
\end{tabular}
\end{center}
6 (c) (iii) Prove the flow found in part (d) (ii) is maximum.\\
6 (c) (iv) Due to maintenance work, the flow through node $E$ is restricted to 9 litres per second.\\[0pt]
Interpret the impact of this restriction on the maximum flow through the system of pipes. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Discrete Q6 [11]}}