GATE CS 1995 Q2.25 — Memory Hierarchy: Cache, Main Memory and Secondary Storage

To find the number of bits in the SET and WORD fields of the main memory address format, we can analyze the cache parameters step-by-step:

1. WORD Field:

   The block size is given as $64$ words.

   $$\text{Number of bits for WORD field} = \log_2(\text{Block size}) = \log_2(64) = 6 \text{ bits}$$

2. Total Cache Blocks:

   The total cache size is $4\text{ K}$ words ($4 \times 1024 = 4096$ words).

   $$\text{Total blocks in cache} = \frac{\text{Cache size}}{\text{Block size}} = \frac{4096}{64} = 64 \text{ blocks}$$

3. SET Field:

   The cache is $4$-way set-associative, meaning there are $4$ blocks per set.

   $$\text{Number of sets} = \frac{\text{Total blocks}}{\text{Blocks per set}} = \frac{64}{4} = 16 \text{ sets}$$

   $$\text{Number of bits for SET field} = \log_2(\text{Number of sets}) = \log_2(16) = 4 \text{ bits}$$

Thus, the number of bits in the SET and WORD fields are $4$ and $6$ respectively.