SSC IMD SA Mock 04 – Question 158

A four-way set-associative cache memory unit with a capacity of \(32\text{ KB}\) is built using a block size of \(1\text{ K}\). The word length is \(8\text{ bits}\). The size of physical address space is \(32\text{ GB}\). The number of bits for the TAG field is:

• 22
• 24
• 25
• 23

Step-by-step calculation:
1. Word length = \(8\text{ bits} = 1\text{ Byte}\), so memory is byte-addressable.
2. Physical Address Space = \(32\text{ GB} = 2^5 \times 2^{30}\text{ Bytes} = 2^{35}\text{ Bytes}\) \(\Rightarrow\) Total Physical Address = \(35\text{ bits}\).
3. Block size = \(1\text{ KB} = 2^{10}\text{ Bytes}\) \(\Rightarrow\) Block/Word offset = \(10\text{ bits}\).
4. Total cache capacity = \(32\text{ KB}\).
5. Number of cache lines/blocks = \(\frac{32\text{ KB}}{1\text{ KB}} = 32\).
6. For a 4-way set-associative cache, lines per set = 4. Number of sets = \(\frac{32}{4} = 8 = 2^3\) \(\Rightarrow\) Set index = \(3\text{ bits}\).
7. TAG bits = Total Address Bits \(- (\text{Set Index Bits} + \text{Block Offset Bits}) = 35 – (3 + 10) = 22\text{ bits}\).