| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Dynamic programming production scheduling |
| Difficulty | Standard +0.3 This is a standard textbook dynamic programming problem with clearly defined states, transitions, and costs. While it requires systematic working through a DP table (4 months, limited storage states), the problem structure is straightforward with no novel insights needed—students follow the standard backward induction algorithm taught in D2. The arithmetic is simple and the constraints are clearly stated, making this easier than average A-level questions which often require more problem-solving creativity. |
| Spec | 7.07a Simplex tableau: initial setup in standard format7.07b Simplex iterations: pivot choice and row operations7.07c Interpret simplex: values of variables, slack, and objective7.07d Simplex terminology: basic feasible solution, basic/non-basic variable7.07e Graphical interpretation: iterations as edges of convex polygon7.07f Algebraic interpretation: explain simplex calculations |
| Month | January | February | March | April |
| Number of robots required | 2 | 2 | 3 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| E.g. | 1M1 1A1 | |
| 2 | ||
| Stage | State | Action |
| April (4) | 0 | 4 |
| 1 | 3 | 0 |
| 2 | 2 | 0 |
| March (3) | 0 | 3 |
| 4 | 1 | \(400+\) |
| 1 | 2 | 0 |
| 3 | 1 | |
| 4 | 2 | \(400+150+300+600 =1450\) |
| 2 | 1 | 0 |
| 2 | 1 | |
| 3 | 2 | |
| Feb (2) | 0 | 2 |
| 3 | 1 | |
| 4 | 2 | \(400\) |
| 1 | 1 | 0 |
| 2 | 1 | |
| 3 | 2 | |
| 2 | 0 | 0 |
| 1 | 1 | |
| 2 | 2 | |
| Jan (2) | 0 | 2 |
| 3 | 1 | |
| 4 | 2 | \(400\) |
| B1 | ||
| 1 | ||
| Month | Jan | Feb |
| Number made | 2 | 3 |
| Total 12 |
| Answer/Working | Marks | Guidance |
|---|---|---|
| **E.g.** | 1M1 1A1 | |
| | 2 | |
| Stage | State | Action | Dest. | Value | | |
| April (4) | 0 | 4 | 0 | $400+$ | $300$ | $= 700^*$ | |
| | 1 | 3 | 0 | | $300$ | $= 300^*$ | |
| | 2 | 2 | 0 | | $300$ | $= 300^*$ | |
| March (3) | 0 | 3 | 0 | | $300+700 = 1000^*$ | 2M1 |
| | | 4 | 1 | $400+$ | $300+450 = 1150$ | |
| | 1 | 2 | 0 | | $300+700 = 1000$ | 2A1ft |
| | | 3 | 1 | | $300+450 = 900^*$ | |
| | | 4 | 2 | $400+150+300+600 =1450$ | | 3A1 |
| | 2 | 1 | 0 | | $300+700 = 1300$ | |
| | | 2 | 1 | | $300+450 = 1050^*$ | | 3M1 |
| | | 3 | 2 | | $300+600 = 1200$ | | 4A1ft |
| Feb (2) | 0 | 2 | 0 | | $300+1000 = 1300$ | | |
| | | 3 | 1 | | $300+ 900 = 1200^*$ | | 5A1ft |
| | | 4 | 2 | $400$ | $+300+1050 =1750$ | | |
| | 1 | 1 | 0 | | $300+1000 = 1450$ | | |
| | | 2 | 1 | | $300+ 900 = 1350^*$ | | 6A1 |
| | | 3 | 2 | | $300+1050 =1500$ | | |
| | 2 | 0 | 0 | | $1000 = 1000^*$ | | | 4M1 7A1 |
| | | 1 | 1 | | $300+ 900 = 1500$ | | 2 |
| | | 2 | 2 | | $300+1050 =1650$ | | |
| Jan (2) | 0 | 2 | 0 | | $300+1200 = 1500^*$ | | |
| | | 3 | 1 | | $300+1350 = 1650$ | | |
| | | 4 | 2 | $400$ | $+300+1300 = 2000$ | | |
| | | | | | | **B1** |
| | | | | | **1** |
| | | Month | Jan | Feb | March | April | |
| | | Number made | 2 | 3 | 3 | 3 | |
| | | **Total 12** |
**Notes for question 8 – see alts too:**
- ALL M marks - Must bring earlier optimal results into calculations. Ignore extra rows. Must have necessary right 'ingredients' (– storage costs, overheads, extra worker costs) at least once per stage.
- 1M1: First stage completed. 3 rows.
- 1A1: CAO condone missing * here. No extra rows.
- 2M1: Second stage completed. Expect 3 states.
- 2A1ft: Any 2 states correct. Ft on * values only No missing/extra rows. (Penalise * errors only once in the qn).
- 3A1: CAO All 3 states correct. No missing rows. (Penalise * errors only once in the question).
- 3M1: 3rd stage completed. Expect 3 states.
- 4A1ft: Any state correct. Ft on * values only. No missing rows. (Penalise * errors only once in the qn).
- 5A1ft: Any 2 states correct. Ft on * values only. No missing rows. (Penalise * errors only once in the qn).
- 6A1: CAO All 3 states correct. No missing/extra rows. (Penalise * errors only once in the question).
- 4M1: 4th stage completed.
- 7A1: CAO Final, state correct. No missing/extra rows. (Penalise * errors only once in the question).
- 1B1: CAO. Must have attempted algorithm, getting at least one M mark.
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# Alternative solutions for Question 8
The document also contains alternative solutions showing different approaches (Special Cases 1-4) based on different interpretations of when storage costs are added, but all follow the same marking structure above.8. A company makes industrial robots.
They can make up to four robots in any one month, but if they make more than three they will have to hire additional labour at a cost of $\pounds 400$ per month.\\
They can store up to two robots at a cost of $\pounds 150$ per robot per month.\\
The overhead costs are $\pounds 300$ in any month in which work is done.\\
Robots are delivered to buyers at the end of each month. There are no robots in stock at the beginning of January and there should be none in stock after the April delivery.
The order book for robots is
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
Month & January & February & March & April \\
\hline
Number of robots required & 2 & 2 & 3 & 4 \\
\hline
\end{tabular}
\end{center}
Use dynamic programming to determine the production schedule which minimises the costs, showing your working in the table provided in the answer book.\\
(Total 12 marks)\\
\hfill \mbox{\textit{Edexcel D2 2012 Q8 [12]}}