Transcript Introduction and Five Components of a Computer

Computer Architecture
Chapter 2
The Role of Performance
Lec2.1
Introduction
° This chapter discusses how to measure, report, and summarize
performance and describes the major factor that determine the
performance of a computer
° A primary reason for examining performance is that hardware
performance is often key to the effectiveness of an entire
system of hardware and software
° Assessing the performance of such a system can be quite
challenging
° Key to understanding underlying organizational motivation
Why is some hardware better than others for different programs?
What factors of system performance are hardware related?
(e.g., Do we need a new machine, or a new operating system?)
How does the machine's instruction set affect performance?
° For example, to improve the performance of a software system, we
may need to understand what factors in the hardware contribute to
the overall performance and the relative importance of these factors
Lec2.2
Which of these airplanes has the best performance?
Airplane
Passengers
Boeing 737-100
Boeing 747
BAD/Sud Concorde
Douglas DC-8-50
101
470
132
146
Range (mi) Speed (mph)
630
4150
4000
8720
598
610
1350
544
°How much faster is the Concorde compared to the 747?
°How much bigger is the 747 than the Douglas DC-8?
Lec2.3
Computer Performance
° Response Time (latency): the time between the start and
completion of a task
— How long does it take for my job to run?
— How long does it take to execute a job?
— How long must I wait for the database query?
° Throughput: the total amount of work done in a given time
— How many jobs can the machine run at once?
— What is the average execution rate?
— How much work is getting done?
° If we upgrade a machine with a new processor what do we increase?
If we add a new machine to the lab what do we increase?
Lec2.4
Two notions of “performance”
Plane
DC to Paris
Speed
Passengers
Throughput
(pmph)
Boeing 747
6.5 hours
610 mph
470
286,700
BAD/Sud
Concorde
3 hours
1350 mph
132
178,200
Which has higher performance?
Lec2.5
Example
• 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 = 0.62 “times faster”
• Boeing is 286,700 pmph / 178,200 pmph = 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
Lec2.6
Execution Time
° Elapsed Time
• counts everything (disk and memory accesses, I/O, etc.)
• a useful number, but often not good for comparison purposes
° CPU time
• doesn't count I/O or time spent running other programs
• can be broken up into system time, and user time
° Our focus: user CPU time
• time spent executing the lines of code that are "in" our
program
Lec2.7
Book's Definition of Performance
° For some program running on machine X,
PerformanceX = 1 / Execution timeX
° "X is n times faster than Y"
PerformanceX / PerformanceY = n
° Problem:
• machine A runs a program in 20 seconds
• machine B runs the same program in 25 seconds
Lec2.8
Clock Cycles
° Almost all computers are constructed using a clock that runs at
a constant rate and determines when events take place in the
hardware
° These discrete time intervals are called clock cycles (or clock
ticks, clock periods, clocks, cycles)
time
° Instead of reporting execution time in seconds, we often use
cycles
° cycle time = time between ticks = seconds per cycle
° clock rate (frequency) = cycles per second (1 Hz. = 1 cycle/sec)
A 200 Mhz. clock has a
time
1
200  10 6
 10 9  5 nanoseconds
cycle
Lec2.9
Now that we understand cycles
° A given program will require
•
some number of instructions (machine instructions)
•
some number of cycles
•
some number of seconds
° We have a vocabulary that relates these quantities:
•
cycle time (seconds per cycle)
•
clock rate (cycles per second)
•
CPI (cycles per instruction)
a floating point intensive application might have a higher CPI
•
MIPS (millions of instructions per second)
this would be higher for a program using simple instructions
Lec2.10
CPI
“Average cycles per instruction”
CPI = (CPU Time * Clock Rate) / Instruction Count
= Clock Cycles / Instruction Count
CPU clock cycles =
 CPI i
*
C
i
Lec2.11
CPI Example
° Suppose we have two implementations of the same
instruction set
architecture (ISA).
For some program,
Machine A has a clock cycle time of 10 ns. and a CPI of 2.0
Machine B has a clock cycle time of 20 ns. and a CPI of 1.2
What machine is faster for this program, and by how much?
° If two machines have the same ISA which of our quantities (e.g.,
clock rate, CPI, execution time, # of instructions, MIPS) will always
be identical?
Lec2.12
# of Instructions Example
° A compiler designer is trying to decide between two code
sequences for a particular machine. Based on the hardware
implementation, there are three different classes of
instructions: Class A, Class B, and Class C, and they
require one, two, and three cycles (respectively).
The first code sequence has 5 instructions: 2 of A, 1 of B,
and 2 of C
The second sequence has 6 instructions: 4 of A, 1 of B, and
1 of C.
Which sequence will be faster? How much?
What is the CPI for each sequence?
Lec2.13
MIPS example
6
° MIPS = Instruction count/(Execution time * 10 )
° Two different compilers are being tested for a 100 MHz.
machine with three different classes of instructions: Class A,
Class B, and Class C, which require one, two, and three cycles
(respectively). Both compilers are used to produce code for a
large piece of software.
The first compiler's code uses 5 million Class A instructions, 1
million Class B instructions, and 1 million Class C instructions.
The second compiler's code uses 10 million Class A
instructions, 1 million Class B instructions, and 1 million Class
C instructions.
° Which sequence will be faster according to MIPS?
° Which sequence will be faster according to execution time?
Lec2.14
Metrics of performance
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)
Lec2.15
Benchmarks
° Performance best determined by running a real application
•
Use programs typical of expected workload
•
Or, typical of expected class of applications
e.g., compilers/editors, scientific applications, graphics, etc.
° Small benchmarks
•
nice for architects and designers
•
easy to standardize
•
can be abused
° SPEC (System Performance Evaluation Cooperative)
•
companies have agreed on a set of real program and inputs
•
can still be abused (Intel’s “other” bug)
•
valuable indicator of performance (and compiler technology)
Lec2.16
SPEC ‘89
° Compiler “enhancements” and performance
800
700
SPEC performance ratio
600
500
400
300
200
100
0
gcc
espresso
spice
doduc
nasa7
li
eqntott
matrix300
fpppp
tomcatv
Benchmark
Compiler
Enhanced compiler
Lec2.17
SPEC95
° 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
Lec2.18
SPEC ‘95
Benchmark
go
m88ksim
gcc
compress
li
ijpeg
perl
vortex
tomcatv
swim
su2cor
hydro2d
mgrid
applu
trub3d
apsi
fpppp
wave5
Description
Artificial intelligence; plays the game of Go
Motorola 88k chip simulator; runs test program
The Gnu C compiler generating SPARC code
Compresses and decompresses file in memory
Lisp interpreter
Graphic compression and decompression
Manipulates strings and prime numbers in the special-purpose programming language Perl
A database program
A mesh generation program
Shallow water model with 513 x 513 grid
quantum physics; Monte Carlo simulation
Astrophysics; Hydrodynamic Naiver Stokes equations
Multigrid solver in 3-D potential field
Parabolic/elliptic partial differential equations
Simulates isotropic, homogeneous turbulence in a cube
Solves problems regarding temperature, wind velocity, and distribution of pollutant
Quantum chemistry
Plasma physics; electromagnetic particle simulation
Lec2.19
SPEC ‘95
Does doubling the clock rate double the performance?
10
10
9
9
8
8
7
7
6
6
SPECfp
SPECint
Can a machine with a slower clock rate have better performance?
5
5
4
4
3
3
2
2
1
1
0
0
50
100
150
Clock rate (MHz)
200
250
Pentium
Pentium Pro
50
100
150
Clock rate (MHz)
200
Pentium
Pentium Pro
Lec2.20
250
Amdahl's Law
Execution Time After Improvement =
Execution Time Unaffected +( Execution Time Affected / Amount of
Improvement )
° Example:
"Suppose a program runs in 100 seconds on a machine, with
multiply responsible for 80 seconds of this time. How much do we
have to improve the speed of multiplication if we want the program
to run 4 times faster?"
How about making it 5 times faster?
Lec2.21