Consider virtual addressing of 32 bits. Assume that the page size is 4 KB. You h

Question

Consider virtual addressing of 32 bits. Assume that the page size is 4 KB. You have 2 GB of RAM. How many entries do you need in the page table? How many bits are needed to index the page table? Assume that Table 1 lists all the virtual pages of process A that are currently in physical memory. If a virtual page is not listed in Table 1, it implies that any address in that range will result a page fault and will be a miss. Translate the virtual addresses below to the physical address using Table 1 and mention which will be hit and which will cause a miss: Translate the following virtual addresses to physical address using Table 1 (if possible): 1546. 20716. 85060. 36320, 2455

Explanation / Answer

a)

Using above formula we can say that there will be 2^(32-12) = 2^20 entries in page table.
No. of bits required to address the 64MB Physical memory = 26.
So there will be 2^(26-12) = 2^14 page frames in the physical memory. And page table needs to store the address of all these 2^14 page frames. Therefore, each page table entry will contain 14 bits address of the page frame and 1 bit for valid-invalid bit.
Since memory is byte addressable. So we take that each page table entry is 16 bits i.e. 2 bytes long.

b)

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