ch4a- 2 ee/cs/cpe 3760 - computer organization seattle pacific university performance what...

Ch4a- 2EE/CS/CPE 3760 - Computer OrganizationSeattle Pacific University

Performance

• What differences do we see in performance?

• Almost all computers operate correctly (within reason)

• Most computers implement useful operations• This is a matter of taste...

• Computers all operate at different speeds

• Speed is the most important performance metric

• The entire point of computer hardware is to “perform”

• Operate correctly

• Implement useful operations

• Do so as fast as possible


Measuring speed

• Raw speed

• Ferrari wins

• Which is faster?

• School Bus: 57 MPH, 40 people

• Ferrari: 170 MPH, 2 people

• Throughput

• Ferrari: 340 passenger-MPH

• School Bus: 2280 passenger-MPH

• Other issues...

• Range, reliability, cost


Performance of computers

• How long does it take to run my favorite program?

• To compare two computers, we compare the execution time of the same program on the two computers

• Faster one wins

• Lower execution time is better

• Batch throughput• Response time


A little background...

• The compiler converts this code into machine-language instructions

• Computer programs are (usually) written in a high-level language (e.g. C)

• The performance of a program depends on:• The number and types of instructions executed• How fast the CPU can execute those instructions

• The CPU interprets machine-language instructions and executes them


Tick-tock• Almost all modern computers are based on a clock

Period

• All events are controlled by and synchronized to a regular clock

• Clocks are just regular periodic waveforms

• Cycle time: time for the waveform to repeat itself

• Also known as the clock period

• Frequency: 1/Period

• Example:• 10ns clock cycle --> period = 10-8 s• Frequency 1/10ns = 1/10-8 s = 108 cycles/sec


Execution time

• Performance can be improved by:

• Decreasing the cycle time• Hardware solution: Use faster technology

• Decreasing the number of cycles for the program• Software: Write a better program• Hardware: Re-design CPU

• Time = cycles * cycle time

• Time = cycles / clock frequency

• Since the cycle time of a computer is constant, we can express time in terms of CPU cycles


Instruction execution time• Every instruction takes time to execute

• Some instructions may take more or less time than others

• The time for an instruction is expressed in terms of clock cycles

Instruction CyclesADD 1MULT 4CMP 1SUB 2

Example:

• The time to run a program depends on:

• How many instructions

• What type of instructions

• 30 ADDs and 4 MULTs --> 46 cycles


Average CPI

• The Cycles-Per-Instruction (CPI) varies depending on what instructions are used

• Take an Average CPI

• Cycles = Number of Instructions * Average CPI

• Average CPI should reflect the mix of instructions in the program

• A large proportion of 4-cycle MULTs should raise the CPI, a large proportion of 1-cycle ADDs should lower it

• The average should be the weighted average

Ch4a- 10

EE/CS/CPE 3760 - Computer OrganizationSeattle Pacific University

Weighting the average

Instruction Cycles %ADD 1 40MULT 4 10CMP 1 20SUB 2 30

Average CPI = 1 * 40% + 4 * 10% + 1 * 20% + 2 * 30% = .4 + .4 + .2 + .6 = 1.6

Average CPI = 1 * 40% + 4 * 10% + 1 * 20% + 2 * 30% = .4 + .4 + .2 + .6 = 1.6

Notice: The average CPI depends on the code we’re executing!Notice: The average CPI depends on the code we’re executing!

Example mix of instructions

Example mix of instructions

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How long?

• Remember, lower is better

• Reducing any one of the three components reduces execution time

• Execution time = Cycles * Cycle Time

• Cycles = Average CPI * Instruction Count

• Execution time = Instruction Count * CPI * Cycle Time

• Cycle time - Reduced through technology change, change in CPU design

• CPI - Reduced through better code, better compiler, change in CPU design

• Instruction count - Reduced through better code, better compiler, change in CPU design

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Examples

Time = CPI x Period x Instructions = 1.10 x 22ns x 4 x 108 = 9.68s

System A: 400,000,000 instr., 22ns clock and a CPI of 1.10.How long does it take to run the program on system A?

System B: 10s to run a program, 20ns clock, 400,000,000 instr.What is the CPI? CPI = Time / (Period x InstrC) = 10s / (20 x 10-9 x 4 x 108) = 1.25

System C: 3,000,000,000 instr., 2GHz clock and a CPI of 1.25.How long does it take to run the program on system C?

Period = 1/Frequency = 1/2GHz = 0.5ns = 0.5 x 10-9

Time = CPI x Period x Instructions = 1.25 x 0.5 x 10-9 x 3 x 109 = 1.875s

Ch4a- 13


ExamplesAssume an add takes 1 cycle, a mult 4 cycles, and a sub 2 cycles

Two different compilers produce the following loops for the same code:

addaddmultsubaddadd

multaddmultsub

A: B:

loop1000000times

loop1000000times

What’s the CPI?CPIA = (4 + 1 + 4 + 2)/4 = 2.75

CPIB = (1 + 1 + 4 + 2 + 1 + 1)/6 = 1.667How long does it take to run each program on a 200MHz CPU?

TimeA = CPIA x PeriodA x InstructionsA = 2.75 x 5ns x 4000000 = .0055sTimeB = CPIB x PeriodB x InstructionsB = 1.667 x 5ns x 6000000 = .0050s

ch4a- 2 ee/cs/cpe 3760 - computer organization seattle pacific university performance what...

Documents

cycle time time

cycle time hardware

number of instructions

wins lower execution

redesign cpu time

average cpi cycles

cycles clock frequency

number of cycles