Answer:
1G
Explanation:
1 'G' as in first generation.
Hope this helped. :)
Answer:
#include <iostream>
using namespace std;
int main() {
int k;
double d;
string s;
cin >> k >> d >> s;
cout << s << " " << d << " " << k << "\n" << k << " " << d << " " << s; }
Explanation:
k is int type variable that stores integer values.
d is double type variable that stores real number.
s is string type variable that stores word.
cin statement is used to take input from user. cin takes an integer, a real number and a word from user. The user first enters an integer value, then a real number and then a small word as input.
cout statement is used to display the output on the screen. cout displays the value of k, d and s which entered by user.
First the values of k, d and s are displayed in reverse order. This means the word is displayed first, then the real number and then the integer separated again by EXACTLY one space from each other. " " used to represent a single space.
Then next line \n is used to produce a new line.
So in the next line values of k, d and s are displayed in original order (the integer , the real, and the word), separated again by EXACTLY one space from each other.
The program along with the output is attached.
Answer:
develop a business plan to describe how to maintain and grow revenues
Explanation:
Answer:
a) CPU utilization for uniprogramming system = 18.2%
b) CPU utilization for multiprogramming system = 41.379%
Explanation:
IO time = total time – CPU time
For JOB 1
CPU time = 3ms ,
total time= 23ms
IO time = 23-3 = 20ms ,
For JOB 2
CPU time =5ms
total = 29ms
IO time = 29-5 = 24ms
For JOB 3
CPU time = 4ms
total = 14ms
IO time = 14-10 =10ms
1.) In uniprogramming system, operations are performed sequentially
CPU utilization=total CPU time / total real time
CPU utilization =(3+5+4) / (23+29+14)
CPU utilization =0.182 or 18.2%
2) In multiprogramming system, jobs wait for the CPU to get free while performing IO operations concurrently
steps followed by the os
1. IO for 1st job at 0ms
2. IO for 2nd job at 0ms
3. IO for 3rd job at 0ms
4. CPU time for 3rd job at 5ms
5. Next IO job for 3rd job at 9ms END at 14ms
6. CPU time for 1st job at 9ms
7. Next IO job for 1st job at 14ms END at 23ms
8. CPU time for 2nd job at 14ms
9. Next IO job for 2nd job at 15ms END at 29ms
CPU time is 3+5+4 =12ms
Total execution time = 29ms
CPU utilization = 12*100/29= 41.379%