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Re: What's the best hardver for PostgreSQL 8.1?

From: Dawid Kuroczko <qnex42(at)gmail(dot)com>
To: pgsql-performance(at)postgresql(dot)org
Subject: Re: What's the best hardver for PostgreSQL 8.1?
Date: 2005-12-27 07:24:45
Message-ID: 758d5e7f0512262324i4dccd6bbs87d80e59eb776a8@mail.gmail.com (view raw or flat)
Thread:
Lists: pgsql-performance
On 12/26/05, David Lang <dlang(at)invendra(dot)net> wrote:
> raid5 writes n+1 blocks not n+n/2 (unless n=2 for a 3-disk raid). you can
> have a 15+1 disk raid5 array for example
>
> however raid1 (and raid10) have to write 2*n blocks to disk. so if you are
> talking about pure I/O needed raid5 wins hands down. (the same 16 drives
> would be a 8+8 array)
>
> what slows down raid 5 is that to modify a block you have to read blocks
> from all your drives to re-calculate the parity. this interleaving of
> reads and writes when all you are logicly doing is writes can really hurt.
> (this is why I asked the question that got us off on this tangent, when
> doing new writes to an array you don't have to read the blocks as they are
> blank, assuming your cacheing is enough so that you can write blocksize*n
> before the system starts actually writing the data)

Not exactly true.

Let's assume you have a 4+1 RAID5 (drives A, B, C, D and E),
and you want to update drive A.  Let's assume the parity
is stored in this particular write on drive E.

One way to write it is:
 write A,
 read A, B, C, D,
 combine A+B+C+D and write it E.
 (4 reads + 2 writes)

The other way to write it is:
 read oldA,
 read old parity oldE
 write newA,
 write E = oldE + (newA-oldA) -- calculate difference between new and
old A, and apply it to old parity, then write
 (2 reads + 2 writes)

The more drives you have, the smarter it is to use the second approach,
unless of course A, B, C and D are available in the cache, which is the
niciest situation.

   Regards,
       Dawid

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