Performance, Cost and Amdahl`s Law

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Performance, Cost and
Amdahl’s Law
Arquitectura de Computadoras
Arturo Díaz Pérez
Centro de Investigación y de Estudios Avanzados del IPN
Laboratorio de Tecnologías de Información
[email protected]
Arquitectura de Computadoras
Performance- 1
Performance
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♦ Purchasing perspective
■ given a collection of machines, which has the
» best performance ?
» least cost ?
» best performance / cost ?
♦ Design perspective
■ faced with design options, which has the
» best performance improvement ?
» least cost ?
» best performance / cost ?
♦ Both require
■ basis for comparison
■ metric for evaluation
♦ Our goal is to understand cost & performance implications of
architectural choices
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Performance- 2
Two notions of “performance”
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Plane
DC to Paris
Speed
Passengers
Throughput
(pmph)
Boeing 747
6.5 hours
610 mph
470
286,700
Concorde
3 hours
1350 mph
132
178,200
Which has higher performance?
° Time to do the task (Execution Time)
– execution time, response time, latency
° Tasks per day, hour, week, sec, ns. .. (Performance)
– throughput, bandwidth
Response time and throughput often are in opposition
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Performance- 3
What is Performance ?
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♦ KEY: A measure of Speed (Rate)
■ Car: miles driven per hour
■ Car wash: cars washed per day
■ Auto plant: cars built per year
♦ Two metrics:
■ Latency (response or execution time)
» time to start to finish of a task
■ Throughput (bandwidth)
» rate of task completion
= rate of task initiation
= 1 / (time between task completions)
♦ Deterministic vs. average
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Definitions
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♦ Performance is in units of things-per-second
■ bigger is better
♦ If we are primarily concerned with response time
■ performance(x) =
1
execution_time(x)
" X is n times faster than Y" means
Performance(X)
n
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=
---------------------Performance(Y)
Performance- 5
Example
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♦ Time of Concorde vs. Boeing 747?
Concord is 1350 mph / 610 mph
= 2.2 times faster
= 6.5 hours / 3 hours
♦ Throughput of Concorde vs. Boeing 747 ?
Concord is 178,200 pmph / 286,700 pmph
Boeing is 286,700 pmph / 178,200 pmph
= 0.62 “times faster”
= 1.60 “times faster”
♦ Boeing is 1.6 times (“60%”) faster in terms of throughput
♦ Concord is 2.2 times (“120%”) faster in terms of flying
time
We will focus primarily on execution time for a single job
Lots of instructions in a program => Instruction throughput important!
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Performance- 6
Relative Performance
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♦ Definition: X is n % faster than Y if
execution rateX execution timeY
n
=
= 1+
100
execution rateY execution timeX
♦ Example: X = 1 minute, Y = 2 minutes
2 minute
100
= 1+
1 minute
100
Thus, X is 100 % faster than Y
♦ Example: Car wash that starts one car per minute and
holds four cars.
■ Latency = four minutes per car
■ Throughput = one car per minute
■ Throughput > 1/Latency due to overlap
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Key idea: pipelining
Performance- 7
Basis of Evaluation
Cons
Pros
• representative
• portable
• widely used
• improvements
useful in reality
• easy to run, early in
design cycle
• identify peak
capability and
potential bottlenecks
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Actual Target Workload
• very specific
• non-portable
• difficult to run, or
measure
• hard to identify cause
Full Application Benchmarks
•less representative
Small “Kernel” Benchmarks
• easy to “fool”
Microbenchmarks
• “peak” may be a long
way from application
performance
Performance- 8
Metrics of performance
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Answers per month
Application
Useful Operations per second
Programming
Language
Compiler
ISA
(millions) of Instructions per second – MIPS
(millions) of (F.P.) operations per second – MFLOP/s
Datapath
Control
Megabytes per second
Function Units
Transistors Wires Pins
Cycles per second (clock rate)
Each metric has a place and a purpose, and each can be misused
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Performance- 9
Aspects of CPU Performance
CPU
CPUtime
time
== Seconds
Seconds
Program
Program
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==Instructions
xx Seconds
Instructions xx Cycles
Cycles
Seconds
Program
Instruction
Cycle
Program
Instruction
Cycle
instr count
CPI
clock rate
Program
Compiler
Instr. Set
Organization
Technology
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Aspects of CPU Performance
CPU
CPUtime
time
== Seconds
Seconds
Program
Program
Program
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==Instructions
xx Seconds
Instructions xx Cycles
Cycles
Seconds
Program
Instruction
Cycle
Program
Instruction
Cycle
instr count
X
CPI
clock rate
Compiler
X
X
Instr. Set
X
X
X
X
X
Organization
Technology
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X
Performance- 11
CPI: “Average cycles per instruction”
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♦ CPI
= Instruction Count / (CPU Time * Clock Rate)
= Instruction Count / Cycles
♦ CPU Time = Cycle Time *
♦ CPU Time =
n
∑ CPI i * I i
i =1
n
∑ CPI i * Fi
i =1
♦ where
Fi =
Ii
Instruction Count
♦ Invest resources where time is spent !
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Performance- 12
Controversial Example
CPU time =
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Instruction
Cycles
Seconds
×
×
Program
Instruction
Cycle
♦ Some have argued:
■ CISC CPU Time = P x 8 x T = 8PT
■ RISC CPU Time = 2P x 2 x T = 4PT
■ RISC CPU Time = (CISC CPU Time)/2
♦ DISCLAIMER:
■ The truth is much, much more complex
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Amdahl's Law
Speedup due to enhancement E:
ExTime w/o E
Speedup(E) = -------------------- =
ExTime w/ E
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Performance 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
then,
ExTime(with E) = ((1-F) + F/S) X ExTime(without E)
Speedup(with E) =
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1
(1-F) + F/S
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Amdahl’s Law
♦ Let
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Speedup =
new rate old latency
=
old rate new latency
♦ Consider an enhancement x that speedups fraction fx of a
task by Sx
Speedupoverall =
=
old latency
new latency
[( 1 - f x ) + f x ] × old latency
( 1 − f x ) × old latency + (f x / S x ) × old latency
♦ Amdahl’s Law gives:
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Speedupoverall =
1
( 1 − f x ) + (f x / S x )
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Amdahl’s Law, cont.
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♦ Example: fx = 95 % and Sx = 1.10
Speedup overall =
1
= 1.094
( 1 − 0.95 ) + ( 0.95 / 110
. )
♦ Example: fx = 5% and Sx = 10
Speedupoverall =
1
= 1.047
( 1 − 0.0.5 ) + ( 0.05 / 10 )
♦ Example: fx = 5% and Sx
Speedupoverall =
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→∞
1
= 1.052
( 1 − 0.05 ) + ( 0.05 / ∞ )
Performance- 16
Amdahl’s Law Corollary
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Since Sx → ∞ implies
Speedup overall →
1
( 1 − f x ) + (f x / ∞ )
For real speedups:
Speedup overall <
1
(1 − f x )
Example:
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fx
1
(1 − f x )
1%
2%
5%
10 %
20 %
50 %
1.01
1.02
1.05
1.11
1.25
2.00
Performance- 17
Standard Example: Load/Store
Machine
Operation
ALU Ops
Loads
Stores
Branches
Frequency
43 %
21 %
12 %
24 %
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Cycle Count
1
1
2
2
♦ Suppose we could make stores execute in 1 cycle, by slowing down
the cycle time by 15 %
■ Should we make this optimization ?
♦ Old CPI = 0.43 + 0.21 + (0.12 + 0.24)x2 = 1.36
♦ New CPI = 0.43 + 0.21 + 0.12 + 0.24x2 = 1.24
New CPU time P × New CPI × 115
. T
=
= 1.05
Old CPU time
P × Old CPI × T
♦ Conclusion: Don’t make the change
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Performance- 18
Example (RISC processor)
Base Machine (Reg / Reg)
Op
Freq Cycles CPI(i)
ALU
50%
1
.5
Load
20%
5
1.0
Store
10%
3
.3
Branch
20%
2
.4
2.2
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% Time
23%
45%
14%
18%
Typical Mix
How much faster would the machine be if a better data cache
reduce the average load time to 2 cycles?
How does this compare with using branch prediction to shave a
cycle off the branch time?
What if two ALU instructions could be executed at once?
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Performance- 19
Evaluating Instruction Sets?
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Design-time metrics:
° Can it be implemented, in how long, at what cost?
° Can it be programmed? Ease of compilation?
Static Metrics:
° How many bytes does the program occupy in memory?
Dynamic Metrics:
° How many instructions are executed?
° How many bytes does the processor fetch to execute the program?
CPI
° How many clocks are required per instruction?
° How "lean" a clock is practical?
Best Metric: Time to execute the program!
Inst. Count
Cycle Time
NOTE: this depends on instructions set, processor organization, and
techniques.
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Performance- 20
Corollary: Make The Common Case
Fast
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♦ All instructions require an instruction fetch, only a fraction
require a data fetch/store.
⇒ Optimize instructions access over data access
♦ Programs exhibit locality
spatial locality
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temporal locality
Performance- 21
Corollary: Make The Common Case
Fast
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♦ Access to small memories is faster
⇒ provide a storage hierarchy such that the most frequent accesses
are the smallest (closest) memories
Memory
Regs.
Disk/Tape
Cache
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Performance- 22
Marketing Metrics
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♦ Clock Frequency
■ 3 Ghz better than 2 Ghz?
♦ Only relevant for comparing processors from the same family
■ The same architecture
■ The same ISA
♦ Machine with different instruction sets ?
■ Intel Pentium vs PowerPC
♦ Program with different instruction mixes ?
♦ Dynamic frequency of instructions
♦ Uncorrelated to performance
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Performance- 23
Marketing Metrics
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♦ MIPS= instruction Count /Time * 106
= Clock Rate / CPI * 106
■
■
■
■
machine with different instruction sets ?
program with different instruction mixes ?
dynamic frequency of instruction
uncorrelated to performance
♦ MFLOPS = FP Operations / Time * 106
■ machine dependent
■ often not where time is spent
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Normalized:
add, sub, compare, mult
divide, sqrt
exp, sin, ...
1
4
8
Performance- 24
Normalized MFLOPS
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♦ Not all machines implement the same FP
operations
■ Cray-1 does not implement Divide
■ Motorola 68882 does SQRT, SIN, and COS
♦ Not all FP operations are the same
■ ADD is much faster than Divide
♦ Normalized MFLOPS
■ Assign a “canonical number of FP operations” to a
program
Normalized MFLOPS =
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Canonical FP operations
time × 10 6
Performance- 25
Metrics of performance
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Answers per month
Application
Useful Operations per second
Programming
Language
Compiler
ISA
(millions) of Instructions per second – MIPS
(millions) of (F.P.) operations per second – MFLOP/s
Datapath
Control
Megabytes per second
Function Units
Transistors Wires Pins
Cycles per second (clock rate)
Each metric has a place and a purpose, and each can be misused
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Performance- 26
Benchmarks
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♦ Real Programs
■ Representative of real workload
■ The only accurate way to characterize performance
■ e.g., gcc, spice, ...
♦ Kernels
■ “Representative” program fragments
■ Time critical excerpts of real programs.
■ e.g., Livermore loops
♦ Toy Benchmarks
■ 10-100 lines
■ e. g. Sieve, Puzzle, Towers
♦ Synthetic Benchmarks
■ attempt to match average frequencies of real workloads
■ e.g. Whetstone, dhrystone
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Benchmarking
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♦ Reproducible results
■ must control outside factors
♦ Important factors
■
■
■
■
■
■
■
■
■
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Program input
Version of program
Version of compiler
Optimization level
Version of operating system
Amount of memory
Number and type of disks
Version of CPU
Cache configuration
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Benchmarking: SPEC
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♦ Limitations of de facto Benchmarks
♦ Dhrystone
■ Synthetic integer benchmark
■ Heavy string emphasis
■ Optimization compilers cause MAJOR
problems
♦ Whetsone
■ Synthetic floating-point benchmark
■ Designed to thwart optimization
♦ Linpack
■ Floating-point kernel
■ DAXPY() = A(I) = B(I) + C * D(I)
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SPEC95
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♦ Standard Performance Evaluation Corporation
♦ Eighteen application benchmarks (with inputs) reflecting a
technical computing workload
♦ Eight integer
■ go, m88ksim, gcc, compress, li, ijpeg, perl, vortex
♦ Ten floating-point intensive
■ tomcatv, swim, su2cor, hydro2d, mgrid, applu, turb3d, apsi, fppp,
wave5
♦ Must run with standard compiler flags
■ eliminate special undocumented incantations that may not even
generate working code for real programs
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Benchmarking: SPEC200
Integer
Gzip
Vpr
Gcc
Mcf
Crafty
Parser
Eon
Perlbmk
Gap
Vortex
Bzip2
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Floating Point
Compression
FPGA circuit placement
and routing
C programming language
compiler
Combinatorial
optimization
Game playing: chess
Word processing
Computer visualization
Perl programming
language
Group theory
Object oriented database
Compression
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Wupwise
Swim
Mgrid
Applu
Mesa
Galgel
Art
Equaqke
Facerec
Ammp
Lucas
Fma3d
Sixtrack
Apsi
Physics: quantum chromadinamics
Shallow water modelling
Multigrid solver: 3D potential field
Partial differential equations
3D Graphics library
Computational fluid dynamics
Image recognition neural networks
Seismic wave propagation
simulation
Image processing: face
recognition
Computational chemistry
Number theory/primality testing
Finite-element crash simulation
Nuclear physics accelerator design
Meteorology: pollutant distribution
Performance- 31
Summarizing Results: A CounterExample
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A car goes 30 MPH for the first then miles and 90 MPH for the
second ten miles. What the car’s average speed over the twenty
miles?
Wrong answer:
Avg Speed =
30 MPH + 90 MPH
= 60 MPH
2
Correct answer:
Avg Speed =
=
=
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total distance
total time
10 miles + 10 miles
(10 miles / 30 MPH) + (10 miles / 90 MPH)
20 miles
= 45 MPH
(1 / 3) hour + (1 / 9 ) hour
Performance- 32
Summarizing Results: Averages
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Use the ARITHMETIC mean for times (cycles per instruction):
1 n
∑ timei
n i =1
Use the HARMONIC mean for rates (MIPS, MFLOPS):
⎛ 1 n 1 ⎞−1
⎜ ∑
⎟
⎝ n i =1 ratei ⎠
Use the GEOMETRIC mean for ratios (normalized numbers):
⎛ 1 n 1 ⎞1 / n
⎜ ∏
⎟
⎝ n i =1 ratei ⎠
Better yet: don’t average normalized numbers
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Performance- 33
Summarizing Results: A Measure of
Time
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♦ Property 1:
A single-number performance measure for a set of
benchmarks expressed in units of time should be directly
proportional to the total (weighted) time consumed by the
benchmarks.
♦ Property 2:
A single-number performance measure for a set of
benchmarks expressed as a rate should be indirectly
proportional to the total (weighted) time consumed by the
benchmarks.
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Performance- 34
Summarizing Results: Which Mean ?
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Ti = Execution time for Benchmark i
Fi = FP Operations for Benchmark i
Ri = Fi / Ti = Rate of Benchmark i
Average Time:
A − mean =
1 n
∑T
n i =1 i
A − mean =
1 n
∑R
n i =1 i
Average Rate:
Violates Property 2: Not proportional to inverse of time. Use Harmonic
mean:
⎛ 1 n 1 ⎞−1 ⎛ 1 n Fi ⎞−1
H − mean = ⎜ ∑ ⎟ = ⎜ ∑ ⎟
⎝ n i =1 Ri ⎠
⎝ n i =1 Ti ⎠
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Performance- 35
Homework 2
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♦ Choose a program to evaluate performance of a PC
■ It can be for Linux or Windows
♦ Choose performance metrics for:
■ Speed of CPU
■ Speed of Main memory
■ Speed of graphics applications
■ Speed of hard disk
♦ Run performance program in two different computers
■ Your assigned PC at the lab
■ Your home computer
♦ Compare results for two computers and stand if one is faster than the
other according each metrics
♦ Three pages long report
■ Describe the performance program (one page)
■ Describe performance tests and metrics (one page)
■ Describe characteristics of both computers, compare results and make
conclusions (one page)
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Homework 2
Computer A
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Computer B
Comparison
CPU speed
Memory speed
Graphics speed
HD speed
Due date: September 19th, 2008.
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Performance- 37