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Secondary Storage Management

Secondary Storage Management. Chapter 13. 13.1 The Memory Hierarchy. The Memory Hierarchy. Computer systems have several different components in which data may be stored. Data capacities & access speeds range over at least seven orders of magnitude

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Secondary Storage Management

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  1. Secondary Storage Management Chapter 13

  2. 13.1 The Memory Hierarchy

  3. The Memory Hierarchy • Computer systems have several different components in which data may be stored. • Data capacities & access speeds range over at least seven orders of magnitude • Devices with smallest capacity also offer the fastest access speed

  4. Description of Levels • Cache • Megabyte or more of Cache storage. • On-board cache : On same chip. • Level-2 cache : On another chip. • Cache data accessed in few nanoseconds. • Data moved from main memory to cache when needed by processor • Volatile

  5. Description of Levels 2. Main Memory • 1 GB or more of main memory. • Instruction execution & Data Manipulation - involves information resident in main memory. • Time to move data from main memory to the processor or cache is in the 10-100 nanosecond range. • Volatile 3. Secondary Storage • Typically a magnetic disk. • Capacity upto 1 TB. • One machine can have several disk units. • Time to transfer a single byte between disk & main memory is around 10 milliseconds.

  6. Description of Levels 4. Tertiary Storage • Holds data volumes measured in terabytes. • Significantly higher read/write times. • Smaller cost per bytes. • Retrieval takes seconds or minutes, but capacities in the petabyte range are possible.

  7. Transfer of Data Between Levels • Data moves between adjacent levels of the hierarchy. • Each level is organized to transfer huge amounts of data to or from the below level • Key technique for speeding up database operations is to arrange data so that when one piece of a disk block is needed, it is likely that other data on the same block will also be needed at about the same time.

  8. Volatile & Non Volatile Storage • A volatile device “forgets” what is stored in it when the power goes off. • Example: Main Memory • A nonvolatile device, on the other hand, is expected to keep its contents intact even for long periods when the device is turned off or there is a power failure. • Example: Secondary & Tertiary Storage Note: No change to the database can be considered final until it has migrated to nonvolatile, secondary storage.

  9. Virtual Memory • Managed by Operating System. • Some memory in main memory & rest on disk. • Transfer between the two is in units of disk blocks (pages). • Not a level of the memory hierarchy

  10. Thank you!

  11. Secondary storage management Section 13.2 CS-257 Database System Principles AvinashAnantharamu (102) 008629907

  12. Index 13.2 Disks 13.2.1 Mechanics of Disks 13.2.2 The Disk Controller 13.2.3 Disk Access Characteristics

  13. Structure of a Disk

  14. Mechanics of Disks Two principal moving pieces of hard drive 1- Head Assembly 2- Disk Assembly Disk Assembly has 1 or more circular platters that rotate around a central spindle. Platters are covered with thin magnetic material

  15. Top View of Disk Surface

  16. Mechanics of Disks • Tracks are concentric circles on a platter. • Tracks are organized into sectors which are segments of circular platter. • Sectors are indivisible as far as errors are concerned. • Blocks are logical data transfer units.

  17. Disk Controller Control the actuator to move head assembly Selecting the surface from which to read or write Transfer bits from desired sector to main memory Possibly, buffering an entire track or more in local memory of the disk controller, hoping that many sectors of this track will be read soon, and additional accesses to the disk can be avoided.

  18. Simple Single Processor Computer

  19. Disk Access characteristics Seek time Rotational latency Transfer time Latency of the disk.

  20. Thank you

  21. 13.3 Accelerating Access to Secondary Storage San Jose State University Spring 2012

  22. 13.3 Accelerating Access to Secondary StorageSection Overview • 13.3.1: The I/O Model of Computation • 13.3.2: Organizing Data by Cylinders • 13.3.3: Using Multiple Disks • 13.3.4: Mirroring Disks • 13.3.5: Disk Scheduling and the Elevator Algorithm • 13.3.6: Prefetching and Large-Scale Buffering

  23. 13.3 Introduction • Average block access is ~10ms. • Disks may be busy. • Requests may outpace access delays, leading to infinite scheduling latency. • There are various strategies to increase disk throughput. • The “I/O Model” is the correct model to determine speed of database operations

  24. 13.3 Introduction (Contd.) • Actions that improve database access speed: • Place blocks closer, within the same cylinder • Increase the number of disks • Mirror disks • Use an improved disk-scheduling algorithm • Use prefetching

  25. 13.3.1 The I/O Model of Computation • If we have a computer running a DBMS that: • Is trying to serve a number of users • Has 1 processor, 1 disk controller, and 1 disk • Each user is accessing different parts of the DB • It can be assumed that: • Time required for disk access is much larger than access to main memory; and as a result: • The number of block accesses is a good approximation of time required by a DB algorithm

  26. 13.3.2 Organizing Data by Cylinders • It is more efficient to store data that might be accessed together in the same or adjacent cylinder(s). • In a relational database, related data should be stored in the same cylinder.

  27. 13.3.3 Using Multiple Disks • If the disk controller supports the addition of multiple disks and has efficient scheduling, using multiple disks can improve performance significantly • By striping a relation across multiple disks, each chunk of data can be retrieved in a parallel fashion, improving performance by up to a factor of n, where n is the total number of disks the data is striped over

  28. 13.3.4 Mirroring Disks • A drawback of striping data across multiple disks is that you increase your chances of disk failure. • To mitigate this risk, some DBMS use a disk mirroring configuration • Disk mirroring makes each disk a copy of the other disks, so that if any disk fails, the data is not lost • Since all the data is in multiple places, access speedup can be increased by more than n since the disk with the head closest to the requested block can be chose • Unfortunately, the writing of disk blocks does not speed up at all. The reason is that the new block must be written to each of the n disks.

  29. 13.3.4 Mirroring Disks

  30. 13.3.5 Disk Scheduling • One way to improve disk throughput is to improve disk scheduling, prioritizing requests such that they are more efficient • The elevator algorithm is a simple yet effective disk scheduling algorithm • The algorithm makes the heads of a disk oscillate back and forth similar to how an elevator goes up and down • The access requests closest to the heads current position are processed first

  31. 13.3.5 Disk Scheduling • When sweeping outward, the direction of head movement changes only after the largest cylinder request has been processed • When sweeping inward, the direction of head movement changes only after the smallest cylinder request has been processed • Example:

  32. 13.3.6 Prefetching and Large-Scale Buffering • In some cases we can anticipate what data will be needed • We can take advantage of this by prefetching data from the disk before the DBMS requests it • Since the data is already in memory, the DBMS receives it instantly

  33. ? Questions ?

  34. Disk Failures Presented by Timothy Chen Spring 2013

  35. Index 13.4 Disk Failures 13.4.1 Intermittent Failures 13.4.2 Organizing Data by Cylinders 13.4.3 Stable Storage 13.4.4 Error- Handling Capabilities of Stable Storage 13.4.5 Recovery from Disk Crashes 13.4.6 Mirroring as a Redundancy Technique 13.4.7 Parity Blocks 13.4.8 An Improving: RAID 5 13.4.9 Coping With Multiple Disk Crashers

  36. Intermittent Failures If we try to read the sector but the correct content of that sector is not delivered to the disk controller Controller will check good and bad sector If the write is correct: Read is performed Good sector and bad sector is known by the read operation

  37. CheckSum Read operation that determine the good or bad status If, on reading, we find that the checksum is not proper for the data bits, then we know the is an error in reading. If the checkum is proper, there is still a small chance that the block was not read correctly, but by using many checksum bits we can make the probability of missing a bad read arbitrarily small.

  38. How CheckSum perform Each sector has some additional bits Set depending on the values of the data bits stored in each sector If the data bit in the not proper we know there is an error reading Odd number of 1: bits have odd parity(01101000) Even number of 1: bit have even parity (111011100) Find Error is the it is one bit parity

  39. Stable Storage Deal with disk error Sectors are paired and each pair X showing left and right copies as Xl and Xr It check the parity bit of left and right by subsituting spare sector of Xl and Xr until the good value is returned

  40. Error-Handling Capabilities of Stable Storage Since it has XL and XR, one of them fail we can still read other one Chance both of them fail are pretty small The write Fail, it happened during power outage

  41. Recover Disk Crash The most serious mode of failure for disks is “head crash” where data permanently destroyed. The way to recover from crash , we use RAID method

  42. Mirroring as a Redundancy Technique it is call Raid 1 Just mirror each disk

  43. Raid 1 graph

  44. Parity Block It often call Raid 4 technical read block from each of the other disks and modulo-2 sum of each column and get redundant disk disk 1: 11110000 disk 2: 10101010 disk 3: 00111000 get redundant disk 4(even 1= 0, odd 1 =1) disk 4: 01100010

  45. Raid 4 graphic

  46. Parity Block- Fail Recovery It can only recover one disk fail If it has more than one like two disk Then it can’t be recover us modulo-2 sum

  47. An Improvement Raid 5

  48. Coping with multiple Disk Crash For more one disk fail Either raid 4 and raid 5 can’t be work So we need raid 6 It is need at least 2 redundant disk

  49. Raid 6

  50. Reference http://www.definethecloud.net/wp-content/uploads/2010/12/325px-RAID_1.svg_.png http://en.wikipedia.org/wiki/RAID

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