Slide 1 Fundamentals of Computer Design CSCE430/830 Computer Architecture Instructor: Hong Jiang...
-
Upload
allan-bracher -
Category
Documents
-
view
218 -
download
1
Transcript of Slide 1 Fundamentals of Computer Design CSCE430/830 Computer Architecture Instructor: Hong Jiang...
Slide 1
Fundamentals of Computer Design
CSCE430/830 Computer Architecture
Instructor: Hong Jiang
Courtesy of Prof. Yifeng Zhu @ U. of Maine
Fall, 2007
Portions of these slides are derived from:Dave Patterson © UCB
Slide 2
Motivations and Introduction
•Phenomenal growth in computer industry/technology:
X2/18mo in 20yr. multi-GFLOPs processors, largely due to–Micro-electronics technology–Computer Design innovations
•We have come a long way in a short time of 60 years since the 1st general purpose computer in 1946:
• Instruction Set Architecture: •An Introduction
Slide 3
Motivations and Introduction
Past (Milestones):– First electronic computer ENIAC in 1946: 18,000 vacuum tubes, 3,000 cubic feet, 20 2-foot 10-digit registers, 5 KIPs (thousand additions per second);
– First microprocessor (a CPU on a single IC chip) Intel 4004 in 1971: 2,300 transistors, 60 KIPs, $200;
– Virtual elimination of assembly language programming reduced the need for object-code compatibility;
– The creation of standardized, vendor-independent operating systems, such as UNIX and its clone, Linux, lowered the cost and risk of bringing out a new architecture
– RISC instruction set architecture paved ways for drastic design innovations that focused on two critical performance techniques: instruction-level parallelism and use of caches
Slide 4
Motivations and Introduction
Present (State of the art): – Microprocessors approaching/surpassing 10 GFLOPS;– A high-end microprocessor (<$10K) today is easily more powerful than a supercomputer (>$10million) ten years ago;
– While technology advancement contributes a sustained annual growth of 35%, innovative computer design accounts for another 25% annual growth rate a factor of 15 in performance gains!
Slide 6
Technology Trend
PCWork-stationMini-
computer
Mainframe
Mini-supercomputer
Supercomputer
Massively Parallel
Processors
1988 Computer Food Chain
Slide 7
Technology Trend
1998 Computer Food Chain
PCWork-station
Mainframe
Supercomputer
Mini-supercomputerClusters
Mini-computer
Now who is eating whom?
Server
Slide 8
Parallel Computing Architectures in Top 500
www.top500.orgNov. 2004
MEMORY
BUS/CROSSBAR
CPU CPU CPU CPU
Symmetric Multiprocessing (SMP)
Massively Parallel Processor (MPP)
CPU M
CPU MCPU M
CPU MPC PCPC
network
cluster
MPP
Cluster
SMPConstellations
SIMD
Single processor
Supercomputer Trends in Top 500
Slide 9
Why Such Changes in 10 years?
• Performance– Technology Advances
» CMOS VLSI dominates older technologies (TTL, ECL) in cost AND performance
– Computer architecture advances improves low-end » RISC, superscalar, RAID, …
• Price: Lower costs due to …– Simpler development
» CMOS VLSI: smaller systems, fewer components– Higher volumes
» CMOS VLSI : same dev. cost 10,000 vs. 10,000,000 units – Lower margins by class of computer, due to fewer services
• Function– Rise of networking/local interconnection technology
Slide 10
Amazing Underlying Technology Change
• In 1965, Gordon Moore sketched out his prediction of the pace of silicon technology.
• Moore's Law: The number of transistors incorporated in a chip will approximately double every 24 months.
• Decades later, Moore's Law remains true.
From Intel
Slide 11
Technology Trends: Moore’s Law
• Gordon Moore (Founder of Intel) observed in 1965 that the number of transistors on a chip doubles about every 24 months.
• In fact, the number of transistors on a chip doubles about every 18 months.
From intel
Slide 12
Technology Trends
Based on SPEED, the CPU has increased dramatically, but memory and disk have increased only a little. This has led to dramatic changed in architecture, Operating Systems, and programming practices.
Slide 13
Technology dramatic change• Processor
– transistor number in a chip: about 55% 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– 10 years: 10Mb 100Mb– 5 years: 100Mb 1 Gb
Slide 15
Computer Architecture Is …
the attributes of a [computing] system as seen by the programmer, i.e., the conceptual structure and functional behavior, as distinct from the organization of the data flows and controls, the logic design, and the physical implementation.
Amdahl, Blaaw, and Brooks, 1964SOFTWARESOFTWARE
Slide 16
Computer Architecture’s 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
• 2010s: Computer Architecture Course:
Self adapting systems? Self organizing structures?DNA Systems/Quantum Computing?
Slide 17
CSCE430/830 Course Focus
Understanding the design techniques, machine structures, technology factors, evaluation methods that will determine the form of computers in the 21st Century
Technology ProgrammingLanguages
OperatingSystems
History
ApplicationsInterface Design
(ISA)
Measurement & Evaluation
Parallelism
Computer Architecture:• Instruction Set Design• Organization• Hardware/Software Boundary Compilers
Slide 18
Computer Engineering Methodology
TechnologyTrends
Evaluate ExistingEvaluate ExistingSystems for Systems for BottlenecksBottlenecks
Benchmarks
Simulate NewSimulate NewDesigns andDesigns and
OrganizationsOrganizations
Workloads
Implement NextImplement NextGeneration SystemGeneration System
ImplementationComplexity
Architecture design is an iterative process: Searching the space of possible designs at all levels of computer systems
Slide 19
Summary
1. Moors’s laws: The number of transistors incorporated in a chip will approximately double every 18 months.
2. CPU speed increases dramatically, but the speed of memory, disk and network increases slowly.
3. Architecture design is an iterative process. Measure performance: Benchmarks
Slide 20
Quantitative Principles• Performance Metrics: How do we
conclude that System-A is “better” than System-B?
• Amdahl’s Law: Relates total speedup of a system to the speedup of some portion of that system.
• Topics: (Sections 1.1, 1.2, 1.5, 1.6)– Metrics for different market segments– Benchmarks to measure performance– Quantitative principles of computer design
Slide 21
Importance of Measurement
Architecture design is an iterative process:• Search the possible design space• Make selections • Evaluate the selections made
Good IdeasGood Ideas
Mediocre IdeasBad Ideas
Cost /PerformanceAnalysis
Good measurement tools are required to accurately evaluate the selection.
Slide 22
Two notions of “performance”
Plane
Boeing 747
BAD/Sud Concodre
Speed
610 mph
1350 mph
DC to Paris
6.5 hours
3 hours
Passengers
470
132
Throughput (pmph)
286,700
178,200
• Time to do the task (Execution Time)
– execution time, response time, latency, etc.
• Tasks per day, hour, week, sec, ns. .. (Performance)
– throughput, bandwidth, etc.
Which has higher performance?
Slide 23
Performance Definitions
• Performance is in units of things-per-second.– bigger is better
• Execution time is the reciprocal of performance.– performance(x) = 1
execution_time(x)
• "X is n times faster than Y" means
execution_time (Y) performance(X)
n = ----------------- = -----------------
execution_time (X) performance(Y)• When is throughput more important than execution
time?• When is execution time more important than
throughput?
Slide 24
Performance Terminology“X is n% faster than Y” means:ExTime(Y) Performance(X) n
-------- = -------------- = 1 + ------
ExTime(X) Performance(Y) 100
n = 100(Performance(X) - Performance(Y))
Performance(Y)
Example: Y takes 15 seconds to complete a task, X takes 10 seconds. What % faster is X than Y?
n = 100(ExTime(Y) - ExTime(X))
ExTime(X)
Slide 25
Suppose that enhancement E accelerates a fraction F of the task by a factor S, and the remainder of the task is unaffected
tEnhancemenWithoutePerformanc
tEnhancemenWithePerformanc
tEnhancemenWithTimeExecution
tEnhancemenWithoutTimeExecutionESpeedup
__
__
___
___)(
Speedup due to enhancement E:
This fraction enhanced
Quantitative Design: Amdahl's Law
Amdahl’s Law gives a quick way to find the speedup from some enhancement.
Slide 26
Quantitative Design: Amdahl's Law
This fraction enhanced
ExTimeold ExTimenew
ExTimenew = ExTimeold x (1 - Fractionenhanced) + Fractionenhanced
Speedupoverall =ExTimeold
ExTimenew
Speedupenhanced
=
1
(1 - Fractionenhanced) + Fractionenhanced
Speedupenhanced
Slide 27
Pictorial Depiction of Amdahl’s Pictorial Depiction of Amdahl’s LawLaw
Before: Execution Time without enhancement E
After: Execution Time with enhancement E:
Enhancement E accelerates fraction F of original execution time by a factor of S
Unaffected fraction: (1- F) Affected fraction: F
Unaffected fraction: (1- F) F/S
Unchanged
Execution Time without enhancement E 1Speedup(E) = --------------------------------------------------------- = ---------------------- Execution Time with enhancement E (1 - F) + F/S
• shown normalized to 1 = (1-F) + F =1
Slide 28
• Floating point (FP) instructions improved to run 2X; but only 10% of actual instructions are FP. Suppose the old execution time is ExTimeold, What are the current execution time and speedup?
Quantitative Design: Amdahl's Law
Speedupoverall = 1
0.95= 1.053
ExTimenew = ExTimeold x (0.9 + .1/2) = 0.95 x ExTimeold
Speedup =ExTimeold
ExTimenew
=
1
(1 - Fractionenhanced) + Fractionenhanced
Speedupenhanced
Speedup =1
(1 - 0.1) + 0.1/2= 1.053
Slide 29
• The clock cycle time is the amount of time for one clock period to elapse (e.g. 5 ns).
• The clock rate is the inverse of the clock cycle time.
• For example, if a computer has a clock cycle time of 5 ns, the clock rate is:
1 ---------------------- = 200 MHz
5 x 10 sec
Computer Clocks • A computer clock runs at a constant rate
and determines when events take placed in hardware.
Clk
clock period
-9
Slide 30
Computing CPU time• The time to execute a given program is
CPU time = CPU clock cycles for a program x clock cycle time
Since clock cycle time and clock rate are reciprocals, thusCPU time = CPU clock cycles for a program / clock rate
• CPI: clock cycles per instruction CPU clock cycle for a program CPI = ------------------------- Instruction count
Slide 31
Computing CPU time• The time to execute a given program is
CPU time = CPU clock cycles for a program x clock cycle time
Since clock cycle time and clock rate are reciprocals, thus
CPU time = CPU clock cycles for a program / clock rate
• The number of CPU clock cycles can be determined by
CPU clock cycles = (instructions/program) x (clock cycles/instruction)
= Instruction count x CPI
which gives
• The units for this are instructions clock cycles secondsseconds = ---------------- x -------------- x -------------- program instruction clock cycle
CPU time = Instruction count x CPI x clock cycle timeCPU time = Instruction count x CPI / clock rate
Slide 32
Example of Computing CPU
time• If a computer has a clock rate of 2
GHz, how long does it take to execute a program with 1,000,000 instructions, if the CPI for the program is 3.5?
Slide 33
Example of Computing CPU
time• If a computer has a clock rate of 2 GHz, how long
does it take to execute a program with 1,000,000 instructions, if the CPI for the program is 3.5?
• Using the equation
CPU time = Instruction count x CPI / clock rate
gives
CPU time = 1000000 x 3.5 / (2 x 109 )
• If a computer’s clock rate increases from 200 MHz to 250 MHz and the other factors remain the same, how many times faster will the computer be?
CPU time old clock rate new 250 MHz-------------- = ------------------ = -------------- = 1.25CPU time new clock rate old 200 MHZ
• What simplifying assumptions did we make?
6
Slide 34
Performance Example
• Two computers M1 and M2 with the same instruction set.
• For a given program, we have
• How many times faster is M2 than M1 for this program?
ExTimeM1 ICM1 x CPIM1 / Clock RateM1
=ExTimeM2 ICM2 x CPIM2 / Clock RateM2
=2.8/50
3.2/75= 1.31
Clock rate
(MHz)
CPI
M1 50 2.8M2 75 3.2
Slide 35
Aspects of CPU Performance
CPU 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 CPICycle Time
Program X
Compiler X (X)
Inst. Set. X X
Organization X X
Technology X
Slide 36
Performance Summary• Two performance metrics execution time and
throughput.
• Amdahl’s Law
• When trying to improve performance, look at what occurs frequently => make the common case fast.
• CPU time:
CPU time = Instruction count x CPI x clock cycle time
CPU time = Instruction count x CPI / clock rate
Execution Time without enhancement E 1Speedup(E) = --------------------------------------------------------- = ---------------------- Execution Time with enhancement E (1 - F) + F/S