Question RE free float and total float
Submitted by rcouprie on Fri, 12/28/2012 - 19:36
Hi,
I'm preparing for my PMP exam in a couple weeks and have been reviewing Oliver Lehmann's collection of 75 questions. I'm a little puzzled by Q18 below.
18. Start dates in the following network logic diagram are defined as early morning, finish dates are evening. If tasks are scheduled to begin at early start date, what is true?
1) Activity B has a free float of 10 d.
2) Activity B has a total float of 10 d.
3) Activity A has a free float of 10 d.
4) Activity A has a total float of 10 d.
The posted answer is option 2, and I agree that statement is true. However, I believe that option 1 is also true: task B can be delayed by a max of 10 days before it impacts on the early start date for task D. Why would option 1 be untrue?
Thanks!
Forums:
sspawar
Sat, 12/29/2012 - 05:12
Permalink
The universal formula of
The universal formula of float (or total float) is = LF-EF = LS-ES (This stands similar in both method, 0start or 1start)
For free float -- you can't slip ES of successor as scheduled.
Now in current example - if activity B, gets LF on 31st day evening, how could you start activity D on 31st day morning. (remember it is example of start with 1 method).
Thus you can say that free float would be 9 days and not 10days.
Formula for 1start method FF = ESs -EFp -1 -Lag
Formula for 0 start method FF = ESs -EFp -Lag
I can understand where you are confusing, while looking for Total float, you probably not looking at LS of activity of D (32nd day morning), and comparing with ES of D (31th day).
(Remember,in question, even if, LS of activity D were not given or may be given 31st day, then and that, still, by default total float or float becomes always = LF-EF = LS-ES =10 days. Because by definition, network end scheduled date will not be affected, and LS and LF are calculated on that basis during backword pass. and here part network is given. End of network is still ahead.)