How many total bits are required for a direct mapped cache with 16KB of data and 4-word blocks assuming a 32-bit address?

To calculate the total number of bits required for a direct-mapped cache, we need to consider the cache size, block size, and address size. Let's break down the calculation step by step:

Cache Size: 16KB

Block Size: 4 words

Address Size: 32 bits

1. Determine the number of blocks in the cache:

? ?Cache Size (in bytes) / Block Size (in bytes)

? ?16KB / (4 words * 4 bytes/word) = 4KB / 16 bytes = 256 blocks

2. Calculate the number of index bits:

? ?Number of blocks = 2^(Number of index bits)

? ?Number of index bits = log2(Number of blocks)

? ?Number of index bits = log2(256) = 8 bits

3. Calculate the number of offset bits:

? ?Offset bits = log2(Block Size in bytes)

? ?Offset bits = log2(4 words * 4 bytes/word) = log2(16) = 4 bits

4. Calculate the number of tag bits:

? ?Tag bits = Address Size - (Number of index bits + Number of offset bits)

? ?Tag bits = 32 bits - (8 bits + 4 bits) = 20 bits

5. Calculate the total number of bits required:

? ?Total bits = (Number of blocks) * (Tag bits + 1 valid bit)

? ?Total bits = 256 blocks * (20 bits + 1 bit) = 256 * 21 = 5376 bits

Therefore, a direct-mapped cache with 16KB of data, 4-word blocks, and a 32-bit address would require a total of 5376 bits.