Re: Disk Mangement Advice
From: Phillip Renouf (PhillipRenouf_at_discussions.microsoft.com)
Date: 01/07/05
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Date: Fri, 7 Jan 2005 08:39:06 -0800
RAID1 doesn't have multiple writes either, it writes to both drives at the
same time.
The issue with RAID5 writes being more intensive than writes to a RAID1
array is that not only does it have to break the file up and write it to
multiple disks, it also has to calculate the parity information and write
that to another disk. There are a couple of other performance related issues
pertaining to RAID5, but this link explains them better than I could:
http://www.storagereview.com/guide2000/ref/hdd/perf/raid/concepts/perfReadWrite.html
I have never ready any recommendations that stated a system drive should be
placed on a RAID5 array. Every one I've ever seen has listed RAID1 for system
and RAID5 for data/application. The exception to that is when an
application/data needs exceptional disk performance and then the
recommendation changes to RAID0+1 (mirrored stripe set without parity). To
each their own, if RAID5 on your system drive works and you don't see
performance issues then go with it. I've never had a problem with performance
when using RAID1.
Also, there is only one RAID5 and that is striping with parity. Striping
without parity is RAID0 and is indeed faster.
Phil
"Phillip Windell" wrote:
> No, the fact that it is 3 drives (or 4 or 5) doesn't not make it slower. In
> fact it is faster because it writes to all of them at same time and does not
> have to write the data twice. The data is also evenly divided across the
> drives so that only a portion of the data need to be writen to each one.
> There is a slight time loss for the Parity but the "cost" of this negligable
> and often exaggerated and is nearly compensated for by the faster
> writetimes. There is also two RAID5's,...there is RAID5 with Parity and
> Raid5 without Parity (Stripped Set). The RAID5 without Parity is the fastest
> RAID of them all.
>
> RAID1 is slower because and entire copy of the data goes to the second drive
> so the data being completely written twice in its entirety creates a "cost".
>
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