Performance and Quantitative Principles

1. ENGS 116 Lecture 2 Vincent H. Berk September 26th, 2008 Reading for today: Chapter 1.1 - 1.4, Amdahl article Reading for Monday: Chapter 1.5 – 1.11, Mazor article Homework for Wednesday: 1.1, 1.3, 1.6, 1.7, 1.13 Performance and Quantitative Principles

2. ENGS 116 Lecture 2 Review • Task of Computer Designers • Determine which attributes are important for a new machine • Design a machine to maximize performance without violating cost/power/functionality constraints • 3 Components of “Architecture” • Instruction set design • Organization • Hardware

3. ENGS 116 Lecture 2 Benchmarking Games • Different configurations used to run the same workload on two systems. • Compiler customized to optimize the workload. • Workload arbitrarily picked to skew results. • Test specification written to be biased toward one machine.

4. ENGS 116 Lecture 2 Design benchmarks for: • Industrial and design • Consumer Electronics • Networking, routers • Office applications • Telecommunications • Weapon systems

5. ENGS 116 Lecture 2 Execution time • Weighted arithmetic mean: sum over execution time of all programs run, times their relative frequencies • Normalized execution time: take a reference machine, set it to 1, then compute normalized execution times for others based on this machine • Geometric mean of normalized execution time (reference computer becomes irrelevant, ratios can arbitrarily be compared)

6. ENGS 116 Lecture 2 Amdahl’s Law Execution time after improvement = Make the common case fast

7. ENGS 116 Lecture 2 Speedup due to enhancement E: Suppose that enhancement E accelerates a fraction F of the task by a factor S, and the remainder of the task is unaffected: ExTime (E) = Speedup (E) = Amdahl’s Law ExTime w/o E Performance w/ E  Speedup(E) = ExTime w/ E Performance w/o E

8. ENGS 116 Lecture 2 Amdahl’s Law

9. ENGS 116 Lecture 2 Example: Floating point instructions improved to run 2X, but only 10% of actual instructions are FP Amdahl’s Law

10. ENGS 116 Lecture 2 All instructions require an instruction fetch, only a fraction require a data fetch/store. Optimize instruction access over data access Programs exhibit locality. Spatial Locality Temporal Locality Access to small memories is faster. Provide a storage hierarchy such that the most frequent accesses are to the smallest (closest) memories. Disk/Tape Registers Cache Memory Corollary: Make The Common Case Fast

11. ENGS 116 Lecture 2 Metrics of Performance Application Answers per month Operations per second Programming Language Compiler Millions of instructions per second: MIPS Millions of FP operations per second: MFLOPS ISA Datapath Megabytes per second Control Function Units Cycles per second (clock rate)‏ Transistors Wires Pins

12. ENGS 116 Lecture 2 Machines with different instruction sets? Programs with different instruction mixes? Dynamic frequency of instructions Uncorrelated with performance Marketing Metrics • Machine dependent • Often not where time is spent

13. ENGS 116 Lecture 2 Instr. Count CPI Clock Rate Program Compiler Instruction Set Organization Technology Aspects of CPU Performance

14. ENGS 116 Lecture 2 Instr. Count CPI Clock Rate Program X Compiler X (X) Instruction Set X X Organization X X Technology X Aspects of CPU Performance

15. ENGS 116 Lecture 2 Average Cycles per Instruction CPI = (CPU Time  Clock Rate) / Instruction Count = Cycles / Instruction Count CPU time = Cycle Time  Instruction Frequency Invest resources where time is spent! Cycles Per Instruction

16. ENGS 116 Lecture 2 Base Machine (Reg / Reg)‏ Op Freq Cycles CPI (i) (% Time)‏ ALU 50% 1 .5 (33%)‏ Load 20% 2 .4 (27%)‏ Store 10% 2 .2 (13%)‏ Branch 20% 2 .4 (27%)‏ 1.5 Typical Mix Example: Calculating CPI

17. ENGS 116 Lecture 2 Want to add register / memory operations - One source operand in memory - One source operand in register - Cycle count of 2 Side effect: Branch cycle count will increase to 3. What fraction of the loads must be eliminated for this to pay off? Base Machine (Reg / Reg)‏ Op Freq Cycles ALU 50% 1 Load 20% 2 Store 10% 2 Branch 20% 2 Example

18. ENGS 116 Lecture 2 Exec Time = Instruction Count  CPI  Clock Op Freq Cycles CPI Freq Cycles CPI ALU .50 1 .5 Load .20 2 .4 Store .10 2 .2 Branch .20 2 .4 Reg/Mem 1.00 1.5 Example Solution

19. ENGS 116 Lecture 2 Exec Time = Instruction Count  CPI  Clock Op Freq Cycles CPI Freq Cycles CPI ALU .50 1 .5 .5 – X 1 .5 – X Load .20 2 .4 .2 – X 2 .4 – 2X Store .10 2 .2 .1 2 .2 Branch .20 2 .4 .2 3 .6 Reg/Mem X 2 2X 1.00 1.5 1 – X (1.7 – X) /(1 – X)‏ CPINew must be normalized to new instruction frequency Example Solution

20. ENGS 116 Lecture 2 Exec Time = Instruction Count  CPI  Clock Op Freq Cycles CPI Freq Cycles CPI ALU .50 1 .5 .5 – X 1 .5 – X Load .20 2 .4 .2 – X 2 .4 – 2X Store .10 2 .2 .1 2 .2 Branch .20 2 .4 .2 3 .6 Reg/Mem X 2 2X 1.00 1.5 1 – X (1.7 – X) / (1 – X)‏ Instr CntOld CPIOld ClockOld = Instr CntNew CPINew ClockNew Example Solution

21. ENGS 116 Lecture 2 Exec Time = Instruction Count  CPI  Clock Op Freq Cycles CPI Freq Cycles CPI ALU .50 1 .5 .5 – X 1 .5 – X Load .20 2 .4 .2 – X 2 .4 – 2X Store .10 2 .2 .1 2 .2 Branch .20 2 .4 .2 3 .6 Reg/Mem X 2 2X 1.00 1.5 1 – X (1.7 – X) / (1 – X)‏ Instr CntOld CPIOld ClockOld = Instr CntNew CPINew ClockNew 1.00  1.5 = (1 – X)  (1.7 – X) / (1 – X)‏ 1.5 = 1.7 – X 0.2 = X ALL loads must be eliminated for this to be a win! Example Solution