The idea, step by step
Minimum under a model
A forward scan finds the earliest completion consistent with the precedence rules and supplied durations. It assumes activities can start as soon as their predecessors finish. If the project needs one machine for two tasks at once, resource limits may make the real schedule longer.
For the display project, the task durations sum to 21 days of work, but parallel work allows completion in 13 elapsed days. Neither figure is wrong: they measure different things. The 13-day result is an ideal dependency-based minimum, not a promise that weather, staffing or material deliveries cannot cause delay.
| 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 |
Shortening the project
To reduce the finish time, shorten a controlling activity or change the process so a controlling dependency is removed legitimately. Then rescan the network. You cannot simply subtract the time saved on one task from the original completion without checking other routes.
In the display project, reducing D from 4 to 3 days reduces the controlling route from 13 to 12 days. At that point A–C–F also takes 12. Reducing D further does not take the project below 12 unless work on that second route is also shortened.
A common trap
Assuming the earliest network finish includes staffing and equipment limits that were never supplied.
Instead: State the parallel-resource assumption. Add constraints or revise durations if real resource conflicts prevent the modelled starts.
See the thinking
Work versus elapsed time
Tasks A (5 hours) and B (8 hours) start together. C (2 hours) follows both. Find the total work and minimum elapsed duration, assuming independent resources.
- Total work is 5+8+2=15 task-hours.
- The parallel tasks finish after max(5,8)=8 hours.
- C adds 2 hours to the elapsed schedule.
Check a proposed reduction
A project has path durations 18 and 16 days. A task only on the 18-day path can be shortened by 4 days. What is the new earliest completion?
- The changed route becomes 14 days.
- The other route remains 16 days.
- The project still waits for both, so take max(14,16).
Check your understanding
The longest route is 30 days and the next longest is 28. A task appearing only on the longest is shortened by 5 days. What finish time results?
Show answer
28 days.
The routes become 25 and 28; the formerly second-longest route now controls the project.
Your turn
6 questions · 16 marks · Work at your own pace.
Read the questions online
- Find the display project's total activity duration and its earliest elapsed completion.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 - Reduce D in the display project from 4 to 3 days. Find the new completion and all critical real-activity paths.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 - Instead reduce D from 4 to 1 day. Find the new project completion.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 - Two independent tasks each last 6 hours but need the same one available operator. A final 1-hour task follows both. Compare the dependency-only minimum with a schedule respecting the operator limit.3 marks
- A project has a 22-day critical path and a 21-day alternative. A one-day reduction is available on a task shared by both routes. What revised completion results?2 marks
- A contractor promises the exact network minimum with no allowance for uncertainty. Give two assumptions that should be checked.2 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- Find the display project's total activity duration and its earliest elapsed completion.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 Sum 4+3+5+4+2+3=21.
The controlling route A–D–E–F takes 13.
21 task-days; 13 elapsed days. - Reduce D in the display project from 4 to 3 days. Find the new completion and all critical real-activity paths.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 The changed D route is 4+3+2+3=12.
The C route remains 4+5+3=12.
12 days; A–D–E–F and A–C–F. - Instead reduce D from 4 to 1 day. Find the new project completion.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 The D route becomes 4+1+2+3=10.
The unchanged A–C–F route still needs 12.
12 days. - Two independent tasks each last 6 hours but need the same one available operator. A final 1-hour task follows both. Compare the dependency-only minimum with a schedule respecting the operator limit.3 marks
Parallel model: max(6,6)+1=7.
One operator must do the two tasks sequentially: 6+6+1=13.
Dependency-only minimum 7 hours; with one operator, 13 hours. - A project has a 22-day critical path and a 21-day alternative. A one-day reduction is available on a task shared by both routes. What revised completion results?2 marks
Both path durations fall by one.
They become 21 and 20; the maximum is 21.
21 days. - A contractor promises the exact network minimum with no allowance for uncertainty. Give two assumptions that should be checked.2 marks
The calculation uses fixed durations and immediate starts.
Supply delays, resource conflicts or rework can invalidate these assumptions.
For example: the activity durations are reliable, and enough independent resources exist for parallel tasks.
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 ↗