The reason that the drag of an activity is the lowest total float of the parallel activity with the lowest total float is that, as one removes days from the duration of the critical path activity, the project duration is reduced UNTIL the float of a parallel activity becomes zero. Thereafter, shortening the CP activity will NOT reduce the project duration because that parallel activity with the least float will have become part of a new critical path (and will therefore have drag!).
One issue is how to define "parallel". Two activities are parallel NOT because they are scheduled to occur simultaneously, but because they are not on the same path. In other words, neither is an ancestor nor descendant of the other.
1. B is parallel with F and H, but NOT with G, which is a descendant of B. Therefore if we start taking days out of B, after we have shortened it from 20D to 5D, the project duration will have come in from 65D to 50D. But now, although H will STILL have 5D left of its 20D of float, all 15 days of F's float will have disappeared and it will be on a parallel critical path. Taking more days out of B will not shorten the project further, but just create float on the original critical path.
2. C is parallel to exactly the same activities as B (F and H) and so WOULD have drag of 5D -- except C's duration is just 5D and so is adding only 5D to the project duration.
3. D is NOT an ancestor or descendant of G, and so is parallel with all of F, H AND G. Therefore D's drag is constrained by G's TF of 5D, as reducing D's duration by just 5D uses up G's float and puts it on a parallel CP even while F and H still have float.
I hope this helps, Kannan.
I still would like to know why Oracle Primavera ignores this fundamental and critical scheduling metric.
Member for
20 years 7 monthsHi, Kannan.The reason that
Hi, Kannan.
The reason that the drag of an activity is the lowest total float of the parallel activity with the lowest total float is that, as one removes days from the duration of the critical path activity, the project duration is reduced UNTIL the float of a parallel activity becomes zero. Thereafter, shortening the CP activity will NOT reduce the project duration because that parallel activity with the least float will have become part of a new critical path (and will therefore have drag!).
One issue is how to define "parallel". Two activities are parallel NOT because they are scheduled to occur simultaneously, but because they are not on the same path. In other words, neither is an ancestor nor descendant of the other.
The Wikipedia page for "critical path drag" and the diagrams on it may help:
[[wysiwyg_imageupload:6886:]]
In the diagram:
1. B is parallel with F and H, but NOT with G, which is a descendant of B. Therefore if we start taking days out of B, after we have shortened it from 20D to 5D, the project duration will have come in from 65D to 50D. But now, although H will STILL have 5D left of its 20D of float, all 15 days of F's float will have disappeared and it will be on a parallel critical path. Taking more days out of B will not shorten the project further, but just create float on the original critical path.
2. C is parallel to exactly the same activities as B (F and H) and so WOULD have drag of 5D -- except C's duration is just 5D and so is adding only 5D to the project duration.
3. D is NOT an ancestor or descendant of G, and so is parallel with all of F, H AND G. Therefore D's drag is constrained by G's TF of 5D, as reducing D's duration by just 5D uses up G's float and puts it on a parallel CP even while F and H still have float.
I hope this helps, Kannan.
I still would like to know why Oracle Primavera ignores this fundamental and critical scheduling metric.
Fraternally in project management,
Steve the Bajan
Member for
17 years 4 monthsHi Stephen,Can you please
Hi Stephen,
Can you please explain the concept that the drag days shall be the lowest total float of parallel activities. If possible with a clear example.
Best Regards
Kannan