2. We have an L3 Cache (write-back). Its configuration is: • Address size= 48 bits • Block size = 64 Bytes. • Total cache size= 8 MB • Associativity = 16-ways Find the total number of bits in the cache. Note: Be sure to consider all required bits and to show your steps to find the solution. 3. Calculate the total number of bits for a write-through cache with the following configuration: • Address Size= 48 bits • Block Size = 64 Bytes • Total cache size = 32 KB • Associativity = 4-ways
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Total cache size = 8 MB = $8 \times 2^{20}$ Bytes Block size = 64 Bytes = $2^6$ Bytes Number of blocks = $\frac{8 \times 2^{20}}{2^6} = 8 \times 2^{14} = 2^3 \times 2^{14} = 2^{17}$ blocks Show more…
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