Performance Analysis

A CPU design change raises clock frequency 1.5× but also raises CPI 1.2× (instruction count unchanged). How many times faster is execution?

  1. A1.25×
  2. B1.5×
  3. C1.8×
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Answer: A. 1.25×

Execution time is instruction count × CPI ÷ frequency, so the new time is 1.2 ÷ 1.5 = 0.8 of the original and the speedup is 1 ÷ 0.8 = 1.25×. Deepening the pipeline to raise frequency often increases CPI, for example through larger branch misprediction penalties. Frequency alone suggests 1.5×, but the real gain is only 1.25×. Performance must be judged by the product of all three factors.

  • A. Correct. 1.5 ÷ 1.2 = 1.25.
  • B. That ignores the CPI increase.
  • C. That multiplies 1.5 by 1.2.
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More Performance Analysis questions

  1. A new compiler optimization reduces executed instructions by 20% but increases average CPI by 10% (frequency unchanged). What is its effect?
  2. A program's instruction mix is 50% integer ALU (CPI 1), 30% loads and stores (CPI 2) and 20% branches (CPI 5). What is the average CPI?
  3. When comparing CPU performance by instructions executed per second (millions of instructions per second), what should be kept in mind?
  4. How is speedup, which expresses a performance improvement, calculated?
  5. A part taking 50% of total time is made 4× faster. What is the overall speedup (Amdahl's law)?