Virtual Memory

Question

Comments

@srestha I multiplied with 16KB page size then applied log₂(). It's fine

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ohh yea thanks

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the question is wrongly framed

Answer 1

Given

Page table entry size = address translation bits + a valid bit + a modified bit + 2 replacement bits

page size = 16KB

physical address = 32 bits

#f frames = 2^32 /2^14 = 2^18

Page table entry size = 18+1+1+2 =22bits

page table size = 22MB

#f pages = page table size / Page table entry size

= 22* 2^20 * 8 bits / 22 bits

= 2^23

logical address = #f bits required to represent pages + offset

= 23+14 =37bits

Comments

Assuming Byte addressable

Page table entry size = 18+1+1+2 = 22bits  It can't be stored exactly and will take 3 bytes to store 22 bits.

Answer 2

yes 37 is correct one

Comments

The calculation you have done implies memory is allocated in bits. Memory is always allocated in words. And here 1 word = 1 byte