GATE CS 2021 Set 2 Q19 — Memory Hierarchy: Cache, Main Memory and Secondary Storage

In a set-associative cache, a $32$-bit physical address is divided into three fields:

[ Tag | Set Index | Block Offset ]

Let's calculate the size of each field based on the given parameters:

1. Block Offset:

   The cache is byte-addressable and the block size is $64$ bytes ($2^6$ bytes).

   $$\text{Block Offset bits} = \log_2(64) = 6\text{ bits}$$
2. Tag Field:

   The width of the tag field is given as $22$ bits.

3. Set Index:

   Since the total address length is $32$ bits:

   $$\text{Set Index bits} = 32 - (\text{Tag bits} + \text{Block Offset bits})$$ $$\text{Set Index bits} = 32 - (22 + 6) = 4\text{ bits}$$

   Therefore, the number of sets in the cache is:

   $$\text{Number of Sets} = 2^4 = 16$$

Now, we calculate the total number of blocks in the cache:

For a $P$-way set-associative cache, the number of blocks per set is $P$ (associativity):

Thus, the associativity of the cache is 2 (i.e., it is a 2-way set-associative cache).