Mathematics Standard • Year 12 • Module 6 • Lesson 8
Multiple Critical Paths — Past-Paper Style
Practise HSC Mathematics Standard 2-style writing on identifying, analysing and crashing networks with more than one critical path.
1. Short-answer questions
1.1 A project has activities: A(3,−), B(4,A), C(4,A), D(2,B), E(2,C), F(1,D,E). List all critical paths and state the project duration. 3 marks Band 3
1.2 A network has: A(2,−), B(5,A), C(4,A), D(2,B), E(3,C), F(1,D,E).
(a) Find all critical paths and the project duration.
(b) Identify any shared critical activities.
(c) If activity C's duration increases by 1 day, what happens to the number of critical paths? 4 marks Band 3-4
1.3 A project has two critical paths sharing no common activities, each 14 days long. Three crash options exist: A on Path 1 ($300/day), B on Path 2 ($500/day), and C on Path 2 ($400/day, max 2 days). The site manager wants to reduce the project to 12 days at minimum cost. Calculate the cheapest crash plan, showing all working. 4 marks Band 4
Stuck on 1.3? With no shared activities, every cut day needs ONE activity crashed on Path 1 AND ONE on Path 2.2. Extended response
2.1 A regional NSW council is upgrading a community centre. The activity network has:
Site demo (D): 3 days, no predecessor
North-wing slab (N): 4 days, predecessor D
South-wing slab (S): 4 days, predecessor D
North-wing frame (Nf): 3 days, predecessor N
South-wing frame (Sf): 3 days, predecessor S
Roofing across both wings (R): 2 days, predecessors Nf and Sf
(a) Carry out a forward scan and state the project duration.
(b) List all critical paths and identify the shared activities.
(c) The council can crash one activity by 1 day for cost: D = $1500, N = $400, S = $400, R = $1200. They want a 1-day cut at minimum cost. Recommend an action and show all working, including a clear conclusion sentence. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct forward scan (ES and EF for all six activities).
• 1 mark — correct project duration (12 days).
Part (b) — 2 marks
• 1 mark — both critical paths listed: D → N → Nf → R and D → S → Sf → R.
• 1 mark — shared activities D and R identified.
Part (c) — 3 marks
• 1 mark — recognises that crashing N alone (or S alone) does NOT reduce duration because the other critical path stays at 12.
• 1 mark — compares the shared-activity options D ($1500) vs R ($1200) against the combined non-shared plan (N + S = $400 + $400 = $800).
• 1 mark — explicit conclusion sentence naming N + S = $800 as the cheapest plan, with clear justification it cuts both critical paths.
Your response:
Stuck on (c)? Compute the cost of every option that cuts BOTH paths by 1 day. The cheapest may be two simultaneous non-shared crashes.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Symmetric network (3 marks)
Sample response. Forward: A(0,3), B(3,7), C(3,7), D(7,9), E(7,9), F(9,10). Paths: A-B-D-F = 3+4+2+1 = 10. A-C-E-F = 3+4+2+1 = 10. Both critical. Project duration = 10 days. Shared: A, F.
Marking notes. 1 mark — forward scan. 1 mark — both paths listed (or numerically equal). 1 mark — duration stated with both paths declared critical.
1.2 — Asymmetric network (4 marks)
(a) Forward: A(0,2), B(2,7), C(2,6), D(7,9), E(6,9), F(9,10). Paths: A-B-D-F = 2+5+2+1 = 10. A-C-E-F = 2+4+3+1 = 10. Both critical. Duration = 10 days.
(b) Shared critical activities: A and F.
(c) If C increases by 1: A-C-E-F = 2+5+3+1 = 11. A-B-D-F = 10 (unchanged). Now only A-C-E-F is critical; duration = 11. Number of critical paths drops from 2 to 1.
Marking notes. (a) 1 — forward scan + paths. 1 — both critical, duration 10. (b) 1 — A, F. (c) 1 — explanation of path becoming sole critical, duration 11.
1.3 — Crash plan with disjoint critical paths (4 marks)
Sample response. No shared activities, so every day cut requires ONE activity-crash on each path.
Per-day cuts needed = 14 − 12 = 2 days.
Path 1: only A is available ($300/day). 2 days × $300 = $600.
Path 2: cheaper of B ($500) or C ($400) — use C (up to 2 days). 2 days × $400 = $800.
Total = $600 + $800 = $1,400.
Marking notes. 1 mark — recognises disjoint paths require independent crashes on both. 1 mark — correctly identifies C as cheaper than B per day on Path 2. 1 mark — correct totals on each path. 1 mark — final sum $1,400 with units.
2.1 — Community centre (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Forward scan and project duration.
D: ES=0, EF=3.
N: ES=3, EF=7. S: ES=3, EF=7.
Nf: ES=7, EF=10. Sf: ES=7, EF=10.
R: ES = max(10, 10) = 10, EF = 12. [forward scan — 1 mark]
Project duration = 12 days. [1 mark]
(b) Critical paths and shared activities.
Paths: D → N → Nf → R = 3+4+3+2 = 12. D → S → Sf → R = 3+4+3+2 = 12.
Both critical. Critical paths: D → N → Nf → R and D → S → Sf → R. [1 mark]
Shared critical activities: D and R. [1 mark]
(c) Cheapest 1-day cut.
Crashing N alone ($400) reduces D-N-Nf-R to 11 but leaves D-S-Sf-R at 12 ⇒ duration still 12. Crashing one non-shared activity alone does NOT cut the project. [1 mark]
Options that genuinely cut 1 day:
— Crash D (shared) by 1: cost = $1,500. Both paths drop to 11.
— Crash R (shared) by 1: cost = $1,200. Both paths drop to 11.
— Crash N + S simultaneously, each by 1: cost = $400 + $400 = $800. Path 1 drops by 1, Path 2 drops by 1 ⇒ both now 11. [comparison of options — 1 mark]
Conclusion: the cheapest 1-day cut is to crash N and S simultaneously by 1 day each, at a total cost of $800. This is cheaper than crashing either shared activity (D = $1,500 or R = $1,200) and is the only option that reduces both critical paths within budget. [1 mark — explicit conclusion]
Total: 7/7.
Band descriptors for marker.
Band 3: Forward scan attempted; only one critical path identified; suggests crashing N or S alone. ≈ 3 marks.
Band 4: Both critical paths found; recognises duration is 12; either misses the shared-vs-disjoint cost comparison or doesn't compute N + S = $800. ≈ 5 marks.
Band 5: Comparison of shared crash vs paired non-shared crash done numerically, but conclusion sentence missing the explicit dollar amount or activity names. ≈ 6 marks.
Band 6: Complete forward scan; both critical paths and shared activities named; cost comparison of all options; explicit conclusion sentence "Crash N and S simultaneously at $800". 7/7.