Edexcel D1 2012 June — Question 6 14 marks

Exam BoardEdexcel
ModuleD1 (Decision Mathematics 1)
Year2012
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCritical Path Analysis
TypeCalculate lower bound for workers
DifficultyModerate -0.5 This is a standard Critical Path Analysis question from D1 covering routine procedures: precedence tables, early/late times, float calculations, and lower bound for workers (sum of durations ÷ critical path length). All parts follow textbook algorithms with no novel problem-solving required, making it slightly easier than average.
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

6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{4ad45e8f-f50a-4125-866b-a6951f85600f-7_624_1461_194_301} \captionsetup{labelformat=empty} \caption{Figure 5}
\end{figure} Figure 5 is the activity network relating to a development project. 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 the precedence table in the answer book.
    (2)
  2. Complete Diagram 1 in the answer book to show the early event times and late event times.
    (4)
  3. Calculate the total float for activity E. You must make the numbers you use in your calculation clear.
    (2)
  4. Calculate a lower bound for the number of workers needed to complete the project in the minimum time. You must show your working.
    (2)
  5. Schedule the activities using the minimum number of workers so that the project is completed in the minimum time.

Question 6:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Precedence table: A–; B–; C–; D–A; E–A; F–BE; G–BE; H–C; I–DF; J–CDFG; K–HB2, 1, 0 (2)
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Early times correct for first set of nodes1M1, 1A1
All early and late times correct2M1, 2A1 (4)
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Total float on \(E = 21 - 5 - 3 = 13\)M1, A1 (2)
Part (d):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{62}{28} = 2.21\), so lower bound is 3 workersM1, A1 (2)
Part (e):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Correct scheduling diagram on 3 workers up to time 30, e.g. Worker 1: A, E, D, I; Worker 2: B, F, G, J; Worker 3: C, H, K1M1, 1A1, 2A1, 3A1 (4)
Question 6:
Part a:
AnswerMarks Guidance
AnswerMarks Guidance
Any 3 rows completed correctlya1B1
All five rows completed correctlya1B2
Part b:
AnswerMarks Guidance
AnswerMarks Guidance
All top boxes complete, values generally increasing left to rightb1M1 Condone one rogue
CAOb1A1
All bottom boxes complete, values generally decreasing R to Lb2M1 Condone one rogue; condone missing 0 or 28 for M only
CAOb2A1
Part c:
AnswerMarks Guidance
AnswerMarks Guidance
Correct calculation, all three numbers correct (ft). Float \(\geq 0\)c1M1
CAOc1A1
Part d:
AnswerMarks Guidance
AnswerMarks Guidance
Attempt to find lower bound: \([52\text{-}72 \div \text{their finish time}]\)d1M1 Accept awrt 2.2
CAO – correct calculation seen or awrt 2.2d1A1 Beware \(28/11\) gives 3 also, so 3 with no working gets M0A0
Part e:
AnswerMarks Guidance
AnswerMarks Guidance
Not a cascade chart; 4 workers used at most; at least 7 activitiese1M1 If in doubt send to review
CHKAB correct: C-14; H-10; K-4; A-5; B-9. A and B completed by late finish times (A by time = 18, B by time = 21)e1A1
3 workers; all 11 activities present (just once); condone one error either precedence or activity length on activities D, E, F, G, I, Je2A1
3 workers; all 11 activities present (just once); no errors on activities D, E, F, G, I, Je3A1
Precedence checks:
- F must not start until after B and E are complete
- G must not start until after B and E are complete
- J must not start until after C, D, F are complete
- I must not start until after D and F are complete
Length checks:
- Length 5: A, I
- Length 4: D
- Length 3: E, G, J
- Length 2: F
- Length 9: B
## Question 6:

### Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Precedence table: A–; B–; C–; D–A; E–A; F–BE; G–BE; H–C; I–DF; J–CDFG; K–H | B2, 1, 0 **(2)** | |

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Early times correct for first set of nodes | 1M1, 1A1 | |
| All early and late times correct | 2M1, 2A1 **(4)** | |

### Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Total float on $E = 21 - 5 - 3 = 13$ | M1, A1 **(2)** | |

### Part (d):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{62}{28} = 2.21$, so lower bound is 3 workers | M1, A1 **(2)** | |

### Part (e):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct scheduling diagram on 3 workers up to time 30, e.g. Worker 1: A, E, D, I; Worker 2: B, F, G, J; Worker 3: C, H, K | 1M1, 1A1, 2A1, 3A1 **(4)** | |

# Question 6:

## Part a:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Any 3 rows completed correctly | a1B1 | |
| All five rows completed correctly | a1B2 | |

## Part b:
| Answer | Marks | Guidance |
|--------|-------|----------|
| All top boxes complete, values generally increasing left to right | b1M1 | Condone one rogue |
| CAO | b1A1 | |
| All bottom boxes complete, values generally decreasing R to L | b2M1 | Condone one rogue; condone missing 0 or 28 for M only |
| CAO | b2A1 | |

## Part c:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct calculation, all three numbers correct (ft). Float $\geq 0$ | c1M1 | |
| CAO | c1A1 | |

## Part d:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to find lower bound: $[52\text{-}72 \div \text{their finish time}]$ | d1M1 | Accept awrt 2.2 |
| CAO – correct calculation seen or awrt 2.2 | d1A1 | Beware $28/11$ gives 3 also, so 3 with no working gets M0A0 |

## Part e:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Not a cascade chart; 4 workers used at most; at least 7 activities | e1M1 | If in doubt send to review |
| CHKAB correct: C-14; H-10; K-4; A-5; B-9. A and B completed by late finish times (A by time = 18, B by time = 21) | e1A1 | |
| 3 workers; all 11 activities present (just once); condone one error either precedence or activity length on activities D, E, F, G, I, J | e2A1 | |
| 3 workers; all 11 activities present (just once); no errors on activities D, E, F, G, I, J | e3A1 | |

**Precedence checks:**
- F must not start until after B and E are complete
- G must not start until after B and E are complete
- J must not start until after C, D, F are complete
- I must not start until after D and F are complete

**Length checks:**
- Length 5: A, I
- Length 4: D
- Length 3: E, G, J
- Length 2: F
- Length 9: B

---
6.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{4ad45e8f-f50a-4125-866b-a6951f85600f-7_624_1461_194_301}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}

Figure 5 is the activity network relating to a development project. 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 the precedence table in the answer book.\\
(2)
\item Complete Diagram 1 in the answer book to show the early event times and late event times.\\
(4)
\item Calculate the total float for activity E. You must make the numbers you use in your calculation clear.\\
(2)
\item Calculate a lower bound for the number of workers needed to complete the project in the minimum time. You must show your working.\\
(2)
\item Schedule the activities using the minimum number of workers so that the project is completed in the minimum time.
\end{enumerate}

\hfill \mbox{\textit{Edexcel D1 2012 Q6 [14]}}