HPE EVA Storage

# EVA8100 array capability statistics

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## EVA8100 array capability statistics

Hello all,
I have recently been asked by a contractor if we can provide the bandwidth and latency for the physical disks in a disk group. I have looked at quite a few HP tech docs but can't find these particular data. Does anyone know where I can get it? The array is an EVA8100. The disk group in question is composed of 56x300GB 15krpm FC drives.
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Honored Contributor
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## Re: EVA8100 array capability statistics

Hi, assuming a typical workload of 60% reads, 40% writes, 56 x 15K drives gives you:

6,412 IOPS, 280 MB/S writes in VRAID 1
4,080 IOPS, 448 MB/S writes in VRAID 5

560 MB/S read in both cases

## Re: EVA8100 array capability statistics

What formula did you use to determine that? Is there a whitepaper that goes into this in more detail?
Honored Contributor

## Re: EVA8100 array capability statistics

No sorry, you cannot have the Excel file that calculates that, it's HP internal use only.

If you can get the exact numbers for % read operations / write operations, and the % of space in RAID 1 / RAID 5, I can use them to calculate the exact numbers for your case.

## Re: EVA8100 array capability statistics

Thanks for the information. I didn't realize that you were an HP employee. :-) I'll get with the contractor to see if I can get the exact number of reads and writes they specify and post that some time tomorrow.
Honored Contributor

## Re: EVA8100 array capability statistics

If you don't take cache hits into account you need to consider that host read I/O's translate 1:1 to Array I/O's, but write I/O's are worse:

RAID1: every host write are 2 array writes (writes to each mirror)
RAID5 sequential: every host write are (1 + (1 devided by <) writes: 4RAID5: 1 + 1/4 = 1,25 IOps
RAID5 random: every host write are 4 array writes (read original data + parity, write new data + new parity)
RAID6 sequential: every host write are (1 + (2 devided by <) writes: 4RAID5: 1 + 2/4 = 1,5 IOps
RAID6 random: every host write are 6 array writes (read original data + parity1 + parity2, write new data + new parity1 + new parity2)

If your host and your array have lots of cache, the actual number of disk IOps can be greatly reduced, depending on the actual activity of this cache

Say a host will do only 250 write IOPs at a time and you are planning for this (no reads). When "Determining the disk load" (section in the guide) use the parity formula.

Raid 5 IOPs = reads + 4* Writes. In our case 0 + 4*250 =1000. Say we are going to use 15k fibre drives, so 1000/180 = 5.555 drives (call it 6). This means we need a 5+1 raid 5 to handle the IOP load from this host.

Correct (if the I/O's are random) !
But if you want to choose between RAID5 and RAID10 and you don't need that much space), you might want to choose for RAID10, since the write penalty is only 2 !

0 + 2*250 = 500
15k drives: 500/180 = 2.778, let's call it 4, so you'd only need a 2+2 RAID10 RG.