A Computer system implements 8 kilobyte pages and a 32-bit physical address…
2015
A Computer system implements 8 kilobyte pages and a 32-bit physical address space. Each page table entry contains a valid bit, a dirty bit three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is _______________ bits?
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Correct answer: 36
Page/offset facts
Page size = 8 KB = 2^13 bytes ⇒ offset = 13 bits
Physical frame number (PFN) bits in PTE
Physical address = 32 bits
Offset = 13 bits ⇒ PFN bits = 32 − 13 = 19
PTE size
Fields: valid (1) + dirty (1) + permissions (3) + PFN (19)
Total = 1 + 1 + 3 + 19 = 24 bits = 3 bytes per PTE
Max page table size ⇒ number of entries
Max page table size = 24 MB = 24 × 2^20 bytes
Number of PTEs = (24 × 2^20) / 3 = 8 × 2^20 = 2^23
So, number of virtual pages = 2^23 ⇒ VPN bits = 23
Virtual address length
Virtual address bits = VPN bits + offset bits = 23 + 13 = 36 bits
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