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Let the page fault service time be 10ms in a computer with average memory access time being 20ns. If one page fault is generated for every 10^6 memory accesses, what is the effective access time for the memory?
  • a)
    21ns
  • b)
    30ns
  • c)
    23ns
  • d)
    35ns
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Let the page fault service time be 10ms in a computer with average mem...
Let P be the page fault rate
Effective Memory Access Time = p * (page fault service time) + (1 - p) * (Memory access time)
= ( 1/(10^6) )* 10 * (10^6) ns + (1 - 1/(10^6)) * 20 ns
= 30 ns (approx)
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Most Upvoted Answer
Let the page fault service time be 10ms in a computer with average mem...


Given data:
- Page fault service time = 10ms
- Average memory access time = 20ns
- Page fault rate = 1 per 10^6 memory accesses

Calculating effective access time:
- Effective access time = (1 - page fault rate) * memory access time + page fault rate * (memory access time + page fault service time)
- Given page fault rate = 1 per 10^6 memory accesses, so page fault rate = 1 / 10^6 = 10^-6

Substitute the values:
- Effective access time = (1 - 10^-6) * 20ns + 10^-6 * (20ns + 10ms)
- Effective access time = 0.999999 * 20ns + 0.000001 * (20ns + 10000000ns)
- Effective access time = 19.99998ns + 10.02ns
- Effective access time = 30.01998ns

Therefore, the effective access time for memory is approximately 30ns. So, option 'B' is correct.
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Let the page fault service time be 10ms in a computer with average memory access time being 20ns. If one page fault is generated for every 10^6 memory accesses, what is the effective access time for the memory?a)21nsb)30nsc)23nsd)35nsCorrect answer is option 'B'. Can you explain this answer?
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