Question

A MIPS processor has a 32-bit address bus with and a cache memory to hold 8K (= 2^13) memory words. If a Direct-mapped cache is used to implement cache with a block size of 8 memory words then find:

• Number of bits for block field = [Select]
• Number of bits for set field= [Select]
• Number of bits for tag field = [Select]
• Capacity (in number of main memory words) =
• Actual size of the cache memory with a valid bit for each set V​

Answer 1

To determine the number of bits for the block field, set field, and tag field, we need to consider the cache organization and the given information.

Given:
- Cache memory size: 8K memory words (2^13)
- Block size: 8 memory words

1. Number of bits for the block field:
The block field is used to identify the specific block within a cache set. Since the block size is 8 memory words, we need log2(8) = 3 bits to represent the block field.

2. Number of bits for the set field:
In a direct-mapped cache, each memory address maps to a specific cache set. The number of cache sets can be calculated by dividing the cache memory size by the product of the block size and the number of sets per block. In this case, we have 8K memory words, a block size of 8 memory words, and a direct-mapped cache, so the number of sets is 8K / (8 * 1) = 1K = 2^10. Therefore, we need log2(2^10) = 10 bits to represent the set field.

3. Number of bits for the tag field:
The tag field is used to identify the memory block stored in the cache. Since the remaining bits after considering the block and set fields are used for the tag field, we can calculate it as the total number of bits minus the block and set field bits. In this case, the total number of bits is 32 (32-bit address bus), so the tag field will be 32 - 3 (block field) - 10 (set field) = 19 bits.

4. Capacity (in number of main memory words):
The capacity of the cache can be calculated by multiplying the number of cache sets by the block size. In this case, the number of cache sets is 2^10 and the block size is 8, so the capacity is 2^10 * 8 = 8K memory words.

5. Actual size of the cache memory with a valid bit for each set (V):
The actual size of the cache memory can be calculated by multiplying the number of cache sets by the block size and the number of bits per memory word. In this case, the number of cache sets is 2^10, the block size is 8, and the number of bits per memory word is 32 (32-bit address bus), so the actual size of the cache memory is 2^10 * 8 * 32 = 256K bits.

To summarize:
- Number of bits for block field = 3
- Number of bits for set field = 10
- Number of bits for tag field = 19
- Capacity (in number of main memory words) = 8K
- Actual size of the cache memory with a valid bit for each set (V) = 256K bits