Question

For a direct-mapped cache design with a 32-bit address, the following bits are used to access the cache: Tag Index 12 11-6 31- 4.a) What is the cache block size (in words)? 4.b) How many entries does the cache have? Offset 5-0 4.c) What is the total number of bits in this cache? ALU Sancion xxx sad add J0000 100100 10010 10 4.d) What is the ratio between the total number of bits required for this cache implementa- tion over the data storage bits?

Answer 1

In a direct-mapped cache design with a 32-bit address, the tag is represented by the most significant bits, the index is represented by the bits in the middle, and the offset is represented by the least significant bits.

Step-by-step explanation:

In a direct-mapped cache design with a 32-bit address, the tag is represented by the most significant bits (31-12), the index is represented by bits 11-6, and the offset is represented by bits 5-0.

a) To find the cache block size, we need to determine the number of bits for the offset. In this case, there are 6 bits for the offset, which means each cache block is 26 = 64 bits or 8 bytes.

b) The number of entries in the cache is determined by the number of bits available for the index. In this case, there are 6 bits for the index, which means there are 26 = 64 entries in the cache.

c) To find the total number of bits in the cache, we sum the number of bits for the tag, index, and offset. In this case, it is 12 bits for the tag + 6 bits for the index + 6 bits for the offset, giving us a total of 24 bits.

d) The ratio between the total number of bits required for this cache implementation over the data storage bits is given by: (tag + index + offset) / offset. In this case, it is (12 + 6 + 6) / 6 = 4.

Answer 2

Final answers

a) Cache block size (in words): 64 words

b) Number of entries in the cache: 4096 entries

c) Total number of bits in this cache: 65536 bits

d) Ratio between total implementation bits and data storage bits: 16:1

Explanation

For a direct-mapped cache design with a 32-bit address, the breakdown of bits into tag, index, and offset allows us to determine crucial aspects of the cache.

a) The cache block size in words is found by considering the offset bits. With 6 offset bits (5-0), the cache block size can address 2^6 words, resulting in 64 words per block.

b) The number of entries in the cache is determined by the index bits. With 6 index bits (11-6), the cache can accommodate 2^6 entries, yielding 4096 entries.

c) Calculating the total number of bits in the cache involves accounting for the tag, index, and offset bits. In this setup, 12 bits for the tag, 6 bits for the index, and 6 bits for the offset sum up to 24 bits per entry. Considering 4096 entries, the total number of bits becomes 24 bits/entry * 4096 entries = 65536 bits.

d) The ratio between the total implementation bits and the data storage bits is derived from the total number of bits (65536 bits) compared to the data storage bits in a single entry (offset bits - 6 bits). The ratio is 65536 bits / (64 words * 6 bits) = 16:1, indicating that the implementation overhead is 16 times the actual data storage bits.