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A computer system has 4 K word cache organized in a block-set associative manner, with 4 blocks per set, 64 words per block. The number of bits in the SET and WORD fields of the main memory address format is

  1. 15, 4
  2. 6, 4
  3. 7, 2
  4. 4, 6
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64 words per set - so $\lg 64 = 6$ bits for WORD.

No. of sets = cache size/(block size * no. of blocks per set)

= 4K/(64 * 4) = 16.

So, no. of set bits = $\lg 16 = 4.$

Answer is - d)4, 6
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