On 2013-06-17 22:52, Stan Hoeppner wrote:
>>> (num_of_disks * 4KB) * stripe_cache_size
>>>
>>> In your case this would be
>>>
>>> (3 * 4KB) * 32768 = 384MB
>>
>> I'm actually seeing a bit more memory difference: 401-402 MB when going
>> from 256 to to 32768, on a mostly idle system, so maybe there's
>> something else coming into play.
>
> 384MB = 402,653,184 bytes
:)
I think that's just a coincidence, but it's possible I'm measuring it
wrong. I just did "free -m" (without --si) immediately before and after
changing the cache size.
stripe_cache_size = 256
---
total used free shared buffers cached
Mem: 16083 13278 2805 0 1387 4028
-/+ buffers/cache: 7862 8221
Swap: 0 0 0
---
stripe_cache_size = 32768
---
total used free shared buffers cached
Mem: 16083 12876 3207 0 1387 4028
-/+ buffers/cache: 7461 8622
Swap: 0 0 0
---
The exact memory usage isn't really that important to me; I just
mentioned it.
> memory_consumed = system_page_size * nr_disks * stripe_cache_size
>
> The current default, 256. On i386/x86-64 platforms with default 4KB
> page size, this consumes 1MB memory per drive. A 12 drive arrays eats
> 12MB. Increase the default to 1024 and you now eat 4MB/drive. A
> default kernel managing a 12 drive md/RAID6 array now eats 48MB just to
> manage the array, 96MB for a 24 drive RAID6. This memory consumption is
> unreasonable for a default kernel.
>
> Defaults do not exist to work optimally with your setup. They exist to
> work reasonably well with all possible setups.
True, and I will grant you that I was not considering low-memory setups.
I wouldn't want the kernel to frivolously consume RAM either. In a choice
between getting the low performance I was seeing vs. spending the RAM,
though, I'd much rather spend the RAM. Now that I know I can tune that,
I'm happy enough; I was just surprised...
Thanks,
Corey
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