1. CS35101 – Computer Architecture Week 9: Understanding Performance Paul Durand ( www.cs.kent.edu/~durand ) [Adapted from M Irwin (www.cse.psu.edu/~mji) ] [Adapted from COD, Patterson & Hennessy, © 2005, UCB]www.cse.psu.edu/~mji

2. Indeed, the cost-performance ratio of the product will depend most heavily on the implementer, just as ease of use depends most heavily on the architect. The Mythical Man-Month, Brooks, pg 46

3. Performance Metrics  Purchasing perspective l given a collection of machines, which has the -best performance ? -least cost ? -best cost/performance?  Design perspective l faced with design options, which has the -best performance improvement ? -least cost ? -best cost/performance?  Both require l basis for comparison l metric for evaluation  Our goal is to understand what factors in the architecture contribute to overall system performance and the relative importance (and cost) of these factors

4. Defining (Speed) Performance  Normally interested in reducing l Response time (aka execution time) – the time between the start and the completion of a task -Important to individual users l Thus, to maximize performance, need to minimize execution time l Throughput – the total amount of work done in a given time -Important to data center managers l Decreasing response time almost always improves throughput performance X = 1 / execution_time X If X is n times faster than Y, then performance X execution_time Y -------------------- = --------------------- = n performance Y execution_time X

5. Performance Problem 1 Consider two machines, A and B A runs a program in 20 seconds B runs the same program in 25 seconds All other things being equal Which machine has the better performance? Perf (A) / Perf (B) = Time (B) / Time (A) = 25 / 20 = 1.25 Perf (B) / Perf (A) = Time (A) / Time (B) = 20 / 25 = 0.80 Machine A has better performance By how much? 25% - Machine A is 25% faster than machine B or, Machine A is 1.25 times as fast as machine B

6. Performance Factors  Want to distinguish elapsed time and the time spent on our task  CPU execution time (CPU time) – time the CPU spends working on a task l Does not include time waiting for I/O or running other programs CPU execution time # CPU clock cycles for a program for a program = x clock cycle time CPU execution time # CPU clock cycles for a program for a program clock rate = -------------------------------------------  Can improve performance by reducing either the length of the clock cycle or the number of clock cycles required for a program or

7. Review: Machine Clock Rate  Clock rate (MHz, GHz) is inverse of clock cycle time (clock period) CC = 1 / CR one clock period 10 nsec clock cycle => 100 MHz clock rate 5 nsec clock cycle => 200 MHz clock rate 2 nsec clock cycle => 500 MHz clock rate 1 nsec clock cycle => 1 GHz clock rate 500 psec clock cycle => 2 GHz clock rate 250 psec clock cycle => 4 GHz clock rate 200 psec clock cycle => 5 GHz clock rate

8. Performance Problem 2 Our favorite program runs in 10 seconds on computer A, which has a 4 GHZ 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. However, this will affect the rest of the CPU design, causing 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? Remember that cpu execution time = # cpu clock cycles / clock rate For machine A: 10 sec = n cpu clock cycles / 4 GHz For machine B : 6 sec = 1.2 n cpu clock cycles / target GHz Dividing equation A by equation B => (10/6) = target / (4 x 1.2) Or, target = 4.8 x 1.6667 = 8 GHz

9. Performance Problem 3 - CPI Suppose we have two implementations of the same instruction set architecture (ISA). For some program with n instructions: Machine A has a clock cycle time of 250 ps and a CPI 0f 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? Machine A: cpu time = n instruc x 2.0 clocks/instruc x 250 ps/clock Machine B: cpu time = n instruc x 1.2 clocks/instruc x 500 ps/clock Perf(A/B) = Time(B/A) = 600/500 = 6/5 = 1.2 So machine A is 1.2 times as fast as machine B Or, machine A is 20% faster than machine B

10. Instructions Have Different ExecutionTimes  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)

11. Clock Cycles per Instruction  Since not all instructions take the same amount of time to execute l One way to think about execution time is that it equals the number of instructions executed multiplied by the average time per instruction  Clock cycles per instruction (CPI) – the average number of clock cycles each instruction takes to execute l A way to compare two different implementations of the same ISA # CPU clock cycles # Instructions Average clock cycles for a program for a program per instruction = x CPI for this instruction class ABC CPI123

12. Effective CPI  Computing the overall effective CPI is done by looking at the different types of instructions and their individual cycle counts and averaging Overall effective CPI =  (CPI i x IC i ) i = 1 n l Where IC i is the count (percentage) of the number of instructions of class i executed l CPI i is the (average) number of clock cycles per instruction for that instruction class l n is the number of instruction classes  The overall effective CPI varies by instruction mix – a measure of the dynamic frequency of instructions across one or many programs

13. THE Performance Equation  Our basic performance equation is then CPU time = Instruction_count x CPI x clock_cycle Instruction_count x CPI clock_rate CPU time = ----------------------------------------------- or  These equations separate the three key factors that affect performance l Can measure the CPU execution time by running the program l The clock rate is usually given l Can measure overall instruction count by using profilers/ simulators without knowing all of the implementation details l CPI varies by instruction type and ISA implementation for which we must know the implementation details

14. Determinates of CPU Performance CPU time = Instruction_count x CPI x clock_cycle Instruction_ count CPIclock_cycle Algorithm Programming language Compiler ISA Processor organization Technology

15. Determinates of CPU Performance CPU time = Instruction_count x CPI x clock_cycle Instruction_ count CPIclock_cycle Algorithm Programming language Compiler ISA Processor organization Technology X XX XX XX X X X X X

16. Performance Problem 4 – Compiler Effect  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? By how much? What is the CPI for each sequence? Instruction Class CPICode Sequence 1 # Instruc clock cycles Code Sequence 2 # Instruc clock cycles A12244 B21212 C32613 10 clocks CPI = 2.0 9 clocks CPI = 1.5

17. A Simple Example – Implementation Effects  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? OpFreqCPI i Freq x CPI i ALU50%1. Load20%5 Store10%3 Branch20%2  =

18. A Simple Example  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? OpFreqCPI i Freq x CPI i ALU50%1 Load20%5 Store10%3 Branch20%2  =.5 1.0.3.4 2.2 CPU time new = 1.6 x IC x CC so 2.2/1.6 means 37.5% faster 1.6.5.4.3.4.5 1.0.3.2 2.0 CPU time new = 2.0 x IC x CC so 2.2/2.0 means 10% faster.25 1.0.3.4 1.95 CPU time new = 1.95 x IC x CC so 2.2/1.95 means 12.8% faster

19. Comparing and Summarizing Performance  Guiding principle in reporting performance measurements is reproducibility – list everything another experimenter would need to duplicate the experiment (version of the operating system, compiler settings, input set used, specific computer configuration (clock rate, cache sizes and speed, memory size and speed, etc.))  How do we summarize the performance for benchmark set with a single number? l The average of execution times that is directly proportional to total execution time is the arithmetic mean (AM) AM = 1/n  Time i i = 1 n l Where Time i is the execution time for the i th program of a total of n programs in the workload l A smaller mean indicates a smaller average execution time and thus improved performance

20. SPEC Benchmarks www.spec.orgwww.spec.org Integer benchmarksFP benchmarks gzipcompressionwupwiseQuantum chromodynamics vprFPGA place & routeswimShallow water model gccGNU C compilermgridMultigrid solver in 3D fields mcfCombinatorial optimizationappluParabolic/elliptic pde craftyChess programmesa3D graphics library parserWord processing programgalgelComputational fluid dynamics eonComputer visualizationartImage recognition (NN) perlbmkperl applicationequakeSeismic wave propagation simulation gapGroup theory interpreterfacerecFacial image recognition vortexObject oriented databaseammpComputational chemistry bzip2compressionlucasPrimality testing twolfCircuit place & routefma3dCrash simulation fem sixtrackNuclear physics accel apsiPollutant distribution

21. SPEC 2000 Does doubling the clock rate double the performance? Can a machine with a slower clock rate have better performance?

22. Other Performance Metrics  Power consumption – especially in the embedded market where battery life is important (and passive cooling) l For power-limited applications, the most important metric is energy efficiency

23. Summary: Evaluating ISAs  Design-time metrics: l Can it be implemented, in how long, at what cost? l Can it be programmed? Ease of compilation?  Static Metrics: l How many bytes does the program occupy in memory?  Dynamic Metrics: l How many instructions are executed? How many bytes does the processor fetch to execute the program? l How many clocks are required per instruction? l How "lean" a clock is practical? Best Metric: Time to execute the program! CPI Inst. CountCycle Time depends on the instructions set, the processor organization, and compilation techniques.