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.