Question

A computer system implements 8 kilobyte pages and a 32-bit physical address space. Each page table entry contains a valid bit, a dirty bit, three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is_____bits.

• A. 36

Answer 1

The correct option is A 36
Page size = 8 KB $\Rightarrow$ 13 bit offset
Number of frames bits = 32 - 13 = 19 bits
Page table entry = Valid + Dirty + Permission bits + Translation (frame bits)
= 1 + 1 + 3 + Frame bits
= 5 + 19 =24 bits
Page table size = 24 Mbytes
Number of pages = Number of page table entries
= $\frac{24Mbytes}{24Bits}=8M=2^{23}pages$
$\therefore$ 23 bits needed for page and 13 bits offset Length of virtual address = 23 + 13 = 36 bits.