Keep in mind that in the case of both RAID-5 and RAID-6 parity is distributed across all disks in the disk group.  As a simple picture for your (4+1) RAID-5 let's visualize the following where "D" represents data and "P" represents parity on any one of the disks in the 5-disk group; Disk # 1  2 3  4 5 D D D D P P D D D D D P D D D D D P D D D D D P D In the case of a single disk failure, all data will still be available to the host/application. In the representation above, let's assume that disk #5 fails. If the host/application has to read data from the first row, since only the "P" parity was lost. all existing "D" data is still available. If the host/application has to read data from the second through the 5 rows, because "D" data was lost due to the drive failure, the remaining "D" data from the row and the associated "P" parity is used to recreate the lost "D" data on the fly... So while the response time experienced may be slightly longer due to the on-the-fly calculation, all user data is still available for host/application access...  In the case of RAID-6 (6+2), the same basic idea will apply, except there will be double "P" parity available... Another simple picture shown below: Disk # 1 2  3  4 5  6 7  8 D D D D D D P P P D D D D D D P P P D D D D D D D P P D D D D D D D P P D D D D D D D P P D D D D D D D P P D D D D D D D P P D In the case of a single disk failure, the same "basic" logic will apply and all data will still be available to the host/application because both "D" data and "P" parity are available for the any lost data to be reconstructed, as required.  In both scenarios, once the failed disk is replaced, the existing "D" and/or "P" parity is used to rebuild what was lost. In some implementations, there is also a "diagonal" approach used to minimize the amount of read operations that need to occur for better overall efficiency, but I will not attempt to explain this approach further... I'm sure there are much better write-ups available on the Internet if you need further details...[IMAGE]
Greetings,  It could be a general discussion and not biased for a specific model. It's all about RAID.  In a VNX if i create a pool with the following disks:  -Extreme performance: 5 disks RG 5(4+1) -performance: 5 disks RG 5(4+1) -capacity: 8 disks RG 6(6+2)  The above configuration has no hot spare configured. In case of RAID 5: one disk failed. Does the host will continue accessing its data(access time will be impacted).or will be facing some sort of data unavailability till replacing the faulty disk and rebuild the RAID.  In case of RAID 6 and same scenario (one disk failed) what is the outcome??   Thanks,
R5,R6 and parity