Slack and changes to activities | Year 12

01 / LEARN

Understand the idea.

An activity's allowable delay

Total float is the delay between an activity's earliest and latest permissible start. It can also be calculated from its latest finish minus earliest finish. The chosen project deadline and consistent scheduling of related work determine the allowance.

For the displayed network, B has $8 - 0 - 3 = 5$ days of float and C has $10 - 4 - 5 = 1$ day. Zero-float activities A, D, E and F control completion. Float is an available delay, not additional working time or extra activity duration.

\begin{aligned} Total float & = L S - E S \\ = L F - E F \\ = L_{j} - E_{i} - d \end{aligned}

Change the network, then recompute

Increasing B from 3 to 7 days leaves E waiting for D until day 8, so the project still finishes at day 13. Increasing B to 9 days makes E wait until day 9 and moves completion to day 14.

After any change, recompute the schedule and floats. Successive non-critical activities can share the same allowance on a branch. Adding their original floats as if each were independent can predict a finish that the combined delays cannot achieve.

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

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Compute float from a pair of events

A task lasts 5 hours, with tail earliest time 4 and head latest time 12. Find its total float.

Worked solution

1
Calculate the earliest finish.
$E F = 4 + 5 = 9 hours$
The task starts no earlier than its tail event and requires five hours.
2
Calculate its total float.
$T F = L F - E F = 12 - 9 = 3 hours$
Finishing up to three hours after its earliest finish still meets its latest finish.
3
Check using the start window.
$\begin{aligned} L S & = 12 - 5 = 7 \\ T F & = 7 - 4 = 3 hours \end{aligned}$
Latest minus earliest start gives the same allowable delay.

Answer

3 hours.

EXAMPLE 02

Two tasks share a branch allowance

A non-critical branch has two successive tasks P and Q. The branch takes 8 days while the critical route takes 11. Each task is reported to have 3 days total float. Can both be extended by 3 days without delaying the project?

Worked solution

1
Find the branch's shared allowance.
$11 - 8 = 3 days$
The branch can grow by only three days before reaching the critical route's duration.
2
Combine the proposed extensions.
$3 + 3 = 6 days$
P and Q are successive, so both extensions add to the same branch duration.
3
Recalculate the branch and project finish.
$\begin{aligned} T_{P Q} & = 8 + 6 = 14 days \\ T & = max (11, 14) = 14 days \end{aligned}$
The branch exceeds its shared allowance and becomes the new controlling route.

Answer

No. Their allowances are shared along the branch; using both extensions delays completion to 14 days.

03 / YOUR TURN

Try it yourself.

Check your understanding

A task has ES=3 and LS=8. It starts 4 days later than its earliest start. With no other changes, has it exceeded its total float?

Show answer & reasoning

No.

No. The float is $8 - 3 = 5$ days. A four-day delay means starting at $3 + 4 = 7$ , which is before the latest start of 8 when no other changes use the allowance.

Keep practising

6 worksheet questions · 17 marks

Open worksheet Worked solutions

Read the questions online

Calculate the total floats of B and C in the display project.
3 marks
Increase B's duration in the display project from 3 to 7 days. Find the project finish.
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
Increase B's duration instead to 9 days. Find the finish and critical real-activity sequence.
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
Activity P has $ES = 6$ , $EF = 10$ , $LS = 9$ and $LF = 13$ . Find its duration and float.
2 marks
A non-critical route takes 12 hours and the project finishes in 16. Two consecutive tasks on that route are both extended by 3 hours. What happens?
3 marks
The project deadline is fixed at day 20, but its earliest completion is day 18. Explain why a backward scan using 20 answers a different question from one using 18.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Calculate the total floats of B and C in the display project.
3 marks
B: $8 - 0 - 3 = 5$ .
C: $10 - 4 - 5 = 1$ .
B: 5 days; C: 1 day.
Increase B's duration in the display project from 3 to 7 days. Find the project finish.
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
E waits for $max (B finish 7, D finish 8) = 8$ .
E finishes 10 and F finishes 13.
13 days.
Increase B's duration instead to 9 days. Find the finish and critical real-activity sequence.
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
E starts at $max (9, 8) = 9$ .
It finishes 11; F finishes 14.
The B branch now controls.
14 days; B–E–F.
Activity P has $ES = 6$ , $EF = 10$ , $LS = 9$ and $LF = 13$ . Find its duration and float.
2 marks
$Duration = 10 - 6 = 4$ (also $13 - 9 = 4$ ).
$Float = 9 - 6 = 3$ .
Duration 4; float 3 time units.
A non-critical route takes 12 hours and the project finishes in 16. Two consecutive tasks on that route are both extended by 3 hours. What happens?
3 marks
Its shared allowance was $16 - 12 = 4$ .
The combined extension is 6, so it exceeds the allowance by 2.
That route becomes 18 hours, delaying the project by 2 hours.
The project deadline is fixed at day 20, but its earliest completion is day 18. Explain why a backward scan using 20 answers a different question from one using 18.
3 marks
A looser finish target creates two days of overall schedule allowance.
State the finish target before interpreting zero or positive float.
The day-20 scan finds latest times that meet the deadline; the day-18 scan finds float that preserves the earliest completion.

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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 ↗

In this lesson

Understand the idea.

An activity's allowable delay

Change the network, then recompute

See the method in action.

Compute float from a pair of events

Calculate the earliest finish.

Calculate its total float.

Check using the start window.

Two tasks share a branch allowance

Find the branch's shared allowance.

Combine the proposed extensions.

Recalculate the branch and project finish.

Try it yourself.

Check your understanding

Keep practising