Consider a computer with a 64MB of RAM and a 32- bit virtual address space. The

Question

Consider a computer with a 64MB of RAM and a 32- bit virtual address space. The offset within a page is 12 bits. a) What is the size, in bytes, of a single virtual memory page on this computer? b) Each page table entry is 4 bytes in size. How much memory must be reserved for an architecture that uses a single level page table? c) If the computer architecture is updated to use a 2-level page table scheme like the diagram below How large of a virtual address range would a single Is1 level page table entry cover?

Explanation / Answer

Given that 12 bit offset. So the memory size of page is, (2**12)*8 bits=4096*8 bits=32768 bits=4096 Bytes (here we are cinsidering size of block is 8 bit).

There are 1048576 pages, if each page table entry occupy 4 Bytes then the total memory reserved is 1048576*4 Bytes=4194304 Bytes=4 MB.

If the page table is updated, then the address range of first level page table is 000h to 3FFh (0000000000b to 1111111111b)

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