Mathematics Standard • Year 12 • Module 6 • Lesson 8
Multiple Critical Paths — Problem Set
Apply multiple-critical-path reasoning to real scheduling decisions in construction, events, software, manufacturing and emergency response.
Problem 1 — Townhouse build with two trade streams
A two-storey townhouse build has these activities (days): Site prep (3, −), Frame ground floor (4, Site prep), Frame upper floor (4, Site prep), Roof (3, Frame upper), Slab finish (3, Frame ground), Final hand-over (2, Roof and Slab finish).
Set up: What are we solving for?
(i) Perform a forward scan and compute the project duration. 2 marks
(ii) List ALL critical paths and identify the shared critical activities. 2 marks
(iii) The project manager wants to cut 1 day. Should they crash Site prep (shared) at $500 or Frame ground floor at $200? Justify. 2 marks
Stuck? Revisit lesson § Shared Activities — crashing a shared activity shortens every path it lies on.Problem 2 — Wedding catering twin-stream
A Sydney caterer is preparing for a 200-guest wedding. Activities (hours): Plan menu (2, −), Buy ingredients (3, Plan menu), Hire equipment (3, Plan menu), Prep food (4, Buy ingredients), Polish glassware (4, Hire equipment), Final plating + service start (1, Prep food and Polish glassware).
Set up: What are we solving for?
(i) Calculate the project duration using a forward scan. 2 marks
(ii) List all critical paths and shared activities. 2 marks
(iii) The caterer can shorten "Prep food" by 1 hour (extra chef, $80) or shorten "Plan menu" by 1 hour (planning meeting, $40). Which option actually reduces the project duration? Justify in one sentence. 2 marks
Stuck? Revisit lesson § Identifying Multiple Paths — a crash only reduces duration if it shortens every critical path.Problem 3 — Software release with twin test tracks
A software team is shipping a release. Activities (days): Architecture (2, −), Backend code (5, Architecture), Frontend code (5, Architecture), Backend tests (3, Backend code), Frontend tests (3, Frontend code), Deploy (1, Backend tests and Frontend tests).
Set up: What are we solving for?
(i) Forward scan, project duration, and ALL critical paths. 3 marks
(ii) The team has $1000 to spend cutting one day. Crash options: Architecture $1000 (shared), Backend code $400 (Path 1 only), Frontend code $400 (Path 2 only). What is the cheapest plan that genuinely cuts 1 day, and at what cost? 3 marks
(iii) One year later the team renames Frontend tests but the durations stay the same. If a near-critical path develops at 10 days while the critical paths are 11, what is the float of the near-critical path? 1 mark
Stuck? Revisit lesson § Near-Critical Paths — float = critical path length − path length.Problem 4 — Country show pavilion build
A regional NSW show is putting up two pavilions side by side. Activities (days): Survey site (1, −), Build pavilion A frame (4, Survey), Build pavilion B frame (4, Survey), Wire pavilion A (3, Frame A), Wire pavilion B (3, Frame B), Cladding both pavilions (2, Wire A and Wire B), Inspection (1, Cladding).
Set up: What are we solving for?
(i) Forward scan, project duration, and critical paths. 2 marks
(ii) The show committee says "this project has too much risk because there are two critical paths". Explain in your own words why multiple critical paths increase project risk. 2 marks
(iii) The committee can pay extra to crash one shared activity by 1 day. Which shared activity would give the biggest schedule reduction? Justify. 2 marks
Stuck? Revisit lesson § Risk Multiplier — every critical path is a separate route to project delay.Problem 5 — Bushfire incident response
An RFS captain is co-ordinating a 3-team response. Activities (hours): Brief teams (1, −), Team 1 contain northern flank (4, Brief), Team 2 contain southern flank (4, Brief), Team 3 evacuate residents (3, Brief), Team 1 mop-up (2, Contain north), Team 2 mop-up (2, Contain south), Final all-clear (1, both mop-ups and Evacuate).
Set up: What are we solving for?
(i) Forward scan and project duration. 2 marks
(ii) List ALL critical paths and the shared activities. 2 marks
(iii) One extra strike team can be assigned to a single mop-up to halve its time (2 hours → 1 hour). Will that reduce total response time? Explain with reference to the two critical paths. 3 marks
Stuck? Revisit lesson § Crashing With Multiple Paths — cutting only one of two equal critical paths leaves the duration controlled by the unshortened path.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Townhouse build
Set up. We are performing a forward scan, finding all critical paths, and choosing the cheaper of two crash options that actually reduces project duration.
(i) Site prep: 0→3. Frame ground: 3→7. Frame upper: 3→7. Roof: 7→10. Slab finish: 7→10. Hand-over: 10→12. Project duration = 12 days.
(ii) Paths: Site prep → Frame upper → Roof → Hand-over = 3+4+3+2 = 12. Site prep → Frame ground → Slab finish → Hand-over = 3+4+3+2 = 12. Both critical. Shared activities: Site prep and Hand-over.
(iii) Site prep ($500) is shared — crashing by 1 day shortens BOTH critical paths and the project drops to 11 days. Frame ground floor ($200) is only on one critical path — crashing it leaves the other critical path at 12, so the project duration stays at 12. Crash Site prep at $500 — it is the only option that actually cuts the project duration.
Problem 2 — Wedding catering
Set up. We are computing project duration, listing all critical paths, then deciding which crash actually shortens the duration.
(i) Plan menu 0→2. Buy 2→5. Hire 2→5. Prep food 5→9. Polish glassware 5→9. Plating 9→10. Project duration = 10 hours.
(ii) Paths: Plan → Buy → Prep → Plating = 2+3+4+1 = 10. Plan → Hire → Polish → Plating = 2+3+4+1 = 10. Both critical. Shared: Plan menu and Plating.
(iii) Plan menu ($40) is shared. Crashing it by 1 hour cuts BOTH paths by 1, so duration → 9 hours. Prep food ($80) is on only one critical path; crashing it leaves Polish glassware's path at 10. Plan menu $40 actually reduces duration — and it is cheaper too.
Problem 3 — Software release
Set up. We are finding both critical paths, choosing the cheapest crash that actually saves a day, and reading float for a near-critical path.
(i) Architecture 0→2. Backend 2→7. Frontend 2→7. Backend tests 7→10. Frontend tests 7→10. Deploy 10→11. Duration = 11 days. Critical paths: Arch → Backend → Backend tests → Deploy and Arch → Frontend → Frontend tests → Deploy. Shared: Architecture and Deploy.
(ii) Architecture $1000 (shared): cuts both paths by 1 ⇒ new duration 10 days. Backend $400 alone: Path 1 → 10, Path 2 stays 11 ⇒ no change. Frontend $400 alone: similarly no change. Backend + Frontend = $800: both paths drop to 10 ⇒ new duration 10 days. Cheapest 1-day cut = $800 (crash Backend + Frontend, one day each). The shared-activity crash costs $1000 — more expensive here. Insight: shared-activity crashing isn't always cheapest; compare cost(shared) to sum of one-per-path crashes.
(iii) Float = critical path − near-critical path = 11 − 10 = 1 day.
Problem 4 — Show pavilion
Set up. We are finding the critical paths, explaining the risk of multiple critical paths, and choosing the highest-impact shared activity to crash.
(i) Survey 0→1. Frame A 1→5. Frame B 1→5. Wire A 5→8. Wire B 5→8. Cladding 8→10. Inspection 10→11. Paths: Survey → Frame A → Wire A → Cladding → Inspection = 1+4+3+2+1 = 11. Survey → Frame B → Wire B → Cladding → Inspection = 1+4+3+2+1 = 11. Duration = 11 days. Two critical paths. Shared: Survey, Cladding, Inspection.
(ii) Each critical path is an independent route to project slippage. With one critical path, you watch one chain. With two, a delay on EITHER chain delays the project — risk doubles. Additionally, the chance of "something going wrong" on at least one of the two paths is higher than on a single path.
(iii) The shared activities are Survey (1 day), Cladding (2 days), Inspection (1 day). Cladding has the most days to potentially crash. Crashing Cladding by 1 day shortens both critical paths simultaneously — duration drops to 10 days. (Survey and Inspection at 1 day each cannot be reduced to zero realistically.)
Problem 5 — Bushfire response
Set up. We are finding the critical paths through a 3-team response, then deciding whether speeding up one mop-up cuts the total response time.
(i) Brief 0→1. Team 1 contain 1→5. Team 2 contain 1→5. Team 3 evacuate 1→4. Team 1 mop-up 5→7. Team 2 mop-up 5→7. All-clear: predecessors are both mop-ups (EF=7) and Evacuate (EF=4) → ES = max(7,7,4) = 7, EF = 8. Duration = 8 hours.
(ii) Paths to All-clear: Brief → Contain N → Mop-up N → All-clear = 1+4+2+1 = 8. Brief → Contain S → Mop-up S → All-clear = 1+4+2+1 = 8. Brief → Evacuate → All-clear = 1+3+1 = 5 (not critical, float = 3). Two critical paths. Shared: Brief and All-clear.
(iii) Halving Team 1's mop-up (2 → 1 hour) shortens only Path North to 7 hours. Path South is still 8 hours. Project duration stays at 8 hours. To get a real reduction, the extra team must speed up BOTH mop-ups, OR crash a shared activity (Brief or All-clear). Operationally: a single mop-up acceleration only helps the residents on the northern flank get an "all-clear" message earlier from that team — the official project all-clear still waits for the slower path.