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Consider a computer system using 2-level paging with TLB.The logical address supported is 32 bits.The page table is divided into 512 pages each of size 1K.The memory access time is 100ns and the TLB access time 15ns .Page table entry size at 1st level is 2 Bytes and that at second level us 4 Bytes each.

What is the $memory$ $overhead$ of storing the top level page table along with the page of the second level page table for a process?





My approach, The last level over head is $2^9*2B=2^{10}B=1KB$

Now for the first level has $2^9*2^{10}entries$ and each have  overhead $4B$ so the overhead of the 1st level should be $2^{21}B$ right?

So total overhead→ $1KB+2048KB=2049KB$ right?

in Operating System by (698 points) | 30 views
@shahin , @debapaul

Got it...I assumed/misread page size as 1KB...but question says number of entries/page as 1K.
You didn't misread it - the question is open to misinterpretation that way.

When you write an essay and give it to someone - you won't say "here is an essay of 2". You will say "here is an essay of 2 pages" or "here is an essay of 2 paragraphs". The units and context are important.

Especially when in computer science, 1K can represent size, count, units anything based on context.
haha....true that!!


The page table is divided into 512 pages each of size 1K

what does this line in the question mean ?

What i understood is that Outer page table contains 512 entries.Each entry of outer page table points to an Inner page table. And each Inner page table has 1K entries in it. Is it right ? 

That's the only way I could interpret it, yes.

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