How lastLogontimeStamp update in DC?

According to MS the lastLogontimeStamp update follow this process:
1. (Assuming the value of the ms-DS-Logon-Time-Sync-Interval is at the default of 14)
2. User logs on to the domain
3. The lastLogontimeStamp attribute value of the user is retrieved
4. 14 – (Random percentage of 5) = X
5. Current date – value of lastLogontimeStamp = Y
6. X = Y – update lastLognTimeStamp
7. X > Y – do not update lastLogontimeStamp


What I'm not sure to understand is "Random percentage of 5". What does it mean?



What I'm not sure to understand is "Random percentage of 5". What does it mean?

14 days minus random percentage of 5 days

Also see the below link for further details.
https://msdn.microsoft.com/fr-fr/library/windows/desktop/ms676824%28v=vs.85%29.aspx

Will.



What I'm not sure to understand is "Random percentage of 5". What does it mean?

14 days minus random percentage of 5 days

Also see the below link for further details.
https://msdn.microsoft.com/fr-fr/library/windows/desktop/ms676824%28v=vs.85%29.aspx

Will.



Is it like this:
14 days minus random percentage of 5 days = 14 days - (X/100*5)



Below is how the last logonTimeStamp is updated.


1. (Assuming the value of the ms-DS-Logon-Time-Sync-Interval is at the default of 14)
2. User logs on to the domain
3. The lastLogontimeStamp attribute value of the user is retrieved
4. 14 - (Random percentage of 5) = X
5. Current date - value of lastLogontimeStamp = Y
6. X = Y - update lastLognTimeStamp
7. X > Y - do not update lastLogontimeStamp

Reference: http://blogs.technet.com/b/askds/archive/2009/04/15/the-lastlogontimestamp-attribute-what-it-was-designed-for-and-how-it-works.aspx
Credit for this above example goes to Warren from the Active Directory Team Blog Post.

Will.



ok so it's like what I said then:
14 days minus random percentage of 5 days = 14 days - (X/100*5)

Where x/100 is the random pourcentage.

Thanks



Share this

Related Posts

There was an error in this gadget