Mathematics Standard • Year 12 • Module 6 • Lesson 5
Critical Path and Float — Problem Set
Apply critical-path analysis to real Australian schedules: an apartment build, a school excursion, a clinical drug trial, a regional theatre production and a council park upgrade.
Problem 1 — Apartment-block fit-out (Parramatta)
An apartment build has the following activities (days):
A(3, —), B(4, A), C(2, A), D(5, B), E(3, C), F(2, D and E), G(1, F).
Set up: What are we solving for?
(i) Perform both scans and tabulate Activity | Duration | ES | EF | LS | LF | Float. 2 marks
(ii) State the critical path and the project duration. 2 marks
(iii) The site manager asks: "How many days can activity C be delayed before it becomes critical?" Answer in one sentence using the float. 1 mark
Stuck? Revisit lesson § Critical Path — critical activities have float 0 and form a connected chain from start to finish.Problem 2 — Year 12 school excursion planning (Sydney)
A teacher schedules tasks (weeks):
Permission slips PS(2, —), Book transport BT(3, PS), Book venue BV(4, PS), Pay deposits PD(1, BT and BV), Risk assessment RA(2, BV), Final brief FB(1, PD and RA).
Set up: What are we solving for?
(i) Perform both scans and produce a clean table with floats. 2 marks
(ii) State the critical path and the project duration. 2 marks
(iii) Can BT (book transport) be delayed by 1 week without delaying the final brief? Justify in one short sentence using your float. 2 marks
Stuck? Revisit lesson § Float — slack = LS − ES. A delay ≤ slack is absorbed without affecting the project.Problem 3 — Clinical drug trial (university hospital, NSW)
A trial coordinator schedules (weeks):
Protocol PT(4, —), Ethics ET(6, PT), Recruit RC(8, ET), Dose phase DP(10, RC), Follow-up FU(4, DP), Lab analysis LA(5, FU), Statistical analysis SA(3, LA), Report RP(2, SA).
Set up: What are we solving for?
(i) Perform both scans and produce a clean table with floats. 2 marks
(ii) State the critical path. Comment in one sentence on why every activity in this project is critical. 2 marks
(iii) The funder asks: "If we shorten LA from 5 weeks to 3 weeks, does the trial finish 2 weeks earlier?" Use your float and critical-path analysis to give a yes/no answer with reasoning. 2 marks
Stuck? Revisit lesson § Critical Path — shortening a critical activity by k shortens the project by k, provided no other path becomes critical first.Problem 4 — Regional theatre production (NSW Central Coast)
A theatre schedules (weeks):
Choose play CP(1, —), Cast CA(2, CP), Rehearse RH(8, CA), Build set BS(6, CP), Costumes CO(5, CA), Promotion PR(4, CP), Tech run TR(1, RH and BS and CO), Opening night ON(1, TR and PR).
Set up: What are we solving for?
(i) Perform both scans and produce a clean table with floats. 2 marks
(ii) State the critical path and the project duration. 2 marks
(iii) Which activity has the largest float and how many weeks of slack does it have? 2 marks
Stuck? Revisit lesson § Float — compute LS − ES for every activity and pick the largest.Problem 5 — Council playground upgrade (regional VIC)
A council schedules a 10-week playground upgrade (weeks):
Design DS(2, —), Tender TE(2, DS), Award contract AC(1, TE), Demo old equipment DM(2, AC), Soft fall surface SF(3, DM), Equipment install EI(4, DM), Fencing FC(2, AC), Landscaping LS(3, EI and SF and FC), Inspection IN(1, LS).
Set up: What are we solving for?
(i) Perform both scans and produce a clean table with floats. 2 marks
(ii) State the critical path and the project duration. 2 marks
(iii) Council asks: "If FC (fencing) is delayed by 4 weeks, will the upgrade finish late?" Use your float to answer with one short sentence. 2 marks
Stuck? Revisit lesson § Float — if delay > slack, the project moves out by (delay − slack).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Apartment-block fit-out
Set up. Both scans, identify critical path, give C's slack.
(i) Forward: A(0,3), B(3,7), C(3,5), D(7,12), E(5,8), F(max(12,8)=12, 14), G(14, 15). Duration = 15.
Backward (G LF=15): G LS=14. F LF=14, 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.
Table (Float = LS − ES):
A | 3 | 0 | 3 | 0 | 3 | 0
B | 4 | 3 | 7 | 3 | 7 | 0
C | 2 | 3 | 5 | 7 | 9 | 4
D | 5 | 7 | 12 | 7 | 12 | 0
E | 3 | 5 | 8 | 9 | 12 | 4
F | 2 | 12 | 14 | 12 | 14 | 0
G | 1 | 14 | 15 | 14 | 15 | 0
(ii) Critical path: A → B → D → F → G, duration 15 days.
(iii) C has float = 4 days. C can be delayed up to 4 days before it becomes critical (delaying by more pushes the project out).
Problem 2 — Excursion planning
Set up. Both scans, critical path, BT delay analysis.
(i) Forward: PS(0,2), BT(2,5), BV(2,6), PD(max(5,6)=6, 7), RA(6,8), FB(max(7,8)=8, 9). Duration = 9.
Backward (FB LF=9): FB LS=8. PD LF=8, LS=7. RA LF=8, LS=6. BT LF=7, LS=4. BV LF=min(7,6)=6, LS=2. PS LF=min(4,2)=2, LS=0.
Floats: PS=0, BT=2, BV=0, PD=1, RA=0, FB=0.
(ii) Critical path: PS → BV → RA → FB, duration 9 weeks.
(iii) BT slack = 2 weeks. A 1-week delay (1 ≤ 2) is absorbed without affecting the final brief. Yes — BT can be delayed by 1 week with no impact.
Problem 3 — Clinical drug trial
Set up. Single-chain pipeline; both scans; LA shortening analysis.
(i) Forward: PT(0,4), ET(4,10), RC(10,18), DP(18,28), FU(28,32), LA(32,37), SA(37,40), RP(40,42). Duration = 42.
Backward: every LF = next LS, so every activity has LS = ES (zero float).
Floats: all 0.
(ii) Critical path: PT → ET → RC → DP → FU → LA → SA → RP. Every activity is critical because the network is a single chain — there is only one path from start to finish, so every activity lies on it.
(iii) LA is critical (float 0). Shortening it by 2 weeks shortens the only path by 2 weeks, and no other path can take over (there is no other path). Yes — the trial finishes 2 weeks earlier, in 40 weeks.
Problem 4 — Theatre production
Set up. Both scans, critical path, largest float.
(i) Forward: CP(0,1), CA(1,3), RH(3,11), BS(1,7), CO(3,8), PR(1,5), TR(max(11,7,8)=11, 12), ON(max(12,5)=12, 13). Duration = 13.
Backward (ON LF=13): ON LS=12. TR LF=12, LS=11. PR LF=12, LS=8. RH LF=11, LS=3. BS LF=11, LS=5. CO LF=11, LS=6. CA LF=min(3,6)=3, LS=1. CP LF=min(1,5,1)=1, LS=0.
Floats: CP=0, CA=0, RH=0, BS=4, CO=3, PR=7, TR=0, ON=0.
(ii) Critical path: CP → CA → RH → TR → ON, duration 1 + 2 + 8 + 1 + 1 = 13 weeks.
(iii) Largest float = PR with 7 weeks slack (it can start as late as week 8 and still finish by week 12 in time for ON).
Problem 5 — Council playground upgrade
Set up. Both scans, critical path, FC delay impact.
(i) Forward: DS(0,2), TE(2,4), AC(4,5), DM(5,7), SF(7,10), EI(7,11), FC(5,7), LS(max(11,10,7)=11, 14), IN(14,15). Duration = 15.
Backward (IN LF=15): IN LS=14. LS LF=14, LS=11. EI LF=11, LS=7. SF LF=11, LS=8. FC LF=11, LS=9. DM LF=min(7,8)=7, LS=5. AC LF=min(5,9)=5, LS=4. TE LF=4, LS=2. DS LF=2, LS=0.
Floats: DS=0, TE=0, AC=0, DM=0, SF=1, EI=0, FC=4, LS=0, IN=0.
(ii) Critical path: DS → TE → AC → DM → EI → LS → IN, duration 2 + 2 + 1 + 2 + 4 + 3 + 1 = 15 weeks.
(iii) FC has float = 4 weeks. A 4-week delay exactly uses up all of FC's slack — FC becomes critical but the project still finishes on time. No — the upgrade does not finish late, but FC now sits on a second critical path; any further FC delay would push the project out.