Understand the idea.
The chain that cannot be delayed
A critical path is a longest-duration directed path from the project start to finish. Its duration sets the earliest project completion because every activity along that chain must occur in sequence. A shorter chain has some allowance before it controls the finish.
With the earliest finish used as the deadline, critical activities have zero total float. Trace connected zero-float arrows from start to finish, including any dummy links; a disconnected list of critical activity labels does not describe a path.
There can be more than one
Two or more complete paths can tie for the longest duration. Each is critical. Shortening work on just one tied path leaves another path controlling the old finish, so a proposed time saving must be checked against every complete route.
The display project has route durations , and days. Thus A–D–E–F is critical. These route durations measure sequential elapsed time, not the sum of all project work.
| 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 |
See the method in action.
Compare complete paths
A project has complete path durations 17, 14, 17 and 11 hours. State the minimum completion time and the number of critical paths.
Worked solution
- 1
Compare all complete path durations.
The project must wait for its longest required start-to-finish chain.
- 2
Count paths attaining that maximum.
A critical path has duration equal to the project's earliest finish; ties are retained.
- 3
State the project conclusion.
Both longest paths can control completion, so neither may be ignored when considering delays or time savings.
17 hours, with two critical paths.
Two parallel routes
A and B start together. A takes 2 days then C takes 3; B takes 4 days then D takes 1. A final task E takes 2 days after C and D. Find the critical paths.
Worked solution
- 1
Find the route through A and C.
A, C and E occur sequentially along this complete route.
- 2
Find the route through B and D.
B and D form the other branch that must finish before E starts.
- 3
Confirm the merge and final completion.
Both branches finish at day 5, so both A–C–E and B–D–E are critical.
A–C–E and B–D–E are both critical; completion is 7 days.
Try it yourself.
Check your understanding
One critical path takes 20 days and another takes 20. If an activity found only on the first is shortened by 2 days, must the project finish at 18?
Show answer & reasoning
No; the other critical path still requires 20 days.
No. The unchanged critical path still takes 20 days, so it prevents an 18-day finish. Shortening an activity exclusive to the other path does not shorten that remaining controlling route.
Keep practising
6 worksheet questions · 15 marks
Read the questions online
- Find the critical path and its duration for the displayed network.3 marks
- Complete paths have durations 9, 15, 12 and 15 weeks. Find the completion time and number of critical paths.2 marks
- A project starts with P (3 hours) and Q (5 hours) in parallel. R (4 hours) follows P and S (2 hours) follows Q. T (1 hour) follows both R and S. Find all critical paths.3 marks
- A path contains a zero-duration dummy. Can it still be critical?2 marks
- A network shows three connected zero-float activities beginning at the project start but not reaching the finish. Is that list enough to name a complete critical path?2 marks
- The critical path has duration 24 hours. A non-critical path takes 21. If an activity appearing only on the non-critical path is extended by 4 hours, what is the revised completion time?3 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- Find the critical path and its duration for the displayed network.3 marks
; ; .
The maximum is 13.
A–D–E–F, 13 days. - Complete paths have durations 9, 15, 12 and 15 weeks. Find the completion time and number of critical paths.2 marks
Take the largest complete-path duration.
Both 15-week routes are controlling.
15 weeks and two critical paths. - A project starts with P (3 hours) and Q (5 hours) in parallel. R (4 hours) follows P and S (2 hours) follows Q. T (1 hour) follows both R and S. Find all critical paths.3 marks
.
.
P–R–T and Q–S–T, both 8 hours. - A path contains a zero-duration dummy. Can it still be critical?2 marks
A critical route is determined by total duration and dependency structure.
A dummy may be needed to connect that controlling route.
Yes. - A network shows three connected zero-float activities beginning at the project start but not reaching the finish. Is that list enough to name a complete critical path?2 marks
A critical path must run continuously from the start event to the finish event.
Continue tracing the controlling arrows.
No. - The critical path has duration 24 hours. A non-critical path takes 21. If an activity appearing only on the non-critical path is extended by 4 hours, what is the revised completion time?3 marks
The changed path becomes .
The new maximum is , so the critical path changes.
25 hours.
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 ↗