Computer Sciences > GATE 2025 SET-2 > Paging
A computer system supports a logical address space of 232 bytes. It uses two-level hierarchical paging with a page size of 4096 bytes. A logical address is divided into a b-bit index to the outer page table, an offset within the page of the inner page table, and an offset within the desired page. Each entry of the inner page table uses eight bytes. All the pages in the system have the same size.
The value of b is ______ (Answer in integer)
The value of b is ______ (Answer in integer)
Correct : 10
Page offset bits:
Page size = 4096 bytes = 212 bytes
Page offset bits = 12
Total logical address bits:
Logical address space = 232 bytes
Total bits = 32
Bits for inner page table index:
Each inner page table fits exactly in one page = 4096 bytes
Each entry of inner page table = 8 bytes
Number of entries in inner page table = 4096 ÷ 8 = 512 = 29
Inner page table index bits = 9
Logical address structure:
Total bits = b (outer index) + 9 (inner index) + 12 (page offset)
32 = b + 9 + 12
32 = b + 21
b = 32 − 21 = 10
∴ The value of b = 10
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