Mathematics Standard • Year 12 • Module 6 • Lesson 10
Cost-Time Trade-offs — Past-Paper Style
Practise HSC Mathematics Standard 2-style writing on cost slopes, indirect costs and minimum-cost duration.
1. Short-answer questions
1.1 An activity has a normal duration of 9 days at a cost of $4,500, and can be crashed to 6 days at a cost of $5,700. Calculate the cost slope per day saved. 2 marks Band 3
1.2 A project's normal duration is 18 days at a direct cost of $12,000. Indirect costs are $400 per day. A single critical activity, A, can be crashed by up to 2 days at a slope of $500/day.
(a) Calculate the total project cost at 18 days.
(b) Calculate the total project cost if A is crashed by 2 days.
(c) Should A be crashed? Justify in one sentence. 4 marks Band 3-4
1.3 A council infrastructure project has normal duration 20 days, direct cost $80,000, indirect $1,000/day. Two critical activities can be crashed: P (slope $600/day, max 3 days) and Q (slope $1,200/day, max 2 days).
(a) Which activity should be crashed first? Justify.
(b) Calculate the total cost at the optimal duration.
(c) Explain in one sentence why Q should not be crashed (or only partially crashed). 4 marks Band 4
2. Extended response
2.1 A NSW concert promoter is upgrading a regional venue. Normal project duration is 24 days. Direct cost is $90,000. Indirect cost (site overhead + insurance) is $700 per day. A penalty of $1,500 per day applies for every day the project runs past day 22 (a contracted opening date). Three critical activities can be crashed:
Stage build — normal 8 days, crash 6 days, cost normal $20,000 / crash $22,000
Wiring — normal 6 days, crash 4 days, cost normal $15,000 / crash $17,800
Acoustic treatment — normal 4 days, crash 3 days, cost normal $10,000 / crash $10,900
(a) Calculate the cost slope for each activity. (b) Calculate the total project cost at the normal 24-day duration (direct + indirect + penalty). (c) Find the minimum-cost duration. Show your working using a small comparison table for durations 24, 23, 22, 21, 20. Write a clear conclusion sentence stating the optimal duration and total cost. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — Stage build slope $1,000/day; Wiring slope $1,400/day.
• 1 mark — Acoustic treatment slope $900/day.
Part (b) — 1 mark
• 1 mark — total at 24 days = $90,000 + 24×$700 + 2×$1,500 = $109,800.
Part (c) — 4 marks
• 1 mark — ranks activities cheapest-to-most-expensive (Acoustic $900, Stage $1,000, Wiring $1,400).
• 1 mark — recognises that indirect+penalty saving = $700+$1,500 = $2,200/day above day 22 (and $700/day below day 22).
• 1 mark — produces a table with cost at 24, 23, 22, 21, 20 with correct totals.
• 1 mark — explicit conclusion sentence stating optimal duration (22 days) and total cost ($106,000).
Your response:
Stuck on (c)? Once you hit day 22 (no more penalty), the only saving from further crashing is $700/day indirect. Compare each slope to $700.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Slope (2 marks)
Sample response. Slope = ($5,700 − $4,500) ÷ (9 − 6) = $1,200 ÷ 3 = $400 per day.
Marking notes. 1 mark — correct numerator and denominator. 1 mark — correct slope with $/day units.
1.2 — Crash decision (4 marks)
(a) Total at 18 days = $12,000 + 18 × $400 = $12,000 + $7,200 = $19,200.
(b) Crash A 2 days: extra = 2 × $500 = $1,000. New duration 16. Total = $12,000 + $1,000 + 16 × $400 = $12,000 + $1,000 + $6,400 = $19,400.
(c) No — crashing A costs $500/day but saves only $400/day in indirect costs (net loss $100/day). The 2-day crash makes the project $200 more expensive.
Marking notes. (a) 1 mark — total at 18. (b) 1 mark — extra direct, 1 mark — new total. (c) 1 mark — recommendation with reason (slope > indirect).
1.3 — Two crashing activities (4 marks)
(a) P slope $600 < Q slope $1,200. P first — lower cost per day saved.
(b) Day 20: $80,000 + 20×$1,000 = $100,000.
Crash P 3 days (max): cost $1,800. Day 17: $80,000 + $1,800 + 17×$1,000 = $98,800.
Further crash Q: slope $1,200 > indirect $1,000 — net loss. Stop.
Optimal duration = 17 days. Total cost = $98,800.
(c) Q's slope ($1,200/day) exceeds the indirect saving ($1,000/day), so each day of Q crashing makes the project $200 more expensive.
Marking notes. (a) 1 mark. (b) 2 marks — optimal duration and total cost. (c) 1 mark — explanation.
2.1 — Concert venue upgrade (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Cost slopes.
Stage build: ($22,000 − $20,000) ÷ (8 − 6) = $2,000 ÷ 2 = $1,000/day.
Wiring: ($17,800 − $15,000) ÷ (6 − 4) = $2,800 ÷ 2 = $1,400/day. [1 mark — Stage and Wiring]
Acoustic: ($10,900 − $10,000) ÷ (4 − 3) = $900 ÷ 1 = $900/day. [1 mark — Acoustic]
(b) Total at 24 days.
Direct $90,000 + indirect 24×$700 = $16,800 + penalty 2×$1,500 = $3,000.
Total = $109,800. [1 mark]
(c) Optimal duration.
Slopes ranked cheapest first: Acoustic $900, Stage $1,000, Wiring $1,400. [1 mark — ranking]
Effective saving per day above day 22: $700 indirect + $1,500 penalty = $2,200/day. Below or at day 22: only $700/day. [1 mark — recognising shift]
Table of total cost (direct + crash + indirect + penalty):
Day 24: $90,000 + 0 + 24×$700 + 2×$1,500 = $109,800.
Day 23 (crash Acoustic 1 day, $900): $90,000 + $900 + 23×$700 + 1×$1,500 = $90,000 + $900 + $16,100 + $1,500 = $108,500.
Day 22 (crash Acoustic 1 + Stage 1, $900 + $1,000 = $1,900): $90,000 + $1,900 + 22×$700 + 0 = $90,000 + $1,900 + $15,400 = $107,300.
Day 21 (Acoustic 1 + Stage 2, $900 + $2,000 = $2,900): $90,000 + $2,900 + 21×$700 = $90,000 + $2,900 + $14,700 = $107,600.
Day 20 (Acoustic 1 + Stage 2 + Wiring 1, $900 + $2,000 + $1,400 = $4,300): $90,000 + $4,300 + 20×$700 = $90,000 + $4,300 + $14,000 = $108,300. [1 mark — table correct]
Conclusion: the optimal duration is 22 days with a total cost of $107,300. Crashing past day 22 loses $700/day in indirect saving against slope costs of $1,000+, so the savings stop once the penalty is eliminated. [1 mark — explicit conclusion with duration and total]
Total: 7/7.
Band descriptors for marker.
Band 3: Calculates slopes but only one cost figure (e.g. only Day 24). ≈ 3 marks.
Band 4: Slopes correct, day-24 total correct, attempts crash but doesn't switch to $700 saving once penalty is gone. ≈ 5 marks.
Band 5: Full table; identifies 22 days as optimal but conclusion sentence missing dollar value. ≈ 6 marks.
Band 6: All slopes; effective-saving change recognised at day 22; complete table; conclusion sentence names both 22 days AND $107,300. 7/7.