RAID

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Exercises Enterprise Digital Infrastructures 2014-2015 1. Consider 12 identical HDD, each with a capacity of 1 T B. All disks have an identical M T T F = 1000 days and M T T R = 2 days. State the total capacity, and the M T T DL (mean time to data loss), if the disks are connected in: (a) RAID 0 (b) RAID 1+0 (6 groups of 2 disks) (c) RAID 0+1 (2 groups of 6 disks) (d) RAID 5 (e) RAID 6 SOLUTIONS See slides in ”L02 - RAIDs.pdf”. RAID 0: striping - distribute data across several disks, improve pe rformance, it does not affect availability of RAID RAID 1: mirroring - replicate data on disks, typically two copies of the data (otherwise the cost is too high), it improve the RAID availability RAID X+Y: first apply technique X to each disk of a group, then apply technique Y to all groups considering them as single disks. (a) No fault tolerance, a fault in any of the 12 disk (see the minimum tim e between several concurrent faults ”L05 - RAID availability” pp 11-14) causes a failur e in the whole RAID. M T T DL= M T T F/12 = 1000 /12 = 83 .33day All disk are used to store data, thus C = 12T B. (b) Mirroring to each disk of the group, data stripped over the 6 gr oups. RAID failure due to: a failure in any disk k followed by a failure in the mirror of disk k before repair of disk k. M T T DL= M T T F2/(12M T T R) = 1000 2/(12 ∗ 2) = 41666 .66day Half of the disks are used to store a copy of the data, thus C = 6T B (c) Data are stripped over the 6 disks of a group, the whole group is mirrored. RAID failure due to: a failure in any disk k followed by a failure in any of the 6 disks of the mirror group before repair of disk k. M T T DL= 2M T T F2/(122M T T R) = 2 ∗ 10002/(122 ∗ 2) = 6944 .44 As in previous case, half of the disks are used to store a copy of the data, thus C = 6T B.…