The idea, step by step
An activity's allowable delay
Total float is the difference between latest and earliest start: LS−ES. Equivalently, for an arrow from event i to event j with duration d, it is Lj−Ei−d. This is how far that activity may be delayed, with other relevant work scheduled consistently, before the chosen project finish is delayed.
In the display project, B has float 8−0−3=5 days and C has float 10−4−5=1 day. A, D, E and F have zero float. Float is not extra activity duration and does not mean a worker is occupied throughout the whole available window.
Change the network, then recompute
If B's duration rises from 3 to 7 days in the display project, E still waits until day 8 for D, so the overall finish remains 13. If B rises to 9 days, E starts at 9 and F finishes at 14. The old float correctly predicts when this branch begins to control.
After a change, recompute times rather than continuing to quote the old floats. Delays on connected non-critical activities may use the same path allowance; adding their separate original floats can overstate the delay that the project can tolerate.
| Activity | Duration | Immediate predecessors |
|---|---|---|
| A | 4 | None |
| B | 3 | None |
| C | 5 | A |
| D | 4 | A |
| E | 2 | B and D |
| F | 3 | C and E |
A common trap
Adding the floats of successive tasks and treating the sum as extra spare time on that path.
Instead: Their total floats can represent the same scheduling allowance. Recalculate the full path or network for combined changes.
See the thinking
Compute float from a pair of events
A task lasts 5 hours, with tail earliest time 4 and head latest time 12. Find its total float.
- Use Lj−Ei−d.
- 12−4−5=3.
- This is the permitted delay before the selected project finish is affected.
Two tasks share a branch allowance
A non-critical branch has two successive tasks P and Q. The branch takes 8 days while the critical route takes 11. Each task is reported to have 3 days total float. Can both be extended by 3 days without delaying the project?
- The branch has only 11−8=3 days of total route allowance.
- Extending both adds 6 days.
- The branch becomes 14 days and exceeds the old finish by 3.
Check your understanding
A task has ES=3 and LS=8. It starts 4 days later than its earliest start. With no other changes, has it exceeded its total float?
Show answer
No.
Total float is 8−3=5 days. A four-day delay stays inside that allowance.
Your turn
6 questions · 17 marks · Work at your own pace.
Read the questions online
- Calculate the total floats of B and C in the display project.3 marks
- Increase B's duration in the display project from 3 to 7 days. Find the project finish.3 marks
A school display project. Times are in days. Activity Duration Immediate predecessors A 4 None B 3 None C 5 A D 4 A E 2 B and D F 3 C and E - Increase B's duration instead to 9 days. Find the finish and critical real-activity sequence.3 marks
A school display project. Times are in days. Activity Duration Immediate predecessors A 4 None B 3 None C 5 A D 4 A E 2 B and D F 3 C and E - Activity P has ES=6, EF=10, LS=9 and LF=13. Find its duration and float.2 marks
- A non-critical route takes 12 hours and the project finishes in 16. Two consecutive tasks on that route are both extended by 3 hours. What happens?3 marks
- The project deadline is fixed at day 20, but its earliest completion is day 18. Explain why a backward scan using 20 answers a different question from one using 18.3 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- Calculate the total floats of B and C in the display project.3 marks
B: 8−0−3=5.
C: 10−4−5=1.
B: 5 days; C: 1 day. - Increase B's duration in the display project from 3 to 7 days. Find the project finish.3 marks
A school display project. Times are in days. Activity Duration Immediate predecessors A 4 None B 3 None C 5 A D 4 A E 2 B and D F 3 C and E E waits for max(B finish 7, D finish 8)=8.
E finishes 10 and F finishes 13.
13 days. - Increase B's duration instead to 9 days. Find the finish and critical real-activity sequence.3 marks
A school display project. Times are in days. Activity Duration Immediate predecessors A 4 None B 3 None C 5 A D 4 A E 2 B and D F 3 C and E E starts at max(9,8)=9.
It finishes 11; F finishes 14.
The B branch now controls.
14 days; B–E–F. - Activity P has ES=6, EF=10, LS=9 and LF=13. Find its duration and float.2 marks
Duration=10−6=4 (also 13−9=4).
Float=9−6=3.
Duration 4; float 3 time units. - A non-critical route takes 12 hours and the project finishes in 16. Two consecutive tasks on that route are both extended by 3 hours. What happens?3 marks
Its shared allowance was 16−12=4.
The combined extension is 6, so it exceeds the allowance by 2.
That route becomes 18 hours, delaying the project by 2 hours. - The project deadline is fixed at day 20, but its earliest completion is day 18. Explain why a backward scan using 20 answers a different question from one using 18.3 marks
A looser finish target creates two days of overall schedule allowance.
State the finish target before interpreting zero or positive float.
The day-20 scan finds latest times that meet the deadline; the day-18 scan finds float that preserves the earliest completion.
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 ↗