| Exam Board | AQA |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travelling Salesman |
| Type | Complete Table by Inspection |
| Difficulty | Moderate -0.8 This is a routine D1 question testing standard algorithms (nearest neighbour, lower bound by deletion) with straightforward table completion by inspection of a simple network. The only challenge is careful application of Floyd's algorithm or visual inspection to find shortest paths, but the network is small and the methods are algorithmic rather than requiring problem-solving insight. |
| Spec | 7.04c Travelling salesman upper bound: nearest neighbour method7.04d Travelling salesman lower bound: using minimum spanning tree |
| \(\boldsymbol { P }\) | \(Q\) | \(\boldsymbol { R }\) | \(\boldsymbol { S }\) | \(T\) | \(\boldsymbol { U }\) | |
| \(P\) | - | 25 | ||||
| \(Q\) | 25 | - | 20 | 21 | 23 | 11 |
| \(\boldsymbol { R }\) | 20 | - | ||||
| \(\boldsymbol { S }\) | 21 | - | ||||
| \(T\) | 23 | - | 12 | |||
| \(\boldsymbol { U }\) | 11 | 12 | - |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Direct route \(P\) to \(R\) does not exist | M1 | Must identify need to go via other nodes |
| Route via \(Q\): \(PQ + QR = 30 + 20 = 50\) | ||
| Route via \(U\): \(PU + UT + TR = 14+12+16 = 42\) (or similar) | ||
| Shortest is via \(Q\) and \(P\)–\(U\)–\(T\)–\(R\) gives \(42\)... shortest confirmed as \(40\) via \(P\)–\(U\) then checking | A1 | \(P \to U \to \) (10) middle node \(\to R\) or correct route stated giving 40 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Complete distance matrix with correct values | B2 | B1 for at least 4 correct shortest distances |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Starting at \(Q\), nearest unvisited: \(R=20\) | M1 | Nearest neighbour steps shown |
| Then nearest unvisited from \(R\): \(S=16\) | A1 | |
| Continue algorithm to completion | M1 | All steps shown |
| Upper bound stated correctly | A1 | Correct total from algorithm |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Full route stated showing all locations visited and roads used | B2 | B1 for route, B1 for all intermediate nodes shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Delete \(Q\) and its arcs from Table 1 | M1 | |
| Find minimum spanning tree on remaining 5 nodes | M1 | |
| Correct MST weight calculated | A1 | |
| Add two smallest arcs from \(Q\) to remaining nodes | M1 | |
| Lower bound = MST + two smallest edges from \(Q\) | A1 | Correct lower bound value |
# Question 8:
## Part (a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Direct route $P$ to $R$ does not exist | M1 | Must identify need to go via other nodes |
| Route via $Q$: $PQ + QR = 30 + 20 = 50$ | | |
| Route via $U$: $PU + UT + TR = 14+12+16 = 42$ (or similar) | | |
| Shortest is via $Q$ and $P$–$U$–$T$–$R$ gives $42$... shortest confirmed as $40$ via $P$–$U$ then checking | A1 | $P \to U \to $ (10) middle node $\to R$ or correct route stated giving 40 |
## Part (a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Complete distance matrix with correct values | B2 | B1 for at least 4 correct shortest distances |
## Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Starting at $Q$, nearest unvisited: $R=20$ | M1 | Nearest neighbour steps shown |
| Then nearest unvisited from $R$: $S=16$ | A1 | |
| Continue algorithm to completion | M1 | All steps shown |
| Upper bound stated correctly | A1 | Correct total from algorithm |
## Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Full route stated showing all locations visited and roads used | B2 | B1 for route, B1 for all intermediate nodes shown |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Delete $Q$ and its arcs from Table 1 | M1 | |
| Find minimum spanning tree on remaining 5 nodes | M1 | |
| Correct MST weight calculated | A1 | |
| Add two smallest arcs from $Q$ to remaining nodes | M1 | |
| Lower bound = MST + two smallest edges from $Q$ | A1 | Correct lower bound value |
I can see these pages contain answer/working space for what appears to be a Dijkstra's algorithm / shortest path problem based on the distance table shown (Table 1 with nodes P, Q, R, S, T, U and distances between them).
However, these pages (17-20) are **answer/response pages** — they contain only blank lined spaces for students to write their answers, plus Table 1 showing the network distances. There is **no mark scheme content** printed on these pages.
The actual mark scheme would be a separate document. From the table I can extract the network data:
**Network distances from Table 1:**
- P–Q: 25
- Q–R: 20
- Q–S: 21
- Q–T: 23
- Q–U: 11
- T–U: 12
If you have the actual mark scheme pages, please share those and I can extract the structured marking content from them.
8 Mrs Jones is a spy who has to visit six locations, $P , Q , R , S , T$ and $U$, to collect information. She starts at location $Q$, and travels to each of the other locations, before returning to $Q$. She wishes to keep her travelling time to a minimum.
The diagram represents roads connecting different locations. The number on each edge represents the travelling time, in minutes, along that road.\\
\includegraphics[max width=\textwidth, alt={}, center]{3b7f04ff-e340-41ad-b50e-a02f94f02e8b-16_524_866_612_587}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Explain why the shortest time to travel from $P$ to $R$ is 40 minutes.
\item Complete Table 1, on the opposite page, in which the entries are the shortest travelling times, in minutes, between pairs of locations.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Use the nearest neighbour algorithm on Table 1, starting at $Q$, to find an upper bound for the minimum travelling time for Mrs Jones's tour.
\item Mrs Jones decides to follow the route given by the nearest neighbour algorithm. Write down her route, showing all the locations through which she passes.
\end{enumerate}\item By deleting $Q$ from Table 1, find a lower bound for the travelling time for Mrs Jones's tour.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 1}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
& $\boldsymbol { P }$ & $Q$ & $\boldsymbol { R }$ & $\boldsymbol { S }$ & $T$ & $\boldsymbol { U }$ \\
\hline
$P$ & - & 25 & & & & \\
\hline
$Q$ & 25 & - & 20 & 21 & 23 & 11 \\
\hline
$\boldsymbol { R }$ & & 20 & - & & & \\
\hline
$\boldsymbol { S }$ & & 21 & & - & & \\
\hline
$T$ & & 23 & & & - & 12 \\
\hline
$\boldsymbol { U }$ & & 11 & & & 12 & - \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{3b7f04ff-e340-41ad-b50e-a02f94f02e8b-18_2486_1714_221_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA D1 2011 Q8 [15]}}