Final answer:
The memory address 0x172f3d4c in a 16-way associative cache of size 1024KB with 32B blocks will have a 16-bit tag, 11-bit index, and a 5-bit offset.
Step-by-step explanation:
To calculate the tag, index, and offset for the memory address 0x172f3d4c in a 16-way associative cache of size 1024KB with 32B blocks, we need to break down the address into these components based on the cache's configuration.
Firstly, let's translate the hexadecimal address 0x172f3d4c into binary. The 32-bit binary equivalent will then be divided into the tag, index, and offset fields.
The size of the offset is determined by the block size. With 32B blocks, we have 5 bits for the offset (25 = 32). Therefore, the least significant 5 bits of the address are used for the offset.
The number of index bits is determined by the number of cache blocks, which can be calculated as the cache size divided by the number of ways and the block size. In a 1024KB cache, there are (1024 * 1024) / (16 * 32) = 2048 blocks, requiring 11 bits for the index (since 211 = 2048).
The remaining bits of the 32-bit address are used for the tag. After accounting for the index and offset bits, we have 32 - 5 (offset) - 11 (index) = 16 bits for the tag.
Now, we can dissociate the binary address into its corresponding parts:
- The tag is the leftmost 16 bits of the address.
- The index is the next 11 bits.
- The offset is the rightmost 5 bits.
Using these calculations, we can then extract the tag, index, and offset from the memory address 0x172f3d4c. The exact binary values will depend on the binary representation of the given hexadecimal address.