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Strategic Cost Management · Network Analysis - PERT, CPM

Project Crashing and Time-Cost Trade-off in CPM

Updated 11 October 2026 · Fact-checked

Project crashing shortens a project by speeding up activities on the critical path at extra direct cost. You compute each activity's cost slope, crash the cheapest critical activity one day at a time, and stop when the slope exceeds the indirect cost saved per day. That point is the optimum duration.

Understand Project Crashing and Time-Cost Trade-off

Every activity in a project has a normal time and a normal cost. If you spend more on extra labour, overtime or machines, you can finish it faster, down to a crash time at a higher crash cost. Below the crash time, no amount of money helps.

Crashing is useful because the project has two kinds of cost. Direct cost (labour, material, equipment on activities) rises as you shorten the project. Indirect cost (supervision, overheads, site rent, penalties) falls as the project gets shorter. Total cost is the sum of the two, and it is usually lowest at some duration between the normal and the fully crashed one.

Only activities on the critical path can shorten the project. Crashing a non-critical activity adds cost and saves no time. So the method is: find the critical path, pick the critical activity with the lowest cost slope (extra cost per day saved), crash it, then re-check the network.

When you crash, a second path may become critical. From then on, you must shorten every critical path together by the same number of days. Sometimes one activity common to all critical paths is cheapest. Sometimes you need one activity from each path.

Keep crashing while the cost of saving a day is less than the indirect cost saved per day. Stop when it is equal or higher. The duration where total cost is lowest is the optimum duration.

Key rules to remember

Cost slope
Cost slope = (Crash cost − Normal cost) ÷ (Normal time − Crash time)
Extra cost per day saved. Assumes cost rises in a straight line between normal and crash points.
Maximum crashing available
Maximum days crashable = Normal time − Crash time
You cannot crash an activity by more than this.
Total project cost
Total cost = Total direct cost + Total indirect cost
Direct cost = normal costs plus crashing cost added. Indirect cost = rate per day × project duration.
Crashing decision rule
Crash one more day if cost of saving that day < indirect cost saved per day
If equal, total cost does not change. Stop when the cost is higher.
Crashing cost of a step
Crashing cost = Days crashed × Cost slope
For parallel critical paths, add the slopes of the activities crashed together.

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

Use this table-style method for any crashing question. Work one day (or one block of days) at a time and recheck the critical path every time.

  1. 1Compute the cost slope of every activity and the maximum days it can be crashed.
  2. 2List all paths with their durations. Mark the critical path (the longest) and note the gap to the next longest path.
  3. 3Find the critical activity with the lowest slope that still has crash capacity. If several paths are critical, find the cheapest way to shorten all of them together.
  4. 4Crash by the smaller of: the remaining crash capacity, or the gap to the next path. Add days × slope to direct cost and update path durations.
  5. 5Compare the cost of saving a day with the indirect cost per day. Continue only if the cost is lower.
  6. 6Repeat until no cheaper crash is possible or all critical paths are fully crashed.
  7. 7Prepare a table of duration, direct cost, indirect cost and total cost. Pick the duration with the lowest total cost and state it with the cost.

Quickest way: Slope-first shortlist

When to use it: Use when the network is small (4-8 activities) and the question asks for the optimum duration or the cheapest way to reach a target time.

  1. Write slope and crash capacity beside each activity before anything else.
  2. List every path with its length in one line each.
  3. Crash the cheapest critical activity by one day. Rewrite path lengths straight away.
  4. When two paths tie, look first for a common activity with a slope lower than the sum of two separate activities.
  5. Stop as soon as the next step costs more than the indirect saving. Then write the cost table for the durations tested only.

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

  • Crashing an activity that is not on the critical path

    It has the lowest slope in the list, so it looks attractive.

    Fix: Choose only from critical activities. Non-critical activities save no project time.

  • Crashing beyond the gap to the next path without recalculating

    Students crash the cheapest activity by its full capacity in one go.

    Fix: Limit each step to the gap to the next longest path. After that, crash all critical paths together.

  • Dividing by the wrong days in the slope

    The crash time and normal time are swapped, or the crash time is used instead of the days saved.

    Fix: Slope = extra cost ÷ (normal time − crash time). Days saved is the denominator.

  • Ignoring indirect cost when finding the optimum

    The student crashes until the project cannot be shortened further.

    Fix: Compare each slope with indirect cost per day, and confirm with a total-cost table.

  • Adding the indirect cost of the normal duration to every row

    The indirect cost is copied down the table.

    Fix: Indirect cost = rate × the duration in that row. Recompute it for every duration.

  • Crashing the cheapest activity on each parallel path when one common activity is cheaper

    Students do not compare combined slopes.

    Fix: Check the common activity's slope against the sum of the slopes of separate activities.

Worked examples

Example 1

A project has five activities. Indirect cost is ₹500 per day.

Activity | Predecessor | Normal days | Crash days | Normal cost | Crash cost
A | none | 4 | 3 | ₹1,000 | ₹1,400
B | none | 6 | 4 | ₹2,000 | ₹2,600
C | A | 5 | 3 | ₹1,500 | ₹2,300
D | B | 4 | 3 | ₹800 | ₹1,100
E | C, D | 3 | 2 | ₹600 | ₹1,000

Find the optimum project duration and the total cost at that duration.

Show the solution
  1. Cost slopes: A = 400 ÷ 1 = ₹400. B = 600 ÷ 2 = ₹300. C = 800 ÷ 2 = ₹400. D = 300 ÷ 1 = ₹300. E = 400 ÷ 1 = ₹400.
  2. Paths: A-C-E = 4 + 5 + 3 = 12 days. B-D-E = 6 + 4 + 3 = 13 days. The critical path is B-D-E, so normal duration is 13 days.
  3. Normal direct cost = 1,000 + 2,000 + 1,500 + 800 + 600 = ₹5,900. Total cost at 13 days = 5,900 + 13 × 500 = 5,900 + 6,500 = ₹12,400.
  4. Step 1: crash B by 1 day (slope ₹300, less than ₹500). Both paths are now 12 days. Direct cost = 5,900 + 300 = ₹6,200. Total = 6,200 + 12 × 500 = ₹12,200.
  5. Step 2: both paths are critical. E is common to both, with slope ₹400 (less than ₹500). Crash E by 1 day. Both paths are 11 days. Direct cost = 6,200 + 400 = ₹6,600. Total = 6,600 + 11 × 500 = ₹12,100.
  6. Step 3: E is fully crashed. To save another day you must crash one activity on each path: the cheapest are A (₹400) plus B or D (₹300) = ₹700, which is more than ₹500. Direct cost at 10 days = 6,600 + 700 = ₹7,300. Total = 7,300 + 5,000 = ₹12,300, which is higher than ₹12,100.
  7. Summary of total cost: 13 days ₹12,400; 12 days ₹12,200; 11 days ₹12,100; 10 days ₹12,300.

Answer: The optimum duration is 11 days, with a total cost of ₹12,100 (direct ₹6,600 + indirect ₹5,500). Crash B by 1 day and E by 1 day.

Example 2

The critical path of a project takes 30 days. The next longest path takes 28 days. Activity X lies only on the critical path. Normal time 10 days at ₹50,000; crash time 7 days at ₹68,000. The cheapest activity on the 28-day path has a slope of ₹5,000 per day. Indirect cost is ₹8,000 per day. How many days should X be crashed, and what is the net saving?

Show the solution
  1. Cost slope of X = (68,000 − 50,000) ÷ (10 − 7) = 18,000 ÷ 3 = ₹6,000 per day.
  2. X can be crashed by up to 3 days. Its slope of ₹6,000 is less than ₹8,000, so each day saves ₹2,000 net.
  3. The gap to the next path is 2 days. Crashing X by 2 days brings the project to 28 days, and both paths become critical.
  4. Extra cost for 2 days = 2 × 6,000 = ₹12,000. Indirect saving = 2 × 8,000 = ₹16,000. Net saving = ₹4,000.
  5. For a 3rd day, both paths must be shortened: X ₹6,000 + the other path ₹5,000 = ₹11,000, which is more than ₹8,000. Do not crash a 3rd day.

Answer: Crash X by 2 days (project duration 28 days). Net saving is ₹4,000.

Exam tips

  • Always show the slope table first. Marks are given for method even if a later step goes wrong.
  • Draw the network and list every path with its length. After each crash, rewrite the path lengths.
  • Check whether the question gives indirect cost per day, a penalty, or a target date. A target-date question asks for the cheapest way to reach it, not the optimum.
  • End the answer with a clear recommendation: the duration, the activities crashed, and the total cost.
  • In MCQs, watch for the cost slope trap: the denominator is the days saved, not the crash time.

Practice questions from Network Analysis - PERT, CPM

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.

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

What is the cost slope in project crashing?

Cost slope is the extra cost of saving one day on an activity. It equals (crash cost − normal cost) ÷ (normal time − crash time). Crash the activities with the lowest slope first.

How do I find the optimum project duration?

Crash step by step, always on the critical path, and stop when the cost of saving a day is not less than the indirect cost per day. Confirm with a table of total cost at each duration. The lowest total cost gives the optimum duration.

What do I do when two paths become critical during crashing?

Shorten both paths by the same number of days. Look for a common activity with a low slope. Otherwise, crash one activity on each path, and add their slopes to get the cost per day.

Can I crash a non-critical activity?

It does not shorten the project, so it only adds cost. Crash a non-critical activity only if it becomes critical after other crashing.