| Exam Board | OCR |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Critical Path Analysis |
| Type | Calculate early and late times |
| Difficulty | Moderate -0.5 This is a standard Critical Path Analysis question requiring calculation of early/late times and identification of critical activities. While it involves multiple steps (forward pass, backward pass, finding critical path), these are routine algorithmic procedures taught directly in D2 with no novel problem-solving required. The question is slightly easier than average A-level standard due to its purely procedural nature. |
| 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 |
| (i) | \(A \bullet\) |
| \(B \bullet\) | \(\bullet J\) |
| \(C \bullet\) | \(\bullet K\) |
| \(D \bullet\) | \(\bullet L\) |
| \(E \bullet\) | \(\bullet M\) |
| \(F \bullet\) | \(\bullet N\) |
| \(J\) | \(K\) | \(L\) | \(M\) | \(N\) | \(O\) | |
| \(A\) | 2 | 5 | 2 | 2 | 5 | 2 |
| \(B\) | 2 | 5 | 2 | 0 | 5 | 5 |
| \(C\) | 5 | 0 | 5 | 5 | 2 | 2 |
| \(D\) | ||||||
| \(E\) | ||||||
| \(F\) |
| Answer | Marks | Guidance |
|---|---|---|
| Activity | Duration | Immediate predecessors |
| A | 6 | - |
| B | 4 | - |
| C | 5 | A |
| D | 1 | A, B |
| E | 5 | A, D |
| F | 4 | D |
| G | 2 | C, E, F |
| Answer | Marks |
|---|---|
| B1 | For predecessors for activities \(A, B\) and \(C\) correct |
| B1 | For predecessors for activities \(D, F\) and \(G\) correct |
| B1, [3] | For predecessors for activity \(E\) correct |
| Answer | Marks |
|---|---|
| - Node at start (0, 0) with 12 | 12 |
| - Node after A with times 6 | 6 and label C |
| - Nodes connected with various times including 7 | 7 |
| - Final node with 14 | 14 |
| Answer | Marks |
|---|---|
| M1 | For carrying out forward pass (no more than one independent error) |
| A1 | For all early event times correct |
| M1 | For carrying out backwards pass (no more than one independent error) |
| A1 | For all late event times correct |
| Answer | Marks | Guidance |
|---|---|---|
| Critical activities: A, D, E, G | B1, B1, [6] | For 14 (cao); For A, D, E, G only (cao); For stating that time increases by 2, or equivalent (cao) |
| Answer | Marks | Guidance |
|---|---|---|
| Becomes 16 (hours) | B1, [1] | For stating that time increases by 2, or equivalent |
| B1 | For a resource histogram with no overhanging cells |
| Answer | Marks | Guidance |
|---|---|---|
| Number of workers required = 3 | M1, A1 | For a reasonable attempt, fit their start times if possible; For a completely correct histogram (cao) |
| B1, [4] | For 3 or follow through their histogram if possible |
(i)
| Activity | Duration | Immediate predecessors |
|----------|----------|------------------------|
| A | 6 | - |
| B | 4 | - |
| C | 5 | A |
| D | 1 | A, B |
| E | 5 | A, D |
| F | 4 | D |
| G | 2 | C, E, F |
**ANSWERED ON INSERT**
| B1 | For predecessors for activities $A, B$ and $C$ correct
| B1 | For predecessors for activities $D, F$ and $G$ correct
| B1, [3] | For predecessors for activity $E$ correct
(ii) **Network diagram with:**
- Node at start (0, 0) with 12|12
- Node after A with times 6|6 and label C
- Nodes connected with various times including 7|7
- Final node with 14|14
- Activity G marked
| M1 | For carrying out forward pass (no more than one independent error)
| A1 | For all early event times correct
| M1 | For carrying out backwards pass (no more than one independent error)
| A1 | For all late event times correct
Minimum completion time = 14 hours
Critical activities: A, D, E, G | B1, B1, [6] | For 14 (cao); For A, D, E, G only (cao); For stating that time increases by 2, or equivalent (cao)
(iii) Increased by 2 (hours)
Becomes 16 (hours) | B1, [1] | For stating that time increases by 2, or equivalent
| B1 | For a resource histogram with no overhanging cells
(iv) **Resource histogram shown with workers on y-axis (0-4) and hours on x-axis (0-14)**
Number of workers required = 3 | M1, A1 | For a reasonable attempt, fit their start times if possible; For a completely correct histogram (cao)
| B1, [4] | For 3 or follow through their histogram if possible
---4 Answer this question on the insert provided.
The diagram shows an activity network for a project. The table lists the durations of the activities (in hours).\\
\includegraphics[max width=\textwidth, alt={}, center]{e879b1f5-edc7-4819-80be-2a90dbf3d451-05_680_1125_424_244}
(ii)
Key:
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e879b1f5-edc7-4819-80be-2a90dbf3d451-10_154_225_1119_1509}
\captionsetup{labelformat=empty}
\caption{Early event Late event time time}
\end{center}
\end{figure}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{e879b1f5-edc7-4819-80be-2a90dbf3d451-10_762_1371_1409_427}
\end{center}
Minimum completion time = $\_\_\_\_$ hours
Critical activities: $\_\_\_\_$\\
(iii) $\_\_\_\_$\\
(iv)\\
\includegraphics[max width=\textwidth, alt={}, center]{e879b1f5-edc7-4819-80be-2a90dbf3d451-11_513_1189_543_520}
Number of workers required = $\_\_\_\_$
\begin{center}
\begin{tabular}{ l l }
(i) & $A \bullet$ \\
$B \bullet$ & $\bullet J$ \\
$C \bullet$ & $\bullet K$ \\
$D \bullet$ & $\bullet L$ \\
$E \bullet$ & $\bullet M$ \\
$F \bullet$ & $\bullet N$ \\
\end{tabular}
\end{center}
(ii) $\_\_\_\_$\\
(iii)
\begin{center}
\begin{tabular}{ c | c c c c c c }
& $J$ & $K$ & $L$ & $M$ & $N$ & $O$ \\
\hline
$A$ & 2 & 5 & 2 & 2 & 5 & 2 \\
$B$ & 2 & 5 & 2 & 0 & 5 & 5 \\
$C$ & 5 & 0 & 5 & 5 & 2 & 2 \\
$D$ & & & & & & \\
$E$ & & & & & & \\
$F$ & & & & & & \\
\end{tabular}
\end{center}
Answer part (iv) in your answer booklet.
\hfill \mbox{\textit{OCR D2 2006 Q4 [14]}}