1. EE898 Lec 3.1 10/22/04 EE898.02 Architecture of Digital Systems Lecture 3: I/O Introduction and A Little Queueing Theory Prof. Seok-Bum Ko

2. EE898 Lec 3.2 10/22/04 Motivation: Who Cares About I/O? CPU Performance: 60% per year I/O system performance limited by mechanical delays (disk I/O) < 10% per year (IO per sec) Amdahl's Law: system speed-up limited by the slowest part! 10% IO & 10x CPU => 5x Performance (lose 50%) 10% IO & 100x CPU => 10x Performance (lose 90%) I/O bottleneck: Diminishing fraction of time in CPU Diminishing value of faster CPUs

3. EE898 Lec 3.3 10/22/04 I/O Systems Processor Cache Memory - I/O Bus Main Memory I/O Controller Disk I/O Controller I/O Controller Graphics Network interrupts

4. EE898 Lec 3.4 10/22/04 Storage Technology Drivers Driven by the prevailing computing paradigm –1950s: migration from batch to on-line processing –1990s: migration to ubiquitous computing »computers in phones, books, cars, video cameras, … »nationwide fiber optical network with wireless tails Effects on storage industry: –Embedded storage »smaller, cheaper, more reliable, lower power –Data utilities »high capacity, hierarchically managed storage

5. EE898 Lec 3.5 10/22/04 Disk Device Terminology Several platters, with information recorded magnetically on both surfaces (usually) Actuator moves head (end of arm,1/surface) over track (“seek”), select surface, wait for sector rotate under head, then read or write –“Cylinder”: all tracks under heads Bits recorded in tracks, which in turn divided into sectors (e.g., 512 Bytes) Platter Outer Track Inner Track Sector Actuator HeadArm

6. EE898 Lec 3.6 10/22/04 Photo of Disk Head, Arm, Actuator Actuator Arm Head Platters (12) { Spindle

7. EE898 Lec 3.7 10/22/04 Disk Device Performance Platter Arm Actuator HeadSector Inner Track Outer Track Disk Latency = Seek Time + Rotation Time + Transfer Time + Controller Overhead Seek Time? depends on no. tracks move arm, seek speed of disk Rotation Time? depends on speed disk rotates, how far sector is from head Transfer Time? depends on data rate (bandwidth) of disk (bit density), size of request Controller Spindle

8. EE898 Lec 3.8 10/22/04 Data Rate: Inner vs. Outer Tracks To keep things simple, originally kept same number of sectors per track –Since outer track longer, lower bits per inch Competition  decided to keep BPI the same for all tracks (“constant bit density”)  More capacity per disk  More of sectors per track towards edge  Since disk spins at constant speed, outer tracks have faster data rate Bandwidth outer track 1.7X inner track! –Inner track highest density, outer track lowest, so not really constant –2.1X length of track outer / inner, 1.7X bits outer / inner

9. EE898 Lec 3.9 10/22/04 Devices: Magnetic Disks Sector Track Cylinder Head Platter Purpose: – Long-term, nonvolatile storage – Large, inexpensive, slow level in the storage hierarchy Characteristics: – Seek Time (~8 ms avg) »positional latency »rotational latency Transfer rate – 10-40 MByte/sec –Blocks Capacity –Gigabytes –Quadruples every 2 years (aerodynamics) 7200 RPM = 120 RPS => 8 ms per rev ave rot. latency = 4 ms 128 sectors per track => 0.25 ms per sector 1 KB per sector => 16 MB / s Response time = Queue + Controller + Seek + Rot + Xfer Service time

10. EE898 Lec 3.10 10/22/04 Disk Performance Model /Trends Capacity + 100%/year (2X / 1.0 yrs) Transfer rate (BW) + 40%/year (2X / 2.0 yrs) Rotation + Seek time – 8%/ year (1/2 in 10 yrs) MB/$ > 100%/year (2X / 1.0 yrs) Fewer chips + areal density

11. EE898 Lec 3.11 10/22/04 State of the Art: Barracuda 180 –181.6 GB, 3.5 inch disk –12 platters, 24 surfaces –24,247 cylinders –7,200 RPM; (4.2 ms avg. latency) –7.4/8.2 ms avg. seek (r/w) –64 to 35 MB/s (internal) –0.1 ms controller time –10.3 watts (idle) source: www.seagate.com Latency = Queuing Time + Controller time + Seek Time + Rotation Time + Size / Bandwidth per access per byte { + Sector Track Cylinder Head Platter Arm Track Buffer

12. EE898 Lec 3.12 10/22/04 Areal Density Bits recorded along a track –Metric is Bits Per Inch (BPI) Number of tracks per surface –Metric is Tracks Per Inch (TPI) Disk Designs Brag about bit density per unit area –Metric is Bits Per Square Inch –Called Areal Density –Areal Density = BPI x TPI

13. EE898 Lec 3.13 10/22/04 Historical Perspective 1956 IBM Ramac — early 1970s Winchester –Developed for mainframe computers, proprietary interfaces –Steady shrink in form factor: 27 in. to 14 in Form factor and capacity drives market, more than performance 1970s: Mainframes  14 inch diameter disks 1980s: Minicomputers,Servers  8”,5 1/4 ” diameter PCs, workstations Late 1980s/Early 1990s: –Mass market disk drives become a reality »industry standards: SCSI, IPI, IDE –Pizzabox PCs  3.5 inch diameter disks –Laptops, notebooks  2.5 inch disks –Palmtops didn’t use disks, so 1.8 inch diameter disks didn’t make it 2000s: –1 inch for cameras, cell phones?

14. EE898 Lec 3.14 10/22/04 Disk History Data density Mbit/sq. in. Capacity of Unit Shown Megabytes 1973: 1. 7 Mbit/sq. in 140 MBytes 1979: 7. 7 Mbit/sq. in 2,300 MBytes source: New York Times, 2/23/98, page C3, “Makers of disk drives crowd even more data into even smaller spaces”

15. EE898 Lec 3.15 10/22/04 Disk History 1989: 63 Mbit/sq. in 60,000 MBytes 1997: 1450 Mbit/sq. in 2300 MBytes source: New York Times, 2/23/98, page C3, “Makers of disk drives crowd even more data into even smaller spaces” 1997: 3090 Mbit/sq. in 8100 MBytes

16. EE898 Lec 3.16 10/22/04 1 inch disk drive! 2000 IBM MicroDrive: – 1.7” x 1.4” x 0.2” –1 GB, 3600 RPM, 5 MB/s, 15 ms seek –Digital camera, PalmPC? 2006 MicroDrive? 9 GB, 50 MB/s! –Assuming it finds a niche in a successful product –Assuming past trends continue

17. EE898 Lec 3.17 10/22/04 Fallacy: Use Data Sheet “Average Seek” Time Manufacturers needed standard for fair comparison (“benchmark”) –Calculate all seeks from all tracks, divide by number of seeks => “average” Real average would be based on how data laid out on disk, where seek in real applications, then measure performance –Usually, tend to seek to tracks nearby, not to random track Rule of Thumb: observed average seek time is typically about 1/4 to 1/3 of quoted seek time (i.e., 3X-4X faster)

18. EE898 Lec 3.18 10/22/04 Fallacy: Use Data Sheet Transfer Rate Manufacturers quote the speed off the data rate off the surface of the disk Sectors contain an error detection and correction field (can be 20% of sector size) plus sector number as well as data There are gaps between sectors on track Rule of Thumb: disks deliver about 3/4 of internal media rate (1.3X slower) for data

19. EE898 Lec 3.19 10/22/04 Tape vs. Disk Longitudinal tape uses same technology as hard disk; tracks its density improvements Disk head flies above surface, tape head lies on surface Disk fixed, tape removable Inherent cost-performance based on geometries: fixed rotating platters with gaps (random access, limited area, 1 media / reader) vs. removable long strips wound on spool (sequential access, "unlimited" length, multiple / reader) Helical Scan (VCR, Camcoder, DAT) Spins head at angle to tape to improve density

20. EE898 Lec 3.20 10/22/04 Current Drawbacks to Tape Tape wear out: –Helical 100s of passes to 1000s for longitudinal Head wear out: –2000 hours for helical Both must be accounted for in economic / reliability model Bits stretch Readers must be compatible with multiple generations of media Long rewind, eject, load, spin-up times; not inherent, just no need in marketplace Designed for archival

21. EE898 Lec 3.21 10/22/04 Library vs. Storage Getting books today as quaint as the way WE learned to program –punch cards, batch processing –wander thru shelves, anticipatory purchasing Cost $1 per book to check out $30 for a catalogue entry 30% of all books never checked out Write only journals? Digital library can transform campuses

22. EE898 Lec 3.22 10/22/04 Use Arrays of Small Disks? 14” 10”5.25”3.5” Disk Array: 1 disk design Conventional: 4 disk designs Low End High End Katz and Patterson asked in 1987: Can smaller disks be used to close gap in performance between disks and CPUs?

23. EE898 Lec 3.23 10/22/04 Advantages of Small Formfactor Disk Drives Low cost/MB High MB/volume High MB/watt Low cost/Actuator Cost and Environmental Efficiencies

24. EE898 Lec 3.24 10/22/04 Redundant Arrays of (Inexpensive) Disks Files are "striped" across multiple disks Redundancy yields high data availability –Availability: service still provided to user, even if some components failed Disks will still fail Contents reconstructed from data redundantly stored in the array  Capacity penalty to store redundant info  Bandwidth penalty to update redundant info

25. EE898 Lec 3.25 10/22/04 RAID Fundamental to RAID is “striping”, a method of concatenating multiple drives into one logical storage unit. RAID-0: not redundant RAID-1: by writing all data to two or more drives. “mirroring” RAID-2: Hamming error correction codes RAID-3: striping data at a byte level across several drives, with parity stored on one drive RAID-4: striping data at a block level across several drives, with parity stored on one drive RAID-5: similar to level 4, but distributes parity among the drives

26. EE898 Lec 3.26 10/22/04 Redundant Arrays of Inexpensive Disks RAID 1: Disk Mirroring/Shadowing Each disk is fully duplicated onto its “mirror” Very high availability can be achieved Bandwidth sacrifice on write: Logical write = two physical writes Reads may be optimized Most expensive solution: 100% capacity overhead ( RAID 2 not interesting, so skip) recovery group

27. EE898 Lec 3.27 10/22/04 Redundant Array of Inexpensive Disks RAID 3: Parity Disk P 10010011 11001101 10010011... logical record 1010001110100011 1100110111001101 1010001110100011 1100110111001101 P contains sum of other disks per stripe mod 2 (“parity”) If disk fails, subtract P from sum of other disks to find missing information Striped physical records

28. EE898 Lec 3.28 10/22/04 RAID 3 Sum computed across recovery group to protect against hard disk failures, stored in P disk Logically, a single high capacity, high transfer rate disk: good for large transfers Wider arrays reduce capacity costs, but decreases availability 33% capacity cost for parity in this configuration

29. EE898 Lec 3.29 10/22/04 Inspiration for RAID 4 RAID 3 relies on parity disk to discover errors on Read But every sector has an error detection field Rely on error detection field to catch errors on read, not on the parity disk Allows independent reads to different disks simultaneously

30. EE898 Lec 3.30 10/22/04 Redundant Arrays of Inexpensive Disks RAID 4: High I/O Rate Parity D0D1D2 D3 P D4D5D6 PD7 D8D9 PD10 D11 D12 PD13 D14 D15 P D16D17 D18 D19 D20D21D22 D23 P.............................. Disk Columns Increasing Logical Disk Address Stripe Insides of 5 disks Example: small read D0 & D5, large write D12-D15 Example: small read D0 & D5, large write D12-D15

31. EE898 Lec 3.31 10/22/04 Inspiration for RAID 5 RAID 4 works well for small reads Small writes (write to one disk): –Option 1: read other data disks, create new sum and write to Parity Disk –Option 2: since P has old sum, compare old data to new data, add the difference to P Small writes are limited by Parity Disk: Write to D0, D5 both also write to P disk D0 D1D2 D3 P D4 D5 D6 P D7

32. EE898 Lec 3.32 10/22/04 Redundant Arrays of Inexpensive Disks RAID 5: High I/O Rate Interleaved Parity Independent writes possible because of interleaved parity Independent writes possible because of interleaved parity D0D1D2 D3 P D4D5D6 P D7 D8D9P D10 D11 D12PD13 D14 D15 PD16D17 D18 D19 D20D21D22 D23 P.............................. Disk Columns Increasing Logical Disk Addresses Example: write to D0, D5 uses disks 0, 1, 3, 4

33. EE898 Lec 3.33 10/22/04 Problems of Disk Arrays: Small Writes D0D1D2 D3 P D0' + + D1D2 D3 P' new data old data old parity XOR (1. Read) (2. Read) (3. Write) (4. Write) RAID-5: Small Write Algorithm 1 Logical Write = 2 Physical Reads + 2 Physical Writes

34. EE898 Lec 3.34 10/22/04 System Availability: Orthogonal RAIDs Array Controller String Controller String Controller String Controller String Controller String Controller String Controller... Data Recovery Group: unit of data redundancy Redundant Support Components: fans, power supplies, controller, cables End to End Data Integrity: internal parity protected data paths

35. EE898 Lec 3.35 10/22/04 System-Level Availability Fully dual redundant I/O Controller Array Controller......... Recovery Group Goal: No Single Points of Failure Goal: No Single Points of Failure host with duplicated paths, higher performance can be obtained when there are no failures

36. EE898 Lec 3.36 10/22/04 Introduction to Queueing Theory More interested in long term, steady state than in startup => Arrivals = Departures Little’s Law: Mean number tasks in system = arrival rate x mean response time –Observed by many, Little was first to prove Applies to any system in equilibrium, as long as nothing in black box is creating or destroying tasks ArrivalsDepartures

37. EE898 Lec 3.37 10/22/04 A Little Queuing Theory: Notation Queuing models assume state of equilibrium: input rate = output rate Notation: r average number of arriving customers/second T ser average time to service a customer (traditionally µ(avg. service rate) = 1/ T ser ) userver utilization (0..1): u = r x T ser T q average time/customer in queue T sys average time/customer in system: T sys = T q + T ser L q average length of queue: L q = r x T q L sys average length of system: L sys = r x T sys Little’s Law: Length server = rate x Time server (Mean number customers = arrival rate x mean service time) ProcIOCDevice Queue server System

38. EE898 Lec 3.38 10/22/04 A Little Queuing Theory Service time completions vs. waiting time for a busy server: randomly arriving event joins a queue of arbitrary length when server is busy, otherwise serviced immediately –Unlimited length queues key simplification A single server queue: combination of a servicing facility that accommodates 1 customer at a time (server) + waiting area (queue): together called a system Server spends a variable amount of time with customers; how do you characterize variability? –Distribution of a random variable: histogram? curve? ProcIOCDevice Queue server System

39. EE898 Lec 3.39 10/22/04 A Little Queuing Theory Server spends variable amount of time with customers –Weighted mean m1 = (f1 x T1 + f2 x T2 +...+ fn x Tn)/F (F=f1 + f2...), where Ti is the time for task i and Fi is the frequency of i –variance = (f1 x T1 2 + f2 x T2 2 +...+ fn x Tn 2 )/F – m1 2 »Must keep track of unit of measure (100 ms 2 vs. 0.1 s 2 ) –Squared coefficient of variance: C 2 = variance/m1 2 »Unitless measure (100 ms 2 vs. 0.1 s 2 ) Exponential distribution C 2 = 1 : most short relative to average, few others long; 90% < 2.3 x average, 63% < average Hypoexponential distribution C 2 90% < 2.0 x average, only 57% < average Hyperexponential distribution C 2 > 1 : further from average C 2 =2.0 => 90% < 2.8 x average, 69% < average ProcIOCDevice Queue server System Avg.

40. EE898 Lec 3.40 10/22/04 A Little Queuing Theory: Variable Service Time Server spends a variable amount of time with customers –Weighted mean m1 = (f1xT1 + f2xT2 +...+ fnxTn)/F (F=f1+f2+...) Usually pick C = 1.0 for simplicity Another useful value is average time must wait for server to complete task: m1(z) –Not just 1/2 x m1 because doesn’t capture variance –Can derive m1(z) = 1/2 x m1 x (1 + C 2 ) –No variance => C 2 = 0 => m1(z) = 1/2 x m1 ProcIOCDevice Queue server System

41. EE898 Lec 3.41 10/22/04 A Little Queuing Theory: Average Wait Time Calculating average wait time in queue T q –If something at server, it takes to complete on average m1(z) –Chance server is busy = u; average delay is u x m1(z) –All customers in line must complete; each avg T s er T q = u x m1(z) + L q x T s er = 1/2 x u x T ser x (1 + C) + L q x T s er T q = 1/2 x u x T s er x (1 + C) + r x T q x T s er T q = 1/2 x u x T s er x (1 + C) + u x T q T q x (1 – u) = T s er x u x (1 + C) /2 T q = T s er x u x (1 + C) / (2 x (1 – u)) Notation: raverage number of arriving customers/second T ser average time to service a customer userver utilization (0..1): u = r x T ser T q average time/customer in queue L q average length of queue:L q = r x T q

42. EE898 Lec 3.42 10/22/04 A Little Queuing Theory: M/G/1 and M/M/1 Assumptions so far: –System in equilibrium, number sources of requests unlimited –Time between two successive arrivals in line are exponentially distrib. –Server can start on next customer immediately after prior finishes –No limit to the queue: works First-In-First-Out "discipline" –Afterward, all customers in line must complete; each avg T ser Described “memoryless” or Markovian request arrival (M for C=1 exponentially random), General service distribution (no restrictions), 1 server: M/G/1 queue When Service times have C = 1, M/M/1 queue T q = T ser x u x (1 + C) /(2 x (1 – u)) = T ser x u / (1 – u) T ser average time to service a customer userver utilization (0..1): u = r x T ser T q average time/customer in queue

43. EE898 Lec 3.43 10/22/04 A Little Queuing Theory: An Example processor sends 10 x 8KB disk I/Os per second, requests & service exponentially distrib., avg. disk service = 20 ms On average, how utilized is the disk? –What is the number of requests in the queue? –What is the average time spent in the queue? –What is the average response time for a disk request? Notation: raverage number of arriving customers/second = 10 T ser average time to service a customer = 20 ms (0.02s) userver utilization (0..1): u = r x T ser = 10/s x.02s = 0.2 T q average time/customer in queue = T ser x u / (1 – u) = 20 x 0.2/(1-0.2) = 20 x 0.25 = 5 ms (0.005s) T sys average time/customer in system: T sys =T q +T ser = 25 ms L q average length of queue:L q = r x T q = 10/s x.005s = 0.05 requests in queue L sys average # tasks in system: L sys = r x T sys = 10/s x.025s = 0.25

44. EE898 Lec 3.44 10/22/04 A Little Queuing Theory: Another Example processor sends 20 x 8KB disk I/Os per sec, requests & service exponentially distrib., avg. disk service = 12 ms On average, how utilized is the disk? –What is the number of requests in the queue? –What is the average time a spent in the queue? –What is the average response time for a disk request? Notation: raverage number of arriving customers/second= 20 T ser average time to service a customer= 12 ms userver utilization (0..1): u = r x T ser = /s x. s = T q average time/customer in queue = T s er x u / (1 – u) = x /( ) = x = ms T sys average time/customer in system: T sys =T q +T ser = 16 ms L q average length of queue:L q = r x T q = /s x s = requests in queue L sys average # tasks in system : L sys = r x T sys = /s x s =

45. EE898 Lec 3.45 10/22/04 A Little Queuing Theory: Another Example processor sends 20 x 8KB disk I/Os per sec, requests & service exponentially distrib., avg. disk service = 12 ms On average, how utilized is the disk? –What is the number of requests in the queue? –What is the average time a spent in the queue? –What is the average response time for a disk request? Notation: raverage number of arriving customers/second= 20 T ser average time to service a customer= 12 ms userver utilization (0..1): u = r x T ser = 20/s x.012s = 0.24 T q average time/customer in queue = T s er x u / (1 – u) = 12 x 0.24/(1-0.24) = 12 x 0.32 = 3.8 ms T sys average time/customer in system: T sys =T q +T ser = 15.8 ms L q average length of queue:L q = r x T q = 20/s x.0038s = 0.076 requests in queue L sys average # tasks in system : L sys = r x T sys = 20/s x.016s = 0.32