Consider a demand paging memory management system with 32-bit logical address, 20-bit physical address, and page size of 2048 bytes. Assuming that the memory is byte addressable, what is the maximum number of entries in the page table?

To find the maximum number of entries in the page table:

1. Given Data:

   • Logical Address Space (LAS) size $= 2^{32}$ bytes (since logical address is 32 bits and memory is byte-addressable).

   • Page size $= 2048\text{ bytes} = 2^{11}\text{ bytes}$.

   • Therefore, the page offset ($d$) requires $11$ bits.

2. Number of Pages:

   • The number of bits for the page number ($p$) is:

     $$p = \text{Logical Address Bits} - \text{Page Offset Bits} = 32 - 11 = 21\text{ bits}$$

   • The maximum number of pages in the logical address space is:

     $$\text{Number of Pages} = \frac{2^{32}\text{ bytes}}{2^{11}\text{ bytes}} = 2^{21}$$

3. Page Table Entries:

   • Each page has a corresponding entry in the page table.

   • Therefore, the maximum number of entries in the page table is $2^{21}$.

Thus, (A) is the correct option.