# Andrew S. Tanenbaum (OS) Edition 4 Exercise 3 Question 24 (Page No. 256)

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A machine has $48-bit$ virtual addresses and $32-bit$ physical addresses. Pages are $8\: KB.$ How many entries are needed for a single-level linear page table?

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Answer: $\mathbf{\;2^{35}\;entries}$

$\because$ Total number of bits in the page number field = $48-13 \;\text{bits} = 35$

$\therefore \text{Number of page table entries } = 2^{35} \;\text{entries}$

edited by
1
Why 12 bits?
0
page size give 8KB =13 bit

so no of entries 2^35
1

@yagnesh

Corrected!

2^35 pages will be there
page size give 8KB =13 bit

bit for no of entries in page table at first level=48-13=35 bit

so no of entries 2^35
page size is 8KB
total bit in page offset =13 (2^13=8KB)
Virtual address space will be divided into two part i.e. Page no + Page offset
Hence,Total pages = 2^(no of bits in page no) = 2^(48-13)=2^35

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