Applied Test Series

Question

A hypothetical system OSXZ_09 implements paging. Given 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.

Note : The memory is byte addressable.

Comments

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

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why u are not consider the reference bit and protection bit  

i think page table entry size is 14+2=16 bit which is 2B so page table size  of innermost page table is 2MB and outermost table is 1kB

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yaa yaa u are correct i totally missed that part…my bad..!! 😣