Edexcel D1 2015 January — Question 1 6 marks

Exam BoardEdexcel
ModuleD1 (Decision Mathematics 1)
Year2015
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMinimum Spanning Trees
TypeApply Prim's algorithm from vertex
DifficultyModerate -0.8 This is a straightforward application of Prim's algorithm from a given starting vertex using a distance table. The procedure is algorithmic and routine for D1 students—systematically selecting minimum weight edges connecting to the growing tree. Part (c) adds minimal challenge by asking about uniqueness. This is easier than average A-level maths as it requires only mechanical application of a learned algorithm with no problem-solving or mathematical insight.
Spec7.04b Minimum spanning tree: Prim's and Kruskal's algorithms

1.
ABCDEFGH
A-981317111210
B9-11211524137
C811-2023171715
D132120-15281118
E17152315-312330
F1124172831-1315
G121317112313-23
H1071518301523-
The table represents a network that shows the time taken, in minutes, to travel by car between eight villages, \(\mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D } , \mathrm { E } , \mathrm { F } , \mathrm { G }\) and H .
  1. Use Prim's algorithm, starting at A , to find a minimum spanning tree for this network. You must list the arcs that form your tree in the order in which you select them.
  2. Draw your minimum spanning tree using the vertices given in Diagram 1 in the answer book and state the weight of the tree.
  3. State whether your minimum spanning tree is unique. Justify your answer.

AnswerMarks Guidance
AnswerMarks Guidance
AC, AB, BH; AF AG; DG, DE or BEM1 A1 A1 (3) First three arcs correctly chosen in order {AC, AB, BH,...} or first four nodes correctly chosen in order {A, C, B, H,...}. If any rejections seen at any point then M1 (max) only. Order of nodes may be seen at the top of the matrix {1, 3, 2, -, -, -, 4} so please check the top of the matrix carefully.
Weight of tree = 73 (mins)B1
No – there are two different MST (with a weight of 73) either with DE or BEB1 (1) CAO – mention of two different MST with either arc BE or DE.
6 marks
Notes for Question 1:
- a1A1: First five arcs correctly chosen in order {AC, AB, BH, AF, AG,...} or all eight nodes correctly chosen in order {A, C, B, H, F, G, D, E}. Order of nodes may be seen at the top of the matrix so for the first two marks accept {1, 3, 2, 7, 8, 5, 6, 4} (do not condone any missing numbers e.g. the number 8 must be above E).
- a2A1: All arcs correct stated and chosen in the correct order. Candidates must be considering arcs for this final mark (do not accept a list of nodes or numbers across the top of the matrix unless the correct list of arcs (in the correct order) is also seen). Allow DE or BE for the final arc but not DE and BE.
- Misread: Starting at a node other than A scores M1 only – must have the first three arcs (or four nodes) correct (and in the correct order).
- b1B1: CAO (condone lack of weights on arcs) – condone, say, a dashed line between B and E if arc DE is in the tree (or vice-versa).
- b2B1: CAO (73) (condone lack of units).
- c1B1: CAO – mention of two different MST with either arc BE or DE.
| Answer | Marks | Guidance |
|--------|-------|----------|
| AC, AB, BH; AF AG; DG, DE or BE | M1 A1 A1 (3) | First three arcs correctly chosen in order {AC, AB, BH,...} or first four nodes correctly chosen in order {A, C, B, H,...}. If any rejections seen at any point then M1 (max) only. Order of nodes may be seen at the top of the matrix {1, 3, 2, -, -, -, 4} so please check the top of the matrix carefully. |
| Weight of tree = 73 (mins) | B1 | |
| No – there are two different MST (with a weight of 73) either with DE or BE | B1 (1) | CAO – mention of two different MST with either arc BE or DE. |
| | 6 marks | |

**Notes for Question 1:**

- a1A1: First five arcs correctly chosen in order {AC, AB, BH, AF, AG,...} or all eight nodes correctly chosen in order {A, C, B, H, F, G, D, E}. Order of nodes may be seen at the top of the matrix so for the first two marks accept {1, 3, 2, 7, 8, 5, 6, 4} (do not condone any missing numbers e.g. the number 8 must be above E).
- a2A1: All arcs correct stated and chosen in the correct order. Candidates must be considering arcs for this final mark (do not accept a list of nodes or numbers across the top of the matrix unless the correct list of arcs (in the correct order) is also seen). Allow DE or BE for the final arc but not DE and BE.
- **Misread**: Starting at a node other than A scores M1 only – must have the first three arcs (or four nodes) correct (and in the correct order).
- b1B1: CAO (condone lack of weights on arcs) – condone, say, a dashed line between B and E if arc DE is in the tree (or vice-versa).
- b2B1: CAO (73) (condone lack of units).
- c1B1: CAO – mention of two different MST with either arc BE or DE.

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1.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
 & A & B & C & D & E & F & G & H \\
\hline
A & - & 9 & 8 & 13 & 17 & 11 & 12 & 10 \\
\hline
B & 9 & - & 11 & 21 & 15 & 24 & 13 & 7 \\
\hline
C & 8 & 11 & - & 20 & 23 & 17 & 17 & 15 \\
\hline
D & 13 & 21 & 20 & - & 15 & 28 & 11 & 18 \\
\hline
E & 17 & 15 & 23 & 15 & - & 31 & 23 & 30 \\
\hline
F & 11 & 24 & 17 & 28 & 31 & - & 13 & 15 \\
\hline
G & 12 & 13 & 17 & 11 & 23 & 13 & - & 23 \\
\hline
H & 10 & 7 & 15 & 18 & 30 & 15 & 23 & - \\
\hline
\end{tabular}
\end{center}

The table represents a network that shows the time taken, in minutes, to travel by car between eight villages, $\mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D } , \mathrm { E } , \mathrm { F } , \mathrm { G }$ and H .
\begin{enumerate}[label=(\alph*)]
\item Use Prim's algorithm, starting at A , to find a minimum spanning tree for this network. You must list the arcs that form your tree in the order in which you select them.
\item Draw your minimum spanning tree using the vertices given in Diagram 1 in the answer book and state the weight of the tree.
\item State whether your minimum spanning tree is unique. Justify your answer.
\end{enumerate}

\hfill \mbox{\textit{Edexcel D1 2015 Q1 [6]}}