Which bits of the address would be used in the tag, index and offset in a two-way set associative cache with 1-word blocks and a total capacity of 512 words
1 answer:
Size of block = 1 word = 8 bytes number of blocks = size of cache / size of word = 512 word /1 word = 512 blocks In a two-way set associative cache, 2 blocks will form 1 set. So number of index = 512/2=256=2^8 Size of offset = 3 bits size of index = 8 bits Size of tag = 64-11=53 bits
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.2 x 16 -> 3.2 F 3
.2 x 16 -> 3.2 F 3
.2 x 16 -> 3.2 F 3
.2 x 16 -> 3.2 F 3
.2 x 16 -> 3.2 F 3
.2 x 16 -> 3.2 F 3
.2 x 16 -> 3.2 F 3
.2 x 16 -> 3.2 F 3
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Explanation:
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