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E20-324 VNX Certification - For bandwidth calculations (Please Help Me)
I'm studing from the exame E20-324 in the pdf ...
From: VNX_USSD_01 ( V-ILT)
Formula:
Parity RAID 5 and 3: Drive MB/s = Read MB/s + Write MB/s * (1 + (1/ (number of user data drives in group)))
Example:
For example, if a RAID 5 group size is 4+1 (four user data drives in group), the read load is 100 MB/s, and
write load is 50 MB/s:
Drive MB/s = 100 MB/s + 40 MB/s * (1 + (1/4))
Drive MB/s = 150 MB/s
My question:
When the Write load is 50 MB/s in question, why is it lowered to 40MB/s in the calculation ?
Is it because .. the 50 MB/s / 5 (total diskarray) * 4 (user data drives in group less the parity drive)
Therefore 50 / 5 * 4 = 40 MB/s ? Is my assumption correct?
Thanks for any help you can offer!
andre_rossouw
62 Posts
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October 19th, 2012 06:00
The formula assumes not only that this is a bandwidth calculation, but that the system is performing full stripe writes. Note that full stripe writes are only relevant for parity protected RAID types.
For full stripe writes, the write penalty is (number of disks in the RAID Group) / (number of data disks in the RAID Group).
Examples: R5 4+1, this is 5/4. ForR6 4+2, this is 6/4.
Now, back to the original question ...
Disk MB/s will be the sum of LUN read MB/s and LUN write MB/s * (write penalty), which in the question is
100 MB/s + 50 MB/s * (5/4) = 100 MB/s + 62.5 MB/s = 162.5 MB/s.
The 40 MB/s must be a typo.
andre_rossouw
62 Posts
2
October 19th, 2012 07:00
Your calculation of the data marked dirty is correct:
Data Marked Dirty = 25 writes x 128KiB extend x 15 minutes x 60 Seconds = 2,880,000 KiB
A COFW deals in 64 KiB chunks, so the data marked dirty must be divided by 64, to give 45,000 COFWs.
Dividing the 64 KiB by 8 KiB, as you did, is only applicable in the case of sequential writes.
andre_rossouw
62 Posts
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October 19th, 2012 07:00
Answered on the forum …
Regards,
André Rossouw
Office (phone) (919) 329 6911
Office (fax) (919) 329 6911
Cell (508) 922 6330
vmauro
10 Posts
0
October 19th, 2012 07:00
Thanks Andre for the speedy response.
I have another question....
I read in the community, the following question/ask, without a response, i'd like to understand .. can you please help me again ?
"Customer requirements call for a SnapView snapshot to be made of a clone prior to resynchronization. The source LUN is 256 GiB, and performs 500 random reads/s and 25 random writes/s of 8 KiB. The clone will be fractured every 15 minutes, and resynchronized after the snapshot is started. How many COFWs will be performed each time the clone is resynchronized?
So I calculated on this but I seem to go wrong some where
this is what I have done:
256GB source LUN so the Extend will be 128GB
Data Marked Dirty = 25 writes x 128KiB extend x 15 minutes x 60 Seconds = 2,880,000
With 8Kib writes it means that you have 64Kib : 8Kib = 8 writes to have a full stripe
So 2,880,000 : 8 = 360,000 COFW"
Thanks in advanced for your help
Regards
vmauro
10 Posts
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October 19th, 2012 09:00
TKS .. I'm very happy