GATE CS 2015 Set 2 — Question 43

Given:
Page size = 8 kilobytes = $2^3 \times 2^{10}$ bytes = $2^{13}$ bytes.
Physical address space = 32-bit.
Maximum size of page table = 24 megabytes = $24 \times 2^{20}$ bytes.From the page size, the page offset (or displacement) is 13 bits, since $2^{13}$ bytes.Since the physical address space is 32-bit, the physical address is 32 bits long.
Physical Address = Page Frame Number (PFN) + Page Offset
$32 = \text{PFN} + 13$
$ext{PFN} = 32 - 13 = 19$ bits.Each page table entry (PTE) contains a valid bit, a dirty bit, three permission bits, and the translation (PFN).
So, the size of one PTE = $1 + 1 + 3 + 19 = 24$ bits.Maximum size of page table = 24 megabytes = $24 \times 2^{20}$ bytes.
Number of PTEs in the page table = $\frac{\text{Maximum size of page table}}{\text{Size of one PTE}}$
Size of one PTE in bytes = $\frac{24 \text{ bits}}{8 \text{ bits/byte}} = 3$ bytes.
Number of PTEs = $\frac{24 \times 2^{20} \text{ bytes}}{3 \text{ bytes}} = 8 \times 2^{20} = 2^3 \times 2^{20} = 2^{23}$.The number of PTEs in the page table corresponds to the number of virtual pages.
Number of virtual pages = $2^{23}$.Virtual Address = Virtual Page Number (VPN) + Page Offset
Number of bits for VPN = 23 bits.
Number of bits for Page Offset = 13 bits.Length of virtual address = VPN bits + Page Offset bits = $23 + 13 = 36$ bits.Thus, the length of the virtual address supported by the system is 36 bits.