edited by
558 views
3 votes
3 votes
Consider a computer with 64 MB physical memory and 32 bit virtual address space. If the page size is 4KB , then the size of page table in MB is _____________    (Assume 2 bits per page table entry for extra information other than that for addressing)
edited by

1 Answer

Best answer
8 votes
8 votes
Answer is 2MB.

We need a page table entry for each possible page (virtual) means for $2^{32}/4 KB = 2^{20}$ possible pages.

Each page table entry must address a physical page whose address can be of $\log (64 MB / 4KB) = 14 $bits.

We are mentioned 2 extra bits per page table entry- so 16 bits = 2 bytes per PTE.

So, size of page table = $2 \times 2^{20} = 2 \; MB.$
selected by
Answer:

Related questions

3 votes
3 votes
1 answer
1
Bikram asked Sep 3, 2016
304 views
Swap space in the disk is used to _____save temporary html pagessave process datastore the super-blockstore device drivers
5 votes
5 votes
1 answer
3
Bikram asked Sep 3, 2016
623 views
Which of the following page replacement schemes is the toughest to implement from a hardware point of view?LRUFIFOMRUAll have equal complexity
2 votes
2 votes
1 answer
4
Bikram asked Sep 3, 2016
452 views
The size of the virtual memory in a computer system depends on the size of the data busmain memoryaddress busRAM