Mathematics Standard • Year 12 • Module 6 • Lesson 6
Crashing and Resource Allocation — Problem Set
Apply crashing and resource-levelling decisions to real Australian projects: a roadworks contract, a hospital wing build, an apartment renovation, a tech-product launch and a building-site labour roster.
Problem 1 — Highway resurfacing contract (NSW)
Two critical activities on a Pacific Highway resurface have these times and costs:
Mill MO: normal 6 days at $24,000; crash 4 days at $30,000.
Asphalt AS: normal 8 days at $40,000; crash 5 days at $52,000.
The contractor needs to reduce project duration by 3 days at minimum extra cost.
Set up: What are we solving for?
(i) Calculate the crash cost per day for both MO and AS. 2 marks
(ii) State the maximum number of days each activity can be crashed by. 1 mark
(iii) Decide which activity (or combination) to crash and by how many days to reduce the project by 3 days at minimum cost. State the total extra cost. 2 marks
Stuck? Revisit lesson § Crashing — pick the cheapest critical activity first, respecting its crash limit.Problem 2 — Hospital wing fit-out (Sydney teaching hospital)
Three critical activities are candidates for crashing:
A (rewire): normal 10 days $20,000; crash 7 days $26,000.
B (plumbing): normal 8 days $16,000; crash 6 days $19,000.
C (drywall): normal 12 days $18,000; crash 9 days $22,500.
Reduce the project by 4 days at minimum extra cost.
Set up: What are we solving for?
(i) Compute the crash cost per day for A, B and C. 2 marks
(ii) State the maximum crash for each activity in days. 1 mark
(iii) Build a crashing schedule that reduces the project by 4 days at minimum cost (state which activities you crash and by how many days). Give the total extra cost. 2 marks
Stuck? Revisit lesson § Worked Example — rank by $/day, then crash from cheapest upwards within each activity's limit.Problem 3 — Apartment renovation (Melbourne) — crash vs accept the delay
An apartment renovation runs 25 days. A critical activity can be crashed:
Tiling TI: normal 8 days $4,000; crash 6 days $5,200.
The owner offers a $700 bonus for finishing 2 days early.
Set up: What are we solving for?
(i) Compute the crash cost per day for TI and the total extra cost to crash by 2 days. 2 marks
(ii) Compare the crash cost for 2 days against the $700 bonus. State whether the contractor should accept the bonus (i.e. crash TI by 2 days). 2 marks
(iii) A non-critical activity (Paint PT, float 3) could be crashed for $400/day. Explain in one short sentence why crashing PT is not a valid way to earn the bonus. 2 marks
Stuck? Revisit lesson § Crashing — only crash critical activities; non-critical crashes do not move the finish date.Problem 4 — Tech product launch (Sydney start-up)
A start-up plans to launch in 30 days but the marketing window closes in 27. Three critical activities are crash candidates:
Backend BE: normal 12 days $8,000; crash 9 days $11,000.
Frontend FE: normal 10 days $7,000; crash 8 days $8,400.
Test TE: normal 5 days $3,000; crash 4 days $3,500.
Reduce duration by 3 days at minimum cost.
Set up: What are we solving for?
(i) Compute the crash cost per day for BE, FE and TE. 2 marks
(ii) State the maximum crash days for each activity. 1 mark
(iii) Build a crashing plan that reduces the launch by 3 days at minimum cost. Give the total extra cost and one short reason for your choice of activities. 2 marks
Stuck? Revisit lesson § Crashing — order by $/day, crash within each activity's limit, recheck path if needed.Problem 5 — Building-site labour roster (regional NSW)
A small site has 6 labourers. The following four activities all run during week 3, each needing labourers:
W1 Concrete pour: needs 4 labourers, critical (no slack).
W2 Steel fix: needs 3 labourers, float = 4 days.
W3 Carpentry: needs 3 labourers, float = 3 days.
W4 Plumbing: needs 2 labourers, critical (no slack).
All four are currently scheduled to start on Monday of week 3.
Set up: What are we solving for?
(i) Calculate the total labour demand if all four activities run simultaneously, and state whether it exceeds the 6-labourer cap. 1 mark
(ii) Identify which activities are critical (cannot be shifted) and which can be shifted (within float). Use this to propose a resource-levelling plan that runs W1 and W4 in week 3 and shifts the others so the total never exceeds 6 labourers. 2 marks
(iii) A site manager suggests shifting W4 (plumbing) by 2 days to fit everything in. Explain in one short sentence why this is not allowed. 2 marks
Stuck on (ii)? W1 + W4 = 6 labourers; that exactly fits. Run W2 and W3 (six labourers between them) in the following days using their float.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Highway resurfacing
Set up. We are choosing the cheapest critical activity to crash by 3 days, possibly splitting the crash across MO and AS.
(i) MO $/day = ($30,000 − $24,000) ÷ (6 − 4) = $6,000 ÷ 2 = $3,000/day. AS $/day = ($52,000 − $40,000) ÷ (8 − 5) = $12,000 ÷ 3 = $4,000/day.
(ii) MO max crash = 6 − 4 = 2 days. AS max crash = 8 − 5 = 3 days.
(iii) MO is cheaper per day. Crash MO by 2 days (max) = 2 × $3,000 = $6,000. Need 1 more day from AS at $4,000. Total extra cost = $6,000 + $4,000 = $10,000. (Alternative: crash AS by 3 days = $12,000 — more expensive.)
Problem 2 — Hospital wing fit-out
Set up. Three critical activities; reduce duration by 4 days at minimum cost.
(i) A $/day = ($26,000 − $20,000) ÷ (10 − 7) = $6,000 ÷ 3 = $2,000/day. B $/day = ($19,000 − $16,000) ÷ (8 − 6) = $3,000 ÷ 2 = $1,500/day. C $/day = ($22,500 − $18,000) ÷ (12 − 9) = $4,500 ÷ 3 = $1,500/day.
(ii) A max = 3 days. B max = 2 days. C max = 3 days.
(iii) B and C tie cheapest at $1,500/day. Crash B by 2 days (max) = $3,000. Need 2 more days. Crash C by 2 days = $3,000. Total extra cost = $6,000. (No need to crash A.) Alternative: crash C by 3 + B by 1 = $4,500 + $1,500 = $6,000 — same cost; either is acceptable.
Problem 3 — Apartment renovation
Set up. Compare 2-day crash cost to a $700 bonus and reject crashing a non-critical activity.
(i) TI $/day = ($5,200 − $4,000) ÷ (8 − 6) = $1,200 ÷ 2 = $600/day. Total cost for 2 days = 2 × $600 = $1,200.
(ii) Crash cost ($1,200) > bonus ($700), so the contractor would lose $500 by accepting the bonus. Do not crash TI — reject the bonus offer.
(iii) PT is non-critical with float 3, so it can absorb 3 days of delay without affecting the finish; crashing PT does not bring the finish forward. Crashing PT wastes money without earning the bonus.
Problem 4 — Tech product launch
Set up. Three critical activities; reduce by 3 days at minimum cost.
(i) BE $/day = ($11,000 − $8,000) ÷ (12 − 9) = $3,000 ÷ 3 = $1,000/day. FE $/day = ($8,400 − $7,000) ÷ (10 − 8) = $1,400 ÷ 2 = $700/day. TE $/day = ($3,500 − $3,000) ÷ (5 − 4) = $500 ÷ 1 = $500/day.
(ii) BE max = 3 days. FE max = 2 days. TE max = 1 day.
(iii) Cheapest first: TE $500 (max 1) = $500. Next: FE $700 (max 2) — crash by 2 = $1,400. Total so far = 3 days for $1,900. Total extra cost = $1,900. (Reason: order by $/day and use the maximum allowed at the cheapest activities first.)
Problem 5 — Building-site labour roster
Set up. Decide if total labour exceeds 6, then use float to level non-critical activities.
(i) Total demand = 4 + 3 + 3 + 2 = 12 labourers > 6 cap. Exceeds cap.
(ii) Critical (cannot shift): W1 (4 labourers) and W4 (2 labourers) — together = 6, exactly fitting the cap. Shift W2 and W3 within their float: e.g. run W1+W4 (6 labourers) Mon-Wed, then run W2 (3 labourers) Thu-Fri, then W3 (3 labourers) the following Mon-Wed of week 4 (within its 3-day float). At no point does total demand exceed 6.
(iii) W4 is critical (no slack); shifting it by 2 days delays the project by 2 days. Resource levelling can only shift non-critical activities within their float.