| Exam Board | Edexcel |
|---|---|
| Module | FD2 (Further Decision 2) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Dynamic programming production scheduling |
| Difficulty | Challenging +1.2 This is a standard dynamic programming problem with clearly defined states, transitions, and costs. While it requires systematic tabulation and careful bookkeeping across multiple months with storage constraints, it follows the textbook DP framework taught in FD2 with no novel insights needed—just methodical application of the algorithm. |
| Spec | 7.05a Critical path analysis: activity on arc networks7.05b Forward and backward pass: earliest/latest times, critical activities7.05c Total float: calculation and interpretation7.05d Latest start and earliest finish: independent and interfering float |
| Month | January | February | March | April | May |
| Number ordered | 1 | 3 | 3 | 5 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| May stage completed: State 2: \(160 = 160^*\); State 1: \(80+35=115^*\); State 0: \(70=70^*\) | M1 | Second stage completed, 6 rows, something in each cell |
| Any two states correct for May | A1 | |
| CAO for May (second) stage | A1 | |
| April stage completed with correct states, actions, destinations; e.g. State 3, Action 2: \(240+70+70=380^*\); State 2, Action 3: \(160+105+250+70=585^*\); State 1, Action 4: \(80+140+250+70=540^*\) | M1 | 6 rows, something in each cell |
| Any two states correct | A1 | |
| CAO for April stage | A1 | |
| March stage completed, 9 rows with correct state, action and destination; e.g. State 3, Action 1: \(240+35+540=815^*\); State 2, Action 2: \(160+70+540=770^*\); State 1, Action 3: \(80+105+250+540=975^*\); State 0, Action 4: \(140+250+540=930^*\) | M1 | Condone at least 8 rows correct |
| Any two states correct (ft optimal values) | A1ft | |
| CAO for March stage | A1 | |
| February stage completed, 13 rows correct; e.g. State 3, Action 2: \(240+70+770=1080^*\); State 2, Action 1: \(160+35+930=1125^*\); State 1, Action 2: \(80+70+930=1080^*\); State 0, Action 3: \(105+250+930=1285^*\) | M1 | Condone at least 11 rows |
| Any two states correct (ft optimal values) | A1ft | |
| Any three states correct (ft optimal values) | A1ft | |
| CAO for February stage | A1 | |
| January stage completed, 4 rows; e.g. State 0, Action 2: \(70+1080=1150^*\) | M1 | 4 rows, something in each cell |
| CAO for January stage | A1 | |
| Correct allocation: January 2, February 2, March 4, April 4, May 2 | B1 | Dependent on all previous M marks |
| Minimum Cost: £1150 | B1 | Dependent on all previous M marks |
# Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| May stage completed: State 2: $160 = 160^*$; State 1: $80+35=115^*$; State 0: $70=70^*$ | M1 | Second stage completed, 6 rows, something in each cell |
| Any two states correct for May | A1 | |
| CAO for May (second) stage | A1 | |
| April stage completed with correct states, actions, destinations; e.g. State 3, Action 2: $240+70+70=380^*$; State 2, Action 3: $160+105+250+70=585^*$; State 1, Action 4: $80+140+250+70=540^*$ | M1 | 6 rows, something in each cell |
| Any two states correct | A1 | |
| CAO for April stage | A1 | |
| March stage completed, 9 rows with correct state, action and destination; e.g. State 3, Action 1: $240+35+540=815^*$; State 2, Action 2: $160+70+540=770^*$; State 1, Action 3: $80+105+250+540=975^*$; State 0, Action 4: $140+250+540=930^*$ | M1 | Condone at least 8 rows correct |
| Any two states correct (ft optimal values) | A1ft | |
| CAO for March stage | A1 | |
| February stage completed, 13 rows correct; e.g. State 3, Action 2: $240+70+770=1080^*$; State 2, Action 1: $160+35+930=1125^*$; State 1, Action 2: $80+70+930=1080^*$; State 0, Action 3: $105+250+930=1285^*$ | M1 | Condone at least 11 rows |
| Any two states correct (ft optimal values) | A1ft | |
| Any three states correct (ft optimal values) | A1ft | |
| CAO for February stage | A1 | |
| January stage completed, 4 rows; e.g. State 0, Action 2: $70+1080=1150^*$ | M1 | 4 rows, something in each cell |
| CAO for January stage | A1 | |
| Correct allocation: January 2, February 2, March 4, April 4, May 2 | B1 | Dependent on all previous M marks |
| Minimum Cost: £1150 | B1 | Dependent on all previous M marks |
**Notes:** All M marks must bring earlier optimal results into calculations. Must have right ingredients (storage costs, overhead costs, additional workers) at least once per stage. Penalise lack of $^*$ only once per question.
---\begin{enumerate}
\item Bernie makes garden sheds. He can build up to four sheds each month.
\end{enumerate}
If he builds more than two sheds in any one month, he must hire an additional worker at a cost of $\pounds 250$ for that month.
In any month in which sheds are made, the overhead costs are $\pounds 35$ for each shed made that month.
A maximum of three sheds can be held in storage at the end of any one month, at a cost of $\pounds 80$ per shed per month.
Sheds must be delivered at the end of the month.\\
The order schedule for sheds is
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
Month & January & February & March & April & May \\
\hline
Number ordered & 1 & 3 & 3 & 5 & 2 \\
\hline
\end{tabular}
\end{center}
There are no sheds in storage at the beginning of January and Bernie plans to have no sheds left in storage after the May delivery.
Use dynamic programming to determine the production schedule that minimises the costs given above. Complete the working in the table provided in the answer book and state the minimum cost.\\
\hfill \mbox{\textit{Edexcel FD2 2022 Q6 [14]}}