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I have the wimic and fsutil output from a SQL server. How do I determine proper alignment?

I'm having trouble putting this into context. I have a decent understanding of disks, how can I understand the rest of the information here?

wmic and fsutil

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2 Answers 2

Server 2008 and R2 will automatically align partitions on 1MB. MS chose this number because it's quite uncommon at this time to have a stripe size that doesn't not work well with a 1MB offset (and nobody's missing the wasted 1MB these days)

For R0,1,1+0 this will be fine unless your stripe size is >1MB.
For R5 this will fine unless your stripe size * (disks -1) is > 1MB.

Allocation Unit should be an integer multiple of stripe size. (Allocation Unit and Cluster Size [in NTFS context only] are the same)

If you create multiple logical disks in the hardware raid adapter, watch that they are also aligned. Most controllers will do this automatically these days.

Edit:
Specifically for you, Disks #0, #2, and #3 are not 1MB aligned, but the rest are. Also Disk #0 Partition #1 is alligned, but Partition #0 is not. These disks were probably initialized by a Server 2003 (or XP) or previous version of Windows. If your stripe size is 32KB, this will not be a problem (on disk #0; 64KB for disks #2 and #3).

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This should be what you are after: http://support.microsoft.com/kb/929491

To verify that an existing partition is aligned, divide the size of the stripe unit by the starting offset of the RAID disk group. Use the following syntax: ((Partition offset) * (Disk sector size)) / (Stripe unit size) Note Disk sector size and stripe unit size must be in bytes or in kilobytes (KB).

Example of alignment calculations in bytes for a 256-KB stripe unit size:
(63 * 512) / 262144 = 0.123046875
(64 * 512) / 262144 = 0.125
(128 * 512) / 262144 = 0.25
(256 * 512) / 262144 = 0.5
(512 * 512) / 262144 = 1
Example of alignment calculations in kilobytes for a 256-KB stripe unit size:
(63 * .5) / 256 = 0.123046875
(64 * .5) / 256 = 0.125
(128 * .5) / 256 = 0.25
(256 * .5) / 256 = 0.5
(512 * .5) / 256 = 1

These examples shows that the partition is not aligned correctly for a 256-KB stripe unit size until the partition is created by using an offset of 512 sectors (512 bytes per sector).

More info on the link including wmic commands

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