Virtual Address size = 40 bit
Now Physical memory size is 16GB = 234 bytes
Therefore physical memory size = 34 bits
Page Size / block size = 4K = 212 bytes
1) As Block size is 212 bytes so, OFFSET is of 12 bits
INDEX = 40-12 = 28 bits
2) The Page table entry generally contains the following fields:-
a) Frame number:
As the offset is 12 bits and physical address is of 34 bits so, frame number will be = 34 - 12 = 22 bits
b) Present/Absent (1 bit)
c) Protection (1 bit)
d) Referencing (1 bit)
e) Caching (1 bit)
f) Dirty (1 bit)
Therefore total bits in a page table entry = 27 bits = 4 bytes (in the whole number of bytes)
3) Number of pages needed = 2INDEX = 228 pages
Therefore the size of page table = 228 * page table entry size = 228 * 22 bytes = 230 bytes = 1 GB
It is stil lresonable size for a 16 GB memory as 1GB can be easily allocated in 16 GB of memory. Still we can use multi-level paging for better utilization.
computer architecture Virtual memory 4 physice memar Assume a 64 bit machine with 40 bit addresses,...
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