GATE CS 2014 Set 2 Q9 — Memory Hierarchy: Cache, Main Memory and Secondary Storage

To find the number of bits in the TAG field, we break down the physical address into three fields: TAG, Set Index, and Block Offset.

1. Physical Address (PA) Size:
   Physical Address Space = $4\text{ GB} = 2^{32}\text{ bytes}$.
   Therefore, the Physical Address length is $32\text{ bits}$.
2. Block Offset Bits:
   Word length = $32\text{ bits} = 4\text{ bytes}$.
   Block size = $8\text{ words} = 8 \times 4\text{ bytes} = 32\text{ bytes} = 2^5\text{ bytes}$.
   Thus, Block Offset = $5\text{ bits}$.
3. Set Index Bits:
   Cache Capacity = $16\text{ KB} = 2^{14}\text{ bytes}$.
   Number of blocks in the cache = $\frac{\text{Cache Capacity}}{\text{Block Size}} = \frac{2^{14}}{2^5} = 2^9 = 512$ blocks.
   Since it is a $4$-way set-associative cache, the number of sets is: $$\text{Number of Sets} = \frac{512}{4} = 128 = 2^7$$ Thus, Set Index = $7\text{ bits}$.
4. Tag Bits: $$\text{Tag Bits} = \text{Physical Address Size} - (\text{Set Index Bits} + \text{Block Offset Bits})$$ $$\text{Tag Bits} = 32 - (7 + 5) = 20\text{ bits}$$