| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2024 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Critical Path Analysis |
| Type | Calculate lower bound for workers |
| Difficulty | Moderate -0.5 This is a standard Critical Path Analysis question covering routine D1 techniques: finding early/late times, calculating float, determining lower bound (sum of activity times ÷ project duration), and drawing a cascade chart. All steps are algorithmic procedures taught directly in the syllabus with no novel problem-solving required, making it slightly easier than average. |
| Spec | 7.05a Critical path analysis: activity on arc networks7.05d Latest start and earliest finish: independent and interfering float |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| All top boxes complete, values generally increasing in direction of arrows (left to right), condone one rogue | M1 | a1M1: All top boxes complete |
| CAO (top boxes) — correct values: 0,4,7,9,12,13,18,24 etc. | A1 | a1A1: CAO top boxes |
| All bottom boxes complete, values generally decreasing in opposite direction of arrows (right to left), condone one rogue. Condone missing 0 and/or 24 at end event for M mark only | M1 | a2M1 |
| CAO (bottom boxes) | A1 (4) | a2A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Activity D has a total float of \(12 - 4 - 5 = 3\) (hours) | B1ft (1) | Correct calculation for their event times for activity D (all three figures must be seen) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Lower bound \(= \dfrac{4+3+7+\ldots+5+12}{24} = \dfrac{67}{24} = 2.791\ldots = 3\) workers | M1 A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gantt chart — correct rows for C, F, G, J (critical activities) | M1 | |
| Correct placement of A, B | A1 | |
| Correct placement of D, E | A1 | |
| Correct placement of H, I, K, L | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Minimum is 4 workers e.g. activities F, H, I and L together with \(12 < \text{time} < 13\) | dM1 A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt to find lower bound: value in interval \([55-79]\) / their finish time, or sum of activities (12 values) / their finish time, or awrt 2.8 | c1M1 | Condone one missing value |
| Answer of 3, requires correct calculation or awrt 2.8 seen | c1A1 | CSO - answer of 3 with no working scores no marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At least eight different activities labelled including at least five floats | d1M1 | A scheduling diagram with no floats scores M0 |
| Critical activities C, F, G, J correct (appearing just once) and three non-critical activities with correct duration and total float | d1A1 | |
| Any five non-critical activities correct | d2A1 | Not dependent on previous A mark |
| Completely correct Gantt chart with exactly twelve activities appearing just once | d3A1 | CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Statement with correct number of workers (4) and correct activities (F, H, I, L) with numerical time \(12 \leq t \leq 13\) | e1depM1 | Mark numerical value only, dependent on M mark in (d) |
| Completely correct statement with both time and activities; time within \(12 < t < 13\) (e.g. 12.5) and activities F, H, I and L | e1A1 | Condone 'days' instead of 'hours'; strict inequalities mean time of exactly 12 is incorrect |
| Answer | Marks | Guidance |
|---|---|---|
| Activity | Duration + Float | Activity |
| A | 0 to 4, F: 4 to 7 | F |
| B | 0 to 3, F: 3 to 13 | G |
| C | 0 to 7, Critical | H |
| D | 4 to 9, F: 9 to 12 | I |
| E | 4 to 6, F: 6 to 18 | J |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| All top boxes complete, values generally increasing in direction of arrows (left to right), condone one rogue | M1 | a1M1: All top boxes complete |
| CAO (top boxes) — correct values: 0,4,7,9,12,13,18,24 etc. | A1 | a1A1: CAO top boxes |
| All bottom boxes complete, values generally decreasing in opposite direction of arrows (right to left), condone one rogue. Condone missing 0 and/or 24 at end event for M mark only | M1 | a2M1 |
| CAO (bottom boxes) | A1 **(4)** | a2A1 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Activity D has a total float of $12 - 4 - 5 = 3$ (hours) | B1ft **(1)** | Correct calculation for their event times for activity D (all three figures must be seen) |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Lower bound $= \dfrac{4+3+7+\ldots+5+12}{24} = \dfrac{67}{24} = 2.791\ldots = 3$ workers | M1 A1 **(2)** | |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gantt chart — correct rows for C, F, G, J (critical activities) | M1 | |
| Correct placement of A, B | A1 | |
| Correct placement of D, E | A1 | |
| Correct placement of H, I, K, L | A1 **(4)** | |
## Part (e)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Minimum is 4 workers e.g. activities F, H, I and L together with $12 < \text{time} < 13$ | dM1 A1 **(2)** | |
**Total: 13 marks**
# Question 1 (Critical Path Analysis):
## Part (c) - Lower Bound:
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to find lower bound: value in interval $[55-79]$ / their finish time, or sum of activities (12 values) / their finish time, or awrt 2.8 | c1M1 | Condone one missing value |
| Answer of 3, requires correct calculation or awrt 2.8 seen | c1A1 | CSO - answer of 3 with no working scores no marks |
## Part (d) - Scheduling Diagram:
| Answer | Mark | Guidance |
|--------|------|----------|
| At least eight different activities labelled including at least five floats | d1M1 | A scheduling diagram with no floats scores M0 |
| Critical activities C, F, G, J correct (appearing just once) and three non-critical activities with correct duration and total float | d1A1 | |
| Any five non-critical activities correct | d2A1 | Not dependent on previous A mark |
| Completely correct Gantt chart with exactly twelve activities appearing just once | d3A1 | CSO |
## Part (e) - Workers:
| Answer | Mark | Guidance |
|--------|------|----------|
| Statement with correct number of workers (4) and correct activities (F, H, I, L) with numerical time $12 \leq t \leq 13$ | e1depM1 | Mark numerical value only, dependent on M mark in (d) |
| Completely correct statement with both time and activities; time within $12 < t < 13$ (e.g. 12.5) and activities F, H, I and L | e1A1 | Condone 'days' instead of 'hours'; strict inequalities mean time of exactly 12 is incorrect |
**Float Table:**
| Activity | Duration + Float | Activity | Duration + Float | Activity | Duration + Float |
|----------|-----------------|----------|-----------------|----------|-----------------|
| A | 0 to 4, F: 4 to 7 | F | 7 to 13, Critical | K | 13 to 18, F: 18 to 24 |
| B | 0 to 3, F: 3 to 13 | G | 13 to 18, Critical | L | 9 to 21, F: 21 to 24 |
| C | 0 to 7, Critical | H | 7 to 13, F: 13 to 18 | | |
| D | 4 to 9, F: 9 to 12 | I | 7 to 13, F: 13 to 18 | | |
| E | 4 to 6, F: 6 to 18 | J | 18 to 24, Critical | | |
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1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4814ebd7-f48a-49cf-8ca2-045d84abd63c-2_679_958_315_568}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A project is modelled by the activity network shown in Figure 1. The activities are represented by the arcs. The number in brackets on each arc gives the time, in hours, to complete the corresponding activity. Each activity requires one worker. The project is to be completed in the shortest possible time using as few workers as possible.
\begin{enumerate}[label=(\alph*)]
\item Complete Diagram 1 in the answer book to show the early event times and the late event times.
\item Calculate the total float for activity D. You must make the numbers used in your calculation clear.
\item Calculate a lower bound for the minimum number of workers required to complete the project in the shortest possible time. You must show your working.
\item Draw a cascade chart for this project on Grid 1 in the answer book.
\item Use your cascade chart to determine the minimum number of workers needed to complete the project in the shortest possible time. You must make specific reference to time and activities. (You do not need to provide a schedule of the activities.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2024 Q1 [13]}}