Consider a direct mapped cache of size 32 KB with block size 32 bytes.The CPU…
2005
Consider a direct mapped cache of size 32 KB with block size 32 bytes.The CPU generates 32-bit addresses.The number of bits needed for cache indexing and the number of tag bits are respectively:
Answer: A. 10, 17 — Step 1: Compute the block offset. Block size = 32 bytes, so block offset = log2(32) = 5 bits. Step 2: Compute the number of cache lines and index bits. Cache…
- A.
10, 17
- B.
10, 22
- C.
15, 17
- D.
5, 17
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Correct answer: A
Step 1: Compute the block offset.
Block size = 32 bytes, so block offset = log2(32) = 5 bits.
Step 2: Compute the number of cache lines and index bits.
Cache size = 32 KB = 32 * 1024 = 32768 bytes.
Number of lines = 32768 / 32 = 1024. Index bits = log2(1024) = 10 bits.
Step 3: Compute tag bits.
Total address bits = 32. Tag bits = 32 - index bits - offset bits = 32 - 10 - 5 = 17 bits.
Final answer: index bits = 10, tag bits = 17.
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