In the last post, I discussed the formal method for determining the critical path of a network using the forward and backward pass method. This also gives you the float of each activity on the network. Recall that the float of any activity on the critical path is by definition equal to 0, the float of any activity on a non-critical path is the number of time units it can be delayed without affecting the successor activity.
There is a shortcut alternative which I found saves a lot of time, which is especially important when taking the PMP or CAPM exam.
Here’s how it works.
1. Draw the network diagram for the problem.
2. List the potential paths for the project and their durations.
3. Whichever path you find as a result of step 2 has the longest duration is the critical path.
4. Label ALL activities on the critical path as having float of 0.
5. Calculate the float of the non-critical paths found as a result of step 2 by taking the duration of the critical path minus the duration of that path.
6. Take the non-critical path with the next highest float above 0.
7. Label all activities on that non-critical path with the float of that path EXCEPT for those that you already labeled as a result of step 4.
8. Taking the non-critical path with the next highest float above the one you found as a result of step 6.
9. Label all activities on that non-critical path with the float of that path EXCEPT for those that you already labeled in a previous step.
10. Continue until all activities are labeled.
Let me illustrate.
Here is a chart with the activities of a project. Answer these questions.
1. What is the critical path of the project?
2. What is the float of each activity?
| Activity | Predecessor | Duration | Float |
| Start | None | 0 | |
| A | Start | 2 | |
| B | A | 9 | |
| C | B | 18 | |
| D | A | 6 | |
| E | D | 10 | |
| F | A | 4 | |
| G | F | 14 | |
| H | E, G | 8 | |
| I | C, H | 9 | |
| J | I | 5 | |
| End | 0 |
If you draw the network diagram, you get the following three paths and their durations.
Path #1: A-D-E-H-I-J , Duration = 2 + 6 + 10 + 8 + 9 + 5 = 40
Path #2: A-F-G-H-I-J, Duration = 2 + 4 + 14 + 8 + 9 + 5 = 42
Path #3: A-B-C-I-J, Duration = 2 + 9 + 18 + 9 + 5 = 43
The longest path is path #3, with duration 43. This is the critical path and all activities have a float of 0.
Let’s look at path #2, which has the next smallest duration of 43, and a float of 43 – 42 = 1. All of the activities on path #2 that are NOT ALREADY accounted for in path #3 (the critical path) are assigned a float of 1. That means F, G, H are all have float of 1.
Finally let’s look at path #1, which has the next smallest duration of 40, and a float of 43 – 40 = 3. All of the activities on path #1 that are NOT ALREADY accounted for in either path #3 or path #2 are therefore assigned a float of 3. That mean D and E have a float of 3.
So the finally answer is the duration of the critical path is 43, and the chart below contains the float of all the activities.
| Activity | Predecessor | Duration | Float |
| Start | None | 0 | N.A. |
| A | Start | 2 | 0 |
| B | A | 9 | 0 |
| C | B | 18 | 0 |
| D | A | 6 | 3 |
| E | D | 10 | 3 |
| F | A | 4 | 1 |
| G | F | 14 | 1 |
| H | E, G | 8 | 1 |
| I | C, H | 9 | 0 |
| J | I | 5 | 0 |
| End | 0 | N.A. |
If you would have done that same problem with the forward and backward pass, it would have taken a LOT longer.
The next post will cover the critical chain method; it sounds like the critical path method, but is a different technique altogether, with terminology involving “buffers” borrowed from lean manufacturing.
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4squareviews 
Much better explanation than the course I am taking. One question – where did the ‘2’ come from? Got a little lost.
Thanks for the comment–I appreciate your interest and attention to my blog. In answer to your question, if by the “2” you meant the duration of activity A, that is just given in the problem. If it’s some other figure you’re referring to, you’ll have to be more specific in order for me to answer your question.
D & E should have a float of 3, not 2 … (43-40 = 3)
Thanks, Devin! The method still works, but somehow my brain stopped working during that problem. You’re right, D & E should have a float of 43-40 = 3. Thanks for correcting my mistake–I’ve gone ahead and corrected the post!