Question: n - way set associative = = n Blocks / Set # of sets = ( Total Storage in Bytes - : # Bytes /

n-way set associative == n Blocks/Set
# of sets =(Total Storage in Bytes -: # Bytes/block)-: (# of Blocks/Set)
Offset bits = Log(base 2) of (# of Bytes/block)
Index bits = Log(base 2) of (# of Sets)
Tag bits =32-(Offset bits + Index bits)
Below is a list of 32-bit memory address references for questions A - B :
12,720,172,8,764,352,760,56,724,176,744,1012
A. Show the final cache contents for a 4-way set associative cache with 8-byte blocks and a total storage size of 128 bytes. Use LRU replacement.
Calculate the number of bits for Tag, Index, Offset, and the overhead ratio.
For each reference identify the index bits, the tag bits, DRAM Block Address, and if it is a hit or a miss.
B. Show the final cache contents for a fully associative cache with 4-byte blocks and a total storage size of 128 bytes. Use LRU replacement.
Calculate the number of bits for Tag, Index, Offset, and the overhead ratio.
For each reference identify the index bits, the tag bits, DRAM Block Address, and if it is a hit or a miss.
n - way set associative = = n Blocks / Set # of

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