For a 64-Kbyte. 2-way set associative cache, calculate the cache miss rate for the following loop. Make the following assumptions: There is no entry in the cache initially; Integers are 4 bytes long and two-dimensional arrays are stored in row-major order. Partially credit may be awarded, if intermediate results are shown int array[32][1024]; // An integer array with 32 rows and 1024 columns for (int i = 0; i less than 32; i++) { for (int j = 0: j < 1024; j++) { array[i][j] = (array[i][j] + array[31 - i][1023-j]) / 2; } } What is the cache miss-rate for 32-byte cache blocks What is the cache miss-rate for 16-byte cache blocks? Can you re-write the program, so that the final program performs the same computations and memory accesses but with fewer cache misses.

a miss ratio is calculated by dividing the total number of content requests by the number of misses.
For instance, if you go back over a period of time and discover that your cache had 11 misses while receiving a total of 48 content requests, you would divide 11 by 48 to obtain a miss ratio of 0.229.
32-bit byte addressing is used by computers. The computer makes use of a 32KB 2-way associative cache. The size of a cache block is 16 bytes.
A request takes longer to process the more cache layers a system must examine.
This causes a higher rate of cache misses, especially when the system must consult the main database to retrieve the required data.
For example, a 64-byte cache line indicates that the memory is separated into separate, 64-byte blocks that are not overlapping.
Because each block comprises 64 bytes, the start address of each block has six address bits that are always zero.

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