Chapter 2. The Role of Performance

Transcription

1 Chapter 2 The Role of Performance 1 Performance Why 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

2 Which of these airplanes has the best performance? Airplane Passengers Range (mi) Speed (mph) Boeing Boeing BAC/Sud Concorde Douglas DC The plane with the highest cruising speed is the Concorde. The plane with the longest range is the DC-8. The plane with the largest capacity is the Boeing 747. How much faster is the Concorde compared to the 747? How much bigger is the 747 than the Douglas DC-8? 3 Airplanes Example (cont.) Airplane Boeing Boeing 747 BAC/Sud Concorde Douglas DC-8-50 Passenger Throughput (Passengers x m.p.h) 228, , ,200 79,424 If we define performance in terms of speed, this leaves two possible definitions: 1. Define the fastest plane as the one with the highest speed, taking a single passenger from one point to another in the least time. The Concorde would clearly be the fastest. 2. Define the fastest plane as the one with the highest speed, taking 450 passengers from one point to another in the least time. The Boeing 747 would clearly be the fastest. Check the above table. 4

3 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? The time between the start and completion of a task. Throughput How many jobs can the machine run at once? What is the average execution rate? How much work is getting done? The total amount of work done in a given time. 5 Example 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? Answer: Case 1: Both response time and throughput are improved. Case 2: No one task gets work done faster, so only throughput increases. But by increasing the number of processors would reduce the waiting time in the queue, therefore response time could improve. 6

4 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 Example: Unix time Command 90.7u 12.9s 2:39 65% 90.7u : User CPU Time (seconds) 12.9s : System CPU Time (seconds) 2:39 : Elapsed Time (159 seconds) 65% : Percentage of Elapsed Time that is CPU Time is ( ) / 159 = 0.65 More than a third of the elapsed time in this example was spent waiting for I/O, running other programs, or both. 8

5 Book's Definition of Performance For some program running on machine X, Performance X = 1 / Execution time X machine X is n times faster than machine Y" Performance X / Performance Y = n Problem: machine A runs a program in 10 seconds machine B runs the same program in 15 seconds, how much faster is A than B? Answer: A is 1.5 times faster than B. 9 Clock Cycles Instead of reporting execution time in seconds, we often use cycles seconds program = cycles program seconds cycle A simple formula relates the most basic metrics (clock cycles and clock cycle time) to CPU time: CPU execution time = CPU clock cycles X Clock cycle time for a program for a program Or, because clock rate and clock cycle time are inverses: CPU execution time = (CPU clock cycles ) / Clock rate for a program 10

6 Clock Cycles (Cont.) 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 time = 5 nanoseconds 11 How to Improve Performance seconds program = cycles program seconds cycle So, hardware designer can improve performance by reducing either the length of the clock cycle or the number of clock cycles required for a program. 12

7 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?" 13 Answer First find the number of clock cycles required for the program on A: CPU time A = (CPU clock cycles A ) / Clock rate A 10 seconds = (CPU clock cycles A ) / 400 x 10 6 cycles/sec CPU clock cycles A = 10 sec X 400 X 10 6 cycles/sec = 4000 X 10 6 cycles CPU time for B can be found using this equation: CPU time B = (1.2 X CPU clock cycles A ) / Clock rate B 6 seconds = (1.2 X 4000 X 10 6 cycles) / Clock rate B Clock rateb = (1.2 X 4000 X 10 6 cycles) / 6 seconds = (800 X 10 6 cycles) / second = 800 MHz Machine B must therefore have twice the clock rate of A to run the program in 6 seconds. 14

8 Cycles and Instructions Since the compiler clearly generated instructions to execute, and the machine had to execute the instructions to run the program, the execution time must depend on the number of instructions in a program. 15 How many cycles are required for a program? Could assume that # of cycles = # of instructions 1st instruction 2nd instruction 3rd instruction 4th 5th 6th... This assumption is incorrect, time different instructions take different amounts of time on different machines. Why? hint: remember that these are machine instructions, not lines of C code 16

9 Different numbers of cycles for different instructions time 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) 17 CPI (clock cycles per instruction) The number of clock cycles required for a program can be written as: CPU clock cycles = Instructions for a program X Average clock cycles per instruction The term clock cycles per instruction, which is the average number of clock cycles each instruction takes to execute, is called CPI. Since different instructions may take different amounts of time depending on what they do, CPI is an average of all instructions executed in the program. 18

10 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 1 ns. and a CPI of 2.0 Machine B has a clock cycle time of 2 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? 19 Answer We know that each machine executes the same number of instructions for the same program; let s call this number I First find the number of processor clock cycles for each machine: CPU clock cycles A = I x 2.0 CPU clock cycles B = I x 1.2 Now we can compute the CPU time for each machine: CPU time A = CPU clock cycles A x Clock cycle time A = I x 2.0 x 1 ns = 2 x I ns CPU time B = I x 1.2 x 2 ns = 2.4 x I ns Clearly, machine A is faster. The amount faster is given by the ratio of the execution times: CPU Performance A = Execution time B = 2.4 x I ns = 1.2 CPU Performance B Execution time A 2 x I ns Machine A is 1.2 times faster than machine B for this program 20

11 Instruction Count Basic Performance Equation in terms of instruction count (the number of instructions executed by the program), CPI, and clock cycle time: or CPU time = Instruction count x CPI x Clock cycle time CPU time = (Instruction count x CPI) / Clock rate Execution Time = Instructions x Clock cycles x Seconds Program Instruction Clock cycle How to find the above performance parameters? CPU clock cycles = Summation (CPI i x C i ) i = 1 to n Where C i is the count of the number of instructions of class i executed, CPI i is the average number of cycles per instruction for that instruction class, and n is the number of instruction classes. 21 Number 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 C The second sequence has 6 instructions: 4 of A, 1 of B, and 1 of C. Which sequence will be faster? What is the CPI for each sequence? 22

12 Answer Sequence 1 executes = 5 instructions. Sequence 2 executes = 6 instructions. So, sequence 1 executes fewer instructions. To find the total number of clock cycles for each sequence: CPU clock cycles = Summation (CPI i x C i ) i = 1 to n CPU clock cycles 1 = (2x1) + (1x2) + (2x3) = 10 cycles CPU clock cycles 2 = (4x1) + (1x2) + (1x3) = 9 cycles So code sequence 2 is faster, even though it actually executes one extra instruction. Since code sequence 2 takes fewer overall clock cycles but has more instructions, it must have a lower CPI. 23 Answer (Cont.) The CPI values can be computed by: CPI = CPU clock cycles Instruction count CPI1 = CPU clock cycles 1 = 10 = 2 Instruction count 1 5 CPI2 = CPU clock cycles 2 = 9 = 1.5 Instruction count 2 6 Note: When comparing two machines, you must look at all three components (instruction count, CPI, and clock cycle time), which combine to form execution time. 24

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 vocabulary 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 25 Performance Performance is determined by execution time Do any of the other variables equal performance? number of cycles to execute program? number of instructions in program? number of cycles per second? average number of cycles per instruction? average number of instructions per second? Common pitfall: thinking one of the variables is indicative of performance when it really isn t. 26

14 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) 27 SPEC 89 Compiler enhancements and performance SPEC performance ratio gcc espresso spice doduc nasa7 li eqntott matrix300 fp pp p to m ca tv Benchmark Compiler Enhanced compiler 28

15 SPEC 95 (8 integer followed by 10 floating-point benchmarks) Benchmark Description go Artificial intelligence; plays the game of Go m88ksim Motorola 88k chip simulator; runs test program gcc The Gnu C compiler generating SPARC code compress Compresses and decompresses file in memory li Lisp interpreter ijpeg Graphic compression and decompression perl Manipulates strings and prime numbers in the special-purpose programming language Perl vortex A database program tomcatv A mesh generation program swim Shallow water model with 513 x 513 grid su2cor quantum physics; Monte Carlo simulation hydro2d Astrophysics; Hydrodynamic Naiver Stokes equations mgrid Multigrid solver in 3-D potential field applu Parabolic/elliptic partial differential equations trub3d Simulates isotropic, homogeneous turbulence in a cube apsi Solves problems regarding temperature, wind velocity, and distribution of pollutant fpppp Quantum chemistry wave5 Plasma physics; electromagnetic particle simulation 29 SPECint 95 SPEC ratio: bigger numeric results indicate faster performance SPECint Clock rate (MHz) Pentium Pentium Pro Pentium Pro is 1.4 to 1.5 times faster than Pentium 30

16 SPECfp 95 SPEC ratio: bigger numeric results indicate faster performance SPECfp Clock rate (MHz) Pentium Pentium Pro Pentium Pro is 1.7 to 1.8 times faster than Pentium 31 Synthetic Benchmarks Synthetic benchmarks are artificial programs that are constructed to try to match the characteristics of a large set of programs. Whetstone and Dhrystone are the most popular synthetic benchmarks. Whetstone was based on measurements of Algol programs in a scientific and engineering environment. It was later converted to Fortran and became popular. Dhrystone was created as benchmark for systems programming environments. It was originally written in Ada and later converted in C, after which it became popular. 32

17 Synthetic Benchmarks Drawbacks One major drawback of synthetic benchmarks is that no user would ever run a synthetic benchmark as an application because these programs do not compute anything a user would find remotely interesting. Synthetic benchmarks are not real programs, they usually do not reflect program behavior, other than the behavior considered when they were created. 33 CPU Performance Does doubling the clock rate double the performance? For a given instruction set architecture, increases in CPU performance can come from three sources: 1. Increases in clock rate 2. Improvements in processor organization that lower the CPI 3. Compiler enhancements that lower the instruction count or generate instructions with a lower average CPI (e.g., by using simpler instructions) 34

18 Comparing and Summarizing Performance Example: Execution times of 2 programs on 2 different machines Program 1 (seconds) Program 2 (seconds) Total time (seconds) Computer A Computer B Which is best answer? 1. A is 10 times faster than B for program B is 10 times faster than A for program B is 9.1 times faster than A for programs 1 & Comparing Performance (Cont.) Answer: Best answer is Performance B = Execution time A = 1001 = 9.1 Performance A Execution time B 110 The average of the execution times that is directly proportional to total execution time is the arithmetic mean (AM): AM = 1/n x Summation Time i i = 1 to n where Time i is the execution time for the ith program of a total of n in the workload. Since it is the mean of execution times, a smaller mean indicates a smaller average execution time and thus improved performance. 36

19 Amdahl's Law The performance enhancement possible with a given improvement is limited by the amount that the improved feature is used (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 5 times faster?" 37 Amdahl's Law (Cont.) Answer: Execution time after improvement = ( seconds) + (80 seconds / n) Since we want the performance to be 5 times faster, the new execution time should be 20 seconds, giving: 20 seconds = (20 seconds) + (80 seconds / n) 0 = (80 seconds / n) That is, there is no amount by which we can enhance multiply to achieve a fivefold increase in performance, if multiply accounts for only 80% of the workload. 38

20 Amdahl's Law (Cont.) Amdahl s law is given in another form that yields the speedup. Speedup is the measure of how a machine performs after some enhancement relative to how it performed previously. Speedup = Performance after improvement Performance before improvement = Execution time before improvement Execution time after improvement 39 MIPS (million instructions per second) MIPS = Instruction count / (Execution time x 10 6 ) Faster machines have a higher MIPS rating. Three problems with using MIPS: 1. MIPS specifies the instruction execution rate but does not take into account the capabilities of the instructions. We cannot compare computers with different instruction sets. 2. MIPS varies between programs on the same computer; thus a machine cannot have a single MIPS rating for all programs. 3. MIPS can vary inversely with performance. 40

21 MIPS Example Two different compilers are being tested for a 500 MHz. machine with three different classes of instructions: Class A, Class B, and Class C, which require one, two, and three cycles (CPI) respectively. Both compilers are used to produce code for a large piece of software. The first compiler's code uses 5 billion Class A instructions, 1 billion Class B instructions, and 1 billion Class C instructions. The second compiler's code uses 10 billion Class A instructions, 1 billion Class B instructions, and 1 billion Class C instructions. Which sequence will be faster according to MIPS? Which sequence will be faster according to execution time? 41 MIPS Example (Answer) First we find the execution time for the 2 different compiles using the following equation: Execution time = CPU clock cycles / Clock rate where CPU clock cycles = Summation (CPI i x C i ) i = 1 to n CPU clock cycles1 = (5x1+1x2+1x3)x10 9 = 10 x 10 9 CPU clock cycles2 = (10x1+1x2+1x3)x10 9 = 15 x 10 9 Now, we find the execution time for the 2 compiles: Execution time1 = (10 x 10 9 ) / (500 x 10 6 ) = 20 seconds Execution time2 = (15 x 10 9 ) / (500 x 10 6 ) = 30 seconds So, we conclude that compiler 1 generates the faster program, according to execution time. 42

22 MIPS Example (Answer Cont.) Now let s compute the MIPS rate for each version of the program, using: MIPS = (Instruction count) / (Execution time x 10 6 ) MIPS1 = ((5+1+1) x 10 9 ) / (20 x 10 6 ) = 350 MIPS2 = ((10+1+1) x 10 9 ) / (30 x 10 6 ) = 400 So, the code from compiler 2 has a higher MIPS rating, but the code from compiler 1 runs faster! From the above example, MIPS fails to give a true picture of performance; even when comparing 2 versions of the same program from 2 different compilers on the same machine 43 MFLOPS as a Performance Metric MFLOPS (megaflops): million floating-point operations per second. MFLOPS = Number of floating-point operations in a program Execution time x 10 6 A floating-point is an addition, subtraction, multiplication, or division operation applied to a number in a single or double precision floating-point representation (key words: float, real, double ). MFLOPS rating is dependent on the program. That is, different programs require the execution of different number of floatingpoint operations. Compilers have a MFLOPS rating near to zero no matter how fast the machine is, because compilers rarely use floating-point arithmetic. 44

23 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 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 45 Performance versus Cost All computer designers must balance performance and cost. Performance is the primary goal and cost is secondary. The cost of a machine is affected not only by the cost of the components, but by the costs of labor to assemble the machine, of research and development overhead, of sales and marketing, and of the profit margin. Computer designs will always be measured by cost and performance, and finding the best balance will always be the art of computer design. 46