Answer:
40, 5 and 3
See explaination for the details
Explanation:
A)
Consider the data.
Assume D-cache is empty
An integer size =4bytes
The cache block size =4bytes
Therefore, the number of D-cache misses for reading the first 40 integers is,
= (40×4)/4
= 160/4
= 40
b)
Consider the data.
Assume D-cache is empty
An integer size =4bytes
The cache block size =32bytes
Therefore, the number of D-cache misses for reading the first 40 integers is,
= (40×4)/32
= 160/32
= 5
c)
Consider the data.
Assume D-cache is empty
An integer size =4bytes
The cache block size =64bytes
Therefore, the number of D-cache misses for reading the first 40 integers is,
= (40×4)/64
= 160/64
= 3