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Year 12 Maths Standard 2 (2027) Critical path analysis

Float Times & the Critical Path

20 practice questions 0 video lessons Theory + worked examples

Master float times and the critical path for NSW Year 12 Mathematics Standard 2. Working from the EST and LST scanned into each event of an activity network, you learn to calculate the float of every activity — the spare time it has before it delays the project — using \(\text{Float}=\text{LST}(\text{head})-\text{EST}(\text{tail})-\text{duration}\).

You will identify critical activities (zero float) and non-critical activities (float greater than zero), trace the critical path from start to finish, find the minimum completion time, and predict the effect of shortening a critical task — a core Standard 2 skill for planning real projects such as building works, events and software releases.

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Theory

Float time and the critical path are core ideas in critical path analysis. This Year 12 Standard 2 (NSW) guide shows how to use the EST and LST at each event to work out the float of every activity, spot the critical (zero-float) activities, trace the critical path, and read off the minimum completion time for a project.

In critical path analysis a project is drawn as an activity-on-edge network: each arrow is an activity labelled with its letter and duration, and each numbered circle is an event that shows its earliest start time (EST) in the lower-left and its latest start time (LST) in the lower-right.

The float time (or slack) of an activity is the amount of time it can be delayed without delaying the whole project. It equals the LST at the activity’s head event minus the EST at its tail event minus its duration.

An activity with zero float is critical — it has no spare time. Joined from the start event to the end event, the zero-float activities form the critical path, and its length is the minimum completion time for the whole project. Activities with a float greater than zero are non-critical.

Critical path in an activity networkEvents show EST and LST; the zero-float activities A, C, E form the red critical path. A 4 B 2 C 3 D 6 E 5 1 0 0 2 4 4 3 7 7 4 12 12
Zero-float activities \(A\), \(C\), \(E\) form the critical path (red).
Float in a project networkActivity D has float 4; the critical path A-C-E is drawn red. A 5 B 3 C 6 D 4 E 4 F 5 1 0 0 2 5 5 3 3 7 4 11 11 5 15 15
Activity \(D\) has a float of \(4\) days; the red path has none.

The float (slack) of an activity is found from the times scanned into its two events:

\[\text{Float} = \text{LST}(\text{head}) - \text{EST}(\text{tail}) - \text{duration}\]
Float=LSTESTduration

An activity is critical exactly when its float is zero:

\[\text{critical} \iff \text{Float} = 0\]
Float=0
Minimum completion time. The critical path is the longest path through the network, and its length is the least time in which the whole project can be finished: \[\text{minimum completion time} = \text{length of the critical path}.\]

How to find float times and the critical path

  1. Read the EST and LST at every event from the split circles (forward scan gives EST in the lower-left, backward scan gives LST in the lower-right).
  2. Float each activity: \(\text{Float}=\text{LST}(\text{head})-\text{EST}(\text{tail})-\text{duration}\).
  3. Mark the critical activities — those with zero float — and join them from the start event to the end event to trace the critical path.
  4. State the result: non-critical activities are those with float \(>0\); the length of the critical path is the minimum completion time.
Example 1 — Float of one activity
The network shows a community garden build (durations in days). Find the float time of activity \(D\).
Solution

Activity \(D\) goes from event \(3\) to event \(4\): read \(\text{EST}(3)=3\) and \(\text{LST}(4)=11\).

Community garden networkActivity D from event 3 to 4; critical path A-C-E in red. A 5 B 3 C 6 D 4 E 4 F 5 1 0 0 2 5 5 3 3 7 4 11 11 5 15 15
\(\text{Float}(D)\)\(=\)\(\text{LST}(\text{head})-\text{EST}(\text{tail})-d\)
\(\)\(=\)\(11-3-4\)
\(\)\(=\)\(4\text{ days}\)
1134=4

Activity \(D\) has \(4\) days of spare time.

Example 2 — Find the critical path
The network shows a cafe fit-out (durations in days). Which activities form the critical path, and how long will it take?
Solution

Float every activity; the zero-float ones are critical.

Cafe fit-out networkCritical path A-D-F in red; length 18 days. A 5 B 4 C 3 D 8 E 6 F 5 1 0 0 2 5 5 3 8 12 4 13 13 5 18 18
\(A\)\(:\)\(5-0-5=0\)
\(D\)\(:\)\(13-5-8=0\)
\(F\)\(:\)\(18-13-5=0\)
\(\text{critical path}\)\(=\)\(A\text{-}D\text{-}F\)
\(\text{length}\)\(=\)\(5+8+5=18\text{ days}\)
A-D-F

The fit-out needs a minimum of 18 days.

Example 3 — Non-critical activities
The network shows a warehouse relocation (durations in days). Which activities are non-critical (float greater than zero)?
Solution

The critical activities have zero float; everything else is non-critical.

Warehouse relocation networkCritical path A-C-F in red; B, D, E, G are non-critical. A 6 B 4 C 5 D 3 E 2 F 7 G 4 1 0 0 2 6 6 3 4 8 4 11 11 5 6 14 6 18 18
\(B\)\(:\)\(8-0-4=4\)
\(D\)\(:\)\(11-4-3=4\)
\(E\)\(:\)\(14-4-2=8\)
\(G\)\(:\)\(18-6-4=8\)
\(\text{non-critical}\)\(=\)\(B,\ D,\ E,\ G\)

\(A\), \(C\) and \(F\) are critical; \(B\), \(D\), \(E\) and \(G\) have spare time, with \(E\) and \(G\) the most (8 days).

Example 4 — Shortening a critical task
For a backyard pool installation the critical path \(A\text{-}C\text{-}E\) gives \(6+7+4=17\) days. Extra crew cut activity \(C\) from \(7\) to \(4\) days. Find the new minimum completion time.
Solution

\(C\) is critical, so re-scan after shortening it and check no other path is longer.

Pool installation networkAfter cutting C to 4 days the critical path A-C-E is 14 days. A 6 B 3 C 4 D 5 E 4 F 3 1 0 0 2 6 6 3 3 5 4 10 10 5 14 14
\(\text{new } A\text{-}C\text{-}E\)\(=\)\(6+4+4=14\)
\(B\text{-}D\text{-}E\)\(=\)\(3+5+4=12\)
\(\text{new completion}\)\(=\)\(14\text{ days}\)
6+4+4=14

Cutting \(C\) by 3 days shortens the project to 14 days.

Common pitfalls

Head vs tail. Use the LST of the head event and the EST of the tail event. Swapping them, or using the EST of the head, gives the wrong float.
Zero float is not "unimportant". A float of zero means the activity is critical — it has no spare time and any delay delays the whole project. It cannot be skipped.
Critical = longest path. The critical path is the longest route through the network (it fixes the minimum time), even though it is made of zero-float activities. Do not confuse "longest path" with "most float".

Frequently asked questions

What is the float time of an activity?

The float, or slack, of an activity is how long it can be delayed without delaying the whole project. An activity with plenty of float can start late or run over a little and still not push back the finish date; an activity with zero float has no spare time at all.

How do you calculate float time?

Float equals the latest start time (LST) at the activity's head event minus the earliest start time (EST) at its tail event minus the activity's duration. Read the EST from the lower-left of the starting circle and the LST from the lower-right of the finishing circle, then subtract the duration.

What does it mean if an activity has zero float?

A zero float means the activity is critical: it has no spare time, so any delay to it delays the entire project. All the critical activities joined together from start to finish make up the critical path.

What is the critical path and how do you find it?

The critical path is the chain of zero-float activities running from the start event to the end event. Find it by working out the float of every activity, marking those with zero float, and joining them up. It is the longest path through the network, and its length is the minimum completion time.

Can a network have more than one critical path?

Yes. If two different chains of zero-float activities both run from start to finish and have the same length, the project has two (or more) critical paths. Every one of them must stay on schedule for the project to finish on time.

What happens if you shorten a critical activity?

Shortening a critical activity can shorten the whole project, because the critical path sets the minimum completion time. But you must re-scan: once it is shortened another path may become the longest, so the project only shortens until a different critical path takes over.