Chapter 2 Performance Text in blue is by

Chapter 2 Performance Text in blue is by N. Guydosh Updated 1/25/04* Ó 1998 Morgan Kaufmann Publishers 1

Performance • • Measure, Report, and Summarize Make intelligent choices See through the marketing hype Key to understanding underlying organizational motivation 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? Ó 1998 Morgan Kaufmann Publishers 2

Which of these airplanes has the best performance? Airplane Passengers Boeing 737 -100 Boeing 747 BAC/Sud Concorde Douglas DC-8 -50 Range (mi) 101 470 132 146 Speed (mph) Thru-put(pass*mph) 630 598 4150 610 4000 1350 8720 544 60, 398 286, 700 178, 200 79, 424 • How much faster is the Concorde compared to the 747? • How much bigger is the 747 than the Douglas DC-8? • What about throughput as a metric? Ó 1998 Morgan Kaufmann Publishers 3

Computer Performance: 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? Ó 1998 Morgan Kaufmann Publishers 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 CPU time, and user CPU time • Elapsed time = system CPU time + waiting time • Our focus: user CPU time – time spent executing the lines of code that are "in" our program – Include both system and user CPU time- difficult to distinguish between system and user CPU time from OS to OS Ó 1998 Morgan Kaufmann Publishers 5

Book's Definition of Performance • For some program running on machine X, Performance. X = 1 / Execution time. X • "X is n times faster than Y" Performance. X / Performance. Y = n • Distinguish between execution time based on elapsed time vs. CPU time • Definitions: – System performance is based on elapsed time on an unloaded system – CPU performance is based on user CPU time Problem: – machine A runs a program in 20 seconds – machine B runs the same program in 25 seconds – PA/PB = (1/20) (1/25) = 25/20 = EB/EA … n = 1. 25 • Ó 1998 Morgan Kaufmann Publishers 6

Clock Cycles • Instead of reporting execution time in seconds, we often use cycles • Clock “ticks” indicate when to start activities (one abstraction): time • • 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 Ó 1998 Morgan Kaufmann Publishers 7

How to Improve Performance equation in terms of time: CPU time for the program = CPU cycles for a program x Clock cycle time CPU time for the program = CPU cycles for a program Clock rate 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. Ó 1998 Morgan Kaufmann Publishers 8

How many cycles are required for a program? . . . 6 th 5 th 4 th 3 rd instruction 2 nd instruction Could assume that # of cycles = # of instructions 1 st instruction • time This assumption is incorrect, In general, different instructions take different amounts of time on different machines. Why? hint: remember that these are machine instructions, not lines of C code Ó 1998 Morgan Kaufmann Publishers 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) Ó 1998 Morgan Kaufmann Publishers 10

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 Can be done by doubling the clock frequency to 800 Mhz – se next slide and text pp. 60 -61 – Question: if doubling the clock rate increased substantially performance, why not do it again …. and again …? Ó 1998 Morgan Kaufmann Publishers 11

Example solution steps CPU_clock_cycles. A = CPU_time. A x Clock_rate. A = 10 sec x 400 x 106 cycles/sec = 4000 x 106 cycles Clock_rate. B = 1. 2 x CPU_clock_cycles. A / CPU_time. B = 1. 2 x 4000 x 106 cycles / 6 sec = 800 x 106 = 800 Mhz – Question: if doubling the clock rate substantially increased performance, why not do it again …. and again …? Ó 1998 Morgan Kaufmann Publishers 12

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 Ó 1998 Morgan Kaufmann Publishers 13

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. Basic performance equation: CPU time = Instruction count x CPI x Clock cycle time CPU time = Instruction count x CPI / Clock rate • Ó 1998 Morgan Kaufmann Publishers 14

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? See next slide and text page 62 • If two machines have the same ISA (instruction set architecture) which of our quantities (e. g. , clock rate, CPI, execution time, # of instructions, MIPS) will always be identical? Ó 1998 Morgan Kaufmann Publishers 15

CPI Example - Solution Assume each machine executes the same number of insructions: I CPU_clock_cycles. A = I x 2 cyc/inst CPU_clock_cycles. B = I x 1. 2 cyc/inst CPU_time. A = CPU_clock_cycles. A x Clock_cycle_time. A = I x 2 cyc/inst x 1 ns = 2 x. I ns CPU_time. B = CPU_clock_cycles. B x Clock_cycle_time. B = I x 1. 2 cyc/inst x 2 ns = 2. 4 x. I ns Machine A is 1. 2 times faster than machine B Ó 1998 Morgan Kaufmann Publishers 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 C The 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? See next slide and text pp. 64 -65 Ó 1998 Morgan Kaufmann Publishers 17

# of Instructions Example - Solution CPU clock cycles = n is the number of instruction classes Ci is the number of instruction types in class i executed Where CPIi is the average number of cycles for instructions in class i Remember sequence 1 has a total of 5 instructions and sequence 2 has a total of 6 instructions CPU_clock_cycles 1 = 2 x 1 + 1 x 2 + 2 x 3 = 10 cycles CPU_clock_cycles 2 = 4 x 1 + 1 x 2 + 1 x 3 = 9 cycles Thus code sequence 2 is faster, even though it actually executes one extra instruction CPI 1 = CPU_clock_cycles 1 / Instruction_count 1 = 10/5 = 2 cycles per inst CPI 2 = CPU_clock_cycles 2 / Instruction_count 2 = 9/6 = 1. 5 cycles per inst Ó 1998 Morgan Kaufmann Publishers 18

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 generated code uses 5 billion Class A instructions, 1 billion Class B instructions, and 1 billion Class C instructions. The second compiler's generated code uses 10 billion Class A instructions, 1 billion Class B instructions, and 1 billion Class C instructions. • • Which generated sequence will be faster according to MIPS? Which generated sequence will be faster according to execution time? • See next slide and text pp. 78 -79 Ó 1998 Morgan Kaufmann Publishers 19

MIPS example - Solution Using: CPU clock cycles = And Execution_time = CPU_clock_cycles / Clock_rate CPU_clock_cycles 1 = (5 x 1 + 1 x 2 + 1 x 3 )x 109 = 10 x 109 CPU_clock_cycles 2 = (10 x 1 + 1 x 2 + 1 x 3 )x 109 = 15 x 109 Execution_time 1 = 10 x 109 / 500 x 106 = 20 sec Execution_time 2 = 15 x 109 / 500 x 106 = 30 sec Compiler 1 generates the faster program MIPS = Instruction_count / (Execution_time x 106) MIPS 1 = (5+1+1)x 109 /20 x 106 = 350 MIPS 2 = (10+1+1)x 109 /30 x 106 = 400 MIPS The code from compiler 2 has a higher MIP rate, but the code from compiler 1 runs faster! Ó 1998 Morgan Kaufmann Publishers 20

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) In an “aggressive” attempt to get optimized code for an Intel processor using a particular benchmark, the compiler designer “over-optimized” output, and the compiler generated erroneous code. The compiler was not available to the public and thus was inconsistent with what the user saw. – valuable indicator of performance (and compiler technology) Ó 1998 Morgan Kaufmann Publishers 21

SPEC ‘ 89 • Compiler “enhancements” and performance The effect of designing compilers around benchmarks. The enhanced compiler was finely tuned to matrix 300 and gave a misleading indication of the overall compiler performance. Ó 1998 Morgan Kaufmann Publishers 22

SPEC ‘ 95 Ó 1998 Morgan Kaufmann Publishers 23

SPEC ‘ 95 Does doubling the clock rate double the performance? Can a machine with a slower clock rate have better performance? Ó 1998 Morgan Kaufmann Publishers 24

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? " 100/4 = 20 +(80/n), n = 16 needed improvement for multiply How about making it 5 times faster? 100/5 = 20 +(80/n ==> 0 = 80/n, not possible Principle: Make the common case fast Ó 1998 Morgan Kaufmann Publishers 25

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? Same as problem 2. 41. Solution : Execute_time_before = 10 sec Execute_time_after = 5 + 5/5 = 6 sec speedup = Execute_time_before / Execute_time_after = 10/6 = 1. 7 • We are looking for a benchmark to show off the new floating-point unit described above (speed FP by 5), 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? Same as problem 2. 42. Solution: Speedup = Execute_time_before / Execute_time_after = 3 let p be the fraction of time spent doing floating point in the bench mark Execute_time_after = 100(1 -p) + (100 p)/5 = 100/3 = 33. 3 p = 83. 3% of the original time to be spent doing floating point in order to achieve this speedup … make the common case fast! Ó 1998 Morgan Kaufmann Publishers 26

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) Ó 1998 Morgan Kaufmann Publishers 27

Summary of Fallacies and Pitfalls (sec. 2. 7) • • • Pitfall: Expecting the improvement of one aspect of a machine to increase performance by an amount proportional to the size of the improvement. – Amdahl’s law “the performance enhancement possible with a given improvement is limited by the amount that the improved feature is used … the law of diminishing returns. Fallacy: Hardware independent metrics predict performance. – Ex: code size – a small program generated may do more looping – Execution time is ultimately what should be used Pitfall: Using MIPS as a performance metric – A lower MIP machine with more powerful instructions may oput perform a higher MIP machine – use with caution – perhaps within a given ISA family of machines (Ex: IBM main frames) Fallacy: Synthetic benchmarks predict performance. – Synthetic or artificial benchmark programs (Ex: Whetstone or Dhrystone) may not match the relevant real applications to be run. Pitfall: Using the arithmetic means of normalized execution times to prdict performance. – Can give inconsistent results – geometric mean is consistent but … Fallacy: The geometric mean of execution time ratiosis proportional to total execution time. – Not so – example p. 81 shows the 2 pgms have same performce, but b is 9. 1 faster than A Ó 1998 Morgan Kaufmann Publishers 28