Questions: CPU Performance Metrics and Amdahl's Law

Question 1

Processor A runs at 4 GHz with an average CPI of 4. Processor B runs at 2 GHz with an average CPI of 1. Both execute the same program with the same instruction count. Which processor is faster?

• A. Processor A, because it has a higher clock frequency
• B. Processor B, because it has a lower CPI
• C. Processor B, which is twice as fast as Processor A
• D. They are equally fast because higher frequency cancels lower CPI

Answer: (C)

CPU time = Instruction count × CPI × Clock cycle time. Processor A's time per instruction is 4 × (1/4 GHz) = 1 ns. Processor B's is 1 × (1/2 GHz) = 0.5 ns — so B is twice as fast. This is the key lesson: a higher clock frequency does not mean better performance if CPI is proportionally higher. The CPU time equation requires considering all three factors together.

Question 2

A program spends 90% of its execution time in a routine that you parallelize, achieving a 10× speedup for that portion. What is the maximum overall speedup of the entire program?

• A. 10×, since you sped up the dominant portion by 10×
• B. 9×, since 90% of the work is 10× faster
• C. Approximately 5.3×, because the remaining 10% now dominates execution time
• D. Approximately 1.1×, because parallelism introduces overhead

Answer: (C)

Amdahl's Law: Speedup = 1 / ((1 − f) + f/S) = 1 / (0.10 + 0.90/10) = 1 / 0.19 ≈ 5.3×. The unoptimized 10% forms a hard ceiling. The tempting wrong answer (10×) assumes the serial fraction disappears — but it doesn't. The remaining 10% now constitutes almost all of execution time, limiting the overall gain to roughly half the naive expectation.

Question 3

A processor with a higher clock frequency usually executes a given program faster than a processor with a lower clock frequency.

• T. True
• F. False

Answer: False

False. CPU time = Instruction count × CPI × Clock cycle time. A higher clock frequency reduces cycle time but says nothing about CPI. If more pipeline stages are added to reach a higher frequency, CPI may increase (more hazards, longer stall penalties), leaving execution time unchanged or worse. Performance depends on all three factors together, which is why raw clock speed comparisons across architectures are misleading.

Question 4

According to Amdahl's Law, if you could speed up exactly 50% of a program to take zero time, the maximum possible speedup for the whole program is 2×.

• T. True
• F. False

Answer: True

True. Amdahl's Law: as S → ∞, Speedup → 1 / (1 − f) = 1 / 0.5 = 2. No matter how fast you make half the program, the other half still takes the same time — and that unoptimized half now constitutes all remaining execution time. This illustrates the fundamental insight: the serial fraction is the hard ceiling on total speedup.

Question 5

Why do benchmark suites like SPEC provide more meaningful performance comparisons than raw MIPS (millions of instructions per second) ratings?

Model answer: MIPS varies by program and ignores instruction complexity — a processor can achieve higher MIPS by executing more but simpler instructions, even if it takes longer to complete the same task. SPEC runs standardized real-world workloads and reports performance ratios relative to a reference machine, measuring actual time to complete meaningful tasks rather than a synthetic throughput metric.

MIPS is flawed in several ways: it varies across programs run on the same processor, it ignores that different instructions do different amounts of work, and it can be gamed by compilers that emit many simple instructions. SPEC addresses this by running representative real applications and reporting the geometric mean of speedup ratios — capturing how the processor actually performs on workloads that matter.