Float versus standard deviation

Submitted by shaikfromriyadh on Sun, 04/29/2012 - 12:50

Question: The sponsor wants the project to be completed in 40 days. The CPI is 1.1, the project critical path duration is 38 days with the standard deviation of 2 days. What is the maximum project float.


a) 8 days


b) 2 days


c) 4 days


d) 0 days


The answer is 4.


Can anyone explain what is the relation ship between float and standard deviation.


Thanks,


Abdul khadar


 

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sspawar

Sun, 04/29/2012 - 14:30

As per normal distribution curve :

STANDARD DEVIATION = 2

It means =   +/- 2n DAYS DURATION CAN VARY FROM THE MOST LIKELY (mean value) =38DAYS(OR ANY VALUE IN DAYS)

Here n is degree of probablity

There is appx 68% chances (n=1) that project may complete in 36days to 40 days. ---- here float is 4 days

There is appx 95% chances (n=2) that project may complete in  34days to 42days ------here float is 8 days

there is apx 99.73% (n=3) chances that project may complete in 32days to 44days - ---here float is 12 days

and sso on ---------

second condition is maximum expected time (sponser has fixed) is  = 40days

hence 1st option (C)  is right one. = 4days

 

or

as per variance formua  = ((P-O)/6)^2

Sigma( st deviation) =  P-O/6  =2DAYS

hence V = 2^2 =  +/- 4 days ( this variance formula is for sigma one (68%) probablity )

This variance is here float for total duration.

projmanpro

Sun, 04/29/2012 - 16:37

sspawar i agree with your explanation dude

shaikfromriyadh

Mon, 04/30/2012 - 07:49

Thank you....

sspawar

Mon, 04/30/2012 - 12:06

BY normal dist curve it will be  - 0 to 6days float for n=1

                                                   0 to 12days float for n =2

By variance formula  =    3^2 = 3x 3  =  0 to 9 days float., variance formula is irrespective of any value of n.

It sounds us the later formula is to be followed in exam / exam point of view, because this is PMBOK formula.

Regards