CS252Graduate Computer Architecture

Lecture 22

I/O Continued

John Kubiatowicz

Electrical Engineering and Computer Sciences

University of California, Berkeley

http://www.eecs.berkeley.edu/~kubitron/cs252

http://www-inst.eecs.berkeley.edu/~cs252

4/18/2007 cs252-S07, Lecture 22 2

Review: A Little Queuing Theory• Assumptions:

– System in equilibrium; No limit to the queue– Time between successive arrivals is random and memoryless

• Parameters that describe our system: : mean number of arriving customers/second– Tser: mean time to service a customer (“m1”)– C: squared coefficient of variance = 2/m12

– μ: service rate = 1/Tser

– u: server utilization (0u1): u = /μ = Tser • Parameters we wish to compute:

– Tq: Time spent in queue– Lq: Length of queue = Tq (by Little’s law)

• Results:– Memoryless service distribution (C = 1):

» Called M/M/1 queue: Tq = Tser x u/(1 – u)– General service distribution (no restrictions), 1 server:

» Called M/G/1 queue: Tq = Tser x ½(1+C) x u/(1 – u))

Arrival Rate

Queue ServerService Rate

μ=1/Tser

4/18/2007 cs252-S07, Lecture 22 3

A Little Queuing Theory: An Example

• Processor access memory over “network”• DRAM service properties

– speed = 9cycles+ 2cycles/word– With 8 word cachelines: Tser= 25cycles, =1/25=.04 ops/cycle– Deterministic Servicetime! (C=0)

• Processor behavior– CPI=1, 40% memory, 7.5% cache misses – Rate: = 1 inst/cycle*.4*.075 = .03 ops/cycle

• Notation: average number of arriving customers/cycle=.03

Tser average time to service a customer= 25 cyclesu server utilization (0..1): u = x Tser= .0325=.75Tq average time/customer in queue = Tser x u x (1 + C) /(2 x (1 – u))

= (25 x 0.75 x ½)/(1 – 0.75) = 37.5 cyclesTsys average time/customer in system: Tsys = Tq +Tser= 62.5 cyclesLq average length of queue:Lq= x Tq

= .03 x 37.5 = 1.125 requests in queueLsys average # tasks in system :Lsys = x Tsys = .03 x 62.5 = 1.875

4/18/2007 cs252-S07, Lecture 22 4

Review: A Three-Bus System

• A small number of backplane buses tap into the processor-memory bus

– Processor-memory bus is only used for processor-memory traffic

– I/O buses are connected to the backplane bus

• Advantage: loading on the processor bus is greatly reduced

Processor Memory

Processor Memory Bus

BusAdaptor

BusAdaptor

BusAdaptor

I/O Bus

BacksideCache bus

I/O BusL2 Cache

4/18/2007 cs252-S07, Lecture 22 5

Main components of Intel Chipset: Pentium 4

• Northbridge:– Handles memory

– Graphics

• Southbridge: I/O– PCI bus

– Disk controllers

– USB controllers

– Audio

– Serial I/O

– Interrupt controller

– Timers

4/18/2007 cs252-S07, Lecture 22 6

DeviceController

readwrite

controlstatus

AddressableMemoryand/orQueuesRegisters

(port 0x20)

HardwareController

Memory MappedRegion: 0x8f008020

BusInterface

How does the processor actually talk to the device?

• CPU interacts with a Controller– Contains a set of registers that

can be read and written– May contain memory for request

queues or bit-mapped images • Regardless of the complexity of the connections and

buses, processor accesses registers in two ways: – I/O instructions: in/out instructions

» Example from the Intel architecture: out 0x21,AL– Memory mapped I/O: load/store instructions

» Registers/memory appear in physical address space» I/O accomplished with load and store instructions

Address+Data

Interrupt Request

Processor Memory Bus

CPU

RegularMemory

InterruptController

BusAdaptor

BusAdaptor

Other Devicesor Buses

4/18/2007 cs252-S07, Lecture 22 7

Example: Memory-Mapped Display Controller• Memory-Mapped:

– Hardware maps control registers and display memory into physical address space

» Addresses set by hardware jumpers or programming at boot time

– Simply writing to display memory (also called the “frame buffer”) changes image on screen

» Addr: 0x8000F000—0x8000FFFF– Writing graphics description to

command-queue area » Say enter a set of triangles that describe

some scene» Addr: 0x80010000—0x8001FFFF

– Writing to the command register may cause on-board graphics hardware to do something

» Say render the above scene» Addr: 0x0007F004

• Can protect with page tables

DisplayMemory

0x8000F000

0x80010000

Physical AddressSpace

Status0x0007F000Command0x0007F004

GraphicsCommand

Queue

0x80020000

4/18/2007 cs252-S07, Lecture 22 8

Case for Storage• Shift in focus from computation to

communication and storage of information – E.g., Cray Research/Thinking Machines vs.

Google/Yahoo– “The Computing Revolution” (1960s to 1980s)

“The Information Age” (1990 to today)

• Storage emphasizes reliability and scalability as well as cost-performance

• What is “Software king” that determines which HW actually features used?

– Operating System for storage– Compiler for processor

• Also has own performance theory—queuing theory—balances throughput vs. response time

4/18/2007 cs252-S07, Lecture 22 9

Hard Disk Drives

IBM/Hitachi Microdrive

Western Digital Drivehttp://www.storagereview.com/guide/

Read/Write HeadSide View

4/18/2007 cs252-S07, Lecture 22 10

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 developments

– 5.25 inch floppy disk formfactor (microcode into mainframe)– Emergence of industry standard disk interfaces

• Early 1980s: PCs and first generation workstations• Mid 1980s: Client/server computing

– Centralized storage on file server» accelerates disk downsizing: 8 inch to 5.25

– Mass market disk drives become a reality» industry standards: SCSI, IPI, IDE» 5.25 inch to 3.5 inch drives for PCs, End of proprietary interfaces

• 1900s: Laptops => 2.5 inch drives• 2000s: Shift to perpendicular recording

– 2006: Seagate introduces 750GB drive– 2007: Seagate promises 1TB drive by second quarter

4/18/2007 cs252-S07, Lecture 22 11

Disk History

Data densityMbit/sq. in.

Capacity ofUnit ShownMegabytes

1973:1. 7 Mbit/sq. in140 MBytes

1979:7. 7 Mbit/sq. in2,300 MBytes

source: New York Times, 2/23/98, page C3, “Makers of disk drives crowd even mroe data into even smaller spaces”

4/18/2007 cs252-S07, Lecture 22 12

Disk History

1989:63 Mbit/sq. in60,000 MBytes

1997:1450 Mbit/sq. in2300 MBytes

source: New York Times, 2/23/98, page C3, “Makers of disk drives crowd even mroe data into even smaller spaces”

1997:3090 Mbit/sq. in8100 MBytes

4/18/2007 cs252-S07, Lecture 22 13

Properties of a Hard Magnetic Disk

• Properties– Independently addressable element: sector

» OS always transfers groups of sectors together—”blocks”– A disk can access directly any given block of information it

contains (random access). Can access any file either sequentially or randomly.

– A disk can be rewritten in place: it is possible to read/modify/write a block from the disk

• Typical numbers (depending on the disk size):– 500 to more than 20,000 tracks per surface– 32 to 800 sectors per track

» A sector is the smallest unit that can be read or written• Zoned bit recording

– Constant bit density: more sectors on outer tracks– Speed varies with track location

Track

Sector

Platters

4/18/2007 cs252-S07, Lecture 22 14

MBits per square inch: DRAM as % of Disk over time

0%

10%

20%

30%

40%

50%

1974 1980 1986 1992 1998

source: New York Times, 2/23/98, page C3, “Makers of disk drives crowd even mroe data into even smaller spaces”

470 v. 3000 Mb/si

9 v. 22 Mb/si

0.2 v. 1.7 Mb/si

4/18/2007 cs252-S07, Lecture 22 15

Nano-layered Disk Heads• Special sensitivity of Disk head comes from “Giant

Magneto-Resistive effect” or (GMR) • IBM is (was) leader in this technology

–Same technology as TMJ-RAM breakthrough

Coil for writing

4/18/2007 cs252-S07, Lecture 22 16

Disk Figure of Merit: 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: Areal Density = BPI x TPI

Year Areal Density1973 2 1979 8 1989 63 1997 3,090 2000 17,100 2006 130,000 2007 164,000

1

10

100

1,000

10,000

100,000

1,000,000

1970 1980 1990 2000 2010

Year

Are

al D

ensi

ty

4/18/2007 cs252-S07, Lecture 22 17

Newest technology: Perpendicular Recording

• In Perpendicular recording:– Bit densities much higher

– Magnetic material placed on top of magnetic underlayer that reflects recording head and effectively doubles recording field

4/18/2007 cs252-S07, Lecture 22 18

Seagate Barracuda

• 750 GB! 130 GB/in2

• 4 platters, 2 heads each• 3.5” platters • Perpendicular recording• 7200 RPM• 4.2ms latency• 100MB/Sec transfer speed• 16MB cache• Technology demonstrations:

– 1TB drive due in next couple of months: 164 GB/in2

– Last year, Seagate demonstrated it could do 421 GB/in2!

4/18/2007 cs252-S07, Lecture 22 19

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

OuterTrack

InnerTrackSector

Actuator

HeadArm

4/18/2007 cs252-S07, Lecture 22 20

Disk Time Example• Disk Parameters:

– Transfer size is 8K bytes– Advertised average seek is 12 ms– Disk spins at 7200 RPM– Transfer rate is 4 MB/sec

• Controller overhead is 2 ms• Assume that disk is idle so no queuing delay• Disk Latency =

Queuing Time + Seek Time + Rotation Time + Xfer Time + Ctrl Time

• What is Average Disk Access Time for a Sector?– Ave seek + ave rot delay + transfer time + controller overhead– 12 ms + 0.5/(7200 RPM/60) + 8 KB/4 MB/s + 2 ms– 12 + 4.15 + 2 + 2 = 20 ms

• Advertised seek time assumes no locality: typically 1/4 to 1/3 advertised seek time: 20 ms => 12 ms

4/18/2007 cs252-S07, Lecture 22 21

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: average number of arriving customers/second = 10

Tser average time to service a customer = 20 ms (0.02s)u server utilization (0..1): u = 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= x Tq

= 10/s x .005s = 0.05 requests in queueLsys average # tasks in system: Lsys = x Tsys = 10/s x .025s = 0.25

4/18/2007 cs252-S07, Lecture 22 22

Alternative Data Storage Technologies: Early 1990s

Cap BPI TPI BPI*TPI Data Xfer Access

Technology (MB) (Million) (KByte/s) Time

Conventional Tape:

Cartridge (.25") 150 12000 104 1.2 92 minutes

IBM 3490 (.5") 800 22860 38 0.9 3000 seconds

Helical Scan Tape:

Video (8mm) 4600 43200 1638 71 492 45 secs

DAT (4mm) 1300 61000 1870 114 183 20 secs

Magnetic & Optical Disk:

Hard Disk (5.25") 1200 33528 1880 63 3000 18 ms

IBM 3390 (10.5") 3800 27940 2235 62 4250 20 ms

Sony MO (5.25") 640 24130 18796 454 88 100 ms

4/18/2007 cs252-S07, Lecture 22 23

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)

• New technology trend: Helical Scan (VCR, Camcoder, DAT) Spins head at angle to tape to improve density

4/18/2007 cs252-S07, Lecture 22 24

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

• Long rewind, eject, load, spin-up times; not inherent, just no need in marketplace (so far)

• Designed for archival

4/18/2007 cs252-S07, Lecture 22 25

Future Disk Size and Performance

• Continued advance in capacity (60%/yr) and bandwidth (40%/yr)

• Slow improvement in seek, rotation (8%/yr)• Time to read whole disk

Year Sequentially Randomly (1 sector/seek)

1990 4 minutes 6 hours2000 12 minutes 1 week(!)2006 56 minutes 3 weeks (SCSI)2006 171 minutes 7 weeks (SATA)

4/18/2007 cs252-S07, Lecture 22 26

Use Arrays of Small Disks?

14”10”5.25”3.5”

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?

4/18/2007 cs252-S07, Lecture 22 27

Replace Small # of Large Disks with Large # of Small Disks! (1988 Disks)

Data Capacity

Volume

Power

Data Rate

I/O Rate

MTTF

Cost

IBM 3390 (K)

20 GBytes

97 cu. ft.

3 KW

15 MB/s

600 I/Os/s

250 KHrs

$250K

IBM 3.5" 0061

320 MBytes

0.1 cu. ft.

11 W

1.5 MB/s

55 I/Os/s

50 KHrs

$2K

x70

23 GBytes

11 cu. ft.

1 KW

120 MB/s

3900 IOs/s

??? Hrs

$150K

Disk Arrays have potential for

large data and I/O rates

high MB per cu. ft., high MB per KW

reliability?

4/18/2007 cs252-S07, Lecture 22 28

Advantages of Small Formfactor Disk Drives

Low cost/MBHigh MB/volumeHigh MB/wattLow cost/Actuator

Cost and Environmental Efficiencies

4/18/2007 cs252-S07, Lecture 22 29

Array Reliability

• Reliability of N disks = Reliability of 1 Disk ÷ N

50,000 Hours ÷ 70 disks = 700 hours

Disk system MTTF: Drops from 6 years to 1 month!

• Arrays (without redundancy) too unreliable to be useful!

Hot spares support reconstruction in parallel with access: very high media availability can be achievedHot spares support reconstruction in parallel with access: very high media availability can be achieved

4/18/2007 cs252-S07, Lecture 22 30

Redundant Arrays of Disks

• Files are "striped" across multiple spindles• Redundancy yields high data availability

Disks will fail

Contents reconstructed from data redundantly stored in the array

Capacity penalty to store it

Bandwidth penalty to update

Mirroring/Shadowing (high capacity cost)

Horizontal Hamming Codes (overkill)

Parity & Reed-Solomon Codes

Failure Prediction (no capacity overhead!)VaxSimPlus — Technique is controversial

Techniques:

4/18/2007 cs252-S07, Lecture 22 31

Redundant Arrays of DisksRAID 1: Disk Mirroring/Shadowing

• Each disk is fully duplicated onto its "shadow" 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

Targeted for high I/O rate , high availability environments

recoverygroup

4/18/2007 cs252-S07, Lecture 22 32

Redundant Arrays of Disks RAID 5+: High I/O Rate Parity

A logical writebecomes fourphysical I/Os

Independent writespossible because ofinterleaved parity

Reed-SolomonCodes ("Q") forprotection duringreconstruction

A logical writebecomes fourphysical I/Os

Independent writespossible because ofinterleaved parity

Reed-SolomonCodes ("Q") forprotection duringreconstruction

D0 D1 D2 D3 P

D4 D5 D6 P D7

D8 D9 P D10 D11

D12 P D13 D14 D15

P D16 D17 D18 D19

D20 D21 D22 D23 P

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.Disk Columns

IncreasingLogical

Disk Addresses

Stripe

StripeUnit

Targeted for mixedapplications

4/18/2007 cs252-S07, Lecture 22 33

Problems of Disk Arrays: Small Writes

D0 D1 D2 D3 PD0'

+

+

D0' D1 D2 D3 P'

newdata

olddata

old parity

XOR

XOR

(1. Read) (2. Read)

(3. Write) (4. Write)

RAID-5: Small Write Algorithm

1 Logical Write = 2 Physical Reads + 2 Physical Writes

4/18/2007 cs252-S07, Lecture 22 34

Subsystem Organization

hostarray

controller

single boarddisk

controller

single boarddisk

controller

single boarddisk

controller

single boarddisk

controller

hostadapter

manages interfaceto host, DMA

control, buffering,parity logic

physical devicecontrol

often piggy-backedin small format devices

striping software off-loaded from host to array controller

no applications modifications

no reduction of host performance

4/18/2007 cs252-S07, Lecture 22 35

System Availability: Orthogonal RAIDs

ArrayController

StringController

StringController

StringController

StringController

StringController

StringController

. . .

. . .

. . .

. . .

. . .

. . .

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

4/18/2007 cs252-S07, Lecture 22 36

System-Level Availability

Fully dual redundantI/O Controller I/O Controller

Array Controller Array Controller

. . .

. . .

. . .

. . . . . .

.

.

.RecoveryGroup

Goal: No SinglePoints ofFailure

Goal: No SinglePoints ofFailure

host host

with duplicated paths, higher performance can beobtained when there are no failures

OceanStore:Global Scale Persistent Storage

Global-Scale Persistent Storage

4/18/2007 cs252-S07, Lecture 22 38

Pac Bell

Sprint

IBMAT&T

CanadianOceanStore

• Service provided by confederation of companies– Monthly fee paid to one service provider– Companies buy and sell capacity from each other

IBM

Utility-based Infrastructure

4/18/2007 cs252-S07, Lecture 22 39

Important P2P Technology(Decentralized Object Location and Routing)

GUID1

DOLR

GUID1GUID2

4/18/2007 cs252-S07, Lecture 22 40

Peer-to-peer systems can be very stable

(May 2003: 1.5 TB over 4 hours)

In JSAC, To appear

4/18/2007 cs252-S07, Lecture 22 41

The Path of an OceanStore UpdateSecond-Tier

Caches

Multicasttrees

Inner-RingServers

Clients

4/18/2007 cs252-S07, Lecture 22 42

Archival Disseminationof Fragments

4/18/2007 cs252-S07, Lecture 22 43

Aside: Why erasure coding?High Durability/overhead ratio!

• Exploit law of large numbers for durability!• 6 month repair, FBLPY:

– Replication: 0.03– Fragmentation: 10-35

Fraction Blocks Lost Per Year (FBLPY)

4/18/2007 cs252-S07, Lecture 22 44

• OceanStore Concepts Applied to Tape-less backup– Self-Replicating, Self-Repairing, Self-Managing

– No need for actual Tape in system » (Although could be there to keep with tradition)

The Berkeley PetaByte Archival Service

4/18/2007 cs252-S07, Lecture 22 45

Summary• Disk industry growing rapidly, improves:

– bandwidth 40%/yr ,

– areal density 60%/year, $/MB faster?

• queue + controller + seek + rotate + transfer

• Advertised average seek time benchmark much greater than average seek time in practice

• Redundancy useful to gain reliability– Redundant disks+controllers+etc (RAID)

– Geographical scale systems (OceanStore)

• Queueing theory: for (c=1):

u

uxCW

1

121

uux

W1