Edexcel D1 2007 June — Question 5 7 marks

Exam BoardEdexcel
ModuleD1 (Decision Mathematics 1)
Year2007
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMinimum Spanning Trees
TypeApply Prim's algorithm in matrix form
DifficultyEasy -1.3 This is a straightforward application of Prim's algorithm with a small 6-vertex network. Part (a) requires mechanical execution of a standard algorithm taught in D1, parts (b) and (c) are trivial bookkeeping, and part (d) tests basic understanding of why Prim's works. No problem-solving or novel insight required—purely procedural recall below typical A-level difficulty.
Spec7.04b Minimum spanning tree: Prim's and Kruskal's algorithms

$$\begin{array}{c|c|c|c|c|c} & M & A & B & C & D & E \\ \hline M & - & 215 & 170 & 290 & 210 & 305 \\ \hline A & 215 & - & 275 & 100 & 217 & 214 \\ \hline B & 170 & 275 & - & 267 & 230 & 200 \\ \hline C & 290 & 100 & 267 & - & 180 & 220 \\ \hline D & 210 & 217 & 230 & 180 & - & 245 \\ \hline E & 305 & 214 & 200 & 220 & 245 & - \end{array}$$ The table shows the cost, in pounds, of linking five automatic alarm sensors, \(A,B,C,D\) and \(E\), and the main reception, \(M\).
  1. Use Prim's algorithm, starting from \(M\), to find a minimum spanning tree for this table of costs. You must list the arcs that form your tree in the order that they are selected. [3]
  2. Draw your tree using the vertices given in Diagram 1 in the answer book. [1]
  3. Find the total weight of your tree. [1]
  4. Explain why it is not necessary to check for cycles when using Prim's algorithm. [2]
(Total 7 marks)

(a)
MB, BE, MD, DC, CA
AnswerMarks
m1, A1, A1 (3)
(b)
[Diagram showing network with vertices E, A, B, C, D and edges labeled with distances]
AnswerMarks
B1' (1)
(c)
\(170 + 2×0 + 210 + 480 + 100 = 860\)
AnswerMarks
B1 (1)
(d)
(A cycle is formed when an arc is used that connects two vertices already connected to such other in the tree)
From algorithm always selects arcs that bring a vertex not in the tree so a cycle can't happen
AnswerMarks
B2, 1, 0 (e) [2]
## (a)
MB, BE, MD, DC, CA

| m1, A1, A1 (3) |

## (b)
[Diagram showing network with vertices E, A, B, C, D and edges labeled with distances]

| B1' (1) |

## (c)
$170 + 2×0 + 210 + 480 + 100 = 860$

| B1 (1) |

## (d)
(A cycle is formed when an arc is used that connects two vertices already connected to such other in the tree)
From algorithm always selects arcs that bring a vertex not in the tree so a cycle can't happen

| B2, 1, 0 (e) [2] |

---
$$\begin{array}{c|c|c|c|c|c}
 & M & A & B & C & D & E \\
\hline
M & - & 215 & 170 & 290 & 210 & 305 \\
\hline
A & 215 & - & 275 & 100 & 217 & 214 \\
\hline
B & 170 & 275 & - & 267 & 230 & 200 \\
\hline
C & 290 & 100 & 267 & - & 180 & 220 \\
\hline
D & 210 & 217 & 230 & 180 & - & 245 \\
\hline
E & 305 & 214 & 200 & 220 & 245 & -
\end{array}$$

The table shows the cost, in pounds, of linking five automatic alarm sensors, $A,B,C,D$ and $E$, and the main reception, $M$.

\begin{enumerate}[label=(\alph*)]
\item Use Prim's algorithm, starting from $M$, to find a minimum spanning tree for this table of costs. You must list the arcs that form your tree in the order that they are selected. [3]

\item Draw your tree using the vertices given in Diagram 1 in the answer book. [1]

\item Find the total weight of your tree. [1]

\item Explain why it is not necessary to check for cycles when using Prim's algorithm. [2]
\end{enumerate}

(Total 7 marks)

\hfill \mbox{\textit{Edexcel D1 2007 Q5 [7]}}