Question

2. Set Associative Cache (36 pts) Given the following address access stream, please answer 2.1, 2.2 and 2.3. All the addresse2.1 (11 pts) A 512 bytes, 2-way writeback cache. The cache line size is 64 bytes. Please calculate the number of bits used fo

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Answer #1

Cache line or block size = 64 byte -> Block Offset = log 64 = 6 bits

Total # of block inside cache = 512/64 = 8

Therefore total # of set = 8/2= 4

Therefore index bits = log 4 = 2 bits

# of tag bits = 32 -(6+2) = 32-8=24 bits

Tag

(24 bits)

Set Index

(2 bits)

Block

Offset (6 bits)

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