GATE CS 2016 Set 1 Q.57 (CS Q.47): Memory Management and Virtual Memory

Consider a computer system with 40-bit virtual addressing and page size of sixteen kilobytes. If the computer system has a one-level page table per process and each page table entry requires 48 bits, then the size of the per-process page table is _____ megabytes.

Your answer

Official answer and solution

Official answer: 384
From: Official GATE 2016 answer key

Solution

Calculate the number of pages in the virtual address space and multiply by the page table entry size.

Given parameters:

Virtual address space $= 40$ bits $\implies 2^{40}$ bytes.
Page size $= 16\text{ KB} = 16 \times 2^{10}\text{ bytes} = 2^4 \times 2^{10} = 2^{14}\text{ bytes}$ .

Calculate the number of pages:

\text{Number of pages} = \frac{\text{Virtual Address Space}}{\text{Page Size}} = \frac{2^{40}}{2^{14}} = 2^{26}\text{ pages}

Calculate page table size:

Each page table entry (PTE) requires $48\text{ bits} = \frac{48}{8} = 6\text{ bytes}$ .
A one-level page table must contain an entry for every virtual page:

\text{Total Page Table Size} = 2^{26} \times 6\text{ bytes} = 6 \times 2^{26}\text{ bytes}

Convert to megabytes ( $1\text{ MB} = 2^{20}\text{ bytes}$ ):

\text{Size in MB} = \frac{6 \times 2^{26}}{2^{20}} = 6 \times 2^6 = 6 \times 64 = 384\text{ MB}

Final answer

384

Common trap

Forgetting to divide 48 bits by 8 to convert to bytes, or confusing 16 KB with $2^{16}$ bytes instead of $2^{14}$ bytes.

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

Source

Paper: GATE 2016 Computer Science and Information Technology (CS) Session 1
Question: 57 of 65, printed as CS Q.47
Section: Computer Science and Information Technology

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