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Performance COE 308 Computer Architecture Prof. Muhamed Mudawar Computer Engineering Department King Fahd University of Petroleum and Minerals
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 2 How can we make intelligent choices about computers? Why is some computer hardware performs better at some programs, but performs less at other programs? How do we measure the performance of a computer? What factors are hardware related? software related? How does machine’s instruction set affect performance? Understanding performance is key to understanding underlying organizational motivation What is Performance?
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 3 Response Time and Throughput Response Time Time between start and completion of a task, as observed by end user Response Time = CPU Time + Waiting Time (I/O, OS scheduling, etc.) Throughput Number of tasks the machine can run in a given period of time Decreasing execution time improves throughput Example: using a faster version of a processor Less time to run a task more tasks can be executed Increasing throughput can also improve response time Example: increasing number of processors in a multiprocessor More tasks can be executed in parallel Execution time of individual sequential tasks is not changed But less waiting time in scheduling queue reduces response time
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 4 For some program running on machine X X is n times faster than Y Book’s Definition of Performance Execution time X 1 Performance X = Performance Y Performance X Execution time X Execution time Y = n=
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 5 Real Elapsed Time Counts everything: Waiting time, Input/output, disk access, OS scheduling, … etc. Useful number, but often not good for comparison purposes Our Focus: CPU Execution Time Time spent while executing the program instructions Doesn't count the waiting time for I/O or OS scheduling Can be measured in seconds, or Can be related to number of CPU clock cycles What do we mean by Execution Time?
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 6 Clock cycle = Clock period = 1 / Clock rate Clock rate = Clock frequency = Cycles per second 1 Hz = 1 cycle/sec1 KHz = 10 3 cycles/sec 1 MHz = 10 6 cycles/sec1 GHz = 10 9 cycles/sec 2 GHz clock has a cycle time = 1/(2×10 9 ) = 0.5 nanosecond (ns) We often use clock cycles to report CPU execution time Clock Cycles Cycle 1Cycle 2Cycle 3 CPU Execution Time = CPU cycles × cycle time Clock rate CPU cycles =
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 7 Improving Performance To improve performance, we need to Reduce number of clock cycles required by a program, or Reduce clock cycle time (increase the clock rate) Example: A program runs in 10 seconds on computer X with 2 GHz clock What is the number of CPU cycles on computer X ? We want to design computer Y to run same program in 6 seconds But computer Y requires 10% more cycles to execute program What is the clock rate for computer Y ? Solution: CPU cycles on computer X = 10 sec × 2 × 10 9 cycles/s = 20 × 10 9 CPU cycles on computer Y = 1.1 × 20 × 10 9 = 22 × 10 9 cycles Clock rate for computer Y = 22 × 10 9 cycles / 6 sec = 3.67 GHz
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 8 Instructions take different number of cycles to execute Multiplication takes more time than addition Floating point operations take longer than integer ones Accessing memory takes more time than accessing registers CPI is an average number of clock cycles per instruction Important point Changing the cycle time often changes the number of cycles required for various instructions (more later) Clock Cycles per Instruction (CPI) 1 I1 cycles I2I3I6I4I5I7 234567891011121314 CPI =14/7 = 2
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 9 To execute, a given program will require … Some number of machine instructions Some number of clock cycles Some number of seconds We can relate CPU clock cycles to instruction count Performance Equation: (related to instruction count) Performance Equation CPU cycles = Instruction Count × CPI Time = Instruction Count × CPI × cycle time
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 10 I-CountCPICycle ProgramX CompilerXX ISAXXX OrganizationXX TechnologyX Understanding Performance Equation Time = Instruction Count × CPI × cycle time
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 11 Suppose we have two implementations of the same ISA For a given program Machine A has a clock cycle time of 250 ps and a CPI of 2.0 Machine B has a clock cycle time of 500 ps and a CPI of 1.2 Which machine is faster for this program, and by how much? Solution: Both computer execute same count of instructions = I CPU execution time (A) = I × 2.0 × 250 ps = 500 × I ps CPU execution time (B) = I × 1.2 × 500 ps = 600 × I ps Computer A is faster than B by a factor = = 1.2 Using the Performance Equation 600 × I 500 × I
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 12 Determining the CPI Different types of instructions have different CPI Let CPI i = clocks per instruction for class i of instructions Let C i = instruction count for class i of instructions Designers often obtain CPI by a detailed simulation Hardware counters are also used for operational CPUs CPU cycles = (CPI i × C i ) i = 1 n ∑ CPI = (CPI i × C i ) i = 1 n ∑ n ∑ CiCi
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 13 Example on Determining the CPI Problem 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 per instruction, respectively. The first code sequence has 5 instructions:2 of A, 1 of B, and 2 of C The second sequence has 6 instructions:4 of A, 1 of B, and 1 of C Compute the CPU cycles for each sequence. Which sequence is faster? What is the CPI for each sequence? Solution CPU cycles (1 st sequence) = (2×1) + (1×2) + (2×3) = 2+2+6 = 10 cycles CPU cycles (2 nd sequence) = (4×1) + (1×2) + (1×3) = 4+2+3 = 9 cycles Second sequence is faster, even though it executes one extra instruction CPI (1 st sequence) = 10/5 = 2CPI (2 nd sequence) = 9/6 = 1.5
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 14 Given: instruction mix of a program on a RISC processor What is average CPI? What is the percent of time used by each instruction class? Class i Freq i CPI i ALU50%1 Load20%5 Store10%3 Branch20%2 How faster would the machine be if load time is 2 cycles? What if two ALU instructions could be executed at once? Second Example on CPI CPI i × Freq i 0.5×1 = 0.5 0.2×5 = 1.0 0.1×3 = 0.3 0.2×2 = 0.4 %Time 0.5/2.2 = 23% 1.0/2.2 = 45% 0.3/2.2 = 14% 0.4/2.2 = 18% Average CPI = 0.5+1.0+0.3+0.4 = 2.2
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 15 MIPS: Millions Instructions Per Second Sometimes used as performance metric Faster machine larger MIPS MIPS specifies instruction execution rate We can also relate execution time to MIPS MIPS as a Performance Measure Instruction Count Execution Time × 10 6 Clock Rate CPI × 10 6 MIPS == Inst Count MIPS × 10 6 Inst Count × CPI Clock Rate Execution Time ==
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 16 Drawbacks of MIPS Three problems using MIPS as a performance metric 1.Does not take into account the capability of instructions Cannot use MIPS to compare computers with different instruction sets because the instruction count will differ 2.MIPS varies between programs on the same computer A computer cannot have a single MIPS rating for all programs 3.MIPS can vary inversely with performance A higher MIPS rating does not always mean better performance Example in next slide shows this anomalous behavior
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 17 Two different compilers are being tested on the same program for a 4 GHz machine with three different classes of instructions: Class A, Class B, and Class C, which require 1, 2, and 3 cycles, respectively. The instruction count produced by the first compiler is 5 billion Class A instructions, 1 billion Class B instructions, and 1 billion Class C instructions. The second compiler produces 10 billion Class A instructions, 1 billion Class B instructions, and 1 billion Class C instructions. Which compiler produces a higher MIPS? Which compiler produces a better execution time? MIPS example
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 18 Solution to MIPS Example First, we find the CPU cycles for both compilers CPU cycles (compiler 1) = (5×1 + 1×2 + 1×3)×10 9 = 10×10 9 CPU cycles (compiler 2) = (10×1 + 1×2 + 1×3)×10 9 = 15×10 9 Next, we find the execution time for both compilers Execution time (compiler 1) = 10×10 9 cycles / 4×10 9 Hz = 2.5 sec Execution time (compiler 2) = 15×10 9 cycles / 4×10 9 Hz = 3.75 sec Compiler1 generates faster program (less execution time) Now, we compute MIPS rate for both compilers MIPS = Instruction Count / (Execution Time × 10 6 ) MIPS (compiler 1) = (5+1+1) × 10 9 / (2.5 × 10 6 ) = 2800 MIPS (compiler 2) = (10+1+1) × 10 9 / (3.75 × 10 6 ) = 3200 So, code from compiler 2 has a higher MIPS rating !!!
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 19 Amdahl’s Law Amdahl's Law is a measure of Speedup How a computer performs after an enhancement E Relative to how it performed previously Enhancement improves a fraction f of execution time by a factor s and the remaining time is unaffected Performance with E Performance before ExTime before ExTime with E Speedup(E) == ExTime with E = ExTime before × (f / s + (1 – f )) Speedup(E) = (f / s + (1 – f )) 1
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 20 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? Solution: suppose we improve multiplication by a factor s 25 sec (4 times faster) = 80 sec / s + 20 sec s = 80 / (25 – 20) = 80 / 5 = 16 Improve the speed of multiplication by s = 16 times How about making the program 5 times faster? 20 sec ( 5 times faster) = 80 sec / s + 20 sec s = 80 / (20 – 20) = ∞ Impossible to make 5 times faster! Example on Amdahl's Law
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 21 Benchmarks Performance best obtained by running a real application Use programs typical of expected workload Representatives of expected classes of applications Examples: compilers, editors, scientific applications, graphics,... SPEC (System Performance Evaluation Corporation) Funded and supported by a number of computer vendors Companies have agreed on a set of real program and inputs Various benchmarks for … CPU performance, graphics, high-performance computing, client- server models, file systems, Web servers, etc. Valuable indicator of performance (and compiler technology)
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 22 The SPEC CPU2000 Benchmarks 12 Integer benchmarks (C and C++)14 FP benchmarks (Fortran 77, 90, and C) Name Description Name Description gzipCompressionwupwiseQuantum chromodynamics vprFPGA placement and routingswimShallow water model gccGNU C compilermgridMultigrid solver in 3D potential field mcfCombinatorial optimizationappluPartial differential equation craftyChess programmesaThree-dimensional graphics library parserWord processing programgalgelComputational fluid dynamics eonComputer visualizationartNeural networks image recognition perlbmkPerl applicationequakeSeismic wave propagation simulation gapGroup theory, interpreterfacerecImage recognition of faces vortexObject-oriented databaseammpComputational chemistry bzip2CompressionlucasPrimality testing twolfPlace and route simulatorfma3dCrash simulation using finite elements sixtrackHigh-energy nuclear physics apsiMeteorology: pollutant distribution Wall clock time is used as metric Benchmarks measure CPU time, because of little I/O
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 23 SPEC 2000 Ratings (Pentium III & 4) SPEC ratio = Execution time is normalized relative to Sun Ultra 5 (300 MHz) SPEC rating = Geometric mean of SPEC ratios Note the relative positions of the CINT and CFP 2000 curves for the Pentium III & 4 Pentium III does better at the integer benchmarks, while Pentium 4 does better at the floating-point benchmarks due to its advanced SSE2 instructions
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 24 Performance and Power Power is a key limitation Battery capacity has improved only slightly over time Need to design power-efficient processors Reduce power by Reducing frequency Reducing voltage Putting components to sleep Energy efficiency Important metric for power-limited applications Defined as performance divided by power consumption
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 25 Performance and Power
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 26 Energy Efficiency Energy efficiency of the Pentium M is highest for the SPEC2000 benchmarks Relative Energy Efficiency Always on / maximum clockLaptop mode / adaptive clockMinimum power / min clock Benchmark and power mode SPECINT 2000SPECFP 2000SPECINT 2000SPECFP 2000SPECINT 2000SPECFP 2000 Pentium M @ 1.6/0.6 GHz Pentium 4-M @ 2.4/1.2 GHz Pentium III-M @ 1.2/0.8 GHz
PerformanceCOE 308 – Computer Architecture© Muhamed Mudawar – slide 27 Performance is specific to a particular program Any measure of performance should reflect execution time Total execution time is a consistent summary of performance For a given ISA, performance improvements come from Increases in clock rate (without increasing the CPI) Improvements in processor organization that lower CPI Compiler enhancements that lower CPI and/or instruction count Algorithm/Language choices that affect instruction count Pitfalls (things you should avoid) Using a subset of the performance equation as a metric Expecting improvement of one aspect of a computer to increase performance proportional to the size of improvement Things to Remember
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