Consider a machine with a byte addressable main memory of 2^20 bytes, block…

Solution:

Step 1 — Determine bit widths:

• Total address bits = 20 (since main memory is 2^20 bytes).

• Block offset = 4 bits (block size 16 bytes = 2^4).

• Index = 12 bits (2^12 cache lines).

Step 2 — Compute tag bits:

• Tag bits = 20 - 12 - 4 = 4 bits (one hex digit).

Step 3 — Map the hex address E201F to tag, index, offset:

• Write the 20-bit address as 5 hex digits: E 2 0 1 F.

• Top hex digit (E) = tag.

• Next three hex digits (2 0 1) = index (cache line address) = 201 (hex).

• Last hex digit (F) = block offset within the cache line.

Answer: tag = E (hex), cache line address = 201 (hex).