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2 votes
2 votes

In a paged memory, the page hit ratio is 0.35. The time required to access a page in secondary memory is 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page is

  1. 100 ns
  2. 68.0 ns
  3. 68.5 ns
  4. 78.5 ns

3 Answers

Best answer
8 votes
8 votes
Answer is -  c)  68.5 ns

Main memory and secondary memory are hierarchical. But as per convention page fault service time includes primary memory access time. Here we are not given page fault service time but secondary memory access time. So,
Effective Memory Access Time = $0.35 \times 10 + 0.65 \times (100+10) = 3.5 + 65+6.5 = 75.$

This is not among the choices. So, lets try excluding primary memory access time for secondary memory access:

Effective Memory Access Time = $0.35 \times 10 + 0.65 \times 100 = 3.5 + 65 = 68.5.$
edited by
2 votes
2 votes
Answer must be d) as not only the time to do address translation of page table but also the time to access a page table from  memory is required
0 votes
0 votes
Answer must be D
Answer:

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