Can anyone with project management knowledge explain how to do a forward and backward pass calculation?
I have an exam coming up and I really can’t wrap my head around this. I would be very happy if someone would be willing to meet in chat and explain it.
Thanks!
Really it’s just common sense, except by the time you have some fancy computer software to "help" you do it, the commn sense gets forgotten!
The first thing you want to know is how quick you can get the project done, i.e. the earlest possible start and end time for every task.
You get that from doing a forwards pass. The earliest time a task can start is when all the tasks that precede it have finished.
There are always some tasks that start at the beginning of the project. They start at time 0. The earliest time they can end is just the length of the task.
After you have "filled in the blanks" for those tasks, there will be some more tasks where you know the end time for all the preceding tasks. If a task depends on three others that end at times 2 7 and 5 say, the earliest that task can start is time 7, and the earliest it can end is 7 + its length.
You keep going till you have the earliest start and end times for all the tasks. Then, you know the earliest end time for the whole project.
Now, you want to know the LATEST time that every task can start, and still meet that end time. So you work backwards from the end. Find the all tasks that finish at the end of the project. Their latest end time is obviously the project end time that you got from the forward pass. Their latest start times are the end time – the length of the task. Then work backwards through the remaining tasks till you get to the project start. If you don’t have at least one task where the latest start time is 0, you made a mistake somewhere.
You now have the earilest start time (from the forwards pass) and the latest end time (from the backwards pass) for each task, so you can see if the tasks are on the crifical path (latest end – earliest start = length of task) and if not, how much "float time" there is for the task.
Really it’s just common sense, except by the time you have some fancy computer software to "help" you do it, the commn sense gets forgotten!
The first thing you want to know is how quick you can get the project done, i.e. the earlest possible start and end time for every task.
You get that from doing a forwards pass. The earliest time a task can start is when all the tasks that precede it have finished.
There are always some tasks that start at the beginning of the project. They start at time 0. The earliest time they can end is just the length of the task.
After you have "filled in the blanks" for those tasks, there will be some more tasks where you know the end time for all the preceding tasks. If a task depends on three others that end at times 2 7 and 5 say, the earliest that task can start is time 7, and the earliest it can end is 7 + its length.
You keep going till you have the earliest start and end times for all the tasks. Then, you know the earliest end time for the whole project.
Now, you want to know the LATEST time that every task can start, and still meet that end time. So you work backwards from the end. Find the all tasks that finish at the end of the project. Their latest end time is obviously the project end time that you got from the forward pass. Their latest start times are the end time – the length of the task. Then work backwards through the remaining tasks till you get to the project start. If you don’t have at least one task where the latest start time is 0, you made a mistake somewhere.
You now have the earilest start time (from the forwards pass) and the latest end time (from the backwards pass) for each task, so you can see if the tasks are on the crifical path (latest end – earliest start = length of task) and if not, how much "float time" there is for the task.
References :