cache block size

Question

Consider a 64 KB cache with 256 sets. Addresses are 32 bits. Tags are 19 bits.

What is associativity (blocks per set) of this cache?

• A. 4
• B. 16
• C. 2
• D. 8

Answer 1

I think its 8.

address size 32 and tag is 19 bit so 

32 - 19 =13 bit (set bit +  offset) 

and there are 256 sets so set bit = 8 

word off = 13-8=5 bit

in set associative: size of cache = no. of sets * associativity * offset  

2¹⁶=2⁸ * associativity * 2⁵= > associativity =2³

Comments

Another way  to solve

Calculate no of blocks=2^16/2^5=2^11

Now associativity=no of blocks/no of sets=2^11/2^8=2^3=8