os slip8

Q.1 Write a C program to accept the number of process and resources and find the need matrix content and display it.
ANSWER-1

#include <stdio.h>
int main() {
int n, m;
printf("Enter processes and resources: ");
scanf("%d %d", &n, &m);
int alloc[n][m], max[n][m], need[n][m];
printf("Enter allocation and max matrices:\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
scanf("%d %d", &alloc[i][j], &max[i][j]);
need[i][j] = max[i][j] - alloc[i][j];}}
printf("Need matrix:\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
printf("%d ", need[i][j]);}
printf("\n");}
return 0;}

Q.2. Write the simulation program to implement demand paging and show the page scheduling and total number of page faults for the following given page reference string. Give input n
=3 as the number of memory frames. Reference String : 12,15,12,18,6,8,11,12,19,12,6,8,12,15,19,8 Implement OPT
ANSWER-2

#include <stdio.h>
int main() {
int n; printf("Enter memory frames: "); scanf("%d", &n);
int ref[] = {12, 15, 12, 18, 6, 8, 11, 12, 19, 12, 6, 8, 12, 15, 19, 8};
int len = sizeof(ref) / sizeof(ref[0]), mem[n], faults = 0;
for (int j = 0; j < n; j++) mem[j] = -1;
printf("Page Scheduling using OPT:\n");
for (int r = 0; r < len; r++) {
int p = ref[r], found = 0, future = -1, fIdx = 0;
for (int j = 0; j < n; j++) {
if (mem[j] == p) {
found = 1;
break;}
for (int k = r + 1; k < len; k++) {
if (mem[j] == ref[k]) {
future = k;
break;}}
if (future == -1) {
fIdx = j;
break;}
if (future > fIdx) {
fIdx = j;}}
if (!found) {
faults++;
if (mem[fIdx] != -1) {
printf("Page %d replaced by Page %d\n", mem[fIdx], p);}
mem[fIdx] = p;
printf("Page %d -> Page Fault\nMemory: ", p);
for (int j = 0; j < n; j++) {
printf("%d ", mem[j]);}
printf("\n");}}
printf("Total Page Faults: %d\n", faults);
return 0;