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In a paged memory, the page hit ratio is 0.35. The time required to access a page in secondary memory is equal to 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page is
- a)3.0 ns
- b)68.0 ns
- c)68.5 ns
- d)78.5 ns
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
In a paged memory, the page hit ratio is 0.35. The time required to ac...
Hit ratio = 0.35
Time (secondary memory) = 100 ns
T(main memory) = 10 ns
Average access time = h(Tm) + (1 - h) (Ts)
= 0.35 x 10 +(0.65) x 100
= 3.5 + 65
= 68.5 ns
Time (secondary memory) = 100 ns
T(main memory) = 10 ns
Average access time = h(Tm) + (1 - h) (Ts)
= 0.35 x 10 +(0.65) x 100
= 3.5 + 65
= 68.5 ns
Hence option (C) is correct
For basics of Memory management lecture click on the following link:
Most Upvoted Answer
In a paged memory, the page hit ratio is 0.35. The time required to ac...
Given:
Page hit ratio = 0.35
Time required to access a page in secondary memory = 100 ns
Time required to access a page in primary memory = 10 ns
To find: Average time required to access a page
Calculation:
We can use the formula for average access time:
Average access time = (page hit time * page hit ratio) + (page fault time * (1 - page hit ratio))
Here,
- Page hit time is the time required to access a page in primary memory, which is 10 ns.
- Page fault time is the time required to transfer a page from secondary memory to primary memory, which is the sum of the time required to access a page in secondary memory (100 ns) and the time required to transfer it to primary memory. We assume this transfer time to be negligible.
Substituting the values in the formula, we get:
Average access time = (10 * 0.35) + (100 * (1 - 0.35))
= 3.5 + 65
= 68.5 ns
Therefore, the average time required to access a page is 68.5 ns.
Final answer: Option (c)
Page hit ratio = 0.35
Time required to access a page in secondary memory = 100 ns
Time required to access a page in primary memory = 10 ns
To find: Average time required to access a page
Calculation:
We can use the formula for average access time:
Average access time = (page hit time * page hit ratio) + (page fault time * (1 - page hit ratio))
Here,
- Page hit time is the time required to access a page in primary memory, which is 10 ns.
- Page fault time is the time required to transfer a page from secondary memory to primary memory, which is the sum of the time required to access a page in secondary memory (100 ns) and the time required to transfer it to primary memory. We assume this transfer time to be negligible.
Substituting the values in the formula, we get:
Average access time = (10 * 0.35) + (100 * (1 - 0.35))
= 3.5 + 65
= 68.5 ns
Therefore, the average time required to access a page is 68.5 ns.
Final answer: Option (c)
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