On a simple paging system with a page table containing 512 entries of 18 bits (i

ID: 3866327 • Letter: O

Question

On a simple paging system with a page table containing 512 entries of 18 bits (including valid/invalid bit) each, and a page size of 1 KB,

a. how many bits in the logical address specify the page number?

b. how many bits in the logical address specify the offset within the page?

c. how many bits are in a logical address?

d. what is the size of the logical address space?

e. how many bits in the physical address specify the page frame number?

f. how many bits in the physical address specify the offset within the page frame?

g. how many bits are in a physical address?

h. what is the size of the physical address space?

i. Draw the diagram of the above system to explain it clearly (similar to the one I drew in the class while working with a similar example).

a. total pages = 512 , so total pages = 2^9, so it requires 9 bits to specify page number

b. total 10 bits to specify offest because page size = 1024 bytes

c. total number of bits in logical address space = 9(page table entry) + 10(page size) = 19

d.logicall address space size = 2^19 bytes

Please ask other questions as separate questions.

Cheers :)

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