Consider a 64-bit architecture machine where physical memory is 128GB. a) If we

ID: 651698 • Letter: C

Question

Consider a 64-bit architecture machine where physical memory is 128GB.

a) If we would like to run processes as big as 256GB how many bits would be required for the logical address?

b) If we are using pages of size 4KB, how many bits are needed for displacement into a page?

c) If a single level page table is used, what is the maximum number of entries in this table?

d) What is the size of this single level page table in terms of 4KB pages?

e) If a two-level page-table is used and the outer page table is an 4KB page, how many entries does it contain, maximally?

f) How many bits of the logical address are used to specify an index into the inner page(page of page table)?

The size of the physical memory is 128GB.

The number of bits required for the logical address= No. of pages * Page size

The number of pages required =256GB/4K=2^6MB.

Page size=4KB.

The number of bits required for the logical address is 2^6 MB * 2^2= 2^18.

The number of bits required for the logical address is 18 bits.

The size of the physical memory is 128GB= 2^37 bytes

The size of a page is 4KB= 2^12 bytes

The number of bits needed for displacement is (2^37)/2^12 = 2^25 bytes

The number of bits required for displacement is 25 bits.

When the single level page table with page size 4KB=2^12 bytes.

The size of phsical memory is 128GB is 2^37 bytes of memory.

The number of bytes per one virtual page is 2^64 addressable bytes/2^12 bytes per page=2^52 page table entries.

The maximum page table size is= 2^52 + 2^2 = 2^54 bytes

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