Applied Course | Mock GATE | Test 1 | Question: 49

Question

How much memory is used for page tables, when there are $10$ process running in the system with the below details?

• Virtual Address space: $1GB$
• Page size: $1KB$

Each page table entry contains a valid bit, dirty bit and the resulting frame number. And the system has a maximum of $2^{14}$ physical pages.

• A. $10$ MB
• B. $20$ MB
• C. $30$ MB
• D. None of these

Answer 1

Given information: Virtual address space is $1GB$ so we need $30$ bits required Virtual Address:
$\begin{array}{|c|c|} \hline \text{VPN 20 bits} & \text{Offset 10 bits} \\ \hline \end{array}$
Page table entry $\Rightarrow$
$\begin{array}{|c|c|c|} \hline \\ \text{Valid bit (1 bit)} & \text{Dirty bit(1 bit)} & \text{Page frame numbers(14 bits)} \\ \hline \end{array}$
$\text{i.e., } 2$ bytes are required for page table entry.
Total memory is used for page table of $10$ process is =
$2^{20} \times 2 \text{ bytes /page table} \times 10 \text{ process } = 20 \text{ MB}$

Comments

There won't be $2^{20}$ page tables or page table entries.

$2^{20}$×2 bytes /page table×10 process =20 MB220×2 bytes /page table×10 process =20 MB

Answer 2

Virtual address space is 1GB so we need 30 bits for Virtual Address. We will need 20 bits (i.e. 30-10) for Virtual Page Number.

Size of Page Table Entry (PTE) = 2B (14b for frame number, 2b for valid and dirt bit)

Number of page table entries in one page =  1 KB/ 2B = 512

On each page we will have 512 entries and we need 9 bits to address their page table entries.

For Page Table lvl 3 (2 bits)

For Page Table lvl 2 (9 bits)

For Page Table lvl 3 (9 bits)

Page Number (10 bits)

So number of pages required for page table = 1 + $2^{2}$ + $2^{2}$ * $2^{9}$ = 2053

Memory required for page tables = 2053 * 10 * 1 KB = 20.05 MB

Comments

Page m page table entry kaise aagyi???

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On each page we will have 512 entries and we need 9 bits to address their page table entries

This statement is totally wrong

memory is byte addressable or word addressable

but according to above statement is has become PTE(page table entry)  addressable