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licensing question

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floppyraid

Technical User
Aug 16, 2009
30
US
Greetings

Let's say that I have two Windows 2003 Standard domain controllers.

1 is doing all of the heavy duty stuff like network shares/active directory, etc.

The other one is only devoted to being an internet content filtration system for all clients on our network.

Would this mean that I would need to purchase 2x the amount of CALs seeing as how all users that would authenticate on one server are also going to be using the other server to get to the net?
 
Can you elaborate on that? Sorry, I don't really understand many of their licensing terms--- and I've read many, many of their pages on it.

What do you mean by 'Windows' CAL? From all of their documentation all I see are 'device' and 'user' CALs. And they make it a point to mention that you cannot get around purchasing additional CALs by placing intermediate devices between the users and the server that somehow multiplex their data stream as if though it were coming from one source.

Let's say I have 50 people that use computers but only 25 actual computers (minus the servers). One of the servers hosts network shares and does dns/dhcp/active directory/etc. The other server runs some sort of filtering software- and all users traffic must go through it to get online.

Would I then need 25 'device' CALs for their communication with the DNS/DHCP/etc server, AND 25 more 'device' CALs for their communication with the filtration server?
 
So the CALs are only valid for a connection between 1 single client and 1 single server?

If I have 2 workstations and 5 Windows 2003 Standard Servers I have to purchase 10 CALs for each workstation to be able to access each of the servers?
 
No. You buy one CAL for each user in your environment, and one Windows CAL for each server. Assuming you have one user for each of the 2 workstations, you'd end up with two user CALs and five server CALs.

Pat Richard MVP
Plan for performance, and capacity takes care of itself. Plan for capacity, and suffer poor performance.
 
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