CS 554: Advanced Database System Notes 02: Hardware

1. CS 554: Advanced Database SystemNotes 02: Hardware Hector Garcia-Molina Notes 2

2. Outline Hardware: Disks Access Times (disk) Optimizations (disk access time) Other Topics: Storage costs Using secondary storage Disk failures Notes 2

3. Hardware DBMS Data Storage Notes 2

4. CPU P Typical Computer Disk Controller ... ... M C Memory Secondary Storage Notes 2

5. Secondary storage Many flavors: - Disk: Floppy (hard, soft) Removable Packs Winchester (most common) SSD disks Optical, CD-ROM… Arrays - Tape: Reel, cartridge Robots Notes 2

6. “Typical Disk:” Platter Head … Terms: Platter, Head, Cylinder, Track, Sector (physical), Block (logical), Gap Notes 2

7. Top View Gap Sector Track Notes 2

8. Block Block Block = group of sectors that form a unit of access One read/write operation will read/write one block Notes 2

9. Disk Access Time block x in memory I want block X How long ? Notes 2

10. Platter Head … Time = Seek Time + Rotational Delay + Transfer Time + Other Seek time: to move head to the desired cylinder (track) Rotational delay: for waiting on the desired sector Transfer time: to transfer data on sectors to memory Notes 2

11. Seek Time Once head moving, the head travels fast 3 or 5x Seek Time x Cylinders Traveled 1 N Takes time to start the head moving Notes 2

12. Average Random Seek Time Start at cylinder i  Go to cylinder j N N  SEEKTIME (i  j) S = N(N-1) j=1 ji i=1 There are N starting cylinders and N-1 cylinders Total: N(N-1) possible values Notes 2

13. Average Random Seek Time N N  SEEKTIME (i  j) S = N(N-1) j=1 ji i=1 “Typical” S: 10 ms  40 ms Notes 2

14. Typical Seek Time • Ranges from • 4ms for high end drives • 15ms for mobile devices • Typical SSD (Solid State): ranges from • 0.08ms • 0.16ms • Source: Wikipedia, "Hard disk drive performance characteristics" Notes 2

15. Rotational Delay Disk platter rotates Head is here Block I Want Notes 2

16. Average Rotational Delay R=0 for SSDs R = 1/2 revolution Typical HDD figures Source: Wikipedia, "Hard disk drive performance characteristics" Notes 2

17. Transfer Rate: # bits transferred/sec • Transfer rates: • HDD: up to 1000 Mbit/sec • 12x Blu-Ray: 432 Mbit/sec • 1xCD: 1.23 Mbits/sec • for SSDs, limited by interfacee.g., SATA 3000 Mbit/s • Transfer time: Amount data transferred Transfer rate Notes 2

18. Other Delays • CPU time to issue I/O • Contention delay for disk controller • Different programs can be using the disk • Contention delay for bus, memory • Different programs can be transferring data These delays are negligible compared to Seek time + rotational delay + transfer time Notes 2

19. So far: One (Random) Block Access • What about: Reading “Next” block? Notes 2

20. If we do things right(e.g., Double Buffer, Stagger Blocks…) Time to get = Block Size + Negligible “next” block Transfer rate - skip gap - switch track - once in a while, next cylinder Notes 2

21. Rule ofRandom I/O: ExpensiveThumb Sequential I/O: Much less Notes 2

22. Cost for Writingsimilar to Reading …. unless we want to verify: need to add (full) rotation + Block size Transfer time Notes 2

23. To Modify a Block? Notes 2

24. To Modify Block: (a) Read Block into Memory (b) Modify block in Memory (c) Write Block [(d) Verify?] To Modify a Block? Notes 2

25. Random Access Time • Hand Drive:Ranges from 2.9 msec (high end server drive) to 12 msec (laptop HDD) • Due to the need to move the heads and wait for the data to rotate under the read/write head Notes 2

26. Data Transfer Rate • Hard Disk:Once the head is positioned, an enterprise HDD can transfer data at about 140 MBytes/sec. • In practice, much lower speeds because…. • Data transfer rate depends also on rotational speed (of the platter) ! Notes 2

27. Reliability • Hard Disk: According to a study performed by CMU for both consumer and enterprise-grade HDDs, their average failure rate is 6 years, and life expectancy is 9–11 years. Notes 2

28. Cost and Capacity • Hard Drive: • In 2013: HDDs of up to 6 TB were available. • In 2014: Cost: around $50 per TeraByte Notes 2

29. Kibibytes • 1 kibibyte = 210 bytes = 1024 bytes. from Wikipedia Notes 2

30. Outline • Hardware: Disks • Access Times • Optimizations • Other Topics • Storage Costs • Using Secondary Storage • Disk Failures here Notes 2

31. Optimizations(in controller or O.S.) • Disk Scheduling Algorithms • e.g., elevator algorithm • Pre-fetch (Double buffering) • Arrays (RAID) • Mirrored Disks Notes 2

32. Disk Scheduling: Elevator Algorithm Situation: Have many read/write requests Question: In which order do you process the requests ? Notes 2

33. Disk Scheduling: Elevator Algorithm • Process requests • for these cylinders 2. Then process requests this way Current cylinder Notes 2

34. Double Buffering Algorithm Problem: You have a File • Sequence of Blocks B1, B2, …, Bn You have a Program that: • Process B1 • Process B2 • Process B3 ... Notes 2

35. Single Buffer Solution (“naïve” solution) (1) Read B1  Buffer (2) Process Data in Buffer (3) Read B2  Buffer (4) Process Data in Buffer ... Notes 2

36. Say P = time to process/block R = time to read in 1 block n = # blocks  R (1) Read B1  Buffer (2) Process Data in Buffer (3) Read B2  Buffer (4) Process Data in Buffer ...  P  R  P Time to process n block = n(P + R) Notes 2

37. A B C D E F G Double Buffering process Memory: Disk: Read block 1 Notes 2

38. B A B C D E F G done Double Buffering process Memory: Disk: A Process block 1 AND read block 2 simultaneously Notes 2

39. A B C D E F G Double Buffering process Memory: Disk: B C A AND read block 3 simultaneously Process block 2 done Notes 2

40. Say P > R P = Processing time/block R = IO time/block n = # blocks What is processing time? Notes 2

41. A B C D E F G Double Buffering process Memory: Disk: Read block 1  R Notes 2

42. B A B C D E F G done Double Buffering Time needed = P (P > R) process Memory: Disk: A AND read block 2  R simultaneously Process block 1  P Notes 2

43. A B C D E F G Double Buffering Time needed = P (P > R) process Memory: Disk: B C A AND read block 3  R simultaneously Process block 2  P done Notes 2

44. Say P  R P = Processing time/block R = IO time/block n = # blocks What is processing time? • Double buffering time = R + nP • Single buffering time = n(R+P) Notes 2

45. Using disk array to accelerate disk access • Why use multiple disks: • Multiple disks  multiple disk heads • Multiple outputs = Increased data rate Notes 2

46. Techniques to deploit multiple disks • Block Striping: • Store blocks of a file over multiple disks • (This technique uses multiple disks as point 2) • Mirror disk: • Store the same data on multiple disks • RAID: • Redundant Array of Independent (inexpensive) Disks Notes 2

47. Block Striping • Blocks of the same file stored on different disks Data blocks of 1 file Notes 2

48. Disk Mirroring • Mirrored disks contain identical content • Read operation: n times as fast • Write operation: about the same as 1 disk logicallyone disk Notes 2

49. Disk Arrays (Even parity) • RAIDs (various flavors) Parity block Data blocks 00 01 00 10 11 logicallyone disk Notes 2

50. Disk Failures • Intermittent read failure • Cause: power fluctuations/failure • Intermittent write failure • Cause: power fluctuation/failure • Media decay  discuss first • Disk surface worn out • Permanent failure  redundancy… • Disk crash Notes 2