CS61C:GreatIdeasinComputerArchitecture

PerformanceIronLaw,Amdahl’sLaw

Instructors:NicholasWeaver&VladimirStojanovichttp://inst.eecs.berkeley.edu/~cs61c/

New-SchoolMachineStructures(It’sabitmorecomplicated!)

• ParallelRequestsAssigned tocomputere.g.,Search“Katz”

• ParallelThreadsAssigned tocoree.g.,Lookup,Ads

• ParallelInstructions>1instruction@onetimee.g.,5pipelined instructions

• ParallelData>1dataitem@one timee.g.,Addof4pairsofwords

• HardwaredescriptionsAllgates@onetime

• ProgrammingLanguages2

SmartPhone

WarehouseScale

Computer

SoftwareHardware

HarnessParallelism&AchieveHighPerformance

LogicGates

Core Core…

Memory(Cache)

Input/Output

Computer

CacheMemory

Core

InstructionUnit(s) FunctionalUnit(s)

A3+B3A2+B2A1+B1A0+B0

Howdoweknow?

WhatisPerformance?

• Latency(orresponsetimeorexecutiontime)– Timetocompleteonetask

• Bandwidth(orthroughput)– Taskscompletedperunittime• Ifyouhavesufficientindependenttasks,youcanalwaysthrowmoremoneyattheproblem:Throughput/$oftenamoreimportantmetricthanjustthroughput

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CloudPerformance:WhyApplicationLatencyMatters

• Keyfigureofmerit:applicationresponsiveness– Longerthedelay,thefewertheuserclicks,thelesstheuserhappiness,andthelowertherevenueperuser

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DefiningCPUPerformance• WhatdoesitmeantosayXisfasterthanY?

• Ferrarivs.SchoolBus?• 2013Ferrari599GTB– 2passengers,quartermilein10secs

• 2013TypeDschoolbus– 50passengers,quartermilein20secs

• ResponseTime (Latency):e.g.,timetotravel¼mile• Throughput (Bandwidth): e.g.,passenger-miin1hour

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DefiningRelativeCPUPerformance• PerformanceX =1/ProgramExecutionTimeX• PerformanceX >PerformanceY =>1/ExecutionTimeX>1/ExecutionTimey=>ExecutionTimeY >ExecutionTimeX

• ComputerXisNtimesfasterthanComputerYPerformanceX /PerformanceY =NorExecutionTimeY /ExecutionTimeX=N

• BustoFerrariperformance:– Program:Transfer1000passengersfor1mile– Bus:3,200sec,Ferrari:40,000sec

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MeasuringCPUPerformance

• Computersuseaclocktodeterminewheneventstakesplacewithinhardware

• Clockcycles: discretetimeintervals– akaclocks,cycles,clockperiods,clockticks

• Clockrateorclockfrequency: clockcyclespersecond(inverseofclockcycletime)

• 3GigaHertz clockrate=>clockcycletime=1/(3x109)seconds

clockcycletime=333picoseconds(ps)

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CPUPerformanceFactors

• TodistinguishbetweenprocessortimeandI/O,CPUtimeistimespentinprocessor

• CPU Time/Program= Clock Cycles/Program

x Clock Cycle Time

• OrCPU Time/Program= Clock Cycles/Program ÷ Clock Rate

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IronLawofPerformancebyEmer andClark

• A programexecutesinstructions• CPU Time/Program

= Clock Cycles/Program x Clock Cycle Time= Instructions/Program

x Average Clock Cycles/Instruction x Clock Cycle Time

• 1st termcalledInstructionCount• 2nd termabbreviatedCPIforaverageClockCyclesPerInstruction

• 3rdtermis1/Clockrate

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RestatingPerformanceEquation• Time= Seconds

ProgramInstructions Clockcycles SecondsProgram Instruction ClockCycle

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WhatAffectsEachComponent?A)InstructionCount,B)CPI,C)ClockRate

AffectsWhat?(click inletterofcomponentnotaffected)

Algorithm

ProgrammingLanguageCompiler

InstructionSet Architecture11

WhatAffectsEachComponent?InstructionCount,CPI,ClockRate

AffectsWhat?Algorithm InstructionCount,

CPIProgrammingLanguage

InstructionCount,CPI

Compiler InstructionCount,CPI

InstructionSetArchitecture

InstructionCount,ClockRate,CPI

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Clickers

• Whichcomputerhasthehighestperformanceforagivenprogram?

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Computer Clockfrequency

Clock cyclesperinstruction

#instructionsperprogram

A 1GHz 2 1000

B 2GHz 5 800

C 500MHz 1.25 400

D 5GHz 10 2000

WorkloadandBenchmark

• Workload: Setofprogramsrunonacomputer– Actualcollectionofapplicationsrunormadefromrealprogramstoapproximatesuchamix

– Specifiesprograms,inputs,andrelativefrequencies• Benchmark:Programselectedforuseincomparingcomputerperformance– Benchmarksformaworkload– Usuallystandardizedsothatmanyusethem

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SPEC(SystemPerformanceEvaluationCooperative)• ComputerVendorcooperativeforbenchmarks,startedin1989

• SPECCPU2006– 12IntegerPrograms– 17Floating-PointPrograms

• Oftenturnintonumberwherebiggerisfaster• SPECratio:referenceexecutiontimeonoldreferencecomputerdividebyexecutiontimeonnewcomputertogetaneffectivespeed-up

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SPECINT2006onAMDBarcelonaDescription

Instruc-tion

Count (B)CPI

Clock cycle

time (ps)

Execu-tion

Time (s)

Refer-ence

Time (s)

SPEC-ratio

Interpreted string processing 2,118 0.75 400 637 9,770 15.3Block-sorting compression 2,389 0.85 400 817 9,650 11.8GNU C compiler 1,050 1.72 400 724 8,050 11.1Combinatorial optimization 336 10.0 400 1,345 9,120 6.8Go game 1,658 1.09 400 721 10,490 14.6Search gene sequence 2,783 0.80 400 890 9,330 10.5Chess game 2,176 0.96 400 837 12,100 14.5Quantum computer simulation 1,623 1.61 400 1,047 20,720 19.8Video compression 3,102 0.80 400 993 22,130 22.3Discrete event simulation library 587 2.94 400 690 6,250 9.1Games/path finding 1,082 1.79 400 773 7,020 9.1XML parsing 1,058 2.70 400 1,143 6,900 6.016

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SummarizingPerformance…

Clickers:Whichsystemisfaster?

System Rate(Task1) Rate(Task2)

A 10 20

B 20 10

A:SystemAB:SystemBC:SameperformanceD:Unanswerablequestion!

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… DependsWho’sSellingSystem Rate(Task1) Rate(Task2)

A 10 20

B 20 10

Average

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15Averagethroughput

System Rate(Task1) Rate(Task2)

A 0.50 2.00

B 1.00 1.00

Average

1.25

1.00ThroughputrelativetoB

System Rate(Task1) Rate(Task2)

A 1.00 1.00

B 2.00 0.50

Average

1.00

1.25ThroughputrelativetoA

SummarizingSPECPerformance

• Variesfrom6xto22xfasterthanreferencecomputer

• Geometricmeanofratios:N-th rootofproductofNratios– GeometricMeangivessamerelativeanswernomatterwhatcomputerisusedasreference

• GeometricMeanforBarcelonais11.7

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Administrivia

• Midterm2intheeveningnextMonday• Project2.1gradesin

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BigIdea:Amdahl’s(Heartbreaking)Law• SpeedupduetoenhancementEis

Speedupw/E= ----------------------Exectimew/oEExectimew/E

• SupposethatenhancementEacceleratesafractionF(F<1)ofthetaskbyafactorS(S>1)andtheremainderofthetaskisunaffected

ExecutionTimew/E=

Speedupw/E=21

ExecutionTimew/oE´[(1-F)+F/S]

1/[(1-F)+F/S]

BigIdea:Amdahl’sLaw

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Speedup=1(1- F)+F

SNon-speed-uppart Speed-uppart

10.5+0.5

2

10.5+0.25

= = 1.33

Example:theexecutiontimeofhalfoftheprogramcanbeacceleratedbyafactorof2.Whatistheprogramspeed-upoverall?

Example#1:Amdahl’sLaw

• Consideranenhancementwhichruns20timesfasterbutwhichisonlyusable25%ofthetime

Speedupw/E=1/(.75+.25/20)=1.31

• Whatifitsusableonly15%ofthetime?Speedupw/E=1/(.85+.15/20)=1.17

• Amdahl’sLawtellsusthattoachievelinearspeedupwith100processors,noneoftheoriginalcomputationcanbescalar!

• Togetaspeedupof90from100processors,thepercentageoftheoriginalprogramthatcouldbescalarwouldhavetobe0.1%orless

Speedupw/E=1/(.001+.999/100)=90.9923

Speedupw/E=1/ [(1-F)+F/S]

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Iftheportionoftheprogramthatcanbeparallelizedissmall,thenthespeedupislimited

Thenon-parallelportionlimitstheperformance

StrongandWeakScaling• Togetgoodspeeduponaparallelprocessorwhilekeepingtheproblemsizefixedisharderthangettinggoodspeedupbyincreasingthesizeoftheproblem.– Strongscaling:whenspeedupcanbeachievedonaparallelprocessorwithoutincreasingthesizeoftheproblem

– Weakscaling:whenspeedupisachievedonaparallelprocessorbyincreasingthesizeoftheproblemproportionallytotheincreaseinthenumberofprocessors

• Loadbalancingisanotherimportantfactor:everyprocessordoingsameamountofwork– Justoneunitwithtwicetheloadofotherscutsspeedupalmostinhalf

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Clickers/PeerInstruction

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Supposeaprogramspends80%ofitstimeinasquarerootroutine.Howmuchmustyouspeedupsquareroottomaketheprogramrun5timesfaster?

A:5B:16C:20D:100E:Noneoftheabove

Speedupw/E=1/ [(1-F)+F/S]

AndInConclusion,…

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• Time(seconds/program)ismeasureofperformanceInstructions Clockcycles SecondsProgram Instruction ClockCycle

• Floating-pointrepresentationsholdapproximationsofrealnumbersinafinitenumberofbits

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