AQA D2 2009 January — Question 5 10 marks

Exam BoardAQA
ModuleD2 (Decision Mathematics 2)
Year2009
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDynamic Programming
TypeDynamic programming maximin route
DifficultyModerate -0.5 This is a standard textbook application of dynamic programming to find a maximin route. The algorithm is straightforward (working backwards from Y, taking maximum of minimums along paths), and the network is small enough to solve systematically. Part (a) requires simple comparison of two routes, while part (b) is routine execution of a taught algorithm with a provided table structure. Easier than average as it's algorithmic rather than requiring problem-solving insight.
Spec7.05e Cascade charts: scheduling and effect of delays

5 [Figure 3, printed on the insert, is provided for use in this question.]
A truck has to transport stones from a quarry, \(Q\), to a builders yard, \(Y\). The network shows the possible roads from \(Q\) to \(Y\). Along each road there are bridges with weight restrictions. The value on each edge indicates the maximum load in tonnes that can be carried by the truck along that particular road. \includegraphics[max width=\textwidth, alt={}, center]{6c407dbf-efe5-49e4-881f-91e7de5c46d9-6_723_1280_589_372} The truck is able to carry a load of up to 20 tonnes. The optimal route, known as the maximin route, is that for which the possible load that the truck can carry is as large as possible.
  1. Explain why the route \(Q A C Y\) is better than the route \(Q B E Y\).
  2. By completing the table on Figure 3, or otherwise, use dynamic programming, working backwards from \(\boldsymbol { Y }\), to find the optimal (maximin) route from \(Q\) to \(Y\). Write down the maximin route and state the maximum possible load that the truck can carry from \(Q\) to \(Y\).

5 [Figure 3, printed on the insert, is provided for use in this question.]\\
A truck has to transport stones from a quarry, $Q$, to a builders yard, $Y$. The network shows the possible roads from $Q$ to $Y$. Along each road there are bridges with weight restrictions. The value on each edge indicates the maximum load in tonnes that can be carried by the truck along that particular road.\\
\includegraphics[max width=\textwidth, alt={}, center]{6c407dbf-efe5-49e4-881f-91e7de5c46d9-6_723_1280_589_372}

The truck is able to carry a load of up to 20 tonnes. The optimal route, known as the maximin route, is that for which the possible load that the truck can carry is as large as possible.
\begin{enumerate}[label=(\alph*)]
\item Explain why the route $Q A C Y$ is better than the route $Q B E Y$.
\item By completing the table on Figure 3, or otherwise, use dynamic programming, working backwards from $\boldsymbol { Y }$, to find the optimal (maximin) route from $Q$ to $Y$. Write down the maximin route and state the maximum possible load that the truck can carry from $Q$ to $Y$.
\end{enumerate}

\hfill \mbox{\textit{AQA D2 2009 Q5 [10]}}