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### PMP®:CAPM® - Critical Path Method (CPM)

The Critical Path Method (CPM) helps in keeping projects on track.
Critical path schedules help in identifying the activities that must be completed on time in order to complete the whole project on time.
They show which tasks can be delayed and for how long without impacting the overall project schedule.
They help in calculating the minimum amount of time it will take to complete the project.
They help in knowing the earliest and latest dates each activity can start on in order to maintain the schedule.
The CPM has 4 key elements...
1.Analyzing the Critical Path
2.Calculation of Early Start & Early Finish
3.Calculation of Late Start & Late Finish
4.Calculation of Float
1. Analyzing the Critical Path
The critical path is the sequence of activities with the longest duration. A delay in any of these activities will result in a delay for the whole project.  The duration of each activity is listed in each node in the diagram. For each path, add the duration of each node to determine its total duration.

There are 2 paths in this sample project. Task 1, Task 2 and Task 4 is one path with total duration 1+3+1=5. Task 1, Task 3 and Task 4 is the second path with total duration 1+2+1=4. So in this case, it is very obvious that the top path is the longest path and hence, it is the critical path.

2. Calculation of Early Start & Early Finish
Let us use the technique called the Forward Pass which is used to determine the earliest date an activity can start and the earliest date it can finish. These dates are valid as long as all prior activities in that path started on their earliest start date and didn't slip.
Starting with the critical path, the Early Start (ES) of the first task is zero. The Early Finish (EF) of an activity is its ES plus its duration. Using our earlier example, Task 1 is the first activity on the critical path: ES = 0, EF = 0 + 1 = 1.
You then move to the next task in the path, in this case, Task 2. Its ES is the previous activity's EF. Task 2 ES = 1. Its EF is calculated the same as before: EF = 1 + 3 = 4.For Task 3, ES = 1 and EF = 1+2=3.
If an activity has more than one predecessor, to calculate its ES you will use the activity with the latest EF that means choose whatever is maximum. Task 2 and Task 3 are merging in Task 4. So here we have to choose the maximum of EF of Task 2 (4) and Task 3 (3) as the ES of Task 4. 4 is the maximum and so Task 4's ES = 4. EF of Task 4 is 4+1=5.

3. Calculation of Late Start & Late Finish
Next we use the Backward Pass technique to determine the latest date an activity can start and the latest date it can finish before it delays the project.
You'll start once again with the critical path, but this time, you'll begin from the last activity in the path. The Late Finish (LF) for the last activity in every path is the same as the last activity's EF in the critical path. The Late Start (LS) is the LF - duration.
In our example, Task 4 is the last activity on the critical path. Its LF is the same as its EF, which is 5. To calculate the LS, subtract its duration from its LF. LS = 5 - 1 = 4.
You then move on to the next activity in the path. Its LF is the previous activity's LS. In our example, the next Activity in the critical path is Task 2. Its LF is equal to Task 4's LS. Task 2 LF = 4. Its LS is calculated the same as earlier, by subtracting its duration from the LF.  Task 2 LS = 4 - 3 = 1.
Similarly, Task 3 Late Finish and Late Start are calculated as 4 and 2 respectively. If an activity has more than one successor to find its Late Finish you have to choose the minimum of their Late Start. For Task 1’s Late finish, you have to choose the minimum of Late Start of Task 2 and Task 3.
The activities in the critical path will have 0 float. Now let us find the float for all the activities. If in a path all the activities have 0 float, then that will be a critical path. There can be more than 1 critical path.
Total float is calculated by LS-ES or LF - EF.
For Task 1, TF = 0-0 = 1-1= 0. For Task 2, TF = 1-1 = 4-4 = 0. For Task 4,TF = 4-4=5-5=0.
For Task 3,  TF = 2-1 = 4-3 = 1.
Using the critical path diagram from the previous section, Tasks 1, 2, and 4 are on the critical path because they have a float of zero.
The next longest path is Tasks 1, 3, and 4. Since Tasks 1 and 4 are also on the critical path, their float will remain as zero. Task 3 has a float of 1 which makes the path Task 1,3 and 4 noncritical.
Free float of an activity = ES of successor - EF of present activity. (ES of successor - in case of multiple successors choose immediate successor,i.e.the least of the ES of successors).
Only Task 3, has Free float of 1 (4-3). Free float cannot be greater than Total float. For Task 1, 2 and 4 since the Total float is 0, Free float also is 0.