Edexcel D1 2018 June — Question 5 14 marks

Exam BoardEdexcel
ModuleD1 (Decision Mathematics 1)
Year2018
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCritical Path Analysis
TypeComplete precedence table from network
DifficultyStandard +0.3 This is a standard D1 critical path analysis question covering routine procedures: reading a network diagram, completing a precedence table, finding early/late times, identifying critical path, calculating float, and basic scheduling. All techniques are textbook exercises requiring careful application of algorithms rather than problem-solving insight. Slightly easier than average due to being methodical rather than conceptually challenging.
Spec7.05a 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 float7.05e Cascade charts: scheduling and effect of delays
5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{5b18e92c-540e-4e89-8d60-d60294f50dda-06_630_1237_189_412} \captionsetup{labelformat=empty} \caption{Figure 6}
\end{figure} A project is modelled by the activity network shown in Figure 6. 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.
  1. Complete the precedence table in the answer book.
  2. Complete Diagram 1 in the answer book to show the early event times and late event times.
  3. State the minimum project completion time and list the critical activities.
  4. Calculate the maximum number of hours by which activity E could be delayed without affecting the shortest possible completion time of the project. You must make the numbers used in your calculation clear.
  5. Calculate a lower bound for the number of workers needed to complete the project in the minimum time. You must show your working. The project is to be completed in the minimum time using as few workers as possible.
  6. Schedule the activities using Grid 1 in the answer book.
    (3) Before the project begins it becomes apparent that activity E will require an additional 6 hours to complete. The project is still to be completed in the shortest possible time and the time to complete all other activities is unchanged.
  7. State the new minimum project completion time and list the new critical activities.
Question 5:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Predecessor table completed correctlyB2, 1, 0 Any 7 of 10 rows correct (allow A and B blank) for B1; CAO (allow A and B blank) for B2
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
All top and bottom boxes completed with values generally increasing left to right (top) and decreasing right to left (bottom)M1 Condone missing 0 or 21; condone one rogue value in top and one in bottom boxes
CAO top boxesA1
CAO bottom boxesA1
Part (c)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Minimum project completion time is 21 hoursB1ft Follow through candidate's value from (b)
Critical activities: B, G, JB1 CAO
Part (d)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
E could be delayed by \(16 - 5 - 6 = 5\) hoursB1 Correct calculation with all three numbers present; answer of 5 with no working scores B0
Part (e)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Lower bound \(= \frac{52}{21} = 2.476\ldots\) so 3 workers requiredB1 Correct calculation seen then 3; answer of 3 with no working scores B0
Part (f)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Scheduling diagram started with 4 workers, at least 9 unique activities placedM1 Not a cascade chart; 4 workers used at most
4 workers, all 10 activities present just once; at most two errorsA1 Duration, time interval and IPA all checked
4 workers, all 10 activities present just once; no errorsA1
Activities and time intervals: A(0–7), B(0–7), C(7–12), D(11–15), E(6–16), F(7–16), G(7–15), H(13–21), I(15–21), J(15–21)
Part (g)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Activities A, E and H are now criticalB1 Correctly stating activities now critical (A, E and H); no extras
Minimum project completion time is now 22 hoursB1 Correctly stating new project completion time (22); no units required 
# Question 5:

## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Predecessor table completed correctly | B2, 1, 0 | Any 7 of 10 rows correct (allow A and B blank) for B1; CAO (allow A and B blank) for B2 |

## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| All top and bottom boxes completed with values generally increasing left to right (top) and decreasing right to left (bottom) | M1 | Condone missing 0 or 21; condone one rogue value in top and one in bottom boxes |
| CAO top boxes | A1 | |
| CAO bottom boxes | A1 | |

## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Minimum project completion time is 21 hours | B1ft | Follow through candidate's value from (b) |
| Critical activities: B, G, J | B1 | CAO |

## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| E could be delayed by $16 - 5 - 6 = 5$ hours | B1 | Correct calculation with all three numbers present; answer of 5 with no working scores B0 |

## Part (e)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Lower bound $= \frac{52}{21} = 2.476\ldots$ so 3 workers required | B1 | Correct calculation seen then 3; answer of 3 with no working scores B0 |

## Part (f)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Scheduling diagram started with 4 workers, at least 9 unique activities placed | M1 | Not a cascade chart; 4 workers used at most |
| 4 workers, all 10 activities present just once; at most two errors | A1 | Duration, time interval and IPA all checked |
| 4 workers, all 10 activities present just once; no errors | A1 | |

Activities and time intervals: A(0–7), B(0–7), C(7–12), D(11–15), E(6–16), F(7–16), G(7–15), H(13–21), I(15–21), J(15–21)

## Part (g)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Activities A, E and H are now critical | B1 | Correctly stating activities now critical (A, E and H); no extras |
| Minimum project completion time is now 22 hours | B1 | Correctly stating new project completion time (22); no units required |

---
5.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{5b18e92c-540e-4e89-8d60-d60294f50dda-06_630_1237_189_412}
\captionsetup{labelformat=empty}
\caption{Figure 6}
\end{center}
\end{figure}

A project is modelled by the activity network shown in Figure 6. 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.
\begin{enumerate}[label=(\alph*)]
\item Complete the precedence table in the answer book.
\item Complete Diagram 1 in the answer book to show the early event times and late event times.
\item State the minimum project completion time and list the critical activities.
\item Calculate the maximum number of hours by which activity E could be delayed without affecting the shortest possible completion time of the project. You must make the numbers used in your calculation clear.
\item Calculate a lower bound for the number of workers needed to complete the project in the minimum time. You must show your working.

The project is to be completed in the minimum time using as few workers as possible.
\item Schedule the activities using Grid 1 in the answer book.\\
(3)

Before the project begins it becomes apparent that activity E will require an additional 6 hours to complete. The project is still to be completed in the shortest possible time and the time to complete all other activities is unchanged.
\item State the new minimum project completion time and list the new critical activities.
\end{enumerate}

\hfill \mbox{\textit{Edexcel D1 2018 Q5 [14]}}
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