| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Dynamic programming production scheduling |
| Difficulty | Moderate -0.5 This is a standard textbook dynamic programming problem with clearly defined states, constraints, and a straightforward backward recursion. While it requires systematic working through multiple stages, the problem structure is typical for A-level Decision Maths with no novel insights needed—just careful application of the DP algorithm to minimize costs given storage and production constraints. |
| Spec | 7.03h Best/worst/typical case: run-time analysis |
| Month | January | February | March | April |
| Number of cabinets required | 3 | 3 | 5 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Storage cost: \(2 \times £50 = £100\) | M1 | |
| Total = \(£300 + £100 = £400\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Systematic backwards dynamic programming from April | M1 | |
| April stage correct | A1 | |
| March stage correct | A1 | |
| February stage correct | A1 | |
| January stage correct | A1 | |
| Optimal production schedule identified | A1 | |
| Minimum total cost stated | A2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Revenue: \(13 \times £2000 = £26000\) | M1 | |
| Material costs: \(13 \times £300 = £3900\) | M1 | |
| Wages: \(4 \times £2000 = £8000\); overhead and storage costs added | ||
| Total profit = Revenue \(-\) all costs | A1 |
# Question 5:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Storage cost: $2 \times £50 = £100$ | M1 | |
| Total = $£300 + £100 = £400$ | A1 | |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Systematic backwards dynamic programming from April | M1 | |
| April stage correct | A1 | |
| March stage correct | A1 | |
| February stage correct | A1 | |
| January stage correct | A1 | |
| Optimal production schedule identified | A1 | |
| Minimum total cost stated | A2 | |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Revenue: $13 \times £2000 = £26000$ | M1 | |
| Material costs: $13 \times £300 = £3900$ | M1 | |
| Wages: $4 \times £2000 = £8000$; overhead and storage costs added | |
| Total profit = Revenue $-$ all costs | A1 | |
---5 [Figure 3, printed on the insert, is provided for use in this question.]\\
A small firm produces high quality cabinets.\\
It can produce up to 4 cabinets each month.\\
Whenever at least one cabinet is made during that month, the overhead costs for that month are $\pounds 300$.
It is possible to hold in stock a maximum of 2 cabinets during any month.\\
The cost of storage is $\pounds 50$ per cabinet per month.\\
The orders for cabinets are shown in the table below. There is no stock at the beginning of January and the firm plans to clear all stock after completing the April orders.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Month & January & February & March & April \\
\hline
Number of cabinets required & 3 & 3 & 5 & 2 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Determine the total cost of storing 2 cabinets and producing 3 cabinets in a given month.
\item By completing the table of values on Figure 3, or otherwise, use dynamic programming, working backwards from April, to find the production schedule which minimises total costs.
\item Each cabinet is sold for $\pounds 2000$ but there is an additional cost of $\pounds 300$ for materials to make each cabinet and $\pounds 2000$ per month in wages. Determine the total profit for the four-month period.
\end{enumerate}
\hfill \mbox{\textit{AQA D2 2008 Q5 [13]}}