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Assume a system with 2-levels of paging with the main memory access time of 100ns, and the cache lookup time of 50ns. Assuming a cache hit rate of 90% and no parallel memory access , what is the average time to read a location from main memory? (in nanoseconds)

the answer mentioned is 180ns and sugested explanation says 3 memory accesses would be required however i got my answer as 60nsec.

can someone please help me with this.
ago in Operating System by (20 points) | 33 views
0
No,I don't think by default we assume page tables in cache...they should mention it explicitly.
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I feel that the answer revolves around hierarchical access of memory also no overlapping of operations.

since the question says cannot be accessed parallely it might even mean that the data cannot even be read parallel that is each required word has to be accessed individually, therefore, they might have multiplied it by 3.

which gives us 60*3 as the answer.this is all that i can conclude from all the responses.
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what is wrong if i compute it this way?

0.9*(50+100) + 0.1*350
+1
Firstly - that is a whole lot of assumptions you are making simply to justify the given answer. Remember - solutions come from concepts. Not concepts from solutions.

Secondly, assuming page tables are cached - 0.9*(50+100) + 0.1*350 - the problem with this is that you are assuming both levels of page table will be available in the cache at all times, or none will be available. Whereas in reality, it is entirely possible for one level of page table to be in cache, and the other level being a cache miss. This solution does not account for that.

Anyway - if you understand the concept, move on to the next one. Don't try to justify the given answer by any means necessary and risk picking up wrong concepts :)
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thankyou for the response,it really did helped :)

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