Edexcel D1 2024 January — Question 1 13 marks

Exam BoardEdexcel
ModuleD1 (Decision Mathematics 1)
Year2024
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCritical Path Analysis
TypeCalculate lower bound for workers
DifficultyModerate -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.
Spec7.05a Critical path analysis: activity on arc networks7.05d Latest start and earliest finish: independent and interfering float

1. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{4814ebd7-f48a-49cf-8ca2-045d84abd63c-2_679_958_315_568} \captionsetup{labelformat=empty} \caption{Figure 1}
\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.
  1. Complete Diagram 1 in the answer book to show the early event times and the late event times.
  2. Calculate the total float for activity D. You must make the numbers used in your calculation clear.
  3. 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.
  4. Draw a cascade chart for this project on Grid 1 in the answer book.
  5. 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.)

Question 1:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
All top boxes complete, values generally increasing in direction of arrows (left to right), condone one rogueM1 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 onlyM1 a2M1
CAO (bottom boxes)A1 (4) a2A1
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks 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)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Lower bound \(= \dfrac{4+3+7+\ldots+5+12}{24} = \dfrac{67}{24} = 2.791\ldots = 3\) workersM1 A1 (2)
Part (d)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Gantt chart — correct rows for C, F, G, J (critical activities)M1
Correct placement of A, BA1
Correct placement of D, EA1
Correct placement of H, I, K, LA1 (4)
Part (e)
AnswerMarks Guidance
Answer/WorkingMarks 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:
AnswerMarks Guidance
AnswerMark 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 2.8c1M1 Condone one missing value
Answer of 3, requires correct calculation or awrt 2.8 seenc1A1 CSO - answer of 3 with no working scores no marks
Part (d) - Scheduling Diagram:
AnswerMarks Guidance
AnswerMark Guidance
At least eight different activities labelled including at least five floatsd1M1 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 floatd1A1
Any five non-critical activities correctd2A1 Not dependent on previous A mark
Completely correct Gantt chart with exactly twelve activities appearing just onced3A1 CSO
Part (e) - Workers:
AnswerMarks Guidance
AnswerMark 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 Le1A1 Condone 'days' instead of 'hours'; strict inequalities mean time of exactly 12 is incorrect
Float Table:
AnswerMarks Guidance
ActivityDuration + Float Activity
A0 to 4, F: 4 to 7 F
B0 to 3, F: 3 to 13 G
C0 to 7, Critical H
D4 to 9, F: 9 to 12 I
E4 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]}}