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Experiemnt 5

The document contains two experiments demonstrating scheduling concepts in process management. Experiment 5A illustrates non-preemptive scheduling with a C program that calculates waiting and turnaround times for processes. Experiment 5B showcases preemptive scheduling, using a time slice to manage process execution and compute average waiting and turnaround times.

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10 views4 pages

Experiemnt 5

The document contains two experiments demonstrating scheduling concepts in process management. Experiment 5A illustrates non-preemptive scheduling with a C program that calculates waiting and turnaround times for processes. Experiment 5B showcases preemptive scheduling, using a time slice to manage process execution and compute average waiting and turnaround times.

Uploaded by

dakshmahajan2022
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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PROCESS MANAGEMENT:SCHEDULING

EXPERIMENT NO - 5A

AIM - Write a program to demonstrate the concept of non-preemptive scheduling .

#include <stdio.h>
int main() {
int bt[20], wt[20], tat[20], i, n;
float wtavg = 0.0, tatavg = 0.0;

printf("\nEnter the number of processes: ");


scanf("%d", &n);

for(i = 0; i < n; i++) {


printf("\nEnter Burst Time for Process %d: ", i + 1);
scanf("%d", &bt[i]);
}

wt[0] = 0;
tat[0] = bt[0];

for(i = 1; i < n; i++) {


wt[i] = wt[i - 1] + bt[i - 1]; tat[i] = tat[i - 1] + bt[i];
}

for(i = 0; i < n; i++) {


wtavg += wt[i];
tatavg += tat[i];
}

printf("\nPROCESS\tBURST TIME\tWAITING TIME\tTURNAROUND TIME\n");

for(i = 0; i < n; i++) {


printf("\nP%d\t\t%d\t\t%d\t\t%d", i + 1, bt[i], wt[i], tat[i]);
}

printf("\nAverage Waiting Time: %.2f", wtavg / n);


printf("\nAverage Turnaround Time: %.2f", tatavg / n);

return 0;}
PROCESS MANAGEMENT:SCHEDULING
EXPERIMENT NO - 5B

AIM - Write a program to demonstrate the concept of preemptive scheduling .

#include <stdio.h>
int main() {
int i, j, n, bu[10], wa[10], tat[10], t, ct[10], max;
float awt = 0.0, att = 0.0, temp = 0;

printf("Enter the number of processes: ");


scanf("%d", &n);

for(i = 0; i < n; i++) {


printf("\nEnter Burst Time for Process %d: ", i + 1);
scanf("%d", &bu[i]);
ct[i] = bu[i];
}

printf("\nEnter the size of time slice: ");


scanf("%d", &t);

max = bu[0];
for(i = 1; i < n; i++) {
if(max < bu[i]) {
max = bu[i];
}
}

for(j = 0; j < (max / t) + 1; j++) {


for(i = 0; i < n; i++) {
if(bu[i] != 0) {
if(bu[i] <= t) {
tat[i] = temp + bu[i];
temp += bu[i];
bu[i] = 0;
} else {
bu[i] -= t;
temp += t;
}
}
}
}
for(i = 0; i < n; i++) {
wa[i] = tat[i] - ct[i]; // Waiting time = Turnaround time - Burst time
att += tat[i];
awt += wa[i]; }

printf("\nThe Average Turnaround Time is: %.2f", att / n);


printf("\nThe Average Waiting Time is: %.2f", awt / n);

printf("\nPROCESS\tBURST TIME\tWAITING TIME\tTURNAROUND TIME\n");

for(i = 0; i < n; i++) {


printf("\nP%d\t\t%d\t\t%d\t\t%d", i + 1, ct[i], wa[i], tat[i]);
}

return 0;
}

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