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. What role does the DBMS have? 1.QUERY PARSING, RELATIONAL OPERATORS 2. OPTIMISATION AND EXECUTION 3. FILES AND ACCESS METHODS 4. BUFFER MANAGEMENT 5. DISK SPACE MANAGEMENT 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. 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 Block Seeking a track Rotational delay (latency) Illustration – Disk, Track, Sector

12. 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 Block Seeking a track Rotational delay (latency) Block transfer Illustration – Disk, Track, Sector …

13. 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 + latency is much greater than transfer 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

22. 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 = 86,507,52015*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) = 73728600 r/min = 10 r/sec 10* 73,728 = 737,2801024/737280 = 0.001388 sec 1.39 msec

23. Hardware parameters and access performance Access time = seek time + latency + transfer time seek time + latency is much greater than transfer 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

24. 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