| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Dynamic programming order sequencing |
| Difficulty | Moderate -0.5 This is a standard textbook dynamic programming problem with a small state space (3 items, 3 stages). The backwards recursion is straightforward and mechanical, requiring only systematic enumeration of paths and comparison of totals. Part (b) is trivial (just change min to max). Easier than average A-level as it requires no mathematical insight, just following the DP algorithm. |
| Spec | 7.03h Best/worst/typical case: run-time analysis |
| \multirow[t]{2}{*}{Month} | \multirow[t]{2}{*}{Already built} | Cost | ||
| \(\boldsymbol { A }\) | B | \(\boldsymbol { C }\) | ||
| 1 | - | 500 | 440 | 475 |
| 2 | A | - | 440 | 490 |
| B | 510 | - | 500 | |
| \(\boldsymbol { C }\) | 520 | 490 | - | |
| 3 | \(\boldsymbol { A }\) and \(\boldsymbol { B }\) | - | - | 520 |
| \(\boldsymbol { A }\) and \(\boldsymbol { C }\) | - | 500 | - | |
| \(\boldsymbol { B }\) and \(\boldsymbol { C }\) | 510 | - | - | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Stage 3 values: \(A\)+\(B\) built → build \(C\): cost 520; \(A\)+\(C\) built → build \(B\): cost 500; \(B\)+\(C\) built → build \(A\): cost 510 | B1 | |
| Stage 2 working: e.g. built \(A\): min of (440+500, 490+510) = 940 or 1000 → choose \(B\) then \(C\), cost 940 | M1 A1 | |
| All stage 2 options evaluated correctly | A1 | |
| Stage 1: min of (500+940, 440+..., 475+...) correctly evaluated | M1 | |
| Optimal order identified with minimum cost stated | A1 | 6 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Repeat working but maximising at each stage | M1 A1 | |
| Correct optimal order for maximum profit identified | A1 A1 | 4 marks total |
# Question 5:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Stage 3 values: $A$+$B$ built → build $C$: cost 520; $A$+$C$ built → build $B$: cost 500; $B$+$C$ built → build $A$: cost 510 | B1 | |
| Stage 2 working: e.g. built $A$: min of (440+500, 490+510) = 940 or 1000 → choose $B$ then $C$, cost 940 | M1 A1 | |
| All stage 2 options evaluated correctly | A1 | |
| Stage 1: min of (500+940, 440+..., 475+...) correctly evaluated | M1 | |
| Optimal order identified with minimum cost stated | A1 | 6 marks total |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Repeat working but maximising at each stage | M1 A1 | |
| Correct optimal order for maximum profit identified | A1 A1 | 4 marks total |
I can see this is an AQA Mathematics Decision 2 (MD02) June 2007 exam paper, but the images provided show the **question paper and insert**, not the mark scheme. I cannot extract mark scheme content from these pages as they don't contain mark scheme information.
To find the official mark scheme for this paper, I can instead work through the solutions based on the questions shown:
---5 [Figure 3, printed on the insert, is provided for use in this question.]\\
A maker of exclusive furniture is planning to build three cabinets $A , B$ and $C$ at the rate of one per month. The order in which they are built is a matter of choice, but the costs will vary because of the materials available and suppliers' costs. The expected costs, in pounds, are given in the table.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multirow[t]{2}{*}{Month} & \multirow[t]{2}{*}{Already built} & \multicolumn{3}{|c|}{Cost} \\
\hline
& & $\boldsymbol { A }$ & B & $\boldsymbol { C }$ \\
\hline
1 & - & 500 & 440 & 475 \\
\hline
2 & A & - & 440 & 490 \\
\hline
& B & 510 & - & 500 \\
\hline
& $\boldsymbol { C }$ & 520 & 490 & - \\
\hline
3 & $\boldsymbol { A }$ and $\boldsymbol { B }$ & - & - & 520 \\
\hline
& $\boldsymbol { A }$ and $\boldsymbol { C }$ & - & 500 & - \\
\hline
& $\boldsymbol { B }$ and $\boldsymbol { C }$ & 510 & - & - \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Use dynamic programming, working backwards from month 3, to determine the order of manufacture that minimises the total cost. You may wish to use Figure 3 for your working.
\item It is discovered that the figures given were actually the profits, not the costs, for each item. Modify your solution to find the order of manufacture that maximises the total profit. You may wish to use the final column of Figure 3 for your working.
\end{enumerate}
\hfill \mbox{\textit{AQA D2 2007 Q5 [10]}}