Edexcel D1 2013 June — Question 7 17 marks

Exam BoardEdexcel
ModuleD1 (Decision Mathematics 1)
Year2013
SessionJune
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCritical Path Analysis
TypeEffect of activity delay/change
DifficultyStandard +0.3 This is a comprehensive critical path analysis question covering standard D1 techniques (early/late times, float, cascade charts, worker bounds). While multi-part with 17 marks, each component is routine application of well-defined algorithms with no novel problem-solving required. The cascade chart completion and comparing lower bounds are textbook exercises, making this slightly easier than average overall.
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 float
7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{5b32eb57-c9cd-46ec-a328-12050148bdf7-8_724_1730_241_167} \captionsetup{labelformat=empty} \caption{Figure 5}
\end{figure} \section*{[The sum of the duration of all activities is 172 days]} A project is modelled by the activity network shown in Figure 5. The activities are represented by the arcs. The number in brackets on each arc gives the time, in days, to complete the activity. Each activity requires one worker. The project is to be completed in the shortest possible time.
  1. Complete Diagram 1 in the answer book to show the early event times and late event times.
  2. Calculate the total float for activity M. You must make the numbers you use in your calculation clear.
  3. For each of the situations below, explain the effect that the delay would have on the project completion date.
    1. A 2 day delay on the early start of activity P.
    2. A 2 day delay on the early start of activity Q .
  4. Calculate a lower bound for the number of workers needed to complete the project in the shortest possible time. Diagram 2 in the answer book shows a partly completed cascade chart for this project.
  5. Complete the cascade chart.
  6. Use your cascade chart to determine a second lower bound on the number of workers needed to complete the project in the shortest possible time. You must make specific reference to times and activities.
  7. State which of the two lower bounds found in (d) and (f) is better. Give a reason for your answer.
    (Total 17 marks)
AnswerMarks Guidance
M1 A1
M1 A1 (4)
Float on M = 42 – 26 – 8 = 8M1 A1 (2)
2 day delay on P – no effect on the project completion date (float on P is 4)B1 (2)
2 day delay on Q – project finishes 2 days late (Q is a critical activity)B1
(172/53 = 3.245, so) a minimum of 4 workers neededB1 (1)
Notes for Question 7:
- a1M1: All top boxes complete, values generally increasing left to right, condone one 'rogue' (if values do not increase from left to right then if one value is ignored and then the values do increase from left to right then this is considered to be only one rogue value)
- a1A1: CAO.
- a2M1: All bottom boxes complete, values generally decreasing right to left, condone one 'rogue'.
- a2A1: CAO
- b1M1: Correct calculation seen – all three numbers correct (ft), float = 0.
- b1A1: Float correct (no ft on this mark)
- c1B1: CAO
- c2B1: CAO
- d1B1: 4 with (or without) working scores this mark
- e1M1: At least six activities added including six floats. Scheduling diagram scores M0.
- e1A1: Six activities including their floats dealt with correctly.
- e2M1: All remaining eleven activities including all eleven floats.
- e2A1: CAO.
Examples for part (f):
Example 1: Activities H, I, J, K and L together with 22 < time < 26 so 5 workers needed.
Example 2: At 10 < time < 14, F, D, E and H must be happening. Activity G must be happening 7 < time < 18 but its duration is 5 so it must also occur at some point in the interval 10 < time < 14 so 5 workers needed.
- f1M1: Example 1: A statement with the correct number of workers (5) and the correct activities (H, I, J, K and L) with some mention of time, or Example 2: A statement with the correct number of workers (5), the correct activities (F,D,E and H) with some mention of time and an indication that G must be happening with the other four activities at some point - give bod but e.g. 'at time 11 F, D, E and H must be happening' is M0). Scheduling the activities only scores M0.
- f1A1: A correct, complete full statement with details of both time and activities. Candidates only need to give a time within the intervals stated.
Please note strict inequalities for the time. Allow e.g. on 'day 23' as equivalent to 22 < time <23.
- g1M1: Must have attempted both parts (d) and (f). Their higher lower bound chosen + attempt at a reason.
- g1A1: CAO plus a correct reason given. Acceptable reasons e.g. the cascade gives a larger value or the bound for the cascade shows that the project cannot be done with fewer workers, etc.
Question 8: Part (e)
AnswerMarks
M1 A1
(any 6 more)
M1 A1
(all 11)(4)
E.g. Activities H, I, J, K and L together with 22 < time < 26 stated. So 5 workers neededM1 A1 (2)
The cascade gives a higher lower bound, so (f) is better.M1 A1 (2)
17 marks 
| | M1 A1 | |
| | | M1 A1 (4) |
| Float on M = 42 – 26 – 8 = 8 | M1 A1 (2) | |
| 2 day delay on P – no effect on the project completion date (float on P is 4) | B1 | (2) |
| 2 day delay on Q – project finishes 2 days late (Q is a critical activity) | B1 | |
| (172/53 = 3.245, so) a minimum of 4 workers needed | B1 (1) | |
| | | |

**Notes for Question 7:**

- a1M1: All top boxes complete, values generally increasing left to right, condone one 'rogue' (if values do not increase from left to right then if one value is ignored and then the values do increase from left to right then this is considered to be only one rogue value)
- a1A1: CAO.
- a2M1: All bottom boxes complete, values generally decreasing right to left, condone one 'rogue'.
- a2A1: CAO
- b1M1: Correct calculation seen – all three numbers correct (ft), float = 0.
- b1A1: Float correct (no ft on this mark)
- c1B1: CAO
- c2B1: CAO
- d1B1: 4 with (or without) working scores this mark
- e1M1: At least six activities added including six floats. Scheduling diagram scores M0.
- e1A1: Six activities including their floats dealt with correctly.
- e2M1: All remaining eleven activities including all eleven floats.
- e2A1: CAO.

**Examples for part (f):**

Example 1: Activities H, I, J, K and L together with 22 < time < 26 so 5 workers needed.

Example 2: At 10 < time < 14, F, D, E and H must be happening. Activity G must be happening 7 < time < 18 but its duration is 5 so it must also occur at some point in the interval 10 < time < 14 so 5 workers needed.

- f1M1: Example 1: A statement with the correct number of workers (5) **and the correct activities (H, I, J, K and L) with some mention of time**, or Example 2: A statement with the correct number of workers (5), the correct activities (F,D,E and H) with some mention of time **and an indication that G must be happening with the other four activities at some point - give bod but e.g. 'at time 11 F, D, E and H must be happening' is M0). Scheduling the activities only scores M0**.
- f1A1: A correct, complete full statement with details of both time **and activities**. Candidates only need to give a time within the intervals stated.

**Please note strict inequalities for the time. Allow e.g. on 'day 23' as equivalent to 22 < time <23.**

- g1M1: **Must have attempted both parts (d) and (f). Their higher lower bound chosen + attempt at a reason.**
- g1A1: CAO plus a correct reason given. Acceptable reasons e.g. the cascade gives a larger value **or** the bound for the cascade shows that the project cannot be done with fewer workers, etc.

---

## Question 8: Part (e)

| | M1 A1 | |
| | (any 6 more) | |
| | M1 A1 | |
| | (all 11) | (4) |
| E.g. Activities H, I, J, K and L together with 22 < time < 26 stated. So 5 workers needed | M1 A1 (2) | |
| The cascade gives a higher lower bound, so (f) is better. | M1 A1 (2) | |
| | 17 marks |
7.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{5b32eb57-c9cd-46ec-a328-12050148bdf7-8_724_1730_241_167}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}

\section*{[The sum of the duration of all activities is 172 days]}
A project is modelled by the activity network shown in Figure 5. The activities are represented by the arcs. The number in brackets on each arc gives the time, in days, to complete the activity. Each activity requires one worker. The project is to be completed in the shortest possible time.
\begin{enumerate}[label=(\alph*)]
\item Complete Diagram 1 in the answer book to show the early event times and late event times.
\item Calculate the total float for activity M. You must make the numbers you use in your calculation clear.
\item For each of the situations below, explain the effect that the delay would have on the project completion date.
\begin{enumerate}[label=(\roman*)]
\item A 2 day delay on the early start of activity P.
\item A 2 day delay on the early start of activity Q .
\end{enumerate}\item Calculate a lower bound for the number of workers needed to complete the project in the shortest possible time.

Diagram 2 in the answer book shows a partly completed cascade chart for this project.
\item Complete the cascade chart.
\item Use your cascade chart to determine a second lower bound on the number of workers needed to complete the project in the shortest possible time. You must make specific reference to times and activities.
\item State which of the two lower bounds found in (d) and (f) is better. Give a reason for your answer.\\
(Total 17 marks)
\end{enumerate}

\hfill \mbox{\textit{Edexcel D1 2013 Q7 [17]}}
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