Operations Management and Strategic Management · Project Management, Monitoring and Control

Project Crashing and Time-Cost Trade-off in CPM

Updated 10 October 2026 · Fact-checked

Project crashing means shortening a project's duration by spending extra money on selected activities. Calculate each activity's cost slope = (crash cost − normal cost) ÷ (normal time − crash time). Then crash critical activities, cheapest slope first, one step at a time, watching for new critical paths, until you reach the target time.

Understand Project Crashing and Time-Cost Trade-off

Every activity in a project has a normal time and a normal cost. This is the usual, most economical way of doing it. If you add resources such as overtime, extra labour or extra machines, you can finish sooner. The shortest possible time is the crash time, and the cost at that time is the crash cost.

Shortening an activity costs money. The extra cost per unit of time saved is the cost slope. A low slope means time is cheap to buy. A high slope means it is expensive.

The project duration depends only on the critical path, the longest path through the network. Crashing an activity that is not on the critical path saves no project time and wastes money. So you always crash critical activities, and among those you pick the one with the lowest cost slope first.

There is a catch. When you shorten the critical path, another path may become equally long. Then you have two critical paths, and you must shorten both together to gain more time. This is why you crash one unit at a time and redraw the path check after each step.

The aim is the minimum extra cost for the required reduction. Total project cost also includes indirect costs such as overheads, which fall as the project gets shorter. The best duration is where direct cost plus indirect cost is lowest.

Key rules to remember

Cost slope: Cost slope = (Crash cost − Normal cost) ÷ (Normal time − Crash time); Assumes cost rises in a straight line between normal and crash points. It is the extra cost per unit of time saved.
Maximum crashing available: Maximum crash = Normal time − Crash time; You cannot shorten an activity beyond this limit.
Cost of crashing an activity: Extra cost = Cost slope × Time units crashed; Add this to the normal project cost to get the new direct cost.
Total project cost: Total cost = Direct cost + Indirect cost; Direct cost rises as you crash. Indirect cost usually falls with duration. Choose the duration with minimum total.
Net saving from a crash step: Net saving = Indirect cost saved − Crashing cost; Keep crashing only while this is positive.

How to solve Project Crashing and Time-Cost Trade-off questions

Use this method for any crashing question. Work in a neat table so you can show step marks.

1Draw the network and find all paths with their durations using normal times. Identify the critical path and the normal project duration.
2Compute the cost slope for every activity. Also note the maximum crash available for each.
3List critical activities in order of lowest cost slope.
4Crash the cheapest critical activity by one unit, or by the amount allowed before another path becomes critical. Do not exceed its maximum crash.
5Recheck all path lengths. If a new critical path appears, crash activities on all critical paths together, choosing the cheapest combination. A common activity on both paths may be cheaper than two separate ones.
6Repeat until the target duration is reached or no critical activity can be crashed further.
7Add total crashing cost to the normal cost. If indirect cost is given, compute total cost at each duration and pick the lowest.
8State the final duration, extra cost and the activities crashed.

Quickest way: Cheapest-critical-first table

When to use it: Use when the question asks for the minimum cost of reducing the project by a given number of days.

Compute slopes first and write them beside each activity.
List every path with its length once. Use it to see which paths are close to critical.
Reduce one day at a time on the cheapest critical activity. Note the cumulative cost after each day.
After each day, update only the path lengths that changed.
Stop when the target is met. Ignore any non-critical activity completely.
Do a quick check: crashed time of each activity must still be at least its crash time.

Common mistakes in Project Crashing and Time-Cost Trade-off

Crashing an activity that is not on the critical path.
Students pick the lowest slope in the whole table.
Fix: Choose the lowest slope only among critical activities. Non-critical crashing gives no time saving.
Dividing by the wrong time difference in the slope.
Students use crash time minus normal time or mix up the order.
Fix: Always use normal time minus crash time in the denominator, and crash cost minus normal cost in the numerator.
Crashing too far and ignoring a second critical path.
Students crash the full allowed amount in one go.
Fix: Crash only until another path becomes critical, then recheck and crash on all critical paths together.
Crashing beyond the crash time limit.
The cheapest activity is reused without tracking how much is left.
Fix: Keep a column for remaining crash days for each activity and stop at zero.
Forgetting to add crashing cost to normal cost for the total.
Students stop once the duration is found.
Fix: Always give the final total direct cost, and include indirect cost when it is given.
Ignoring indirect costs when asked for the optimum duration.
Students keep crashing until the target without testing savings.
Fix: Compare each day's crashing cost with the indirect cost saved per day. Stop when crashing costs more than it saves.

Worked examples

Example 1

A project has activities: A (1-2), B (1-3), C (2-4), D (3-4). Normal time/crash time in days: A 6/4, B 5/4, C 7/5, D 8/6. Normal cost/crash cost in ₹: A 6,000/8,000, B 4,000/4,600, C 9,000/10,000, D 10,000/11,600. Find the normal duration and the minimum extra cost to complete the project in 11 days.

Show the solution

Paths: 1-2-4 (A + C) = 6 + 7 = 13 days. 1-3-4 (B + D) = 5 + 8 = 13 days. Both paths are critical, normal duration is 13 days.
Slopes: A = (8,000 − 6,000) ÷ (6 − 4) = ₹1,000 per day. B = (4,600 − 4,000) ÷ (5 − 4) = ₹600 per day. C = (10,000 − 9,000) ÷ (7 − 5) = ₹500 per day. D = (11,600 − 10,000) ÷ (8 − 6) = ₹800 per day.
We need a 2 day reduction. Both paths are critical, so each path needs 2 days saved.
Path A-C: cheapest is C at ₹500 per day, maximum 2 days. Crash C by 2 days: cost ₹1,000. Path becomes 11 days.
Path B-D: cheapest is B at ₹600 but maximum 1 day. Crash B by 1 day: ₹600. Then crash D by 1 day: ₹800. Total ₹1,400. Path becomes 11 days.
Total extra cost = ₹1,000 + ₹1,400 = ₹2,400.

Answer: Normal duration is 13 days. The minimum extra cost to finish in 11 days is ₹2,400, by crashing C by 2 days, B by 1 day and D by 1 day.

Example 2

A project has a single critical path A-B-C of 20 days with crashing details: A (normal 8 days, crash 6 days, slope ₹300 per day), B (normal 7 days, crash 6 days, slope ₹200 per day), C (normal 5 days, crash 3 days, slope ₹400 per day). Another path A-D-C is 18 days. Indirect cost is ₹350 per day. Normal direct cost is ₹60,000. Find the best project duration and the total cost at that duration.

Show the solution

Critical path A-B-C is 20 days. Path A-D-C is 18 days.
Indirect cost at 20 days = 20 × ₹350 = ₹7,000. Total cost = ₹60,000 + ₹7,000 = ₹67,000.
Day 1: crash the cheapest critical activity, B at ₹200 (1 day available). Duration 19. Cost ₹200 vs indirect saving ₹350. Net saving ₹150. Continue.
Day 2: B is exhausted. Next cheapest is A at ₹300. Path A-B-C becomes 18 days. A is common to A-D-C so that path also falls to 17 days. Duration is now 18. Cost ₹300 vs saving ₹350. Net saving ₹50. Continue.
Day 3: A-B-C is the only critical path at 18 days. Cheapest left is A at ₹300 (1 day left). Crash A again. Path A-B-C becomes 17 days. A-D-C becomes 16 days. Duration 17. Cost ₹300 vs saving ₹350. Net saving ₹50. Continue.
Day 4: A is exhausted. Next is C at ₹400. Saving is only ₹350, so crashing costs more than it saves. Stop.
Total crashing cost = ₹200 + ₹300 + ₹300 = ₹800. Duration is 17 days.
Direct cost = ₹60,000 + ₹800 = ₹60,800. Indirect cost = 17 × ₹350 = ₹5,950. Total cost = ₹66,750.

Answer: The best duration is 17 days. Total cost is ₹66,750, compared with ₹67,000 at the normal duration of 20 days.

Exam tips

Always show the slope table and the path lengths. Step marks come from these even if the final figure slips.
Write a short crashing table with columns for day, activity crashed, cost and new duration. It keeps the working clear and quick.
When two critical paths appear, check whether one common activity can serve both before paying for two separate ones.
If indirect costs are given, the question is about the optimum duration. Compare crashing cost with the saving each day.
For MCQs, compute the cost slope first. Many questions ask only for the slope or the cost of one crash step.

Practice questions from Project Management, Monitoring and Control

Project Crashing and Time-Cost Trade-off in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

CMA FinalProject Crashing and Time-Cost Trade-off

Project Crashing and Time-Cost Trade-off: frequently asked questions

What is the difference between normal cost and crash cost?

Normal cost is the cost of completing an activity in its normal time. Crash cost is the higher cost of completing it in the shortest possible time. The difference between them, spread over the time saved, gives the cost slope.

How do you calculate cost slope in crashing?

Cost slope = (crash cost − normal cost) ÷ (normal time − crash time). It tells you the extra rupees paid for each day saved. Lower slopes are crashed first, provided the activity is critical.

Why can we not crash non-critical activities?

Project duration is set by the critical path alone. Shortening an activity with float does not reduce the longest path, so it adds cost without saving any time.

What happens when a second path becomes critical during crashing?

You now need to shorten both critical paths to reduce the project further. Choose the cheapest combination, which may be a common activity on both paths or one activity on each.