performance power
play

Performance, Power CS301 Prof Szajda Performance Metrics (How do - PowerPoint PPT Presentation

Performance, Power CS301 Prof Szajda Performance Metrics (How do we compare two machines?) What to Measure? Which airplane has the best performance? 3 Performance One size does not fit all Depends on application domain


  1. Performance, Power CS301 Prof Szajda

  2. Performance Metrics (How do we compare two machines?)

  3. What to Measure? Which airplane has the best performance? � 3

  4. Performance • One size does not fit all • Depends on application domain � Scientific computing � Graphics � Databases � General-Purpose desktop � Beware of designing to benchmark! • Depends on technology characteristics � DRAM speed and capacity, chip size, etc.

  5. Which Metric Do We Use? • Response or execution time � Di fg erence between start and end time � Individual user cares most about this • Throughput � Total amount of work done in given time � Frequently used for servers and clusters • How are these a fg ected by � Replacing processor with faster version? � Adding more processors?

  6. Execution Time • Shorter execution time is better • Allows comparison between 2 machines

  7. Relative Performance • “X is n times faster than Y” • Example: � Machine A takes 10s to run program � Machine B takes 15s to run same program � What is the performance ratio?

  8. Di fg erent Time Values Execution time • � Wall-clock, response, or elapsed time � Includes everything (processing,I/O, OS overhead, etc)! � Determines system performance CPU time • � Time spent executing code for this task only � Does not include I/O or time-sharing � Comprises user CPU time and system CPU time � Di fg erent programs are a fg ected di fg erently by CPU and system performance � man time � 90.7u 12.9s 2:39 65% � User: 90.7 sec � System: 12.9 sec � Elapsed time: 2 min 39 sec

  9. Clock Cycles • Instead of expressing time in seconds, use clock cycles • Clock � Determines when events take place � Runs at constant rate (ex. 1 GHz) � Easy to convert between clock rate and seconds � Clock rate = 1 / Clock Cycle time � 500 MHz = 1 / (2 ns) � 1 ns = 10 -9 s

  10. 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 × 10 9 Hz Chapter 1 — Computer Abstractions and Technology —

  11. An Aside Chapter 1 — Computer Abstractions and Technology —

  12. 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 —

  13. 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 —

  14. 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 —

  15. 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 —

  16. Application Characteristics • Determine the mix of di fg erent instruction types � Integer arithmetic � Logical operations � Floating point arithmetic � Loads and stores • Di fg erent applications have di fg erent CPI because of di fg erent instruction mixes

  17. CPI in More Detail � If different instruction classes take different numbers of cycles � Weighted average CPI Relative frequency Chapter 1 — Computer Abstractions and Technology —

  18. 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 � Sequence 2: IC = 6 � Clock Cycles 
 � Clock Cycles 
 = 2 × 1 + 1 × 2 + 2 × 3 
 = 4 × 1 + 1 × 2 + 1 × 3 
 = 10 = 9 � Avg. CPI = 10/5 = 2.0 � Avg. CPI = 9/6 = 1.5 Chapter 1 — Computer Abstractions and Technology —

  19. 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 —

  20. Amdahl’s Law • How much speedup do you get from an enhancement? Speedup = Execution time w/o enhancement Execution time w/ enhancement • Based on � Fraction of time enhancement used � Improvement in enhanced mode fraction enh Exec new = Exec old × ((1-fraction enh ) + ) Speedup enh

  21. §1.10 Fallacies and Pitfalls 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 —

  22. Review Question • Your machine has a clock rate of 2.4GHz. How long is the clock cycle?

  23. Review Questions • Suppose you are given the following: � Machine A � 1 GHz � Average CPI = 1.6 � Instructions = 1.7 Billion � Machine B � 3.3 GHz � Average CPI = 6.1 � Instructions = 2 Billion • Which machine is faster? By how much?

  24. Review Questions • What is the average CPI for a machine with the following CPIs on an application with the following instruction frequency? Frequen Type CPI cy Arithmeti 0.45 1 c Memory 0.3 8 Control 0.2 3 Mult/Div 0.05 5

  25. Review Questions • What factors must be included when comparing the relative performance of two machines?

  26. Amdahl’s Law fraction enh Exec new = Exec old × ((1-fraction enh ) + ) Speedup enh • Suppose you have an enhancement that makes a functional unit 10x faster. • Speedup if used 5% of the time? • Speedup if used 40% of the time?

  27. Review Questions • What is the equation for execution time? • What does Amdahl’s Law say?

  28. Benchmarks • Programs specifically used to measure performance • Hope is that it is representative of how computer will be used • Examples � SPEC Integer and Floating Point � MediaBench � MineBench � TPC

  29. 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 —

  30. CINT2006 for Intel Core i7 920 Chapter 1 — Computer Abstractions and Technology —

  31. §1.7 The Power Wall Recent Concern: Power � In CMOS IC technology × 40 5V → 1V × 1000 Chapter 1 — Computer Abstractions and Technology —

  32. Tricks to Increase Power • Attach large cooling devices • Turn o fg parts of chips not used in given clock cycle � Can increase power to 300 watts... � ...But these and other ways all prohibitively expensive for desktop computers. So... � 32

  33. More Recent Approaches: 
 Chip Multiprocessors • Reasons for change � Limited opportunities to improve single thread performance � Power � On-chip communication latencies

  34. §1.8 The Sea Change: The Switch to Multiprocessors Uniprocessor Performance Constrained by power, instruction-level parallelism, memory latency Chapter 1 — Computer Abstractions and Technology —

  35. 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 —

  36. §1.9 Concluding Remarks 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 —

Recommend


More recommend