A computer system with a word length of 32 bits has a 16 MB byte- addressable…
2020
A computer system with a word length of 32 bits has a 16 MB byte- addressable main memory and a 64 KB, 4-way set associative cache memory with a block size of 256 bytes. Consider the following four physical addresses represented in hexadecimal notation.
\(A1 = 0x42C8A4, \ \ \ \ \ \ \ \ A2 = 0x546888, \ \ \ \ \ \ \ \ A3 = 0x6A289C, \ \ \ \ \ \ \ \ A4 = 0x5E4880\)
Which one of the following is TRUE ?
- A.
\(A1\)and\(A4\)are mapped to different cache sets. - B.
\(A2\)and\(A3\)are mapped to the same cache set. - C.
\(A3\)and\(A4\)are mapped to the same cache set. - D.
\(A1\)and\(A3\)are mapped to the same cache set.
Attempted by 131 students.
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Correct answer: B
Key insight: with a 256-byte block and a 64 K B (64 KB) cache that is 4-way set associative, the block offset is 8 bits and there are 64 sets (6 index bits). To find the set: block number = address >> 8, set = block_number & 0x3F.
Compute block numbers (address >> 8): A1 = 0x42C8, A2 = 0x5468, A3 = 0x6A28, A4 = 0x5E48.
Compute set indices (block_number & 0x3F): A1 -> 0x08 (8), A2 -> 0x28 (40), A3 -> 0x28 (40), A4 -> 0x08 (8).
Conclusion: Addresses 0x546888 and 0x6A289C map to the same cache set (set 40), so the correct statement is that A2 and A3 are mapped to the same cache set.
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