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GATE CSE 2021 Set 2 | Question: 19

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Consider a set-associative cache of size $\text{2KB (1KB} =2^{10}$ bytes$\text{)}$ with cache block size of $64$ bytes. Assume that the cache is byte-addressable and a $32$ -bit address is used for accessing the cache. If the width of the tag field is $22$ bits, the associativity of the cache is _________
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Calculate Tag size in Direct Mapping = MM Size/CM size

MM = 2^32

CM = 2^11

DM TAG = 2^32/2^11 = 2^21 = 21 bit

SET Associative tag  = DM tag + log base 2 N-way

22 = 21 + log base 2 N-way

N-way = 2

that's the answer.

 
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