Discussion
A computer with a 32-bit address uses two-level page table. Virtual addresses are split into a 9-bit top level page table field, an 11-bit second level page table field and an offset. How large are the pages ?
Answer: B
The pages are 4 KB.
The virtual address is split into the top-level index, the second-level index and the offset within the page. The offset bits fix the page size.
\[ \begin{aligned} \text{offset} &= 32 - 9 - 11 = 12\ \text{bits} \\ \text{page} &= 2^{12} = 4096\ \text{bytes} \\ &= 4\ \text{KB} \end{aligned} \]
2 KB would need an 11-bit offset and 8 KB a 13-bit offset. The trap is to use one of the table fields as the offset; 11 bits gives 2 KB.
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