1. I/O Chapter 8

2. Outline Introduction - 8.1 Disk Storage and Dependability – 8.2 Buses and other connectors – 8.4 I/O performance measures – 8.6

3. Input / Ouput devices Communicate between human and computer –keyboard, mouse, printer, game controllers, … Store more than what is on processor –hard drive, thumb drive, … Enhanced functionality –music, video, …

4. Constraints Users intolerant of lost data Unknown (non-standard) devices Slow devices (relatively speaking)

5. Goals Dependability Expandability

6. Taxonomy Behavior – –Input (read once) –Output (write once, never read) –Storage (read / write / carries state) Partner –Human or machine on other side? Data rate (speed) –peak data transfer rate

7. Measures of Performance Response time –latency – time a user must wait for task Bandwidth –I/O operations per unit time –Data transferred per unit time

8. Outline Introduction - 8.1 Disk Storage and Dependability – 8.2 Buses and other connectors – 8.4 I/O performance measures – 8.6

9. Anatomy of a Disk Drive

10. Vocabulary Head – the device that reads data from a disk Each disk is divided into ________ _______ called _________ Each track is made up of _________ cylinder – volume of all _______ that lie under the heads at a given point on all surfaces nonvolatile – data that remains even when ______ is removed Concentric circles tracks sectors tracks power

11. Vocabulary seek – the act of positioning the _____ over the correct ________ rotational delay or latency – average latency to rotate disk to put the ______ over the correct _______ transfer time – time required to _________ a block of data disk controller – controls disk accesses head track head sector send / receive

12. Example 1 - Performance What is the average time to read or write a 512-byte sector for a typical disk rotating at 10,000 RPM? The advertised average seek time is 6 ms, the transfer rate is 50 MB/sec, and the controller overhead is 0.2 ms. Assume that the disk is idle, so that there is no waiting time.

13. seek time: 6ms rotational delay: ½ rev * 1min / 10,000 rev transfer time: 512 B * 1 sec / 50*1024*1024 B controller overhead = 0.2ms 6ms + 3ms + 0.01ms + 0.2ms = 9.21ms

14. Reliability Reliability – measure of a continuously working system Availability – how often, on average, the system is working properly MTTF – Mean Time to Failure MTTR – Mean Time to Repair MTBF – Mean time between failures

15. Availability Availability – how often, on average, the system is working properly Availability = MTTF / (MTTF + MTTR)

16. Improving MTTF Fault avoidance Fault tolerance Fault forecasting preventing fault occurrence by construction using redundancy to continue executing in the presence of faults (usually hardware faults) predicting the presence and creation of faults (hardware & software faults)

17. RAID - Redundant Arrays of Inexpensive Disks Shift from one large disk to several small disks Cheaper, smaller, faster Inherently less reliable Provide redundancy to counteract lower reliability

18. RAID 0 No redundancy!!! Only a performance increase Striping (interleaving) – allocation of logically sequential blocks to separate disks to increase performance Parallel access controlled by disk controller – computer knows nothing about it.

19. RAID 1 mirroring – write the identical data to multiple disks Requires twice as many disks as RAID 0 If a disk fails, use the backup copy, move to a working set of mirrored space.

20. RAID 3 Bit-interleaved parity Store only enough data to recover original Group N blocks Add one bit of parity – xor of all bits. Lost data can be reconstructed by looking at the rest of the bits in the group.

21. On a write Read all blocks of data in parity group Calculate new parity Write new block Write new parity

22. RAID 4 More efficient parity update On write: –Read old data –xor with new data –adjust parity –Write parity, Write new data

23. RAID 5 Rotate parity blocks around system Spread out writing (since parity always written)

24. Summary RAID 1 and RAID 5 most common 80% of server disks use RAID Repair: –hot swapping – replace disks with power on –Standby spares – spares included in system for immediate reconstruction of data

25. Outline Introduction - 8.1 Disk Storage and Dependability – 8.2 Buses and other connectors – 8.4 I/O performance measures – 8.6

26. Connecting I/O Devices Much slower than processor / memory Support lots of heterogeneous devices

27. Bus control lines – send / receive commands data lines – transfer data processor-memory bus – fast, small bus connecting DRAM to processor I/O bus – slow, long bus connecting many devices to system through a controller.

28. Synchrony Synchronous – clock in control line, fixed protocol is relative to clock. Asynchronous – no clock – must coordinate through hand-shaking to determine when data is ready to send / receive. split transaction protocol – bus is released between data request and data response

29. hand-shaking Series of steps used to coordinate bus transfers. Both parties must acknowledge they are ready before moving to next step. Control lines: ReadReq: proc/device wants to read DataRdy: dev/proc is ready to send data Ack: acknowledge ReadReq or DataRdy

30. Handshaking protocol ReadReq Data Ack DataRdy AddressData 1 2 3 5 6 7 4: Response0: Request

31. Outline Introduction - 8.1 Disk Storage and Dependability – 8.2 Buses and other connectors – 8.4 I/O performance measures – 8.6

32. Metric Units Memory: GB = 2^30 I/O: GB = 10^9 Be careful when reading specs For this class, we will pretend that all use base-2 units.

33. Measuring Performance Throughput more important than latency Large database operations (TP – Transaction Processing)

34. Example 1 Execution time = 100 seconds 90 seconds CPU time, 10 sec I/O time CPU time improves by 50%/yr for 5 years I/O does not improve How much faster is program after 5 years? What percentage of new time is I/O?

35. New CPU time: 90 / (1.5^5) = 12 seconds CPU improvement: 90/12 = 7.5 speedup Overall improvement: 100/22 = 4.5 speedup Now I/O is 10/22 = 45% of total time

36. Example 2 System A: –.005 sec per I/O op –4 overlapping I/O ops at a time System B: –.002 sec per I/O op –no overlapping I/O ops Which has the higher throughput?

37. Throughput: 4 ops / 0.005 sec = 800 ops/sec Throughput: 1 op / 0.002 sec = 500 ops / sec