The idea, step by step
Forward scan: all predecessors must be ready
Start the first event at time 0. For every incoming arrow, add its duration to the earliest time at its tail. The head event must wait for all incoming activities, so use the largest candidate. Continue in a direction that ensures each needed tail time is already known.
In the display network, event 4 has candidates 4+4=8 from D and 3+0=3 from the dummy. Therefore its earliest time is 8. Event 5 has candidates 4+5=9 and 8+2=10, so its earliest time is 10. The finish event occurs at 10+3=13.
Backward scan: protect every successor
Set the final event's latest time equal to the required project finish. For the ordinary no-delay calculation, use the earliest possible project finish. Work backwards: subtract each outgoing duration from the latest time at that arrow's head. Choose the smallest candidate so none of the successors is held up.
For event 2, the candidates are 10−5=5 through C and 8−4=4 through D. The latest permitted event time is 4, not 5. Event 3 can occur as late as 8 because its only outgoing arrow is the zero-duration dummy. Check that every latest time is at least its corresponding earliest time.
| Event | Earliest time | Latest time |
|---|---|---|
| 1 | 0 | 0 |
| 2 | 4 | 4 |
| 3 | 3 | 8 |
| 4 | 8 | 8 |
| 5 | 10 | 10 |
| 6 | 13 | 13 |
A common trap
Using the smallest incoming value in a forward scan because it is the fastest arrival.
Instead: A merge waits for every predecessor. Use max forwards; use min backwards to protect every following activity.
See the thinking
A merge in the forward scan
Event 4 can be reached from event 2 at time 6 by an activity of duration 5, or from event 3 at time 8 by an activity of duration 2. Find its earliest time.
- The first candidate is 6+5=11.
- The second is 8+2=10.
- Both arrivals are required; choose the larger.
A split in the backward scan
An event has outgoing arrows of durations 4 and 7. Their head events have latest times 18 and 20. Find the event's latest time.
- Through the first arrow it must occur by 18−4=14.
- Through the second it must occur by 20−7=13.
- Choose 13 so both requirements are met.
Check your understanding
An event receives candidates 12, 15 and 15 in the forward scan. What earliest time is recorded?
Show answer
15.
The slower incoming branches control the event; equal largest values may signal more than one controlling branch.
Your turn
6 questions · 16 marks · Work at your own pace.
Read the questions online
- Perform a complete forward scan on the displayed network.4 marks
- Perform the backward scan for the displayed network using project finish 13.4 marks
- Find the earliest time at a merge when incoming tail times and durations are (2,7), (6,2) and (4,6).2 marks
- Find a latest event time when outgoing head times and durations are (15,4), (19,6) and (14,2).2 marks
- A forward scan gives an event earliest time 9. A claimed backward scan gives latest time 7 for a finish target equal to the earliest project finish. What does this indicate?2 marks
- Why does adding all incoming durations at a merge usually give the wrong earliest time?2 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- Perform a complete forward scan on the displayed network.4 marks
E1=0; E2=4; E3=3.
E4=max(4+4,3+0)=8.
E5=max(4+5,8+2)=10; E6=10+3=13.
Events 1–6: 0, 4, 3, 8, 10, 13. - Perform the backward scan for the displayed network using project finish 13.4 marks
L6=13; L5=10; L4=8; L3=8.
L2=min(10−5,8−4)=4.
L1=min(4−4,8−3)=0.
Events 1–6 latest: 0, 4, 8, 8, 10, 13. - Find the earliest time at a merge when incoming tail times and durations are (2,7), (6,2) and (4,6).2 marks
Candidates are 9, 8 and 10.
Their maximum is 10.
10. - Find a latest event time when outgoing head times and durations are (15,4), (19,6) and (14,2).2 marks
Candidates are 11, 13 and 12.
Their minimum is 11.
11. - A forward scan gives an event earliest time 9. A claimed backward scan gives latest time 7 for a finish target equal to the earliest project finish. What does this indicate?2 marks
The event cannot both require at least 9 and be required by 7.
Recheck durations, directions and the final-time assumption.
The scans or the model contain an error. - Why does adding all incoming durations at a merge usually give the wrong earliest time?2 marks
Each candidate already measures elapsed time from the common start.
The event waits for the latest arrival, not the sum of arrivals.
It assumes parallel branches occurred one after another.
Curriculum reference & lesson scope
SACE Board of South Australia
Stage 2 General Mathematics: discrete models, project scheduling and assignment. Activity-on-arrow conventions are used for project networks; technology may verify calculations but does not replace interpreting the model.
Official curriculum source ↗