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.