| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2001 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Route Inspection |
| Type | Basic Chinese Postman (closed route) |
| Difficulty | Moderate -0.3 This is a standard route inspection (Chinese Postman) problem requiring identification of odd vertices, pairing them optimally, and finding the minimum extra distance. While it involves multiple steps, it's a direct application of a well-defined algorithm with no novel problem-solving required. The 7 marks suggest moderate length but the procedure is algorithmic and routine for D1 students who have learned the method. |
| Spec | 7.04e Route inspection: Chinese postman, pairing odd nodes |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | B1 cae | Table showing vertices A, B, C, D, E, F with valencies 3, 2, 4, 3, 3, 3 |
| M1 A1 | Possible pairings shortest route total: (A,F) and (D,E): \(\max(60,90) = 150\) | |
| M1 A1 | (A,E) and (D,F): \(\max(190,70) = 240\) | |
| M1 A1 | (A,D) and (E,F): \(\max(120,110) = 230\) | |
| A1(5) | So repeat AF and DE | |
| (b) | M1 A1(2) | Total length = Total weight of edges + 150 = 690 + 150 = 840 m |
(a) | B1 cae | Table showing vertices A, B, C, D, E, F with valencies 3, 2, 4, 3, 3, 3 |
| M1 A1 | Possible pairings shortest route total: (A,F) and (D,E): $\max(60,90) = 150$ |
| M1 A1 | (A,E) and (D,F): $\max(190,70) = 240$ |
| M1 A1 | (A,D) and (E,F): $\max(120,110) = 230$ |
| A1(5) | So repeat AF and DE |
(b) | M1 A1(2) | Total length = Total weight of edges + 150 = 690 + 150 = 840 m |
---\includegraphics{figure_1}
\begin{enumerate}[label=(\alph*)]
\item Using an appropriate algorithm, obtain a suitable route starting and finishing at A. [5 marks]
\item Calculate the total length of this route. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2001 Q3 [7]}}