shortest job first scheduling non-preemptive| Code

#include<stdio.h> int main() { system("Title CPU Scheduling SJF-Non preemptive "); int i,n,j,min,line[10],line1[10],p[10]={1,2,3,4,5,6,7,8,9,10}; int B_T[1000],A_V_T[1000],W_T[1000],temp,T_A_T[10]; int bt=0,ta=0,sum=0,k=1; float AVR_W_T=0,AVR_T_A_T=0,T_A_T_SUM=0,W_T_SUM=0; printf("Enter the Number of processes : "); scanf("%d",&n); for(i=0;i<n;i++) { printf("\nEnter The Arrival Time Of Process %d : ",i+1); scanf(" %d",&A_V_T[i]); printf("Enter The Burst Time Of Process%d : ",i+1); scanf(" %d",&B_T[i]); } for(i=0;i<n;i++) { for(j=0;j<n;j++) { if(A_V_T[i]<A_V_T[j]) { temp=p[j]; p[j]=p[i]; p[i]=temp; temp=A_V_T[j]; A_V_T[j]=A_V_T[i]; A_V_T[i]=temp; temp=B_T[j]; B_T[j]=B_T[i]; B_T[i]=temp; } } } for(j=0;j<n;j++) { bt=bt+B_T[j]; min=B_T[k]; for(i=k;i<n;i++) { if (bt>=A_V_T[i] && B_T[i]<min) { temp=p[k]; p[k]=p[i]; p[i]=temp; temp=A_V_T[k]; A_V_T[k]=A_V_T[i]; A_V_T[i]=temp; temp=B_T[k]; B_T[k]=B_T[i]; B_T[i]=temp; } } k++; } W_T[0]=0; for(i=1;i<n;i++) { sum=sum+B_T[i-1]; W_T[i]=sum-A_V_T[i]; line[i]=W_T[i]; W_T_SUM=W_T_SUM+W_T[i]; } AVR_W_T=(W_T_SUM/n); for(i=0;i<n;i++) { ta=ta+B_T[i]; T_A_T[i]=ta-A_V_T[i]; line1[i]=T_A_T[i]; T_A_T_SUM=T_A_T_SUM+T_A_T[i]; } AVR_T_A_T=(T_A_T_SUM/n); printf("\nProcess Name\tBurst Time\t Arrival Time\t Waiting Time\t Turn Around Time" ); for(i=0;i<n;i++) { printf("\np%d\t\t%d\t\t %d\t\t %d\t\t %d",p[i],B_T[i],A_V_T[i],W_T[i],T_A_T[i]); } printf("\n\nAVERAGE WAITING TIME : %.2f",AVR_W_T); printf("\nAVERAGE TURN AROUND TIME : %.2f\n",AVR_T_A_T); printf("\n"); for(i=0;i<T_A_T_SUM/2;i++) { printf(" "); } printf("GANTT CHART\n\n"); line[0]=0; for(i=0;i<n;i++) { for(j=0;j<(line[i]+line1[i])/2;j++) { printf(" "); } printf("P%d",p[i]); } printf("\n"); printf("0"); for(i=0;i<n; i++) { for(j=0;j<line1[i];j++) { printf("-"); } printf("%d",line1[i]); } printf("\n"); getch(); return 0; }

1 Response

This code was Excellent because of this I understand how SJF in a non-primitive way works.

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