Question

Consider a system with 48-bit address that supports paging AND segmentation. The page size is 8KB and each page table entry (PTE) is 4B. A process can have up to 256 segments and each segment table entry (STE) is 8B. How large in bytes is the largest possible page table?

Answer 1

Logical address size = 48 bits => Virtual address space = $\small 2^{48}\;bytes$

Number of segments = 256 = $\small 2^8\;bytes$ => Segment Number = 8 bits

log2 (28) =

Size of segment = Logical address space / #segments = $\small \frac{2^{48}\;bytes}{2^8} = 2^{40}\;bytes$

Each segment will be divided into pages where page size = 8 KB = 8*1024 bytes = $\small 2^{13}\;bytes$

• Number of pages = segment size / page size = $\small \frac{2^{40}}{2^{13}} = 2^{27}$

• Page offset = $\small log_2(8*1024\;bytes) = log_2(2^{13})$ = 13 bits
• Page Number = $\small log_2(\#pages) = log_2(2^{27})$ = 27 bits

Page table entry size = 4 bytes

Size of page table

= Number of entries in page table * page table entry size

= $\small 2^{27}*4 \;bytes = 2^{29}\;bytes = \bold{512\;MB}$