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Princeton COS 318 - Storage Devices

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Slide 1Today’s TopicsA Typical Magnetic Disk ControllerDisk CachingDisk Arm and HeadMechanical Component of A Disk DriveDisk SectorsDisks Were LargeThey Are Now Much SmallerAreal Density vs. Moore’s Law50 Years Later (Mark Kryder at SNW 2006)Sample Disk Specs (from Seagate)Disk PerformanceMore on PerformanceFIFO (FCFS) orderSSTF (Shortest Seek Time First)Elevator (SCAN)C-SCAN (Circular SCAN)DiscussionsRAID (Redundant Array of Independent Disks)Synopsis of RAID LevelsRAID Level 6 and BeyondDealing with Disk FailuresNext Generation: FLASHSome Current FLASH ParametersWhat’s Wrong With FLASH?Where Is Flash GoingSummaryCOS 318: Operating SystemsStorage Devices2Today’s TopicsMagnetic disksMagnetic disk performanceDisk arrays (RAID)Flash memory3A Typical Magnetic Disk ControllerExternal connectionIDE/ATA, SATASCSI, SCSI-2, Ultra SCSI, Ultra-160 SCSI, Ultra-320 SCSIFibre channelCacheBuffer data between disk and interfaceControllerRead/write operationCache replacement Failure detection and recoveryDRAMcacheInterfaceControllerExternal connectionDisk4Disk CachingMethodUse DRAM to cache recently accessed blocks•Most disk has 16MB•Some of the RAM space stores “firmware” (an embedded OS)Blocks are replaced usually in an LRU orderPros of disk cachingGood for reads if accesses have localityConsCostNeed to deal with reliable writesDisk Arm and HeadDisk armA disk arm carries disk headsDisk headMounted on an actuatorRead and write on disk surfaceRead/write operationDisk controller receives a command with <track#, sector#>Seek the right cylinder (tracks) Wait until the right sector comesPerform read/write6Mechanical Component of A Disk DriveTracksConcentric rings around disk surface, bits laid out serially along each trackCylinderA track of the platter, 1000-5000 cylinders per zone, 1 spare per zoneSectorsEach track is split into arc of track (min unit of transfer)7Disk SectorsWhere do they come from?Formatting processLogical maps to physical What is a sector?Header (ID, defect flag, …)Real space (e.g. 512 bytes)Trailer (ECC code)What about errors?Detect errors in a sectorCorrect them with ECCIf not recoverable, replace it with a spareSkip bad sectors in the futureHdrSector…512 bytes ECCi i+1 i+2defectdefect8Disks Were LargeFirst Disk: IBM 305 RAMAC (1956)5MB capacity50 disks, each 24”9They Are Now Much SmallerForm factor: .5-1” 4” 5.7”Storage: 120-750GB Form factor: .4-.7”  2.7”  3.9”Storage: 60-200GB Form factor: .2-.4”  2.1”  3.4”Storage: 1GB-8GB10Areal Density vs. Moore’s Law (Mark Kryder at SNW 2006)1150 Years Later (Mark Kryder at SNW 2006)IBM RAMAC (1956)Seagate Momentus(2006)DifferenceCapacity 5MB 160GB 32,000Areal Density 2K bits/in2130 Gbits/in265,000,000Disks 50 @ 24” diameter 2 @ 2.5” diameter 1 / 2,300Price/MB $1,000 $0.01 1 / 3,200,000Spindle Speed1,200 RPM 5,400 RPM 5Seek Time 600 ms 10 ms 1 / 60Data Rate 10 KB/s 44 MB/s 4,400Power 5000 W 2 W 1 / 2,500Weight ~ 1 ton 4 oz 1 / 9,00012Sample Disk Specs (from Seagate)Cheetah 15k.5 Barracuda ES.2CapacityFormatted capacity (GB) 300 1000Discs 4 4Heads 8 8Sector size (bytes) 512 512PerformanceExternal interface Ultra320 SCSI, FC, S. SCSI SATASpindle speed (RPM) 15,000 7,200Average latency (msec) 2.0 4.16Seek time, read/write (ms) 3.5/4.0 8.5/9.5Track-to-track read/write (ms) 0.2-0.4 0.8/1.0External transfer (MB/sec) 300-400 350Internal transfer rate (MB/sec) 89-150 105 (sustained)Cache size (MB) 16 32ReliabilityRecoverable read errors 1 per 1012 bits read 1 per 1010 bits readNon-recoverable read errors 1 per 1016 bits read 1 per 1015 bits readDisk PerformanceSeekPosition heads over cylinder, typically 3.5-9.5 msRotational delayWait for a sector to rotate underneath the headsTypically 8  4 ms (7,200 – 15,000RPM) or ½ rotation takes 4 - 2msTransfer bytes Transfer bandwidth is typically 40-125 Mbytes/secPerformance of transfer 1 KbytesSeek (4 ms) + half rotational delay (2ms) + transfer (0.013 ms)Total time is 6.01 ms or 167 Kbytes/sec (1/360 of 60MB/sec)!14More on PerformanceWhat transfer size can get 90% of the disk bandwidth?Assume Disk BW = 60MB/sec, ½ rotation = 2ms, ½ seek = 4msBW * 90% = size / (size/BW + rotation + seek)size = BW * (rotation + seek) * 0.9 / 0.1 = 60MB * 0.006 * 0.9 / 0.1 = 3.24MBSeek and rotational times dominate the cost of small accessesDisk transfer bandwidth are wastedNeed algorithms to reduce seek timeSpeed depends on which sectors to accessBlock Size (Kbytes) % of Disk Transfer Bandwidth1Kbytes 0.28%1Mbytes 73.99%3.24Mbytes 90%15FIFO (FCFS) orderMethodFirst come first serveProsFairness among requestsIn the order applications expectConsArrival may be on random spots on the disk (long seeks)Wild swings can happen0 19998, 183, 37, 122, 14, 124, 65, 675316SSTF (Shortest Seek Time First)MethodPick the one closest on diskRotational delay is in calculationProsTry to minimize seek timeConsStarvationQuestionCan we avoid the starvation?0 19998, 183, 37, 122, 14, 124, 65, 67(65, 67, 37, 14, 98, 122, 124, 183)5317Elevator (SCAN)MethodTake the closest request in the direction of travelReal implementations do not go to the end (called LOOK)ProsBounded time for each requestConsRequest at the other end will take a while0 19998, 183, 37, 122, 14, 124, 65, 67(37, 14, 65, 67, 98, 122, 124, 183)5318C-SCAN (Circular SCAN)MethodLike SCANBut, wrap aroundReal implementation doesn’t go to the end (C-LOOK)ProsUniform service timeConsDo nothing on the return0 19998, 183, 37, 122, 14, 124, 65, 67(65, 67, 98, 122, 124, 183, 14, 37)5319DiscussionsWhich is your favorite?FIFOSSTFSCANC-SCANDisk I/O request bufferingWhere would you buffer requests?How long would you buffer requests?20RAID (Redundant Array of Independent Disks)Main ideaStore the error correcting codes on other disksGeneral error correcting codes are too powerfulUse XORs or single parityUpon any failure, one can recover the entire block from the spare disk (or any disk) using XORsProsReliabilityHigh bandwidthConsThe controller is complexD1 D2 D3 D4 PRAID controllerP = D1  D2  D3  D4D3 = D1  D2  P  D421Synopsis of RAID


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Princeton COS 318 - Storage Devices

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