Computer Science, asked by qwertyuiop35, 1 year ago

Consider a system using 2-level paging and the virtual address is 38 bits. The most significant 10 bits are used to index the page directory and next 16 bits index the page table. Each entry in both levels is 4 Bytes. What is the maximum size of a page table in KB?​

Answers

Answered by KGFKING
0

Answer:

what you question I can't understand can you give image for it.

Answered by cooldudeniku1
0

Answer:

Explanation:

the most significant bit is 10 bits so the size of page table2 is 2^10 and page table entry is 4 bytes so no. of page entries is 2^10 / 2^2 = 2^8 = 256 page tables  , 256 pages table enteries in page table2 it's means that 256 pages are present in level 1 , level 1 + level 2 => 256 +1 = 257 page table required

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