A computer system has 4 k word cache organised 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
| A. | 15, 4 |
| B. | 6, 4 |
| C. | 7, 2 |
| D. | 4, 6 |
Answer : D Explanation : There are 64 words in a block. So 4K cache has ( 4 x 1024)/64 = 64 blocks. Since 1 set has 4 blocks, there are 16 sets. 16 sets needs 4 bits for representation. In a set there are 4 blocks, which needs 2 bits. Each block has 64 words. So, the word field has 6 bits. |
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Option: A Explanation : Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. |
