1. CpE 442 Introduction to Computer Architecture The Role of Performance
Instructor: H. H. Ammar

2. Overview of Today’s Lecture: The Role of Performance
Review from Last Lecture
Definition and Measures of Performance
Summarizing Performance and Performance Pitfalls

3. Review: What is "Computer Architecture"
° Co-ordination of levels of abstraction
Application
Operating
Compiler
System
Instruction Set
Architecture
Instr. Set Proc.
I/O system
Digital Design
Circuit Design
° Under a set of rapidly changing Forces

4. Review: Levels of Representation
temp = v[k];
v[k] = v[k+1];
v[k+1] = temp;
High Level Language Program
Compiler
lw $15, 0($2)
lw $16, 4($2)
sw $16, 0($2)
sw $15, 4($2)
Assembly Language Program
Assembler
Machine Language Program
Machine Interpretation
Control Signal Specification

5. Review: Levels of Organization
SPARCstation 20
Computer
SPARC
Processor
Memory
Devices
Control
Input
That is, any computer, no matter how primitive or advance, can be divided into five parts:
1. The input devices bring the data from the outside world into the computer.
2. These data are kept in the computer’s memory until ...
3. The datapath request and process them.
4. The operation of the datapath is controlled by the computer’s controller.
All the work done by the computer will NOT do us any good unless we can get the data back to the outside world.
5. Getting the data back to the outside world is the job of the output devices.
The most COMMON way to connect these 5 components together is to use a network of busses.
Datapath
Output

6. Review: Summary from Last Lecture
All computers consist of five components
Processor: (1) datapath and (2) control
(3) Memory
(4) Input devices and (5) Output devices
Not all “memory” are created equally
Cache: fast (expensive) memory are placed closer to the processor
Main memory: less expensive memory--we can have more
Input and output (I/O) devices has the messiest organization
Wide range of speed: graphics vs. keyboard
Wide range of requirements: speed, standard, cost ... etc.
Least amount of research (so far)
Let me summarize what I have said so far.
The most important thing I want you to remember is that: all computers, no matter how complicated or expensive, can be divided into five components:
(1) The datapath and (2) control that make up the processor.
(3) The memory system that supplies data to the processor.
And last but not least, the (4) input and (5) output devices that get data in and out of the computer.
One thing about memory is that Not all “memory” are created equally.
Some memory are faster but more expensive and we place them closer to the processor and call them “cache.”
The main memory can be slower than the cache so we usually use less expensive parts so we can have more of them.
Finally as you can see from the last few slides, the input and output devices usually has the messiest organization. There are several reasons for it:
(1) First of all, I/O devices can have a wide range of speed.
(2) Then I/O devices also have a wide range of requirements.
(s) Finally to make matters worse, historically I/O has attracted the least amount of research interest. But hopefully this is changing.
In this class, you will learn about all these five components and we will try to make this as enjoyable as possible. So have fun.

7. Processor Performance

8. Metrics of performance
Answers per month
Operations per second
Application
Programming
Language
Compiler
(millions) of Instructions per second – MIPS
(millions) of (F.P.) operations per second – MFLOP/s
ISA
Datapath
Megabytes per second
Control
Function Units
Cycles per second (clock rate)
Transistors
Wires
Pins

9. Relating Processor Metrics
CPU execution time = CPU clock cycles/pgm X clock cycle time
or CPU execution time = CPU clock cycles/pgm ÷ clock rate
CPU clock cycles/pgm = Instructions/pgm X CPI the avg. clock cycles per instruction
or CPI = CPU clock cycles/pgm ÷ Instructions/pgm
CPI tells us something about the Instruction Set Architecture, the Implementation of that architecture, and the program measured

10. Aspects of CPU Performance
CPU time = Seconds = Instructions x Cycles x Seconds
Program Program Instruction Cycle
instr. count CPI clock rate
Program
Compiler
Instr. Set Arch.
Organization
Technology

11. Aspects of CPU Performance
CPU time = Seconds = Instructions x Cycles x Seconds
Program Program Instruction Cycle
instr count CPI clock rate
Program X (x)
Compiler X (x)
Instr. Set X X
Organization X X
Technology X

12. Organizational Trade-offs
Application
Programming
Language
Compiler
ISA
Instruction Mix
Datapath
CPI
Control
Function Units
Transistors
Wires
Pins
Cycle Time

13. CPI “Average cycles per instruction” "instruction frequency"
CPI = (CPU Time * Clock Rate) / Instruction Count
= Clock Cycles / Instruction Count n
CPU time = ClockCycleTime * S CPI * I i i
i = 1 n
"instruction frequency"
CPI = S CPI * F where F = I i i i i
i = 1
Instruction Count
Invest Resources where time is Spent!

14. Base Machine (Reg / Reg) Op Freq(Fi) CPI(i) % Time ALU 50% 1 .5 33%
Example
Base Machine (Reg / Reg)
Op Freq(Fi) CPI(i) % Time
ALU 50% %
Load 20% %
Store 10% %
Branch 20% %
1.5
Typical Mix
The CPI = 1.5 cycles per instruction
Assignment 1: Turn in the solution of the following problems
from the text book By Thursday September 4,
Chapter 2, Exercises Section, problems number
2.1, 2.2, 2.3, 2.4, 2.10, 2.11, 2.12, 2.13, and 2.15

15. Assume a program of 1 million instructions, Compare the performance of
Base Machine (B) with the above CPI, 1 GHZ clock, and
Enhanced Machine (E) with GHZ and a one cycle increase for L/S
And branch instructions
Enhanced Machine (Reg / Reg)
Op Freq CPI(i) % Time
ALU 50% %
Load 20% %
Store 10% %
Branch20% %
2.0

16. Perf. of machine X = 1 / exec. Time of prog on machine X
Perf. of E / Perf. of B = exec. Time of B / exec. Time of E = 1.5 * 1 / 2 * 0.75
= 1
Performance of B is similar to that of E,
No gain in performance

17. Marketing Metrics MIPS = Instruction Count / (Time * 10^6)
= Clock Rate / (CPI * 10^6)
machines with different instruction sets ?
programs with different instruction mixes ?
dynamic frequency of instructions
uncorrelated with performance
MFLOP/S= FP Operations / (Time * 10^6)
machine dependent
often not where time is spent

18. Example showing why MIPS can fail Compare performance with Compilers 1 and 2 for a given program on a given machine Instruction Count in Billion for instruction classes A B C Compiler Compiler clock cycles Clock cycles using compiler1 = 10 Billion Clock cycles using compiler2 = 15 Billion assuming 1GHZ clock CPU Time 1 = 10 secs CPU Time 2 = 15 secs yet the MIPS rating is MIPS 1 = (instr. Count/cpu time in sec x 10^6) = 700 MIPS 2 = 800

19. Why Do Benchmarks? How we evaluate differences Different systems
Changes to a single system
Provide a target
Benchmarks should represent large class of important programs
Improving benchmark performance should help many programs
For better or worse, benchmarks shape a field
Good ones accelerate progress
good target for development
Bad benchmarks hurt progress
help real programs v. sell machines/papers?
Inventions that help real programs don’t help benchmark

20. Programs to Evaluate Processor Performance
(Toy) Benchmarks
line
e.g.,: sieve, puzzle, quicksort
Synthetic Benchmarks
attempt to match average frequencies of real workloads
e.g., Whetstone, dhrystone
Kernels
Time critical excerpts Real programs
e.g., gcc, spice

21. Successful Benchmark: SPEC
EE Times + 5 companies band together to perform Systems Performance Evaluation Committee (SPEC) in 1988: Sun, MIPS, HP, Apollo, DEC
Create standard list of programs, inputs, reporting: some real programs, includes OS calls, some I/O

22. SPEC first round
First round 1989; 10 programs, single number to summarize performance
One program: 99% of time in single line of code
New front-end compiler could improve dramatically

23. SPEC second round, SPEC95
8 integer benchmarks in C and 10 floating pt benchmarks in Fortran

24. Amdahl's Law Speedup due to enhancement E:
ExTime w/o E Performance w/ E
Speedup(E) = =
ExTime 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) = ExTime(without E) ÷ ((1-F) + F/S) X ExTime(without E)
<= 1/(1-F) speed up is bounded by this factor

25. Performance Evaluation Summary
CPU time = Seconds = Instructions x Cycles x Seconds
Program Program Instruction Cycle
Time is the measure of computer performance!
Good products created when have:
Good benchmarks
Good ways to summarize performance
If not good benchmarks and summary, then choice between improving product for real programs vs. improving product to get more sales=> sales almost always wins
Remember Amdahl’s Law: Speedup is limited by unimproved part of program