1. COMS 361 Computer Organization
Title: Performance
Date: 9/06/2004
Lecture Number: 5

2. Announcements
Homework 2
Due 9/14/04 (1 week)

3. Review Performance Know the criteria for measuring We choose
Execution time as our measure

4. Review System clock Clock rate (cycles/second)
Clock period (seconds/cycle)
Common clock rate:
1 Giga Hz

5. Outline Orders of magnitude Performance MIPS, MOPS, FLOPS
The big and small of it
Performance
Relationship between
Execution time and clock rate
Execution time and the number of instructions
Instruction and number of clocks
MIPS, MOPS, FLOPS
SPEC Benchmarks

6. Orders of Magnitude The small ones The BIG ones thousandths 10-3
milli, m
millionths
10-6
micro, u
billionths
10-9
nano, n
trillionths
10-12
pico, p
thousands
103
kilo, K
millions
106
mega, M
billions
109
giga, G
trillions
1012
tera, T

7. Execution time vs clock/rate
Relation between execution time and clock rate
Tells what happens to execution time as clock rate changes
Increase performance
Reduce the number of clocks cycles to execute the program
Reduce the clock period (increase the clock rate)

8. Example Don't Panic, can easily work this out from basic principles
Our favorite program runs in 10 seconds on computer A, which has a 400 MHz. clock. We are trying to help a computer designer build a new machine B, that will run this program in 6 seconds. The designer can use new (or perhaps more expensive) technology to substantially increase the clock rate, but has informed us that this increase will affect the rest of the CPU design, causing machine B to require 1.2 times as many clock cycles as machine A for the same program. What clock rate should we tell the designer to target?
Don't Panic, can easily work this out from basic principles

9. Solution What are we given? What do we want to determine?
CPU time(A) = 10sec
Clock Rate(A) = 400MHz
CPU time(B) = 6sec
Clock cycles(B) = 1.2 * clock cycles(A)
What do we want to determine?
The clock rate(B)
What do we need an equation to relate?
CPU time
Clock rate

10. Solution Rearrange to solve for the unknown
We need a relationship between machine A and B

11. Solution Plug in the known values
Clock rate of machine B must be twice that of machine A
Example showed
Giving machine B a faster clock required 20% more clock cycles to execute the same program
Perhaps faster instructions do less work, more are needed

12. Execution time vs # of instructions
Execution time is proportional to the number of clocks required by the program
The number of clocks to execute a program is proportional to the number of instructions executed

13. Execution time vs # of instructions
Various instructions may take different numbers of clock cycles to execute
lw or sw compared to an add
Memory instructions vs on processor instructions
Consider the average number of clocks per instruction

14. Execution time vs # of instructions
CPI is the average number of clocks per instruction
Useful in comparing different implementations of the same ISA
Checking units

15. Execution time vs # of instructions
Same equation in terms of clock rate instead of clock period
checking units

16. Execution time vs # of instructions
Execution time in terms of three parameters:
Number of instructions per program
Number of clocks per instruction
Number of clocks per second

17. Execution time vs # of instructions
Execution time decreases (performance increases)
Reduce the number of instructions per program
Compiler writer or programmer has some control here
Make instructions that perform many operations simultaneously
Reduce CPI
Make instructions simple so they do not require many clocks
May make the number of instructions in the program go up
Increase the clock rate
May increase CPI unless instructions are simple
May increase the number of instructions per program

18. Typical values CPU execution time Clock rate Instruction Count CPI
Determined (measured) for different programs
msec, sec, min
Clock rate
Specified by the manufacturer
500MHz, 1GHz, 2.66MHz
Instruction Count
Depends on the compiler and program
Can be determined
CPI
Usually not available
Varies from one ISA to another

19. Typical values CPI Look at CPI for different instruction classes
Let n be the number of instruction classes
C(i) be the number of instructions in class i
CPI(i) be the number of clocks per instruction for class I
CPI depends on the number of cycles for each instruction class and the percentage of instructions from each class

20. Example Given Which sequence executes the fastest?
CPI(A) = 1, CPI(B) = 2, CPI(C) = 3
Two code sequences (programs) with the following distribution of instructions
1) Num(A) = 2, Num(B) = 1, Num(C) = 2
2) Num(A) = 4, Num(B) = 1, Num(C) = 1
More instructions of class A, less instructions from class C
Which sequence executes the fastest?
5 instructions for seq. 1 vs 6 instructions for seq. 2

21. Example How many clock cycles are required for each sequence?
CPU clock cycles(1) = 2*1 + 1*2 + 2*3 = 10 cycles
CPU clock cycles(2) = 4*1 + 1*2 + 4*3 = 9 cycles
Sequence 2 executes faster than sequence 1 even though there are more instructions in sequence 2!

22. Example What are the CPI’s for the 2 sequences?
CPI(2) < CPI(1) makes sense
Sequence 2 has more instructions to execute
Requires less clock cycles