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Performance Computer Architecture – CS401 Erkay Savas Sabanci University 7/17/2015 Erkay Savas 1 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 hardware are related to system overall performance? • How does the machine's instruction set affect performance? 7/17/2015 Erkay Savas 2 Airplane Analogy • Which of these airplanes has the best performance? Airplane Boeing 777 Passenger throughput Passenger Range Speed Capacity (miles) (m.p.h) (passenger x m.p.h) 228750 375 4630 610 Boeing 747 470 4150 610 268700 Airbus A 3xx 656 8400 600 393600 Concorde 132 4000 1350 178200 Douglas DC-8-50 146 8720 544 79424 7/17/2015 Erkay Savas 3 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? 7/17/2015 Erkay Savas 4 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 7/17/2015 Erkay Savas 5 CPU Time • Unix time command 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 7/17/2015 Erkay Savas 6 A Definition of Performance • For some program running on machine X PerformanceX = 1/Execution_timeX • The machine X is said to be “n times faster” than the machine Y if 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? 7/17/2015 Erkay Savas 7 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. 7/17/2015 Erkay Savas 8 Clock Cycles • Clock “ticks” indicate when to start activities time Start of events often the rising edge of the clock • Instead of reporting execution time in seconds, we often use cycles seconds cycles seconds program program cycle 7/17/2015 Erkay Savas 9 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 7/17/2015 Erkay Savas 10 CPI • CPI Clocks 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 7/17/2015 Erkay Savas 11 Improving Performance seconds cycles seconds program program cycles So, to improve performance 1. Increase the clock frequency (i.e. decrease the clock period) 2. Reduce the number of the clock cycles per program (IC CPI) 7/17/2015 Erkay Savas 12 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) 7/17/2015 Erkay Savas 13 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. 7/17/2015 Erkay Savas 14 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 write CPU clock cycles for a program = IC CPI • Then CPU execution time = (IC CPI)/Clock rate 7/17/2015 Erkay Savas 15 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? 7/17/2015 Erkay Savas 16 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 variables is indicative of performance when it really isn’t. 7/17/2015 Erkay Savas 17 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): 1. 2106 of A, 106 of B, and 2106 of C (IC = 5106) 2. 4106 of A, 106 of B, and 106 of C (IC = 6106) – Which code sequence is faster? 7/17/2015 Erkay Savas 18 MIPS • Millions Instructions Per Second = MIPS MIPS MIPS MIPS = = = = IC/(Execution_time 106) IC/(#of clocks cycle time 106) (IC clock rate)/(IC CPI 106) clock rate/(CPI 106) • A faster machine has a higher MIPS Execution_time = IC/(MIPS 106) 7/17/2015 Erkay Savas 19 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? 7/17/2015 Erkay Savas 20 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. 7/17/2015 Erkay Savas 21 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 7/17/2015 Erkay Savas 22 Overview • • A given program will require 1. Some number of instructions 2. Some number of clock cycles 3. 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 7/17/2015 Erkay Savas 23 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 7/17/2015 Erkay Savas 24 Question • Assuming two machines have the same ISA, which of the following quantities are identical? – – – – – Clock rate CPI Execution time # of instructions MIPS 7/17/2015 Erkay Savas 25 Program Performance HW or SW component Algorithm Affects what? How? IC, possibly CPI Programming IC, CPI Language Compiler IC, CPI ISA 7/17/2015 IC, clock rate, CPI Erkay Savas 26 Benchmarks • Programs specifically chosen to measure performance – must reflect typical workload of the user • Benchmark types – – – – Real applications Small benchmarks Benchmark suites Synthetic benchmarks 7/17/2015 Erkay Savas 27 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 7/17/2015 Erkay Savas 28 Comparing & Summarizing Performance Computer A Computer B Program 1 1s 100 s Program 2 1000 s 100 s Total time 1001 s 200 s • 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: 1 n AM Timei n i 1 7/17/2015 Erkay Savas 29 Arithmetic Mean • If each program, in the workload, do not run equal times, then we have to use weighted arithmetic mean 1 n AM wi Timei n i 1 • Suppose that the program 1 runs 10 times as often as the program 2. Which machine is faster? Computer A Computer B Program 2 (seconds) weight 10 1 1 1000 100 100 Weighted AM - ? ? Program 1 (seconds) 7/17/2015 Erkay Savas 30