Larger disks would give more capacity that you don’t need and faster disks would provide performance above and beyond what was requested. This may be good depending on your confidence in the performance requirements.
random I/O
RAID10: write penalty = 2, read = 1; available space = number of disks devided by 2
RAID5: write penalty = 4, read = 1; available space = number of disks minus 1 disk
RAID6: write penalty = 6, read = 1; available space = number of disks minus 2 disks
Always count all the drives involved, since the write penalty takes care of that.
An app does 1000 IOps, where the read / write ratio is 3 / 1, so 3 times as many reads as writes. These 1000 IOps are 750 reads and 250 writes
Backend IOps:
RAID10: 750 + 2 x 250 = 1250; you'll need 1250/180 15k = 7, so at least 8 drives or 1250/130 10k = at least 10 drives
RAID5: 750 + 4 x 250 = 1750; you'll need 1750/180 15k = at least 10 drives or 1750/130 10k = at least 14 drives
RAID6: 750 + 6 x 250 = 2250; you'll need 2250/180 15k = at least 13 drives or 2250/130 10k = at least 18 drives
Not digested ???
Let me take it this way now...
As for the IOPS per drive here is what is used as industry standard:
FC 10K= 150 IOPS
FC 15K= 200 IOPS
SSD = 400 IOPS
SATA = 80 IOPS
Flash: 3500 IOPS
SAS 15k: 180 IOPS
NLSAS: 90 IOPS
These are just rules of thumb used to size environments.
No comments:
Post a Comment