A computer system supports a logical address space of 232 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 ______.

The correct answer is 11.

To determine the value of b, we can break down the structure of the logical address step-by-step:

1. Logical Address Size:
The computer system supports a logical address space of 2³² bytes. Therefore, the length of the logical address is 32 bits.

2. Offset within the Desired Page:
The page size is given as 4096 bytes.

$$\text{Page Size}=4096\text{bytes}=2^{12}\text{bytes}$$

This means 12 bits are required for the page offset (the offset within the desired page).

3. Index of the Inner Page Table:
All pages in the system, including those storing the page tables, have the same size of 4096 bytes. The inner page table must fit within a single page of size 4096 bytes.
Each entry in the inner page table uses 8 bytes. We can calculate the number of entries in the inner page table as follows:

$$\text{Number of entries in inner page table}=\frac{\text{Page Size}}{\text{Size of an entry}}=\frac{4096}{8}=512=2^9\text{entries}$$

To index these 512 entries in the inner page table, we need 9 bits. This corresponds to the offset within the page of the inner page table.

4. Calculating the value of b:
The 32-bit logical address is divided into:
- b bits (outer page table index)
- 9 bits (inner page table index)
- 12 bits (page offset)

$$b+9+12=32$$

$$b+21=32$$

$$b=32-21=11$$

Thus, the value of the outer page table index size b is 11.