bfeist Posted July 18, 2011 Share Posted July 18, 2011 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 Link to comment
Joe L. Posted July 19, 2011 Share Posted July 19, 2011 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. Link to comment
jumperalex Posted July 19, 2011 Share Posted July 19, 2011 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 Link to comment
bfeist Posted July 19, 2011 Author Share Posted July 19, 2011 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 Link to comment
mcs Posted July 19, 2011 Share Posted July 19, 2011 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 ... Link to comment
Joe L. Posted July 19, 2011 Share Posted July 19, 2011 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) Link to comment
jumperalex Posted July 19, 2011 Share Posted July 19, 2011 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 Link to comment
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