Mathematics Standard • Year 12 • Module 6 • Lesson 5
Critical Path and Float — Past-Paper Style
Practise HSC Mathematics Standard 2-style writing on critical path and float: identify critical activities, calculate slack and write a structured analysis on delays.
1. Short-answer questions
1.1 A project has scan results: 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). (ES, EF, LS, LF.) Calculate the float for each activity, name the critical path and state the project duration. 3 marks Band 3
1.2 Activity C in the project above has float = 1 day. The project manager is considering delaying C by 2 days. State the impact on the project finish date and explain in one short sentence why. 3 marks Band 3-4
1.3 A project has activities A(2, —), B(3, —), C(4, A), D(3, A and B), E(2, C and D).
(a) Perform both scans and calculate the float for each activity.
(b) State every critical path and give the project duration. 4 marks Band 4
2. Extended response
2.1 A community pool refurbishment in regional NSW has the following precedence table (days):
Drain pool DP(2, —)
Resurface RS(5, DP)
Tile TI(4, RS)
Pump service PS(3, DP)
Filter replace FR(2, PS)
Refill RF(3, TI and FR)
Inspect IN(1, RF)
(a) Perform both scans and produce a six-column table including float (Activity | Duration | ES | EF | LS | LF | Float).
(b) State the critical path and the project duration.
(c) The council asks: "If we delay PS (pump service) by 4 days, will the pool reopen late? If yes, by how many days?" Use your float and the path-sums to give a clear yes/no answer with the exact delay (if any). 7 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — correct forward scan.
• 1 mark — correct backward scan.
• 1 mark — clean six-column table including the float column.
Part (b) — 2 marks
• 1 mark — critical path named explicitly.
• 1 mark — correct project duration with units.
Part (c) — 2 marks
• 1 mark — compares the 4-day delay to PS's float (or recomputes the PS path).
• 1 mark — clear sentence stating whether the pool reopens late and by how many days, with reasoning.
Your response:
Stuck on (c)? If PS's slack is k days, a delay of d days delays the project by max(0, d − k). Apply this directly.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Floats and critical path from scan values (3 marks)
Sample response. Floats (LS − ES): A=0, B=0, C=1, D=0, E=1, F=0. Critical activities are A, B, D, F. Critical path: A → B → D → F, duration 3 + 3 + 3 + 3 = 12 days.
Marking notes. 1 mark — all six floats correct. 1 mark — critical path named explicitly. 1 mark — duration with units. A response that only gives "12 days" with no floats scores 1/3.
1.2 — Delay C by 2 days (3 marks)
Sample response. C's float is 1 day. A delay of 2 days exceeds the float by 1 day, so C becomes critical and pushes the project finish back by 1 day. The new project duration is 13 days. Reason: any delay beyond the float uses up the cushion and then propagates directly to the finish date.
Marking notes. 1 mark — recognises C's float = 1. 1 mark — applies "delay − float = project slippage". 1 mark — explicit conclusion (1-day late, new duration 13 days). A "yes — project is late" with no number scores 1/3.
1.3 — Two start activities, find every critical path (4 marks)
(a) Sample response. 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.
(b) Sample response. Project duration = 8 days. Two critical paths: A → C → E (2+4+2 = 8) and B → D → E (3+3+2 = 8).
Marking notes. 1 mark — forward scan correct (MAX at D and at E). 1 mark — backward scan correct (MIN at A). 1 mark — floats all zero stated. 1 mark — both critical paths named with the project duration. A response that only names one critical path scores 3/4.
2.1 — Community pool refurbishment (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Scans and table.
Forward:
DP(0, 2)
RS(2, 7)
TI(7, 11)
PS(2, 5)
FR(5, 7)
RF(max(11, 7) = 11, 14)
IN(14, 15)
Project duration = EF(IN) = 15 days. [1 mark — forward scan correct.]
Backward (IN LF=15):
IN: LF=15, LS=14
RF: LF=14, LS=11
TI: LF=11, LS=7
FR: LF=11, LS=9
RS: LF=7, LS=2
PS: LF=9, LS=6
DP: LF=min(LS(RS), LS(PS)) = min(2, 6) = 2, LS=0
[1 mark — backward scan correct including MIN at DP.]
Act | Dur | ES | EF | LS | LF | Float
DP | 2 | 0 | 2 | 0 | 2 | 0
RS | 5 | 2 | 7 | 2 | 7 | 0
TI | 4 | 7 | 11 | 7 | 11 | 0
PS | 3 | 2 | 5 | 6 | 9 | 4
FR | 2 | 5 | 7 | 9 | 11 | 4
RF | 3 | 11 | 14 | 11 | 14 | 0
IN | 1 | 14 | 15 | 14 | 15 | 0
[1 mark — clean six-column table including float column.]
(b) Critical path and duration.
Critical path: DP → RS → TI → RF → IN. [1 mark.]
Project duration = 15 days. [1 mark.]
(c) Delay PS by 4 days.
PS has float = 4 days (LS = 6, ES = 2). A 4-day delay exactly uses PS's slack. Recomputing the PS path: DP(2) + PS(3 + 4 delay = 7) + FR(2) + RF(3) + IN(1) = 15. This equals the original 15-day critical-path duration. [1 mark — compares delay to PS float and recomputes path-sum.]
Conclusion: No — the pool will not reopen late, but the PS path now ties at 15 days, so PS becomes critical. Any further delay on PS (or on FR) would push the reopen date out by the additional days. [1 mark — clear yes/no with explicit reasoning.]
Total: 7/7.
Band descriptors for marker.
Band 3: Forward scan correct and duration stated; backward scan partial; no float column. ≈ 3 marks.
Band 4: Both scans correct; table given but no float column; critical path named. ≈ 5 marks.
Band 5: Complete six-column table with float; critical path named; (c) says "no" with float ≥ delay reasoning but does not note PS becomes critical. ≈ 6 marks.
Band 6: Complete — full table, critical path, AND (c) notes that PS now sits on a tied second critical path so future delays propagate directly. 7/7.