Discussion
Computer A having clock cycle time of 250 ps and cycle per instruction of 2.0 for some programs, and computer B having clock cycle time of 500 ps and a cycle per instruction of 1.2 for same program. Which one is faster for this program
Answer: A
Correct answer: Computer A
Compare the time per instruction, which is the clock cycle time multiplied by the cycles per instruction:
\[ A: 250\ \text{ps}\times 2.0=500\ \text{ps} \qquad B: 500\ \text{ps}\times 1.2=600\ \text{ps} \]
A takes 500 ps per instruction against B's 600, so A is the faster machine by a factor of 600/500 = 1.2. Note that B has the better CPI and A the better clock; neither figure decides the question by itself.
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