Year 12 General Mathematics: Project crashing
Back to Graph theory tutorialsThis lesson builds on critical path analysis. Make sure you can identify critical activities and calculate float before continuing.
Crashing means shortening a project by paying to reduce the duration of selected activities. The goal is not simply to crash the cheapest activity. You must shorten an activity on every current critical path, stay within each activity's crash limit, and check whether the saving in project time is worth the extra direct cost.
Crash cost per day
The cost slope compares an activity's normal and fully crashed plans:
The denominator is the maximum number of days that can be removed. Unless stated otherwise, assume the cost rises linearly, so this slope is the additional cost of removing one day.
Worked project
A project has two start-to-finish paths:
- A–C–F: 4 + 8 + 5 = 17 days
- B–D–F: 6 + 4 + 5 = 15 days
The normal project duration is 17 days and the initial critical path is A–C–F.
| Activity | Normal time | Crash time | Maximum reduction | Extra cost/day |
|---|---|---|---|---|
| A | 4 | 2 | 2 days | $300 |
| B | 6 | 4 | 2 days | $350 |
| C | 8 | 5 | 3 days | $200 |
| D | 4 | 3 | 1 day | $250 |
| F | 5 | 4 | 1 day | $500 |
Iteration 1: reduce 17 days to 16
On the critical path A–C–F, activity C is cheapest at $200 per day. Crash C by one day.
- A–C–F: 4 + 7 + 5 = 16
- B–D–F: 6 + 4 + 5 = 15
Project duration: 16 days. Added cost: $200.
Iteration 2: reduce 16 days to 15
Crash C once more for another $200.
- A–C–F: 4 + 6 + 5 = 15
- B–D–F: 6 + 4 + 5 = 15
Iteration 3: reduce both critical paths
Activity F is shared by both paths. Crashing it by one day shortens both at once and costs $500. The alternative is to crash one activity on each branch; the cheapest such pair is A and D, costing $300 + $250 = $550. Therefore crash F.
- A–C–F: 4 + 6 + 4 = 14
- B–D–F: 6 + 4 + 4 = 14
The project has been shortened by three days for an additional direct cost of $200 + $200 + $500 = $900.
Should the project be crashed further?
Suppose normal direct cost is $20,000 and indirect costs such as supervision and equipment hire are $550 per project day.
| Duration | Added crash cost | Direct cost | Indirect cost | Total cost |
|---|---|---|---|---|
| 17 days | $0 | $20,000 | 17 × $550 = $9,350 | $29,350 |
| 16 days | $200 | $20,200 | $8,800 | $29,000 |
| 15 days | $400 | $20,400 | $8,250 | $28,650 |
| 14 days | $900 | $20,900 | $7,700 | $28,600 |
At 14 days, F cannot be shortened again. Both critical branches must now be crashed. The cheapest combination is A and D for $550, exactly equal to one day's indirect saving. If there is no separate early-finish bonus, crashing further does not reduce total cost.
A reliable exam method
- Calculate every path length and identify all critical paths.
- Calculate each activity's crash cost per day and maximum reduction.
- Choose the cheapest feasible activity on the critical path.
- If several critical paths exist, crash a shared activity or one activity on every critical path.
- Reduce by one day—or only until another path becomes critical.
- Recalculate all path lengths and repeat.
- Stop at the target duration, the minimum possible duration, or the minimum total cost.
Common mistakes
- Crashing a non-critical activity. Its float absorbs the reduction and the finish date does not change.
- Using total crash cost instead of additional cost per day.
- Reducing an activity beyond its stated crash limit.
- Continuing with the original critical path without recalculating after each change.
- When two paths are critical, crashing only one of them.
- Assuming the shortest possible duration is also the lowest-cost duration.