Question about size of disk failure protection space

[Christopher McCormick]

Hello all,
I am a new member, so please let me know if there are better resources to answer these questions. I have all the usual documentation, the best practices guide, and so on.

The disk failure protection level on our SAN has a requested level and an actual level of "single". All of our disks have a hardware capacity of 146.8 GB, and an addressable capacity of 136.73 GB. I wasn't sure which value would be used for the disk failure protection, but I figured probably the latter.

However, though experimentation, it seems that the SAN is aetting aside 273.44 GB of space, which is equivalent to *two* of our disks.

Can someone tell me why this is so?

Thanks for your help,
Chris McCormick

[Uwe Zessin]

Hello Chris,

"146.8 GB" is hardware capacity on 'base 10' and
"136.73 GB" is software capacity on 'base 2':

146.8/1024*1000/1024*1000/1024*1000 = 136.718

The vendors specify 'hardware capacity', because that looks like a larger value and thus more impressive ("my disk is larger than yours" ;-)

It is correct that protection level= single (1) reserves *two* times the size of the largest disk in a disk group. This is necessary for VRAID-1 data, because the mirrored data is stored on an even and odd member disk.

Create a disk group with an odd number of disks and fill it with VRAID-1 only virtual disks. Check all physical disk drive's occupancy and you will see that one disk is unused.

[Christopher McCormick]

Thank you Uwe.

Yes, the hardware/software capacity thing I understand. I was assuming that most likely when documents said "the size of the largest disk," they meant the addressable capacity. But somewhere along the line, I got the idea that the space reserved was 1 * protection level * size of largest disk. Not sure where I thought I read that.

Thanks very much,
Chris

[Uwe Zessin · Accepted solution]

Chris,
it does not have anything with 'addressable' or not. You cannot say that 146.8 - 136.73 = 10.07 GB are 'unaddressable' - HW GB and SW GB are simpliy different representations.

It is like: 10(10) = A(16) = 12(8) = 1010(2)

[Christopher McCormick]

Wow. You know, I have had a fundamental misunderstanding on this, that somehow persisted even through graduate school.

I always assumed that hard drives were manufactured to certain sizes that were convenient for manufacture (cheaper to use platters of a cartain size, etc.) or better for marketing purposes (making a drive of size 200 GB is better than making a 195 GB or 205 GB rive), but which did not map neatly to a base 2 addressable memory space. I did wonder how something that was in between powers of 2 could have "lost" space in base 2, but I never thought much beyond that, because it was right in front of my eyes that there was missing space.

I always thought it wasteful that they didn't just make the drives so that they were exactly the right size to have it all be neatly addressable.

Little did I know that they first set the "marketing" (base 10) size, then figure out what base 2 capacity they need to have it be that size in base 10.

I have learned something new here. Thanks for helping sweep away my ignorance on this.

[Sheldon Smith]

Going back to the original question, as Uwe said, because of VRAID-1, the EVA stores mirrored data on an even and odd member disk. Since no assumption can be made about the VRAID level of the virtual disks in existance on a particular array, all virtual disks have to be assumed worst case, which is VRAID-1.
As far as I've been able to gather, a PSEG of host data is stored using matching PSEGs on two (numerically?) adjacent disks. If a spindle goes bad, the data needs to be moved. Problem is, the mirror PSEG *also* needs to be moved. Hence, single disk failure protection requires reserving TWO spindles worth of storage space.
Hope this helps (and I am sure that if I got any part wrong, we will hear about it).