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
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Answer:
<em>C++</em>
///////////////////////////////////////////////////////////////////////////////////////////
#include <iostream>
using namespace std;
//////////////////////////////////////////////////////////////////
class QuadraticEquation {
int a, b, c;
public:
QuadraticEquation(int a, int b, int c) {
this->a = a;
this->b = b;
this->c = c;
}
////////////////////////////////////////
int getA() {
return a;
}
int getB() {
return b;
}
int getC() {
return c;
}
////////////////////////////////////////
// returns the discriminant, which is b2-4ac
int getDiscriminant() {
return (b*2)-(4*a*c);
}
int getRoot1() {
if (getDiscriminant() < 0)
return 0;
else {
// Please specify how to calculate the two roots.
return 1;
}
}
int getRoot2() {
if (getDiscriminant() < 0)
return 0;
else {
// Please specify how to calculate the two roots.
return -1;
}
}
};
//////////////////////////////////////////////////////////////////
int main() {
return 0;
}
Answer:
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