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May 14th, 2009 14:00
Disk alignment with ESX and Windows 2003 guest
I'm trying to figure out disk alignment for ESX and windows with DMX storage. I found a white paper from VMware that tells me I need to align both the ESX host and the Windows guest. PowerLink tells me to align at the guest if I'm presenting raw volumes or align at the VMFS for VMDK files. Does anyone know who is correct?
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RRR
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May 15th, 2009 02:00
And ESX host is just like a physical Windows host and all drives should be aligned before installing the OS.
Withing ESX each and every guest should also be aligned, whether or not this is RDM or VMFS based, that doesn't matter. However: the VMFS itself will be aligned already when you create this filesystem after you installed ESX. But when you create a VMDK for a guest, you should align the disk from within the guest. And on guest's c drives this can be done as follows: present the c drive to another guest as h drive for example and align it. After that disconnect this drive and present it to the new host for which it's going to be the c drive.
Once again: it doesn't matter whether or not this is RDM or VMFS based. Always align !! And on DMX you should use "create partition primary align=64".
xe2sdc
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May 15th, 2009 03:00
VmWare is talking from a higher pint of view. And from VmWare's point of view I ask you to align both VMDK itself (the virtual disk, inside the virtual host) and the whole VMFS (as EMC suggests).
RRR
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May 15th, 2009 04:00
HankDorsett
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May 15th, 2009 07:00
Thanks for the answers.
xe2sdc
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September 21st, 2009 06:00
Thank you...
Netwizrm
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September 21st, 2009 13:00
diskpart(5.2.3790)
create partition primary align=128
diskpar for W2K
This is because a Symm, cache slot boundary, which is one track covers 64 Blocks ¿32Kb¿
A RAID5 set Meta boundary, which is 4 tracks covers 256 Blocks ¿128Kb¿
For CLARiiON LUN¿s presented to a Window OS host the same command is
create partition primary align=64
When we discuss Storage on EMC array we view things from a Block perspective as opposed to Windows which always assumes disk sectors this reports a max of 255 heads, which counts from 0, = 254 as the last head.
And if the application that you are going to host on that volume is MS-SQL, then you have to format the drive with a 64k allocation unit size.
I understand all this stuff might became null and void with the introduction of Windows 7
AranH1
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September 21st, 2009 14:00
Also Microsoft has changed their recommendations for Disk Alignment. Their recommendation is now to create the alignment at 2048 since this will cover most array stripe sizes. See http://support.microsoft.com/kb/929491 for more details. You do not need to align the partition in Windows 2008 and later as the partition will automatically be aligned at 2048.
Now if the array manufacturer has a specific recommendation for an alignment offset to use in place of the Microsoft general recommendation that would probably be the best one to use as it will be array specific, but most likely the Microsoft general recommendation will be sufficient.
RRR
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September 22nd, 2009 06:00
So you align twice, although the ESX part isn't needed, since it's automatically aligned.
dynamox
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September 22nd, 2009 06:00
RRR
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AranH1
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AranH1
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AranH1
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dynamox
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RRR
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September 22nd, 2009 07:00