
A.
Given Information:
The definition for the code is mentioned below:
//Traverse through the grid
for (i =0; i < 16; i++)
{
for (j = 0; j < 16; j++)
{
//add values of x into grid
total_x += grid[j][i].x;
//add values of y into grid
total_y += grid[j][i].y;
}
}
B.
Given Information:
The definition for the code is mentioned below:
//Traverse through the grid
for (i =0; i < 16; i++)
{
for (j = 0; j < 16; j++)
{
//add values of x into grid
total_x += grid[j][i].x;
//add values of y into grid
total_y += grid[j][i].y;
}
}
C.
Explanation of Solution
Miss rate:
- The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.
- Hence, each iteration will have one hit and one miss.
- This means one will have 256 hits and 256 misses...
D.
Explanation of Solution
New Miss Rate:
If the cache were twice as big the n it could hold the entire grid array and the only misses would be the initial cold miss...

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