CIMA E2 Syllabus C. Managing projects - EET - Notes ^{6 / 16}

Earliest event time (EET) is entered in the upper-right quadrant of each event circle.

We work from left to right to find the EET. It is the earliest time in which we can finish all the activities that need be done.

Using our ‘Example Above’ (EA), we will calculate EET for Events 2, 3, 4 and 5 in the following way:

The EET for Event 2 is simply 5 weeks as only activity A needs to be completed.

The EET for Event 3 is 9 weeks because I would have to do (Ac A) 5 + (Ac B) 4 weeks.

The EET for Event 4 is 10 weeks because I would have to do (Ac A) 5 + (Ac C) 5 weeks.

The EET for Event 5 is 8 weeks because I would have to do (Ac A) 5 + (Ac D) 3 weeks. 

On the diagram, it would look like this:

What would the EET for Event 6 be?

I could either go:

Activities ABE: 5+4+5 = 14

Activities ACF: 5+5+6 = 16

Activities ADG: 5+3+7 = 15

I must finish ALL of the Events, the earliest i can finish of these activities is via the longest of all of the paths mentioned above which is 16, therefore the EET for Event 6 is 16.

For Event 7, it will be 16 + 7 for Activity H, therefore the EET for Event 7 is 23 which is also the total duration for the project.