Array Raid adice for win2k3

[acl03]

I have a bunch of new drives to add to my 2003 system. Since the Smart Array 6400 has a 2TB logical volume limit, any recommendations on how to set up the new drives?

Is it better practice to add more 2TB logical drives to my existing array, or create new arrays for them? Any reasons for or against each method? Thanks!



Thanks,
Andrew

 

[tecknic]

Have you ever consider a san? 4T for a single server is a lot of data. your SA 6400 come with two channel. I would use the second if you are to bust the 2T.

It's depend on how you applications can handle multiple drives. Expand is pretty cool if you can do it. however using the second channel may increase time respond performance.

 

[acl03]

Well we have our existing array running on both channels to increase performance. We would continue to do so on any new drive we create.



Thanks,
Andrew

 

[w33mhz]

Depends on what you are hosting on your Win2k3 server. If this is going to be a file server adding on isn't a bad idea. But if you are using it for a database or exchange server you may want to split them up and possibly setup a RAID 0 + 1 for preformance. For a file server you would want a RAID 5 solution. The reason for the RAID 0 + 1 solution for a database/exchange solutions is that you will get latancy while writing your parity on all the drives with a RAID 5, a RAID 0 + 1 gives you the best disk preformance. If you lose a drive in your 1st RAID 0 array, the mirror will take over so you get the best of both worlds with that solution. The down side is that if you lose a drive in both RAID 0 array's your screwed. If you wish to lose a little on preformance and gain in protection you could do a RAID 1 + 0 this way you could loss more than 1 drive and be fine.

 

[acl03]

Oh this is a simple fileserver....

Just dont know whether to make new arrays, or add logical drives to my existing.



Thanks,
Andrew

 

[w33mhz]

Well I would just create another Array. You may start getting a preformance lag writing to all those drives.