Consider a machine with a byte addressable main memory of 216 bytes block size…
2020
Consider a machine with a byte addressable main memory of 216 bytes block size of 8 bytes. Assume that a direct mapped cache consisting of 32 lines used with this machine. How many bits will be there in Tag, line and word field of format of main memory addresses?
Answer: A. 8,5,3 — Solution: Total main memory size = 2^16 bytes, so address length = 16 bits. Block size = 8 bytes, so block offset (word) = log2(8) = 3 bits. Cache has 32…
- A.
8,5,3
- B.
8,6,2
- C.
7,5,4
- D.
7,6,3
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Correct answer: A
Solution:
Total main memory size = 2^16 bytes, so address length = 16 bits.
Block size = 8 bytes, so block offset (word) = log2(8) = 3 bits.
Cache has 32 lines, so index (line) = log2(32) = 5 bits.
Tag bits = total address bits - index bits - offset bits = 16 - 5 - 3 = 8 bits.
Therefore the fields (Tag, line, word) are 8, 5, 3.
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