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physical memory 32 bit

cache memory 256 KB = 2^18

4 way set associative

cache block size 16B = 2^4

so block offset bit =4

no of blocks = 2^14

no of sets = 2^14 / 4 = 2^12

set offset bits = 12 bits

so tag bits = 32 - 12 - 4 =16 bits.

size of comparator=no of tag bits

size of comparator needed is 16 bits only.

(all options are wrong,basically they wanted to know the tag memory size)

no of tag bits in set associative cache is  = 16

size of  tag bits  is = 16* 2^14 = 2^18.

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