Q.30 ) Design 1 : Fully Associative Cache Memory
Word Size = 16 bits = 2 Bytes
$\text{Cache Size} = \frac{256KB}{2B} = 128K \text{words}$
Block Size = 1K words
# Cache lines = 128K / 1K = 128
Therefore, # Tag Comparators required = # Cache Lines = 128
Q. 31 ) Design 2 : 8 Way Set Associative Cache Memory
Block Size = 64 words
# Cache Lines = 128K / 64 = 2K
#Sets = 2K / 8 = $2^8$
Therefore, tag size = 32 - 8 - 6 = 18 bits
Tag Memory Size = #Cache lines * tag size = 2K * 18 bits = 4.5 KBytes
