Mathematics Standard • Year 12 • Module 6 • Lesson 5

Critical Path and Float — Skill Drill

Build fluency in calculating float (slack), identifying critical activities and tracing the critical path through a network using ES, EF, LS, LF.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Write the float formula in two equivalent forms.

Float = LS − ____ = LF − ____.

Q1.2 An activity is critical when its float equals ____.

This is equivalent to ES = ____ (and EF = ____).

Q1.3 The critical path is the ____________ path from start to finish; every activity on it has float ____.

Q1.4 True or false: a project can have more than one critical path.

Stuck? Revisit lesson § Key Ideas — Float, Critical activity, Critical path, Multiple critical paths.

2. Worked example — calculating float and tracing the critical path

Follow each line of working. Every step has a reason on the right.

Problem. A project has scan values: A(0, 3, 0, 3), B(3, 7, 3, 7), C(3, 5, 5, 7), D(7, 12, 7, 12), E(12, 14, 12, 14). (Format: ES, EF, LS, LF.) Find the float for each activity and the critical path.

Step 1 — Apply Float = LS − ES for every activity.

A: 0 − 0 = 0
B: 3 − 3 = 0
C: 5 − 3 = 2
D: 7 − 7 = 0
E: 12 − 12 = 0

Reason: float says how much an activity can slip without pushing the whole project back.

Step 2 — Mark zero-float activities as critical.

Critical: A, B, D, E. Not critical: C (float 2).

Step 3 — Trace the critical path through the critical activities from start to finish.

A → B → D → E

Reason: only critical activities can lie on a critical path, and they must form a connected chain from start to finish.

Step 4 — Check by summing the path-duration.

Duration of A-B-D-E = 3 + 4 + 5 + 2 = 14 = project duration. OK.

Conclusion. Floats: A=0, B=0, C=2, D=0, E=0. Critical path: A → B → D → E, duration 14. C has 2 days slack.

3. Faded example — fill in the missing floats and critical path

A project has scans P(0,2,0,2), Q(2,5,2,5), R(2,6,4,8), S(5,7,5,7), T(6,9,8,11), U(7,10,7,10). (ES, EF, LS, LF.) Fill in the floats and the critical path. 4 marks

Step 1 — Float = LS − ES.

P: __ − __ = __ Q: __ − __ = __ R: __ − __ = __
S: __ − __ = __ T: __ − __ = __ U: __ − __ = __

Step 2 — Mark zero-float as critical.

Critical activities: ______________________________________.

Step 3 — Trace the critical path from start to finish.

Critical path: ______________________________________. Duration: ____.

Stuck? Revisit lesson § Worked Example — critical activities have ES = LS (equivalently EF = LF).

4. Graduated practice — float and critical-path problems

Show your working. For each network, calculate float for every activity and identify the critical path.

Foundation — single-activity float (4 questions)

Q	Problem	Answer
4.1 1	Activity W has ES = 4, LS = 4. Find the float and decide if W is critical.
4.2 1	Activity X has EF = 9, LF = 13. Find the float and decide if X is critical.
4.3 1	Activity Y has ES = 2, EF = 7, LS = 5, LF = 10. Find the float.
4.4 1	True or false: every activity on the critical path has float > 0.

Standard — typical HSC difficulty (6 questions)

For each scan table, compute Float = LS − ES for every activity and name the critical path.

4.5 Scans: A(0,4,0,4), B(4,7,4,7), C(4,6,5,7), D(7,11,7,11). Find floats and the critical path. 2 marks

4.6 Scans: A(0,3,0,3), B(3,6,3,6), C(3,7,4,8), D(6,9,6,9), E(7,10,8,11), F(9,12,9,12). Find floats. Is there one critical path or more than one? 2 marks

4.7 A project has critical activities A, C, E and non-critical activities B, D. Activity B has float 3 days. State the maximum delay B can absorb without delaying the project, and explain in one short sentence what happens if B is delayed by 4 days. 2 marks

4.8 A project has duration 20 days. You shorten one critical activity by 3 days. The new project duration is 18 days (not 17). Explain in one short sentence what likely happened. 2 marks

4.9 Scans: A(0,2,0,2), B(2,5,2,5), C(2,6,2,6), D(5,7,5,7), E(6,9,6,9), F(7,10,7,10), G(9,11,9,11). Find every critical activity. How many critical paths exist? 2 marks

4.10 Activity Z has float = 0 in the original scan. Explain in one or two short sentences why Z must be on the critical path. 2 marks

Extension — full pipeline (2 questions)

4.11 Activities: A(3, —), B(4, A), C(2, A), D(5, B), E(3, C), F(2, D and E). Perform both scans, calculate float for every activity, and state the critical path with its duration. 3 marks

4.12 Activities: A(2, —), B(3, —), C(4, A), D(3, A and B), E(2, C and D). Perform both scans, calculate float for every activity, and identify every critical path. 3 marks

Stuck on 4.12? Look at every path from a start activity to E. The longest is the critical path; if two are equally longest, both are critical.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For Q4.1 (ES = LS = 4) I got float = 0 and recognised W as critical.

For Q4.2 (EF = 9, LF = 13) I got float = 4 and recognised X as NOT critical.

For Q4.4 I said FALSE — critical activities have float = 0, not > 0.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Float formula

Float = LS − ES = LF − EF.

Q1.2 — Critical activity

Float = 0. Equivalently ES = LS (and EF = LF).

Q1.3 — Critical path

The critical path is the longest path from start to finish; every activity on it has float = 0.

Q1.4 — Multiple critical paths

True. If two or more paths tie for the longest length, all of them are critical.

Q3 — Faded example (P, Q, R, S, T, U)

Floats: P = 0, Q = 0, R = 2, S = 0, T = 2, U = 0. Critical activities: P, Q, S, U. Critical path: P → Q → S → U, duration 2 + 3 + 2 + 3 = 10.

Q4.1 — Float for ES = LS = 4

Float = LS − ES = 4 − 4 = 0. W is critical.

Q4.2 — Float for EF = 9, LF = 13

Float = LF − EF = 13 − 9 = 4. X is not critical.

Q4.3 — Float for Y (ES=2, LS=5)

Float = 5 − 2 = 3. (Check: LF − EF = 10 − 7 = 3. Consistent.)

Q4.4 — Critical path floats

False. Activities on the critical path have float exactly equal to 0, not > 0.

Q4.5 — ABCD critical path

Floats: A=0, B=0, C=1, D=0. Critical activities: A, B, D. Critical path: A → B → D, duration 4 + 3 + 4 = 11.

Q4.6 — A-F critical path

Floats: A=0, B=0, C=1, D=0, E=1, F=0. Critical activities: A, B, D, F. Critical path: A → B → D → F, duration 3 + 3 + 3 + 3 = 12. Only one critical path (C and E both have float 1 so neither is on the critical path).

Q4.7 — B with float 3

Maximum delay = 3 days. If B is delayed by 4 days, it exceeds its float by 1 day, becomes critical, and pushes the project finish back by 1 day.

Q4.8 — Critical activity crash didn't shorten by 3

A different path became critical after the crash. The 3-day saving on the original critical path was "wasted" beyond the point where another path caught up, so the project only shortened by 2 days (and a second path is now critical).

Q4.9 — Multiple critical paths

Floats: every activity = 0 (LS = ES for all). All activities are critical. There are multiple critical paths: A → B → D → F = 2+3+2+3 = 10 and A → C → E → G = 2+4+3+2 = 11. (Tie at 11 if you check the full path-sum; here the longer one is A → C → E → G.) Note: with the scan data given (every activity has LS = ES), all activities lie on some critical path; in a realistic table you would trace the actual longest path.

Q4.10 — Why float = 0 implies critical path

Float = 0 means ES = LS, so the activity has no flexibility. The path through Z must have length equal to the project duration (otherwise Z would have positive float). Therefore Z lies on a longest path from start to finish — a critical path.

Q4.11 — A-F full pipeline

Forward: A(0,3), B(3,7), C(3,5), D(7,12), E(5,8), F(max(12,8)=12, 14). Duration = 14.
Backward (F LF=14): F LS=12. D LF=12, LS=7. E LF=12, LS=9. B LF=7, LS=3. C LF=9, LS=7. A LF=min(3,7)=3, LS=0.
Floats: A=0, B=0, C=4, D=0, E=4, F=0. Critical path = A → B → D → F, duration 14.

Q4.12 — Two start activities, full pipeline

Forward: A(0,2), B(0,3), C(2,6), D(max(2,3)=3, 6), E(max(6,6)=6, 8). Duration = 8.
Backward (E LF=8): E LS=6. C LF=6, LS=2. D LF=6, LS=3. A LF=min(2,3)=2, LS=0. B LF=3, LS=0.
Floats: A=0, B=0, C=0, D=0, E=0. Two critical paths: A → C → E (2+4+2 = 8) and B → D → E (3+3+2 = 8). Both have duration 8 = project duration.