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Warning: AG size is a multiple of stripe width?

To: xfs@xxxxxxxxxxx
Subject: Warning: AG size is a multiple of stripe width?
From: Gim Leong Chin <chingimleong@xxxxxxxxxxxx>
Date: Fri, 17 Jun 2011 19:36:12 +0800 (SGT)
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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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