Question

Given the following information: \[ \text{Logical address space} = 2^{32}, \quad \text{Page size} = 2084 \, \text{bytes}, \quad \text{PTE size} = 8 \, \text{bytes}, \quad \text{2-level paging system}. \] Calculate the number of bits required to search in the outer level page table.

• A. \( 2^8 \)
• B. \( 2^9 \)
• C. \( 2^{10} \)
• D. \( 2^7 \)

Hint:

In paging systems, the total number of address bits is split between the page offset and the page table indices. Calculate the bits required for each component based on the page size and address space.

Answer 1

The Correct Option is B

Step 1: Calculate the number of bits for the page offset.

Given the page size is 2084 bytes, we calculate the number of bits required for the page offset as: \[ \text{Page offset bits} = \log_2(\text{Page size}) = \log_2(2084) \approx 11 \]

Step 2: Calculate the number of bits for the outer page table index.

In a 2-level paging system, the total number of bits required to address a location in memory is the sum of the bits for the page offset and the bits for both the inner and outer page table indices.

We are given the total logical address space is \( 2^{32} \), so: \[ \text{Total address bits} = 32 \]

Since the page offset requires 11 bits, we have: \[ \text{Remaining bits for page table} = 32 - 11 = 21 \]

This 21 bits will be split between the outer and inner page table indices. Given the page table entry (PTE) size is 8 bytes, the number of bits required for the outer level page table index is: \[ \text{Outer level page table index bits} = \frac{21}{2} = 9 \]

Thus, the number of bits required to search in the outer level page table is \( 2^9 \).

Thus, the correct answer is \( 2^9 \).