Consider a machine with a byte addressable main memory of $2^{20}$ bytes, block size of 16 bytes and a direct mapped cache having $2^{12}$ cache lines. Let the addresses of two consecutive bytes in main memory be $(\text{E}201\text{F})_{16}$ and $(\text{E}2020)_{16}$. What are the tag and cache line address (in hex) for main memory address $(\text{E}201\text{F})_{16}$?

1. Physical Address Size: Main memory size $= 2^{20}\text{ bytes} \implies 20\text{ bits}$.

2. Block Offset (Word Offset): Block size $= 16\text{ bytes} = 2^4\text{ bytes} \implies 4\text{ bits}$.

3. Cache Line (Index) Bits: Number of cache lines $= 2^{12} \implies 12\text{ bits}$.

4. Tag Bits:

   $$\text{Tag bits} = 20 - 12 - 4 = 4\text{ bits}$$

5. Address Division for $(\text{E}201\text{F})_{16}$:

   • In hexadecimal, 20 bits correspond to 5 hex digits: E 2 0 1 F.

   • Tag (first 4 bits / 1 hex digit): E

   • Cache Line Index (next 12 bits / 3 hex digits): 201

   • Block Offset (last 4 bits / 1 hex digit): F

Therefore, the tag is E and the cache line address is 201.