GATE Overflow Test Series | Operating Systems | Test 2 | Question: 15

Question

Consider a machine with $8\; GB$ physical memory and a $34$-bit virtual address space. If the page size s $4\; KB,$ the approximate size of the page table assuming $3$ auxiliary bits (for protection, page replacement information etc.) on average for each page table entry will be _______

• A. $\text{16 MB}$
• B. $\text{8 MB}$
• C. $\text{2 MB}$
• D. $\text{12 MB}$

Answer 1

For $34-$bit virtual address and page size of $2^{12},$ number of pages $ = 2^{34 -12} = 2^{22}.$

With $8GB = 2^{33}B$ physical memory, number of address bits for $2^{12}$ size pages $ = 33 - 12 = 21. $

With, $3$ auxiliary bits we need $24$ bits for each page table entry which equals $3$ bytes.

So, for $2^{22}$ pages, we will need $3 \times 2^{22} = 12 MB$ for the page table.