Determine the number of cache sets (S), tag bits (t), set index bits (s), and block offset bits (b) for a 40964096-byte cache using 3232-bit memory addresses, 88-byte cache blocks and a 88-way associative design. The cache has

Complete Question:

Determine the number of cache sets (S), tag bits (t), set  index bits (s), and block offset bits (b) for a 4096-byte cache using 32-bit memory addresses, 8-byte cache blocks and a 8-way associative design. The cache has :

Cache size = 1024 bytes, sets t = 26.8, tag bits, s = 3.2, set index bit =2

Check below for explanations

Explanation:

Cache size = 4096 bytes = 2¹² bytes

Memory address bit = 32

Block size = 8 bytes = 2³ bytes

Cache line = (cache size)/(Block size)

Cache line =

Cache line = 2⁹

Block offset = 3 (From 2³)

Tag = (Memory address bit - block offset - Cache line bit)

Tag = (32 - 3 - 9)

Tag = 20

Total number of sets = 2⁹ = 512

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explanation:

d. long term memory