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COMPUTER ARCHITECTURE

Prof. Basel MahafzahLesson 2 - The role of performance - Part 2

Copyright Universit Telematica Internazionale UNINETTUNO

The Role ofPerformanceThe Role of

PerformancePart 2Part 2

OutlineOutline

Clock Cycles perInstruction (CPI)Clock Cycles perInstruction (CPI)

Definition of PerformanceDefinition of Performance

Instruction CountInstruction Count

Cycles and InstructionsCycles and Instructions

Since the compiler clearlygenerated instructions toexecute, and the computerhad to execute the instructionsto run the program

Since the compiler clearlygenerated instructions toexecute, and the computerhad to execute the instructionsto run the program

The execution time mustdepend on the numberof instructions in a program

The execution time mustdepend on the numberof instructions in a program

Cycles and InstructionsCycles and Instructions

How many cyclesare required for

a program?

How many cyclesare required for

a program?

Could we assume that # ofcycles = # of instructionsCould we assume that # ofcycles = # of instructions

1 s

t i n s

t r u c t

i o n

1 s

t i n s

t r u c

t i o n

2 n

d i

n s

t r u c

t i o n

2 n

d i

n s

t r u c

t i o n

3 r

d i

n s

t r u c

t i o n

3 r

d i

n s

t r u c

t i o n

4 t h 4 t h

5 t h 5 t h

6 t h 6 t h

. . . . . .

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If the two computers have thesame ISA which of our quantities(e.g., clock rate, CPI, executiontime, # of instructions, MIPS)will always be identical?

If the two computers have thesame ISA which of our quantities( e.g. , clock rate , CPI , executiontime , # of instructions , MIPS )will always be identical?

Answer Answer

lets call this number Ilets call this number I

We know that each computerexecutes the same numberof instructions for the sameprogram

We know that each computerexecutes the same numberof instructions for the sameprogram

First find the number ofprocessor clock cycles for eachcomputer:CPU clock cycles A = I x 2.0CPU clock cycles B = I x 1.2

First find the number ofprocessor clock cycles for eachcomputer:CPU clock cycles A = I x 2.0CPU clock cycles B = I x 1.2

Now we can compute the CPUtime for each computer:Now we can compute the CPUtime for each computer:

CPU time A = CPU clock cycles A x Clock cycle time A= I x 2.0 x 1 ns = 2 x I nsCPU time B = I x 1.2 x 2 ns

= 2.4 x I ns

CPU time A = CPU clock cycles A x Clock cycle time A= I x 2.0 x 1 ns = 2 x I nsCPU time B = I x 1.2 x 2 ns

= 2.4 x I ns

Clearly, computer A is faster.The amount faster is given bythe ratio of the execution times:

Clearly, computer A is faster .The amount faster is given bythe ratio of the execution times:

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CPU Performance A / CPUPerformance B =Execution Time B / Execution Time A= 2.4 x I ns / 2 x I ns = 1.2

CPU Performance A / CPUPerformance B =Execution Time B / Execution Time A= 2.4 x I ns / 2 x I ns = 1.2

Computer A is 1.2 timesfaster than Computer B forthis program

Computer A is 1.2 timesfaster than Computer B forthis program

Final AnswerFinal Answer

Definitionof Performance

(Continues)

Definitionof Performance

(Continues)Instruction CountInstruction Count

Basic performance equationin terms of instruction count(the number of instructionsexecuted by the program), CPI,and clock cycle time:

Basic performance equationin terms of instruction count(the number of instructionsexecuted by the program) , CPI,and clock cycle time: Inst.Count CPI

CR Inst.Count CPI

CR

CC =Clock CyclesCR =Clock Rate

CC =Clock CyclesCR =Clock Rate

Inst.Count CPI CC TimeInst.Count CPI CC Time

Or CPU time =Or CPU time =

CPU time =CPU time =

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Execution Time =Execution Time =

I C C S P I C CI C C S P I C C

Where I=InstructionsP= ProgramCC= Clock cyclesS= Seconds

Where I =InstructionsP= ProgramCC= Clock cyclesS= Seconds

CPU clock cycles =CPU clock cycles =

n

i ii=1

CPI C n

i ii=1

CPI C

How to find the performanceparameters (such as CC)?How to find the performanceparameters (such as CC) ?

CPI i is the average number ofcycles per instruction for thatinstruction class, and n is thenumber of instruction classes

CPI i is the average number ofcycles per instruction for thatinstruction class, and n is thenumber of instruction classes

Where C i is the count of thenumber of instructions of class iexecuted

Where Ci is the count of thenumber of instructions of class iexecuted

A compiler designer is tryingto decide between two codesequences for a particularmachine

A compiler designer is tryingto decide between two codesequences for a particularmachine

ExampleExample

Example ContinuesExample Continues

Based on the hardwareimplementation, there are threedifferent classes of instructions:Class A, Class B, and Class C, andthey require one, two, and threecycles (respectively)

Based on the hardwareimplementation, there are three

different classes of instructions:Class A , Class B , and Class C , andthey require one , two , and threecycles (respectively)

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The first code sequence has 5instructions: 2 of A, 1 of B,and 2 of C

The first code sequence has 5instructions: 2 of A, 1 of B,and 2 of C

The second sequence has 6instructions: 4 of A, 1 of B,and 1 of C

The second sequence has 6instructions: 4 of A, 1 of B,and 1 of C

Which sequence willbe faster?Which sequence willbe faster?

What is the CPI for eachsequence?What is the CPI for eachsequence?

Answer AnswerSequence 1 executes2 + 1 + 2 = 5 instructionsSequence 1 executes2 + 1 + 2 = 5 instructions

Sequence 2 executes4 + 1 + 1 = 6 instructionsSequence 2 executes4 + 1 + 1 = 6 instructions

So, sequence 1 executesfewer instructionsSo, sequence 1 executesfewer instructions

Answer Continues Answer Continues

CPU clock cycles =CPU clock cycles =

To find the total number ofclock cycles for each sequence:To find the total number ofclock cycles for each sequence:

n

i ii=1

CPI C n

i ii=1

CPI CCPU clock cycles 2 = (4x1) +(1x2) + (1x3) = 9 cyclesCPU clock cycles 2 = (4x1) +(1x2) + (1x3) = 9 cycles

CPU clock cycles 1 = (2x1) +(1x2) + (2x3) = 10 cyclesCPU clock cycles 1 = (2x1) +(1x2) + (2x3) = 10 cycles

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So code sequence 2 is faster,even though it actuallyexecutes one extra instruction

So code sequence 2 is faster ,even though it actuallyexecutes one extra instruction

Since code sequence 2 takesfewer overall clock cycles buthas more instructions, it musthave a lower CPI

Since code sequence 2 takesfewer overall clock cycles buthas more instructions, it musthave a lower CPI

The CPI values can becomputed by:The CPI values can becomputed by:

CPI =CPI = CPU CCInst. Count

CPU CCInst. Count

CPI 1 =CPU clock cycles 1 = 10 = 2Instruction count 1 5

CPI 1 =CPU clock cycles 1 = 10 = 2Instruction count 1 5

CPI 2 =CPU clock cycles 2 = 9 = 1.5Instruction count 2 6

CPI 2 =CPU clock cycles 2 = 9 = 1.5Instruction count 2 6

In SummaryIn SummaryNote: When comparing twomachines, you must look at allthree components (instructioncount, CPI, and clock cycle time),which combine to formexecution time

Note: When comparing two

machines, you must look at allthree components (instructioncount, CPI, and clock cycle time) ,which combine to formexecution time

Now that weunderstand cyclesNow that weunderstand cycles

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some number of secondssome number of seconds

A given program will require A given program will require

some number of instructions(machine instructions)some number of instructions(machine instructions)

some number of cyclessome number of cycles

We have a vocabularythat relates these quantities:We have a vocabularythat relates these quantities:

Cycle Time (seconds per cycle)Cycle Time (seconds per cycle)

Inst. Count (inst. per program)Inst. Count (inst. per program)

Clock Rate (cycles per second)Clock Rate (cycles per second)

CPI (cycles per instruction)CPI (cycles per instruction)

A floating point intensiveapplication might havea higher CPI

A floating point intensiveapplication might havea higher CPI

PerformancePerformance

Performance is determinedby execution timePerformance is determinedby execution time

Do any of the other variablesequal performance?Do any of the other variablesequal performance?

number of cycles to executeprogram?number of cycles to executeprogram?

number of instructionsin program?number of instructionsin program?

number of cycles per second?number of cycles per second?

average number of cyclesper instruction?average number of cyclesper instruction?

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average number of instructionsper second?average number of instructionsper second?

Common pitfall: thinking oneof the variables is indicative ofperformance when it really isnt

Common pitfall: thinking oneof the variables is indicative ofperformance when it really isnt

BenchmarksBenchmarks

Performance best determinedby running a real applicationPerformance best determinedby running a real application

Use programs typical ofexpected workloadUse programs typical ofexpected workload

Or, typical of expected classof applicationse.g., compilers/editors,scientific applications, graphics,etc.

Or, typical of expected classof applicationse.g., compilers/editors,scientific applications, graphics,etc.

Small benchmarksSmall benchmarks

nice for architectsand designersnice for architectsand designers

easy to standardizeeasy to standardize

can be abusedcan be abused

SPEC (System PerformanceEvaluation Cooperative)SPEC (System PerformanceEvaluation Cooperative)

companies have agreed on a setof real program and inputscompanies have agreed on a setof real program and inputs

can still be abusedcan still be abused

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Benchmarks are valuableindicator of performance(and compiler technology)

Benchmarks are valuableindicator of performance(and compiler technology)

SPEC 89SPEC 89800800

700700

600600

500500

400400

300300

200200

100100

00

Compiler enhancements and performance

Compiler enhancements and performance

gccgcc espressoespresso spicespice doducdoduc nasa7nasa7 lili eqntotteqntott matrix300matrix300

fppppfpppp tomcatvtomcatv

Benchmark Benchmark compilercompiler Enhanced compilerEnhanced compiler

Benchmark Description

go Artificial intelligence; plays the game of Gom88ksim Motorola 88k chip simulator; runs test programgcc The Gnu C compiler generating SPARC codecompress Compresses and decompresses file in memoryli Lisp interpreter ijpeg Graphic compression and decompressionperl Manipulates strings and prime numbers in the special-purpose programming language Perlvortex A database program

Benchmark Description

go Artificial intelligence; plays the game of Gom88ksim Motorola 88k chip simulator; runs test programgcc The Gnu C compiler generating SPARC codecompress Compresses and decompresses file in memoryli Lisp interpreter ijpeg Graphic compression and decompressionperl Manipulates strings and prime numbers in the special-purpose programming language Perlvortex A database program

SPEC 95 (8 integer benchmarks)SPEC 95 (8 integer benchmarks)

tomcatv A mesh generation programswim Shallow water model with 513 x 513 gridsu2cor quantum physics; Monte Carlo simulationhydro2d Astrophysics; Hydrodynamic Naiver Stokes equationsmgrid Multigrid solver in 3-D potential fieldapplu Parabolic/elliptic partial differential equationstrub3d Simulates isotropic, homogeneous turbulence in a cubeapsi Solves problems regarding temperature, wind velocity, and distribution of pollutantfppp p Q uant um c hem is trywave5 Plasma physics; electromagnetic particle simulation

tomcatv A mesh generation programswim Shallow water model with 513 x 513 gridsu2cor quantum physics; Monte Carlo simulationhydro2d Astrophysics; Hydrodynamic Naiver Stokes equationsmgrid Multigrid solver in 3-D potential fieldapplu Parabolic/elliptic partial differential equationstrub3d Simulates isotropic, homogeneous turbulence in a cubeapsi Solves problems regarding temperature, wind velocity, and distribution of pollutantfpppp Quant um c hem is tr ywave5 Plasma physics; electromagnetic particle simulation

SPEC 95 (10 floating-pointbenchmarks)

SPEC 95 (10 floating-pointbenchmarks)

Benchmark DescriptionBenchmark Description

SPECint 95SPECint 95

SPEC ratio: bigger numeric resultsindicate Faster performanceSPEC ratio: bigger numeric resultsindicate Faster performance

Pentium Prois 1.4 to 1.5times fasterthan Pentium


1

Clock rate (MHz)Clock rate (MHz)5050 100100 150150 200200 250250

1-1-22

00

3-3-445-5-667-7-889-9-

1010

SPECfp 95SPECfp 95


Pentium Prois 1.7 to 1.8times fasterthan Pentium 5050 100100 150150 200200 250250

Clock rate (MHz)Clock rate (MHz)

11-1-22

00

3-3-445-5-667-7-889-9-

1010

SPEC ratio: bigger numeric resultsindicate Faster performanceSPEC ratio: bigger numeric resultsindicate Faster performance

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