os slip2

Q.1 Create a child process using fork(), display parent and child process id. Child process will display the message “Hello World” and the parent process should display “Hi”.
ANSWER-1

#include <stdio.h>
#include <unistd.h>
int main() {
pid_t child_pid = fork();
if (child_pid == 0) {
printf("Hello World\n");
} else if (child_pid > 0) {
printf("Hi\n");
} else {
perror("Fork failed");
return 1;}
return 0;}

2.Write the simulation program using SJF (non-preemptive). The arrival time and first CPU bursts of different jobs should be input to the system. Assume the fixed I/O waiting time (2
units). The next CPU burst should be generated using random function. The output should give the Gantt chart, Turnaround Time and Waiting time for each process and average times
ANSWER-2

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define MAX_PROCESSES 10
typedef struct {
int arrival, burst, turnaround, waiting;
} Process;
int cmpArrival(const void *a, const void *b) { return ((Process *)a)->arrival - ((Process *)b)->arrival; }
int main() {
int n; scanf("%d", &n);
Process p[MAX_PROCESSES] = {0};
srand(time(0));
for (int i = 0; i < n; i++) p[i].arrival = i, p[i].burst = rand() % 10 + 1;
qsort(p, n, sizeof(Process), cmpArrival);
printf("Gantt Chart:\n");
float tTAT = 0, tWT = 0, t = 0;
for (int i = 0; i < n; i++) {
printf("P%d ", i + 1);
p[i].turnaround = t + p[i].burst - p[i].arrival;
p[i].waiting = p[i].turnaround - p[i].burst;
t += p[i].burst + 2;
tTAT += p[i].turnaround;
tWT += p[i].waiting;}
printf("\nProcess\tTurnaround\tWaiting\n");
for (int i = 0; i < n; i++) printf("P%d\t%d\t\t%d\n", i + 1, p[i].turnaround, p[i].waiting);
printf("Avg TAT: %.2f\nAvg WT: %.2f\n", tTAT / n, tWT / n);}