[Date Prev][Date Next]   [Thread Prev][Thread Next]   [Thread Index] [Date Index] [Author Index]

questions regarding file-system optimization for sortware-RAID array


I created a RAID1 array of two physical HDD's with chunk size of 64KiB under Debian "wheezy" using mdadm. As a next step, I would like to create an ext3(or ext4) file-system to this RAID1 array using mke2fs utility. According to RAID-related tutorials, I should create the file-system like this:

# mkfs.ext3 -v -L myarray -m 0.5 -b 4096 -E stride=16,stripe-width=32 /dev/md0


1) According to manual of mke2fs, value of the "stride" has to be the RAID chunk size in clusters. As I use chunk size of 64KiB, then I have to use "stride" value of 16(16*4096=65536). Why is it important for file-system to know the size of chunk used in RAID array? I know it improves the I/O performance, but why is this so?

2) If the "stride" size in my case is 16, then the "stripe_width=" is 32 because there are two drives in the array which contain the actual data. Manual page of the mke2fs explain this option as "This allows the block allocator to prevent read-modify-write of the parity in a RAID stripe if possible when the data is written.". How to understand this? What is this "read-modify-write" behavior? Could somebody explain this with an example?


[Date Prev][Date Next]   [Thread Prev][Thread Next]   [Thread Index] [Date Index] [Author Index]