Suppose we have a byte-addressable computer using a direct mapped caching with 20-bit main memory addresses and 64 blocks of cache. If each block contains 16 bytes, determine: Size of the offset field: type your answer. bits, Size of the block field: type your answer. bits, Size of the tag field: type your answer. bits

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A: For a 4-way set associative cache, the length of the TAG field is given below:

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A: Given: char arr[5]; Address of arr=100. Size of one char = 1 byte. To Find: Address of arr[4]

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A: Block size = 32 words  Block Offset = 5 bits Total # of blocks inside cache = 16 So index bits = 4…

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A: The following solution is

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A: The answer is here

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A: Solution is in Step 2.

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A: The answer is

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A: ANSWER:-

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Q: re there in a set? ANSWER: cache has? ANSWER: address format is => | Tag: blocks sets bits | Set:…

A:

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A:

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A:

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A: The correct option is (c) Answer 1 and Answer 2 are given below choose one from them which is easy…

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A:

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A: Actually, the answer has given below:

Q: Suppose we have a byte-addressable computer using fully associative o 20-bit main memory addresses…

A: Actually, 1 byte =8 bits. cache memory is a fast access memory.