Performance, Power, Die Yield CS301 Prof Szajda
Administrative HW #1 assigned w Due Wednesday, 9/3 at 5:00 pm
Performance Metrics (How do we compare two machines?)
What to Measure? Which airplane has the best performance? 4
Performance One size does not fit all Depends on application domain w Scientific computing w Graphics w Databases w General-Purpose desktop w Beware of designing to benchmark! Depends on technology characteristics w DRAM speed and capacity, chip size, etc.
Which Metric Do We Use? Response or execution time w Difference between start and end time w Individual user cares most about this Throughput w Total amount of work done in given time w Frequently used for servers and clusters How are these affected by w Replacing processor with faster version? w Adding more processors?
Execution Time Shorter execution time is better Allows comparison between 2 machines
Relative Performance X is n times faster than Y Example: w Machine A takes 10s to run program w Machine B takes 15s to run same program w What is the performance ratio?
Different Time Values Execution time w Wall-clock, response, or elapsed time Includes everything (processing,i/o, OS overhead, etc)! w Determines system performance CPU time w Time spent executing code for this task only Does not include I/O or time-sharing w Comprises user CPU time and system CPU time Difference programs are affected differently by CPU and system performance w man time 90.7u 12.9s 2:39 65% User: 90.7 sec System: 12.9 sec Elapsed time: 2 min 39 sec
Clock Cycles Instead of expressing time in seconds, use clock cycles Clock w Determines when events take place w Runs at constant rate (ex. 1 GHz) w Easy to convert between clock rate and seconds Clock rate = 1 / Clock Cycle 500 MHz = 1 / (2 ns) 1 ns = 10-9 s
CPU Clocking Operation of digital hardware governed by a constant-rate clock Clock period Clock (cycles) Data transfer and computation Update state Clock period: duration of a clock cycle e.g., 250ps = 0.25ns = 250 10 12 s Clock frequency (rate): cycles per second e.g., 4.0GHz = 4000MHz = 4.0 109 Hz Chapter 1 Computer Abstractions and Technology
CPU Time Performance improved by Reducing number of clock cycles Increasing clock rate Hardware designer must often trade off clock rate against cycle count Chapter 1 Computer Abstractions and Technology
CPU Time Example Computer A: 2GHz clock, 10s CPU time Designing Computer B Aim for 6s CPU time Can do faster clock, but causes 1.2 clock cycles How fast must Computer B clock be? Chapter 1 Computer Abstractions and Technology
Instruction Count and CPI Instruction Count for a program Determined by program, ISA and compiler Average cycles per instruction Determined by CPU hardware If different instructions have different CPI Average CPI affected by instruction mix Chapter 1 Computer Abstractions and Technology
CPI Example Computer A: Cycle Time = 250ps, CPI = 2.0 Computer B: Cycle Time = 500ps, CPI = 1.2 Same ISA Which is faster, and by how much? A is faster by this much Chapter 1 Computer Abstractions and Technology
Application Characteristics Determine the mix of different instruction types w Integer arithmetic w Logical operations w Floating point arithmetic w Loads and stores Different applications have different CPI because of different instruction mixes
CPI in More Detail If different instruction classes take different numbers of cycles Weighted average CPI Relative frequency Chapter 1 Computer Abstractions and Technology
CPI Example Alternative compiled code sequences using instructions in classes A, B, C Class A B C CPI for class 1 2 3 IC in sequence 1 2 1 2 IC in sequence 2 4 1 1 Sequence 1: IC = 5 Clock Cycles = 2 1 + 1 2 + 2 3 = 10 Avg. CPI = 10/5 = 2.0 Sequence 2: IC = 6 Clock Cycles = 4 1 + 1 2 + 1 3 = 9 Avg. CPI = 9/6 = 1.5 Chapter 1 Computer Abstractions and Technology
Performance Summary The BIG Picture Performance depends on Algorithm: affects IC, possibly CPI Programming language: affects IC, CPI Compiler: affects IC, CPI Instruction set architecture: affects IC, CPI, T c Chapter 1 Computer Abstractions and Technology
Amdahl s Law How much speedup do you get from an enhancement? Speedup = Execution time w/o enhancement Execution time w/ enhancement Based on w Fraction of time enhancement used w Improvement in enhanced mode Exec new = Exec old ((1-fraction enh ) + fraction enh ) Speedup enh
Pitfall: Amdahl s Law Improving an aspect of a computer and expecting a proportional improvement in overall performance 1.10 Fallacies and Pitfalls Example: multiply accounts for 80s/100s How much improvement in multiply performance to get 5 overall? Can t be done! Corollary: make the common case fast Chapter 1 Computer Abstractions and Technology
Review Question Your machine has a clock rate of 2.4GHz. How long is the clock cycle?
Review Questions Suppose you are given the following: w Machine A 1 GHz Average CPI = 1.6 Instructions = 1.7 Billion w Machine B 3.3 GHz Average CPI = 6.1 Instructions = 2 Billion Which machine is faster? By how much?
Review Questions What is the average CPI for a machine with the following CPIs on an application with the following instruction frequency? Type Frequency CPI Arithme(c 0.45 1 Memory 0.3 8 Control 0.2 3 Mult/Div 0.05 5
Review Questions What factors must be included when comparing the relative performance of two machines?
Amdahl s Law Exec new = Exec old ((1-fraction enh ) + fraction enh ) Speedup enh Suppose you have an enhancement that makes function 10x faster. Speedup if used 5% of the time? Speedup if used 40% of the time?
Review Questions What is the equation for execution time? What does Amdahl s Law say?
Benchmarks Programs specifically used to measure performance Hope is that it is representative of how computer will be used Examples w SPEC Integer and Floating Point w MediaBench w MineBench w TPC
SPEC CPU Benchmark Programs used to measure performance Supposedly typical of actual workload Standard Performance Evaluation Corp (SPEC) Develops benchmarks for CPU, I/O, Web, SPEC CPU2006 Elapsed time to execute a selection of programs Negligible I/O, so focuses on CPU performance Normalize relative to reference machine Summarize as geometric mean of performance ratios CINT2006 (integer) and CFP2006 (floating-point) Chapter 1 Computer Abstractions and Technology
CINT2006 for Intel Core i7 920 Chapter 1 Computer Abstractions and Technology
Recent Concern: Power Trends 1.7 The Power Wall In CMOS IC technology 30 5V 1V 1000 Chapter 1 Computer Abstractions and Technology
Tricks to Increase Power Attach large cooling devices Turn off parts of chips not used in given clock cycle w Can increase power to 300 watts... w...but these and other ways all prohibitively expensive for desktop computers. So... 32
More Recent Approaches: Chip Multiprocessors Reasons for change w Limited opportunities to improve single thread performance w Power w On-chip communication latencies
Tapering Processor Performance
Uniprocessor Performance 1.8 The Sea Change: The Switch to Multiprocessors Constrained by power, instruction-level parallelism, memory latency Chapter 1 Computer Abstractions and Technology
Multiprocessors Multicore microprocessors More than one processor per chip Requires explicitly parallel programming Compare with instruction level parallelism Hardware executes multiple instructions at once Hidden from the programmer Hard to do Programming for performance Load balancing Optimizing communication and synchronization Chapter 1 Computer Abstractions and Technology
Concluding Remarks Cost/performance is improving Due to underlying technology development Hierarchical layers of abstraction In both hardware and software Instruction set architecture The hardware/software interface Execution time: the best performance measure Power is a limiting factor Use parallelism to improve performance 1.9 Concluding Remarks Chapter 1 Computer Abstractions and Technology