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Performance Computer Architecture – CS401 Erkay Savas Sabanci University

This article discusses the concept of performance in computer architecture, including how to measure performance, performance metrics, evaluation, and factors that affect hardware performance. It also explores how the machine's instruction set impacts performance.

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Performance Computer Architecture – CS401 Erkay Savas Sabanci University

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  1. Performance Computer Architecture – CS401 Erkay Savas Sabanci University Erkay Savas

  2. Performance • What is performance? • How to measure performance? • Performance metrics • Performance evaluation • Why some hardware perform better than others for different programs? • What factors in hardwareare related to performance? • How does the machine's instruction set affect performance? Erkay Savas

  3. Airplane Passenger Capacity Range (miles) Speed (m.p.h) Passenger throughput (passenger x m.p.h) 228750 Boeing 777 375 4630 610 268700 Boeing 747 470 4150 610 393600 Airbus A 3xx 656 8400 600 178200 Concorde 132 4000 1350 79424 Douglas DC-8-50 146 8720 544 Airplane Analogy • Which of these airplanes has the best performance? Erkay Savas

  4. Computer Performance • Response time (latency) • How long does it take for my job to run? • How long does it take to execute a program? • How long must I wait for a database query? • Throughput • How many jobs can the machine run at once? • What is the average execution rate? • How much work is getting done? • If we upgrade a machine with a new processor what do we increase? • If we add a new machine what do we increase? Erkay Savas

  5. Which Time to Measure? • Elapsed Time (Wall clock time, response time) • Counts everything (disk and memory access, I/O, operating system overhead, work on other processes) • Useful but not always good for comparison purposes • CPU (execution) time • The time CPU spends computing for the user task • Not include time spent waiting for I/O, running other programs • user CPU time CPU time spent within the program, • system CPU time CPU time spent in the operating system performing tasks on behalf of the program Erkay Savas

  6. CPU Time • Unix timecommand reflects this breakdown by returning the following when prompted: 90.7u 12.9s 2:39 65% Interpretation: • User CPU time is 90.7 s • System CPU time is 12.9s • Elapsed time is 159 s ( 90.7+12.9) • CPU time is 65% of total elapsed time Erkay Savas

  7. A Definition of Performance • For some program running on machine X PerformanceX = 1/Execution_timeX • The machine X is said to be “ntimes faster” than the machine Yif PerformanceX/PerformanceY = n Execution_timeY/Execution_timeX= n • Example:Machine A runs a program in 10 seconds and machine B runs the same program in 15 seconds, how much faster is A than B? Erkay Savas

  8. Metrics of Performance • “Time to execute a program” is the ultimate metric in determining the performance • However, it is convenient to inspect other metrics as well when we examine the details of a machine. • Computers use a clock that runs at a constant rate and determines when an event takes place in hardware. • These discrete time intervals are called clock cycles(or ticks, clock ticks, clock periods). • Clock rate (frequency) is the inverse of clock period. Erkay Savas

  9. time Start of events often the rising edge of the clock Clock Cycles • Clock “ticks” indicate when to start activities • Instead of reporting execution time in seconds, we often use cycles Erkay Savas

  10. Clock Cycle • cycle time (CT) = time between ticks = seconds per cycle • Cycle Count (CC): the number of clock cycles to execute a program • clock rate (frequency) = cycles per second (1 Hz = 1 cycle/sec) • A 200 MHz clock has a 1/(200·106) = ? nanosecond cycle time • A 4 GHz clock has a 1/(4· 109) = ? nanosecond cycle time Erkay Savas

  11. CPI • CPIClocks Per Instruction • Number of cycles spent on an instruction on average. • CC = IC  CPI • Hard to compute. • It is useful when comparing the performances of two machines with the same ISA. (Why?) • Example:two machines with the same ISA. For a certain program we have • Machine A:CPI = 2.0 • Machine B:CPI = 1.2 • Which machine is faster? • What if machine A uses 250 ps and machine B 500 ps cycle time Erkay Savas

  12. Improving Performance So, to improve performance • Increase the clock frequency (i.e. decrease the clock period) • Reduce the number of the clock cycles per program (IC  CPI) Erkay Savas

  13. Instruction  Cycle ? • No ! • The number of cycles per instruction depends on the implementations of the instructions in hardware • The number differs for each processor (even with the same ISA) Erkay Savas

  14. The Reason • Operations take different number of cycles • Multiplication takes longer than addition • Floating point operations take longer than integer operations • The access time to a register is much shorter than access to the main memory. Erkay Savas

  15. Simple Formulae for CPU Time • CPU execution time = CPU clock cycles for a program  Clock cycle time (CC  CT) • CPU execution time = CPU clock cycles for a program/Clock rate • We can writeCPU clock cycles for a program =IC  CPI • ThenCPU execution time = (IC  CPI)/Clock rate Erkay Savas

  16. Example • Computer A of 800 MHz • It runs our favorite program in 15 s • Our goal • Design computer B with the same ISA • It will run the same program in 8 s. • We will use a new technology • can increase the clock rate; • however, it will also increase CPI by 1.25. • What clock rate should we aim to use? Erkay Savas

  17. Performance • Performance is determined by execution time (CPU time) • We have also other indicators • # of cycles to execute program • # of instructions in program (IC) • # of cycles per second • average # of cycles per instruction (CPI) • average # of instructions per second • Common pitfall: thinking one of the above is indicative of performance when it really isn’t. Erkay Savas

  18. Number of Instructions Example • A compiler designer has the following two alternatives to generate a certain piece of code with instructions A(1 cycle) , B (2 cycles), and C(3 cycles): • 2106 of A, 106 of B, and 2106 of C (IC = 5106) • 4106 of A, 106 of B, and 106 of C (IC = 6106) • Which code sequence is faster? Erkay Savas

  19. MIPS • Millions Instructions Per Second = MIPS = IC/(Execution_time  106) MIPS = IC/(CC cycle time  106) MIPS = (IC  clock rate)/(IC  CPI  106) MIPS = clock rate/(CPI  106) • A faster machine has a higher MIPS Execution_time = IC/(MIPS  106) Erkay Savas

  20. A MIPS Example • A computer with 500 MHz clock • Three different classes of instructions: • A (1 cycle), B (2 cycles), C (3 cycles) • Two compilers used to produce code for a large piece of software. • Compiler 1: • 5 billion A, 1 billion B, and 1 billion C instructions. • Compiler 2: • 10 billion A, 1 billion B, and 1 billion C instructions. • Which sequence will be faster according to execution time? • Which sequence will be faster according to MIPS? Erkay Savas

  21. Problems of MIPS • MIPS specifies instruction execution rate • MIPS does not take into account the capabilities of the instructions • Thus, it is impossible to compare computers with different ISA using MIPS. • MIPS is not constant, even on a single machine, depends on the application. • As we saw in the previous example, MIPS can vary inversely with performance. Erkay Savas

  22. CPI example • CPI • Machine A: CPI = 10/7 = 1.43 • Machine B: CPI = 15/12 = 1.25 • CPU time • CPU time = (IC  CPI) / clock rate • Let us assume both machines use 200 MHz clock Erkay Savas

  23. Overview • A given program will require • Some number of instructions • Some number of clock cycles • Some number of seconds • Vocabulary • Cycle time: (micro or nano) seconds per cycle • Clock rate (frequency): cycles per second • CPI: clock per instruction • MIPS: millions of instruction per second • MFLOPS: millions of floating point operations per second Erkay Savas

  24. Performance • Performance is ultimately determined by execution time • Is any of the following metrics good to measure performance by itself? Why? • # of cycles to execute a program • # of instructions in a program • # of cycles per second • Average # of cycles per instruction • Average # number of instructions per second Erkay Savas

  25. Question • Assuming two machines have the same ISA, which of the following quantities are identical? • Clock rate • CPI • Execution time • # of instructions • MIPS Erkay Savas

  26. HW or SW component Affects what? How? Program Performance Algorithm IC, possibly CPI Programming Language IC, CPI Compiler IC, CPI ISA IC, clock rate, CPI Erkay Savas

  27. Benchmarks • Programs specifically chosen to measure performance • must reflect typical workload of the user • Benchmark types • Real applications • Small benchmarks • Benchmark suites • Synthetic benchmarks Erkay Savas

  28. Real Applications • Workload: Set of programs a typical user runs day in and day out. • To use these real applications for metrics is a direct way of comparing the execution time of the workload on two machines. • Using real applications for metrics has certain restrictions: • They are usually big • Takes time to port to different machines • Takes considerable time to execute • Hard to observe the outcome of a certain improvement technique Erkay Savas

  29. Comparing & Summarizing Performance • A is 100 times faster than B for program 1 • B is 10 times faster than A for program 2 • For total performance, arithmetic mean is used: Erkay Savas

  30. Arithmetic Mean • If each program, in the workload, do not run equal times, then we have to use weighted arithmetic mean • Suppose that the program 1 runs 10 times as often as the program 2. Which machine is faster? Erkay Savas

  31. Small Benchmarks • Small code segments which are common in many applications • For example, loops with certain instruction mix • for (j = 0; j<8; j++) S = S + Aj  Bi-j • Good for architects and designers • Since small code segments are easy to compile and simulate even by hand, designers use these kind of benchmarks while working on a novel machine • Can be abused by compiler designers by introducing special-purpose optimizations targeted at specific benchmark. Erkay Savas

  32. Benchmark Suites • SPEC (Standard Performance Evaluation Corporation) • non-profit organization that aims to produce "fair, impartial and meaningful benchmarks for computers” • Began in 1989 - SPEC89 (CPU intensive) • companies agreed on a set of real programs and inputs which they hope reflect a typical user’s workload best. • valuable indicator of performance • can still be abused • Updates are required as the applications and their workload change by time Erkay Savas

  33. SPEC Benchmark Sets • CPU Performance (SPEC CPU2006) • Graphics (SPECviewperf) • High-performance computing (HPC2002, MPI2007, OMP2001) • Java server applications (jAppServer2004) • a multi-tier benchmark for measuring the performance of Java 2 Enterprise Edition (J2EE) technology-based application servers. • Mail systems (MAIL2001, SPECimap2003) • Network File systems (SFS97_R1 (3.0)) • Web servers (SPEC WEB99, SPEC WEB99 SSL) • More information: http://www.spec.org/ Erkay Savas

  34. SPECInt

  35. SPECfp

  36. SPEC CPU2006 – Summarizing • SPEC ratio: the execution time measurements are normalized by dividing the measured execution time by the execution time on a reference machine • Sun Microsystems Fire V20z, which has anAMD Opteron 252 CPU, running at 2600 MHz. • 164.gzip benchmark executes in 90.4 s. • The reference time for this benchmark is 1400 s, • benchmark is 1400/90.4 × 100 = 1548 (a unitless value) • Performances of different programs in the suites are summarized using “geometric mean” of SPEC ratios. Erkay Savas

  37. Pentium III & Pentium 4 Erkay Savas

  38. Comparing Pentium III and Pentium 4 Implementation efficiency? Erkay Savas

  39. SPEC WEB99 Erkay Savas

  40. Power Consumption Concerns • Performance studied at different levels: • Maximum power • Intermediate level that conserves battery life • Minimum power that maximizes battery life • Intel Mobile Pentium & Pentium M: two available clock rates • Maximum • Reduced clock rate • Pentium M @ 1.6/0.6 GHz • Pentium 4-M @ 2.4/1.2 GHz • Pentium III-M @ 1.2/0.8 GHz Erkay Savas

  41. Three Intel Mobile Processors Erkay Savas

  42. Energy Efficiency Erkay Savas

  43. Synthetic Benchmarks • Artificial programs constructed to try to match the characteristics of a large set of program. • Goal: Create a single benchmark program where the execution frequency of instructions in the benchmark simulates the instruction frequency in a large set of benchmarks. • Examples: • Dhrystone, Whetstone • They are not real programs • Compiler and hardware optimizations can inflate the improvement far beyond what the same optimization would do with real programs Erkay Savas

  44. Amdahl’s Law in Computing • Improving one aspect of a machine by a factor of n does not improve the overall performance by the same amount. • Speedup = (Performance after imp.) / (Performance before imp.) • Speedup = (Execution time before imp.)/(Execution time after imp.) • Execution Time After Improvement = Execution Time Unaffected +(Execution Time Affected/n) Erkay Savas

  45. Amdahl’s Law • Example: Suppose a program runs in 100 s on a machine, with multiplication responsible for 80 s of this time. • How much do we have to improve the speed of multiplication if we want the program to run 4 times faster? • Can we improve the performance by a factor 5? Erkay Savas

  46. Amdahl’s Law • The performance enhancement possible due to a given improvement is limited by the amount that the improved feature is used. • In previous example, it makes sense to improve multiplication since it takes 80% of all execution time. • But after certain improvement is done, the further effort to optimize the multiplication more will yield insignificant improvement. • Law of Diminishing Returns • A corollary to Amdahl’s Law is to make a common case faster. Erkay Savas

  47. Examples • Suppose we enhance a machine making all floating-point instructions run five times faster. If the execution time of some benchmark before the floating-point enhancement is 10 seconds, what will the speedup be if half of the 10 seconds is spent executing floating-point instructions? • We are looking for a benchmark to show off the new floating-point unit described above, and want the overall benchmark to show a speedup of 3. One benchmark we are considering runs for 90 seconds with the old floating-point hardware. How much of the execution time would floating-point instructions have to account for in this program in order to yield our desired speedup on this benchmark? Erkay Savas

  48. Remember • Total execution time is a consistent summary of performance • Execution Time = (IC  CPI)/f • For a given architecture, performance increases come from: • increases in clock rate (without too much adverse CPI effects) • improvements in processor organization that lower CPI • compiler enhancements that lower CPI and/or IC Erkay Savas

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