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Prof. Seok-Bum Ko

EE898.02 Architecture of Digital Systems Lecture 3: I/O Introduction and A Little Queueing Theory. Prof. Seok-Bum Ko. 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)

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Prof. Seok-Bum Ko

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  1. EE898.02Architecture of Digital SystemsLecture 3: I/O Introduction and A Little Queueing Theory Prof. Seok-Bum Ko

  2. 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. I/O Systems interrupts Processor Cache Memory - I/O Bus Main Memory I/O Controller I/O Controller I/O Controller Graphics Disk Disk Network

  4. 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. Inner Track Outer Track Sector Head Arm Platter Actuator Disk Device Terminology • Several platters, with information recorded magnetically on both surfaces (usually) • Bits recorded in tracks, which in turn divided into sectors (e.g., 512 Bytes) • 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

  6. { Platters (12) Photo of Disk Head, Arm, Actuator Spindle Arm Head Actuator

  7. Disk Device Performance Inner Track Outer Track Sector Head Controller Arm Spindle • 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 Platter Actuator

  8. 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. Response time = Queue + Controller + Seek + Rot + Xfer Service time Devices: Magnetic Disks Track Sector • 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) Cylinder Platter Head 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

  10. 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. Latency = Queuing Time + Controller time + Seek Time + Rotation Time + Size / Bandwidth { per access + per byte State of the Art: Barracuda 180 Track • 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) Sector Cylinder Track Buffer Arm Platter Head source: www.seagate.com

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

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

  23. Advantages of Small Formfactor Disk Drives Low cost/MB High MB/volume High MB/watt Low cost/Actuator Cost and Environmental Efficiencies

  24. 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. 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. 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 Arrivals Departures

  27. System server Queue Proc IOC Device A Little Queuing Theory: Notation • Queuing models assume state of equilibrium: input rate = output rate • Notation: r average number of arriving customers/secondTser average time to service a customer (traditionally µ(avg. service rate) = 1/ Tser)u server utilization (0..1): u = r x Tser Tq average time/customer in queue Tsys average time/customer in system: Tsys = Tq + TserLq average length of queue: Lq = r x Tq Lsys average length of system: Lsys = r x Tsys • Little’s Law: Lengthserver = rate x Timeserver (Mean number customers = arrival rate x mean service time)

  28. System server Queue Proc IOC Device 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?

  29. System server Queue Proc IOC Device 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 T12 + f2 x T22 +...+ fn x Tn2)/F – m12 • Must keep track of unit of measure (100 ms2 vs. 0.1 s2 ) • Squared coefficient of variance: C2 = variance/m12 • Unitless measure (100 ms2 vs. 0.1 s2) • Exponential distribution C2 = 1: most short relative to average, few others long; 90% < 2.3 x average, 63% < average • Hypoexponential distributionC2 < 1: most close to average, C2=0.5 => 90% < 2.0 x average, only 57% < average • Hyperexponential distributionC2 > 1: further from average C2 =2.0 => 90% < 2.8 x average, 69% < average Avg.

  30. System server Queue Proc IOC Device 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 + C2) • No variance => C2 = 0 => m1(z) = 1/2 x m1

  31. A Little Queuing Theory:Average Wait Time • Calculating average wait time in queue Tq • 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 Tser Tq = uxm1(z) + Lq x Ts er= 1/2 x ux Tser x (1 + C) + Lq x Ts er Tq = 1/2 x uxTs er x (1 + C) + r x Tq x Ts er Tq = 1/2 x uxTs er x (1 + C) + u x TqTqx (1 – u) = Ts er x u x (1 + C) /2Tq = Ts er x u x (1 + C) / (2 x (1 – u)) • Notation: r average number of arriving customers/secondTser average time to service a customeru server utilization (0..1): u = r x TserTq average time/customer in queueLq average length of queue:Lq= r x Tq

  32. 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 Tser • 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 queueTq = Tser x u x (1 + C) /(2 x (1 – u)) = Tser x u / (1 – u) Tser average time to service a customeru server utilization (0..1): u = r x TserTq average time/customer in queue

  33. 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: r average number of arriving customers/second = 10Tser average time to service a customer = 20 ms (0.02s)u server utilization (0..1): u = r x Tser= 10/s x .02s = 0.2Tq average time/customer in queue = Tser x u / (1 – u) = 20 x 0.2/(1-0.2) = 20 x 0.25 = 5 ms (0 .005s)Tsys average time/customer in system: Tsys =Tq +Tser= 25 msLq average length of queue:Lq= r x Tq= 10/s x .005s = 0.05 requests in queueLsys average # tasks in system: Lsys = r x Tsys = 10/s x .025s = 0.25

  34. 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: r average number of arriving customers/second= 20Tser average time to service a customer= 12 msu server utilization (0..1): u = r x Tser= /s x .s = Tq average time/customer in queue = Ts er x u / (1 – u) = x /() = x = msTsys average time/customer in system: Tsys =Tq +Tser= 16 msLq average length of queue:Lq= r x Tq= /s x s = requests in queueLsys average # tasks in system : Lsys = r x Tsys = /s xs =

  35. 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: r average number of arriving customers/second= 20Tser average time to service a customer= 12 msu server utilization (0..1): u = r x Tser= 20/s x .012s = 0.24Tq average time/customer in queue = Ts er x u / (1 – u) = 12 x 0.24/(1-0.24) = 12 x 0.32 = 3.8 msTsys average time/customer in system: Tsys =Tq +Tser= 15.8 msLq average length of queue:Lq= r x Tq= 20/s x .0038s = 0.076 requests in queueLsys average # tasks in system : Lsys = r x Tsys = 20/s x .016s = 0.32

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