What even is this? Some tasks can't start until others finish. The critical path is the longest chain of tasks, it sets the minimum project time. Float is the spare time a non-critical task has before it starts to delay the project.
If you only do one thingThe critical path = the longest path through the project = the minimum time to finish. Tasks on it have zero float: any delay to them delays everything.
The 10-minute version
critical path = longest path = minimum project time · float = LST − EST
List the paths, add the durations, the longest one is critical. Its tasks have zero float. Then do Practice Q1.
📺 Watch it explained
Four short ways in. The 🎬 cards are waiting on Nat's videos, and each has a ready-to-read film script tucked underneath. 🖨️ Prefer printable notes? 4 styles here.
1 · The two big ideas
Critical paththe LONGEST path through the project = the minimum project time
Floatfloat = latest start − earliest start (LST − EST)
The longest path is the bottleneck: until it's done, the project isn't done. Tasks on it can't slip at all, they have zero float. Only tasks off the critical path have spare time.
2 · Finding it: forward and backward scans
1
Forward scan → EST (earliest start time). Work left to right, adding durations. Where paths join, take the largest arriving time (you must wait for the slowest one).
2
Backward scan → LST (latest start time). Work right to left from the finish, subtracting durations. Where paths split, take the smallest.
3
Float = LST − EST for each task. A float of 0 means the task is on the critical path.
Critical = longest, not shortest. And critical tasks have the least slack (zero float), not the most.
3 · Worked example
🛠️ A 4-task project
Task A (3 wks) must finish before B (4 wks) and C (2 wks). Both B and C must finish before D (5 wks). Find the critical path, the project time, and the float of each task.
Gold = critical (A, B, D). The two paths: A-B-D = 12, A-C-D = 10.
Step 1, longest path: A-B-D = 3 + 4 + 5 = 12; A-C-D = 3 + 2 + 5 = 10. The longer one wins, so the critical path is A → B → D and the project takes 12 weeks.
| Task | Dur | EST | LST | Float = LST − EST |
|---|---|---|---|---|
| A | 3 | 0 | 0 | 0 ← critical |
| B | 4 | 3 | 3 | 0 ← critical |
| C | 2 | 3 | 5 | 2 |
| D | 5 | 7 | 7 | 0 ← critical |
Critical path A → B → D, project time 12 weeks. Only C has float (2 weeks): it could start as late as week 5 instead of week 3 without delaying the finish.
4 · Practice questions
1. The paths through a project are P-Q-S = 9, P-R-S = 13, P-R-T = 11. What is the critical path and the minimum project time?Tap to reveal
The longest path is P-R-S = 13, so that's the critical path and the project takes 13 units. (Critical = longest.)
2. A task has EST = 4 and LST = 9. What is its float, and is it on the critical path?Tap to reveal
Float = LST − EST = 9 − 4 = 5. Since the float is not zero, it is not on the critical path, it has 5 units of slack.
3. In the worked example, by how long could task C be delayed before it pushes the whole project out?Tap to reveal
C has a float of 2 weeks. It can slip up to 2 weeks with no effect on the 12-week finish. Beyond that, the project is delayed.
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Practise in Sparky's Workshop
The networks escape room covers critical path, float, MST and more.