Final answer:
In a fully associative mapping, the number of bits in the tag field of the MAR would be 18 bits. In a direct mapping, the number of bits in the tag field would be 15 bits. In a 2-way associative mapping, the number of bits in the tag field would be 16 bits.
Step-by-step explanation:
In a fully associative mapping, the cache uses a tag field to check if the requested memory address is present in the cache. Since the cache has a line size of 8 words, and each word is 32 bits (4 bytes) long, the tag field in the MAR would need to be able to identify each line in the cache. Therefore, the number of bits in the tag field would be equal to the log base 2 of the number of lines in the cache, which is equal to log2(2Mbyte/8 words)= log2(256K) = 18 bits.
In a direct mapping, the cache is divided into equally sized blocks, and each block is associated with a specific portion of main memory. With a line size of 8 words, each block in the cache would correspond to a block of 8 words in main memory. The number of bits in the MAR required for the block offset field would be log base 2 of the line size, which is equal to 3 bits in this case. The remaining bits in the MAR would be used for the tag field, which would be equal to the log base 2 of the number of blocks in the cache. Therefore, the number of bits in the tag field would be log2(2Mbyte/8 words/8 blocks) = log2(32K) = 15 bits.
In a 2-way associative mapping, the cache is divided into sets, and each set contains two blocks. With a line size of 4 words, each set in the cache would contain two blocks, where each block corresponds to a block of 4 words in main memory. The number of bits in the MAR required for the block offset field would be log base 2 of the line size, which is equal to 2 bits in this case. The remaining bits in the MAR would be used for the tag field, which would be equal to the log base 2 of the number of sets in the cache. Therefore, the number of bits in the tag field would be log2(2Mbyte/4 words/2 sets) = log2(256K) = 16 bits.