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A CPU has 32 KB cache, 2-way set associative, 16-byte block size. How many bits are required to use index each set in this cache?

User Zack Brown
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1 Answer

7 votes

Answer:

10 bits

Step-by-step explanation:

Cache size = 32 KB

We were also given that it is 2 way set associative

Block size = 16 bytes = 2^4 = 4 bits for offset

Number of blocks = cache size / block size = (32*2^10) / 16

=2^11

Hence, index = number of blocks / 2 (since 2 way associative)

=(2^11)/2 = 2^10 = 10 bits for index

User Halxinate
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