CPSC 321 Computer Architecture Summer 2005 Lecture 1 Introduction and Five Components of a Computer Praveen Bhojwani Adapted from CS 152 Spring 2002 UC.

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CPSC 321 Computer Architecture Summer 2005 Lecture 1 Introduction and Five Components of a Computer Praveen Bhojwani Adapted from CS 152 Spring 2002 UC Berkeley Copyright (C) 2001 UCB

Course Information Instructor –Praveen Bhojwani 514B, HRBB tel: Office Hours: to be announced on the class website TA –Ping Luo

Course Information [contd…] Grading: Projects, Labs, Exams –Labs20% –Mid Term 20% –Finals25% –Projects30% –Quizzes 5% Labs –MIPS (Assembly Programming), Verilog (HDL) Projects –Project 1: MIPS –Projects 2 & 3: Verilog (Datapath Design)

Course Information [contd…] Books –Computer Organization and Design: The Hardware/Software Interface, Second Edition, David A. Patterson and John L. Hennessy, 2 nd edition, Morgran Kaufmann Publishers. –Digital Design M. Morris Mano, 3 rd Edition, Prentice Hall –Check the course webpage for other materials and links

Course Information [contd…] Course Webpage – CS Accounts –Use your CS accounts to turnin and check any regarding course

Course Overview Arithmetic Single/multicycle Datapaths Computer Arithmetic Datapaths

Course Overview [contd…] IFetchDcdExecMemWB IFetchDcdExecMemWB IFetchDcdExecMemWB IFetchDcdExecMemWB PipeliningMemory Systems Performance Memory

What’s in 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) °HP PA-RISC(v1.1, v2.0) °Sun Sparc(v8, v9) °SGI MIPS(MIPS I, II, III, IV, V) °Intel(8086,80286,80386, ,Pentium, MMX,...)

The MIPS R3000 ISA (Summary) Instruction Categories –Load/Store –Computational –Jump and Branch –Floating Point coprocessor –Memory Management –Special R0 - R31 PC HI LO OP rs rt rdsafunct rs rt immediate jump target 3 Instruction Formats: all 32 bits wide

CPSC 321 “What” is Computer Architecture ? I/O systemInstr. Set Proc. Compiler Operating System Application Digital Design Circuit 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 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 ??? Computer Architecture Technology Programming Languages Operating Systems 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!!!

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!

°In ~1985 the single-chip processor (32-bit) and the single-board computer emerged °In the timeframe, these may well look like mainframes compared single-chip computer (maybe 2 chips) DRAM YearSize Kb Kb Mb Mb Mb Mb Mb Gb Microprocessor Logic Density DRAM chip capacity Technology Trends

Smaller feature sizes – higher speed, density ECE/CS 752; copyright J. E. Smith, 2002 (Univ. of Wisconsin)

Technology Trends Number of transistors doubles every 18 months (amended to 24 months) ECE/CS 752; copyright J. E. Smith, 2002 (Univ. of Wisconsin)

Levels of Representation High 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) ALUOP[0:3] <= InstReg[9:11] & MASK

Execution Cycle Instruction 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

Performance Metrics Response Time –Delay between start end end time of a task Throughput –Numbers of tasks per given time

Examples (Throughput/Performance) Replacing the processor with a faster version ? Adding additional processor to a system ?

Measuring Performance Wall-clock time –or- 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

Amdahl’s Law Pitfall: Expecting the improvement of one aspect of a machine to increase performance by an amount proportional to the size of improvement

Amhdahl’s Law [contd…] A program runs in 100 seconds on a machine, with 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 five times faster ? Execution Time After improvement = (exec time affected by improvement/amount of improvement) + exec time unaffected exec time after improvement = (80 seconds / n) + (100 – 80 seconds) We want performance to be 5 times faster => 20 seconds = 80/n seconds / n + 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

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 Faster”