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Operating System :: Memory Management

10. Consider a machine with 64 MB physical memory and a 32-bit virtual address space. If the page size s 4KB, what is the approximate size of the page table?

Answer: Option C

Explanation :

Number of pages = 232 / 4KB = 220 as we need to map every possible virtual address.

So, we need 220 entries in the page table. Physical memory being 64 MB, a physical address must be 26 bits and a page (of size 4KB) address needs 26-12 = 14 address bits. So, each page table entry must be at least 14 bits.

So, total size of page table = 220 * 14 bits ~ 2 MB

Asked In ::  Cisco  

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