ECE 321Computer Architecture

Chapter 1

Computer Abstractions and Technology

Course Overview

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3 4 Arithmetic

Single/multicycleDatapaths

Computer Arithmetic

Datapaths

Course Overview [contd…]IFetchDcd Exec Mem WB

IFetchDcd Exec Mem WB

IFetchDcd Exec Mem WB

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Pipelining

Memory Systems

Performance

Memory

What You Will Learn

• How programs are translated into the machine language– And how the hardware executes them

• The hardware/software interface

• What determines program performance– And how it can be improved

• How hardware designers improve performance

• What is parallel processing

What’s In It For Me ?

• In-depth understanding of the inner-workings of modern computers, their evolution, and trade-offs present at the hardware/software boundary.– Insight into fast/slow operations that are easy/hard to

implementation hardware

• Experience with the design process in the context of a large complex (hardware) design.– Functional Spec --> Control & Datapath --> Physical

implementation– Modern CAD tools

Computer Architecture - Definition

• Computer Architecture = ISA + MO

• Instruction Set Architecture– What the executable can “see” as underlying hardware– Logical View

• Machine Organization– How the hardware implements ISA ?– Physical View

Computer Architecture – Changing Definition• 1950s to 1960s: Computer Architecture Course:

–Computer Arithmetic

• 1970s to mid 1980s: Computer Architecture Course: –Instruction Set Design, especially ISA appropriate for compilers

• 1990s: Computer Architecture Course: Design of CPU, memory system, I/O system, Multiprocessors,

Networks

• 2000s: Computer Architecture Course: –Non Von-Neumann architectures, Reconfiguration

• DNA Computing, Quantum Computing ????

Some Examples …° Digital Alpha (v1, v3) 1992-97

° HP PA-RISC (v1.1, v2.0) 1986-96

° Sun SPARC (v8, v9) 1987-95

° SGI MIPS (MIPS I, II, III, IV, V)1986-96

° IA-16/32 (8086,286,386, 486, 1978-1999 Pentium, MMX, SSE, …)

° IA-64 (Itanium) 1996-now

° AMD64/EMT64 2002-now

° IBM POWER (PowerPC,…) 1990-now

° Many dead processor architectures live on in microcontrollers

Generations of Computer• Vacuum tube - 1946-1957• Transistor - 1958-1964• Small scale integration - 1965 on

– Up to 100 devices on a chip• Medium scale integration - to 1971

– 100-3,000 devices on a chip• Large scale integration - 1971-1977

– 3,000 - 100,000 devices on a chip• Very large scale integration - 1978 to date

– 100,000 - 100,000,000 devices on a chip• Ultra large scale integration

– Over 100,000,000 devices on a chip

The MIPS R3000 ISA (Summary)

• Instruction Categories

– Load/Store– Computational– Jump and Branch– Floating Point

• coprocessor– Memory Management– Special

R0 - R31

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ECE 321

“What” is Computer Architecture ?

I/O systemInstr. Set Proc.

Compiler

OperatingSystem

Application

Digital DesignCircuit Design

Instruction Set Architecture

Firmware

• Coordination of many levels of abstraction• Under a rapidly changing set of forces• Design, Measurement, and Evaluation

Datapath & Control

Layout

Impact of Changing ISA

• Early 1990’s Apple switched instruction set architecture of the Macintosh– From Motorola 68000-based machines

– To PowerPC architecture

• Intel 80x86 Family: many implementations of same architecture– program written in 1978 for 8086 can be run

on latest Pentium chip

Factors Affecting ISA ???

ComputerArchitecture

Technology ProgrammingLanguages

OperatingSystems

History

Applications

Cleverness

ISA: Critical Interface

instruction set

software

hardware

Examples: 80x86 50,000,000 vs. MIPS 5500,000 ???

The Big Picture

Control

Datapath

Memory

Processor

Input

Output

Since 1946 all computers have had 5 components!!!

Example Organization• TI SuperSPARCtm TMS390Z50 in Sun SPARCstation20

Floating-point Unit

Integer Unit

InstCache

RefMMU

DataCache

StoreBuffer

Bus Interface

SuperSPARC

L2$

CC

MBus Module

MBus

L64852 MBus controlM-S Adapter

SBus

DRAM Controller

SBusDMA

SCSIEthernet

STDIO

serialkbdmouseaudioRTC

FloppySBusCards

Moore’s Law

• Increased density of components on chip• Gordon Moore - cofounder of Intel• Number of transistors on a chip will double 18-24

months• Since 1970’s development has slowed a little

– Number of transistors doubles every 24 months

• Cost of a chip has remained almost unchanged• Higher packing density means shorter electrical paths,

giving higher performance• Smaller size gives increased flexibility• Reduced power and cooling requirements• Fewer interconnections increases reliability

Technology Trends

• Processor– logic capacity: about 30% per year– clock rate: about 20% per year

• Memory– DRAM capacity: about 60% per year (4x every 3 years)– Memory speed: about 10% per year– Cost per bit: improves about 25% per year

• Disk– capacity: about 60% per year– Total use of data: 100% per 9 months!

• Network Bandwidth– Bandwidth increasing more than 100% per year!

i4004

i8086

i80386

Pentium

i80486

i80286

SU MIPS

R3010

R4400

R10000

1000

10000

100000

1000000

10000000

100000000

1965 1970 1975 1980 1985 1990 1995 2000 2005

Tra

nsis

tors

i80x86

M68K

MIPS

Alpha

° In ~1985 the single-chip processor (32-bit) and the single-board computer emerged

° In ~2002 started having multiple processor cores on a chip (IBM POWER4)

DRAMYear Size1980 64 Kb1983 256 Kb1986 1 Mb1989 4 Mb1992 16 Mb1996 64 Mb1999 256 Mb2002 1 Gb

uP-Name

Microprocessor Logic DensityDRAM chip capacity

Technology Trends

Technology Trends

Smaller feature sizes – higher speed, density

Technology Trends

Number of transistors doubles every 18 months

(amended to 24 months)

Levels of RepresentationHigh Level Language

Program

Assembly Language Program

Machine Language Program

Control Signal Specification

Compiler

Assembler

Machine Interpretation

temp = v[k];

v[k] = v[k+1];

v[k+1] = temp;

• lw $15, 0($2)• lw $16, 4($2)• sw $16, 0($2)• sw $15, 4($2)

0000 1001 1100 0110 1010 1111 0101 10001010 1111 0101 1000 0000 1001 1100 0110 1100 0110 1010 1111 0101 1000 0000 1001 0101 1000 0000 1001 1100 0110 1010 1111

ALUOP[0:3] <= InstReg[9:11] & MASK

Execution CycleInstruction

Fetch

Instruction

Decode

Operand

Fetch

Execute

Result

Store

Next

Instruction

Obtain instruction from program storage

Determine required actions and instruction size

Locate and obtain operand data

Compute result value or status

Deposit results in storage for later use

Determine successor instruction

The Role of Performance

Understanding Performance

• Algorithm– Determines number of operations executed

• Programming language, compiler, architecture– Determine number of machine instructions executed

per operation

• Processor and memory system– Determine how fast instructions are executed

• I/O system (including OS)– Determines how fast I/O operations are executed

Example of Performance Measure

Performance Metrics

• Response Time– Delay between start and end time of a task

• Throughput– Numbers of tasks per given time

• New: Power/Energy– Energy per task, power

CPU Clocking• Operation of digital hardware governed by a

constant-rate clock

Clock (cycles)

Data transferand computation

Update state

Clock period

Clock period: duration of a clock cycle e.g., 250ps = 0.25ns = 250×10–12s

Clock frequency (rate): cycles per second e.g., 4.0GHz = 4000MHz = 4.0×109Hz

Examples (Throughput/Performance)

• Replace the processor with a faster version?– 3.8 GHz instead of 3.2 GHz

• Add an additional processor to a system?– Core Duo instead of P4

Measuring Performance

• Wall clock time vs. Total execution time

• CPU Time– User Time– System Time

Try using time command on UNIX system

Relating the Metrics

• Performance = 1/Execution Time

• CPU Execution Time = CPU clock cycles for program x Clock cycle time

• CPU clock cycles = Instructions for a program x Average clock cycles per Instruction

Performance Summary

• Performance depends on– Algorithm: affects IC, possibly CPI– Programming language: affects IC, CPI– Compiler: affects IC, CPI

– Instruction set architecture: affects IC, CPI, Tc

The BIG Picture

cycle Clock

Seconds

nInstructio

cycles Clock

Program

nsInstructioTime CPU

SPEC CPU Benchmark• Programs used to measure performance

– Supposedly typical of actual workload• Standard Performance Evaluation Corp (SPEC)

– Develops benchmarks for CPU, I/O, Web, …

• SPEC CPU2006– Elapsed time to execute a selection of programs

• Negligible I/O, so focuses on CPU performance– Normalize relative to reference machine– Summarize as geometric mean of performance ratios

• CINT2006 (integer) and CFP2006 (floating-point)

n

n

1iiratio time Execution

CINT2006 for Opteron X4 2356

Name Description IC×109 CPI Tc (ns) Exec time Ref time SPECratio

perl Interpreted string processing 2,118 0.75 0.40 637 9,777 15.3

bzip2 Block-sorting compression 2,389 0.85 0.40 817 9,650 11.8

gcc GNU C Compiler 1,050 1.72 0.47 24 8,050 11.1

mcf Combinatorial optimization 336 10.00 0.40 1,345 9,120 6.8

go Go game (AI) 1,658 1.09 0.40 721 10,490 14.6

hmmer Search gene sequence 2,783 0.80 0.40 890 9,330 10.5

sjeng Chess game (AI) 2,176 0.96 0.48 37 12,100 14.5

libquantum Quantum computer simulation 1,623 1.61 0.40 1,047 20,720 19.8

h264avc Video compression 3,102 0.80 0.40 993 22,130 22.3

omnetpp Discrete event simulation 587 2.94 0.40 690 6,250 9.1

astar Games/path finding 1,082 1.79 0.40 773 7,020 9.1

xalancbmk XML parsing 1,058 2.70 0.40 1,143 6,900 6.0

Geometric mean 11.7

High cache miss rates

Amdahl’s Law

• Pitfall: Expecting the improvement of one aspect of a machine to increase overall performance by an amount proportional to the size of improvement

Amhdahl’s Law [contd…]• A program runs in 100 seconds on a machine• Multiply operations responsible for 80 seconds of this time. • How much do I have to improve the speed of multiplication if I want

my program to run 5 times faster ?

• Execution Time After improvement = (exec time affected by improvement/amount of improvement) + exec time unaffectedexec time after improvement = (80 seconds / n) + (100 – 80 seconds)

We want performance to be 5 times faster =>

20 seconds = 80/n seconds + 20 seconds

0 = 80 / n !!!!

Amdahl’s Law [contd…]

• Opportunity for improvement is affected by how much time the event consumes

• Make the common case fast

• Very high speedup requires making nearly every case fast

• Focus on overall performance, not one aspect

Summary • Computer Architecture = Instruction Set Architure + Machine

Organization• All computers consist of five components

– Processor: (1) datapath and (2) control– (3) Memory– (4) Input devices and (5) Output devices

• Not all “memory” are created equally– Cache: fast (expensive) memory are placed closer to the

processor– Main memory: less expensive memory--we can have more

• Interfaces are where the problems are - between functional units and between the computer and the outside world

• Need to design against constraints of performance, power, area and cost

Summary

• Performance “eye of the beholder”

Seconds/program =

(Instructions/Pgm)x(Clk Cycles/Instructions)x(Seconds/Clk cycles)

• Amdahl’s Law “Make the Common Case Fast”

Homework

• Chapter 1

• 1.3, 1.4, 1.10, 1.15, 1.16 (first 4 parts of each question)

• Due Next Tuesday