1 / 34

Chapter 2: Performance

Chapter 2: Performance. Measure, Report, and Summarize Make intelligent choices See through the marketing hype

Samuel
Télécharger la présentation

Chapter 2: Performance

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 2: Performance • Measure, Report, and Summarize • Make intelligent choices • See through the marketing hype • Key to understanding underlying organizational motivationWhy is some hardware better than others for different programs?What factors of system performance are hardware related? (e.g., Do we need a new machine, or a new operating system?)How does the machine's instruction set affect performance?

  2. Which of these airplanes has the best performance? • How much faster is the Concorde compared to the 747? • How much bigger is the 747 than the Douglas DC-8? Airplane Passengers Range (mi) Speed (mph) Boeing 737-100 101 630 598 Boeing 747 470 4150 610 BAC/Sud Concorde 132 4000 1350 Douglas DC-8-50 146 8720 544

  3. Definitions • Performance is in units of things-per-second • bigger is better • If we are primarily concerned with response time • performance(x) = 1 execution_time(x) " X is n times faster than Y" means Performance(X) n = ---------------------- Performance(Y)

  4. Example • Time of Concorde vs. Boeing 747? • Concord is 1350 mph / 610 mph = 2.2 times faster • = 6.5 hours / 3 hours • Throughput of Concorde vs. Boeing 747 ? • Concord is 178,200 pmph / 286,700 pmph = 0.62 “times faster” • Boeing is 286,700 pmph / 178,200 pmph = 1.6 “times faster” • Boeing is 1.6 times (“60%”)faster in terms of throughput • Concord is 2.2 times (“120%”) faster in terms of flying time • We will focus primarily on execution time for a single job

  5. Computer Performance: TIME, TIME, TIME • Response Time (latency) — How long does it take for my job to run? — How long does it take to execute a job? — How long must I wait for the 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 to the lab what do we increase?

  6. Execution Time • Elapsed Time • counts everything (disk and memory accesses, I/O , etc.) • a useful number, but often not good for comparison purposes • CPU time • doesn't count I/O or time spent running other programs • can be broken up into system time, and user time • Our focus: user CPU time • time spent executing the lines of code that are "in" our program

  7. Book's Definition of Performance • For some program running on machine X, PerformanceX = 1 / Execution timeX • "X is n times faster than Y" PerformanceX / PerformanceY = n • Problem: • machine A runs a program in 20 seconds • machine B runs the same program in 25 seconds

  8. time Clock Cycles • Instead of reporting execution time in seconds, we often use cycles • Clock “ticks” indicate when to start activities (one abstraction): • cycle time = time between ticks = seconds per cycle • clock rate (frequency) = cycles per second (1 Hz. = 1 cycle/sec)A 200 Mhz. clock has a cycle time

  9. How to Improve Performance So, to improve performance (everything else being equal) you can either________ the # of required cycles for a program, or________ the clock cycle time or, said another way, ________ the clock rate.

  10. 1st instruction 2nd instruction 3rd instruction ... 4th 5th 6th How many cycles are required for a program? • Could assume that # of cycles = # of instructions time This assumption is incorrect, different instructions take different amounts of time on different machines.Why?hint: remember that these are machine instructions, not lines of C code

  11. Different numbers of cycles for different instructions • Multiplication takes more time than addition • Floating point operations take longer than integer ones • Accessing memory takes more time than accessing registers • Important point: changing the cycle time often changes the number of cycles required for various instructions (more later) time

  12. Example • Our favorite program runs in 10 seconds on computer A, which has a 400 Mhz. clock. We are trying to help a computer designer build a new machine B, that will run this program in 6 seconds. The designer can use new (or perhaps more expensive) technology to substantially increase the clock rate, but has informed us that this increase will affect the rest of the CPU design, causing machine B to require 1.2 times as many clock cycles as machine A for the same program. What clock rate should we tell the designer to target?" • Don't Panic, can easily work this out from basic principles

  13. Now that we understand cycles • A given program will require • some number of instructions (machine instructions) • some number of cycles • some number of seconds • We have a vocubulary that relates these quantities: • cycle time (seconds per cycle) • clock rate (cycles per second) • CPI (cycles per instruction) a floating point intensive application might have a higher CPI • MIPS (millions of instructions per second)this would be higher for a program using simple instructions

  14. Performance • Performance is determined by execution time • Do any of the other variables equal performance? • # of cycles to execute program? • # of instructions in program? • # of cycles per second? • average # of cycles per instruction? • average # of instructions per second? • Common pitfall: thinking one of the variables is indicative of performance when it really isn’t.

  15. CPI Example • Suppose we have two implementations of the same instruction set architecture (ISA). For some program,Machine A has a clock cycle time of 10 ns. and a CPI of 2.0 Machine B has a clock cycle time of 20 ns. and a CPI of 1.2 What machine is faster for this program, and by how much? • If two machines have the same ISA which of our quantities (e.g., clock rate, CPI, execution time, # of instructions, MIPS) will always be identical?

  16. # of Instructions Example • A compiler designer is trying to decide between two code sequences for a particular machine. Based on the hardware implementation, there are three different classes of instructions: Class A, Class B, and Class C, and they require one, two, and three cycles (respectively). The first code sequence has 5 instructions: 2 of A, 1 of B, and 2 of CThe second sequence has 6 instructions: 4 of A, 1 of B, and 1 of C.Which sequence will be faster? How much?What is the CPI for each sequence?

  17. MIPS example • Two different compilers are being tested for a 100 MHz. machine with three different classes of instructions: Class A, Class B, and Class C, which require one, two, and three cycles (respectively). Both compilers are used to produce code for a large piece of software.The first compiler's code uses 5 million Class A instructions, 1 million Class B instructions, and 1 million Class C instructions.The second compiler's code uses 10 million Class A instructions, 1 million Class B instructions, and 1 million Class C instructions. • Which sequence will be faster according to MIPS? • Which sequence will be faster according to execution time?

  18. Benchmarks • Performance best determined by running a real application • Use programs typical of expected workload • Or, typical of expected class of applications e.g., compilers/editors, scientific applications, graphics, etc. • Small benchmarks • nice for architects and designers • easy to standardize • can be abused • SPEC (System Performance Evaluation Cooperative) • companies have agreed on a set of real program and inputs • can still be abused (Intel’s “other” bug) • valuable indicator of performance (and compiler technology)

  19. Basis of Evaluation Cons Pros • very specific • non-portable • difficult to run, or • measure • hard to identify cause • representative Actual Target Workload • portable • widely used • improvements useful in reality • less representative Full Application Benchmarks • easy to “fool” Small “Kernel” Benchmarks • easy to run, early in design cycle • “peak” may be a long way from application performance • identify peak capability and potential bottlenecks Microbenchmarks

  20. SPEC ‘89 • Compiler “enhancements” and performance

  21. SPEC95 • Eighteen application benchmarks (with inputs) reflecting a technical computing workload • Eight integer • go, m88ksim, gcc, compress, li, ijpeg, perl, vortex • Ten floating-point intensive • tomcatv, swim, su2cor, hydro2d, mgrid, applu, turb3d, apsi, fppp, wave5 • Must run with standard compiler flags • eliminate special undocumented incantations that may not even generate working code for real programs

  22. SPEC ‘95

  23. SPEC CPU2000

  24. SPEC ‘95 Does doubling the clock rate double the performance? Can a machine with a slower clock rate have better performance?

  25. SpecCPU2000 Benchmark results • Advanced Micro Devices AMD Athlon (TM) XP 2700+ 878 913 • Advanced Micro Devices, AMD Athlon (TM) XP 2800+ 898 933 • Advanced Micro Devices, AMD Athlon (TM) XP 3000+ 960 995 • IBM Corporation pSeries 630 (1450 MHz, 1 CPU) 884 910 • Intel Corporation (2.67 GHz, Pentium 4 processor) 998 1005 • Intel Corporation (2.8 GHz, Pentium 4 processor) 1032 1040 • Intel Corporation (3.06 GHz, Pentium 4 processor 1099 1107 • Sun Microsystems Sun Blade 2000 (1.015GHz) 516 576 • Sun Microsystems Sun Blade Model 2050 537 610

  26. More SPEC CPU2000 Results • This data is current as of Wed Mar 19 20:01:27 PST 2003 • SPECint ® 2000 • 1099.0 3067.0 MHz Pentium 4 HT Intel D850EMVR motherboard (3.06 GHz, Pentium 4 processor wi • 1085.0 3066.0 MHz Pentium 4 Dell Precision WorkStation 350 (3.06 GHz P4) • 967.0 2800.0 MHz Xeon Fujitsu Siemens Computers CELSIUS R610 • 960.0 2167.0 MHz Athlon ASUS A7N8X Deluxe Motherboard, AMD Athlon (TM) XP 3000+ • 909.0 1450.0 MHz POWER4+ IBM Corporation IBM eServer pSeries 650 Model 6M2 (1450 MHz, 1 CPU) • 898.0 2250.0 MHz Athlon XP ASUS A7N8X (REV 1.02) Motherboard, AMD Athlon (TM) XP 2800+ • 845.0 1250.0 MHz Alpha Hewlett-Packard Company hp AlphaServer ES45 68/1250 • 822.0 1300.0 MHz POWER4 IBM Corporation IBM eServer pSeries 655 Model 651 (1300 MHz, 1 CPU) • 810.0 1000.0 MHz Itanium 2 Hewlett-Packard Company hp server rx2600 (1000 MHz, Itanium 2) • 751.0 2133.0 MHz Athlon MSI K7D Master Motherboard, AMD Athlon (TM) MP 2600+ • 747.0 1350.0 MHz SPARC64 V Fujitsu Limited PRIMEPOWER900 (1350MHz) • 737.0 2000.0 MHz Athlon MP Asus A7M266-D Motherboard, AMD Athlon (TM) MP 2400+ • 648.0 1400.0 MHz Pentium III Dell PowerEdge 1500SC (1.4 GHz PIII) • 642.0 875.0 MHz PA-RISC Hewlett-Packard Company hp workstation c3750 • 569.0 750.0 MHz PA-RISC 8700 Hewlett-Packard Company hp workstation j6700 • 537.0 1050.0 MHz UltraSPARC III Cu Sun Microsystems Sun Blade Model 2050 • 512.0 810.0 MHz SPARC64 GP Fujitsu Limited PRIMEPOWER650 (810MHz) • 483.0 600.0 MHz R14000 SGI SGI Origin 3200 1X 600MHz R14k • 471.0 600.0 MHz R14000A SGI SGI Origin 300 1X 600MHz R14000A • 439.0 900.0 MHz UltraSPARC III Sun Microsystems Sun Blade 1000 Model 1900 • 437.0 1000.0 MHz Pentium III Xeon Dell PowerEdge 4400 (1.0 GHz PIII Xeon) • 431.0 750.0 MHz RS64 IV IBM Corporation IBM eServer pSeries 620 Model 6F0 (750 MHz) • 417.0 552.0 MHz PA-RISC 8600 Hewlett-Packard Company hp visualize j6000 UNIX workstation • 379.0 800.0 MHz Itanium Hewlett-Packard Company hp server rx4610 • 333.0 450.0 MHz POWER3 II IBM Corporation RS/6000 44P-170 (450 MHz) • 328.0 400.0 MHz R12000 SGI SGI Origin 3200 400MHz R12k • 313.0 400.0 MHz PA-RISC 8500 Hewlett-Packard Company hp visualize b2000 UNIX workstation • 264.0 500.0 MHz PowerPC RS64 III IBM Corporation RS/6000 Model 7026-M80 (1 CPU) • 225.0 480.0 MHz UltraSPARC II Sun Microsystems Sun Enterprise 450 • 168.0 340.0 MHz RS64 II IBM Corporation RISC System/6000 H70 (1 CPU) • 165.0 500.0 MHz UltraSPARC IIe Sun Microsystems Sun Blade 100 • 99.4 250.0 MHzP owerPC 604e IBM Corporation RS/6000 43P-150 (250MHz)

  27. Metrics of performance Answers per month Useful Operations per second Application Programming Language Compiler (millions) of Instructions per second – MIPS (millions) of (F.P.) operations per second – MFLOP/s ISA Datapath Megabytes per second Control Function Units Cycles per second (clock rate) Transistors Wires Pins Each metric has a place and a purpose, and each can be misused

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

  29. CPI Invest Resources where time is Spent! “Average cycles per instruction” • CPI = (CPU Time * Clock Rate) / Instruction Count • = Clock Cycles / Instruction Count n CPU time = ClockCycleTime * CPI * I i i i = 1 n CPI =  CPI * F where F = I i i i i i = 1 Instruction Count "instruction frequency"

  30. Example (RISC processor) Base Machine (Reg / Reg) Op Freq Cycles CPI(i) % Time ALU 50% 1 .5 23% Load 20% 5 1.0 45% Store 10% 3 .3 14% Branch 20% 2 .4 18% 2.2 Typical Mix How much faster would the machine be if a better data cache reduced the average load time to 2 cycles? How does this compare with using branch prediction to shave a cycle off the branch time? What if two ALU instructions could be executed at once?

  31. Amdahl's Law Execution Time After Improvement = Execution Time Unaffected +( Execution Time Affected / Amount of Improvement ) • Example: "Suppose a program runs in 100 seconds on a machine, with multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster?" How about making it 5 times faster? • Principle: Make the common case fast

  32. 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) Š ((1-F) + F/S) X ExTime(without E) Speedup(with E) Š 1 (1-F) + F/S

  33. Example • 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 100 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?

  34. Remember • Performance is specific to a particular program/s • Total execution time is a consistent summary of performance • For a given architecture performance increases come from: • increases in clock rate (without adverse CPI affects) • improvements in processor organization that lower CPI • compiler enhancements that lower CPI and/or instruction count • Pitfall: expecting improvement in one aspect of a machine’s performance to affect the total performance • You should not always believe everything you read! Read carefully! (see newspaper articles, e.g., Exercise 2.37)

More Related