Mathematics Standard • Year 12 • Module 6 • Lesson 9
Scheduling with Constraints — Past-Paper Style
Practise HSC Mathematics Standard 2-style writing on lag, lead and resource constraints — multi-mark short answers and one structured extended response.
1. Short-answer questions
1.1 A network has: A(3,−), B(4, A lag 2), C(2, B). Compute ES and EF for each activity and state the project duration. 3 marks Band 3
1.2 A network has: A(5,−), B(3, A), C(4, A lead 2), D(2, B and C). (a) Compute ES and EF for all four activities. (b) State the project duration. (c) On which days do A and C overlap? 4 marks Band 3-4
1.3 A renovation requires Paint (3 days), then a 2-day drying lag, then Hang fixtures (1 day). A separate Inspection (1 day) is a successor of Hang fixtures with NO lead. The builder claims the lag can be removed by using fast-drying paint, saving 2 days.
(a) State the project duration with the 2-day lag.
(b) State the project duration if the lag is removed.
(c) In one sentence, explain whether removing a lag is the same as "crashing" an activity. 4 marks Band 4
2. Extended response
2.1 A regional NSW developer is building a small medical centre. The activity network has:
Site prep (Sp): 3 days, no predecessor
Slab pour (Sl): 2 days, predecessor Sp
Cure (lag of 4 days after Sl)
Frame (F): 5 days, starts after cure
Plumbing rough (P): 3 days, predecessor F
Electrical rough (E): 3 days, predecessor F with LEAD 1 day (E starts 1 day before F finishes)
Final fitout (Fit): 2 days, predecessors P and E
(a) Compute ES and EF for each activity, treating the cure as a 4-day lag.
(b) Identify whether E and F overlap, and if so, by how many days.
(c) State the project duration and identify the critical path. The developer asks: "If we used quick-cure concrete to drop the lag to 2 days, would this shorten the project, and by how much?" Provide the new duration and write a clear conclusion sentence. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — Sp, Sl, F correctly computed including the 4-day curing lag between Sl and F.
• 1 mark — P correctly computed.
• 1 mark — E correctly computed using the 1-day LEAD (E starts 1 day before F finishes), and Fit computed using max of EF(P), EF(E).
Part (b) — 1 mark
• 1 mark — overlap of E and F correctly identified as 1 day (the lead).
Part (c) — 3 marks
• 1 mark — original duration correctly stated.
• 1 mark — new duration with 2-day cure correctly computed (saving 2 days).
• 1 mark — explicit conclusion sentence stating the project shortens by 2 days, naming both old and new totals.
Your response:
Stuck on (c)? Recompute Sp + Sl + (lag) + F + P + Fit. Compare lag = 4 vs lag = 2.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Lag in a small chain (3 marks)
Sample response. A: ES = 0, EF = 3. B: ES = 3 + 2 = 5, EF = 5 + 4 = 9. C: ES = 9, EF = 9 + 2 = 11. Duration = 11 days.
Marking notes. 1 mark — A. 1 mark — B with lag applied correctly (ES = EF(A) + lag). 1 mark — C and duration.
1.2 — Lead and merge (4 marks)
(a) A: 0→5. B: 5→8. C: ES = 5 − 2 = 3, EF = 3 + 4 = 7. D: ES = max(8, 7) = 8, EF = 10.
(b) Duration = 10 days.
(c) A runs 0–5; C runs 3–7. Overlap on days 3 to 5 (2 days).
Marking notes. (a) 1 mark — A, B. 1 mark — C with lead. 1 mark — D merge. (b) 1 mark — duration and overlap stated correctly.
1.3 — Lag vs crashing (4 marks)
(a) Paint 0→3. Lag 2 (wait until day 5). Hang fixtures 5→6. Inspection 6→7. Duration = 7 days.
(b) Lag removed: Paint 0→3. Hang fixtures 3→4. Inspection 4→5. Duration = 5 days. (Saves 2 days.)
(c) No — removing a lag is not the same as crashing. Crashing speeds up an activity using extra resources; removing a lag eliminates a mandatory waiting period (here, by switching to a fast-drying material). They both reduce duration but through different mechanisms.
Marking notes. (a) 1 mark — duration 7. (b) 1 mark — duration 5. (c) 2 marks — explains "remove lag = change process / material" vs "crash = add resources to existing activity".
2.1 — Medical centre (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) ES and EF for each activity.
Sp: ES=0, EF=3.
Sl: ES=3, EF=5.
Cure lag = 4 days, so F cannot start until day 5 + 4 = 9.
F: ES=9, EF=14. [Sp, Sl, F including lag — 1 mark]
P: ES=EF(F)=14, EF=14+3=17. [P — 1 mark]
E with lead 1: ES=EF(F) − 1 = 14 − 1 = 13. EF=13+3=16.
Fit: ES = max(EF(P), EF(E)) = max(17, 16) = 17. EF=17+2=19. [E with lead + Fit merge — 1 mark]
(b) Overlap of E and F.
E starts on day 13 (during F, which runs 9–14). E and F overlap on day 13 to day 14 — overlap = 1 day. [1 mark]
(c) Critical path, original duration, and effect of shorter cure.
Original duration = 19 days. Critical path is Sp → Sl → (cure lag) → F → P → Fit. [1 mark — original duration]
With 2-day cure (lag = 2 instead of 4):
Sp: 0→3. Sl: 3→5. F starts at 5 + 2 = 7, ends at 12. P: 12→15. E (lead 1): 11→14. Fit: max(15, 14) = 15 → 17.
New duration = 17 days. Saves 19 − 17 = 2 days. [1 mark — correct new duration]
Conclusion: Using quick-cure concrete shortens the project from 19 days to 17 days — a saving of 2 days. The reduction equals the change in the curing lag (4 → 2) because the lag sits on the critical path. [1 mark — explicit conclusion with both totals named]
Total: 7/7.
Band descriptors for marker.
Band 3: Computes Sp, Sl, F but forgets the 4-day lag, getting an undershoot for the duration. ≈ 3 marks.
Band 4: Lag applied correctly; lead missed (E treated as starting at EF(F)); duration computed but new duration not attempted. ≈ 5 marks.
Band 5: Full forward scan with lag and lead correct; original and new duration computed; conclusion missing or vague. ≈ 6 marks.
Band 6: Complete forward scan, overlap of 1 day stated, both durations (19 and 17) explicitly named with a clear concluding sentence about the 2-day saving. 7/7.