The idea, step by step
Four times for each activity
For an activity of duration d from event i to event j, its earliest start is the earliest time at i. Its earliest finish is ES+d. Its latest finish is the latest time at j, and its latest start is LF−d. An activity may have a range of allowed start times even though its duration is fixed.
The earliest head-event time is not automatically the activity's earliest finish: the head event may also wait for another, slower incoming activity. Keep the activity's own finish calculation separate from the merge time.
| Event | Earliest time | Latest time |
|---|---|---|
| 1 | 0 | 0 |
| 2 | 4 | 4 |
| 3 | 3 | 8 |
| 4 | 8 | 8 |
| 5 | 10 | 10 |
| 6 | 13 | 13 |
Read a time window
Activity B in the display network has ES=0, EF=3, LF=8 and LS=5. It can therefore start anywhere from day 0 to day 5 without delaying the day-13 completion, provided the rest of the schedule follows the model. Activity C has ES=4 and EF=9, but the head event occurs earliest at 10 because E finishes later.
Use a precise interpretation: 'B may start at day 5 and finish at day 8' is clearer than 'B has five days spare somewhere'. These calculations assume durations are reliable and resources allow the listed parallel work.
A common trap
Taking an activity's earliest finish directly from the earliest time at its head event.
Instead: Calculate EF as the activity's own ES plus duration. The head event can occur later because of another incoming activity.
See the thinking
Use event values
An activity takes 6 hours. Its tail has earliest time 3 and its head has latest time 14. Find ES, EF, LS and LF.
- ES is 3.
- EF=3+6=9.
- LF is 14, so LS=14−6=8.
Do not copy the head's earliest time
Activities P and Q finish at the same event. P starts at 2 and lasts 4 hours; Q starts at 5 and lasts 3 hours. Find P's earliest finish and the event's earliest time.
- P's earliest finish is 2+4=6.
- Q's earliest finish is 5+3=8.
- The shared event waits until 8.
Check your understanding
An activity has ES=7, duration=4 and LF=13. What is its latest start?
Show answer
9.
It needs four time units before the latest finish, so LS=13−4=9.
Your turn
6 questions · 15 marks · Work at your own pace.
Read the questions online
- Use the display project to find ES, EF, LS and LF for B.3 marks
Forward and backward scans for the display network, with completion at day 13. Event Earliest time Latest time 1 0 0 2 4 4 3 3 8 4 8 8 5 10 10 6 13 13 - Find all four activity times for C in the display project.3 marks
Forward and backward scans for the display network, with completion at day 13. Event Earliest time Latest time 1 0 0 2 4 4 3 3 8 4 8 8 5 10 10 6 13 13 - Find all four times for E in the display project.2 marks
Forward and backward scans for the display network, with completion at day 13. Event Earliest time Latest time 1 0 0 2 4 4 3 3 8 4 8 8 5 10 10 6 13 13 - A task lasts 8 minutes. Its ES is 12 and LS is 15. Find EF and LF.2 marks
- An activity runs 2→5 for 3 days. E2=6, E5=12 and L5=14. Find its earliest finish and explain why it differs from E5.3 marks
- A claimed schedule gives ES=5, EF=11, LS=9 and LF=14 for one fixed-duration activity. Identify the inconsistency.2 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- Use the display project to find ES, EF, LS and LF for B.3 marks
Forward and backward scans for the display network, with completion at day 13. Event Earliest time Latest time 1 0 0 2 4 4 3 3 8 4 8 8 5 10 10 6 13 13 B is 1→3 of duration 3.
ES=E1=0; EF=3; LF=L3=8; LS=8−3=5.
0, 3, 5 and 8 days respectively. - Find all four activity times for C in the display project.3 marks
Forward and backward scans for the display network, with completion at day 13. Event Earliest time Latest time 1 0 0 2 4 4 3 3 8 4 8 8 5 10 10 6 13 13 C is 2→5 of duration 5.
ES=4; EF=4+5=9; LF=10; LS=10−5=5.
ES=4, EF=9, LS=5, LF=10 days. - Find all four times for E in the display project.2 marks
Forward and backward scans for the display network, with completion at day 13. Event Earliest time Latest time 1 0 0 2 4 4 3 3 8 4 8 8 5 10 10 6 13 13 E is 4→5 of duration 2.
Use E4=8 and L5=10.
ES=8, EF=10, LS=8, LF=10 days. - A task lasts 8 minutes. Its ES is 12 and LS is 15. Find EF and LF.2 marks
Add the same fixed duration to each start.
12+8=20; 15+8=23.
EF=20 and LF=23 minutes. - An activity runs 2→5 for 3 days. E2=6, E5=12 and L5=14. Find its earliest finish and explain why it differs from E5.3 marks
The task itself finishes at 6+3=9.
An event combines all incoming requirements.
EF=9 days; event 5 waits for another incoming activity until day 12. - A claimed schedule gives ES=5, EF=11, LS=9 and LF=14 for one fixed-duration activity. Identify the inconsistency.2 marks
11−5=6.
14−9=5; the same activity cannot have both durations in this model.
The duration would be 6 from ES/EF but 5 from LS/LF.
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 ↗