Question

Solve this problem with details. Thank you

For a direct mapped Cache dekgn wh a 3bit adre Ta D cohta the cache bleck sDe in bords)? How man entr doei the Cache have? ratto betoeen o age

Answer 1

1. Size of cache block = 2^{(number of bits in offset)} words = 2⁰ = 1 word. Hence size of cache is 1 word.

2. Total number of cache blocks = 2^{(number of bits in index field)} = 2¹⁰ = 1024 block

Size of each block = 1 word

Total number of entries in cache = (number of cache blocks)*(number of words in each block) = 1024*1 = 1024 words  

3. Cache size = (Total number of blocks)*(size of each block)

Assuming 1 word = 4 bytes

Cache size = 1024*1 words = 1024 words = 4096 Bytes

4. Total size required to implement cache = (number of bits in tag) * (total number of blocks) = 22*1024 bits = 22*1024/8 Bytes = 2816 Bytes

Required ratio = size required to implement cache/ size of cache = 2816/4096 = 11/16

Hence the required ratio is 11:16.

Please comment for any clarification.