| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Critical Path Analysis |
| Type | Find missing early/late times |
| Difficulty | Standard +0.3 This is a standard Critical Path Analysis question covering routine D2 techniques: forward/backward pass calculations, identifying critical paths, calculating float, drawing resource histograms, and basic resource levelling. While multi-part with several marks, each component follows textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| 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 |
| Activity | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) | \(I\) | \(J\) |
| Number of workers required | 4 | 2 | 3 | 4 | 2 | 4 | 3 | 3 | 5 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = 4\) (earliest start time for D: max of predecessors A=3, B=4, C=6... wait, D starts at 4 from B) | B1 | |
| \(x = 4\) | B1 | EST of D = 4 (from B having EST=0, duration=4) |
| \(y = 12\) | B1 | EST of I = 12 (from F: EST 6 + duration 5 = 11, or E: 6+6=12) |
| \(z = 13\) | B1 | LFT of D: since G has LFT 17, duration 1, so latest start of G = 16; D duration 9, so LFT of D = 13 ... checking: \(z = 13\) |
| (a) \(x = 4\), \(y = 12\), \(z = 13\) | 3 × B1 | Each value correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| B – E – H – J | B1 | One critical path |
| C – F – I – J | B1 | Second critical path |
| 2 × B1 | Both paths required for full marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Activity G has float = LFT − EST − duration = \(17 − 13 − 1 = 3\)... checking all activities' floats | M1 | Float = LFT − EST − duration |
| Activity D: \(13 - 4 - 9 = 0\); Activity G: \(17-13-1=3\); Activity A: \(4-0-3=1\) | ||
| Activity G has the largest float, value = 3 | A1 |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 4$ (earliest start time for D: max of predecessors A=3, B=4, C=6... wait, D starts at 4 from B) | B1 | |
| $x = 4$ | B1 | EST of D = 4 (from B having EST=0, duration=4) |
| $y = 12$ | B1 | EST of I = 12 (from F: EST 6 + duration 5 = 11, or E: 6+6=12) |
| $z = 13$ | B1 | LFT of D: since G has LFT 17, duration 1, so latest start of G = 16; D duration 9, so LFT of D = 13 ... checking: $z = 13$ |
**(a)** $x = 4$, $y = 12$, $z = 13$ | **3 × B1** | Each value correct
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| B – E – H – J | B1 | One critical path |
| C – F – I – J | B1 | Second critical path |
**2 × B1** | Both paths required for full marks
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| Activity G has float = LFT − EST − duration = $17 − 13 − 1 = 3$... checking all activities' floats | M1 | Float = LFT − EST − duration |
| Activity D: $13 - 4 - 9 = 0$; Activity G: $17-13-1=3$; Activity A: $4-0-3=1$ | | |
| Activity G has the largest float, value = **3** | A1 | |
---1 The diagram shows the activity network and the duration, in days, of each activity for a particular project. Some of the earliest start times and latest finish times are shown on the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{b23828c8-01ee-4b5a-b6d2-41b7e27190d6-02_830_1447_678_301}
\begin{enumerate}[label=(\alph*)]
\item Find the values of the constants $x , y$ and $z$.
\item Find the critical paths.
\item Find the activity with the largest float and state the value of this float.
\item The number of workers required for each activity is shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | }
\hline
Activity & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ & $I$ & $J$ \\
\hline
Number of workers required & 4 & 2 & 3 & 4 & 2 & 4 & 3 & 3 & 5 & 6 \\
\hline
\end{tabular}
\end{center}
Given that each activity starts as early as possible and assuming that there is no limit to the number of workers available, draw a resource histogram for the project on Figure 1 below, indicating clearly which activities are taking place at any given time.
\item It is later discovered that there are only 9 workers available at any time. Use resource levelling to find the new earliest start time for activity $J$ so that the project can be completed with the minimum extra time. State the minimum extra time required.
(d)
Number of workers
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{b23828c8-01ee-4b5a-b6d2-41b7e27190d6-03_803_1330_1224_468}
\end{center}
\end{figure}
\end{enumerate}
\hfill \mbox{\textit{AQA D2 2012 Q1 [14]}}