COMPUTER ARCHITECTURE
MEMORY SYSTEMS
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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16 bits
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8 bits
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12 bits
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32 bits
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Detailed explanation-1: -There are 2^32 = 4, 294, 967, 296 total bits in a 32-bit address. Therefore, the required number of bits for a direct-mapped cache with 16 KB of data and 4-word blocks is 2^32 + 4 = 4, 294, 967, 296.
Detailed explanation-2: -The number of addressable units = 2s+w words or bytes. The block size (cache line width not including tag) = 2w words or bytes. The number of blocks in main memory = 2s (i.e., all the bits that are not in w) The number of lines in cache = m = 2. More items
Detailed explanation-3: -Calculate the size of each address in m bits. If main memory has 2048 bytes, then we have 2048=2^m unique addresses. Calculate bit offset n from the number of bytes in a block. 64 bytes/8 blocks = 8 bytes per block. Calculate the set index s. Finally, we know the number of tag bits is T=m-s-n. 13-Mar-2021