5.5 Time–Cost Trade-Off and Crashing
Key Takeaways
- Crashing shortens project duration by adding resources to critical activities; the cost slope = (crash cost − normal cost) ÷ (normal duration − crash duration), expressed in $/day.
- Always crash the critical activity with the lowest cost slope first, and only crash activities on the critical path(s).
- Total project cost = direct costs (rise as you crash) + indirect costs (fall as duration shortens); the optimum duration minimizes the sum.
- As you crash, a parallel path can become critical; once multiple critical paths exist you must crash an activity on every critical path simultaneously (or a shared activity).
- Stop crashing when the marginal cost to save one more day exceeds the daily indirect-cost savings (or a liquidated-damages/bonus rate), not when duration hits an arbitrary target.
Time–Cost Trade-Off and Crashing
Why this topic matters for the PE Construction exam: “Crashing” problems are among the most reliably tested scheduling calculations on the PE Construction exam. You will be given normal and crash durations and costs for activities and asked which activity to shorten, how much it costs, and what the least-cost project duration is. The method is systematic, and the arithmetic is straightforward once you know the sequence.
Crashing is the deliberate shortening of a project's duration by adding resources — overtime, extra crews, larger equipment, or a second shift — to one or more activities. Because those extra resources cost money, crashing is a time–cost trade-off: you spend more direct dollars to buy fewer days.
The Cost Slope
Each activity has a normal point (least-cost, longest acceptable duration) and a crash point (shortest achievable duration, highest cost). The rate at which cost rises per day saved is the cost slope:
The cost slope tells you how many dollars it costs to shorten that specific activity by one day. It is always positive (crashing costs money) and is the single most important number in a crashing problem.
Worked example — computing a slope
An activity has a normal duration of 10 days at a cost of $8,000 and can be crashed to 6 days at a cost of $12,000.
Each day this activity is shortened (down to a floor of 6 days) adds $1,000 to direct cost.
The Crashing Procedure
To shorten a project at the least direct cost, follow this sequence:
- Identify the critical path and its duration from the normal CPM schedule.
- Compute the cost slope for every activity that can be crashed.
- Crash the critical activity with the lowest cost slope first, because it buys a day for the fewest dollars. Never crash a non-critical activity — it costs money without shortening the project.
- Crash one day at a time (or to the next event) and re-check the network. A crashed activity reaches its crash limit (it cannot go below its crash duration), and — more importantly — a parallel path may become critical.
- When two or more paths are critical, you must shorten all of them by the same amount to reduce the project. That means crashing one activity on each critical path, or a single shared activity that lies on all critical paths (often the cheapest move).
- Stop when further crashing is impossible or no longer economical.
Direct vs. Indirect Costs — the Least-Cost Duration
Crashing decisions are not only about direct cost. Two cost streams move in opposite directions:
- Direct costs (labor, material, equipment for the activities) increase as you crash — you are paying the cost slopes.
- Indirect costs (site overhead, supervision, general conditions, equipment rental, financing) decrease as the project gets shorter, because they accrue per unit of time.
| Component | Effect of crashing | Reason |
|---|---|---|
| Direct cost | Increases | Paying cost slopes to add resources |
| Indirect cost | Decreases | Fewer days of overhead/general conditions |
| Total cost | U-shaped | Sum of the two; has a minimum |
The total project cost is the sum of direct and indirect costs. Plotted against duration, it is U-shaped, and its minimum defines the least-cost (optimum) duration. If a contract adds liquidated damages for lateness or an early-completion bonus, treat those as additional daily indirect-cost drivers: keep crashing as long as the daily saving (indirect savings + avoided damages, or earned bonus) exceeds the cheapest available cost slope.
Worked example — when to stop
Suppose indirect costs run $1,500/day. The cheapest remaining way to save a day has a cost slope of $1,000/day. Crashing that day saves $1,500 in overhead while adding $1,000 in direct cost — a net $500 benefit, so you should crash it. Once the cheapest available slope rises above $1,500/day, an additional crashed day costs more than it saves, and you stop. The optimum duration is reached exactly where the marginal cost slope crosses the indirect-cost rate.
Exam Traps
- Only crash critical activities. Spending to crash a non-critical activity wastes money and does not shorten the project.
- Watch for a new critical path. After a few days of crashing, a parallel path can tie the critical duration; from then on you must crash both paths together.
- Respect crash limits. An activity cannot be shortened below its crash duration no matter how much you spend.
- Least-cost duration ≠ fully-crashed duration. The economic optimum usually stops before everything is crashed to its minimum.
An activity has a normal duration of 12 days costing $20,000 and a crash duration of 8 days costing $28,000. What is its cost slope?
When crashing a project to reduce its duration at the lowest direct cost, which activity should be crashed first?
Indirect (overhead) costs are $2,000 per day. The cheapest remaining critical activity that can still be crashed has a cost slope of $2,500 per day. What is the correct economic decision?
After crashing the original critical path by three days, a second, parallel path now has the same duration and is also critical. To shorten the project one more day at least cost, what must you do?