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RAID 5 usable space 1
- Thread starter Billk43
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It usually equates to roughly one drives worth depending on the configuration and number of disks in the Array. If you work on the principle that the configuration usually allows 3/4 of each disk for data storage and 1/4 of the disk for RAID data, you can't go far wrong. This can be changed in SmartStart (during the creation of the Array) but I can't comment on that as I never bother.
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As long as you only have a single stripe set in your RAID it is only one drive. For example we have a RAID configuration with 14 72GB drives. One of the drives is configured as a hotspare (optional). So the RAID Set has 13 drives. We have 814GB of usable space. (Unformatted capacity of 864GB
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1
- #4
AdmanOK
MIS
- Mar 5, 2001
- 412
When calculating the usable space on a RAID5 set you always use the formula
(n - 1) * s
where n = the number of drives and s = the size of the smallest drive in the set (it is recommended that you use all the same drive, but it is actually not a requirement). If you have a RAID controller that is capable of having a hot spare and you take advantage of this feature then your forumula would be
(n - 2) * s
this is because you lose the capacity of one drive for the parity that makes RAID5 fault tolerant, and the capacity of a second drive because it is sitting there doing nothing. It will wait for one of the drives in the RAID set to fail and then immediately take it's place so that you regain your redundancy. Without a hot spare you have no redundancy till you replace the failed drive. Even with a hot spare you do not have redundancy till the RAID replication has taken place which can take many hours on a large array. I hope this answers your question.
(n - 1) * s
where n = the number of drives and s = the size of the smallest drive in the set (it is recommended that you use all the same drive, but it is actually not a requirement). If you have a RAID controller that is capable of having a hot spare and you take advantage of this feature then your forumula would be
(n - 2) * s
this is because you lose the capacity of one drive for the parity that makes RAID5 fault tolerant, and the capacity of a second drive because it is sitting there doing nothing. It will wait for one of the drives in the RAID set to fail and then immediately take it's place so that you regain your redundancy. Without a hot spare you have no redundancy till you replace the failed drive. Even with a hot spare you do not have redundancy till the RAID replication has taken place which can take many hours on a large array. I hope this answers your question.
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