Disk StorageCOMP375 1Disk StorageDisk StorageCOMP375 Computer Architecture dO i tiand OrganizationGoals• Understand how information is stored on a di k d idisk drive.• Be able to calculate how long it will take to read data from a disk.• Understand how data is stored on a CD and how this impacts its useand how this impacts its use.Disk Drives• There are several types of rotating media storage devices.– Hard drives• 36 GB to 2 TB– Floppy disk• 1.4 MB–CD ROM• 600 MB to 750 MB–DVD• 4.7 GB to 8.5 GBHard Drive Storage• A hard drive contains several flat platters d ith th i ti lik th tcovered with a smooth iron coating like that of a tape cassette.• Data can be stored by magnetizing small areas of the iron coating.•The disk surface is polished to be very flat•The disk surface is polished to be very flat.• The disk head floats over the platter spinning at 4.3 million inches/sec or 70 mph.Disk StorageCOMP375 2Disk Read/Write Head© C. HamacherDisk Tracks• Data is written on the disk drive in concentric circles called tracks.•A track is composed of blocks of data•A track is composed of blocks of data called sectors. Each sector has a header including address and checksum.• An arm containing the read/write head can move closer or farther from the center of the disk.• All of the tracks on both sides of all platters that can be read without moving the heads is called a cylinder.Cylinder, Heads and Sectors Terminology• Sector or Block – the smallest unit that b d itt Oft 512 b tcan be read or written. Often 512 bytes.• Track – all blocks that form a ring on a disk surface that can be read without moving the head.•Cylinder–all tracks on all surfaces one•Cylinder–all tracks on all surfaces, one on top of another, that can be read without moving the head.Disk StorageCOMP375 3Disk OperationTo read (or write) data to the disk:• The arm containing the read/write heads must be moved to the proper radius from the center.• The system must wait for the data to rotate under the read head.•The data is read as it passes under the•The data is read as it passes under the read head.• The data is checked and then passed to the I/O controller.Disk Drive Side View© C. HamacherIf a disk has 4 platters and 16K tracks per surface, how many tracks are on the disk?20% 20% 20%20%20%1. 42. 16K3. 32K4 16K 32K 64K 128K4. 64K5. 128KDisk Performance ParametersDisk read or write involves three factors1.Seek time1.Seek time– time it takes to position the head at the desired track2. Rotational delay or rotational latency– time its takes for the beginning of the sector ththhdto reach the head3. Transfer time– time required for the data to move under the headDisk StorageCOMP375 4Seek Time• Seek Time is fixed by the design of the disk.• Manufacturers will usually tell you the– average seek time– maximum seek time (from center to edge)– time to seek to the next adjacent track.Rotational Delay• Best case is when the data comes under the head just as it is needed (delay is zero).Wt i jti ditdh t•Worst case is you just missed it and have to wait a whole revolution. If you know the rotational speed, you can calculate the time per revolution.1000ms/sec*60sec/min• Average is half a revolution or above/2 ms/revrev/min1000ms/sec60sec/min=How many sectors?• Files are stored in sectors or blocks on the di kdisk.• The number of bytes in a sector varies per disk but is often 512 bytes/sector or ½K / sector Transfer Time• The transfer time is determined by how long it t k th d t t t l d th h dit takes the data to travel under the head.• The fraction of sectors on the track that are being read times the rotation time gives the transfer time.dimetransfer_tms/rev*vsectors/readsectors_re=Disk StorageCOMP375 5Rotational Speed 7,200 RPMARdSkTi89Average Read Seek Time8.9 msTrack-To-Track Seek Time 2.0 ms (average)Full Stroke Seek 21.0 ms (average)Cylinders 16,383Number of Heads (Physical) 6Sectors Per Track 63Bytes Per Sector 512Performance ExampleHow long does it take to read a 512 byte block from the disk?Rtti TiRotation TimeAverage Seek time8.9 msAverage Rotational Delay8.33ms / 2 = 4.17 msn7200rev/mi1000ms/sec*60sec/min8.33ms =DelayTransfer timeTotal13.2 ms0.13ms8.33ms*rev63sectors/1sector=Performance ExampleHow long does it take to read two 512 byte blocks from the disk?Rtti TiRotation TimeAverage Seek time8.9 msAverage Rotational Delay8.33ms / 2 = 4.17 msn7200rev/mi1000ms/sec*60sec/min8.33ms =DelayTransfer timeTotal13.3 ms0.26ms8.33ms*rev63sectors/2sector=Performance Notes• The largest component of the time to read a block is the seek time.Si i th dikf t d th•Spinning the disk faster reduces the rotational delay and transfer time.• It is very advantageous to be able to read data without moving the heads.•Disk fragmentation causes a significantDisk fragmentation causes a significant reduction in speed.– 2 consecutive blocks takes 13.3 ms– 2 randomly located blocks takes 26.4 msDisk StorageCOMP375 6How long does it take to read one block of data on the average?20% 20% 20%20%20%Rotational Speed 10,000 RPMAverage Seek Time 4 5 msAverage Seek Time 4.5 msBytes / sector 512Sectors / track 631. 7.6 ms2.10.6 ms 7.6 ms 10.6 ms 13.2 ms 4.5 ms 8.3*1042 seconds2.10.6 ms3. 13.2 ms4. 4.5 ms5. 8.3*1042secondsLogical vs. Physical• Many disks present the OS with a logical layout that is different from the physicallayout that is different from the physical layout.• Most modern disks use Logical Block Addressing (LBA) to hide the physical layout.• LBA represents the disk as a sequential list of blocks.CD ROMEarly Laserdisk and current DVDCD ROM• A CD contains an aluminum layer sandwiched between layers of clear plastic•An infrared laser is used to read the CDAn infrared laser is used to read the CD.• Bits are recorded as pits or spots in the aluminum. These affect the reflectance of the laser light.• Each pit is approximately 100 nm deep by 500nm wide and varies from 850nm to500nm wide, and varies from 850nm to 3.5 μm in length. • The data on a CD is written as one long 5.38 km spiralDisk StorageCOMP375 7CD Data FormatA CD contains 2352 byte blocks with:• 2048 bytes of data• 16 byte header containing the address• 288 bytes of error correcting codesMusic CDs• 16 bit samples• 44,100 samples per second CD Performance• The “X” of CD speed claims represents the b f ti f t th CD i thnumber of times faster the CD