A direct-mapped cache holds 64KB of useful data (not including tag or control bits). Assuming that the block size is 32-byte and
1 answer:
Answer:
A) Number of bits for byte = 6 bits
B) number of bits for index = 17 bits
C) number of bits for tag = 15 bits
Explanation:
Given data :
cache size = 64 kB
block size = 32 -byte
block address = 32 -bit
number of blocks in cache memory
cache size / block size = 64 kb / 32 b = 2^11 hence the number of blocks in cache memory = 11 bits = block offset
A) Number of bits for byte
= 6 bits
B) number of bits for index
block offset + byte number
= 11 + 6 = 17 bits
c ) number of bits for tag
= 32 - number of bits for index
= 32 - 17 = 15 bits
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Explanation:
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