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The size of the physical address space of a processor is 2P bytes. The word length is 2W bytes. The capacity of cache memory is 2N bytes. The size of each cache block is 2M words. For a K-way set-associative cache memory, the length (in number of bits) of the tag field is
  • a)
    P − N − log2K
  • b)
    P − N + log2K
  • c)
    P − N − M − W − log2K
  • d)
    P − N − M − W + log2K
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The size of the physical address space of a processor is 2Pbytes. The ...
Physical Address Space = 2P Bytes. Word Length is 2W bytes, which means each word is of size 2W bytes. Cache memory size = 2N Bytes and Tag Size = 2X Bytes. Physical address is P - W bits Number of blocks in cache = 2(N-W-M) It is a K-way set associative cache memory, each set in cache will have K-blocks. So, Number of sets = 2(N-W-M)/ K SET bits will be N-W-M-logk Offset bits will be M We know, TAG bits = Main memory bits - SET bits - offset bits So, TAG bits(x) = P - W - (N-M-W-logk)- M       = P - W - N + M + W + logk - M       x = P - N + logk Option (B) is Correct.
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The size of the physical address space of a processor is 2Pbytes. The ...
Understanding Cache Memory and Tag Field Calculation
To determine the length of the tag field in a K-way set-associative cache, we need to analyze the components of the memory architecture.
Key Parameters:
- Physical Address Space: 2^P bytes
- Word Length: 2^W bytes
- Cache Memory Capacity: 2^N bytes
- Cache Block Size: 2^M words
Calculating Cache Parameters:
1. Total Cache Lines:
- Cache size in bytes divided by block size in bytes gives the number of cache lines.
- Cache size = 2^N bytes
- Block size = 2^M words * 2^W bytes/word = 2^(M + W) bytes
- Total lines = 2^N / 2^(M + W) = 2^(N - M - W)
2. Number of Sets:
- For a K-way set-associative cache, the number of sets is total lines divided by K.
- Number of sets = (2^(N - M - W)) / K = 2^(N - M - W - log2(K))
3. Address Breakdown:
- The physical address consists of the tag, index, and block offset.
- The block offset is determined by the block size: log2(block size) = M + W.
- The index consists of the number of sets: log2(number of sets) = N - M - W - log2(K).
Tag Field Calculation:
- The tag field length can be derived from the physical address size:
- Tag length = P (total address bits) - Index length - Block offset length
- Tag length = P - (N - M - W - log2(K)) - (M + W)
- Simplifying gives us: Tag length = P - N + log2(K).
Conclusion:
- The correct answer for the length of the tag field in the K-way set-associative cache is:
P - N + log2(K)
Thus, the correct option is B.
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The size of the physical address space of a processor is 2Pbytes. The word length is 2Wbytes. The capacity of cache memory is 2Nbytes. The size of each cache block is 2Mwords. For a K-way set-associative cache memory, the length (in number of bits) of the tag field isa)P − N − log2Kb)P − N + log2Kc)P − N − M − W − log2Kd)P − N − M − W + log2KCorrect answer is option 'B'. Can you explain this answer?
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