Complete Question:
Determine the number of cache sets (S), tag bits (t), set index bits (s), and block offset bits (b) for a 4096-byte cache using 32-bit memory addresses, 8-byte cache blocks and a 8-way associative design. The cache has :
Cache size = 1024 bytes, sets t = 26.8, tag bits, s = 3.2, set index bit =2
Answer:
Check below for explanations
Explanation:
Cache size = 4096 bytes = 2¹² bytes
Memory address bit = 32
Block size = 8 bytes = 2³ bytes
Cache line = (cache size)/(Block size)
Cache line =
Cache line = 2⁹
Block offset = 3 (From 2³)
Tag = (Memory address bit - block offset - Cache line bit)
Tag = (32 - 3 - 9)
Tag = 20
Total number of sets = 2⁹ = 512
Answer:
A computer ___Task__ is a text file that a computer can understand
Answer:
Java code is given below
Explanation:
import java.util.Random;
class Die{
private int sides;
public Die(){
sides = 6;
}
public Die(int s){
sides = s;
}
public int Roll(){
Random r = new Random();
return r.nextInt(6)+1;
}
}
class DieRoll{
public static void main(String[] args) {
Die die = new Die();
int arr[] = new int[6];
for(int i=0; i<6; i++)
arr[i] = 0;
for(int i=0; i<100; i++){
int r = die.Roll();
arr[r-1]++;
}
for(int i=0; i<6; i++)
System.out.println((i+1)+" was rolled "+arr[i]+" times.");
}
}
Answer:if you are looking for video and audio data its mp4
Explanation:
there is no explaination