OCR MEI D2 2010 June — Question 2

Exam BoardOCR MEI
ModuleD2 (Decision Mathematics 2)
Year2010
SessionJune
TopicSign Change & Interval Methods
TypeFloyd's Algorithm Application

2 The network is a representation of a show garden. The weights on the arcs give the times in minutes to walk between the six features represented by the vertices, where paths exist.
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  1. Why might it be that the time taken to walk from vertex \(\mathbf { 2 }\) to vertex \(\mathbf { 3 }\) via vertex \(\mathbf { 4 }\) is less than the time taken by the direct route, i.e. the route from \(\mathbf { 2 }\) to \(\mathbf { 3 }\) which does not pass through any other vertices? The matrices shown below are the results of the first iteration of Floyd's algorithm when applied to the network. \begin{center} \begin{tabular}{ | c | c | c | c | c | c | c | } \cline { 2 - 7 } \multicolumn{1}{c|}{} & \(\mathbf { 1 }\) & \(\mathbf { 2 }\) & \(\mathbf { 3 }\) & \(\mathbf { 4 }\) & \(\mathbf { 5 }\) & \(\mathbf { 6 }\)
    \hline \(\mathbf { 1 }\) & \(\infty\) & 15 & \(\infty\) & \(\infty\) & 7 & 8
    \hline \(\mathbf { 2 }\) & 15 & 30 & 6 & 2 & 6 & 23
    \hline