Computer Architecture. Chapter 1 Part 2 Performance Measures

Transcription

1 Computer Architecture Chapter 1 Part 2 Performance Measures 1

2 Topics Designing for Performance Performance Measures 2

3 Designing for Performance (1) Support-Demand Cycle Computer Performance Demands Supports (Motivates) Application Requirement 3

4 Designing for Performance (2) Performance balance The rate at which performance is changing in the various technology areas (processor, buses, memory, peripherals) differs greatly from one type of elements to another. New applications and new peripheral devices constantly change the nature of the demand on the system. 4

5 Speeding it up (Processor) Branch prediction Data flow analysis Speculative execution 5

6 Performance Mismatch Processor speed increased Memory capacity increased Memory speed lags behind processor speed 6

7 DRAM and Processor Characteristics 7

8 Solutions Increase #bits retrieved at one time by: Make DRAM wider rather than deeper Change DRAM interface to make it more efficient by including cache on the DRAM chip. Reduce frequency of memory access More complex cache and cache on chip Increase interconnection between processors and memory to structure data: High speed buses Hierarchy of buses 8

9 Performance Measurement (1) Performance Execution time (latency): Time between the start and the completion of an event Performance 1/(Execution time) Throughput (bandwidth) Total amount of work done in a given time Machine X is n% faster than Machine Y: 9

10 Performance Measurement (2) Example: Machine A runs a program in 10 seconds, Machine B runs the same program in 15 seconds, A is % faster than B. 1 n Executiontime 100 Executiontime n 50 B A

11 Performance Measurement (3) Improve performance Increase performance Improve execution time Decrease execution time Question: Can we improve performance 10 times faster by using a 10-time-faster machine? 11

12 Amdahl s Law (1) The performance improvement to be gained from using some faster mode of execution is limited by the fraction of the time the faster mode can be used. It defines the speedup can be gained by using a particular enhancement. 12

13 Amdahl s Law (2) Speedup = Performance for entire task using the enhancement when possible Performance for entire task w/o using the enhancement = Execution time for entire task w/o using the enhancement Execution time for entire task using the enhancement when possible 13

14 Execution time new = Execution time old x Amdahl s Law (3) where f E : fraction of enhancement s E : improvement gained by the enhancement mode Speedup = 14

15 Amdahl s Law (4) Example: An enhancement run 10 times faster than the original machine, but it is usable 40% of the time, then the speedup =. 15

16 Amdahl s Law (5) Sol :f E = 0.4 s E = 10 Speedup = 1/((1-0.4) + 0.4/10) =

17 Amdahl s Law (6) Extreme Cases Speedup (1 f 1 E ) f s E E f E = 0 Speedup = 1 f E = 1 Speedup = s E 17

18 CPU Performance (1) All computers are constructed using a clock running at a constant rate CPU time is the time the CPU spends computing a program and does not include time spent waiting for I/O or running other programs Clock cycle time = 1/ clock rate 18

19 Distinct time events ticks clock ticks clocks cycles clock periods clock cycle Computer designers refer to the time of clock period by its duration (e.g. 1 ns) or by its rate (e.g. 1 GHZ) ms (milli-second) = 10 3 sec, s (micro-second) = 10 6 sec, ns (nano-second)= 10 9 sec Hz = 1/sec, KHz = 10 3 Hz, MHz = 10 6 Hz, GHz = 10 9 Hz 19

20 CPU Performance (2) CPI = clock cycle per instruction CPU time for a program = CPU clock cycles for a program x clock cycle time CPU clock cycles for a program clock rate CPI Instruction Count clock rate 20

21 A program is comprised of a number of instructions, I Measured in: instructions/program The average instruction takes a number of cycles per instruction (CPI) to be completed. Measured in: cycles/instruction, CPI CPU has a fixed clock cycle time C = 1/clock rate Measured in: seconds/cycle CPU execution time is the product of the above three parameters as follows: 21

22 CPU Performance (3) CPI x Instruction Count x 1/(clock rate) = CPU time BUT, not every instruction takes the same number of clock cycles to execute. Take the average. 22

23 CPI CPU Performance (4) n: number of different instructions in a program CPI i : CPI of instruction i f i : frequency of instruction i in a program Example: Operations frequency clock cycle ADD 60% 1 LOAD 40% 2 CPI overall = 23

24 24

25 MIPS (1) MIPS: Million Instruction Per Second MIPS 6 Instruction Count 10 Execution time Instruction Count 1 Execution time 10 clock rate 1 6 CPI

26 Advantage: Easy to understand. Disadvantages MIPS (2) The MIPS value is dependent on the instruction set used and thus any comparison between computers with different instruction sets is not valid. MIPS value varies between programs on the same computer MIPS can vary inversely to performance 26

27 Other Measurements MFLOPS: Mega (Millions) of floating point (FP) operations per second GFLOPS: Giga (Billions) FP operations per sec. Cost 27

28 Exercise If computer A has a clock rate of 200 MHz and executes a 30,000,000,000 instruction program in 2 minutes, and computer B has a clock rate of 300 MHz and executes the same program in 3 minutes, a) what are the CPIs of computers A and B? b) Assume that A and B have the same instruction set architecture. For this program, which computer is faster and by what percent? 28

29 Solution a) CPI = (execution_time x clock_rate)/instructions CPI_A = ((2 x 60) x (200 x 10^6))/(3 x 10^10)= 0.8 CPI_B = ((3 x 60) x (300 x 10^6))/(3 x 10^10)= 1.8 b) Assume that A and B have the same instruction set architecture. For this program, which computer is faster and by what percent? (cpu_ex_time_b/cpu_ex_time_a) = 3/2 = 1.5 Therefore, computer A is 1.5 times as fast a computer B, or computer A is faster than computer B by 50% percent. 29

30 Exercise If a program currently takes 100 seconds to execute and loads and stores account for 20% of the execution times, a)how long will the program take if loads and stores are made 30% faster? b)what is the maximum speedup that can be achieved by improving the performance of loads and stores? 30

31 a) For this, you can use Ahmdahl's law or you can reason it out step by step. Doing it step by step gives (1) Before the improvement loads take 20 seconds (2) If loads and stores are made 30 percent faster they will take 20/1.3 = seconds, which corresponds to seconds less (3) Thus, the final program will take = Note: In step 2, the performance improves by 30 percent or performance_new = performance_old x 1.3 since ex_time = 1/performance This gives 1/ex_time_new = 1.3/ex_time_old ex_time_new = ex_time_old/1.3 31

32 Using Amdahl's law gives EX_TIME_NEW = EX_OLD*(1 - FRAC_EN + FRAC_EN/SPEEDEP_EN) For this problem EX_TIME_NEW = 100*( /1.3) = b) The maximum speedup is achieved if loads and stores take no time at all. In this case, the program runs in 80 seconds and the overall speedup is SPEEDUP = EX_TIME_OLD/EX_TIME_NEW = 100/80 =