Consider a computer with a 4-ways set-associative mapped cache of the…
2008
Consider a computer with a 4-ways set-associative mapped cache of the following characteristics: a total of 1 MB of main memory, a word size of 1 byte, a block size of 128 words and a cache size of 8 KB. While accessing the memory location 0C795H by the CPU, the contents of the TAG field of the corresponding cache line is
- A.
000011000
- B.
110001111
- C.
00011000
- D.
110010101
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Correct answer: A
Solution: compute tag bits for address 0C795H.
Address size: 1 MB = 2^20 bytes → 20-bit physical addresses.
Block offset: block size = 128 words → offset = log2(128) = 7 bits.
Index bits: cache size = 8 KB = 8192 bytes. With 4-way associativity and 128-byte blocks, number of sets = 8192 / (128 * 4) = 16 → index = log2(16) = 4 bits.
Tag bits: tag = 20 - 7 - 4 = 9 bits.
Convert the address 0C795H to 20-bit binary: 0 C 7 9 5 → 0000 1100 0111 1001 0101 → 00001100011110010101.
Extract the most significant 9 bits (tag): 000011000.
Therefore the TAG field of the corresponding cache line is 000011000.
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