Edexcel FD1 2023 June — Question 6 9 marks

Exam BoardEdexcel
ModuleFD1 (Further Decision 1)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCritical Path Analysis
TypeDraw cascade/Gantt chart
DifficultyStandard +0.3 This is a standard Critical Path Analysis question requiring routine application of well-defined algorithms (drawing activity networks, identifying critical paths, calculating floats). While it has multiple parts and requires careful bookkeeping, it involves no novel problem-solving or conceptual insight—just methodical application of Decision Maths techniques that are directly taught and practiced.
Spec7.05a Critical path analysis: activity on arc networks

6. The precedence table below shows the twelve activities required to complete a project.
ActivityImmediately preceding activities
A-
B-
C-
DA
EA, B
FD, E
GA, B, C
HF, G
ID, E
JD, E
KF, G, I, J
LI
  1. Draw the activity network described in the precedence table, using activity on arc. Your activity network must contain the minimum number of dummies only.
    (5) \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{6ccce35f-4e62-4b6b-acf6-f9b3e18d4b52-11_654_1358_153_356} \captionsetup{labelformat=empty} \caption{Figure 6}
    \end{figure} Figure 6 shows a partially completed cascade chart for the project. The non-critical activities F, J and K are not shown in Figure 6. The time taken to complete each activity is given in hours and the project is to be completed in the minimum possible time.
  2. State the critical activities. Given that the total float of activity F is 2 hours,
  3. state the duration of activity F . The duration of activity J is \(x\) hours, and the duration of activity K is \(y\) hours, where \(x > 0\) and \(y > 0\)
    1. State, in terms of \(y\), the maximum possible total float for activity K.
    2. State, in terms of \(x\) and \(y\), the total float for activity J .

Question 6:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Network drawn with at least nine labelled activities, one start, at least two dummies placedM1 AO2.1
Activities A, B, C, D, E and the dummy (+ arrow) at the end of A dealt with correctlyA1 AO1.1b
Activities G, F, I and J and the dummy (+ arrow) at the beginning of G dealt with correctlyA1 AO1.1b
Activities H and L dealt with correctlyA1 AO1.1b
CSO – Final two dummies + arrows and activity K dealt with correctly, all arrows present for every activity with one finish and no additional dummiesA1 AO1.1b
IPA Table:
AnswerMarks Guidance
ActivityA B
IPA- -
Note: Activity on node is M0. Condone lack of/incorrect numbered events. Additional unnecessary correct dummies penalised only on final A mark.
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Critical activities: B, E, I and LB1 AO2.2a — CAO (B, E, I and L only)
Part (c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Earliest start for F is 12 and latest finish is 18, therefore if total float is 2 the duration of F is 4 (hours)B1 AO2.2a — CAO (4)
Part (d)
AnswerMarks Guidance
Answer/WorkingMark Guidance
(i) Maximum possible total float for activity K is \(5 - y\)B1 AO3.1b — CAO \((5-y)\)
(ii) Total float for activity J is \(12 - x - y\)B1 AO2.2a — CAO \((12-x-y)\)
# Question 6:

## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Network drawn with at least nine labelled activities, one start, at least two dummies placed | M1 | AO2.1 |
| Activities A, B, C, D, E and the dummy (+ arrow) at the end of A dealt with correctly | A1 | AO1.1b |
| Activities G, F, I and J and the dummy (+ arrow) at the beginning of G dealt with correctly | A1 | AO1.1b |
| Activities H and L dealt with correctly | A1 | AO1.1b |
| CSO – Final two dummies + arrows and activity K dealt with correctly, all arrows present for every activity with one finish and no additional dummies | A1 | AO1.1b |

**IPA Table:**
| Activity | A | B | C | D | E | F | G | H | I | J | K | L |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| IPA | - | - | - | A | A,B | D,E | A,B,C | F,G | D,E | D,E | F,G,I,J | I |

**Note:** Activity on node is M0. Condone lack of/incorrect numbered events. Additional unnecessary correct dummies penalised only on final A mark.

## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Critical activities: B, E, I and L | B1 | AO2.2a — CAO (B, E, I and L only) |

## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Earliest start for F is 12 and latest finish is 18, therefore if total float is 2 the duration of F is 4 (hours) | B1 | AO2.2a — CAO (4) |

## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) Maximum possible total float for activity K is $5 - y$ | B1 | AO3.1b — CAO $(5-y)$ |
| (ii) Total float for activity J is $12 - x - y$ | B1 | AO2.2a — CAO $(12-x-y)$ |

---
6. The precedence table below shows the twelve activities required to complete a project.

\begin{center}
\begin{tabular}{|l|l|}
\hline
Activity & Immediately preceding activities \\
\hline
A & - \\
\hline
B & - \\
\hline
C & - \\
\hline
D & A \\
\hline
E & A, B \\
\hline
F & D, E \\
\hline
G & A, B, C \\
\hline
H & F, G \\
\hline
I & D, E \\
\hline
J & D, E \\
\hline
K & F, G, I, J \\
\hline
L & I \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Draw the activity network described in the precedence table, using activity on arc. Your activity network must contain the minimum number of dummies only.\\
(5)

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{6ccce35f-4e62-4b6b-acf6-f9b3e18d4b52-11_654_1358_153_356}
\captionsetup{labelformat=empty}
\caption{Figure 6}
\end{center}
\end{figure}

Figure 6 shows a partially completed cascade chart for the project. The non-critical activities F, J and K are not shown in Figure 6.

The time taken to complete each activity is given in hours and the project is to be completed in the minimum possible time.
\item State the critical activities.

Given that the total float of activity F is 2 hours,
\item state the duration of activity F .

The duration of activity J is $x$ hours, and the duration of activity K is $y$ hours, where $x > 0$ and $y > 0$
\item \begin{enumerate}[label=(\roman*)]
\item State, in terms of $y$, the maximum possible total float for activity K.
\item State, in terms of $x$ and $y$, the total float for activity J .
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{Edexcel FD1 2023 Q6 [9]}}