CpE 442 Introduction to Computer Architecture. The Role of Performance

Transcription

1 CpE 442 Introduction to Computer Architecture The Role of Performance Instructor: H. H. Ammar CpE442 Lec2.1

2 Overview of Today s Lecture: The Role of Performance Review from Last Lecture Definition and Measures of Performance Benchmarks Summarizing Performance and Performance Pitfalls CpE442 Lec2.2

3 Review: What is "Computer Architecture" Co-ordination of levels of abstraction Application Compiler Instr. Set Proc. Operating System I/O system Instruction Set Architecture Digital Design Circuit Design Under a set of rapidly changing Forces CpE442 Lec2.3

4 Review: Levels of Representation High Level Language Program Compiler Assembly Language Program Assembler Machine Language Program temp = v[k]; v[k] = v[k+1]; v[k+1] = temp; lw $15, 0($2) lw $16, 4($2) sw $16, 0($2) sw $15, 4($2) Control Signal Specification Machine Interpretation CpE442 Lec2.4

5 Review: Levels of Organization SPARCstation 20 Computer SPARC Processor Control Memory Devices Input Datapath Output CpE442 Lec2.5

6 Computer Architecture Simulation Tools 1. The HASE Architecture Simulation Environment 2. The New Compiler Technology simulation (shown in class) 3. MIPS Assembly Language Simulators a. SPIM A MIPS32 Simulator b. MARS (MIPS Assembler and Runtime Simulator) CpE442 Lec2.6

7 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) CpE442 Lec2.7

8 Overview of Today s Lecture: The Role of Performance Review from Last Lecture Definition and Measures of Performance Benchmarks Summarizing Performance and Performance Pitfalls CpE442 Lec2.8

9 Metrics of performance Application Response time, Answers per month Operations per second Programming Language Compiler ISA (millions) of Instructions per second MIPS (millions) of (F.P.) operations per second MFLOP/s Datapath Control Function Units Transistors Wires Pins Megabytes per second Cycles per second (clock rate) CpE442 Lec2.9

10 Relating Processor Metrics the execution time of a given program on a given CPU architecture CPU execution time = CPU clock cycles/pgm X clock cycle time or CPU execution time = CPU clock cycles/pgm clock rate Define CPI = the avg. clock cycles per instruction, CPI tells us something about the Instruction Set Architecture, the Implementation of that architecture, and the program being measured CPU clock cycles/pgm = Instructions/pgm X CPI or CPI = CPU clock cycles/pgm Instructions/pgm CpE442 Lec2.10

11 Aspects of CPU Performance, CPU time = Seconds = Instructions x Cycles x Seconds Program Program Instruction Cycle Program instr. count CPI clock rate Compiler Instr. Set Arch. Organization Technology CpE442 Lec2.11

12 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 CpE442 Lec2.12

13 Figures from a Simulator for the following code segment comparing two compilers for (i=0;i<3;i++) { in_a(i)++; int_b(i)++; flt_d(i) = flt_d(i) + flt_c(i); } CpE442 Lec2.13

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18 Organizational Trade-offs Application Programming Language Compiler ISA Datapath Control Function Units Transistors Wires Pins Instruction Mix Single-Cycle Processor Design CPI CPI=1, large cycle time-slow clock Multi-cycle Processor Design CPI > 1, smaller cycle time- Faster Cycle Time clock CpE442 Lec2.18

19 CPI Average cycles per instruction CPI = (CPU Time * Clock Rate) / Instruction Count = Clock Cycles / Instruction Count The performance equation can be written as follows using instruction classes and the instruction count I and CPI for each class i n S CPU time = ClockCycleTime * CPI * I i = 1 i i n "instruction frequency" S CPI = CPI i * F where F = I i i i i = 1 Instruction Count See example next slide Invest Resources where time is Spent! CpE442 Lec2.19

20 Example Base Machine (Reg / Reg) Op Freq(Fi) CPI(i) % Time ALU 50% % Load 20% % Store 10% % Branch 20% % Typical Mix 1.5 The CPI = 1.5 cycles per instruction CpE442 Lec2.20

21 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 CpE442 Lec2.21

22 Comparing the performance of two machines Speedup due to enhancement E: ExTime w/o E Performance w/ E Speedup(E) = = ExTime w/ E Performance w/o E = 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 CpE442 Lec2.22

23 Rate Metrics MIPS (Million Instructions Per Second), and MFLPOS (Miilions Floating Point Operations Per Second) MIPS = Instruction Count / (CPU 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 CpE442 Lec2.23 often not where time is spent

24 Example showing why MIPS can fail Compare performance with Compilers 1 and 2 for a given program on a given machine Instruction Count in Billions for instruction classes A B C Compiler 1 Instruction Count Compiler 2 Instruction Count CPI for each class Clock cycles using compiler1 = 10 Billion Clock cycles using compiler2 = 15 Billion assuming 1GHZ clock CPU Time 1 = 5x1+1x2 +1x3 = 10 secs CPU Time 2 = 10x1 + 1x2 + 1x3 = 15 secs yet the MIPS rating is MIPS 1 = (instr. Count/cpu time in sec x 10^6) = (5+1+1)/10 * 1000 = 700 MIPS 2 = 12/15 * 1000 = 800 giving the impression that 2 have a higher rate of executing instructions than 1 CpE442 Lec2.24

25 Overview of Today s Lecture: The Role of Performance Review from Last Lecture Definition and Measures of Performance Benchmarks Summarizing Performance and Performance Pitfalls CpE442 Lec2.25

26 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 CpE442 Lec2.26

27 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 CpE442 Lec2.27

28 Successful Benchmark: SPEC EE Times + 5 companies band together to form the Systems Performance Evaluation Committee (SPEC): Sun, MIPS, HP, Apollo, DEC Create standard list of programs, inputs, reporting: some real programs, includes OS calls, some I/O CpE442 Lec2.28

29 SPEC second round, SPEC95 8 integer benchmarks in C and 10 floating pt benchmarks in Fortran CpE442 Lec2.29

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32 Overview of Today s Lecture: The Role of Performance Review from Last Lecture Definition and Measures of Performance Benchmarks Summarizing Performance and Performance Pitfalls CpE442 Lec2.32

33 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 CpE442 Lec2.33

34 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 programs HW 1, Submit via ecampus CpE442 Lec2.34