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IOPS calculation
How do I calculate the average IOPS in a (2+2) RAID 10 environment over 4 15K disks.
According specs: max. 380 (avg. 190) I/Os per second ST3450857FC 15K.7 450 GB
Is it 190x4=760 IOPS? What is my writepenalty in this configuration?
Am I right when I say 1+(parity/#hdd) = 1+(2/4) = 1,5 write penalty ???
DaveZ1
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January 11th, 2011 10:00
we use a best practice (conservative all-around estimate) of 180 IOPS per drive, for 15 K drives. We can figure the best practice host IOPS rate (assuming random IO) if you know the approximate read/write mix.
To determine the host-deliverable IOPS from the drives in RAID 1 use this equation:
(h = host IOPS, r = read%, w = write%)
h = diskIOPS / (r+2w)
so in your case, with 4 * 180 IOPS, you have 720 disk IOPS available. To get host IOPS you need to know your read/write rate. If it is 60/40 read/write, it is
h = 720 /(0.6 + 2*0.4)
h = 512 IOPS
Note the above equation is derived from the equation for disk IOPS from host load (h = host IOPS, r = read%, w = write%) which for RAID 1 and RAID 1/0 is as follows:
r*h + 2*(w*h) = diskIOPS
example r = 60% and w = 40%, for 10000 host IOPS
.6*10000 + 2 * 0.4 * 10000= diskIOPS
6000 + 2 * 4000= diskIOPS
14000 = diskIOPS
teovmy
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January 11th, 2011 12:00
Excellent explained, thank you.
In your example with RAID 10 I will use my figure to see if I understand it.
So If I use your example with my figures (500 IOPS total) it should look like this.
0,6*500 + 2 * (0,4*500) = DiskIOPS
300 + 2 * 200 = DiskIOPS
700 = DiskIOPS
Am I correct?
And we talking about backend IOPS and not frontend. Correct?
teovmy
211 Posts
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January 11th, 2011 13:00
And you are more then welcome here
DaveZ1
75 Posts
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January 11th, 2011 13:00
just to be explicit:
500 = host IOPS, or "Front-end"
700 = disk IOPS, or "back-end"
in your example
teovmy
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January 11th, 2011 13:00
Thanks for your support and quick feedback. You seem to manage this matter well.
It is absolutely clear to me now.
Of course there is still caching involved. And I believe there are some calculations to take into this example as well regarding this or not?
DaveZ1
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January 11th, 2011 13:00
You are welcome. I don't always have the time to haunt the boards, but try to drop in regularly.
DaveZ1
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January 11th, 2011 13:00
yes that is the correct series of steps to determine back-end load (diskIOPS in my equation) for a given host load. You would use the other equation to determine the potential host load for a given back-end load.
This all assumes of course that you have a random, small block (smaller than the RAID stripe) workload. It also does not work in caching effects. A really accurate estimation requires some programming and knowledge of system optimizations and bottlenecks. But this approach is good for determining longterm sustainable load.
RRR
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January 12th, 2011 02:00
I'd suggest the Clariion Performance Workshop. You'll get the same explanation there plus a whole lot of background info !
teovmy
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January 12th, 2011 02:00
@Rob,
As you know I followed this course. But some how I deleted the info from my mind
Full = Full
RRR
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January 12th, 2011 03:00
I knew that.... I just wanted to tease you. This is important and you should not forget it anymore. Ever (for the time being).
teovmy
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January 12th, 2011 04:00
I thought so
But David explained it very well to me. And sometimes people need a second shot before they understand it.
teovmy
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January 25th, 2011 07:00
Ok thanks. And what about the formula. Is this correct?
teovmy
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January 25th, 2011 07:00
To be sure:
Is the formula in a RAID5(4+1) environment:
r*h +(w*h*4) = diskIOPS
And when using a RAID 5(8+1) is the write Penalty 8?
dynamox
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January 25th, 2011 07:00
it's still 4, unless it's a full stripe write.
dynamox
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January 25th, 2011 08:00
r*h + 4*(w*h)