Mathematics Standard • Year 12 • Module 6 • Lesson 12
Module Review: Critical Path Analysis — Problem Set
Apply the full Module 6 toolkit to integrated real-world scenarios — every concept from lessons 1 to 11 woven together.
Problem 1 — Granny-flat addition
A homeowner is adding a granny flat. Activities (days): A. Council approval (5, −), B. Site clearance (2, A), C. Slab pour (3, B), Cure lag (2 days after C), D. Frame (5, after cure), E. Roof (4, D), F. Plumbing (3, D), G. Final inspection (1, E and F).
Set up: What are we solving for?
(i) Compute ES and EF for all activities, treating the cure as a 2-day lag. State the project duration. 3 marks
(ii) Identify the critical path and floats on any non-critical activities. 2 marks
(iii) Sketch a Gantt chart with all activities, solid bars for ES → EF, lighter extensions for float. 3 marks
Stuck? Revisit the M6 summary panel — apply lag, scans, critical path, then sketch the Gantt.Problem 2 — Tech conference setup
A Sydney tech conference is setting up its main expo. Activities (hours): A. Carpet install (4, −), B. Booth frames (5, A), C. Power runs (3, A), D. AV equipment install (4, B and C), E. Signage (3, B), F. Demo computers setup (2, D), G. Floor check (1, E and F).
Set up: What are we solving for?
(i) Full forward and backward scans. State project duration and critical path. 3 marks
(ii) The venue offers a $400 fee to crash one activity by 1 hour. Crash candidates: B ($400), D ($400), F ($400). Which one shortens the project the most, and by how much? 3 marks
(iii) Justify your choice in (ii) with reference to which crash candidate is on the critical path. 2 marks
Stuck? Crashing a non-critical activity changes nothing.Problem 3 — School musical production
A school is preparing a musical. Activities (weeks): A. Auditions (2, −), B. Cast rehearsals (6, A), C. Build set (4, A), D. Costumes (3, A), E. Tech rehearsals (2, B and C), F. Dress rehearsals (1, D and E), G. Opening night (0, F — milestone).
Set up: What are we solving for?
(i) Forward and backward scans, critical path, and project duration. 3 marks
(ii) The producer wants the show ready 1 week earlier. Crash options: B 6→5 at $300/wk, C 4→3 at $200/wk. Find the minimum-cost plan and explain which activities you would crash. 3 marks
(iii) If two critical paths share activities, crashing a shared activity is more efficient. Identify whether this project has multiple critical paths or one. 2 marks
Stuck on (ii)? Crash an activity that IS on the critical path. Costumes (D) is not on critical path? Check first.Problem 4 — Council park redevelopment
A council is redeveloping a small park. Activities (days): A. Demolition (4, −), B. Earthworks (5, A), C. Drainage (3, A), D. Concrete paths (4, B and C), E. Playground install (3, D), F. Turf laying (2, D), G. Inspection (1, E and F).
Direct cost = $40,000. Indirect overheads = $500/day. The park must reopen within 18 days or a $1,000/day penalty applies.
Set up: What are we solving for?
(i) Full scans, critical path and project duration. 3 marks
(ii) Calculate the total cost at the normal duration (direct + indirect + penalty if any). 2 marks
(iii) Crash options: B 5→3 at $700/day, D 4→3 at $500. Find the minimum-cost duration and total cost. 3 marks
Stuck? Compare savings/day above day 18 ($1,500) vs at/below day 18 ($500).Problem 5 — Backyard pool install
A pool company is installing a fibreglass pool. Activities (days): A. Site survey (1, −), B. Excavation (3, A), C. Crane lift pool shell (1, B), D. Plumbing connection (2, C), Cure lag (3 days after D for concrete surround), E. Pebbledash + tile (2, after cure), F. Fence install (3, E), G. Final inspection (1, F).
Set up: What are we solving for?
(i) Compute ES and EF using the 3-day cure lag. State project duration. 3 marks
(ii) The owner asks: "Can we cut 2 days off?" Cure cannot be crashed. Crash candidates: B 3→2 at $300, F 3→2 at $200. State the impact. 3 marks
(iii) Sketch a Gantt chart highlighting the cure lag as a non-work period between D and E. 2 marks
Stuck? Show the cure as a 3-day gap on the Gantt — no bar, just a wait.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Granny flat
Set up. We do scans treating the 2-day cure as a lag, then identify critical path and sketch the Gantt.
(i) A(0,5), B(5,7), C(7,10), cure 2 days (Frame cannot start until 12), D(12,17), E(17,21), F(17,20), G(max(21,20)=21, 22). Duration = 22 days.
(ii) Backward: G(22,21), E(21,17), F(21,18), D(17,12), C(12,7), B(7,5), A(5,0). Floats: F = 18 − 17 = 1, others 0. Critical path: A → B → C → cure → D → E → G. F has float 1.
(iii) Gantt: A 0–5 solid, B 5–7 solid, C 7–10 solid, [cure gap 10–12, no bar], D 12–17 solid, E 17–21 solid, F 17–20 solid + extension 20–21, G 21–22 solid.
Problem 2 — Tech conference
Set up. We do scans, identify critical path, then pick the most impactful single 1-hour crash.
(i) A(0,4), B(4,9), C(4,7), D(max(9,7)=9, 13), E(9,12), F(13,15), G(max(12,15)=15, 16). Backward: G(16,15), F(15,13), E(15,12), D(13,9), B(9,4), C(9,6), A(min(4,6)=4, 0). Floats: A=B=D=F=G=0, C=2, E=3. Critical: A → B → D → F → G (16 hours).
(ii) Candidates: B (critical, crash to 4 → duration 15). D (critical, crash to 3 → duration 15). F (critical, crash to 1 → duration 15). All three give a 1-hour reduction. Same impact.
(iii) All three are on the critical path, so each saves exactly 1 hour. Any of B, D, F at $400. If preferring fewer single-day exposures, choose the one whose original duration is largest (B at 5 hours) to keep the schedule less compressed.
Problem 3 — School musical
Set up. We do scans, choose the cheapest crash on the critical path, then identify whether there are multiple critical paths.
(i) A(0,2), B(2,8), C(2,6), D(2,5), E(max(8,6)=8, 10), F(max(5,10)=10, 11), G(11). Backward: G(11,11), F(11,10), E(10,8), D(11,8), B(8,2), C(8,4), A(min(2,4,8)=2, 0). Floats: A=B=E=F=G=0, C=2, D=3. Critical: A → B → E → F → G (11 weeks).
(ii) Need 1 week. B is on critical (crash 1 week $300, duration 10). C is NOT critical (float 2) — crashing C wastes money. Crash B by 1 week at $300. (Crashing C does NOT save time.)
(iii) Only one critical path: A → B → E → F → G. C (path A-C-E-F-G length 2+4+2+1+0 = 9) and D (path A-D-F-G length 2+3+1+0 = 6) are not critical. Single critical path.
Problem 4 — Park redevelopment
Set up. We do scans, compute normal cost (with penalty), then evaluate crashing.
(i) A(0,4), B(4,9), C(4,7), D(max(9,7)=9, 13), E(13,16), F(13,15), G(max(16,15)=16, 17). Backward: G(17,16), E(16,13), F(16,14), D(13,9), B(9,4), C(9,6), A(4,0). Floats: A=B=D=E=G=0, C=2, F=1. Critical: A → B → D → E → G (17 days).
(ii) 17 < 18, so NO penalty. Total = $40,000 + 17×$500 = $40,000 + $8,500 = $48,500.
(iii) Crash options on critical path: B slope $700/day > $500 indirect saving (no penalty), so each crash day LOSES $200. D slope $500 = $500 saving = break-even. Don't crash. Optimal duration = 17 days. Total cost = $48,500. (No improvement available below $500/day savings without penalty.)
Problem 5 — Backyard pool
Set up. We do scans with cure lag, then evaluate 2-day crash, then sketch.
(i) A(0,1), B(1,4), C(4,5), D(5,7), cure 3 days (E cannot start until 10), E(10,12), F(12,15), G(15,16). Duration = 16 days.
(ii) Need 2 days. B is critical (currently 3 days). F is critical (currently 3 days). Crashing B 1 day saves 1 day = $300. Crashing F 1 day saves 1 day = $200. Total crash cost = $500 for 2-day reduction. New duration = 14 days. (Cure cannot be crashed, but the work activities can.)
(iii) Gantt sketch:
A: solid 0–1
B: solid 1–4
C: solid 4–5
D: solid 5–7
[cure gap 7 → 10: NO bar, just a labelled wait]
E: solid 10–12
F: solid 12–15
G: solid 15–16
The cure period appears as a 3-day blank on the timeline, with all activities shifting right by 3 days from D's finish onward.