Answer:
see explaination
Explanation:
#include <iostream>
#include <string>
using namespace std;
class LinkedList{
class Node{
public :
int data;
Node* next;
Node(int data){
this->data = data;
next = NULL;
}
};
public :
Node *head;
LinkedList(){
this->head = NULL;
}
void insert(int d){
Node* new_node = new Node(d);
new_node->next = head;
head = new_node;
}
// sort the list with selection sort algorithm.
// Pick the smallest element in the unsorted array and place in the first element in the unsorted.
void sort_list(){
if (head == NULL){
return;
}
Node* current = head;
while (current->next != NULL){
Node* min_node = current;
Node* traverse = current->next;
while(traverse != NULL){
if(traverse->data < min_node->data){
min_node = traverse;
}
traverse = traverse->next;
}
int temp = current->data;
current->data = min_node->data;
min_node->data = temp;
current = current->next;
}
}
void print_list(){
Node* current = head;
while(current !=NULL){
cout<<current->data<<" ";
current = current->next;
}
cout<<"\n";
}
};
int main(){
LinkedList ll;
for(int i=0;i<10;i++){
ll.insert(i);
}
ll.print_list();
cout<<"*******************************************\n";
ll.sort_list();
ll.print_list();
cout<<"*******************************************\n";
}
Explanation:
A.)
we have two machines M1 and M2
cpi stands for clocks per instruction.
to get cpi for machine 1:
= we multiply frequencies with their corresponding M1 cycles and add everything up
50/100 x 1 = 0.5
20/100 x 2 = 0.4
30/100 x 3 = 0.9
CPI for M1 = 0.5 + 0.4 + 0.9 = 1.8
We find CPI for machine 2
we use the same formula we used for 1 above
50/100 x 2 = 1
20/100 x 3 = 0.6
30/100 x 4 = 1.2
CPI for m2 = 1 + 0.6 + 1.2 = 2.8
B.)
CPU execution time for m1 and m2
this is calculated by using the formula;
I * CPI/clock cycle time
execution time for A:
= I * 1.8/60X10⁶
= I x 30 nsec
execution time b:
I x 2.8/80x10⁶
= I x 35 nsec
<span>To reduce chances of system failure, engineers may use process control. This is a system which involves activities that ensures the whole process variables can be predicted and controlled to prevent failure. This involves the use of safety valves and sensors.</span>
