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The size of the physical address space of a processor is $2^{\text{P}}$ bytes. The word length is $2^{\text{W}}$ bytes. The capacity of cache memory is $2^{\text{N}}$ bytes. The size of each cache block $2^{\text{M}}$ words. For a $\text {K-Way}$ set associative cache memory, the length (in number of bits) of the tag fields is:

  1. $ \text {P- N- M- W +} \log \text{K}$

  2. $ \text {P- N- M- W } - \log_{2} \text{K}$

  3. $\text {P- N} + \log_{2} \text{K}$

  4. $\text {P- N} - \log_{2} \text{K}$

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Answer : C

P – N + log2(K)