Mathematics Standard • Year 12 • Module 6 • Lesson 10
Cost-Time Trade-offs — Skill Drill
Build fluency in calculating cost slopes, ranking activities for crashing, and finding the minimum total cost duration when indirect costs decline with time.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Cost slope formula: Cost slope = (Crash cost − ____________) ÷ (____________ − Crash time).
Q1.2 When choosing which activity to crash first, you prioritise the activity with the (highest / lowest) cost slope: ____________.
Q1.3 If indirect costs are zero, the optimal project duration is the ____________ duration (no crashing).
2. Worked example — cost slope and one-day crash decision
Each step explains exactly what to calculate and why.
Problem. Activity Z: normal 10 days at $5,000; crash 7 days at $6,200. (a) Calculate the cost slope. (b) If indirect cost is $500/day, is it worth crashing Z by 1 day?
Step 1 — Apply the cost slope formula.
Slope = ($6,200 − $5,000) ÷ (10 − 7) = $1,200 ÷ 3 = $400 per day saved.
Reason: the slope is rise over run — extra direct cost divided by days saved.
Step 2 — Compare to indirect saving per day.
Indirect saving per day = $500. Crash cost per day = $400.
Reason: crashing makes sense whenever direct cost added < indirect cost saved.
Step 3 — Net change per day saved.
Net saving = $500 − $400 = $100/day.
Conclusion. Yes — crash Z by 1 day. The project saves $100 net per day of crashing. Keep crashing until you hit Z's crash limit (3 days max) or indirect cost drops below the slope.
3. Faded example — fill in the missing steps
Activity W: normal 8 days at $3,000; crash 6 days at $3,800. Indirect cost = $250/day. Fill in each blank line. 4 marks
Step 1 — Cost slope:
Slope = ($3,800 − $____) ÷ (____ − 6) = $____ ÷ ____ = $____ per day.
Step 2 — Compare to indirect saving:
Crash cost per day = $____. Indirect saving = $____.
Step 3 — Is crashing worthwhile?
Net per day = $____ − $____ = $____.
Conclusion. Crash W: (Yes / No, choose one) ____________. Net change for 2 days of crashing = $____.
4. Graduated practice — Cost-slope skills
Foundation — single-activity slopes (4 questions)
| Q | Problem | Answer |
|---|---|---|
| 4.1 1 | Normal 10 days $4,000; crash 8 days $4,600. Cost slope? | |
| 4.2 1 | Normal 6 days $2,000; crash 4 days $2,800. Cost slope? | |
| 4.3 1 | Activity X cost slope = $300/day; indirect = $250/day. Worth crashing? (Yes / No) | |
| 4.4 1 | Activity Y cost slope = $200/day; indirect = $400/day. Worth crashing? (Yes / No) |
Standard — multi-activity ranking (6 questions)
Show each slope, then state which activity to crash first.
4.5 Activities A (normal 7, crash 5, +$800) and B (normal 5, crash 4, +$600). Calculate both slopes and state which is cheaper to crash. 2 marks
4.6 Three activities: P (slope $300/day), Q (slope $250/day), R (slope $200/day). Rank them in the order they should be crashed first. 2 marks
4.7 Project normal 18 days at $12,000. Indirect $400/day. Activity A slope $500/day, max 2-day crash. Calculate total cost at 18 days, then total cost if A is crashed by 2 days. 3 marks
4.8 Project normal 20 days $10,000. Indirect $300/day. Activity X slope $400/day, max 3 days. Activity Y slope $250/day, max 2 days. Reduce to 16 days at minimum cost. State which crashes and the total cost. 3 marks
4.9 Project normal cost $8,000. Indirect $500/day. Activity Z has slope $300/day. How many days should Z be crashed if max crash is 4 days? Justify in one sentence. 2 marks
4.10 A penalty clause adds $1,000 for every day the project runs past day 15. The project's normal duration is 17 days. Cost slope of the cheapest critical activity is $400/day. Should the manager crash to day 15? Show working. 2 marks
Extension — find the optimum (2 questions)
4.11 Project normal 22 days $15,000. Indirect $500/day. A: slope $500/day max 2 days. B: slope $800/day max 1 day. C: slope $600/day max 1 day. Find the optimal duration and minimum total cost. Show your work in a small table. 3 marks
4.12 A project has a curve of total cost vs duration: it drops from 25 days (high indirect) down to a minimum around 21 days, then rises again as crashing becomes expensive. Explain in 2 sentences what the rising portion represents and when you would stop crashing. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Slope formula
Slope = (Crash cost − Normal cost) ÷ (Normal time − Crash time).
Q1.2 — Crash priority
Lowest slope first (cheapest per day saved).
Q1.3 — Zero indirect costs
Normal duration — without indirect costs there is no financial reason to crash.
Q3 — Faded example
Step 1: Slope = ($3,800 − $3,000) ÷ (8 − 6) = $800 ÷ 2 = $400 per day.
Step 2: Crash cost/day = $400. Indirect saving = $250.
Step 3: Net = $250 − $400 = −$150/day (loses money).
Conclusion: No, don't crash. 2-day crash would cost net = 2 × −$150 = −$300 (loss of $300).
Q4.1 — Slope
($4,600 − $4,000) ÷ (10 − 8) = $600 ÷ 2 = $300/day.
Q4.2 — Slope
($2,800 − $2,000) ÷ (6 − 4) = $800 ÷ 2 = $400/day.
Q4.3 — Crash decision X
Slope $300 > indirect $250: crashing costs more than it saves. No.
Q4.4 — Crash decision Y
Slope $200 < indirect $400: crashing saves $200 net per day. Yes.
Q4.5 — A vs B
A: $800 ÷ 2 = $400/day. B: $600 ÷ 1 = $600/day. A is cheaper to crash.
Q4.6 — Rank crash order
Lowest first: R ($200), Q ($250), P ($300).
Q4.7 — 18-day project + 2-day crash
At 18 days: $12,000 + 18 × $400 = $12,000 + $7,200 = $19,200.
Crash A by 2: extra = 2 × $500 = $1,000. Duration = 16. Indirect = 16 × $400 = $6,400. Total = $12,000 + $1,000 + $6,400 = $19,400.
Net change: $19,400 − $19,200 = +$200 (worse). Don't crash.
Q4.8 — Reduce 20 → 16 (4 days)
Y is cheaper ($250/day) vs X ($400/day). Y max = 2 days. So crash Y by 2 days ($500), then X by 2 days ($800).
Total crash cost = $500 + $800 = $1,300.
Indirect at 16 days = 16 × $300 = $4,800.
Total at 16 days = $10,000 + $1,300 + $4,800 = $16,100.
Q4.9 — Z slope $300, indirect $500
Net saving per day = $500 − $300 = $200/day. Worth crashing for all 4 days. Crash Z by 4 days (its maximum). Saves 4 × $200 = $800 net.
Q4.10 — Penalty clause
Without crashing: project runs 17 days. Penalty = 2 days × $1,000 = $2,000.
Crash to day 15 (2 days): cost = 2 × $400 = $800. Penalty = $0.
Saving = $2,000 − $800 = $1,200. Yes — crash to day 15.
Q4.11 — Optimal duration
Crash order (cheapest first): A ($500), C ($600), B ($800). All slopes > indirect of $500 means crashing never saves (slope ≥ $500). Crash A: cost $500/day, indirect saving $500/day, net $0. Break-even.
Table:
Duration 22: total = $15,000 + 22 × $500 = $26,000.
Crash A 1 day → 21: $15,000 + $500 + 21×$500 = $26,000 (same).
Crash A 2 days → 20: $15,000 + $1,000 + 20×$500 = $26,000.
Crash A 2 + C 1 → 19: $15,000 + $1,000 + $600 + 19×$500 = $26,100 (worse).
Optimal duration = 20, 21 or 22 days, total cost = $26,000 (any duration from 22 down to 20 is equally good; pick 20 for fastest delivery at no extra cost).
Q4.12 — Cost curve rising portion
The rising portion shows where crash costs exceed indirect savings — each extra day saved costs more than the saving. Stop crashing at the lowest point of the curve: when the next-cheapest crash slope > indirect cost per day.