Question

In a page segmented system, a virtual address consists of 32 bits of which 12 bits are a displacement, 11 bits are segment number, and 9 bits are page number. What is the page size?

• A. \( 2^{12} \) bytes
• B. \( 2^{23} \) bytes
• C. \( 2^{11} \) bytes
• D. \( 2^{32} \) bytes

Hint:

The page size is determined by the number of bits allocated to the displacement or offset within the page.

Answer 1

The Correct Option is A

In this question: - The 12 bits are for displacement, and the displacement indicates the size of the page. -
Therefore, the page size is \( 2^{12} \) bytes, because 12 bits can address \( 2^{12} \) locations within a page.
Thus, the correct answer is \( 2^{12} \) bytes.