Copied SSD alignment incorrect

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Copied SSD alignment incorrect

Postby GordonM » Tue Dec 11, 2018 5:36 pm

I have been trying to use IFW to image from a Windows 10 hard drive to a SSD using IFWs Copy function. There are two partitions on the source drive, the small System Reserved partition and the main Windows System partition. The target SSD is large enough to accommodate the two partitions on the hard drive.

I have largely used IFW's default settings but made sure that Align Partitions on 2048 sectors is checked. As I understand it, for best SSD operation, partition starting offsets should be divisible by 4,096 with an integer result. For the System Reserved partition, the Starting Offset is 1,048,576, which is exactly divisible by 4,096 with a result of 256. For the larger partition, the Starting Offset is 2,138,572,800. Dividing this by 4,096 gives a non-integer result (522,112.5). Given that I specified that I have checked Align Partitions on 2,048 sectors, I am wondering why the starting offset doesn't seem to be correct for the larger partition. Maybe I am not understanding things correctly. Could somebody please help?

GordonM
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Re: Copied SSD alignment incorrect

Postby Bob Coleman » Tue Dec 11, 2018 6:55 pm

I'm probably not understanding this either, but I would think that "Align Partitions on 2048 sectors" would cause the starting offset to be evenly divisible by 2048, not 4096. I admit that I don't thoroughly understand this though.
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Re: Copied SSD alignment incorrect

Postby GordonM » Tue Dec 11, 2018 7:37 pm

Thanks Bob. My understanding is that SSDs (maybe not all, but most) are physically organized in 4,096 byte sectors. If the starting offset is not divisible by 4,096, a write (or read) of 4,096 bytes could span two sectors making writing and reading slower and with more wear on the SSD.
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Re: Copied SSD alignment incorrect

Postby TeraByte Support » Tue Dec 11, 2018 11:33 pm

It would be aligned to 2048 sectors, not 4096 sectors. 2048 sectors is
1MiB alignment using standard sector sizes and is aligned to 4096 bytes.

Latest versions use the align on 1MiB option instead of 2048 sectors.
although you can customize the alignment to whatever you want.


"GordonM" wrote in message news:16107@public.image...

I have been trying to use IFW to image from a Windows 10 hard drive to a SSD
using IFWs Copy function. There are two partitions on the source drive, the
small System Reserved partition and the main Windows System partition. The
target SSD is large enough to accommodate the two partitions on the hard
drive.

I have largely used IFW's default settings but made sure that Align
Partitions on 2048 sectors is checked. As I understand it, for best SSD
operation, partition starting offsets should be divisible by 4,096 with an
integer result. For the System Reserved partition, the Starting Offset is
1,048,576, which is exactly divisible by 4,096 with a result of 256. For
the larger partition, the Starting Offset is 2,138,572,800. Dividing this
by 4,096 gives a non-integer result (522,112.5). Given that I specified
that I have checked Align Partitions on 2,048 sectors, I am wondering why
the starting offset doesn't seem to be correct for the larger partition.
Maybe I am not understanding things correctly. Could somebody please help?

GordonM

TeraByte Support
 
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Re: Copied SSD alignment incorrect

Postby GordonM » Wed Dec 12, 2018 7:27 am

Thank you. That explains it.
GordonM
 
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Re: Copied SSD alignment incorrect

Postby GordonM » Wed Dec 12, 2018 12:55 pm

On second thoughts, I don't think this does explain the issue to me. ".....2048 sectors is 1MiB alignment using standard sector sizes and is aligned to 4096 bytes ...".

"Latest versions use the align on "1MiB option instead of 2048 sectors".

These two statements seem to say the same thing except I don't understand the "instead of" in "1Mib option instead of 2048 sectors". The first statement says 2048 sectors is 1MiB. Are you saying that this is just different terminology for the same thing, 1MiB versus 2,048 sectors. If this is so I still don't understand why I am not getting an integer when I divide by 4,096, which all the references that I have seen say that this is what should be the result for a properly aligned SSD. Should I be forgetting about dividing by 4,096 as long as the starting point of the partition is exactly divisible by 1MiB with an integer result.

So the main question is, if dividing the Starting Offset number by 4,096 with an integer result is important, how do I achieve this with IFW.

Thank you
GordonM
 
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Re: Copied SSD alignment incorrect

Postby Fracso » Thu Dec 13, 2018 9:45 am

If the sector size is 512 bytes, 2048 sectors is equivalent to 1 MiB, since
2048*512 = 1048576 bytes = 1 MiB (1024*1024).
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Re: Copied SSD alignment incorrect

Postby GordonM » Thu Dec 13, 2018 12:50 pm

Yes, I cannot disagree with that but the point is that the partition Starting Offset does not divide by 4,096 with an integer result. I have now tried another imaging application and this resulted in a different starting offset, which did divide by 4,096 with an integer result. I feel sure that IFW should be able to do this but I haven't found out how.

Thank you
GordonM
 
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Re: Copied SSD alignment incorrect

Postby Brian K » Thu Dec 13, 2018 1:08 pm

GordonM wrote:
>. For
> the larger partition, the Starting Offset is 2,138,572,800.

Is this sectors or bytes?

Have a look in PartInfo. What is the "Start" for each partition?
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Re: Copied SSD alignment incorrect

Postby TeraByte Support(PP) » Thu Dec 13, 2018 1:42 pm

Having 4K (4096 bytes) in a sector is not the same thing as 4096 sectors.

Assuming normal 512 byte sectors, by dividing by 4,096 sectors you are checking alignment at 2MiB. This means that a partition can start on a 1MiB alignment and show a 0.5 remainder (1MiB).

2,138,572,800 / 4096 = 522,112.5
2,138,572,800 / 2048 = 1,044,225

Both of these are aligned as far as the SSD is concerned. One just has a little more wasted space. As another example, on most aligned drives, the first partition begins on sector 2048 (not 4096). On drives that expose 4K sectors, 256 sectors would be the normal starting point (1MiB).
Paul Purviance
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