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A CASE FOR REDUNDANT ARRAYS OF INEXPENSIVE DISKS (RAID)HighlightsOriginal MotivationToday’s MotivationRAID LEVEL 0RAID levels 0 and 1RAID LEVEL 1RAID LEVEL 2RAID levels 2 and 3RAID LEVEL 3How parity works?Recovering from a disk failureHow RAID level 3 works (I)How RAID level 3 works (II)How RAID level 3 works (III)RAID LEVEL 4 (I)RAID LEVEL 4 (II)RAID levels 4 and 5RAID LEVEL 5The small write problemFirst solutionSecond solutionOther RAID organizations (I)Other RAID organizations (II)What about flash drives?CONCLUSION (I)CONCLUSION (II)A review questionThe answersA CASE FOR REDUNDANT ARRAYS OF INEXPENSIVE DISKS (RAID)D. A. Patterson, G. A. Gibson, R. H. KatzUniversity of California, Berkeley•The six RAID organizations•Why RAID 1, 3, 5 and 6 are the most interesting•The small write problem occurring with RAID 5 and 6HighlightsWARNING: Skip the reliability and availability analyses: they are not correctOriginal Motivation•Replacing large and expensive mainframe hard drives (IBM 3310) by several cheaper Winchester disk drives•Will work but introduce a data reliability problem:–Assume MTTF of a disk drive is 30,000 hours–MTTF for a set of n drives is 30,000/n•n = 10 means MTTF of 3,000 hoursToday’s Motivation•“Cheap” SCSI hard drives are now big enough for most applications•We use RAID today for –Increasing disk throughput by allowing parallel access–Eliminating the need to make disk backups•Disks are too big to be backed up in an efficient fashionRAID LEVEL 0•No replication•Advantages:–Simple to implement–No overhead•Disadvantage:–If array has n disks failure rate is n times the failure rate of a single diskRAID levels 0 and 1RAID level 0 RAID level 1MirrorsRAID LEVEL 1•Mirroring–Two copies of each disk block•Advantages:–Simple to implement–Fault-tolerant•Disadvantage:–Requires twice the disk capacity of normal file systemsRAID LEVEL 2•Instead of duplicating the data blocks we use an error correction code•Very bad idea because disk drives either work correctly or do not work at all–Only possible errors are omission errors–We need an omission correction code•A parity bit is enough to correct a single omissionRAID levels 2 and 3RAID level 2 RAID level 3Check disksParity diskRAID LEVEL 3•Requires N+1 disk drives–N drives contain data (1/N of each data block)•Block b[k] now partitioned into N fragments b[k,1], b[k,2], ... b[k,N]–Parity drive contains exclusive or of these N fragmentsp[k] = b[k,1]  b[k,2]  ...  b[k,N]How parity works?•Truth table for XOR (same as parity)A B AB0 0 00 1 11 0 11 1 0Recovering from a disk failure•Small RAID level 3 array with data disks D0 and D1 and parity disk P can tolerate failure of either D0 or D1D0 D1 P0 0 00 1 11 0 11 1 0D1P=D0 D0P=D10 00 11 01 1How RAID level 3 works (I)•Assume we have N + 1 disks•Each block is partitioned into N equal chunksBlockChunk Chunk Chunk ChunkN = 4 inexampleHow RAID level 3 works (II)•XOR data chunks to compute the parity chunk   Parity•Each chunk is written into a separate diskParityHow RAID level 3 works (III)•Each read/write involves all disks in RAID array–Cannot do two or more reads/writes in parallel–Performance of array not netter than that of a single diskRAID LEVEL 4 (I)•Requires N+1 disk drives–N drives contain data•Individual blocks, not chunks–Blocks with same disk address form a stripex x xx ?RAID LEVEL 4 (II)•Parity drive contains exclusive or of the N blocks in stripep[k] = b[k]  b[k+1]  ...  b[k+N-1]•Parity block now reflects contents of several blocks!•Can now do parallel reads/writesRAID levels 4 and 5RAID level 4 RAID level 5BottleneckRAID LEVEL 5•Single parity drive of RAID level 4 is involved in every write –Will limit parallelism•RAID-5 distribute the parity blocks among the N+1 drives–Much betterThe small write problem•Specific to RAID 5•Happens when we want to update a single block–Block belongs to a stripe–How can we compute the new value of the parity block...b[k+1]p[k]b[k+2]b[k]First solution•Read values of N-1 other blocks in stripe•Recomputep[k] = b[k]  b[k+1]  ...  b[k+N-1]•Solution requires–N-1 reads–2 writes (new block and new parity block)Second solution•Assume we want to update block b[m]•Read old values of b[m] and parity block p[k]•Computep[k] = new b[m]  old b[m]  old p[k]•Solution requires–2 reads (old values of block and parity block)–2 writes (new block and new parity block)Other RAID organizations (I)•RAID 6:–Two check disks–Tolerates two disk failures–More complex updatesOther RAID organizations (II)•RAID 10:–Also known as RAID 1 + 0–Data are striped (as in RAID 0 or RAID 5) over pairs of mirrored disks (RAID 1)RAID 0RAID 1 RAID 1 RAID 1 RAID 1What about flash drives?•Having no moving parts should mean fewer failures?–Failures still happen–Flash drives age as they are written to–Irrecoverable red errors occur (at least as frequently as in magnetic disks?)•Pure Storage uses a proprietary 3D-RAID organization for their SSD storesCONCLUSION (I)•RAID original purpose was to take advantage of Winchester drives that were smaller and cheaper than conventional disk drives–Replace a single drive by an array of smaller drives•Current purpose is to build fault-tolerant file systems that do not need backupsCONCLUSION (II)•Low cost of disk drives made RAID level 1 attractive for small installations•Otherwise pick–RAID level 6 for higher protection•Can tolerate one disk failure and irrecoverable read errorsA review question•Consider an array consisting of four 750 GB disks•What is the storage capacity of the array if we organize it–As a RAID level 0 array?–As a RAID level 1 array?–As a RAID level 5 array?The answers•Consider an array consisting of four 750 GB disks•What is the storage capacity of the array if we organize it–As a RAID level 0 array? 3 TB–As a RAID level 1 array? 1.5 TB–As a RAID level 5 array? 2.25

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