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Topic: How competently to saw throughput of a disk subsystem on three

Colleagues, I welcome!
There was a following problem:
There is chassis HP Proliant GEN10 on which it is cocked Hyper-V 2012R2.
The controler is used built in, P816i-a SR Gen10, disks - 10  8 7200 SAS, are organized in RAID10, one logical volume. (Under system - the disks).
Files of virtual machines - lie on this volume.
As guest - 1 Windows Server 2008 Server and 2 Windows Server 2012R2 on which it is installed (according to) SQL SERVER Standard 2005, SQL SERVER Standard 2014SP2, SQL SERVER Standard 2016 SP1 (so was historically added). On one server it is impossible to allocate them, since applications I am not able to work normally with .
How competently to restrict a pass-band for each of virtual machines that they did not devour all disk?
For applications which are serviced by servers, loading "chaotic modification on 1 record of a heap of tables 1000 of times in a second" is characteristic, i.e. these are classical OLTP. But loading not constant, but convulsively-periodic. Happen a storm when all 3 sql th are loaded, happens - the full calm.
I so understand that it is necessary to restrict productivity of the virtual machine from below, but not to restrict on top.
And how many iops thus it is necessary to deliver?
I used here this piece http://www.team.ru/RAID-IOPS-Calculator.php approximately to estimate, on what I should hope.
There in accuracy of such equipment is not present, but I think that for an estimation it is possible to use.
For RAID10 it turned out the order 1200 iops.
I gather  productivity of host-system SQLIO, but here matter is not in exact digits, and, more likely, in a technique.
I.e. here digit and received, and further that that? Where a horse ?
To cut to each of the virtual servers from below 400, on top 0 and all?
Whether it is necessary to pay attention to something? Adjustments , the average size of record in processed tables, an amount of operations, line of the bald?
Prompt,  ? It not absolutely my subject!

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Re: How competently to saw throughput of a disk subsystem on three

uaggster;
Correctly I understand that producing 400IOPS for each server you hope that as a result they at peak loading give 1200 on RAID and type all will be ok. And it thus that at you as a matter of fact SATA disks (judging by 7200)?

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Re: How competently to saw throughput of a disk subsystem on three

uaggster wrote:

loading "chaotic modification on 1 record of a heap of tables 1000 of times in a second" is characteristic, i.e. these are classical OLTP.

it is necessary to concretize, imho, 3 servers differently if to approach "in a forehead" that turn out is multiplied on 1000 iops = 3000 iops by record in peak it much for 10 disks 7200RPM SAS, and can also by a heap of tables all it to increase... (For 10 15000)

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Re: How competently to saw throughput of a disk subsystem on three

The registrar;
Did not add: for 10 hdd 15000rpm raid10 I would be guided on 1000 iops at a profile of loading of 20 % read 80 % write (OLTP)

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Re: How competently to saw throughput of a disk subsystem on three

uaggster;
From disks 7.2 rpm SAS expect 100 IOps from everyone on real loading not from a cache.
And though 100500  on them  - more they do not give, because cannot.
From SATA 7.2 they differ only the interface.

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Re: How competently to saw throughput of a disk subsystem on three

KRS544 wrote:

uaggster;
Correctly I understand that producing 400IOPS for each server you hope that as a result they at peak loading give 1200 on RAID and type all will be ok. And it thus that at you as a matter of fact SATA disks (judging by 7200)?

Well, in general that yes.
Disks SAS, here such approximately:
Hard disk 8TB SAS 12Gb/s HGST HUH721008AL5204 (the link I will not result that did not consider for advertizing of indirect resources).

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Re: How competently to saw throughput of a disk subsystem on three

a_shats wrote:

uaggster;
From disks 7.2 rpm SAS expect 100 IOps from everyone on real loading not from a cache.
And though 100500  on them  - more they do not give, because cannot.
From SATA 7.2 they differ only the interface.

I understand it.
It is necessary for me, that in case of a storm, all it did not shut up absolutely, and continued to work with the maximum productivity which can be squeezed out of a disk subsystem.
I.e. yes, productivity of a disk subsystem for normal operation of all three servers - does not suffice.
But what it is necessary to make to squeeze out a maximum of an existing situation?

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Re: How competently to saw throughput of a disk subsystem on three

As not  - but  to buy the new server.))
And so - to write the report addressed to chiefs, that leaking iron does not provide normal functioning of applications. After that to put all all is identical", and on all questions about productivity - to send to the heads.

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Re: How competently to saw throughput of a disk subsystem on three

uaggster;
How much I remember, this chassis (DL380 Gen10 with 8 3.5")  already several baskets under 2.5" disks.
Here to take one or two depending on a variant of drives of this of a basket, and  there - if is money, to steam SSD in RAID1 and if money it is not enough - then hards 10 rpm of a piece 4 in RAID10. And here on it to suppose SQL, and 8 NearLine SAS disks to use on their direct assignment - under  and .

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Re: How competently to saw throughput of a disk subsystem on three

a_shats wrote:

uaggster;
How much I remember, this chassis (DL380 Gen10 with 8 3.5")  already several baskets under 2.5" disks.
Here to take one or two depending on a variant of drives of this of a basket, and  there - if is money, to steam SSD in RAID1 and if money it is not enough - then hards 10 rpm of a piece 4 in RAID10. And here on it to suppose SQL, and 8 NearLine SAS disks to use on their direct assignment - under  and .

Yes does not help. There 4 bases on 6-8 Tb. They do not get simply that you describe. Therefore - that is, that is.
And it is good, though so is.

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Re: How competently to saw throughput of a disk subsystem on three

Well though tempdb on these disks carry out also that a profit will be

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Re: How competently to saw throughput of a disk subsystem on three

uaggster wrote:

Files of virtual machines - lie on this volume.

If to make a pass-through-disk instead of VHD, possibly, it will be easier.

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Re: How competently to saw throughput of a disk subsystem on three

To try to select 2 disks in separate  and for them to suppose broad gulls from three  - record serial, host drives not so will twitch.
To add a couple of disks on 10 - they are on 1,8 and to suppose there TempDB as a variant to deliver SSD, to add the license on  and in such mode to launch.

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Re: How competently to saw throughput of a disk subsystem on three

The best experts demand selection of physical disks for virtual machines under SQL.