Simple Example of the Alternative way to determine the CP and Calculate the Slack
1:20 – Precedence Information to build the precedence diagram
1:35 – Drawing the Precedence Diagram
2:55 – Identification of the paths through the network and calculating the duration
3:55 – Identification of the CP and activities on the CP
4:30 – Calculating the slack of the nob-CP activities
5:55 – Classical calculation in Fast Forward
6:30 – Comparing the two methods
7:10 – Conclusions
In the classical method, we draw the network diagram, perform the forward and backward pass to determine ES, EF, LS, LF, and the slack, and finally determine the CP.
In the alternative method, we still draw the network diagram but instead of doing the forward and backward pass, we identify the different paths through the network and calculate their length:
ACF = 15
ADF = 14
BEG = 8
And from the definition of the CP, we know that the CP is the longest path through the network, in this case, ACF with a length of 15.
The next step is to calculate the slack of all activities and we know that the slack of all activities on the CP = 0. This means that A, C, and F have zero slack.
Let’s look at the other paths and activities:
A and F are also on the CP, their slack = 0.
The remaining activity D is on a path with length 15 and the CP has a length of 15
The slack of D = 15 – 14 = 1
All three activities B, E, and G are on the same path with length 8 and we have to compare this with the CP.
The slack of B, E and G is then equal to 15 – 8 = 7
When we calculate the schedule using the classical method, we find the same values
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