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Chapter 4 of 10

Crashing & Resource Allocation

In the BOI SO Civil Engineer syllabus under Construction Management · 2 parts

📑 Contents (16 sections)

Part 1 of 2

Project Crashing — Time-Cost Trade-off

Last reviewed 16 Sept 2026 · 6 min read

Project costs

Cost Meaning Behaviour when duration is reduced
Direct costs Costs directly attributable to activities — materials, labour, equipment for each activity Increase (overtime, extra shifts, more equipment, costlier methods)
Indirect costs Overheads — site supervision, office establishment, insurance, interest on capital, general administration, security; also penalties for delay (or bonuses) Decrease (roughly proportional to duration)

Total cost = direct cost + indirect cost — reducing the duration raises direct cost but lowers indirect cost, so the total cost curve has a minimum at the optimum duration.

Normal and crash points

  • Normal time () — duration under normal working conditions at the lowest direct cost (normal cost ).
  • Crash time () — minimum possible duration with maximum resources, at crash cost .
  • Reducing an activity below its crash time is impossible (or only increases cost without saving time).
FormulaCost slope

The extra direct cost per unit time saved (assuming a linear time–cost relationship between normal and crash points).

Part 2 of 2

Resource Allocation & Resource Levelling

Last reviewed 16 Sept 2026 · 4 min read

Resources

Construction resources are often summarised as the 5 Ms: Manpower, Materials, Machinery, Money and Management (with space/time also treated as resources).

  • Renewable resources — available again each period (labour, equipment per day).
  • Non-renewable (consumable) resources — used up (materials, money).

Resource planning

  1. Determine resource requirements of each activity (e.g. 4 masons, 1 mixer).
  2. Schedule activities (early start schedule from CPM).
  3. Aggregate resources per time period → resource histogram (resource loading diagram).
  4. Compare with availability; level or allocate as needed.

Resource histogram

A bar diagram of resource units required versus time. Irregular histograms with high peaks and valleys cause hiring and firing, idle time, overtime and poor productivity.

Resource levelling (smoothing)

Resource levelling reshuffles non-critical activities within their floats so that resource demand becomes as uniform as possible, without changing the project duration.

  • Critical activities are not moved.
  • Activities are shifted (start delayed) using free float first, then total float (which may affect successors).
  • Objective measures: reduce peak demand, reduce fluctuations, minimise the sum of squares of daily resource demand (the total resource-days remain constant, so a smaller sum of squares means a more uniform profile).

Resource allocation (resource-constrained scheduling)

When resources are limited (a maximum available per period), activities are scheduled so that demand never exceeds availability, even if the project duration increases.

Heuristic priority rules for deciding which activity receives resources first:

  • Least total float (minimum slack) first.
  • Earliest late start / earliest late finish.
  • Shortest duration first.
  • Activities with most resources or on the critical path first.

Levelling vs allocation

Aspect Resource levelling (smoothing) Resource allocation (constrained)
Project duration Fixed May increase
Resource limit Not fixed; aim is uniformity Fixed maximum availability
Activities moved Only within floats Any, as required
Objective Reduce fluctuations and peaks Keep within resource limits

Worked example — levelling

Worked ExampleLevelling within float
Activity Duration (days) Workers/day Predecessor Early start schedule
A 3 4 — days 1–3 (critical)
B 2 3 — days 1–2 (total float 4)
C 3 2 A days 4–6 (critical)

Project duration = 6 days; total resource-days = 12 + 6 + 6 = 24.

Early start histogram: days 1–2: 4 + 3 = 7; day 3: 4; days 4–6: 2 Peak = 7; sum of squares

Levelled — shift B to days 4–5 (using its float): days 1–3: 4; days 4–5: 2 + 3 = 5; day 6: 2 Peak = 5; sum of squares

The project still finishes on day 6, peak demand falls from 7 to 5 and the profile is smoother (lower sum of squares).

Worked ExampleResource constraint

In the same project, only 4 workers are available per day. What is the minimum duration?

Solution. A (4 workers) and B (3 workers) cannot run together. A is critical, so schedule A on days 1–3. C (2 workers) and B (3 workers) together need 5 > 4. Schedule C on days 4–6 and B on days 7–8, or B on days 4–5 and C on days 6–8. Minimum duration = 8 days — the resource limit extends the project by 2 days.

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