| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Critical Path Analysis |
| Type | Schedule with limited workers - create schedule/chart |
| Difficulty | Standard +0.3 This is a standard D1 critical path analysis question covering routine techniques: finding early/late times, calculating float, determining lower bound for workers, and creating a schedule diagram. While it requires multiple steps and careful bookkeeping, all methods are algorithmic and directly taught. The worker constraint adds mild complexity but follows textbook procedures. Slightly easier than average A-level due to being purely procedural with no problem-solving insight required. |
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Activity | I.P.A | Activity |
| A (8) | - | H (5) |
| B (7) | - | I (9) |
| C (9) | - | J (11) |
| D (9) | A | K (5) |
| E (5) | A | L (4) |
| F (8) | B C | M (6) |
| G (7) | B C |
**Question 3 Notes:**
- a1M1: All top boxes complete, values generally increasing left to right, condone one rogue
- a1A1: CAO
- a2M1: Bottom boxes complete, values generally decreasing right to left, condone missing 0 or 37 for the M mark only
- a2A1: CAO
- b1M1: Correct calculation seen. All three numbers correct (ft)
- b1A1: Float correct (no follow through on this mark)
- c1M1: Attempt to find lower bound. [$82 - 104 / $ their finish time] accept awrt 2.5
- c1A1: CAO – correct calculation seen or awrt 2.5, then 3. (Beware 37/13 gives 3 also, so 3 with no working gets M0A0.)
- d1M1: Not a cascade chart. 4 workers used at most. At least 8 new (10 in total) activities placed
- d1A1: The critical activities (F I K M) and B correct. F – 8; I – 9; K – 5; M – 6; B – 7. B completed by 9 (its late finish time)
Now check the last 6 activities – the last two marks are for D, E, G, H, J and L only
First check that there are only three workers and that all 11 new (13 in total) activities are present (just once).
Then check precedences (see table below) – each row of the table could give rise to 1 error only in precedences
Finally check the length of each activity (see number in brackets in the activity column in the table below)
| Activity | I.P.A | Activity | I.P.A |
|---|---|---|---|
| A (8) | - | H (5) | C |
| B (7) | - | I (9) | E F |
| C (9) | - | J (11) | G H |
| D (9) | A | K (5) | D I |
| E (5) | A | L (4) | D I |
| F (8) | B C | M (6) | E F J K |
| G (7) | B C | | |
- d2M1: 3 workers. All 11 new (13 in total) activities present (just once). Condone one error either precedence, or activity length, on activities D, E, G, H, J and L.
- d2A1: 3 workers. All 11 new (13 in total) activities present (just once). No errors on activities D, E, G, H, J and L.3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1493d74b-e9ef-4c9a-91f6-877c1eaa74e2-04_549_1347_258_360}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A project is modelled by the activity network shown in Figure 3. The activities are represented by the arcs. The number in brackets on each arc gives the time, in days, to complete the activity. Each activity requires one worker. The project is to be completed in the shortest possible time.
\begin{enumerate}[label=(\alph*)]
\item Complete Diagram 1 in the answer book to show the early event times and late event times.
\item Calculate the total float for activity H. You must make the numbers you use in your calculation clear.
\item Calculate a lower bound for the number of workers needed to complete the project in the shortest possible time. Show your calculation.
Diagram 2 in the answer book shows a partly completed scheduling diagram for this project.
\item Complete the scheduling diagram, using the minimum number of workers, so that the project is completed in the minimum time.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2013 Q3 [12]}}