virtual address 32 bit
block size 16 B
Each block contain 2 elements
Data cache size 64 KB
Number of blocks=212
but each block contain 2 elements.
So, no. of blocks will be 211
So, for the machine total size of the blocks $2^{11}\times 2^{4}=2^{15}$
Now , each element size $2^{3}$ B
So, no. of elements could be present in total block=$2^{12}$
So, 1st time array contain 0 to 1024 elements.
2nd time array can contain 1 to 1024 elements
3rd time array can contain 2 to 1024 elements
4th time array can contain 3 to 1024 elements
So, now all blocks are full.
No, next time block access it will replace 1st element by Arr[4][0]