1. Introduction
What if you want to shorten the schedule of your project without having to change the scope? To do so, you can use one of the schedule compression techniques, which change one of the other project constraints, either cost or risk.
| Technique | Description | Increases | |
| 1. | Crashing | Shortens schedule by adding resources to activities | Cost |
| 2. | Fast-Tracking | Activities done in sequence are now performed in parallel | Risk |
The thing to remember about each of these techniques is that they are done on critical path activities. If they were done on non-critical path activities, it would reduce the duration of those activities, but would not necessarily reduce the duration of the project itself.
Crashing is done by adding resources to a group of activities for the least amount of additional cost required to shorten the duration of those activities by the given amount. Having people work overtime, or hiring additional workers on a temporary basis to do an activity are examples of crashing a project.
Fast tracking is done by taking activities that are in sequence and putting them in parallel to a certain extent. This increases the risk because two activities are going on once for at least part of the work.
2. Crashing exam question example
Given the chart below, what is the cost of completing the project so that it only takes 7 days to complete?
| Task | Duration | Predecessor | Cost | Crash Cost | Maximum Crash Days |
| A |
3 |
None | 1500 | 300/day | 1 |
| B |
5 |
A | 1000 | 200/day | 2 |
| C |
4 |
A | 2400 | 300/day | 2 |
| D |
3 |
B | 1100 | 100/day | 1 |
| E |
2 |
C | 500 | 100/day | 1 |
Step 1: Figure out the critical path.
If you draw the network, the possible paths through the network from start to finish are A-B-D = 3+ 5 + 3 = 11 days or A-C-E = 3 + 4 + 2 = 9 days.
Step 2: Figure out the various possible options you have of different combinations of activities to crash in order to achieve the given result.
Therefore A-B-D is the critical path and A, B, and D are the activities you must concentrate on therefore to reduce to 7 days. However, if the critical path A-B-D is reduced by 4 days, then the durations of the paths will be A-B-D = 7 days and A-C-E will be 9 days, and it will become the new critical path. In order to shorten A-C-E from 9 days to 7 days, you will also have to reduce its duration by 2 days as well.
With the pathway A-B-D, you can shorten A by 1, B by 2, and D by 1 for a total of 4 days. With A-C-E, since you have already shortened A, you must either shorten a combination of C or E by 2 days in order to reduce it by 2 days. Since C can be shortened by 2, and E by 1, there are two ways of doing this:
Option 1: Shorten C by 2 days
Option 2: Shorten C by 1 day, and E by 1.
Step 3: Compare the costs of the different options.
| Task | Duration | Predecessor | Cost | Crash Cost | Maximum Crash Days |
Option 1 |
Option 2 |
| A |
3 |
None | 1500 | 300/day | 1 | 300 | 300 |
| B |
5 |
A | 1000 | 200/day | 2 | 400 | 400 |
| C |
4 |
A | 2400 | 300/day | 2 | 600 | 300 |
| D |
3 |
B | 1100 | 100/day | 1 | 100 | 100 |
| E |
2 |
C | 500 | 100/day | 1 | 100 | |
| Subtotal | 6500 | 1400 | 1200 | ||||
| Total | 7900 | 7700 |
The cost of accomplishing the activities A through E without crashing are 6500. With crash option 1, you are adding 1400 of cost, and with crash option 2, you are adding 1200 of cost. Since 1200 is the option with the least amount of additional cost, you should choose this for a total of 6500 + 1200 = 7700.
Note: be carefu; of the wording of these questions. Some questions will ask you how much it will cost to crash the project, meaning what is the cost of reducing the duration of the project. Other questions, like this one, will ask about the total cost of the project, which will equal the regular cost plus the cost to crash the project.
This is the last of the tools & techniques, except for the project management information software (Microsoft Project, Primavera, etc.). The next post will discuss the last of the schedule management processes, this one being the process used to monitor and control the schedule.
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4squareviews 
Regarding crashing & fast-trackin non-critical path tasks, I think you menat to say, “If they were done on non-critical path activities, it would increase (not reduce) the float of those activities.” Crashing or fast-tracking non-critical task shortens the duration, which would increase the float (e.g., a task with a duration of 10 and float of 2 before crashing might have a float of 3 or even more after crashing, due to reduced duration)
The word “float” is in error. What I meant to say was “if they [crashing & fast-tracking] were done on non-critical path activities, it would reduce the duration of those activities, but would not necessarily reduce the duration of the project itself.” I wanted to contrast between reducing the duration of an activity and reducing the duration of the entire project. This necessarily happens with an activity on the critical path, but not necessarily when an activity is on a non-critical path.
Thanks again, Devin, for your close reading of the blog!
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Thanks for this example! Very helpful. I do have a question, I came to a different pair of solutions. After reducing the A-B-D critical path from 11 to 7 days, the new critical path A-C-E is no longer 9 days long as reducing A by one day means A-C-E is now 8 days long. So we only need to reduce the new critical path by one day instead of two as in your example. We can choose either C($400) or E ($100) to reduce. My totals are $7600 or $7400. Do you agree?
Thanks for this example! Very helpful. I do have a question, I came to a different pair of solutions. After reducing the A-B-D critical path from 11 to 7 days, the new critical path A-C-E is no longer 9 days long as reducing A by one day means A-C-E is now 8 days long. So we only need to reduce the new critical path by one day instead of two as in your example. We can choose either C($300) or E ($100) to reduce. My totals are $7600 or $7400. Do you agree?
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Hello, I’m not sure if I well understood. If the purpose is to reduce it to 7 days and if we already shortened A by 1, why do we still have to shorten C- E by 2 and not by 1? (i.e. if A takes 2 days, C takes 4 days, E takes 1 day, the total project would still take 7 days as the A- B-D crashed version).
In your example, I think with A-C-E, since we have already shortened A (by 1 day), we must either shorten a combination of C or E by 1 day in order to reduce it by 1 day. (not 2 days) because A-C-E = 3 + 4 + 2 = 9 days,right? I mean we have two options:
Option 1: Shorten C by 1 day
Option 2: Shorten E by 1 day.
in fact we will have 2 critical path.
I had the same conclusion. Perhaps we had to crash A twice because A-C-E would be the same duration as the original stated critical path, A-B-D? Thus, this diagram would be without a critical path?
If so, this is a tricky question.