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
