LAS = 4 GB, PAS = 64 MB, Page size = 4 KB. A protection bit and a reference bit is also present in a page table entry. Then the size of the innermost page table is ____ B.
now 4GB =4*2^30...2^32 bytes is LAS and LA=22 bits.
PS=4KB=2^12 bytes
Foffset/Poffset=12bits
Pno=32bits-12bits=20bits
*imp that PTE is not given so find out the Fno(frame number)
PAS=64MB=2^26bytes….PA=26bits and Fno=26-12=14bits
now PTE=14+2=16 bits
not PTS=numb of pages * PTE
=2^20*16bits (1 byte=8 bits)
=2^20*(16/8)bytes
=2MB
NOW 2MB > 1KB so we need MULTILEVEL PAGING but they only asked about inner page table