Edexcel D2 2016 June — Question 2 8 marks

Exam BoardEdexcel
ModuleD2 (Decision Mathematics 2)
Year2016
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNetwork Flows
TypeList saturated arcs
DifficultyEasy -1.2 This is a straightforward network flows question requiring basic identification of saturated arcs (where flow equals capacity) and routine application of standard algorithms. Part (a) is pure observation, parts (b-c) are reading values, part (d) requires finding an augmenting path by inspection (only 3 units), and part (e) applies the max-flow min-cut theorem. All techniques are standard D2 content with no novel problem-solving required.
Spec7.04f Network problems: choosing appropriate algorithm

2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a12f9520-85c0-43f6-a104-2502011d5657-3_780_1155_246_461} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a capacitated, directed network of pipes. The number on each arc represents the capacity of the corresponding pipe. The numbers in circles represent an initial flow.
  1. List the saturated arcs.
  2. State the value of the initial flow.
  3. State the capacities of the cuts \(C _ { 1 }\) and \(C _ { 2 }\)
  4. By inspection, find a flow-augmenting route to increase the flow by three units. You must state your route.
  5. Prove that the new flow is maximal.

Question 2:
(a)
AnswerMarks Guidance
Saturated arcs: \(S \to C\) (22), \(S \to B\) (20), \(C \to F\) (30), \(B \to C\) (8)B1 B1 One mark per two correct arcs
(b)
AnswerMarks
Initial flow = 75B1
(c)
AnswerMarks
\(C_1\): arcs \(S\to A\), \(S\to B\), \(S\to C\); capacity = \(25+20+22 = 67\)B1
\(C_2\): arcs \(D\to T\), \(E\to F\), \(C\to F\)... capacity = \(16+21+30 = ...\); \(C_2 = 78\)B1
(d)
AnswerMarks
Route: \(S \to A \to E \to F \to T\) (increase by 3)B1
(e)
AnswerMarks
Max flow = min cut; identify cut with capacity = new flow value (78)M1
State cut and its capacity equals the flow, therefore maximalA1
## Question 2:

**(a)**
Saturated arcs: $S \to C$ (22), $S \to B$ (20), $C \to F$ (30), $B \to C$ (8) | B1 B1 | One mark per two correct arcs

**(b)**
Initial flow = **75** | B1 |

**(c)**
$C_1$: arcs $S\to A$, $S\to B$, $S\to C$; capacity = $25+20+22 = 67$ | B1 |
$C_2$: arcs $D\to T$, $E\to F$, $C\to F$... capacity = $16+21+30 = ...$; $C_2 = 78$ | B1 |

**(d)**
Route: $S \to A \to E \to F \to T$ (increase by 3) | B1 |

**(e)**
Max flow = min cut; identify cut with capacity = new flow value (78) | M1 |
State cut and its capacity equals the flow, therefore maximal | A1 |

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2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{a12f9520-85c0-43f6-a104-2502011d5657-3_780_1155_246_461}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a capacitated, directed network of pipes. The number on each arc represents the capacity of the corresponding pipe. The numbers in circles represent an initial flow.
\begin{enumerate}[label=(\alph*)]
\item List the saturated arcs.
\item State the value of the initial flow.
\item State the capacities of the cuts $C _ { 1 }$ and $C _ { 2 }$
\item By inspection, find a flow-augmenting route to increase the flow by three units. You must state your route.
\item Prove that the new flow is maximal.
\end{enumerate}

\hfill \mbox{\textit{Edexcel D2 2016 Q2 [8]}}