WM CSCI 424 - Review: Major Components of a Computer

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Review: Major Components of a ComputerMagnetic DiskMagnetic Disk CharacteristicSlide 5Magnetic Disk Examples (www.seagate.com)Disk Latency & Bandwidth MilestonesLatency & Bandwidth ImprovementsAside: Media Bandwidth/Latency DemandsDependability, Reliability, AvailabilityRAIDs: Disk ArraysRAID: Level 0 (No Redundancy; Striping)RAID: Level 1 (Redundancy via Mirroring)RAID: Level 0+1 (Striping with Mirroring)Slide 17RAID: Level 4 (Block-Interleaved Parity)Small WritesRAID: Level 5 (Distributed Block-Interleaved Parity)Distributing Parity BlocksSummarys.1Review: Major Components of a Computer ProcessorControlDatapathMemoryDevicesInputOutputCacheMain MemorySecondary Memory(Disk)s.2Magnetic DiskPurposeLong term, nonvolatile storageLowest level in the memory hierarchy-slow, large, inexpensiveGeneral structureA rotating platter coated with a magnetic surfaceA moveable read/write head to access the information on the diskTypical numbers1 to 4 (1 or 2 surface) platters per disk of 1” to 5.25” in diameter (3.5” dominate in 2004)Rotational speeds of 5,400 to 15,000 RPM10,000 to 50,000 tracks per surface-cylinder - all the tracks under the head at a given point on all surfaces100 to 500 sectors per track-the smallest unit that can be read/written (typically 512B)TrackSectors.3Magnetic Disk CharacteristicDisk read/write components1. Seek time: position the head over the proper track (3 to 14 ms avg)-due to locality of disk references the actual average seek time may be only 25% to 33% of the advertised number2. Rotational latency: wait for the desired sector to rotate under the head (½ of 1/RPM converted to ms)-0.5/5400RPM = 5.6ms to 0.5/15000RPM = 2.0ms3. Transfer time: transfer a block of bits (one or more sectors) under the head to the disk controller’s cache (30 to 80 MB/s are typical disk transfer rates)-the disk controller’s “cache” takes advantage of spatial locality in disk accesses–cache transfer rates are much faster (e.g., 320 MB/s)4. Controller time: the overhead the disk controller imposes in performing a disk I/O access (typically < .2 ms)SectorTrackCylinderHeadPlatterController+Caches.5Typical Disk Access Time If the measured average seek time is 25% of the advertised average seek time, thenAvg disk read/write = 6.0ms + 0.5/(10000RPM/(60sec/minute) )+ 0.5KB/(50MB/sec) + 0.2ms = 6.0 + 3.0 + 0.01 + 0.2 = 9.21msAvg disk read/write = 1.5 + 3.0 + 0.01 + 0.2 = 4.71msThe average time to read or write a 512B sector for a disk rotating at 10,000RPM with average seek time of 6ms, a 50MB/sec transfer rate, and a 0.2ms controller overheadThe rotational latency is usually the largest component of the access times.6Magnetic Disk Examples (www.seagate.com)Characteristic Seagate ST37 Seagate ST32 Seagate ST94Disk diameter (inches) 3.5 3.5 2.5Capacity (GB) 73.4 200 40# of surfaces (heads) 8 4 2Rotation speed (RPM) 15,000 7,200 5,400Transfer rate (MB/sec) 57-86 32-58 34Minimum seek (ms) 0.2r-0.4w 1.0r-1.2w 1.5r-2.0wAverage seek (ms) 3.6r-3.9w 8.5r-9.5w 12r-14wMTTF (hours@25oC)1,200,000 600,000 330,000Dimensions (inches) 1”x4”x5.8” 1”x4”x5.8” 0.4”x2.7”x3.9”GB/cu.inch 3 9 10Power: op/idle/sb (watts) 20?/12/- 12/8/1 2.4/1/0.4GB/watt 4 16 17Weight (pounds) 1.9 1.4 0.2s.7Disk Latency & Bandwidth MilestonesDisk latency is one average seek time plus the rotational latency.Disk bandwidth is the peak transfer time of formatted data from the media (not from the cache).CDC Wren SG ST41 SG ST15 SG ST39 SG ST37RSpeed (RPM) 3600 5400 7200 10000 15000Year 1983 1990 1994 1998 2003Capacity (Gbytes) 0.03 1.4 4.3 9.1 73.4Diameter (inches) 5.25 5.25 3.5 3.0 2.5Interface ST-412 SCSI SCSI SCSI SCSIBandwidth (MB/s) 0.6 4 9 24 86Latency (msec) 48.3 17.1 12.7 8.8 5.7Patterson, CACM Vol 47, #10, 2004s.8Latency & Bandwidth ImprovementsIn the time that the disk bandwidth doubles the latency improves by a factor of only 1.2 to 1.40204060801001983 1990 1994 1998 2003Year of IntroductionBandwidth (MB/s)Latency (msec)s.9Aside: Media Bandwidth/Latency DemandsBandwidth requirementsHigh quality video-Digital data = (30 frames/s) × (640 x 480 pixels) × (24-b color/pixel) = 221 Mb/s (27.625 MB/s)High quality audio-Digital data = (44,100 audio samples/s) × (16-b audio samples) × (2 audio channels for stereo) = 1.4 Mb/s (0.175 MB/s)Compression reduces the bandwidth requirements considerablyLatency issuesHow sensitive is your eye (ear) to variations in video (audio) rates?How can you ensure a constant rate of delivery?How important is synchronizing the audio and video streams?-15 to 20 ms early to 30 to 40 ms late is tolerables.10Dependability, Reliability, AvailabilityReliability – measured by the mean time to failure (MTTF). Service interruption is measured by mean time to repair (MTTR)Availability – a measure of service accomplishmentAvailability = MTTF/(MTTF + MTTR)To increase MTTF, either improve the quality of the components or design the system to continue operating in the presence of faulty components1. Fault avoidance: preventing fault occurrence by construction2. Fault tolerance: using redundancy to correct or bypass faulty components (hardware)Fault detection versus fault correctionPermanent faults versus transient faultss.11RAIDs: Disk ArraysArrays of small and inexpensive disksIncrease potential throughput by having many disk drives-Data is spread over multiple disk-Multiple accesses are made to several disks at a timeReliability is lower than a single diskBut availability can be improved by adding redundant disks (RAID)Lost information can be reconstructed from redundant informationMTTR: mean time to repair is in the order of hoursMTTF: mean time to failure of disks is tens of years Redundant Array of Inexpensive Diskss.12RAID: Level 0 (No Redundancy; Striping)Multiple smaller disks as opposed to one big diskSpreading the blocks over multiple disks – striping – means that multiple blocks can be accessed in parallel increasing the performance -A 4 disk system gives four times the throughput of a 1 disk systemSame cost as one big disk – assuming 4 small disks cost the same as one big diskNo redundancy, so what if one disk fails?Failure of one or more disks is more likely as the number of disks in the system increasesblk1 blk3blk2 blk4s.13RAID: Level


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