# Critical Path – An End-To-End Example Of Finding Critical Path And Float

Critical path– For the purpose of PMP exam critical path is at most important. Also many PMP aspirants find the mathematical calculations of critical path or logic difficult to understand. For these reasons I decided to write a post on end-to-end critical path example to calculate critical path, float in a project.

For the purpose of this post, as it an end to end example for critical path, I would not go into detail of each and every definition or concept of critical path and other related concepts. So incase if you are not familiar, please read through the below blog posts, before start reading this one.

## What is PERT and CPM?

Also, the basis for the all concepts pertaining to critical path is calculating the activity duration. PERT is a technique to calculate the activity duration. Read through program evolution and review technique (PERT), if you are not familiar with this topic.

Coming to Critical Path Method (CPM), the following example will explain on how to calculate critical path and float values in a project example with solution.

So jump into the topic of calculating the critical path and different type of floats right away. ## How to Calculate Critical Path, Float/Total Float and Free Float Example with Solution

For the purpose of this example consider the following network diagram ## How to Calculate Critical Path and Float – Steps

1. List Down All The Paths And Find The Critical Path
2. Find The Early Start And Early Finish
3. Find The Late Start And Late Finish
4. Total Float – How To Find It?
5. Free Float – How To Find It?

## List Down All The Paths And Find The Critical Path

The paths in the above project schedule network diagram are as follows

Path 1 : START=>A=>B=>C=>Finish

Path 2 : START=>A=>B=>F=>Finish

Path 3 : START=>D=>E=>H=>Finish

Path 4 : START=>D=>G=>H=>Finish

Now let us find the length of each of the paths

Total duration of Path 1 =2+3+4=9

and the total duration of Path 2 =2+3+7=12

Total duration of Path 3 =5+6+9=20

Total duration of Path 4 =5+8+9=22

So path 4 is the longest path among all other paths. Hence path 4 is the critical path.

Now if you find subtract the length of each path from critical path you will find the float.

For example float of the activities in path 1=13, path 2=10 and path 3=2.

## Find Out The Early Start And Early Finish

At this stage the network diagram looks as follows. Let us start first with the critical path and find out the early start and early finish using forward pass.

### Longest path is Path 4

So the activities in critical path are D, G and H.

#### Activity D

ES of activity D =1 as this is the first activity in the path.

EF=ES + Duration – 1 = 1+5-1=5

#### Activity G

ES = EF of predecessor activity + 1=5+1 =6

EF= ES + Duration – 1 = 6+8-1=13

#### Activity H

ES = EF of predecessor activity + 1= 13+1 =14

EF= ES + Duration – 1 = 14+9-1=22

### Next Longest Path is Path 3

So the activities in path3 are D, E and H.

#### Activity D

Already calculated as part of critical path.

#### Activity E

ES = EF of predecessor activity + 1=5+1 =6

EF= ES + Duration – 1 = 6+6-1=11

#### Activity H

Activity H has two predecessors. They are E and G.

Remember we learned that ES of H should be the maximum of EF of E and G. In this case it is 13.

So ES of H = 13+1

EF = 14+9-1=22

### Next Longest Path is Path 2

So the activities in path2 are A, B and F.

#### Activity A

Already calculated as part of the other paths.

ES of activity A =1 as this is the first activity in the path.

EF=ES + Duration – 1 = 1+2-1=2

#### Activity B

Already calculated as part of the other paths.

ES = EF of predecessor activity + 1= 2+1 =3

EF= ES + Duration – 1 = 3+3-1=5

#### Activity F

ES = EF of predecessor activity + 1= 5+1 =6

EF= ES + Duration – 1 = 6+7-1=12

### Next Longest Path is Path 1 (This is the last path left)

So the activities in path2 are A, B and C.

#### Activity A

ES of activity A =1 as this is the first activity in the path.

EF=ES + Duration – 1 = 1+2-1=2

#### Activity B

ES = EF of predecessor activity + 1= 2+1 =3

EF= ES + Duration – 1 = 3+3-1=5

#### Activity C

ES = EF of predecessor activity + 1= 5+1 =6

EF= ES + Duration – 1 = 6+4-1=9

So we have covered all the paths and find out early start and early finish using the forward pass.

## Find Out The Late Start And Late Finish

Let us start first with the critical path and find out the late start and late finish using backward pass.

### Longest path is Path 4

So the activities in critical path are D, G and H.

#### Activity H

Late finish of activity H is same as early finish of activity H, as this is on critical path.

Late start of activity H is same as late start of activity H, as this is on critical path.

#### Activity G

Late finish of activity G is same as early finish of activity G, as this is on critical path.

Late start of activity G is same as late start of activity G, as this is on critical path.

#### Activity D

Late finish of activity D is same as early finish of activity D, as this is on critical path.

Late start of activity D is same as late start of activity D, as this is on critical path.

### Next Longest Path is Path 3

So the activities in path3 are D, E and H. Since we are traversing through backward pass, let us traverse through activities in reverse order. That is H, E and D.

#### Activity H

Already calculated as part of critical path.

#### Activity E

LF= EF of predecessor activity-1 = 14-1=13

LS = LF-Duration+1=13-6+1 =8

#### Activity D

Already calculated as part of critical path.

LF of D should be taken as minimum of LS of E and G.  Although D has two successors namely E and G, Since D is on critical path, so the LF is already the minimum value.

LF = 6-1 = 5

LS = 5-5+1=1

### Next Longest Path is Path 2

So the activities in path2 are A, B and F. Since we are traversing through backward pass, let us traverse through activities in reverse order. That is F, B and A.

#### Activity F

LF=same as EF of end critical path activity (H) = 22

LS of activity C =LF-duration+1 = 22-7+1=16

#### Activity B

EF = ES of predecessor activity -1

Remember here we have to take minimum of ES of C and F. But we are traversing the backward pass in descending order of duration of the paths, then it should be automatically considered. If there is any confusion hold this, and find out the LF and LS for activity C and decide the LF for activity B.

LF = 16-1 =15

LS=15-3+1=13

#### Activity A

LF = EF of predecessor activity – 1= 13+1 =12

LS= LF – Duration + 1 = 12-2+1=11

### Next Longest Path is Path 1 (This is the last path left)

So the activities in path1 are A, B and C. Since we are traversing through backward pass, let us traverse through activities in reverse order. That is C, B and A.

#### Activity C

LF of activity C is same as EF of the end critical path activity H.

LF =22

LS = 22-4+1=19

#### Activity B

Already calculated as part of the other paths.

Remember we have LF of B as the minimum of LS of C and F-1.

So LF = 16-1 =15

LS = 15-3+1=13

#### Activity A

Already calculated as part of previous paths.

So we have covered all the paths and find out late start and late finish using the backward pass.

Now the completed network diagram looks as follows. ## Total Float – How To Find it?

Since we have the network diagram completed with the values of early start, early finish, late start and late finish now we can calculate total float.

As a matter of fact the total float of critical path activities is zero by definition.

Total float of activity E = 2,

and the total float of activity F = 10

Total float of activity C = 13

Total float of activity B = 10

Finally total float of activity A = 10

## Free Float – How To Find It?

Now free float make sense when  a successor node has multiple predecessors.

Free float of E=14-11=3

## Conclusion

For the purpose of this post, we have taken a bit more complex example to calculate critical path, total float and free float.

Hope this article helps in learning the mathematical calculations of schedule management, which is the main objective of this article.