Question

A computer system supports a logical address space of \( 2^{32} \) bytes. It uses two-level hierarchical paging with a page size of 4096 bytes. A logical address is divided into a \( b \)-bit index to the outer page table, an offset within the page of the inner page table, and an offset within the desired page. Each entry of the inner page table uses eight bytes. All the pages in the system have the same size. The value of \( b \) is ___________ . (Answer in integer)

Hint:

To determine the number of bits required for indexing, break the logical address into page offset, inner table index, and outer table index.

Answer 1

Correct Answer: 11

We are given: - Logical address space size = \( 2^{32} \) bytes - Page size = \( 4096 = 2^{12} \) bytes - Each inner page table entry = \( 8 \) bytes - Two-level hierarchical paging Step 1: Break Down the Logical Address A logical address consists of: 1. Offset within the page: Since the page size is \( 2^{12} \), the offset requires \( 12 \) bits. 2. Inner page table index: Each inner page table entry is \( 8 \) bytes, meaning an inner page table can hold \( 2^{12} / 8 = 2^9 \) entries. This requires \( 9 \) bits. 3. Outer page table index: The remaining bits are used to index the outer page table. Step 2: Compute the Outer Page Table Index \( b \) The total logical address is \( 32 \) bits, and it is divided as: \[ b + 9 + 12 = 32 \] Solving for \( b \): \[ b = 32 - 9 - 12 = 11 \] Thus, the answer is \( 11 \).