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CIS775: Computer Architecture

Chapter 1: Fundamentals of Computer Design

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Course Objectives

• To evaluate the issues involved in choosing and designing instruction set.

• To learn concepts behind advanced pipelining techniques.

• To understand the “hitting the memory wall” problem and the current state-of-art in memory system design.

• To understand the qualitative and quantitative tradeoffs in the design of modern computer systems

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What is Computer Architecture?

• Functional operation of the individual HW units within a computer system, and the flow of information and control among them.

TechnologyProgrammingLanguageInterface

Interface Design(ISA)

Measurement & Evaluation

Parallelism

Computer Architecture:

Applications OS

Hardware Organization

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Computer Architecture Topics

Instruction Set Architecture

Pipelining, Hazard Resolution,Superscalar, Reordering, Prediction, Speculation,Vector, DSP

Addressing,Protection,Exception Handling

L1 Cache

L2 Cache

DRAM

Disks, WORM, Tape

Coherence,Bandwidth,Latency

Emerging TechnologiesInterleaving Memories

RAID

VLSI

Input/Output and Storage

MemoryHierarchy

Pipelining and Instruction Level Parallelism

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Computer Architecture Topics

M

Interconnection NetworkS

PMPMPMP° ° °

Topologies,Routing,Bandwidth,Latency,Reliability

Network Interfaces

Shared Memory,Message Passing,Data Parallelism

Processor-Memory-Switch

MultiprocessorsNetworks and Interconnections

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Measurement and Evaluation

Design

Analysis

Architecture is an iterative process:• Searching the space of possible designs• At all levels of computer systems

Creativity

Good IdeasGood Ideas

Mediocre IdeasBad Ideas

Cost /PerformanceAnalysis

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Issues for a Computer Designer• Functional Requirements Analysis (Target)

– Scientific Computing – HiPerf floating pt.– Business – transactional support/decimal arith.– General Purpose –balanced performance for a range of tasks

• Level of software compatibility– PL level

• Flexible, Need new compiler, portability an issue

– Binary level (x86 architecture)• Little flexibility, Portability requirements minimal

• OS requirements– Address space issues, memory management, protection

• Conformance to Standards– Languages, OS, Networks, I/O, IEEE floating pt.

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Computer Systems: Technology Trends

• 1988– Supercomputers

– Massively Parallel Processors

– Mini-supercomputers

– Minicomputers

– Workstations

– PC’s

• 2002– Powerful PC’s and

SMP Workstations

– Network of SMP Workstations

– Mainframes

– Supercomputers

– Embedded Computers

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Why Such Change in 10 years?• Performance

– Technology Advances• CMOS (complementary metal oxide semiconductor) VLSI dominates older

technologies like TTL (transistor transistor logic) in cost AND performance

– Computer architecture advances improves low-end • RISC, pipelining, superscalar, RAID, …

• Price: Lower costs due to …– Simpler development

• CMOS VLSI: smaller systems, fewer components

– Higher volumes– Lower margins by class of computer, due to fewer services

• Function :Rise of networking/local interconnection technology

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Growth in Microprocessor Performance

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Six Generations of DRAMs

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Updated Technology Trends(Summary)

Capacity Speed (latency)

Logic 4x in 4 years 2x in 3 years

DRAM 4x in 3 years 2x in 10 years

Disk 4x in 2 years 2x in 10 years

Network (bandwidth) 10x in 5 years

• Updates during your study period??

BS (4 yrs)

MS (2 yrs)

PhD (5 yrs)

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Cost of Microprocessors

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DAP.S98 1

IC cost = Die cost + Testing cost + Packaging cost

Final test yield

Die cost = Wafer cost

Dies per Wafer * Die yield

Dies per wafer = š * ( Wafer_diam / 2)2 – š * Wafer_diam – Test dies

Die Area ¦ 2 * Die Area

Die Yield = Wafer yield * 1 +Defects_per_unit_area * Die_Area

Integrated Circuits Costs

Die Cost goes roughly with die area4

{

}

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Performance Trends(Summary)

• Workstation performance (measured in Spec Marks) improves roughly 50% per year (2X every 18 months)

• Improvement in cost performance estimated at 70% per year

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Computer Engineering Methodology

Evaluate ExistingEvaluate ExistingSystems for Systems for BottlenecksBottlenecks

Simulate NewSimulate NewDesigns andDesigns and

OrganizationsOrganizations

Implement NextImplement NextGeneration SystemGeneration System

TechnologyTrends

Benchmarks

Workloads

ImplementationComplexity

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How to Quantify Performance?

• Time to run the task (ExTime)– Execution time, response time, latency

• Tasks per day, hour, week, sec, ns … (Performance)– Throughput, bandwidth

Plane

Boeing 747

BAD/Sud Concodre

Speed

610 mph

1350 mph

DC to Paris

6.5 hours

3 hours

Passengers

470

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Throughput (pmph)

286,700

178,200

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The Bottom Line: Performance and Cost or Cost

and Performance?"X is n times faster than Y" means

ExTime(Y) Performance(X)

--------- = ---------------ExTime(X) Performance(Y)

• Speed of Concorde vs. Boeing 747

• Throughput of Boeing 747 vs. Concorde

• Cost is also an important parameter in the equation which is why concordes are being put to pasture!

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Measurement Tools

• Benchmarks, Traces, Mixes• Hardware: Cost, delay, area, power estimation• Simulation (many levels)

– ISA, RT, Gate, Circuit

• Queuing Theory• Rules of Thumb• Fundamental “Laws”/Principles• Understanding the limitations of any

measurement tool is crucial.

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Metrics of Performance

Compiler

Programming Language

Application

DatapathControl

Transistors Wires Pins

ISA

Function Units

(millions) of Instructions per second: MIPS(millions) of (FP) operations per second: MFLOP/s

Cycles per second (clock rate)

Megabytes per second

Answers per monthOperations per second

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Cases of Benchmark Engineering• The motivation is to tune the system to the benchmark to achieve peak

performance.• At the architecture level

– Specialized instructions

• At the compiler level (compiler flags)– Blocking in Spec89 factor of 9 speedup– Incorrect compiler optimizations/reordering.– Would work fine on benchmark but not on other programs

• I/O level– Spec92 spreadsheet program (sp)– Companies noticed that the produced output was always out put to a file (so they stored

the results in a memory buffer) and then expunged at the end (which was not measured).– One company eliminated the I/O all together.

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After putting in a blazing performance on the benchmark test, Sun issued a glowing press release claiming that it hadoutperformed Windows NT systems on the test. Pendragon president Ivan Phillips cried foul, saying the resultsweren't representative of real-world Java performance and that Sun had gone so far as to duplicate the test's code within Sun'sJust-In-Time compiler. That's cheating, says Phillips, who claims that benchmark tests and real-world applications aren'tthe same thing.

Did Sun issue a denial or a mea culpa? Initially, Sun neither denied optimizing for the benchmark test nor apologized forit. "If the test results are not representative of real-world Java applications, then that's a problem with the benchmark,"Sun's Brian Croll said.

After taking a beating in the press, though, Sun retreated and issued an apology for the optimization.[Excerpted from PC Online 1997]

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Issues with Benchmark Engineering

• Motivated by the bottom dollar, good performance on classic suites more customers, better sales.

• Benchmark Engineering Limits the longevity of benchmark suites

• Technology and Applications Limits the longevity of benchmark suites.

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SPEC: System Performance Evaluation Cooperative

• First Round 1989– 10 programs yielding a single number (“SPECmarks”)

• Second Round 1992– SPECInt92 (6 integer programs) and SPECfp92 (14 floating point

programs)• Compiler Flags unlimited. March 93 • new set of programs: SPECint95 (8 integer programs) and SPECfp95 (10

floating point)

– “benchmarks useful for 3 years”– Single flag setting for all programs: SPECint_base95,

SPECfp_base95 – SPEC CPU2000 (11 integer benchmarks – CINT2000, and 14

floating-point benchmarks – CFP2000

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SPEC 2000 (CINT 2000)Results

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SPEC 2000 (CFP 2000)Results

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Reporting Performance Results

• Reproducability Apply them on publicly available

benchmarks. Pecking/Picking order– Real Programs– Real Kernels– Toy Benchmarks– Synthetic Benchmarks

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How to Summarize Performance

• Arithmetic mean (weighted arithmetic mean) tracks execution time: sum(Ti)/n or sum(Wi*Ti)

• Harmonic mean (weighted harmonic mean) of rates (e.g., MFLOPS) tracks execution time: n/sum(1/Ri) or 1/sum(Wi/Ri)

• Normalized execution time is handy for scaling performance (e.g., X times faster than SPARCstation 10)

• But do not take the arithmetic mean of normalized execution time, use the geometric mean = (Product(Ri)^1/n)

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Performance Evaluation• “For better or worse, benchmarks shape a field”• Good products created when have:

– Good benchmarks– Good ways to summarize performance

• Given sales is a function in part of performance relative to competition, investment in improving product as reported by performance summary

• If benchmarks/summary inadequate, then choose between improving product for real programs vs. improving product to get more sales;Sales almost always wins!

• Execution time is the measure of computer performance!

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Simulations

• When are simulations useful?

• What are its limitations, I.e. what real world phenomenon does it not account for?

• The larger the simulation trace, the less tractable the post-processing analysis.

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Queueing Theory

• What are the distributions of arrival rates and values for other parameters?

• Are they realistic?

• What happens when the parameters or distributions are changed?

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Quantitative Principles of Computer Design

• Make the Common Case Fast– Amdahl’s Law

• CPU Performance Equation– Clock cycle time– CPI– Instruction Count

• Principles of Locality• Take advantage of Parallelism

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DAP.S98 32

Amdahl's LawSpeedup due to enhancement E:

ExTime w/o E Performance w/ E

Speedup(E) = ------------- = -------------------

ExTime w/ E Performance w/o E

Suppose that enhancement E accelerates a fraction F of the task by a factor S, and the remainder of the task is unaffected

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Amdahl’s Law

ExTimenew = ExTimeold x (1 - Fractionenhanced) + Fractionenhanced

Speedupoverall =ExTimeold

ExTimenew

Speedupenhanced

=

1

(1 - Fractionenhanced) + Fractionenhanced

Speedupenhanced

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Amdahl’s Law

• Floating point instructions improved to run 2X; but only 10% of actual instructions are FP

Speedupoverall =

ExTimenew =

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CPU Performance EquationCPU time = Seconds = Instructions x Cycles x

Seconds

Program Program Instruction Cycle

CPU time = Seconds = Instructions x Cycles x Seconds

Program Program Instruction Cycle

Inst Count CPI Clock RateProgram X

Compiler X (X)

Inst. Set. X X

Organization X X

Technology X

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Cycles Per Instruction

CPU time = CycleTime * CPI * Ii = 1

n

i i

CPI = CPI * F where F = I i = 1

n

i i i i

Instruction Count

“Instruction Frequency”

Invest Resources where time is Spent!

CPI = (CPU Time * Clock Rate) / Instruction Count = Cycles / Instruction Count

“Average Cycles per Instruction”

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Example: Calculating CPI

Typical Mix

Base Machine (Reg / Reg)

Op Freq Cycles CPI(i) (% Time)

ALU 50% 1 .5 (33%)

Load 20% 2 .4 (27%)

Store 10% 2 .2 (13%)

Branch 20% 2 .4 (27%)

1.5

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Chapter Summary, #1• Designing to Last through Trends

Capacity Speed

Logic 2x in 3 years 2x in 3 years

DRAM 4x in 3 years 2x in 10 years

Disk 4x in 3 years 2x in 10 years

• 6yrs to graduate => 16X CPU speed, DRAM/Disk size

• Time to run the task– Execution time, response time, latency

• Tasks per day, hour, week, sec, ns, …– Throughput, bandwidth

• “X is n times faster than Y” means ExTime(Y) Performance(X)

--------- = --------------

ExTime(X) Performance(Y)

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Chapter Summary, #2

• Amdahl’s Law:

• CPI Law:

• Execution time is the REAL measure of computer performance!

• Good products created when have:– Good benchmarks, good ways to summarize performance

• Die Cost goes roughly with die area4

Speedupoverall =ExTimeold

ExTimenew

=

1

(1 - Fractionenhanced) + Fractionenhanced

Speedupenhanced

CPU time = Seconds = Instructions x Cycles x Seconds

Program Program Instruction Cycle

CPU time = Seconds = Instructions x Cycles x Seconds

Program Program Instruction Cycle

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Food for thought

• Two companies reports results on two benchmarks one on a Fortran benchmark suite and the other on a C++ benchmark suite.

• Company A’s product outperforms Company B’s on the Fortran suite, the reverse holds true for the C++ suite. Assume the performance differences are similar in both cases.

• Do you have enough information to compare the two products. What information will you need?

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Food for Thought II• In the CISC vs. RISC debate a key argument of the

RISC movement was that because of its simplicity, RISC would always remain ahead.

• If there were enough transistors to implement a CISC on chip, then those same transistors could implement a pipelined RISC

• If there was enough to allow for a pipelined CISC there would be enough to have an on-chip cache for RISC. And so on.

• After 20 years of this debate what do you think?• Hint: Think of commercial PC’s, Moore’s law and

some of the data in the first chapter of the book (and on these slides)

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Amdahl’s Law (answer)

• Floating point instructions improved to run 2X; but only 10% of actual instructions are FP

Speedupoverall = 1

0.95= 1.053

ExTimenew = ExTimeold x (0.9 + .1/2) = 0.95 x ExTimeold