Arquitecturas para Procesamiento de Señal
Capítulo 1.
Abstracciones, Tecnología y Rendimiento de los
Procesadores
Chapter 1.
Computer Abstractions and Technology
Escuela Politécnica Superior
Universidad Autónoma de Madrid
Profesores:
Sergio López Buedo
Francisco J. Gómez Arribas
Outline
n
n
n
The Computer Revolution
n Moore´s Law
Abstractions
n Instruction Set Architecture
n RISC vs CIS
Performace
n
n
n
CPU performance: Execution Time, CPI.
Amdahl´s law
Measuring and evaluating performance: MIPS, SPEC
Outline
n
n
n
The Computer Revolution
n Moore´s Law
Abstractions
n Instruction Set Architecture
n RISC vs CISC
Performace
n
n
n
CPU performance: Execution Time, CPI.
Amdahl´s law
Measuring and evaluating performance: MIPS, SPEC
Historical Perspective
New Architectures: Eight Great Ideas
n
Design for Moore’s Law
n
Use abstraction to simplify design
n
Make the common case fast (Amdahl)
n
Performance via parallelism
n
Performance via pipelining
n
Performance via prediction
n
Hierarchy of memories
n
Dependability via redundancy
Chapter 1 — Computer Abstractions and Technology — 5
Evolution of Processors in Computers
Multi-everywhere: MT, CMP
1000000
100000
Multi-Threaded, Multi-Core
10000
Multi Threaded
Speculative, OOO
MIPS
1000
Era of
Thread &
Processor
Level
Parallelism
Super Scalar
100
10
8086
286
386
486
Era of
Pipelined
Architecture
1
Era of
Instruction
Level
Parallelism
Special Purpose
HW
0,1
0,01
1970
1980
1990
2000
2010
Chapter 1 — Computer Abstractions and Technology — 6
The Computer Revolution
n
Progress in computer technology
n
Underpinned by Moore’s Law
Moore's law
The number of transistors that can be fit
on a computer chip will double every 18
months, resulting in periodic increases in
computing power.
_
Progress in computer technology
Moore's law
The number of transistors that can be fit
on a computer chip will double every 1824 months, resulting in periodic increases
in computing power.
Parallelism and Moore’s Law
Chapter 1 — Computer Abstractions and Technology — 9
Abstractions
n
Abstraction helps us deal with complexity
n
n
Hide lower-level detail
Instruction set architecture (ISA)
A set of hardware-implemented instructions, the
symbolic name and the binary code format of each
instruction.
n Is the structure of a computer that a machine
language programmer (or a compiler) must
understand to write a correct (timing independent)
program for that machine.
n
n
The hardware/software interface.
Chapter 1 — Computer Abstractions and Technology — 10
Abstractions: ISA examples
Chapter 1 — Computer Abstractions and Technology — 11
Understanding Performance
n
Algorithm
n
n
Programming language, compiler, architecture
n
n
Determine number of machine instructions executed
per operation
Processor and memory system
n
n
Determines number of operations executed
Determine how fast instructions are executed
I/O system (including OS)
n
Determines how fast I/O operations are executed
Chapter 1 — Computer Abstractions and Technology — 12
n
Which airplane has the best performance?
Boeing 777
Boeing 777
Boeing 747
Boeing 747
BAC/Sud
Concorde
BAC/Sud
Concorde
Douglas
DC-8-50
Douglas DC8-50
0
100
200
300
400
0
500
Boeing 777
Boeing 777
Boeing 747
Boeing 747
BAC/Sud
Concorde
BAC/Sud
Concorde
Douglas
DC-8-50
Douglas DC8-50
500
1000
Cruising Speed (mph)
4000
6000
8000 10000
Cruising Range (miles)
Passenger Capacity
0
2000
§1.4 Performance
Defining Performance
1500
0
100000 200000 300000 400000
Passengers x mph
Chapter 1 — Computer Abstractions and Technology — 13
Response Time and Throughput
n
Response time
n
n
How long it takes to do a task
Throughput
n
Total work done per unit time
n
n
e.g., tasks/transactions/… per hour
How are response time and throughput affected
by
Replacing the processor with a faster version?
n Adding more processors?
n
n
We’ll focus on response time for now…
Chapter 1 — Computer Abstractions and Technology — 14
Relative Performance
Define Performance = 1/Execution Time
n “X is n time faster than Y”
n
Performanc e X Performanc e Y
= Execution time Y Execution time X = n
n
Example: time taken to run a program
10s on A, 15s on B
n Execution TimeB / Execution TimeA
= 15s / 10s = 1.5
n So A is 1.5 times faster than B
n
Chapter 1 — Computer Abstractions and Technology — 15
CPU Clocking
n
Operation of digital hardware
governed by a constant-rate clock
Clock (cycles)
Data transfer
and computation
Update state
n
e.g., 250ps = 0.25ns = 250×10–12s
Clock frequency (rate): cycles per second
n
10n
80
24
2
1
Clock period: duration of a clock cycle
n
2n
2 70
2 60
2 50
2 40
2 30
2 20
2 10
Clock period
n
Please remember
prefixes!
e.g., 4.0GHz = 4000MHz = 4.0×109Hz
2−10
2−20
2−30
2−40
2−50
2−60
2−70
2−80
10
10 21
10 18
10 15
10 12
10 9
10 6
10 3
10 2
10 1
1
10−1
10−2
10−3
10−6
10−9
10−12
10−15
10−18
10−21
10−24
Prefix Simb
yotta
zetta
exa
peta
tera
giga
mega
kilo
hecto
deca
-deci
centi
mili
micro
nano
pico
femto
atto
zepto
yocto
Y
Z
E
P
T
G
M
k
h
da
-d
c
m
µ
n
p
f
a
z
y
Chapter 1 — Computer Abstractions and Technology — 16
CPU Time (Tcpu)
CPU Time = CPU Clock Cycles ´ Clock Cycle Time
= (# cycles ´ period )
CPU Clock Cycles # cycles
=
=(
)
Clock Rate
freq
n
Performance improved by
Reducing number of clock cycles
n Increasing clock rate
n Hardware designer must often trade off clock
rate against cycle count
n
Chapter 1 — Computer Abstractions and Technology — 17
Instruction Count and CPI
Clock Cycles = Instruction Count ´ Cycles per Instruction
CPU Time = Instruction Count ´ CPI ´ Clock Cycle Time
Instruction Count ´ CPI
=
Clock Rate
n
Instruction Count for a program
n
n
Determined by program, ISA and compiler
Average cycles per instruction
Determined by CPU hardware
n If different instructions have different CPI
n
n
Average CPI affected by instruction mix
Chapter 1 — Computer Abstractions and Technology — 18
CPI Example
Computer A: Cycle Time = 250ps, CPI = 2.0
n Computer B: Cycle Time = 500ps, CPI = 1.2
n Same ISA
n Which is faster, and by how much?
n
CPU Time
A
= Instruction Count ´ CPI ´ Cycle Time
A
A
= I ´ 2.0 ´ 250ps = I ´ 500ps
A is faster…
CPU Time = Instruction Count ´ CPI ´ Cycle Time
B
B
B
= I ´ 1.2 ´ 500ps = I ´ 600ps
CPU Time
B = I ´ 600ps = 1.2
CPU Time
I ´ 500ps
A
…by this much
Chapter 1 — Computer Abstractions and Technology — 19
CPI in More Detail
n
If different instruction classes take different
numbers of cycles
n
Clock Cycles = å (CPIi ´ Instruction Count i )
i=1
n
Weighted average CPI
n
Clock Cycles
Instructio n Count i ö
æ
CPI =
= å ç CPIi ´
÷
Instructio n Count i=1 è
Instructio n Count ø
Relative frequency
Chapter 1 — Computer Abstractions and Technology — 20
CPI Example
n
Alternative compiled code sequences using
instructions in classes A, B, C
IC stand for:
Instruction
Count
n
Class
A
B
C
CPI for class
1
2
3
IC in sequence 1
2
1
2
IC in sequence 2
4
1
1
Sequence 1: IC = 5
Clock Cycles
= 2×1 + 1×2 + 2×3
= 10
n Avg. CPI = 10/5 = 2.0
n
n
Sequence 2: IC = 6
Clock Cycles
= 4×1 + 1×2 + 1×3
=9
n Avg. CPI = 9/6 = 1.5
n
Chapter 1 — Computer Abstractions and Technology — 21
Performance Summary
Instructions Clock cycles Seconds
CPU Time =
´
´
Program
Instruction Clock cycle
n
Performance depends on
Algorithm: affects IC, possibly CPI
n Programming language: affects IC, CPI
n Compiler: affects IC, CPI
n Instruction set architecture: affects IC, CPI, Tc
n
Chapter 1 — Computer Abstractions and Technology — 22
SpeedUp and Amdahl’s Law
n
Speedup is the improvement achieved in
execution time.
Execution timeold
SpeedupOverall =
Execution timenew
n
Amdahl´s Law is used to find the speedup
to an overall system when only part of the
system is improved.
SpeedupOverall =
1
(1− Fractionenhanced ) +
Fractionenhanced
Speedupenhanced
Chapter 1 — Computer Abstractions and Technology — 23
Amdahl´s law: 1 improvement
Told
Time unaffected
Time affected
Fraction 1−Fe
Fraction Fe
Time unaffected
Time improved
Tnew
SpeedupOverall =
1
Fe
(1− Fe ) +
Se
Time enhanced
Se times
Fe: Fraction enhanced
Se: Speedup enhanced.
Amount of improvement
Amdahl´s law
§ Performance is limited to the non-speedup
portion of the program.
Execution time after improvement = Execution time of
unaffected part + (Execution time of affected part /
Amount of Improvement)
§ Corollary of Amdahl’s law: Make the
common case fast.
Amdahl´s law generalization
Told
Time unaffected
Time to improve
Time to improve
mejorar
Fraction Fe,1
Fracción 1−Fe,1−Fe,2
T1 enhanced
Time unaffected
Tnew
Speedupoverall =
Fraction Fe2
T2 enhanced
Amount of improvement
Se,2 times
Amount of improvement
Se1 times
1
n
n
Fe,i
1− ∑ Fe,i + ∑
i=1
i=1 Se,i
Amdahl´s law: An exercise
Example: Using the Amdahl’s law. Calculate the speed up of the
system in both cases:
Case 1: x10 in
FP Operation
ARITHMETIC LOGIC UNIT (ALU)
INT
FP SUM
FP MUL
FP DIV
Instruction (%)
70
15
10
5
Old CPI
2
40
60
100
New CPI improved
2
4
6
10
ARITHMETIC LOGIC UNIT (ALU)
Case 2: FP with
different improvements
INT
FP SUM
FP MUL
FP DIV
Instruction (%)
70
15
10
5
Old CPI
2
40
60
100
New CPI improved
2
2
15
10
Pitfall: MIPS as a Performance Metric
n
MIPS: Millions of Instructions Per Second
n
Doesn’t account for
Differences in ISAs between computers
n Differences in complexity between instructions
n
MIPS =
Instructio n count
Execution time ´ 10 6
Instructio n count
Clock rate
=
=
6
Instructio n count ´ CPI
CPI
´
10
6
´ 10
Clock rate
n
CPI varies between programs on a given CPU
Chapter 1 — Computer Abstractions and Technology — 28
MFLOPs as a Performance Metric
n
MFLOPS: Millions of Floating-point
Operations Per Second
Floating point instruction count
MFLOPS =
Execution time ×106
The MFLOPS measure, as described by [ref1] , is widely employed because of
its efficiency and effectiveness in terms of simplicity and its indicative abilities.
See www.top500.org
[1] Evaluating the Mflops Measure Ran Giladi , Journal IEEE Micro archive
Vol 16 Issue 4, 1996, pp 69-75
Chapter 1 — Computer Abstractions and Technology — 29
MIPS vs MFLOPs: An exercise
Example: Compare the performance @ 100 MHz of M1-CISC and
M2-RISC. (Also using MIPS and MFLOPs metrics) :
IC M1-cisc
IC M2-risc
CPI
Alu INT
6
5x10
6
10x10
1
Mem
6
10
6
10
2
FP
6
10
6
10
3
•
The performance of M1-CISC =1,5 x M2-RISC ( 50% better)
•
M1 MIPS < M2 MIPS
•
M1 MFLOPs > M2 MFLOPs
SPEC CPU Benchmark
n
Programs used to measure performance
n
n
Standard Performance Evaluation Corp (SPEC)
n
n
Supposedly typical of actual workload
Develops benchmarks for CPU, I/O, Web, …
SPEC CPU2006
n
Elapsed time to execute a selection of programs
n
Negligible I/O, so focuses on CPU performance
Normalize relative to reference machine
n Summarize as geometric mean of performance ratios
n
n
CINT2006 (integer) and CFP2006 (floating-point)
n
n
Õ Execution time ratio
i
i=1
Chapter 1 — Computer Abstractions and Technology — 31
CINT2006 for Intel Core i7 920
Chapter 1 — Computer Abstractions and Technology — 32
Material de apoyo adicional
n
Material Adicional
Fórmulas y relaciones útiles para los ejercicios
n Ejemplos
n Trampas (Pitfalls)
n
Info Adicional: Fórmulas y relaciones útiles para los ejercicios
TCPU = NI x CPI x TCICLO =
Rendimiento(X) = 1 / TEJEC(X)
= NroCiclosReloj / FreqReloj
n=
X es n veces más rápido que Y
significa:
Una mejora en el
sistema se mide con la:
Rendimiento(X) TEJEC (Y)
=
Rendimiento(Y) TEJEC (X)
TEJ (sin mejora) R(con m.)
Aceleración =
=
TEJ (con mejora) R(sin m.)
La ley de Amdahl mide la mejora del
rendimiento global de un sistema. Donde:
Acc global =
Fm: Fracción de tiempo mejorada y Accm: Aceleración mejorada
Accglobal =
1
(1- Fm ) +
Fm
Acc m
1
Fm,i
1 - å Fm,i + å
i =1
i =1 Accm ,i
MIPS = NI / (TCPU x 106) = NI / (NI x CPI x TCiclo x 106) = Frec/ (CPI x 106)
La ley de Amdahl para varias mejoras:
n
n
CPU Time Example
Computer A: 2GHz clock, 10s CPU time
n Designing Computer B
n
n
n
n
Aim for 6s CPU time
Can do faster clock, but causes 1.2 × clock cycles
How fast must Computer B clock be?
Clock CyclesB 1.2 ´ Clock CyclesA
Clock RateB =
=
CPU TimeB
6s
Clock CyclesA = CPU Time A ´ Clock Rate A
= 10s ´ 2GHz = 20 ´ 10 9
1.2 ´ 20 ´ 10 9 24 ´ 10 9
Clock RateB =
=
= 4GHz
6s
6s
Chapter 1 — Computer Abstractions and Technology — 35
Amdahl´s law: example 1
§ Suppose we enhance a machine making all floatingpoint instructions run five times faster. If the
execution time of some benchmark before the
floating-point enhancement is 12 seconds, what will
the speedup be if half of the 12 seconds is spent
executing floating-point instructions?
Time = 6 (other instr) + 6 (FP instr) / 5 = 7.2 seconds.
Speedup = 12 sec / 7.2 sec = 1.67
1
1
1
SpeedupOverall =
=
=
=1.67
Fe
0.5 0.6
(1− Fe ) +
(1− 0.5) +
Se
5
Chapter 1 — Computer Abstractions and Technology — 36
Amdahl´s law: example 2
§ We are looking for a benchmark to show off the new
floating-point unit described in the previous example,
and we want the overall benchmark to show a speedup
of 3. One benchmark we are considering runs for 100
seconds with the old floating-point hardware. How
much of the execution time would floating-point
instructions have to account for in this program in order
to yield our desired speedup on this benchmark?
Speedup = 3 = 100 / (Time_FP / 5 + 100 – Time_FP)
Time_FP= 83.33 seconds
1
1
5
SpeedupOverall =
; 3=
; 3=
; Fe = 0,83
Fe
Fe
5 − 4Fe
(1− Fe ) +
(1− Fe ) +
Se
5
Chapter 1 — Computer Abstractions and Technology — 37
n
Improving an aspect of a computer and
expecting a proportional improvement in
overall performance
Taffected
Timproved =
+ Tunaffected
improvemen t factor
n
Example: multiply accounts for 80s/100s
n
n
§1.10 Fallacies and Pitfalls
Pitfall: Amdahl’s Law
How much improvement in multiply performance to
get 5× overall?
80
n Can’t be done!
20 =
+ 20
n
Corollary: make the common case fast
Chapter 1 — Computer Abstractions and Technology — 38