| Exam Board | Edexcel |
|---|---|
| Module | FD1 (Further Decision 1) |
| Year | 2019 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Shortest Path |
| Type | Dijkstra with directed arcs |
| Difficulty | Moderate -0.5 This is a standard Floyd's algorithm question with directed arcs, requiring routine application of a well-defined algorithm taught in FD1. While it involves multiple iterations and table updates, the procedure is mechanical and follows textbook steps. Part (c) tests basic interpretation of route tables. The directed arcs add minimal complexity since Floyd's algorithm handles them naturally. Slightly easier than average due to the algorithmic nature requiring execution rather than problem-solving insight. |
| Spec | 7.04a Shortest path: Dijkstra's algorithm |
| \cline { 2 - 6 } \multicolumn{1}{c|}{} | A | B | C | D | E |
| A | - | 12 | 7 | 6 | 3 |
| B | 15 | - | 22 | 21 | 18 |
| C | 7 | 5 | - | 4 | 7 |
| D | 11 | 9 | 4 | - | 3 |
| E | 14 | 12 | 7 | 3 | - |
| \cline { 2 - 6 } \multicolumn{1}{c|}{} | A | B | C | D | E |
| A | A | ||||
| B | A | B | |||
| C | A | B | C | ||
| D | C | C | C | D | |
| E | D | D | D | D | E |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct distance table with \(\infty\) entries | B1 | 1.1b |
| Correct route table (all entries B) | B1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| No change in first row/column; at least two values correctly reduced in distance table; two letters correctly changed in route table | M1 | 1.1b |
| All values correct (condone dashes in EA and EB) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| No change from 1st iteration to 2nd iteration (ft from candidate's first iteration) | A1ft | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| No change in third row/column; at least two values correctly reduced; two route table values correctly changed | M1 | 1.1b |
| All values correct (note entry in row B column D for route table could be A) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Row E, column A is D so route from E to A is via D | B1 | 2.4 |
| Row D, column A is C so shortest path from D to A is via C | dB1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| EDCA | B1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| NNA: \(D - E - C - B - A - D\) | B1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3 + 7 + 5 + 15 + 6 = 36\) miles | B1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(D - E - D - C - B - A - E - D\) | B1 | 3.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. cycle \(A - E - D - C - B - A\) has length 30 miles \(< 36\) miles, so Mabintou's route is not optimal | B1 | 2.4 |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct distance table with $\infty$ entries | B1 | 1.1b |
| Correct route table (all entries B) | B1 | 1.1b |
## Part (b) — 1st iteration
| Answer/Working | Mark | Guidance |
|---|---|---|
| No change in first row/column; at least two values correctly reduced in distance table; two letters correctly changed in route table | M1 | 1.1b |
| All values correct (condone dashes in EA and EB) | A1 | 1.1b |
## Part (b) — 2nd iteration
| Answer/Working | Mark | Guidance |
|---|---|---|
| No change from 1st iteration to 2nd iteration (ft from candidate's first iteration) | A1ft | 1.1b |
## Part (b) — 3rd iteration
| Answer/Working | Mark | Guidance |
|---|---|---|
| No change in third row/column; at least two values correctly reduced; two route table values correctly changed | M1 | 1.1b |
| All values correct (note entry in row B column D for route table could be A) | A1 | 1.1b |
## Part (c)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Row E, column A is D so route from E to A is via D | B1 | 2.4 |
| Row D, column A is C so shortest path from D to A is via C | dB1 | 2.4 |
## Part (c)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| EDCA | B1 | 2.2a |
## Part (d)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| NNA: $D - E - C - B - A - D$ | B1 | 1.1b |
## Part (d)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3 + 7 + 5 + 15 + 6 = 36$ miles | B1 | 1.1b |
## Part (d)(iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $D - E - D - C - B - A - E - D$ | B1 | 3.2a |
## Part (d)(iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. cycle $A - E - D - C - B - A$ has length 30 miles $< 36$ miles, so Mabintou's route is not optimal | B1 | 2.4 |
---
3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{162f9d72-84a4-4b1a-93cf-b7eeb7f957ae-04_666_940_173_534}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The network in Figure 2 shows the direct roads linking five villages, $\mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D }$ and E .\\
The number on each arc represents the length, in miles, of the corresponding road.\\
The roads from A to E and from C to B are one-way, as indicated by the arrows.
\begin{enumerate}[label=(\alph*)]
\item Complete the initial distance and route tables for the network provided in the answer book.\\
(2)
\item Perform the first three iterations of Floyd's algorithm. You should show the distance table and the route table after each of the three iterations.
After five iterations of Floyd's algorithm the final distance table and partially completed final route table are shown below.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Distance table}
\begin{tabular}{ | c | c | c | c | c | c | }
\cline { 2 - 6 }
\multicolumn{1}{c|}{} & A & B & C & D & E \\
\hline
A & - & 12 & 7 & 6 & 3 \\
\hline
B & 15 & - & 22 & 21 & 18 \\
\hline
C & 7 & 5 & - & 4 & 7 \\
\hline
D & 11 & 9 & 4 & - & 3 \\
\hline
E & 14 & 12 & 7 & 3 & - \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Route table}
\begin{tabular}{ | c | c | c | c | c | c | }
\cline { 2 - 6 }
\multicolumn{1}{c|}{} & A & B & C & D & E \\
\hline
A & A & & & & \\
\hline
B & A & B & & & \\
\hline
C & A & B & C & & \\
\hline
D & C & C & C & D & \\
\hline
E & D & D & D & D & E \\
\hline
\end{tabular}
\end{center}
\end{table}
\item \begin{enumerate}[label=(\roman*)]
\item Explain how the partially completed final route table can be used to find the shortest route from E to A.
\item State this route.
Mabintou decides to use the distance table to try to find the shortest cycle that passes through each vertex. Starting at D, she applies the nearest neighbour algorithm to the final distance table.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item State the cycle obtained using the nearest neighbour algorithm.
\item State the length of this cycle.
\item Interpret the cycle in terms of the actual villages visited.
\item Prove that Mabintou's cycle is not optimal.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel FD1 2019 Q3 [14]}}