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For the preceding simple implementation, this execution order would be nonideal for the input matrix; however, applying a loop interchange optimization would create a nonideal order for the output matrix. Because loop interchange is not sufficient to improve its performance, it must be blocked instead. (a) What should be the minimum size of the cache to take advantage of blocked execution

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batolisis

Answer:

hi your question lacks the necessary matrices attached to the answer is the complete question

1024 bytes

Explanation:

A) The minimum size of the cache to take advantage of blocked execution

 The minimum size of the cache is approximately 1 kilo bytes

There are 128 elements( 64 * 2 ) in the preceding simple implementation and this because there are two matrices and every matrix contains 64 elements .

note:  8 bytes is been occupied by every element therefore the minimum size of the cache to take advantage of blocked execution

= number of elements * number of bytes

= 128 * 8 = 1024 bytes ≈ 1 kilobytes

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