Mathematics Standard • Year 12 • Module 6 • Lesson 10
Cost-Time Trade-offs — Problem Set
Apply cost-slope and indirect-cost reasoning to real construction, festival, manufacturing and council budget decisions.
Problem 1 — Townhouse build with daily site overhead
A Sydney builder has normal project duration of 22 days at a direct cost of $50,000. Site overheads (scaffold hire, council fees, supervisor) total $600 per day. Three critical activities can be crashed:
Framing — normal 10 days $8,000, crash 8 days $9,200
Roofing — normal 6 days $6,000, crash 5 days $6,500
Cladding — normal 6 days $5,000, crash 4 days $5,800
Set up: What are we solving for?
(i) Calculate the cost slope (extra cost per day saved) of each activity. 2 marks
(ii) Rank the three activities cheapest-to-most-expensive per day of crash. 1 mark
(iii) The builder wants to cut 3 days from the project. State which days to take from which activities, the extra direct cost, the indirect cost saving, and the net change in total project cost. 3 marks
Stuck? Revisit lesson § Cost Slope — calculate all slopes first, then choose the cheapest day-by-day.Problem 2 — Wedding venue penalty contract
A wedding venue has booked a band for a fixed date 14 days from project start. Reno work has normal duration 17 days, direct cost $12,000. Site overhead $200/day. If the venue is not ready by day 14, a $1,500 per day penalty applies.
Crash options on the critical path:
Painting — slope $300/day, max 2 days
Cabinetry — slope $700/day, max 2 days
Flooring — slope $500/day, max 1 day
Set up: What are we solving for?
(i) Calculate total cost at the normal 17-day schedule (direct + overhead + penalty). 2 marks
(ii) Find the minimum-cost crash plan to deliver the project on day 14 (no penalty). Show your crash choices, costs, and the new total. 3 marks
(iii) Compare the two totals and recommend which to choose, with a clear conclusion sentence. 2 marks
Stuck? Revisit lesson § Optimal Duration — compare normal schedule (with penalty) against crashed schedule (extra direct cost, no penalty).Problem 3 — Festival stage rental crunch
A festival's main-stage build has normal duration 8 days, direct cost $22,000. Rental on the staging equipment is $1,200 per day. Three crashing options:
Truss assembly — slope $400/day, max 1 day
Lighting rig — slope $900/day, max 2 days
Sound system — slope $1,500/day, max 1 day
Set up: What are we solving for?
(i) Which activities have a positive net saving when crashed (slope < $1,200/day indirect)? 2 marks
(ii) Find the optimal duration that minimises total cost (direct + indirect + crash). Show working in a small table comparing 8, 7, 6 and 5 days. 3 marks
(iii) State the optimal duration and minimum total cost in a clear conclusion sentence. 2 marks
Stuck? Revisit lesson § Optimal Duration — crash one day at a time and re-total cost; stop where total starts rising.Problem 4 — Country council bridge repair
A NSW council is repairing a bridge. Normal duration 30 days, direct cost $200,000. While the bridge is closed, drivers detour an extra 20 km — the council assigns a social cost of $1,000 per day of closure. Crash options:
Excavation — normal 8 days $40,000, crash 6 days $44,000
Concrete — normal 10 days $60,000, crash 8 days $66,000
Resurfacing — normal 6 days $30,000, crash 4 days $34,000
Set up: What are we solving for?
(i) Calculate slopes for each activity. 2 marks
(ii) Find the optimal duration to minimise total social + direct cost. Show the cost at 30, 28, 26, 24 days. 3 marks
(iii) State the optimal duration and write a recommendation to council in one sentence. 2 marks
Stuck? Revisit lesson § Optimal Duration — keep crashing while slope < $1,000/day social cost.Problem 5 — Software release with launch deadline
A team has normal 25-day release timeline at $40,000 direct cost. Cloud hosting and team costs accumulate $800/day. A competitor is launching on day 20 — every day later than day 20 the team loses $1,200 of market-share value.
Crash options on the critical path:
Backend coding — slope $500/day, max 2 days
QA testing — slope $300/day, max 3 days
Deployment — slope $1,000/day, max 1 day
Set up: What are we solving for?
(i) Total effective cost saving per day if the project lands on or before day 20. 1 mark
(ii) Find the minimum-cost crash plan to deliver on day 20 (cutting 5 days). State which activities to crash and by how much, and the total extra direct cost. 3 marks
(iii) Compare total cost at 25 days (with 5 days market-share loss) vs at 20 days. Recommend the right plan in one sentence. 3 marks
Stuck? Revisit lesson § Optimal Duration — compare two total-cost figures (25-day with penalty vs 20-day with crashes).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Townhouse build
Set up. We calculate cost slopes, rank them, then pick the cheapest 3 days of crash.
(i) Framing: ($9,200 − $8,000) ÷ 2 = $600/day. Roofing: ($6,500 − $6,000) ÷ 1 = $500/day. Cladding: ($5,800 − $5,000) ÷ 2 = $400/day.
(ii) Cheapest first: Cladding ($400), Roofing ($500), Framing ($600).
(iii) Cut 3 days = Cladding 2 days ($800) + Roofing 1 day ($500) = $1,300 extra direct cost. Indirect saving = 3 × $600 = $1,800. Net saving = $1,800 − $1,300 = $500 cheaper. Crash plan is worthwhile.
Problem 2 — Wedding venue
Set up. We compare staying at 17 days (3 days of penalty) vs crashing 3 days to hit day 14 (no penalty).
(i) At 17 days: direct $12,000 + overhead 17×$200 = $3,400 + penalty 3×$1,500 = $4,500. Total = $19,900.
(ii) Need 3-day cut. Cheapest first: Painting 2 days × $300 = $600. Then Flooring 1 day × $500 = $500. Total crash cost = $1,100.
At 14 days: direct $12,000 + crash $1,100 + overhead 14×$200 = $2,800 + penalty $0. Total = $15,900.
(iii) Crashed plan saves $19,900 − $15,900 = $4,000. Recommendation: crash 3 days to deliver on day 14, saving $4,000 overall.
Problem 3 — Festival stage rental
Set up. We identify which activities are worth crashing (slope < $1,200), then table the total cost at each duration.
(i) Truss $400 < $1,200 (worth crashing). Lighting $900 < $1,200 (worth crashing). Sound $1,500 > $1,200 (NOT worth crashing).
(ii) Table:
Day 8: $22,000 + 8×$1,200 = $31,600.
Day 7 (crash Truss 1 day, $400): $22,000 + $400 + 7×$1,200 = $30,800.
Day 6 (crash Truss + Lighting 1, $400 + $900 = $1,300): $22,000 + $1,300 + 6×$1,200 = $30,500.
Day 5 (Truss 1 + Lighting 2 = $400 + $1,800 = $2,200): $22,000 + $2,200 + 5×$1,200 = $30,200.
(iii) Optimal duration = 5 days, total cost = $30,200. Crash Truss by 1 day and Lighting by 2 days; do not crash Sound (slope $1,500 exceeds the indirect $1,200/day saving).
Problem 4 — Bridge repair
Set up. We compute slopes, then table the total social + direct cost at each duration.
(i) Excavation: $4,000 ÷ 2 = $2,000/day. Concrete: $6,000 ÷ 2 = $3,000/day. Resurfacing: $4,000 ÷ 2 = $2,000/day.
(ii) All slopes ≥ $2,000 > indirect saving $1,000/day. No crashing is worthwhile. Day 30: $200,000 + 30×$1,000 = $230,000. Day 28 (crash Excav 2 days, $4,000): $200,000 + $4,000 + 28×$1,000 = $232,000. Day 26 (Excav 2 + Resurfacing 2 = $8,000): $200,000 + $8,000 + 26×$1,000 = $234,000. Day 24 (Excav 2 + Resurfacing 2 + Concrete 2 = $14,000): $200,000 + $14,000 + 24×$1,000 = $238,000. Costs rise as you crash.
(iii) Optimal duration = 30 days (normal). Recommendation to council: do not crash — every crash slope exceeds the $1,000/day social cost, so the cheapest plan is the standard 30-day schedule at $230,000.
Problem 5 — Software release
Set up. We compute effective saving per day saved after day 20, then check whether a 5-day crash plan beats the no-crash plan with 5 days of market-share loss.
(i) Effective saving per day = $800 (cloud + team) + $1,200 (market share) = $2,000/day for any day saved between 25 and 20. For days saved below 20, only the $800 indirect saving applies.
(ii) Need 5-day cut. Cheapest first: QA testing 3 days × $300 = $900. Backend 2 days × $500 = $1,000. (Deployment $1,000 not needed.) Total crash cost = $1,900.
(iii) Day 25 total: $40,000 + 25×$800 + 5×$1,200 = $40,000 + $20,000 + $6,000 = $66,000. Day 20 total: $40,000 + $1,900 + 20×$800 + 0 = $40,000 + $1,900 + $16,000 = $57,900. Crash to day 20 — saves $66,000 − $57,900 = $8,100.