From: Gim Leong Chin <chingimleong@yahoo.com.sg>
To: xfs@oss.sgi.com
Subject: Warning: AG size is a multiple of stripe width?
Date: Fri, 17 Jun 2011 19:36:12 +0800 (SGT) [thread overview]
Message-ID: <385532.69322.qm@web77719.mail.sg1.yahoo.com> (raw)
Hi,
I have a Sun workstation with eight Cheetah 15K.5 SAS 300 GB on RAID 1E (RAID 10) on LSI SAS3081E-R.
I am installing SLED 11 SP1 on it and I thought I will do a thorough optimization right down to the partition boundaries.
Since the default for XFS is to create four aggregation groups, and with the reasoning that Cheetah can do double the seeks of normal 7200 RPM drives, I have four aggregation groups per drive for a total of 16 for 70 GB /dev/sda2 partition, and eight per drive for total of 32 for /dev/sda3 partition (1011 GB).
I have aligned the partition start and end with the stripe width boundaries.
The stripe size is 64 kB, stripe width is 4*64 kB = 256 kB, in terms of 512 byte sectors:
70 GB /
No Start End Number
1 512 67109375 32 GB = 67108864 sectors = 131072 stripe sets
2 67109376 213910015 70 GB = 146800640 sectors = 286720 stripe sets
3 213910016 2335932415 Left = 2335932416 - 213910016 = 2122022400 sectors = 4144575 stripe sets
When I do the following:
mkfs.xfs -f -b size=4k -d agcount=16,su=64k,sw=4 -i size=256,align=1,attr=2 -l version=2,su=64k,lazy-count=1 -n version=2 -s size=512 -L / /dev/sda2
Warning: AG size is a multiple of stripe width. This can cause performance problems by aligning all AGs on the same disk. To avoid this, run mkfs with an AG size that is one stripe unit smaller, for example 1146864
agcount=16 agsize=1146880 blks
bsize=4096
sunit=16 swidth=64 blks
mkfs.xfs -f -b size=4k -d agcount=32,su=64k,sw=4 -i size=256,align=1,attr=2 -l version=2,su=64k,lazy-count=1 -n version=2 -s size=512 -L /home /dev/sda3
Warning: AG size is a multiple of stripe width. This can cause performance problems by aligning all AGs on the same disk. To avoid this, run mkfs with an AG size that is one stripe unit smaller, for example 8289136
agcount=32 agsize=8289152 blks
bsize=4096
sunit=16 swidth=64 blks
I am really puzzled since I thought all I am doing is distributing 4 aggregation groups per drive for sda2 and 8 per drive for sda3.
What have I done wrong and what is the flaw with my understanding?
Thank you!
Chin Gim Leong
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next reply other threads:[~2011-06-17 11:36 UTC|newest]
Thread overview: 3+ messages / expand[flat|nested] mbox.gz Atom feed top
2011-06-17 11:36 Gim Leong Chin [this message]
2011-06-17 15:07 ` Warning: AG size is a multiple of stripe width? Eric Sandeen
2011-06-20 1:21 ` Lachlan McIlroy
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