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Database Systems

Database Systems. Disk Management Concepts. WHY DO DISKS NEED MANAGING?. logical information  physical representation bigger databases, larger records, more complex structures in the logical schema, unusual data types more points at which bottleneck develop, and performance degrades

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Database Systems

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  1. Database Systems Disk Management Concepts

  2. WHY DO DISKS NEED MANAGING? • logical information  physical representation • bigger databases, larger records, more complex structures in the logical schema, unusual data types • more points at which bottleneck develop, and performance degrades • behaviour of the data population over time • behaviour of the user population over time • interaction between

  3. QUERY PARSING, RELATIONAL OPERATORS • 2. OPTIMISATION AND EXECUTION • 3. FILES AND ACCESS METHODS • 4. BUFFER MANAGEMENT • 5. DISK SPACE MANAGEMENT What role does the DBMS have? QUERY

  4. Logical Layers Construct a strategy for a given Query, e.g.: SELECT S.Semester, count(*) FROM Section S WHERE S.Year > 91 GROUP BY S.Semester HAVING COUNT (*) > all (Select count(*) from section Group by Section)

  5. Query Optimiser, Relational Operators • Phase 1: Convert into relational algebra expression using Select, Join, Project; e.g., includes S.Semester in the Projection list • Phase 2: Consider alternate access plans - without index, with index - and choose an execution plan.

  6. Physical Layers • File Level Management Pages - Collection of data records, index records Keeps track of pages within a file Organizes the information within a page • Buffer Management Partitions available main memory into pages Bring pages from disk to main memory Uses routines in DSM • Disk Space Management Management of space on disk

  7. DBMS & STORAGE DEVICE PROPERTIES Why not put everything in Primary Storage? • fast access • but (often) limited storage capacity • even if PS large, there are other reasons… • volatility • stability • cost per unit stored • consistency and integrity • locking and concurrency

  8. DBMS, STORAGE DEVICE PROPERTIES Secondary Storage • tape • magnetic disk • drum • optical disk, CD-ROM, • database-oriented hardware devices • other

  9. Secondary Storage • Hardware parameters and access performance • Disk Capacity • number of surfaces • number of read/write heads • number of tracks (30 to 16000) • sectors, blocks, pages • capacity of a track in blocks (512..2048) • capacity of a cylinder

  10. Illustration – Disk, Track, Sector • Disk, sector, track, block • Seeking a track Seek time Is a time needed to move R-W head from one position to another – the desired one. In estimations only the average seek time is considered The average seek time is provided in the manufacturer disk specification

  11. Illustration – Disk, Track, Sector • Block • Seeking a track • Rotational delay (latency) Rotational delay is a time needed for rotation of the disk resulting in the positioning of of the desired block under the R-W head. In estimations only the average rotational delay is considered The average rotational delay is calculated as ½ time of one full rotation

  12. Illustration – Disk, Track, Sector … • Block • Seeking a track • Rotational delay (latency) • Block transfer Block transfer time is a time of moving one block under the R-W head. Block transfer time (btt) is proportional to the relative size of a block in the track and to the time of one rotation. Eg. If a block occupies 1/100 part of a track and one rotation of a track needs 50msec then btt is 50*1/100 = 0.5 msec

  13. Illustration – Disk, Track, Sector … • Disk pack

  14. Hardware parameters and access performance • Disk Organisation Conceptual: Select Semester from Section … Logical: Read Byte I of Record n of File f Physical: Read block m of track t of cylinder c of disk d and transfer to buffer b Transfer of a block (or cluster of blocks): smallest unit of transfer Total transfer time is combined from seek times, rotational delays and block transfer times: • Access time = seek time +latency+transfer time seek time +latencyis much greater thantransfer time

  15. Illustration - Cylinder • Cylinder, Contiguous blocks • No seek time is necessary for each block if the desired blocks are located on one cylinder • One rotational delay is to be used in time evaluation if blocks are contiguous

  16. Physical Disk Structure

  17. Examples - Exercises • A. Double sided disk has on each side 44 Tracks, each track has 64 blocks, and usable size of each block is 1024 bytes. Find usable capacity of the disk 2*44*64*1024 = 5,767,168 • B. Disk pack consists of 15 disks with the above parameters. Find full capacity of the disk pack and the capacity of a cylinder 5,767,168 * 15 = 86,507,520 15*2*64*1024 = 1,966,080 • C. The size of gaps between blocks (interblock gap) are of 128 bytes. Assuming rotational speed 600 revs/min calculate the time for the transfer of one block ( do not include seek and latency). 64 *(1024+128) = 73728 600 r/min = 10 r/sec 10* 73,728 = 737,280 1024/737280 = 0.001388 sec 1.39 msec

  18. Hardware parameters and access performance • Access time = seek time +latency+transfer time • seek time +latencyis much greater thantransfer time • Hence • Cache storage to capitalise on pages already fetched • Advantage of storing a file in clusters of contiguous blocks • Advantage of storing a file on one cylinder

  19. Hardware parameters and access performance Time taken to accomplish a file read thus depends on: • whether the block with the desired record is in cache; • seek: number of tracks to traverse • latency: disk revolution speed • block transfer rate • buffering • file organisation

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