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Sections 13.1 – 13.3

Secondary storage management. Sections 13.1 – 13.3. Sanuja Dabade & Eilbroun Benjamin CS 257 – Dr. TY Lin. Presentation Outline. 13.1 The Memory Hierarchy 13.1.1 The Memory Hierarchy 13.1.2 Transfer of Data Between Levels 13.1.3 Volatile and Nonvolatile Storage 13.1.4 Virtual Memory

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Sections 13.1 – 13.3

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  1. Secondary storage management Sections 13.1 – 13.3 SanujaDabade & Eilbroun Benjamin CS 257 – Dr. TY Lin

  2. Presentation Outline • 13.1 The Memory Hierarchy • 13.1.1 The Memory Hierarchy • 13.1.2 Transfer of Data Between Levels • 13.1.3 Volatile and Nonvolatile Storage • 13.1.4 Virtual Memory • 13.2 Disks • 13.2.1 Mechanics of Disks • 13.2.2 The Disk Controller • 13.2.3 Disk Access Characteristics

  3. Presentation Outline (con’t) • 13.3 Accelerating Access to Secondary Storage • 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

  4. 13.1.1 Memory Hierarchy • Several components for data storage having different data capacities available • Cost per byte to store data also varies • Device with smallest capacity offer the fastest speed with highest cost per bit

  5. Memory Hierarchy Diagram Programs, DBMS Main Memory DBMS’s Tertiary Storage As Visual Memory Disk File System Main Memory Cache

  6. 13.1.1 Memory Hierarchy • Cache • Lowest level of the hierarchy • Data items are copies of certain locations of main memory • Sometimes, values in cache are changed and corresponding changes to main memory are delayed • Machine looks for instructions as well as data for those instructions in the cache • Holds limited amount of data

  7. 13.1.1 Memory Hierarchy (con’t) • No need to update the data in main memory immediately in a single processor computer • In multiple processors data is updated immediately to main memory….called as write through

  8. Main Memory • Everything happens in the computer i.e. instruction execution, data manipulation, as working on information that is resident in main memory • Main memories are random access….one can obtain any byte in the same amount of time

  9. Secondary storage • Used to store data and programs when they are not being processed • More permanent than main memory, as data and programs are retained when the power is turned off • E.g. magnetic disks, hard disks

  10. Tertiary Storage • Holds data volumes in terabytes • Used for databases much larger than what can be stored on disk

  11. 13.1.2 Transfer of Data Between levels • Data moves between adjacent levels of the hierarchy • At the secondary or tertiary levels accessing the desired data or finding the desired place to store the data takes a lot of time • Disk is organized into bocks • Entire blocks are moved to and from memory called a buffer

  12. 13.1.2 Transfer of Data Between level (cont’d) • A key technique for speeding up database operations is to arrange the data so that when one piece of data block is needed it is likely that other data on the same block will be needed at the same time • Same idea applies to other hierarchy levels

  13. 13.1.3 Volatile and Non Volatile Storage • A volatile device forgets what data is stored on it after power off • Non volatile holds data for longer period even when device is turned off • All the secondary and tertiary devices are non volatile and main memory is volatile

  14. 13.1.4 Virtual Memory • Typical software executes in virtual memory • Address space is typically 32 bit or 232 bytes or 4GB • Transfer between memory and disk is in terms of blocks

  15. 13.2.1 Mechanism of Disk • Mechanisms of Disks • Use of secondary storage is one of the important characteristic of DBMS • Consists of 2 moving pieces of a disk • 1. Disk assembly • 2. Head assembly • Disk assembly consists of 1 or more platters • Platters rotate around a central spindle • Bits are stored on upper and lower surfaces of platters

  16. 13.2.1 Mechanism of Disk • Disk is organized into tracks • The track that are at fixed radius from center form one cylinder • Tracks are organized into sectors • Tracks are the segments of circle separated by gap

  17. 13.2.2 Disk Controller • One or more disks are controlled by disk controllers • Disks controllers are capable of • Controlling the mechanical actuator that moves the head assembly • Selecting the sector from among all those in the cylinder at which heads are positioned • Transferring bits between desired sector and main memory • Possible buffering an entire track

  18. 13.2.3 Disk Access Characteristics • Accessing (reading/writing) a block requires 3 steps • Disk controller positions the head assembly at the cylinder containing the track on which the block is located. It is a ‘seek time’ • The disk controller waits while the first sector of the block moves under the head. This is a ‘rotational latency’ • All the sectors and the gaps between them pass the head, while disk controller reads or writes data in these sectors. This is a ‘transfer time’

  19. 13.3 Accelerating Access to Secondary Storage • Several approaches for more-efficiently accessing data in secondary storage: • Place blocks that are together in the same cylinder. • Divide the data among multiple disks. • Mirror disks. • Use disk-scheduling algorithms. • Prefetch blocks into main memory. • Scheduling Latency – added delay in accessing data caused by a disk scheduling algorithm. • Throughput – the number of disk accesses per second that the system can accommodate.

  20. 13.3.1 The I/O Model of Computation • The number of block accesses (Disk I/O’s) is a good time approximation for the algorithm. • This should be minimized. • Ex 13.3: You want to have an index on R to identify the block on which the desired tuple appears, but not where on the block it resides. • For Megatron 747 (M747) example, it takes 11ms to read a 16k block. • A standard microprocessor can execute millions of instruction in 11ms, making any delay in searching for the desired tuple negligible.

  21. 13.3.2 Organizing Data by Cylinders • If we read all blocks on a single track or cylinder consecutively, then we can neglect all but first seek time and first rotational latency. • Ex 13.4: We request 1024 blocks of M747. • If data is randomly distributed, average latency is 10.76ms by Ex 13.2, making total latency 11s. • If all blocks are consecutively stored on 1 cylinder: • 6.46ms + 8.33ms * 16 = 139ms (1 average seek) (time per rotation) (# rotations)

  22. 13.3.3 Using Multiple Disks • If we have n disks, read/write performance will increase by a factor of n. • Striping – distributing a relation across multiple disks following this pattern: • Data on disk R1: R1, R1+n, R1+2n,… • Data on disk R2: R2, R2+n, R2+2n,… … • Data on disk Rn: Rn, Rn+n, Rn+2n, … • Ex 13.5: We request 1024 blocks with n = 4. • 6.46ms + (8.33ms * (16/4)) = 39.8ms (1 average seek) (time per rotation) (# rotations)

  23. 13.3.4 Mirroring Disks • Mirroring Disks – having 2 or more disks hold identical copied of data. • Benefit 1: If n disks are mirrors of each other, the system can survive a crash by n-1 disks. • Benefit 2: If we have n disks, read performance increases by a factor of n. • Performance increases further by having the controller select the disk which has its head closest to desired data block for each read.

  24. 13.3.5 Disk Scheduling and the Elevator Problem • Disk controller will run this algorithm to select which of several requests to process first. • Pseudo code: • requests[] // array of all non-processed data requests • upon receiving new data request: • requests[].add(new request) • while(requests[] is not empty) • move head to next location • if(head location is at data in requests[]) • retrieve data • remove data from requests[] • if(head reaches end) • reverse head direction

  25. 13.3.5 Disk Scheduling and the Elevator Problem (con’t) Events: Head starting point Request data at 8000 Request data at 24000 Request data at 56000 Get data at 8000 Request data at 16000 Get data at 24000 Request data at 64000 Get data at 56000 Request Data at 40000 Get data at 64000 Get data at 40000 Get data at 16000 64000 56000 48000 40000 32000 24000 16000 8000

  26. 13.3.5 Disk Scheduling and the Elevator Problem (con’t) Elevator Algorithm FIFO Algorithm

  27. 13.3.6 Prefetching and Large-Scale Buffering • If at the application level, we can predict the order blocks will be requested, we can load them into main memory before they are needed.

  28. Disk Failures Xiaqing He ID: 204 Dr. Lin

  29. Content 1)Focus on : “How to recover from disk crashes” common term RAID “redundancy array of independent disks” 2)Several schemes to recover from disk crashes: • Mirroring—RAID level 1; • Parity checks--RAID 4; • Improvement--RAID 5; • RAID 6;

  30. 1) Mirroring • The simplest scheme to recovery from Disk Crashes • How does Mirror work? -- making two or more copied of the data on different disks • Benefit: -- save data in case of one disk will fail; -- divide data on several disks and let access to several blocks at once

  31. 1) Mirroring (con’t) • For mirroring, when the data can be lost? -- the only way data can be lost if there is a second (mirror/redundant) disk crash while the first (data) disk crash is being repaired. • Possibility: Suppose: • One disk: mean time to failure = 10 years; • One of the two disk: average of mean time to failure = 5 years; • The process of replacing the failed disk= 3 hours=1/2920 year; So: • the possibility of the mirror disk will fail=1/10 * 1/2,920 =1/29,200; • The possibility of data loss by mirroring: 1/5 * 1/29,200 = 1/146,000

  32. 2)Parity Blocks • why changes? -- disadvantages of Mirroring: uses so many redundant disks • What’s new? -- RAID level 4: uses only one redundant disk • How this one redundant disk works? -- modulo-2 sum; -- the jth bit of the redundant disk is the modulo-2 sum of the jth bits of all the data disks. • Example

  33. 2)Parity Blocks(con’t)___Example Data disks: • Disk1: 11110000 • Disk2: 10101010 • Disk3: 00111000 Redundant disk: • Disk4: 01100010

  34. 2)RAID 4 (con’t) • Reading -- Similar with reading blocks from any disk; • Writing 1)change the data disk; 2)change the corresponding block of the redundant disk; • Why? -- hold the parity checks for the corresponding blocks of all the data disks

  35. 2)RAID 4 (con’t) _ writing For a total N data disks: 1) naïve way: • read N data disks and compute the modulo-2 sum of the corresponding blocks; • rewrite the redundant disk according to modulo-2 sum of the data disks; 2) better way: • Take modulo-2 sum of the old and new version of the data block which was rewritten; • Change the position of the redundant disk which was 1’s in the modulo-2 sum;

  36. 2)RAID 4 (con’t) • Data disks: • Disk1: 11110000 • Disk2: 10101010  01100110 • Disk3: 00111000 • to do: • Modulo-2 sum of the old and new version of disk 2: 11001100 • So, we need to change the positions 1,2,5,6 of the redundant disk. • Redundant disk: • Disk4: 01100010  10101110

  37. 2)RAID 4 (con’t) _failure recovery • Redundant disk crash: -- swap a new one and recomputed data from all the data disks; • One of Data disks crash: -- swap a new one; -- recomputed data from the other disks including data disks and redundant disk; • How to recomputed? (same rule, that’s why there will be some improvement) -- take modulo-2 sum of all the corresponding bits of all the other disks

  38. 3) An Improvement: RAID 5 • Why need a improvement? -- Shortcoming of RAID level 4: suffers from a bottleneck defect (when updating data disk need to read and write the redundant disk); • Principle of RAID level 5 (RAID 5): -- treat each disk as the redundant disk for some of the blocks; • Why it is feasible? The rule of failure recovery for redundant disk and data disk is the same: “take modulo-2 sum of all the corresponding bits of all the other disks” So, there is no need to retreat one as redundant disk and others as data disks

  39. 3) RAID 5 (con’t) • How to recognize which blocks of each disk treat this disk as redundant disk? -- if there are n+1 disks which were labeled from 0 to N, then we can treat the ith cylinder of disk J as redundant if J is the remainder when I is divided by n+1; • Example;

  40. 3) RAID 5 (con’t)_example N=3; • The first disk, labeled as 0 : 4,8,12…; • The second disk, labeled as 1 : 1,5,9…; • The third disk, labeled as 2 : 2,6,10…; • ………. Suppose all the 4 disks are equally likely to be written, for one of the 4 disks, the possibility of being written: • 1/4 + 3 /4 * 1/3 =1/2 • If N=m => 1/m +(m-1)/m * 1/(m-1) = 2/m

  41. 4) Coping with multiple disk crashes • RAID 6 – deal with any number of disk crashes if using enough redundant disks • Example a system of seven disks ( four data disks_numer 1-4 and 3 redundant disks_ number 5-7); • How to set up this 3*7 matrix ? (why is 3? – there are 3 redundant disks) 1)every column values three 1’s and 0’s except for all three 0’s; 2) column of the redundant disk has single 1’s; 3) column of the data disk has at least two 1’s;

  42. 4) Coping with multiple disk crashes (con’t) • Reading: • read form the data disks and ignore the redundant disk • Writing: • Change the data disk • change the corresponding bits of all the redundant disks

  43. 4) Coping with multiple disk crashes (con’t) • In those system which has 4 data disks and 3 redundant disk, how they can correct up to 2 disk crashes? • Suppose disk a and b failed: • find some row r (in 3*7 matrix)in which the column for a and b are different (suppose a is 0’s and b is 1’s); • Compute the correct b by taking modulo-2 sum of the corresponding bits from all the other disks other than b which have 1’s in row r; • After getting the correct b, Compute the correct a with all other disks available; • Example

  44. 4) Coping with multiple disk crashes (con’t)_example 3*7 matrix data disk redundant disk disk number 1 2 3 4 5 6 7

  45. 4) Coping with multiple disk crashes (con’t)_example First block of all the disks disk contents 1) 11110000 2) 10101010 3) 00111000 4) 01000001 5) 01100010 6) 00011011 7) 10001001

  46. 4) Coping with multiple disk crashes (con’t)_example Two disks crashes; disk contents 1) 11110000 2) ????????? 3) 00111000 4) 01000001 5) ????????? 6) 00011011 7) 10001001

  47. 4) Coping with multiple disk crashes (con’t)_example In that 3*7 matrix, find in row 2, disk 2 and 5 have different value and disk 2’s value is 1 and 5’s value is 0. so: compute the first block of disk 2 by modulo-2 sum of all the corresponding bits of disk 1,4,6; then compute thefirst block of disk 2 by modulo-2 sum of all the corresponding bits of disk 1,2,3; 1) 11110000 2) ????????? => 00001111 3) 00111000 4) 01000001 5) ????????? => 01100010 6) 00011011 7) 10001001

  48. 13.5 Arranging data on disk Meghna Jain ID-205 CS257 ‏Prof: Dr. T.Y.Lin

  49. Data elements are represented as records, which stores in consecutive bytes in same same disk block. Basic layout techniques of storing data : Fixed-Length Records Allocation criteria - data should start at word boundary. Fixed Length record header 1. A pointer to record schema. 2. The length of the record. 3. Timestamps to indicate last modified or last read.

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