GATE Overflow Test Series | Operating Systems | Test 2 | Question: 20

Question

If the virtual address space supported is $2^{64}\:\text{bits},$ the page size is $4\:\text{Kbyte,}$ the size of the physical memory is $2\:\text{GB,}$ the size of a PTE is four bytes for standard page table and eight bytes for inverted page table, and the addressing is at the byte level, the sizes of the page tables (assuming single level) required for both standard and inverted page tables respectively will be

• A. $2^{54}\;\text{bytes},\;2^{22}\;\text{bytes}$
• B. $2^{52}\;\text{bytes},\;2^{32}\;\text{bytes}$
• C. $2^{51}\;\text{bytes},\;2^{22}\;\text{bytes}$
• D. $2^{50}\;\text{bytes},\;256\;\text{bytes}$

Answer 1

Number of pages $= 2^{64}/(8 \times 2^{12}) = 2^{49}.$ (virtual address space is in bits)

PTE size being $4$ bytes page table size will be $2^{49} \times 4 = 2^{51}$ bytes.

For inverted page table number of entries $ = 2GB/4KB = 2^{19}.$

Each PTE being eight bytes for inverted page table, size of inverted page table $ = 2^{19} \times 8 = 2^{22}$ bytes.

Comments

@gatecse

 

Why you are multiplying 8 in the page size ?

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Yes , no. of pages should be 2^52 for standard page table

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@gatecse

#of pages in standard=$\frac{2^{64}}{2^{12}}= 2^{52}$

and standard page size should be =$2^{52}*2^{2}=2^{54}\, byte$

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Virtual address space is given in bits.