Hi all,

Please see the attached image. Disks parity - Disk6 are on motherboard SATA connections, Disk7 - Disk9 are on a AOC-SASLP-MV8. Does anyone know why the AOC-SASLP-MV8 would be causing double reads on those drives? Parity checks take upwards of 24 hours as well.

 

Any help would be greatly appreciated.

 

Ben

Hi all,

Please see the attached image. Disks parity - Disk6 are on motherboard SATA connections, Disk7 - Disk9 are on a AOC-SASLP-MV8. Does anyone know why the AOC-SASLP-MV8 would be causing double reads on those drives? Parity checks take upwards of 24 hours as well.

 

Any help would be greatly appreciated.

 

Ben

probably because the buffer size used by the driver for the AOC-SASLP-MV8 is half the size of that used by the driver for the chipset on the motheboard.  It therefore needs to read two "buffers" worth to access the same amount of data as the other disk controller gets in a single read. 

Ben

probably because the buffer size used by the driver for the AOC-SASLP-MV8 is half the size of that used by the driver for the chipset on the motheboard.  It therefore needs to read two "buffers" worth to access the same amount of data as the other disk controller gets in a single read. 

 

So you are saying:

1byte read operation = 1 read = 2byte read operation

 

as opposed to

 

1byte read = 1 read

2bytes read = 2 reads

Hi all,

Please see the attached image. Disks parity - Disk6 are on motherboard SATA connections, Disk7 - Disk9 are on a AOC-SASLP-MV8. Does anyone know why the AOC-SASLP-MV8 would be causing double reads on those drives? Parity checks take upwards of 24 hours as well.

 

Any help would be greatly appreciated.

 

Ben

probably because the buffer size used by the driver for the AOC-SASLP-MV8 is half the size of that used by the driver for the chipset on the motheboard.  It therefore needs to read two "buffers" worth to access the same amount of data as the other disk controller gets in a single read. 

 

Thanks for the help. Can I infer from your response that this means that it's normal?

 

Thanks,

Ben

 

So you are saying:

1byte read operation = 1 read = 2byte read operation

 

as opposed to

 

1byte read = 1 read

2bytes read = 2 reads

 

In other words:

 

With a 1 byte buffer

  1byte read = 1 read operation

  2bytes read = 2 read operations

  3bytes read = 3 read operations

  4bytes read = 4 read operations

  ...

 

With a 2 byte buffer

  1byte read = 1 read operation

  2bytes read = 1 read operation

  3bytes read = 2 read operations

  4bytes read = 2 read operations

  ...

yes, and one probably has a 512 byte buffer and the other has a 1024 byte buffer. 

(or one 1024 and the other 2048, or something like that)

Right ... so those numbers are counts of read operations/events/requests/whatever, not bytes read.  I always wondered that, now I know   Now back on topic