Learning Curve…

PM0010 : if the optimistic estimate of an activity is 12 days & pessimistic estimate is 18 days. What is the variance of this activity?

Posted on: February 23, 2012

PM0010  : if the optimistic estimate of an activity is 12 days & pessimistic estimate is 18 days. What is the variance of this activity?

Answer:-

Estimated time of the project = (to + 4* tm + tp)

If three standard deviations are chosen for the optimistic and pessimistic times,then there will be six standard deviations between them. In this case the varianceof each activity completion time is given by

Variance= [(tp – to) / 6]2

 to=12

tp=18

 

Variance = [(18-12)/6]2

            = [6/6]2

=[1]2

= 2

Variance = 2

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Author

Learning days (Calendar)

February 2012
M T W T F S S
« Jan   Mar »
 12345
6789101112
13141516171819
20212223242526
272829  

Knowledge Bank (Archives)

I am on Twitter

Blog Stats

  • 498,419 hits

Enter your email address to subscribe to this blog and receive notifications of new posts by email.

Join 43 other followers

%d bloggers like this: