Is my OCZ SSD aligned correctly?
Solution 1
The default alignment in order to fully ensure compatibility with both 4K sectors and SSD cell boundaries (typically 128 or 256 or 512K) would be 1MByte into the disk.
2048 (sectors) * 512 = 1048576 (or exactly 1MByte)
So to me your disk looks to be properly aligned, both with your flash-erase cell size and the page write size.
Remember that sector counts start at 0, so while it claims to be at sector 2048 the actual sector your partition is starting at is the 2049th sector of the disk, which places it in the first sector after the first two NAND-flash cells and likewise that sector is also the first sector past the 256th 4KByte "page".
Solution 2
The current anecdotes suggest to have 512KB alignment (or greater integer value of log2n), therefore you want to make sure all your start sectors are evenly divisible by 1024 sectors (i.e 512Kbytes).
/dev/sda1 * 2048
2048 / 1024 = 2 OK
/dev/sda2 73244670
73244670 / 1024 = 71527.99
So this if off by 2 sectors. However since it is an extended partition, I think this is correct, as long as you start the real disk partition at sector 73244672
. 1 sector must be reserved because the logical partition must be written at the start of the extended partition. Just make sure you check when creating the file system.
/dev/sda5 73244672
73244670 / 1024 = 71528 OK
Related videos on Youtube
jonsca
Updated on September 18, 2022Comments
-
jonsca almost 2 years
I have an OCZ Agility 2 SSD with 40 GB of space. I use it as a system drive in Debian Linux ("Squeeze") and in my opinion it's really fast. I've read a lot on aligning partitions and file systems, and I'm not sure if I succeeded in aligning the partitions correctly. Maybe the SSD could be even faster?
I use ext4 and here is the output of
fdisk -cul
:Disk /dev/sda: 40.0 GB, 40018599936 bytes 255 heads, 63 sectors/track, 4865 cylinders, total 78161328 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: [...] Device Boot Start End Blocks Id System /dev/sda1 * 2048 73242623 36620288 83 Linux /dev/sda2 73244670 78159871 2457601 5 Extended /dev/sda5 73244672 78159871 2457600 82 Linux swap / Solaris
My partitions were created just by the Debian Squeeze setup assistant. So I didn't care about the details of partitioning. But now I think that maybe the installer didn't align it correctly? Actually,
2048
looks good to me (better than odd values like 63 or something like that) but I've no idea.According to some "SSD Alignment Calculator" I found on the web, the OCZ SSDs have a NAND Erase Block Size of 512kB and their NAND Page Size is 4kB.
2048 is divisible by 4 and 512. So are the partitions aligned correctly?
-
cmcginty about 13 yearsYou say the default alignment would be 1M for fully compatibility, but 512K also fits your criteria.
-
cmcginty about 13 yearsI read an article that said 2048 is suggested, because that is what Vista/Win7 uses.