Edexcel FD1 2019 June — Question 4 9 marks

Exam BoardEdexcel
ModuleFD1 (Further Decision 1)
Year2019
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCritical Path Analysis
TypeFind missing early/late times
DifficultyStandard +0.3 This is a standard Critical Path Analysis question requiring systematic application of forward/backward pass algorithms and float calculations. Part (a) uses the float formula (a routine calculation), parts (b-d) involve standard CPA procedures taught in FD1. While it requires careful bookkeeping across multiple steps, it demands no novel insight—just methodical application of well-practiced techniques, making it slightly easier than average.
Spec7.05b Forward and backward pass: earliest/latest times, critical activities7.05c Total float: calculation and interpretation
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{162f9d72-84a4-4b1a-93cf-b7eeb7f957ae-05_1004_1797_205_134} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} The network in Figure 3 shows the activities that need to be undertaken to complete a project. Each activity is represented by an arc and the duration of the activity, in days, is shown in brackets. The early event times and late event times are to be shown at each vertex and one late event time has been completed for you. The total float of activity H is 7 days.
  1. Explain, with detailed reasoning, why \(x = 11\)
  2. Determine the missing early event times and late event times, and hence complete Diagram 1 in your answer book. Each activity requires one worker and the project must be completed in the shortest possible time using as few workers as possible.
  3. Calculate a lower bound for the number of workers needed to complete the project in the shortest possible time.
  4. Schedule the activities using Grid 1 in the answer book.
Question 4:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Early event time at end of activity C is 7; float on activity H is \(25 - 7 - x\)B1 3.1a
Float on H is 7, so \(25 - 7 - x = 7\), giving \(x = 25 - 7 - 7 = 11\)dB1 2.4
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Correct network structure with activity labels and durationsM1 2.1
Early event times all correctA1 1.1b
Late event times all correctA1 1.1b
Part (c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{95}{32} = 2.968\ldots = 3\) workersB1 2.2a
Part (d)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Correct Gantt chart structure showing activities scheduled appropriatelyM1 2.1
All activities placed correctlyA1 1.1b
Resource levelling correctA1 1.1b
Question 4:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Float on H is \(25 - 7 - x\); early event time at end of C is 7, or early event time at start of H is 7 and total float for H is \(25 - 7 - x\) (or \(25 - x - 7\), but not just \(18 - x\))B1 Must mention early event time at end of C is 7; no reason for why early event time at end of C is 7 required
Correct explanation for \(x = 11\); equate \(25 - 7 - x\) to 7 (allow \(18 - x = 7\)) hence \(x = 11\)dB1 Dependent on previous B mark; must show where 18 comes from
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
All top boxes and all bottom boxes completed; values generally increasing left to right (top boxes) and decreasing right to left (bottom boxes)M1 Condone missing 0s at source node or 32 in bottom box at sink node for M only; condone one rogue value in top boxes and one in bottom boxes
CAO - Top boxes (including zero at source node)A1
CAO - Bottom boxes (including zero at sink node)A1
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Correct calculation seen, answer of 3B1 Answer of 3 with no working scores B0
Part (d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Not a cascade chart; 4 workers used at most; at least 10 different activities placedM1
4 workers; all 13 activities present (just once); condone at most two errorsA1 If activity appears for two different workers simultaneously this is A0; activity can give rise to at most three errors: one on duration, one on time interval, one on IPA
4 workers; all 13 activities present (just once); no errorsA1 
# Question 4:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Early event time at end of activity C is 7; float on activity H is $25 - 7 - x$ | B1 | 3.1a |
| Float on H is 7, so $25 - 7 - x = 7$, giving $x = 25 - 7 - 7 = 11$ | dB1 | 2.4 |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct network structure with activity labels and durations | M1 | 2.1 |
| Early event times all correct | A1 | 1.1b |
| Late event times all correct | A1 | 1.1b |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{95}{32} = 2.968\ldots = 3$ workers | B1 | 2.2a |

## Part (d)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct Gantt chart structure showing activities scheduled appropriately | M1 | 2.1 |
| All activities placed correctly | A1 | 1.1b |
| Resource levelling correct | A1 | 1.1b |

# Question 4:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Float on H is $25 - 7 - x$; early event time at end of C is 7, or early event time at start of H is 7 **and** total float for H is $25 - 7 - x$ (or $25 - x - 7$, but not just $18 - x$) | B1 | Must mention early event time at end of C is 7; no reason for why early event time at end of C is 7 required |
| Correct explanation for $x = 11$; equate $25 - 7 - x$ to 7 (allow $18 - x = 7$) hence $x = 11$ | dB1 | Dependent on previous B mark; must show where 18 comes from |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| All top boxes and all bottom boxes completed; values generally increasing left to right (top boxes) and decreasing right to left (bottom boxes) | M1 | Condone missing 0s at source node or 32 in bottom box at sink node for M only; condone one rogue value in top boxes and one in bottom boxes |
| CAO - Top boxes (including zero at source node) | A1 | |
| CAO - Bottom boxes (including zero at sink node) | A1 | |

## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct calculation seen, answer of 3 | B1 | Answer of 3 with no working scores B0 |

## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Not a cascade chart; 4 workers used at most; at least 10 different activities placed | M1 | |
| 4 workers; all 13 activities present (just once); condone at most two errors | A1 | If activity appears for two different workers simultaneously this is A0; activity can give rise to at most three errors: one on duration, one on time interval, one on IPA |
| 4 workers; all 13 activities present (just once); no errors | A1 | |

---
4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{162f9d72-84a4-4b1a-93cf-b7eeb7f957ae-05_1004_1797_205_134}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

The network in Figure 3 shows the activities that need to be undertaken to complete a project. Each activity is represented by an arc and the duration of the activity, in days, is shown in brackets. The early event times and late event times are to be shown at each vertex and one late event time has been completed for you.

The total float of activity H is 7 days.
\begin{enumerate}[label=(\alph*)]
\item Explain, with detailed reasoning, why $x = 11$
\item Determine the missing early event times and late event times, and hence complete Diagram 1 in your answer book.

Each activity requires one worker and the project must be completed in the shortest possible time using as few workers as possible.
\item Calculate a lower bound for the number of workers needed to complete the project in the shortest possible time.
\item Schedule the activities using Grid 1 in the answer book.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FD1 2019 Q4 [9]}}
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