After you have drawn a network diagram and understand the flow the work needs to follow, you can begin to compute the duration of the project. How long will it take to complete all of the work? To answer this question, you will need to add the duration of the activities along each path of the network. This will also allow you to identify the critical path.
The critical path:
By computing the critical path for the project, you will also compute the desired duration. Using the network diagram and the estimates provided, you can determine when each activity can begin and finish.
Computing the critical path:
Consider the following network diagram.
On each node representing an activity, the following attributes have been identified:
|ES –||Early start. The earliest time an activity can start.|
|EF –||Early finish. The earliest time an activity can finish.|
|LF –||Late finish. The latest time an activity can finish without impacting the project schedule.|
|LS –||Late start. The latest time an activity can start without impacting the project schedule.|
|DU –||Duration. The number of work periods (days) required for completion of an activity.|
To calculate the ES and EF times of each activity:
Activity B can start as soon as Activity A is complete, so the ES for Activity B is equal to the EF of Activity A. With the ES of 10, adding the duration of 15 results in the EF of 25.
Continuing through the path:
Note that there is an 8-day lag between activities F and G.
Also note the impact on the ES for an activity with multiple precedents.
Completing the computations shows that the earliest we can complete the project activities is 66 days based on the current network diagram.
The LS will identify the latest time the activity can start without causing a delay in completing the entire project. The LF indicates the latest the activity can finish without delaying the project schedule. To calculate the LF and LS times:
Since we have already identified that we can complete the project in 66 days, this becomes our target time for completing the project. Beginning with the last activity, Activity H, we set the LF equal to the activity's EF. This makes sense because, if Activity H finishes any later than day 66, the project will obviously be delayed.
Subtracting the duration of the activity from the LF results in the LS (66 - 5 = 61.) Moving from right to left through the network, if the latest Activity H can start is time 61.
To complete the calculations for the remaining activities:
Slack, also known as float or total float:
In the network diagram with times shown above, slack is represented as TF. Only activities C, D, and E have slack. Any one of these activities can be delayed by up to 11 days without delaying the project past 66 days. Alternatively, all three activities can be delayed, so long as the total number of days does not exceed 11, and the project will still achieve the 66 day schedule.
As indicated above, any activity in the network with slack equal to 0 is considered to be on the critical path. In the network diagram with times shown above, activities A, B, F, G, and H have TF = 0. Should any of the activities be delayed by even one day, the project schedule will be delayed.
Identifying the critical path:
Jim Muller has been working as a Project/Program Manager for over 20 years. He has managed projects for IT/IT as well as on the business side. Projects ranged from $100 million IS development program, mergers and acquisitions, the launch of business products, and physical relocations of business units. Jim has also worked on the development of internal PM Methodologies, implemented a Project Management Office, and continually provided coaching and mentoring for project management staff. Jim has provided project management training for companies as well as teaching at the university level.