| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2003 |
| Session | June |
| Marks | 18 |
| 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, actions, and costs. The table structure is provided, and students follow a mechanical backward induction process. While it requires careful bookkeeping across multiple stages, it involves routine application of the DP algorithm without novel insight or complex optimization reasoning. |
| 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 interpretation |
| Month | August | September | October | November |
| Number of cycles required | 3 | 3 | 5 | 2 |
| Stage | Demand | State | Action | Destination | Value |
| \multirow[t]{3}{*}{1 (Nov)} | \multirow[t]{3}{*}{2} | 0 (in stock) | (make) 2 | 0 | 200 |
| 1 (in stock) | (make) 1 | 0 | 240 | ||
| 2 (in stock) | (make) 0 | 0 | 80 | ||
| \multirow[t]{2}{*}{2 (Oct)} | \multirow[t]{2}{*}{5} | 1 | 4 | 0 | \(590 + 200 = 790\) |
| 2 | 3 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | Total cost = \(2 \times 40 + 350 + 200 = £630\) | M1 A1 |
| (b) | Stage table showing: October 2 (demand 5, state 1), with actions and destinations leading to values of 790, 480, 870, 520, 560, 750; September 3 (demand 3, state 0) with actions 4, destination 1, value 1340; August 4 (demand 3, state 0) with actions 3, 4, destinations 0, 1, values 1540, 1580 | M1 A1 M1 A1 M1 A1 ft |
| Month | August | September |
| Make | 3 | 4 |
| cost = £1540 | A1 ft | 3 |
| (c) | Profit per cycle = \(13 \times 1400 = 18200\). Cost of Kim's time = £2000. Cost of production = £1540. ∴ Total profit = \(18200 - 3540 = £14660\) | B1 M1 A1 ft |
(a) | Total cost = $2 \times 40 + 350 + 200 = £630$ | M1 A1 | 2 |
(b) | Stage table showing: October 2 (demand 5, state 1), with actions and destinations leading to values of 790, 480, 870, 520, 560, 750; September 3 (demand 3, state 0) with actions 4, destination 1, value 1340; August 4 (demand 3, state 0) with actions 3, 4, destinations 0, 1, values 1540, 1580 | M1 A1 M1 A1 M1 A1 ft | 4 |
| Month | August | September | October | November |
|---|---|---|---|---|
| Make | 3 | 4 | 4 | 2 |
cost = £1540 | A1 ft | 3 |
(c) | Profit per cycle = $13 \times 1400 = 18200$. Cost of Kim's time = £2000. Cost of production = £1540. ∴ Total profit = $18200 - 3540 = £14660$ | B1 M1 A1 ft | 3 |
6. Kris produces custom made racing cycles. She can produce up to four cycles each month, but if she wishes to produce more than three in any one month she has to hire additional help at a cost of $\pounds 350$ for that month. In any month when cycles are produced, the overhead costs are $\pounds 200$. A maximum of 3 cycles can be held in stock in any one month, at a cost of $\pounds 40$ per cycle per month. Cycles must be delivered at the end of the month. The order book for cycles is
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
Month & August & September & October & November \\
\hline
Number of cycles required & 3 & 3 & 5 & 2 \\
\hline
\end{tabular}
\end{center}
Disregarding the cost of parts and Kris' time,
\begin{enumerate}[label=(\alph*)]
\item determine the total cost of storing 2 cycles and producing 4 cycles in a given month, making your calculations clear.
There is no stock at the beginning of August and Kris plans to have no stock after the November delivery.
\item Use dynamic programming to determine the production schedule which minimises the costs, showing your working in the table below.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
Stage & Demand & State & Action & Destination & Value \\
\hline
\multirow[t]{3}{*}{1 (Nov)} & \multirow[t]{3}{*}{2} & 0 (in stock) & (make) 2 & 0 & 200 \\
\hline
& & 1 (in stock) & (make) 1 & 0 & 240 \\
\hline
& & 2 (in stock) & (make) 0 & 0 & 80 \\
\hline
\multirow[t]{2}{*}{2 (Oct)} & \multirow[t]{2}{*}{5} & 1 & 4 & 0 & $590 + 200 = 790$ \\
\hline
& & 2 & 3 & 0 & \\
\hline
\end{tabular}
\end{center}
The fixed cost of parts is $\pounds 600$ per cycle and of Kris' time is $\pounds 500$ per month. She sells the cycles for $\pounds 2000$ each.
\item Determine her total profit for the four month period.\\
(3)\\
(Total 18 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 2003 Q6 [18]}}