Calculate the total number of bits required for the cache listed above, assuming a 32-bit address. Given that total size, fi nd 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 fi rst cache.

Respuesta :

Answer:

Number of bits in a cache = 2588672 bits

Cache with 16- word block of equal size or greater =4308992

Explanation:

To calculate the number of bits required for the cache containing 32- bit address.

formula to calculate the bits is :

Number of bits in a cache[tex](2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))[/tex]

total bits = [tex]2^15 (1+14+(32*2^1))[/tex] = 2588672 bits

Cache with 16- word block of equal size or greater

total bits =[tex]2^13(1 +13+(32*2^4))[/tex] =4308992