Mathematics Standard • Year 12 • Module 6 • Lesson 11
Mixed Project Problems — Past-Paper Style
Practise HSC Mathematics Standard 2-style writing on multi-step project problems combining scans, critical path and crashing.
1. Short-answer questions
1.1 A project has: A(3,−), B(4,A), C(2,A), D(5,B), E(3,C), F(2,D,E). (a) Carry out a forward scan. (b) State the project duration. 3 marks Band 3
1.2 Same network as 1.1. (a) Carry out a backward scan. (b) State the critical path and the floats of non-critical activities. 3 marks Band 3-4
1.3 Same network as 1.1. Crash options: D normal 5 → crash 4 days at +$500; F normal 2 → crash 1 day at +$300.
(a) Calculate the new project duration if both crashes are applied.
(b) State the total crash cost.
(c) Indirect cost is $200/day. Was the crash worthwhile? Justify in one sentence. 4 marks Band 4
2. Extended response
2.1 A construction company is fitting out a NSW retail premises.
Activity table (days):
A. Strip-out (3, no predecessor)
B. Slab repair (4, A)
C. Wall framing (2, A)
D. Plumbing (5, B)
E. Electrical (3, C)
F. Drywall (2, D and E)
G. Fit-out (4, F)
Crash options: D 5→3 at +$1,000 total; F 2→1 at +$400; G 4→3 at +$300. Indirect cost = $300/day. The shop must open within 14 days or a $500/day late-opening penalty applies.
(a) Carry out forward and backward scans for all 7 activities. State the project duration and the critical path. (b) Calculate the total cost at the normal project duration (direct + indirect + any penalty). Use direct cost $25,000. (c) Find the minimum-cost duration. Show your working in a table comparing the normal duration plus three crash options. End with an explicit conclusion sentence naming the optimal duration and the minimum total cost. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — correct forward scan (ES, EF for all 7 activities).
• 1 mark — correct backward scan (LS, LF) and critical path stated as A → B → D → F → G.
• 1 mark — project duration stated as 18 days.
Part (b) — 2 marks
• 1 mark — indirect cost computed at 18 × $300 = $5,400; penalty 4 × $500 = $2,000.
• 1 mark — total cost stated as $32,400 with all three components shown.
Part (c) — 2 marks
• 1 mark — table of total costs at progressively crashed durations including 16, 15, 14 days, with cheapest crashes selected (G $300, F $400, D $500/day).
• 1 mark — explicit conclusion sentence naming the optimal duration (14 days) and total cost ($30,900).
Your response:
Stuck on (c)? Above day 14, penalty bites — savings per day = $300 + $500 = $800. At/below day 14, only $300/day saving.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Forward scan (3 marks)
Sample response.
A: ES=0, EF=3.
B: ES=3, EF=7.
C: ES=3, EF=5.
D: ES=7, EF=12.
E: ES=5, EF=8.
F: ES = max(12, 8) = 12, EF = 14.
Project duration = 14 days.
Marking notes. 1 mark — A, B, C correct. 1 mark — D, E, F correct (including max at F). 1 mark — duration stated.
1.2 — Backward scan + critical path (3 marks)
Sample response.
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.
Floats: A=0, B=0, C=4, D=0, E=4, F=0. Critical path: A → B → D → F. C float = 4, E float = 4.
Marking notes. 1 mark — backward scan all six. 1 mark — A: LF = min(3,7) = 3. 1 mark — critical path and floats stated correctly.
1.3 — Crashing D + F (4 marks)
(a) Both D and F critical. D 5→4 reduces critical by 1. F 2→1 reduces by another 1. New duration = 12 days.
(b) Crash cost = $500 + $300 = $800.
(c) Indirect saving = 2 × $200 = $400. Net = $800 − $400 = +$400. No — the crash makes the project $400 more expensive overall. Worth it only if there's an external benefit (deadline, penalty avoided) not stated here.
Marking notes. (a) 1 mark — new duration. (b) 1 mark — crash cost. (c) 2 marks — explicit comparison and conclusion sentence.
2.1 — Retail fit-out (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Scans + critical path.
Forward: A(0,3), B(3,7), C(3,5), D(7,12), E(5,8), F(max(12,8)=12, 14), G(14,18). [1 mark]
Backward: G(18,14), F(14,12), D(12,7), E(12,9), B(7,3), C(9,7), A(min(3,7)=3, 0). Floats: A=B=D=F=G=0; C=4, E=4. Critical path: A → B → D → F → G. [1 mark]
Project duration = 18 days. [1 mark]
(b) Cost at normal duration.
Direct $25,000. Indirect = 18 × $300 = $5,400. Days past 14 = 4 days. Penalty = 4 × $500 = $2,000. [1 mark — indirect and penalty]
Total cost at 18 days = $25,000 + $5,400 + $2,000 = $32,400. [1 mark]
(c) Optimal duration.
Slopes for critical activities: D $500/day (max 2 days), F $400/day (max 1 day), G $300/day (max 1 day). Cheapest first: G ($300), F ($400), D ($500).
Effective saving per day above day 14: $300 indirect + $500 penalty = $800/day. At/below day 14: $300/day only.
Table (direct $25,000 + crash + indirect + penalty):
Day 18 (no crash): $25,000 + 0 + $5,400 + $2,000 = $32,400.
Day 17 (crash G 1, $300): $25,000 + $300 + 17×$300 + 3×$500 = $25,000 + $300 + $5,100 + $1,500 = $31,900.
Day 16 (G 1 + F 1, $300 + $400 = $700): $25,000 + $700 + 16×$300 + 2×$500 = $25,000 + $700 + $4,800 + $1,000 = $31,500.
Day 15 (G 1 + F 1 + D 1, $700 + $500 = $1,200): $25,000 + $1,200 + 15×$300 + 1×$500 = $25,000 + $1,200 + $4,500 + $500 = $31,200.
Day 14 (G 1 + F 1 + D 2, $1,200 + $500 = $1,700): $25,000 + $1,700 + 14×$300 + 0 = $25,000 + $1,700 + $4,200 = $30,900.
Day 13 — only D has crashing capacity left (already used 2 of 2; F used 1 of 1; G used 1 of 1). Cannot crash further. [1 mark — table]
Conclusion: the optimal duration is 14 days at a total cost of $30,900. Crashing G, F, and D fully (with no penalty remaining) is cheaper than the normal 18-day schedule by $1,500. [1 mark — explicit conclusion with duration and total]
Total: 7/7.
Band descriptors for marker.
Band 3: Forward scan and critical path correct; cost calculation omits penalty or duration. ≈ 3 marks.
Band 4: Scans correct; total at 18 correct; attempts crashing but doesn't switch the "saving per day" calculation at day 14. ≈ 5 marks.
Band 5: Full table; identifies 14 days as optimal but conclusion sentence vague or lacks dollar value. ≈ 6 marks.
Band 6: All scans, total at 18, table with $300+$500 above day 14 and $300 below; explicit conclusion sentence naming 14 days AND $30,900. 7/7.