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Below are listed parameters for different direct-mapped cache designs.

Cache Data Size: 32 KiB
Cache Block Size: 2 words
Cache Access Time: 1 cycle
Word: 4 bytes.
Calculate the total number of bits required for the cache listed above, assuming a 32-bit address. Given that total size, find the total size of the closest direct-mapped cache with 16-word blocks of equal size or greater. Explain why the second cache, despite its larger data size, might provide slower performance than the first cache.

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To calculate the number of bits required for the cache, we can use the following formula:

Number of bits = Cache data size / (Cache block size x Word size)

Plugging in the values, we get:

Number of bits = 32 KiB / (2 x 4 bytes) = 4096 bits

To find the total size of the closest direct-mapped cache with 16-word blocks of equal size or greater, we can use the following formula:

Total size = (Number of blocks) x (Block size)

Since the cache in the first part had 32 KiB of data, we can calculate the number of blocks as follows:

Number of blocks = Cache data size / Cache block size = 32 KiB / 2 words = 16,384 blocks

Therefore, the total size of the closest direct-mapped cache with 16-word blocks or greater would be:

Total size = 16,384 blocks x 16 words = 262,144 words

The second cache might provide slower performance than the first cache because it has a larger block size. A larger block size can lead to an increased number of cache misses, which can slow down the system. Additionally, the second cache has a larger data size, which can lead to longer access times.
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User Lukasz Muzyka
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