hard disk selection in clariion

The 73GB disks are supported by CX4, but not as the so called Vault disks, so the lowest capacity drives you can order are the 146GB drives. This is exactly what we do for each CX4 we install.

For best performance on your application please work with your application experts. They can (should) usually tell you what they need for transaction log performance and database performance.

RAID10 is usually very good, but also very expensive as all data is available twice. But on the other hand on RAID10 random reads are very fast, since data can come from 1 of 2 sources. For random writes RAID10 is also very good as your write penalty is only 2, where on RAID5 this penalty is 4, so you need more disks to be able to handle the IOps.

10k drives deliver about 130 IOps.

15k drives can deliver about 180 IOps.

These are rule of thumbs are they are usually very pessimistic. Depending on the IO size IOps can be higher or lower, but the values I named are mostly the ones we use for calculations.

For example:

If you know your application needs 1000 IOps or so on the log disks (100% random writes) and 2400 IOps or so on the database disks (50% writes and 50% reads), the math would be:

1000 random write IOps for the logs would need 1000/180 x 2 (write penalty)= 6 disks x 2 = 12 disks on RAID10 using 15k drives

          using 10k drives this is: 1000/150 x 2 = 14 drives in RAID10

     If you would use RAID 5 for this:

          using 15k drives this is: 1000/180 x 4 = 24 drives

          using 10k drives this is: 1000/150 x 4 = 28 drives

These numbers are the minimum number of req'd drives. If you use RAID10 the number you'll eventually need to use is an even number. If you use RAID5 you could think of 6 Raid Groups with 4 drives each (3+1) in the 24 drive configuration or 7 Raid Groups with 4 drives each (3+1) in the 28 drive configuration.

Finally you need to take the needed amount of space into account. The higher number of the 2 is the one you need to go with, but for the logs usually simply take the smallest available drives since space isn't an issue.

For the database the math is somewhat similar:

2400 IOps where 50% is writes ans 50% is reads

The total amount of IOps needed on RAID10 = (2 x (2400 x 1/2)) + (2400x 1/2) = 3600

The total amount of IOps needed on RAID5 = (4 x (2400 x 1/2)) + (2400x 1/2) = 6000

     If you would use RAID10 for this:

          using 15k drives this is: 3600/180 = 20 drives

          using 10k drives this is: 3600/150 = 24 drives

     If you would use RAID 5 for this:

          using 15k drives this is: 6000/180 = 34 drives

          using 10k drives this is: 6000/150 = 40 drives

These numbers are the minimum number of req'd drives. If you use RAID10 the number you'll eventually need to use is an even number. If you use RAID5 you could think of 7 Raid Groups with 5 drives each (4+1) in the 34 drive configuration. Another thing you need to consider whether or not you want to share all drives for all LUN's you need. To do this you create a number of equally sized Raid Group, let's say 5 for example. Suppose you need 6 LUN's for your database. Now you create 6 striped META LUN's, each of the size you need where each META is devided in 5 equally sized LUN's each of which residing on a RG. This way each META uses all disks.

Does this make any sence ?