Question 3 - A-Level Maths

Edexcel FD2 AS 2018 June — Question 3 10 marks

Exam Board	Edexcel
Module	FD2 AS (Further Decision 2 AS)
Year	2018
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Network Flows
Type	Calculate cut capacity
Difficulty	Standard +0.3 This is a standard network flows question testing routine application of cut capacity calculations and max-flow min-cut theorem. Parts (a)-(d) involve straightforward reading of capacities from a diagram and applying well-rehearsed algorithms, while part (e) requires simple case analysis based on the bottleneck capacity. The question is slightly easier than average A-level as it's methodical application of a decision maths algorithm with no novel problem-solving required.
Spec	7.04f Network problems: choosing appropriate algorithm

3. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{905f2578-e4b2-4d4d-8455-298170fd824b-4_781_1159_365_551} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 models the flow of fluid through a system of pipes from a source, S , to a sink, T . The weights on the arcs show the capacities of the corresponding pipes in litres per minute. Two cuts \(C _ { 1 }\) and \(C _ { 2 }\) are shown.

Find the capacity of
1. cut \(C _ { 1 }\)
2. cut \(C _ { 2 }\)
Using only the capacities of cuts \(C _ { 1 }\) and \(C _ { 2 }\) state what can be deduced about the maximum possible flow through the system.
On Diagram 1 in the answer book, show how a flow of 120 litres per minute from S to T can be achieved. You do not need to apply the labelling procedure to find this flow.
Prove that 120 litres per minute is the maximum possible flow through the system. A new pipe is planned from S to A . Let the capacity of this pipe be \(x\) litres per minute.
Find, in terms of \(x\) where necessary, the maximum possible flow through the new system.

Show mark scheme Show mark scheme source

Question 3:

Part (a)

Answer	Marks	Guidance
Answer	Mark	Guidance
(i) 170	B1	cao
(ii) 145	B1	cao

Part (b)

Answer	Marks	Guidance
Answer	Mark	Guidance
Deduces maximum possible flow is \(\leq 145\) litres per minute	B1ft	Deduced from least value in (a); must include 'less than or equal to'

Part (c)

Answer	Marks	Guidance
Answer	Mark	Guidance
Valid flow diagram shown with flows summing correctly	M1	Deduces flow out of SB = 120; 'flow in = flow out' at all but one node; one number only on each arc (condone blank for arc FE)
Correct valid flow through network	A1	Check flow in = flow out at each vertex

Part (d)

Answer	Marks	Guidance
Answer	Mark	Guidance
Cut through arcs BA, ED, ET, EF (twice), CF	B1	Finds correct cut through saturated arcs directed S to T
Maximum flow = minimum cut; Flow = 120, Cut = 120 therefore flow of 120 is optimal	B1	Must state 'maximum flow = minimum cut'; dependent on correct cut and correct flow in (c)

Part (e)

Answer	Marks	Guidance
Answer	Mark	Guidance
\(0 < x \leq 25\), flow is \(120 + x\)	M1, A1	Understanding that flow differs depending on values of \(x\); critical value \(x=25\)
\(x > 25\), flow is \(145\)	A1	Correct argument for inequalities for when each flow is valid

# Question 3:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| (i) 170 | B1 | cao |
| (ii) 145 | B1 | cao |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Deduces maximum possible flow is $\leq 145$ litres per minute | B1ft | Deduced from least value in (a); must include 'less than or equal to' |

## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| Valid flow diagram shown with flows summing correctly | M1 | Deduces flow out of SB = 120; 'flow in = flow out' at all but one node; one number only on each arc (condone blank for arc FE) |
| Correct valid flow through network | A1 | Check flow in = flow out at each vertex |

## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| Cut through arcs BA, ED, ET, EF (twice), CF | B1 | Finds correct cut through saturated arcs directed S to T |
| Maximum flow = minimum cut; Flow = 120, Cut = 120 therefore flow of 120 is optimal | B1 | Must state 'maximum flow = minimum cut'; dependent on correct cut and correct flow in (c) |

## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $0 < x \leq 25$, flow is $120 + x$ | M1, A1 | Understanding that flow differs depending on values of $x$; critical value $x=25$ |
| $x > 25$, flow is $145$ | A1 | Correct argument for inequalities for when each flow is valid |

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Show LaTeX source

3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{905f2578-e4b2-4d4d-8455-298170fd824b-4_781_1159_365_551}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 models the flow of fluid through a system of pipes from a source, S , to a sink, T . The weights on the arcs show the capacities of the corresponding pipes in litres per minute. Two cuts $C _ { 1 }$ and $C _ { 2 }$ are shown.
\begin{enumerate}[label=(\alph*)]
\item Find the capacity of
\begin{enumerate}[label=(\roman*)]
\item cut $C _ { 1 }$
\item cut $C _ { 2 }$
\end{enumerate}\item Using only the capacities of cuts $C _ { 1 }$ and $C _ { 2 }$ state what can be deduced about the maximum possible flow through the system.
\item On Diagram 1 in the answer book, show how a flow of 120 litres per minute from S to T can be achieved. You do not need to apply the labelling procedure to find this flow.
\item Prove that 120 litres per minute is the maximum possible flow through the system.

A new pipe is planned from S to A . Let the capacity of this pipe be $x$ litres per minute.
\item Find, in terms of $x$ where necessary, the maximum possible flow through the new system.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FD2 AS 2018 Q3 [10]}}

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