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How to calculate the earliest start and earliest finish in critical path method?

Please provide me with a step-by-step process on how to calculate the ES and EF in CPM.
asked 10 years ago by anonymous

2 Answers

This is the first stage of the two stage process which maps put all the timings.  What you do is this:

Add up all the times as you go from start to finish, so if the first task has a duration of, say 4 weeks and the second one takes, say, 7 weeks, then the earliest start date of the first one is after zero weeks (so you put a zero in the top left corner of it's box) and the earliest start of the second one is after 4 weeks, so you put in a 4.  On the third task you put 11 because that's the earliest you could start it, you can't do it sooner than that.  It's just adding up. The largest total from the various routes is the Critical Path.

Things get interesting when you then do the backward pass, ie the same but from the finish, taking the numbers off the critical path total.  So if the total came to 50 weeks and the last task was 4 weeks long you would put 44 in the bottom left corner - it must be started by then, so that's the latest start.

On the critical path the earliest and latest starts are the same, but on the floating tasks they will be different.  The difference between them is how much float you have for that task - how much it can move and still not hold up the project.

The earliest and latest finish (put in the right hand corners) are the same as the starts but with the duration added.  So in my example that I started with, EF is 4 for the first task, and 11 for the second.  and during the backward pass the LF for the last task is 50.

All this easy if you are shown it and then try it on a real diagram rather than just in words....
answered 9 years ago by anonymous
Let's assume your project consists of 4 tasks: Task A, Task B, Task C, and Task D.

Task A (first task for the project, takes 2 days) starts on January 10th 2011, its early finish is January 12th 2011 (morning), and its late finish is January 14th  2011.

Task B has Task A as its predecessor, and its early start (ES) is the early finish of Task A (since it is the predecessor), so it will start on January 12th 2011. Task B takes 5 days to complete, which makes its Early Finish (EF) January the 19th (morning).

Task C has Task A as its predecessor, and Task C's duration is 1 day. The early start of Task C is January the 12th, and its early finish is January 13th.

Task D (takes 3 days) has both task B and task C as predecessors, in this case, the ES is the longest EF of the two predecessor tasks, which is Task B, so Task D early start is January 18th, and EF is January 21st (morning).

The critical path is Task A Duration + Task B Duration + Task D Duration = 2 + 5 + 3 = 10
answered 9 years ago by humblepm (17,390 points)

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