GATE CS 2015 Set 1 Q.22 (CS Q.12): Memory Management and Virtual Memory

Consider a system with byte-addressable memory, 32-bit logical addresses, $4\text{ kilobyte}$ page size and page table entries of $4\text{ bytes}$ each. The size of the page table in the system in megabytes is _____.

Your answer

Official answer and solution

Official answer: 4
From: Official GATE 2015 answer key

Solution

Calculate the number of virtual pages and multiply by the page table entry size to determine total page table size.

Given parameters:

Logical address space: $32\text{ bits} \implies 2^{32}\text{ bytes}$ .
Page size: $4\text{ KB} = 4 \times 2^{10}\text{ bytes} = 2^{12}\text{ bytes}$ .
Page Table Entry (PTE) size: $4\text{ bytes}$ .

Number of pages in the logical address space:

\\text{Number of pages} = \frac{2^{32}}{2^{12}} = 2^{20}\text{ pages}.

Total size of the single-level page table:

\\text{Page Table Size} = 2^{20}\text{ entries} \times 4\text{ bytes/entry} = 4 \times 2^{20}\text{ bytes}.

Since $1\text{ MB} = 2^{20}\text{ bytes}$ :

\\text{Page Table Size} = 4\text{ MB}.

In megabytes, the answer is $4$ .

Final answer

Common trap

Dividing physical memory size instead of logical address space to calculate the number of page table entries.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 1
Question: 22 of 65, printed as CS Q.12
Section: Computer Science and Information Technology

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