A school project can have a task that is allowed to finish three days later without delaying the final presentation, yet that task may have zero days of free float. The reason is that free float protects the earliest start of the next task, while total float protects the overall completion date or applicable schedule deadline. They answer different questions.
Free float is how long an activity can slip without delaying the early start of its immediate successor. Total float is how long it can slip without delaying the project’s calculated end (or the relevant constrained finish). In a simple finish-to-start project network, free float = the next activity’s earliest start minus this activity’s earliest finish; total float = latest start minus earliest start. Do not substitute one for the other.
- Draw the actual task dependencies first; a calendar list alone is insufficient.
- For each task, calculate earliest start (ES) and earliest finish (EF).
- Use a backward pass to calculate latest start (LS) and latest finish (LF).
- Total float = LS − ES; free float compares an activity’s EF to its successor’s ES.
- A task can have total float of three days but free float of zero days.
- Treat all durations and outcomes in the example below as illustrative, not as a real pupil’s project record.
The learning problem: what does a delay actually affect?
Children and adults often answer “this is not on the critical path, so we can delay it” without asking what happens next. Suppose the photography team needs one day to take pictures and the editing team must then use those pictures. Delaying photography may immediately delay photo editing, even when the final exhibition remains on schedule because that entire branch has some spare time.
This is a strong diagnostic question: “Could the task move later without affecting any other team’s earliest start?” If the student says yes merely because the final deadline remains safe, they have confused free float with total float.
Build the network before doing arithmetic
Consider an invented school exhibition. Assume every duration is in uninterrupted whole working days, tasks use simple finish-to-start dependencies, the project starts at time 0, and no team has an extra resource constraint. There are five tasks, called A to E:
- A — write the display text: 3 days, starting at project time 0.
- B — take photographs: 1 day, also available from time 0.
- C — edit the display text: 2 days, after A.
- D — select photographs: 1 day, after B.
- E — assemble the final presentation: 1 day, after both C and D.
The two branches are A → C → E and B → D → E. The assembly task cannot begin until both incoming branches are finished. A chain drawing is useful: the arrows represent actual prerequisites, not simply tasks placed next to each other on a calendar.
Worked example: earliest starts and finishes
Forward pass for the text branch
A begins at 0 and takes 3 days, so A has ES 0, EF 3. C cannot start until A finishes, so C has ES 3, EF 5. These are time points, not dates on a specific school calendar.
Forward pass for the photograph branch
B has ES 0 and EF 1. D starts when B finishes, so D has ES 1 and EF 2. These activities are finished earlier than the text branch.
The joining task determines the earliest project completion
E needs both C and D. Therefore E’s earliest start is the larger predecessor finish: max(5, 2) = 5. It lasts one day, so E has EF 6. Under the stated assumptions the exhibition’s earliest completion is time 6. The longer dependency path A → C → E takes 3 + 2 + 1 = 6 days.
Now calculate the latest permitted times
Work backward from completion at time 6. E must start at time 5 to finish at 6, so E has LS 5, LF 6. Both C and D must finish by time 5, just before E starts.
C lasts 2 days, hence C has LS 3 and LF 5. A must finish before C starts at 3; A’s LS is 0 and LF is 3. That text branch has no room to move if the project must finish at time 6.
D may finish as late as time 5. Because D takes one day, D has LS 4 and LF 5. B must finish before D’s latest start of 4, so B has LS 3 and LF 4. The photo branch therefore has room to shift.
Find the two kinds of float without confusing them
Task B: zero free float, three days total float
Total float for B = LS − ES = 3 − 0 = 3 days. Its successor D originally starts at time 1; B originally finishes at time 1. Therefore free float for B = 1 − 1 = 0 days. B cannot slip even one day without moving D’s earliest start, though B and D can move together by up to three days without affecting the final assembly.
Task D: three days free float and three days total float
D finishes at time 2, and its successor E cannot start before time 5 because the text branch is still in progress. Therefore free float for D = 5 − 2 = 3 days. Total float is also LS − ES = 4 − 1 = 3 days. Both numbers happen to match here, but that will not be true for every activity.
Tasks A, C and E: zero float in this simplified network
A’s LS = ES = 0, C’s LS = ES = 3, and E’s LS = ES = 5, giving zero total float. These are on the longest dependency path. A delay to any one of them, without any compensating adjustment, pushes the earliest project finish later.
Did you know? A delay can inconvenience a team without missing the final deadline
A delayed photograph shoot from time 0–1 to time 2–3 moves the photo selection from time 1–2 to time 3–4. It inconveniences the selection task and changes that team’s earliest schedule. Yet E can still start at time 5, so the project can still finish at time 6. That is the practical difference between the two floats.
This distinction improves communication. “We have three days of total float” does not mean “nobody else will be affected.” It means the final date may remain unchanged under the stated network and scheduling conditions.
Misconceptions to correct in project-based learning
- “Every non-critical activity can be delayed freely.” A delay may move successors even while total float remains.
- “Total float means free time for every student.” Float is a schedule property, not a promise about human availability.
- “A Gantt chart alone proves the critical path.” Dependencies and durations must be specified.
- “Every delay creates an equally large deadline delay.” That depends on available float, network structure and constraints.
- “Critical path cannot change.” Revised task durations, newly discovered dependencies or resource constraints can change it.
- “Zero free float means zero total float.” Task B above is a direct counterexample.
Guided-to-independent practice
Level 1: one straightforward chain
A task finishes at time 4 and its sole successor starts at time 4. Free float = 4 − 4 = 0. Ask the learner why: any delay in that task moves the successor’s earliest start.
Level 2: a visible gap
A task finishes at time 2 and its only successor’s earliest start is time 6. Free float = 6 − 2 = 4 days. Under the simplified dependency model, that task may slip four days without delaying the successor.
Level 3: two successors
A task finishes at time 3. One successor can start at time 5 and another at time 7. The free float is controlled by the earliest successor: min(5,7) − 3 = 2 days. Using 7 − 3 would incorrectly overlook the successor starting at 5.
Level 4: recompute a whole project
Return to the exhibition example. Now suppose the photo selection takes four days rather than one. B takes 1 day and D takes 4, so the photo branch B → D → E lasts 1 + 4 + 1 = 6 days, equal to the text branch. The earliest finish remains 6, but both branches are now critical in this simplified schedule. The student should explain why float changed although the final date did not.
A practical five-minute project planning routine
Parents can ask students to write each task on one card, with its duration and required predecessor. Teachers can ask teams to draw arrows, calculate the earliest complete date and describe the consequence of moving one task. The goal is to make causes and constraints visible before arguing over whose work is “late.”
Track progress with a short rubric: correct dependency arrows; correct earliest times; correct latest times; correct free/total float distinction; explanation of which successor or deadline is affected. A student who earns five checks on a fresh network has shown stronger transfer than one who copied the completed example.
Keep duration estimates separate from observations. If a task was forecast to take one day but actually took two, update the plan and mark the new information clearly. Do not rewrite the original estimate as though it had always been known.
Limitations: where school examples stop and real schedules begin
Formal schedules can contain imposed dates, leads and lags, different working calendars, shared resources, unexpected rework and non-finish-to-start links. Some project-management systems identify critical activities differently under constraints. The short formulas work for this clearly stated toy network; professional scheduling needs the full model.
The Project Management Institute’s guidance on critical path and float provides the definitions used here, including the distinction between free and total float. It is important not to generalise the numerical illustration into an institutional rule for every type of project.
Frequently asked questions
Is free float ever greater than total float?
In the standard simple dependency calculation illustrated here, free float does not exceed total float. If software presents surprising values, inspect constraints, calendars and calculation settings before assuming a mistake.
Why did task B have no free float but three days of total float?
Because D was ready to start as soon as B ended. Moving B also moves D. But the photo branch could move together into the waiting time before E without delaying the overall finish.
Can a project have two critical paths?
Yes. In our modified example, both branches take six days including the final assembly. A delay on either branch can affect the earliest finish without an offsetting adjustment.
What should I do first when a school project runs late?
Verify which task has really changed, update its remaining duration and dependencies, then recalculate the earliest finish. Avoid turning one missed estimate into unsupported blame.
Continue learning: explain before you schedule
The broader skill is careful reasoning from documented information. Compare How to Read a Graph with a Broken Horizontal Axis for preserving meaningful time information. Browse the eduKate Education Hub and the immutable Mathematics tutoring reference for diagnosis-first reasoning, progressive practice and independent checking.
The core aim is not simply to build a project schedule. It is to teach a student to explain what an apparent delay actually affects. That skill is useful in group homework, family planning, research and later work.