Mathematics Standard • Year 12 • Module 6 • Lesson 6
Crashing and Resource Allocation — Skill Drill
Build fluency in the crash-cost-per-day formula, choosing which activities to crash, and applying float-based resource levelling — one calculation and one rule at a time.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the formula.
Crash cost per day = (Crash cost − ____________) ÷ (____________ − Crash duration)
Q1.2 Fill in the rule. "Only crash ____________ activities — crashing non-critical activities ____________ money without shortening the project."
Q1.3 Define each term in one short phrase.
Resource allocation: ____________________________________________.
Resource levelling: ____________________________________________.
Q1.4 True or false: a critical activity can be shifted within its float to smooth resource demand.
2. Worked example — choose the cheapest crash
Follow each line of working. Every step has a reason on the right.
Problem. Two critical activities on the project's critical path:
A: normal 8 days at $2,000; crash 5 days at $3,500.
B: normal 6 days at $1,500; crash 4 days at $2,500.
Reduce the project duration by 2 days at the minimum extra cost.
Step 1 — Crash cost per day for each activity.
A: ($3,500 − $2,000) ÷ (8 − 5) = $1,500 ÷ 3 = $500/day
B: ($2,500 − $1,500) ÷ (6 − 4) = $1,000 ÷ 2 = $500/day
Reason: cost rises linearly with each day shaved off, so divide total extra cost by total days saved.
Step 2 — Check both are critical.
Both A and B are stated to be critical, so crashing either will shorten the project (provided no other path becomes critical first).
Step 3 — Pick the cheapest critical activity, then crash by 2 days within its crash limit.
Both A and B cost $500/day — tie. Pick A. A's max crash = 8 − 5 = 3 days; we need 2 days, OK.
Crash A by 2 days. Extra cost = 2 × $500 = $1,000.
Conclusion. Crash A by 2 days at extra cost $1,000. (Crashing B by 2 days also costs $1,000 — either is acceptable.)
3. Faded example — fill in the crash analysis
Two critical activities:
X: normal 10 days at $8,000; crash 7 days at $9,500.
Y: normal 6 days at $5,000; crash 5 days at $5,700.
Reduce project duration by 3 days at minimum extra cost. Fill in the blanks. 4 marks
Step 1 — Crash cost per day:
X: ($____________ − $____________) ÷ (____ − ____) = $____________ ÷ ____ = $____________/day
Y: ($____________ − $____________) ÷ (____ − ____) = $____________ ÷ ____ = $____________/day
Step 2 — Which is cheaper to crash per day? ____________.
Step 3 — Max days we can crash that activity: ____ − ____ = ____ days.
Step 4 — Crash the cheaper activity by ____ days (within its limit).
Extra cost = ____ × $____________/day = $____________.
Conclusion. Crash ____ by ____ days at extra cost $____________. No need to crash the other activity.
4. Graduated practice — crash costs and resource problems
Show your working in the space below each part. For every crash calculation, write the formula line and the substitution.
Foundation — single-activity calculations (4 questions)
| Q | Problem | Answer |
|---|---|---|
| 4.1 1 | Normal 12 days at $5,000; crash 9 days at $6,500. Crash cost per day? | |
| 4.2 1 | Normal 10 days at $4,000; crash 8 days at $5,000. Crash cost per day? | |
| 4.3 1 | An activity has crash cost per day = $300 and can be crashed by up to 4 days. What is the maximum extra cost if you crash it fully? | |
| 4.4 1 | True or false: it is worth crashing a non-critical activity to shorten the project. |
Standard — typical HSC difficulty (6 questions)
Show every step. For "which to crash" questions, state the rule "crash the cheapest critical activity first" explicitly.
4.5 Two critical activities:
A: normal 8 days $400 extra to crash to 6 days.
B: normal 5 days $300 extra to crash to 4 days.
Reduce project by 2 days at minimum cost. 2 marks
4.6 Three critical activities with crash cost per day: A = $200/day, B = $350/day, C = $150/day. You need to reduce project by 3 days, and each activity can be crashed by at most 2 days. Find the minimum total extra cost. 2 marks
4.7 A non-critical activity has float 5 days and crash cost per day $200. The project needs to finish 2 days earlier. Should you crash this activity? Give a one-sentence reason. 2 marks
4.8 Why must you re-check the critical path after each round of crashing? Answer in one short sentence. 2 marks
4.9 Four activities each need 3 workers. Two are critical (no slack), two have 4 days slack. You have 8 workers in total. Explain in one or two sentences how resource levelling can keep you within 8 workers without delaying the project. 2 marks
4.10 A project has three critical paths sharing one common activity, X. X's crash cost is $400/day. An alternative activity Y is on only one critical path and costs $250/day. Which should you crash to reduce project duration by 1 day, and why? 2 marks
Extension — multi-step crashing (2 questions)
4.11 A project has critical activities:
P: normal 8 days $3,000; crash 6 days $3,800.
Q: normal 5 days $2,000; crash 3 days $3,000.
R: normal 4 days $1,500; crash 3 days $2,200.
Reduce project by 3 days at minimum total cost. Show your strategy (which to crash and by how much) and the total extra cost. 3 marks
4.12 A critical activity has normal 12 days $6,000 and can be crashed to 8 days for $7,600. If we crash by only 2 days (12 → 10), what extra cost is incurred? State the per-day cost first and then the 2-day cost. 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 — Crash cost per day formula
Crash cost per day = (Crash cost − Normal cost) ÷ (Normal duration − Crash duration).
Q1.2 — Crashing rule
"Only crash critical activities — crashing non-critical activities wastes money without shortening the project."
Q1.3 — Definitions
Resource allocation: assigning limited resources (people, equipment, budget) to activities so that resource limits are respected.
Resource levelling: adjusting non-critical activity start times within their float to smooth peak resource demand.
Q1.4 — Shifting critical activities
False. Critical activities have float = 0, so they cannot be shifted without delaying the project. Only non-critical activities can be shifted within their float.
Q3 — Faded crash analysis (X and Y)
Step 1: X $/day = ($9,500 − $8,000) ÷ (10 − 7) = $1,500 ÷ 3 = $500/day. Y $/day = ($5,700 − $5,000) ÷ (6 − 5) = $700 ÷ 1 = $700/day.
Step 2: X is cheaper.
Step 3: X max crash = 10 − 7 = 3 days.
Step 4: Crash X by 3 days.
Extra cost = 3 × $500 = $1,500.
Conclusion: Crash X by 3 days at extra cost $1,500. No need to crash Y.
Q4.1 — Crash cost per day (normal 12 / crash 9)
($6,500 − $5,000) ÷ (12 − 9) = $1,500 ÷ 3 = $500/day.
Q4.2 — Crash cost per day (normal 10 / crash 8)
($5,000 − $4,000) ÷ (10 − 8) = $1,000 ÷ 2 = $500/day.
Q4.3 — Max extra cost if crashed fully
4 × $300 = $1,200.
Q4.4 — Crashing non-critical
False. Non-critical activities don't control the project duration, so shortening them costs money but doesn't bring the finish forward.
Q4.5 — Cheapest crash for 2 days
A $/day = $400 ÷ 2 = $200/day. B $/day = $300 ÷ 1 = $300/day. A is cheaper. A can be crashed by 2 days (8 → 6) at 2 × $200 = $400. (Crashing B by 1 + A by 1 = $300 + $200 = $500 — more expensive.)
Q4.6 — Three critical activities (capped at 2 days each)
Order by $/day: C $150 < A $200 < B $350. Crash C by 2 days = $300; crash A by 1 day = $200. Total = $500. (Crashing only A and B leaves us paying more, e.g. A 2 days + B 1 day = $400 + $350 = $750.)
Q4.7 — Crash a non-critical activity?
No. Crashing a non-critical activity does not shorten the project duration; the 2 days must come from critical activities.
Q4.8 — Re-checking critical path
Crashing a critical activity might shorten its path enough that a previously non-critical path becomes the new longest path, so we must re-identify the critical path before the next crash.
Q4.9 — Resource levelling with 8 workers
Run the two critical activities (6 workers) at their scheduled times. The two non-critical activities (6 workers) cannot run at the same time as the critical pair without exceeding 8 workers. Use their 4-day float to delay one or both non-critical activities so they start after the critical activities finish, keeping the simultaneous worker count at or below 8.
Q4.10 — Shared critical activity
Crash X. X is shared by all three critical paths, so crashing it by 1 day shortens every critical path by 1 day, reducing the project by 1 day for a single payment of $400. Crashing Y only shortens one of the three critical paths; the other two would still hold the project at the old duration.
Q4.11 — Three-activity multi-crash
$/day: P = ($3,800 − $3,000) ÷ (8 − 6) = $800 ÷ 2 = $400/day (max 2 days). Q = ($3,000 − $2,000) ÷ (5 − 3) = $1,000 ÷ 2 = $500/day (max 2 days). R = ($2,200 − $1,500) ÷ (4 − 3) = $700 ÷ 1 = $700/day (max 1 day).
Cheapest first: crash P by 2 days = $800. Need 1 more day. Next cheapest is Q at $500/day. Crash Q by 1 day = $500.
Total extra cost = $800 + $500 = $1,300. (Note: the assumption is that P, Q, R remain critical after each crash — in practice, recheck the critical path after each step.)
Q4.12 — Partial crash
$/day = ($7,600 − $6,000) ÷ (12 − 8) = $1,600 ÷ 4 = $400/day. Crashing by 2 days (12 → 10): extra cost = 2 × $400 = $800.