Consider a memory system with 1M bytes of main memory and 16K bytes of cache memory. Assume that the processor generates 20-bit memory address, and the cache block size is 16 bytes. If the cache uses direct mapping, how many bits will be required to store all the $\textit{tag}$ values? [Assume memory is byte addressable, 1K=$2^{10}$, 1M=$2^{20}$.]

Given parameters:

• Main memory size $= 1\text{ MB} = 2^{20}\text{ bytes}$, which corresponds to a 20-bit physical address.

• Cache memory size $= 16\text{ KB} = 16 \times 2^{10}\text{ bytes} = 2^{14}\text{ bytes}$.

• Block size (line size) $= 16\text{ bytes} = 2^4\text{ bytes}$.

1. Block/Word Offset bits:

   $$\text{Offset bits} = \log_2(16) = 4\text{ bits}$$

2. Number of lines in the cache:

   $$\text{Number of lines} = \frac{\text{Cache size}}{\text{Block size}} = \frac{16\text{ KB}}{16\text{ B}} = 1024 = 2^{10}\text{ lines}$$

3. Cache line index bits (for direct mapping):

   $$\text{Index bits} = \log_2(2^{10}) = 10\text{ bits}$$

4. Tag bits per line:

   $$\text{Tag bits} = \text{Address bits} - (\text{Index bits} + \text{Block offset bits}) = 20 - (10 + 4) = 6\text{ bits}$$

5. Total bits required to store all tag values in the cache:

   $$\text{Total tag memory} = \text{Number of lines} \times \text{Tag bits per line} = 2^{10} \times 6 = 6 \times 2^{10}\text{ bits}$$

Therefore, option (A) is the correct answer.