1. Gordon Moore Gordon Moore, cofounder of Intel 1965: 2 x trans. per chip/year After 1970: 2 x trans. per chip/1.5year 摩爾定律

2. Growth in CPU transistor count

3. Consequences of Moore ’ s law Cost of a chip remains unchanged during the growth of in density => cost down Electrical path length is shortened => increase operating speed Computer becomes smaller Reduction in power More circuitry on each chip => fewer inter- chip connections => more reliable

4. Chap.4 The Role of Performance Jen-Chang Liu, Spring 2005

5. Hardware performance is often key to the effectiveness of an entire system of hardware and software. What do we mean by saying one computer has better performance than another?

6. Example: performance of airplanes

7. Performance of a hardware system What do we mean by better performance? Fast speed ? Response time (execution time): the time between the start and completion of a task 完成 工作所需的時間 Throughput : the total amount of work done in a given time 單位時間完成的工作 Ex. multi-user system

8. Performance measure Performance X 1 Execution time x = * Relative performance: Performance A Performance B = n = Machine A is n times faster than B Execution time B Execution time A Ex. machine A runs a program in 10 sec., machine B runs a program in 15 sec., Performance A Performance B = 1.5 Execution time B Execution time A = = 15 10 Quantitative relation of performance and execution time on machine x:

9. Problem with previous definition of performance The definition of execution time How about multiple tasks run concurrently? Use which programs to evaluate the performance of a computer ?

10. Execution Time ? The total time to complete a task – response time, elapsed time In a timeshared system, such as Unix, a processor work on several programs Including disk access, memory access, I/O, OS overhead … 執行時間的定義 使用者觀點 Program A swap Prog. BI/O Program A Response time for A

11. CPU time CPU execution time Does not include waiting for I/O, running other programs CPU exec. time = user CPU time + system CPU time user CPU time CPU time spent in the program system CPU time CPU time spent in the OS about our program 不含 I/O, 執行其他程式時間

12. Example : CPU time Unix command : time 90.7u 12.9s 2:39 65% user CPU system CPU elapsed time 90.7+12.9 159 = 0.65 We will discuss CPU performance, i.e. user CPU time in the following discussion

13. Unit of time Seconds Clock cycle Ex. Clock cycle time = 2ns Clock rate = 1 2x10 -6 = 500 MHz CPU time for a program CPU clock cycles for a program =x Clock cycle time Instructions for a program =x Average clock cycle per instruction x Clock cycle time (CPI)

14. Example 1 Machine A,B has the same ISA, for the same program Machine A: clock cycle = 1ns, CPI = 2 Machine B: clock cycle = 2ns, CPI = 1.2 CPU time A = Inst. count x CPI x clock cycle time = I x 2 x 1 = 2I CPU time B = I x 1.2 x 2 = 2.4 I Performance A Performance B Execution time B Execution time A = = 2.4I 2I = 1.2 A is 1.2 times faster than B

15. Quiz 5/9 Program P runs in 10s on computer A, which has 4GHz clock We want run program P in 6s. We design computer B with faster clock rate, but it requires 1.2 times as many clock cycles as computer A. What clock rate should we use in computer B?

16. Example 2 Instruction class CPI A B C 1 2 3 Code 1: 2 1 2 Code 2: 4 1 1 Compiler generate 2 different code sequences A B C CPU clock cycle 1 = 2x1 + 1x2 + 2x3 = 10 cycles CPU clock cycle 2 = 4x1 + 1x2 + 1x3 = 9 cycles Total inst. 5 6 faster? faster

17. Short conclusion Computer Performance software hardware Response time CPU time I/O, other prog.s Instruction count CPI Clock cycle length How to optimize them in a hardware design?

18. Problem with previous definition of performance The definition of execution time How about multiple tasks run concurrently? Use which programs to evaluate the performance of a computer ?

19. Choose programs to evaluate performance Benchmarks: programs chosen to measure performance SPEC (System Performance Evaluation Cooperative) suit of benchmarks Started in 1989 http://open.specbench.org/ SPEC95 in textbook is retired … SPECx contains a set of benchmark programs

20. SPEC – money …

21. SPEC95 benchmarks Integer benchmarks written in C floating-pt benchmarks written in Fortran 77

22. Summarize performance Which is faster? Computer AComputer B Program 1(sec)110 Program 2(sec)1000100 Total time(sec)1001110 Performance B Performance A Execution time A Execution time B = = 1001 110 = 9.1 * Assume the programs occur in equal probability.

23. SPEC ratio The execution time of a benchmark program is normalized (compared to a baseline system) SPECint95, SPECfp95 SPEC ratio = Exec. Time on Sun SPARCstation 10/40 Exec. Time on the measured machine SPECint95 = geometric mean of SPEC ratios

24. Example: SPECint95 for Pentium and Pentium Pro 1 1 Performance improvement 2 2 Clock rate x2 SPECint x 1.7 ?

25. Amdahl’s law in computing CPU time for a program CPU clock cycles for a program =x Clock rate 1 Clock rate => CPU time2 2 * Improvement of one aspect of a machine does not increase performance by the same ratio 部分的改進 * Ex. The bottleneck in the memory system does not improve Exec. time after improve. = Exec. time affected by improve. Amount of improvement Exec. time unaffected + as in previous example

26. Example: Amdahl’s law A program takes 100s to run 20% multiplication, 50% memory op., 30% others What ’ s the speed up for Multiply speed 4 Memory access 2 100 20/4 + 50 + 30 =1.18 100 20 + 50/2 + 30 =1.33

27. MIPS as a measurement (not good … ) MIPS = Million Instructions Per Second High MIPS => faster ? MIPS= Instruction count Execution time x 10 6 Pitfalls: MIPS cannot be used to compare computers with different instruction sets => inst. count differs MIPS varies between programs on the same computer => no single MIPS for a machine

28. Example: MIPS ? Example: 500 MHz machine Code 1 Code 2 Inst. Count(x10 9 ) for each inst. class A BC 5 1 1 10 1 1 2 compilers for the same source program: Instruction class CPI A B C 1 2 3

29. Example: MIPS? MIPS 1 = Inst. count Exec timex10 6 = (5+1+1)x10 9 20x10 6 =350 MIPS 2 = (10+1+1)x10 9 30x10 6 =400 Exec. time 1 < Exec. time 2 MIPS 1 < MIPS 2 Exec. Time 1 = (5x1+1x2+1x3)x10 9 cycles 500x10 6 cycles/sec = 20 sec. Exec. Time 2 = (10x1+1x2+1x3)x10 9 cycles 500x10 6 cycles/sec = 30 sec.