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Performance Lecture notes from MKP, H. H. Lee and S. Yalamanchili.

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Presentation on theme: "Performance Lecture notes from MKP, H. H. Lee and S. Yalamanchili."— Presentation transcript:

1 Performance Lecture notes from MKP, H. H. Lee and S. Yalamanchili

2 Reading Section 1.6 Practice Problems: Module 3 – 20, 21, 28

3 Goals Provide a simple model of performance for processor and memory (later) architectures Simple model for reasoning about the impact of compiler, architecture and technology properties A model for understanding performance limits

4 Understanding Performance
Morgan Kaufmann Publishers May 9, 2019 Understanding Performance Algorithm Determines number of operations executed Programming language, compiler, architecture Determine number of machine instructions executed per operation Processor and memory system Determine how fast instructions are executed I/O system (including OS) Determines how fast I/O operations are executed Instruction Set Architecture Chapter 1 — Computer Abstractions and Technology

5 Morgan Kaufmann Publishers
May 9, 2019 Metrics Response time (latency) How long it takes to do a task Throughput Total work done per unit time e.g., tasks/transactions/… per hour Trading throughput vs. latency Energy/Power Measure of work being performed Increases with clock frequency/voltage Determines temperature Is affected by temperature At the moment our focus is here We will then move here (pipelining) And finish here Chapter 1 — Computer Abstractions and Technology

6 Morgan Kaufmann Publishers
May 9, 2019 Relative Performance “X is n time faster than Y” Example: time taken to run a program 10s on A, 15s on B Execution TimeB / Execution TimeA = 15s / 10s = 1.5 So A is 1.5 times faster than B Speedup Chapter 1 — Computer Abstractions and Technology

7 Morgan Kaufmann Publishers
May 9, 2019 CPU Clocking Operation of digital hardware governed by a constant-rate clock Our cycle time Clock period Clock (cycles) Data transfer and computation Update state Clock period: duration of a clock cycle e.g., 250ps = 0.25ns = 250×10–12s Clock frequency (rate): cycles per second e.g., 4.0GHz = 4000MHz = 4.0×109Hz Chapter 1 — Computer Abstractions and Technology

8 Morgan Kaufmann Publishers
May 9, 2019 CPU Time AKA Core 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

9 Morgan Kaufmann Publishers
May 9, 2019 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

10 Instruction Count and CPI
Morgan Kaufmann Publishers May 9, 2019 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

11 Cycles and Instructions
time Multiplication takes more time than addition Floating point operations take longer than integer ones Accessing memory takes (in general) more time than accessing registers Important point: changing the cycle time often changes the number of cycles required for various instructions (more later)

12 Program Execution time
Number of instruction classes ~= Instruction_count * CPIavg * clock_cycle_time technology algorithms/compiler architecture Relative frequency

13 Morgan Kaufmann Publishers
May 9, 2019 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

14 Morgan Kaufmann Publishers
May 9, 2019 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 IC in sequence 2 4 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 Example: Chapter 1 — Computer Abstractions and Technology

15 Morgan Kaufmann Publishers
May 9, 2019 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, Tc Chapter 1 — Computer Abstractions and Technology

16 Benchmark Suites Report performance metrics for execution on target platforms Designed to assess how well the platforms function in specific domains Examples Media Bench - Multimedia EEMBC – Embedded systems Rodinia, Parboil: For GPU Systems SPECWeb, SPECJbb – Enterprise systems Many more……

17 Morgan Kaufmann Publishers
May 9, 2019 Pitfall: Amdahl’s Law Improving an aspect of a computer and expecting a proportional improvement in overall performance 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

18 Amdahl’s Law f (1 - f) (1 - f) f / P
Speed-up = Exec_timeold / Exec_timenew = Performance improvement from using faster mode is limited by the fraction the faster mode can be applied. affected f (1 - f) Told (1 - f) Tnew f / P

19 Morgan Kaufmann Publishers
May 9, 2019 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 Chapter 1 — Computer Abstractions and Technology

20 Study Guide Practice problems provided on the class website

21 Glossary Amdahl’s Law Benchmarks CPI (cycles per instruction) CPU Time
Execution Time Latency Throughput


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