1 CSE141 - Carro Measuring and Discussing Computer Performance 2 CSE141 - Carro Before that, some announcements! Tas office hours: Ryan: M W , 2:15-3:15, R3349A Jessica: Tu Th, 1:00-2:00 Fritz: F , 1:30-3:30 Misaki (tutor): to be defined There are 15 seats available in 141, and 21 in 141L; I will accept the remaining 8 students (total of 108). Now you should: drop from the waiting list; add into the class. Welcome! Web page on air, please look at it frequently 3 CSE141 - Carro Measure, Report, and Summarize Make intelligent choices See through the marketing hype 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? Performance
Transcript
1CSE141 - Carro
Measuring and DiscussingComputer Performance
2CSE141 - Carro
Before that, some announcements!
Tas office hours:Ryan: M W , 2:15-3:15, R3349AJessica: Tu Th, 1:00-2:00Fritz: F , 1:30-3:30Misaki (tutor): to be defined
There are 15 seats available in 141, and 21 in 141L;I will accept the remaining 8 students (total of 108).Now you should:
drop from the waiting list;add into the class.
Welcome!
Web page on air, please look at it frequently
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Measure, Report, and SummarizeMake intelligent choicesSee through the marketing hypeKey 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?
Performance
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Which of these airplanes has the best performance?
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?
Computer Performance: TIME, TIME, TIME
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Elapsed Timecounts everything (disk and memory accesses, I/O , etc.)a useful number, but often not good for comparison purposes
CPU timedoesn’t count I/O or time spent running other programscan be broken up into system time, and user time
Our focus: user CPU timetime spent executing the lines of code that are "in" our program
Execution Time
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Relative Performance
can be confusingA runs in 12 secondsB runs in 20 seconds
A/B = .6 , so A is 40% faster, or 1.4X faster, or Bis 40% slowerB/A = 1.67, so A is 67% faster, or 1.67X faster, orB is 67% slower
needs a precise definition
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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 secondsmachine B runs the same program in 25 seconds
Book’s Definition of Performance
PerformanceA/PerformanceB=ExecutionTimeB/ExecutionTimeA= n = 20/12 = 1.67
A is 1.67X faster than B!
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Clock Cycles
Instead of reporting execution time in seconds, we often use cycles
cycle time = time between ticks = seconds per cycleclock rate (frequency) = cycles per second (1 Hz = 1 cycle/sec)
A 200 Mhz clock has a cycle time
time
seconds
program=
cycles
program
seconds
cycle
1
200 106 109 = 5 nanoseconds
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To make it clear:
A certain program needs 6000 cycles, and the clock is running at200MHz. How long will it take to complete the program?
Time for program = cycles the program needs*second/cycle = 6000*1/200e6=30s!
Piece of cake!
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So, to improve performance (everything else being equal) you can either
________ the # of required cycles for a program, or
________ the clock cycle time or, said another way,
________ the clock rate.
How to Improve Performance
seconds
program=
cycles
program
seconds
cycle
decrease
decrease
increase
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Could assume that # of cycles = # of instructions
This assumption is incorrect,
different instructions take different amounts of time on different machines.
Why?
time
1st i
nstr
uctio
n
2nd
inst
ruct
ion
3rd
inst
ruct
ion
4th
5th
6th ...
How many cycles are required for a program?
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Multiplication takes more time than addition
Floating point operations take longer than integer ones
Accessing memory takes more time than accessing registers
Important point: changing the cycle time often changes the number ofcycles required for various instructions (more later)
time
Different numbers of cycles for different instructions
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Brief review: what is time
CPU Execution Time = CPU clock cycles * Clock cycle time
Every conventional processor has a clock with an associated clockcycle time or clock rateEvery program runs in an integral number of clock cycles
MHz = millions of cycles/secondX MHz = 1000/X nanoseconds cycle time
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How many clock cycles?
Number of CPU cycles = Instructions executed *Average Clock Cycles per Instruction (CPI)
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All Together Now
CPU ExecutionTime
InstructionCount
CPI Clock CycleTime= X X
instructionscycles/instruction seconds/cycle
seconds
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Our favorite program runs in 10 seconds on computer A, which has a400 Mhz. clock. We are trying to help a computer designer build a newmachine B, that will run this program in 6 seconds. The designer can usenew (or perhaps more expensive) technology to substantially increase theclock rate, but has informed us that this increase will affect the rest of theCPU design, causing machine B to require 1.2 times as many clock cycles asmachine A for the same program. What clock rate should we tell thedesigner to target?"
Don’t Panic, can easily work this out from basic principles
10s = X * 1/400MHz; program takes 4000X106 cyclesB takes 1.2*4000X106 = 4800X106 cyclessince B requires 6s, 4800X106 cycles/6s = 800X106 cycles/s,
OR: 800MHz! See that we most double the clock frequency so that Bis 10/6=1.67 faster than A.
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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
Now that we understand cycles
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Performance
Performance is determined by execution timeDo any of the other variables equal performance?
# of cycles to execute program?# of instructions in program?# of cycles per second?average # of cycles per instruction?average # of instructions per second?
Common pitfall: thinking one of the variables is indicative of
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Suppose we have two implementations of the same instruction setarchitecture (ISA).
For some program,
Machine A has a clock cycle time of 10 ns. and a CPI of 2.0Machine 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?
same programs,different machines,same ISASame programs,different machines
different same
differentdifferent
differentdifferent
same
Somewhat
different
similar
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A compiler designer is trying to decide between two code sequencesfor 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 CThe 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?
# of Instructions Example
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Solution
Instructionclass
CPI for theinstruction class
A 1
B 2
C 3
Codesequence
A B C
1 2 1 2
2 4 1 1
CPU_CK_CYCLES1 = 2*1 + 1*2 + 2*3 = 10 cycles
CPU_CK_CYCLES2 = 4*1 + 1*2 + 1*3 = 9 cycles
Instruction count for sequences
=5 inst
=6 inst
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Two different compilers are being tested for a 500 MHz. machine withthree different classes of instructions: Class A, Class B, and ClassC, which require one, two, and three cycles (respectively). Bothcompilers are used to produce code for a large piece of software.
The first compiler’s code uses 5 million Class A instructions, 1million Class B instructions, and 1 million Class C instructions.
The second compiler’s code uses 10 million Class A instructions, 1million 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?
MIPS example
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MIPS example II
Code from A B C
Compiler 1 5 1 1
Compiler 2 10 1 1
Instruction count (billions) for instruction class
go Artificial intelligence; plays the game of Gom88ksim Motorola 88k chip simulator; runs test programgcc The Gnu C compiler generating SPARC codecompress Compresses and decompresses file in memoryli Lisp interpreterijpeg Graphic compression and decompressionperl Manipulates strings and prime numbers in the special-purpose programming language Perlvortex A database programtomcatv A mesh generation programswim Shallow water model with 513 x 513 gridsu2cor quantum physics; Monte Carlo simulationhydro2d Astrophysics; Hydrodynamic Naiver Stokes equationsmgrid Multigrid solver in 3-D potential fieldapplu Parabolic/elliptic partial differential equationstrub3d Simulates isotropic, homogeneous turbulence in a cubeapsi Solves problems regarding temperature, wind velocity, and distribution of pollutantfpppp Quantum chemistrywave5 Plasma physics; electromagnetic particle simulation
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Does doubling the clock rate double the performance?
Can a machine with a slower clock rate have better performance?
Clock rate (MHz)
SP
EC
int
2
0
4
6
8
3
1
5
7
9
10
200 25015010050
Pentium
Pentium Pro
PentiumClock rate (MHz)
SP
EC
fp
Pentium Pro
2
0
4
6
8
3
1
5
7
9
10
200 25015010050
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Execution Time After Improvement =
Execution Time Unaffected +( Execution Time Affected / Amount of Improvement )
Example:
"Suppose a program runs in 100 seconds on a machine, withmultiply responsible for 80 seconds of this time. How much do we have toimprove the speed of multiplication if we want the program to run 4 timesfaster?"
Amdahl’s Law (or: common sense as math)
100s/4 = (100-80) + 80/x
X=80/5=16, or multiplication should improve by a factor of 5
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Amdahl’s Law II
How about making the program run 5 times faster?
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Suppose we enhance a machine making all floating-point instructions runfive times faster. If the execution time of some benchmark before thefloating-point enhancement is 10 seconds, what will the speedup be if half ofthe 10 seconds is spent executing floating-point instructions?
We are looking for a benchmark to show off the new floating-point unitdescribed above, and want the overall benchmark to show a speedup of 3.One benchmark we are considering runs for 100 seconds with the oldfloating-point hardware. How much of the execution time would floating-point instructions have to account for in this program in order to yield ourdesired speedup on this benchmark?
Example
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Performance is specific to a particular program/s
Total execution time is a consistent summary of performance
For a given architecture performance increases come from:
increases in clock rate (without adverse CPI affects)improvements in processor organization that lower CPIcompiler enhancements that lower CPI and/or instruction count
