engineering computer science MCQ #124

A CPU has 32KB cache with a hit ratio of 0.8. If cache access time is 2ns and main memory access time is 20ns, what is the effective access time?

engineering computer science MCQ #124

  1. Question 1

    Q1. A CPU has 32KB cache with a hit ratio of 0.8. If cache access time is 2ns and main memory access time is 20ns, what is the effective access time?

    • A) 4ns
    • B) 6ns
    • C) 8ns
    • D) 10ns

    Answer: 6ns

    Explanation: Effective access time = (hit ratio * cache access time) + ((1 - hit ratio) * (cache access time + main memory access time)) = (0.8 * 2) + (0.2 * (2 + 20)) = 1.6 + 4.4 = 6ns.