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In a paged memory, the page hit ratio is 0.40. The time required to access a page in secondary memory is equal to 120 ns. The time required to access a page in primary memory is 15 ns. The average time required to access a page is ____

  1. 105
  2. 68
  3. 75
  4. 78
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4 Answers

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Page hit means we access the page from the main memory and page miss means we access it from the secondary memory:

$.40 * 15+.60*120 = 78$

Hence, answer is $78 \, ns$
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Average Time Required to Access Page=0.40*15+0.60*120=6+72=78

Option (4) is correct.
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PAGE HIT = page present in ram or main memory

PAGE MISS = page fault so page swapped from the secondary memory

page hit ratio is 0.4, so

0.4*15 + 0.6* (15+120) = 87ns

because we look for a page in secondary memory when the page is not present in the main memory, therefore, we need to add main memory lookup time.
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question is not clear about page table but solving without considering page table wil be appropriate 

Average Access Page time = 0.4 x15 + 0.6(120+15)  = 6 + 81 => 87ns

with page table

Average Access Page time = 0.4 (15+15) + 0.6(120+15+15)  = 12+90 => 112ns

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