The last post discussed the overall technique of schedule network analysis, which actually consists of a group of techniques. The first one of these is the critical path method, which is the subject of this post. If you are taking the PMP exam, you will get a question that will ask you to figure out the *critical path *for a series of activities. This is the shortest possible duration of the project, which is figured out by showing which of the paths created by the network diagram is actually the longest or *critical* one.

The critical path method does the following:

- It starts out with the following information given in the question.

- activities (usually listed with letters of the alphabet A, B, , C, etc.)
- information on predecessors and successors for each activity, and the
- duration of each activity

The procedure of using a forward and backward pass to calculate the early start, early finish, late start, and late finish of the activities is the standard way to find out which activities are on the *critical path*. The difference between either the late start and the early start, or the late finish and the early finish, give the *float *of each activity. If the float is 0, then this means that the activity cannot be delayed without delaying the successor activity. If the float is some positive number, say 3, this would mean that the activity can be delayed 3 time units (days, weeks, or whatever your network diagram is using as units of time) before the successor activity is delayed.

To determine how long a project will take, you need to find out the *critical path*, that is, the sequence of activities in the network diagram that is the longest. Other paths along the network will yield sequences of activities that are shorter than the critical path, and they are shorter by an amount equal to the *float*. This means that activities that have *float* could be delayed by a certain amount without affecting the schedule. Activities along the critical path have a float of zero. This means that any delay along the critical path *will* affect the schedule.

Here’s an outline of the critical path methodology.

a. You create a network diagram of all the activities.

b. You label each activity with the duration derived from process 6.4 Estimate Activity Durations.

c. You do a *forward pass* to determine the early start and early finish date of all activities, from the start of the project to the end of the project.

d. Once at the end of the project, you do a *backward* pass to determine the late start and late finish date of all activities, from the end of the project to the start of the project.

e. For each activity, you use the results of c and d to calculate the *float* of each activity.

f. All activities that have 0 float are on the *critical*

*path* for that project.

Let’s take a look at the methodlogy in general.

Step 1. For each activity, create a matrix which will contain the duration, the early start, the early finish, the late start, late finish, and float for a particular activity.

Activity Number | |

Duration | |

Early Start (ES) | Early Finish (EF) |

Late Start (LS) | Late Finish (LF) |

Float |

Here are the meanings of the numbers in the boxes:

Activity Number: you can label them A through Z, or 1 through N, just as long as each activity has a unique identifier.

Duration: this is the number that you should get as an output of the 6.4 Estimate Activity Durations process.

Early Start (ES): The Early Start is the number you begin the analysis with to do the *forward pass*. It is defined as 0 for the first activity in the project. The Early Start for subsequent activities is calculated in one of two different ways, which will be demonstrated below.

Early Finish (EF): This is the next number you go to in the *forward pass* analysis. It is taken by adding the number in the ES box plus the number in the Duration box.

Late Finish (LF): The Late Finish is the number you begin the analysis with to do the *backward pass*. It is defined to be equal to the number in the Early Finish box for the last activity in the project. The Late Finish for preceding activities is calculated in one of two different ways, which will be demonstrated below.

Late Start (LS): This is the next number you go to in the *backward pass* analysis. It is taken by subtracting the number in the Duration box from the number in the LF box.

Float: Once ES, ES, LF, and LS are determined, the float is calculated by either LS – ES or LF – EF. Just remember that a piece of wood will float to the top of the water, so the float is calculated by taking the bottom number and then going upward and subtracting the number that’s on the top of it.

Step 2.

For activity A, the first activity in the project, ES = 0.

A | |

0 | |

Step 3.

Then EF for activity A is simply ES + duration. Let’s say activity A takes 5 days. Then EF = 0 + 5 = 5.

A | |

5 | |

0 | 5 |

Step 4.

The forward pass for activity A is complete. Let’s go on to activity B.

Since activity B has only one predecessor, activity B, the ES for activity B is simply equal to the EF of activity A, which was 5.

B | |

3 | |

5 | |

Then the EF for activity B is taken by adding the ES of to the duration of activity B or 3, giving EF = 5 + 3 = 8.

There’s one more situation that we have to discuss and that is if an activity has *more than one predecessor.
*

Let’s assume the durations for each activity are as follows:

Activity | Duration |

A | 5 |

B | 3 |

C | 6 |

Assume Activity A and Activity B are both done concurrently at the start of the project, and both need to be done in order for Activity C to start. Well, before we do the formal forward pass analysis, what does logic tell us. Activity A takes 5 days; Activity B takes 3 days. Both activity A and B have to be done before Activity C can take place. In this case the start date of the project is considered to be 0. Can Activity C take place on day 3, when activity B is done? No, because Activity A isn’t completed yet, and you need BOTH A and B to be done. The earliest possible start date for Activity C will be day 5, because only on that date will both A and B be done.

So this illustrates the other way of calculating ES for an activity B. If there are multiple predecessors, then the ES is equal to the LARGEST of the ES of the predecessor activities.

Step 5.

Now, let’s assume we are at the end of the project at activity Z.

Z | |

5 | |

95 | 100 |

EL = ES + duration gives us EL = 95 + 5 + 100. So the project will take 100 days according to our forward pass calculation.

Now, we have the *backward pass*.

We start this out by stating as a principle that the late finish or LF date for the last activity in the project is equal to the EF date.

Z | |

5 | |

95 | 100 |

100 | |

Then, of course, the late start date or LS = LF – duration = 100 – 5 = 95.

Z | |

5 | |

95 | 100 |

95 | 100 |

Step 6.

Now we go in the reverse direction towards the beginning of the network diagram, this time filling out the bottom LS and LF boxes for each activity.

If the activity has one successor, then the LF for the predecessor activity equals the LS for the successor of activity. But if there are more than one predecessor activity, then here’s what you do. For the forward pass, you take the highest EF of all predecessors.

For the backward pass, you take the lowest LS of all successors. Let’s see how this works.

Let’s assume the forward pass is done on A, B, and C. We do the backward analysis and we get to the following point. What is the LF of activity A?

A | B | C | |||

5 | 3 | 4 | |||

0 | 5 | 5 | 8 | 5 | 9 |

6 | 9 | 5 | 9 | ||

Well, activity B and activity C are both successors of A. In this case, activity B has an LS of 6 and activity C has an LS of 5. The earliest LS is therefore 5, and so LS of activity A is 5.

A | B | C | |||

5 | 3 | 4 | |||

0 | 5 | 5 | 8 | 5 | 9 |

0 | 5 | 6 | 9 | 5 | 9 |

Step 7.

What is the float? Take LF – EF (or LS – ES) for each of the activities.

A | B | C | |||

5 | 3 | 4 | |||

0 | 5 | 5 | 8 | 5 | 9 |

0 | 5 | 6 | 9 | 5 | 9 |

0 | 1 | 0 |

So the float of B is 1, and the float of A and C are 0. Therefore A and C are on the *critical path, *and activity B is on a *non-critical path*. If you needed to compress the schedule, you would do it with activities A and C that are on the critical path. If you shortened B by one day, however, it would not shorten the overall schedule at all, but would just reduce the float from 1 to 0.

3. Exam questions

Oh, PMI just loves giving questions on the critical path, and they can be time consuming. When taking a test with 200 questions, every minute counts, so how can you answer questions on the critical path without having to go through this long and time-consuming process? HINT: Most critical path questions can be answered without the forward and backward pass methodology. How? Read the next post to find out!

NOTE: I am going to have to edit this post because the diagrams showing the forward and backward pass did not come out as I had created them and they may be confusing. If you are reading this note, be patient as I correct the post!

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