存储技术英文 1.2课件

1、Storage System EnvironmentModule 1.2 Storage System EnvironmentUpon completion of this module, you will be able to:List components of storage system environmentHost, connectivity and storageList physical and logical components of hostsDescribe key connectivity optionsDescribe the physical disk struc

2、tureDiscuss factors affecting disk drive performance2Storage System EnvironmentLesson: Components of Storage System EnvironmentUpon completion of this lesson, you will be able to:Describe the three components of storage system environmentHost, Connectivity and StorageDetail Host physical and logical

3、 componentsDescribe interface protocolPCI, IDE/ATA and SCSIDescribe storage optionsTape, optical and disk drives3Storage System EnvironmentHost: Logical Components HostDBMSHBAHBAHBAApplicationsVolume ManagerOperating SystemFile SystemDevice Drivers5Storage System EnvironmentLogical Components of the

4、 HostApplication Interface between user and the host Three-tiered architecture Application UI, computing logic and underlying databasesApplication data access can be classifies as:Block-level access: Data stored and retrieved in blocks, specifying the LBAFile-level access: Data stored and retrieved

5、by specifying the name and path of filesOperating systemResides between the applications and the hardwareControls the environment6Storage System EnvironmentLogical Components of the Host (Cont)Device DriversEnables operating system to recognize the device Provides API to access and control devicesHa

6、rdware dependent and operating system specificFile SystemFile is a collection of related records or data stored as a unit File system is hierarchical structure of filesExamples: FAT 32, NTFS, UNIX FS and EXT2/37Storage System EnvironmentFile System: Metadata ExamplesUNIX (UFS) File type and permissi

7、onsNumber of linksOwner and group IDsNumber of bytes in the fileLast file accessLast file modificationWindows (NTFS)Time stamp and link countFile nameAccess rightsFile dataIndex informationVolume information 8Storage System EnvironmentFile Systems: Journaling and Logging (cont)Improves data integrit

8、y and system restart time over non-journaling file systems.Uses a separate area called a log or journal.May hold all data to be writtenMay hold only metadataDisadvantage - slower than other file systems.Each file system update requires at least 1 extra write to the log10Storage System EnvironmentVol

9、ume GroupsOne or more Physical Volumes form a Volume GroupLVM manages Volume Groups as a single entityPhysical Volumes can be added and removed from a Volume Group as necessary Physical Volumes are typically divided into contiguous equal-sized disk blocks A host will always have at least one disk gr

10、oup for the Operating SystemApplication and Operating System data maintained in separate volume groupsLogical Disk BlockVolume GroupPhysical Disk BlockPhysical Volume 1Physical Volume 2Physical Volume 3Logical VolumeLogical Volume12Storage System EnvironmentLinux LVM架构图14Storage System EnvironmentLi

11、nux LVM架构图 （cont.）15Storage System EnvironmentHow Files are Moved to and from Storage123456Consisting ofMapped by LVM toTeacher (User)Course File(s)File System FilesFile System BlocksLVM Logical ExtentsDisk Physical ExtentsDisk SectorsConfigures/ManagesResiding inReside inMapped by a file system toM

12、anaged by disk storage subsystem16Storage System EnvironmentConnectivityInterconnection between hosts or between a host and any storage devicesPhysical Components of Connectivity are:Bus, port and cableCPUHBAPortCableBUSDisk17Storage System EnvironmentBus Technology SerialSerial Bi-directionalParall

13、el18Storage System EnvironmentConnectivity ProtocolProtocol = a defined format for communication between sending and receiving devicesTightly connected entities such as central processor to RAM, or storage buffers to controllers (example PCI)Directly attached entities connected at moderate distances

14、 such as host to storage (example IDE/ATA)Network connected entities such as networked hosts, NAS or SAN (example SCSI or FC)Tightly ConnectedEntities DirectlyAttachedEntitiesNetwork ConnectedEntities20Storage System EnvironmentPopular Connectivity Options: PCIPCI is used for local bus system within

15、 a computerIt is an interconnection between microprocessor and attached devicesHas Plug and Play functionalityPCI is 32/64 bitThroughput is 133 MB/secPCI Express Enhanced version of PCI bus with higher throughput and clock speed21Storage System EnvironmentPopular Connectivity Options: IDE/ATAIntegra

16、ted Device Electronics (IDE) / Advanced Technology Attachment (ATA)Most popular interface used with modern hard disksGood performance at low costInexpensive storage interconnectUsed for internal connectivitySerial Advanced Technology Attachment (SATA)Serial version of the IDE /ATA specification Hot-

17、pluggableEnhanced version of bus provides upto 6Gb/s (revision 3.0)23Storage System EnvironmentPATA24Storage System EnvironmentPopular Connectivity Options: SCSIParallel SCSI (Small computer system interface)Most popular hard disk interface for serversHigher cost than IDE/ATASupports multiple simult

18、aneous data accessUsed primarily in “higher end” environments SCSI Ultra provides data transfer speeds of 320 MB/sSerial SCSISupports data transfer rate of 3 Gb/s (SAS 300)26Storage System EnvironmentSCSI-Bus Interface ConnectorSCA (Single Connector Attached) W (Wide) N (Narrow)27Storage System Envi

19、ronmentSCSI - Pros and ConsPros:Fast transfer speeds, up to 320 megabytes per secondReliable, durable componentsCan connect many devices with a single bus, more than just HDsSCSI host cards can be put in almost any systemFull backwards compatibilityCons:Configuration and setup specific to one comput

20、erUnlike IDE, few BIOS support the standardOverwhelming number of variations in the standard, hardware, and connectorsNo common software interfaces and protocol28Storage System EnvironmentStorage: Medias and OptionsMagnetic TapeLow cost solution for long term data storageLimitationsSequential data a

21、ccess, Single application access at a time, Physical wear and tear and Storage/retrieval overheadsOptical DisksPopularly used as distribution medium in small, single-user computing environmentsWrite once and read many (WORM): CD-ROM, DVD-ROMLimited in capacity and speedDisk DriveMost popular storage

22、 medium with large storage capacityRandom read/write accessIdeal for performance intensive online application30Storage System EnvironmentLesson SummaryKey points covered in this lesson:Host componentsPhysical and LogicalConnectivity optionsPCI, IDE/ATA, SCSIStorage optionsTape, optical and disk driv

23、e31Storage System EnvironmentLesson: Disk DriveUpon completion of this lesson, you will be able to:List and discuss various disk drive componentsPlatter, spindle, read/write head and actuator arm assemblyDiscuss disk drive geometryDescribe CHS and LBA addressing schemeDisk drive performanceSeek time

24、, rotational latency and transfer rateLaws governing disk drive performanceEnterprise flash drive32Storage System EnvironmentDisk Drive ComponentsInterfaceControllerPower Connector HDA33Storage System EnvironmentPhysical Disk StructureSectorTrackPlatterSectorTrackCylinderSpindle34Storage System Envi

25、ronmentLogical Block AddressingPhysical Address= CHSCylinder 2Head 0Sector 10Block 48Block 16Block 32Logical Block Address= Block#Block 0Block 8(Upper Surface)(Lower Surface)35Storage System EnvironmentPlatter Geometry and Zoned-Bit RecordingPlatter Without ZonesSectorTrackPlatter With Zones36Storag

26、e System EnvironmentDisk Drive PerformanceElectromechanical deviceImpacts the overall performance of the storage systemDisk Service TimeTime taken by a disk to complete an I/O requestSeek TimeRotational Latency Data Transfer Rate37Storage System EnvironmentDisk Drive Performance: Seek TimeTime taken

27、 to position the read/write head Lower the seek time, the faster the I/O operationSeek time specifications include:Full strokeAverageTrack-to-track38Storage System EnvironmentDisk Drive Performance: Rotational Speed/LatencyThe time taken by platter to rotate and position the data under the R/W headD

28、epends on the rotation speed of the spindleAverage rotational latency One-half of the time taken for a full rotationAppx. 5.5 ms for 5400-rpm driveAppx. 2.0 ms for 15000-rpm drive39Storage System EnvironmentDisk Drive Performance: Data Transfer RateAverage amount of data per unit timeInternal Transf

29、er RateSpeed at which data moves from a track to disk internal bufferExternal Transfer RateThe advertised speed of the interfaceInterfaceBufferHBADisk DriveInternal transfer rate measured hereExternal transfer rate measured hereHead Disk Assembly40Storage System EnvironmentDisk Design ProblemSuppose

30、 you are asked to design a rotating disk where the number of bits per track is constant. You know that the number of bits per track is determined by the circumference of the innermost track, which you can assume is also the circumference of the hole. Thus, if you make the hole in the center of the d

31、isk larger, the number of bits per track increases, but the total number of tracks decreases. If you let r denote the radius of the platter, and xr the radius of the hole, what value of x maximizes the capacity of the disk?41Storage System EnvironmentBest Read Time & Random Read Time ProblemSuppose

32、that 3 MB file consisting of 1024-byte logical blocks is stored on a disk drive with the following characteristics:For each case below, suppose that a program reads the logical blocks of the file sequentially, one after the other, and that the time of position the head over the first block is Tavg s

33、eek + Tavg rotationBest case: Estimate the optimal time (in ms) required to read the file over all possible mappings of logical blocks to disk sectors.Random cases: Estimate the time (in ms) required to read the file if blocks are mapped randomly to disk sectors.42Storage System EnvironmentAnatomy o

34、f a Commercial DiskSeagate Cheetah15K.4 geometry43Storage System EnvironmentAnatomy of a Commercial Disk (cont.)Seagate Cheetah15K.4 performance44Storage System EnvironmentSeagate Cheetah 15K.4 Zone MapSource: DIXtrac automatic disk drive characterization tool. Data for zone 14 is not available.45St

35、orage System EnvironmentSeagate Cheetah 15K.4 Zone Map (cont.)More sectors are packed into the outer zones than the inner zones.Each zone has more sectors than logical blocks.The spare sectors form a pool of spare cylinders.The notion of a logical block not only provides a simpler interface to the o

36、perating system, it also provides a level of indirection that enables the disk to be more robust.Use the zone map in the previous slide to determine the number of spare cylinders in the following zones.Zone 0Zone 846Storage System EnvironmentSATA vs SAS drives in 3.5-in form factor in 2006Power cons

37、umed by the disk motor:Power Diameter4.6 RPM2.8 Number of platters Power Consumption47Storage System EnvironmentFundamental Laws Governing Disk PerformanceLittles LawDescribes the relationship between the number of requests in a queue and the response time. N = a R“N” is the total number of requests

38、 in the system“a” is the arrival rate “R” is the average response timeUtilization lawDefines the I/O controller utilizationU = a RS“U” is the I/O controller utilization“RS“ is the service time126543I/O ControllerProcessed I/O RequestArrivalI/O Queue 48Storage System EnvironmentLittle LawMean number

39、of tasks in system = Arrival rate Mean response timeMean number of tasks in system = Timeaccumulated / TimeobserveTimeaccumulated / Timeobserve = Timeaccumulated / Numbertasks Numbertasks/ TimeobserveMean response time = Timeaccumulated / NumbertasksArrival rate = Numbertasks/ Timeobserve49Storage S

40、ystem EnvironmentSmall Example on Little LawImagine a small shop with a single counter and an area for browsing, where only one person can be at the counter at a time, and no one leaves without buying something. So the system is roughly:Entrance Browsing Counter ExitThis is a stable system, so the r

41、ate at which people enter the store is the rate at which they arrive at the counter and the rate at which they exit as well. We call this the arrival rate. By contrast, an arrival rate exceeding an exit rate would represent an unstable system, and cause the store to overflow eventually.Littles Law t

42、ells us that the average number of customers in the store is the arrival rate times the average time that a customer spends in the store.Assume customers arrive at the rate of 10 per hour and stay an average of 0.5 hour. This means we should find the average number of customers in the store at any t

43、ime to be 5.50Storage System EnvironmentSmall Example on Little Law (cont.)Now suppose the store is considering doing more advertising to raise the arrival rate to 20 per hour. The store must either be prepared to host an average of 10 occupants or must reduce the time each customer spends in the st

44、ore to 0.25 hour. We can apply Littles Law to systems within the shop. For example, the counter and its queue. Assume we notice that there are on average 2 customers in the queue and at the counter. We know the arrival rate is 10 per hour, so customers must be spending 0.2 hour on average checking o

45、ut.We can even apply Littles Law to the counter itself. The average number of people at the counter would be in the range (0,1) since no more than one person can be at the counter at a time. In that case, the average number of people at the counter is also known as the counters utilization.51Storage

46、 System EnvironmentUtilization vs. Response timeConsider a disk I/O system in which an I/O request arrives at a rate of 100 I/Os per second. The service time, RS, is 4 ms. Utilization of I/O controller (U = a Rs) Total response time (R= Rs /1U) Time spent by a request in a queue Rq = Rs U /(1U) Aver

47、age queue size = U2 / (1U)Calculate the same with service time is doubled0%100%UtilizationKnee of curve: disks at about 70% utilizationLow Queue Size70%52Storage System EnvironmentApplication Requirements and Disk Performance Exercise:Consider an application that requires 1TB of storage capacity and

48、 performs 4900 IOPS Application I/O size is 4KBAs it is business critical application, response time must be within acceptable rangeSpecification of available disk drive:Drive capacity = 73 GB 15000 RPM5 ms average seek time40 MB/sec transfer rateCalculate the number of disks required?53Storage Syst

49、em EnvironmentSolutionCalculate time required to perform one I/O Seek time + (rotational delay)/speed in RPM + (block size/transfer rate)Therefore, 5 ms + 0.5 /15000 + 4K/40MB = 7.1 msecCalculate max. number of IOPS a disk can perform1 / 7.1 ms = 140 IOPSFor acceptable response time disk controller

50、utilization must be less than 70%Therefore, 140 X 0.7 = 98 IOPS To meet application Performance requirement we need 4900/98 i.e. 50 diskCapacity requirement we need 1TB/ 73 GB i.e. 14 diskDisk required = max (capacity, performance) 54Storage System EnvironmentEnterprise Flash Drives: A New Generatio

51、n DrivesConventional disk driveMechanical Delay associated with conventional driveSeek time Rotational latencyMore power consumption due to mechanical operationsLow Mean Time Between FailureEnterprise flash driveHighest possible throughput per driveNo Spinning magnetic mediaNo Mechanical movement wh

52、ich causes seek and latencySolid State enables consistent I/O performanceVery low latency per I/OEnergy efficient storage designLower power requirement per GB of storageLower power requirement per IOPS55Storage System EnvironmentEnterprise Flash Drives OverviewDrive is based on Flash Solid State memory technologyHigh performance and low latency Non volatile memoryUses single layer cell (SLC) or Multi Level cell (MLC) to store dataEnterprise Flash Drives use a 4Gb FC interface56Storage System EnvironmentEnterprise Flash Drives BenefitsFaster performanceUp to 30 times greater IOPS (benchmark