Computer Engineering Made Easy: February 2012

Wednesday, 15 February 2012

Semester 4 Operating System : Pre Emptive SJF java program / code

Given below is Java program for Pre Emptive SJF.

//Aim:To implement preemptive SJF scheduling   
//Program:
import java.util.*;
class PR
{
int bt;
int at;
PR()
{
bt=0;
at=0;
}
PR(int x,int y)
{
bt=x;
at=y;
}
}
class PESJF
{
static int sum(PR p[],int n)
{
int s=0;
for(int i=0;i<n;++i)
s+=p[i].bt;
if(s==0)
return 1;
else
return 0;
}
/* Assume that the none of the processes terminate before all the processes have arrived */
public static void main(String args[])
{
PR p[];
Scanner scr=new Scanner(System.in);
System.out.println("Enter the number of processes");
int n=scr.nextInt();
p=new PR[n];
int i=0,j=0;
int bt,at;
for(i=0;i<n;++i)
{
System.out.println("Enter the burst time and arrival time of a process");
bt=scr.nextInt();
at=scr.nextInt();
p[i]=new PR(bt,at);
}
PR temp;
//  Sorting processes in increasing order of arrival time
for(i=0;i<=n-2;++i)
{
for(j=i;j<=n-1;++j)
{
if(p[i].at>p[j].at)
{
temp=p[i];
p[i]=p[j];
p[j]=temp;
}
}
}
int time=p[n-1].at;
int tt[]=new int[n];
int wt[]=new int[n];
int b[]=new int[n];
for(i=0;i<n;++i)
{
System.out.println("Process "+(i+1)+" Burst time= "+p[i].bt+" Arrival time= "+p[i].at);
b[i]=p[i].bt;
}
int mat=p[n-1].at;
int pn=0;
i=0;
j=0;
p1:while(j<mat)
{
while(j<p[i+1].at)
{
if(p[pn].bt>0)
{
p[pn].bt--;
++j;
}
else
{
System.out.println("Process "+(pn+1)+" executed.Turn around time= "+b[pn]+" Wait time=0");
++i;
for(int y=0;y<i;++y)
{
for(int z=y;z<i-1;++z)
{
if((p[y].bt<p[z].bt)&&p[y].bt>0)
pn=y;
}
}
continue p1;
}}
if(i<n-1)
if(p[i+1].bt<p[pn].bt)
pn=(i+1);
++i;
}
System.out.println("After arrival of all processes");
for(int g=0;g<n;++g)
System.out.println("Burst time of process "+(g+1)+"left = "+ p[g].bt+" ");
System.out.println("Time elapsed= "+i);
time=i;
int k=0;
int min=0;
do
{
min=0;
if(p[0].bt==0)
for(i=0;i<n;++i)
{
if(p[i].bt!=0)
{
min=i;
break;
}
}
for(i=0;i<n;++i)
{
if((p[i].bt<p[min].bt)&&p[i].bt>0)
min=i;
}
while(p[min].bt>0)
{
++time;
p[min].bt--;
}
tt[min]=(time-p[min].at);
wt[min]=(tt[min]-b[min]);
System.out.println("Process "+(min+1)+" executed.Turn around time= "+tt[min]+" Wait time= "+wt[min]);
++k;
}
while(sum(p,n)!=1);
double att=0,awt=0;
double s1=0,s2=0;
for(i=0;i<n;++i)
{
s1+=tt[i];
s2+=wt[i];
}
att=s1/n;
awt=s2/n;
System.out.println("Average Turn around time= "+att+" Average wait time= "+awt);
}
}
/* OUTPUT
Enter the number of processes
4
Enter the burst time and arrival time of a process
8
0
Enter the burst time and arrival time of a process
4
1
Enter the burst time and arrival time of a process
9
2
Enter the burst time and arrival time of a process
5
3
Process 1 Burst time= 8 Arrival time= 0
Process 2 Burst time= 4 Arrival time= 1
Process 3 Burst time= 9 Arrival time= 2
Process 4 Burst time= 5 Arrival time= 3
After arrival of all processes
Burst time of process 1left = 7
Burst time of process 2left = 2
Burst time of process 3left = 9
Burst time of process 4left = 5
Time elapsed= 3
Process 2 executed.Turn around time= 4 Wait time= 0
Process 4 executed.Turn around time= 7 Wait time= 2
Process 1 executed.Turn around time= 17 Wait time= 9
Process 3 executed.Turn around time= 24 Wait time= 15
Average Turn around time= 13.0 Average wait time= 6.5    */

Sunday, 12 February 2012

Sem 4 AOAD : Java Code / Program for MAXMIN algorithm

Given below is the java code for MAXMIN concept.

import java.util.*;
class Stack 
{
int top;
int s[]=new int[50];
Stack()
{
top=-1;
}

void push(int n)
{
s[++top]=n;
}
int pop()
{
return s[top--];
}
void display()
{
for(int i=0;i<=top;++i)
System.out.print(" "+s[i]);
}
}
class MAXMIN2
{
static int a[];
static int i,j,max,min;
static int flag=0;
Scanner scr=new Scanner(System.in);
MAXMIN2()
{
System.out.println("Enter number of elements in array");
int n=scr.nextInt();
a=new int[n];
System.out.println("Enter the elements");
for(int i=0;i<n;++i)
a[i]=scr.nextInt();
}
Stack s1=new Stack();
Stack s2=new Stack();
void max_min(int i,int j)
{
if(i==j)
{
   //if(flag==0)
   //{
   max=a[j];
   min=a[i];
   s1.push(max);
   s2.push(min);
   //}
   /*else
   {
   nmax=a[j];
   nmin=a[i];
   
   
   }  */
}
else
if(i==(j-1))
{
     //if(flag==0)
     //{
      min=(a[i]<a[j])?a[i]:a[j];
      max=(a[i]>a[j])?a[i]:a[j];
     s1.push(max);
	 s2.push(min);
	 //}
    /*else
    {
     nmin=(a[i]<a[j])?a[i]:a[j];
     nmax=(a[i]>a[j])?a[i]:a[j];
    }  */
 }
  else
 {
    int m=(i+j)/2;
   max_min(i,m);
   max_min(m+1,j);
   int lmax,lmin,rmax,rmin;
   rmax=s1.pop();
   rmin=s2.pop();
   lmax=s1.pop();
   lmin=s2.pop();
   if(rmax>lmax)
   s1.push(rmax);
   else
   s1.push(lmax);
  if(rmin<lmin)
    s2.push(rmin);
  else
    s2.push(lmin);  
}
}
void display1()
{
int low=s2.pop();
int high=s1.pop();
System.out.println("max value in the array is= "+high+" min value in the array= "+low);
}
}
class MainMAXMIN
{
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
System.out.println("Please enter the length of array");
int l=scr.nextInt();
MAXMIN2 m=new MAXMIN2();
m.max_min(0,(l-1));
m.display1();
/*
System.out.println("max value in the array is= "++" min value in the array= "+min);
System.out.println("max value in the array is= "+max+" min value in the array= "+min);*/
}
}

Wednesday, 8 February 2012

Semester 4 Operating System (OS) : Optimal page replacement policy Java program / Code

Given below is the Java Code for Optimal page replacement.

//Aim:To implement optimal page replacement policy  
//Program:
import java.util.*;
class PR3
{
static int check2(int p[],int pf[],int count[],int j,int s)
{int k;
for(int i=0;i<s;++i)
count[i]=0;
for(k=0;k<s;++k)
{
loop:for(int i=j-1;;--i)
{
++count[k];
if(p[i]==pf[k])
break loop;
}
//System.out.println("count = "+count[k]);
}
int max=count[0];
int pos=0;
for(int l=0;l<s;++l)
{
if(count[l]>max)
{
max=count[l];
pos=l;
}
}
return pos;
}
static int check(int p[],int pf[],int count[],int j,int s)
{
int k;
for(int i=0;i<s;++i)
count[i]=0;int found=0;
for(k=0;k<s;++k)
{
loop:for(int i=j+1;i<p.length;++i)
{
++count[k];
if(p[i]==pf[k])
{
++found;
break loop;
}
}
}
int pos;
if(found<(s-1))
{
System.out.println();
System.out.println("More than 1 used pages not required for furthur operation.\nSwapping LRU  page.");
pos=check2(p,pf,count,j,s);
}
else
{
 int max=count[0];
 pos=0;
for(int l=0;l<s;++l)
{
if(count[l]>max)
{
max=count[l];
pos=l;
}
}
}
return pos;
}
static boolean check1(int p[],int x,int n)
{
for(int i=0;i<n;++i)
{
if(p[i]==x)
return true;
}
return false;
}
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
System.out.println("Enter the number of page entries");
int n=scr.nextInt();
int fault=0,hit=0;
int p[]=new int[n];
System.out.println("Enter the pages ");
for(int i=0;i<n;++i)
p[i]=scr.nextInt();
System.out.print("The page table entries are ");
for(int i=0;i<n;++i)
System.out.print(" "+p[i]);
System.out.println();
int s=3;
System.out.println("Do you wish to set a the number of page frames?(1=yes/0=no)");
int ch=scr.nextInt();

if(ch==1)
 {
 System.out.println("Enter the number of page frames ");
 s=scr.nextInt();
 System.out.println("The number of page frames set to "+s);
 }
else
 System.out.println("The default number of page frames are "+s);
int count[]=new int[s];
int pf[]=new int[s];
int t=0;
int incr=0;
for(int i=0;i<s || (t!=s);++i)
{
if(check1(pf,p[i],t))
{
++incr;
System.out.println();
for(int y=0;y<t;++y)
System.out.print(pf[y]+" ");
System.out.print(" Hit");
++hit;
}
else
{++incr;
++t;
pf[t-1]=p[i];
System.out.println();
for(int y=0;y<t;++y)
System.out.print(pf[y]+" ");
System.out.print(" F");
++fault;
}
}
for(int j=incr;j<n;++j)
{
if(check1(pf,p[j],s))
{
++hit;
System.out.println();
for(int k=0;k<s;++k)
System.out.print(pf[k]+" ");
System.out.print(" H");
}
else
{
int choice=check(p,pf,count,j,s);
++fault;
pf[choice]=p[j];
System.out.println();
for(int k=0;k<s;++k)
System.out.print(pf[k]+" ");
System.out.print(" F");
}
}
System.out.println();
System.out.println("Page faults= "+fault+" Page hits= "+hit);
}
}
/* OUTPUT
Enter the number of page entries
20
Enter the pages
  7 0 1 2 0 3 0 4 2 3 0 3 2 1 2 0 1 7 0 1
The page table entries are  7 0 1 2 0 3 0 4 2 3 0 3 2 1 2 0 1 7 0 1
Do you wish to set a the number of page frames?(1=yes/0=no)
0
The default number of page frames are 3

7  F
7 0  F
7 0 1  F
2 0 1  F
2 0 1  H
2 0 3  F
2 0 3  H
2 4 3  F
2 4 3  H
2 4 3  H
2 0 3  F
2 0 3  H
2 0 3  H
2 0 1  F
2 0 1  H
2 0 1  H
2 0 1  H
7 0 1  F
7 0 1  H
7 0 1  H
Page faults= 9 Page hits= 11



Enter the number of page entries
20
Enter the pages
  7 0 1 2 0 3 0 4 2 3 0 3 2 1 2 0 1 7 0 1
The page table entries are  7 0 1 2 0 3 0 4 2 3 0 3 2 1 2 0 1 7 0 1
Do you wish to set a the number of page frames?(1=yes/0=no)
1
Enter the number of page frames
4
The number of page frames set to 4

7  F
7 0  F
7 0 1  F
7 0 1 2  F
7 0 1 2  H
3 0 1 2  F
3 0 1 2  H
3 0 4 2  F
3 0 4 2  H
3 0 4 2  H
3 0 4 2  H
3 0 4 2  H
3 0 4 2  H
1 0 4 2  F
1 0 4 2  H
1 0 4 2  H
1 0 4 2  H
7 0 4 2  F
7 0 4 2  H
1 0 4 2  F
Page faults= 9 Page hits= 11   */

Sem 4 Operating System (OS) : LRU page replacement Java Code / Program

Given below is the Java program for LRU page replacement policy.

import java.util.*;
class LRU
{
static int check1(int p[],int pf[],int n,int s)
{
int count[]=new int[s];
for(int i=0;i<s;++i)
{
int j=n;
while(j>=0 && p[j]!=pf[i])
{
++count[i];
--j;
}
}

int max=count[0];
int pos=0;
for(int i=1;i<s;++i)
{
if(count[i]>max)
{
max=count[i];
pos=i;
}
}
return pos;
}
static boolean check(int p[],int n,int x)
{
for(int i=0;i<n;++i)
{
if(p[i]==x)
return true;
}
return false;
}
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
System.out.println("Enter the number of page entries");
int n=scr.nextInt();
int fault=0,hit=0;
int p[]=new int[n];
System.out.println("Enter the pages ");
for(int i=0;i<n;++i)
p[i]=scr.nextInt();
System.out.print("The page table entries are ");
for(int i=0;i<n;++i)
System.out.print(" "+p[i]);
System.out.println();
int s=3;
System.out.println("Do you wish to set a the number of page frames?(1=yes/0=no)");
int ch=scr.nextInt();

if(ch==1)
 {
 System.out.println("Enter the number of page frames ");
 s=scr.nextInt();
 System.out.println("The number of page frames set to "+s);
 }
else
 System.out.println("The default number of page frames are "+s);

int pf[]=new int[s];
for(int i=0;i<s;++i)
pf[i]=-1;
int d=0;
int choice=0;
for(int i=0;i<n;++i)
{
if(check(pf,s,p[i]))
{
++hit;
for(int j=0;j<s && pf[j]!=-1;++j)
System.out.print(pf[j]+" ");
System.out.print("H\n");
}
else
{
++fault;
if(fault<=s)
{
pf[d]=p[i];
++d;
}
else
{
choice=check1(p,pf,i-1,s);
pf[choice]=p[i];
}
for(int j=0;j<s && pf[j]!=-1;++j)
System.out.print(pf[j]+" ");
System.out.print("F\n");
}
}
System.out.println("Number of page Hits= "+hit+"\nNumber of page Faults= "+fault);
}
}

Tuesday, 7 February 2012

Sem 4 Operating System (OS) : FIFO Page replacement Java code / program

Given below is the Java program for FIFO page replacement algorithm.You can also use the DOWNLOAD BOX to download the '.java' file.

import java.util.*;
class FIFO
{
static boolean check(int p[],int n,int x)
{
for(int i=0;i<n;++i)
{
if(p[i]==x)
return true;
}
return false;
}
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
System.out.println("Enter the number of page entries");
int n=scr.nextInt();
int fault=0,hit=0;
int p[]=new int[n];
System.out.println("Enter the pages ");
for(int i=0;i<n;++i)
p[i]=scr.nextInt();
System.out.print("The page table entries are ");
for(int i=0;i<n;++i)
System.out.print(" "+p[i]);
System.out.println();
int s=3;
System.out.println("Do you wish to set a the number of page frames?(1=yes/0=no)");
int ch=scr.nextInt();

if(ch==1)
 {
 System.out.println("Enter the number of page frames ");
 s=scr.nextInt();
 System.out.println("The number of page frames set to "+s);
 }
else
 System.out.println("The default number of page frames are "+s);

int pf[]=new int[s];
for(int i=0;i<s;++i)
pf[i]=-1;
int d=0;

for(int i=0;i<n;++i)
{
if(check(pf,s,p[i]))
{
++hit;
for(int j=0;j<s && pf[j]!=-1;++j)
System.out.print(pf[j]+" ");
System.out.print("H\n");
}
else
{
++fault;
if(d>s-1)
d=0;
pf[d]=p[i];
++d;
for(int j=0;j<s && pf[j]!=-1;++j)
System.out.print(pf[j]+" ");
System.out.print("F\n");
}
}
System.out.println("Number of page Hits= "+hit+"\nNumber of page Faults= "+fault);
}
}

Sem 4 Operating System (OS) : Round Robin process scheduling Java Code / Program

Given below is the Java code for Round Robin Process Scheduling algorithm.

import java.util.*;
class Process
{
Scanner scr=new Scanner(System.in);
int p[];
int bt,tt,wt;
Process(int i)
{
System.out.println("Enter the burst time of process "+(i+1));
bt=scr.nextInt();
}
}
class RR1
{
static boolean check(Process p[],int n)
{
int s=0;
for(int i=0;i<n;++i)
s+=p[i].bt;
if(s<=0)
return false;
else
return true;
}
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
System.out.println("Enter the number of processes");
int n =scr.nextInt();
Process p[]=new Process[n];
for(int i=0;i<n;++i)
p[i]=new Process(i);
System.out.println("The quantum size is 5 ms ");
int burst[]=new int[n];

for(int i=0;i<n;++i)
burst[i]=p[i].bt;
for(int i=0;i<n;++i)
System.out.println("The  burst time of process "+(i+1)+" is "+burst[i]);
int time=0;
int temp;
int flag[]=new int[n];
for(int i=0;i<n;++i)
flag[i]=0;
do
{

loop:for(int k=0;k<n;++k)
{
//System.out.println(" Flag value is "+flag[k]);
if(flag[k]==0)
{
temp=p[k].bt;
if(p[k].bt>5)
{
p[k].bt-=5;
time+=5;
//System.out.println("The time value "+time);
}
else
{
p[k].bt=0;
time+=temp;
p[k].tt=time;
flag[k]=1;
//System.out.println("The process "+(k+1)+" ends");
//System.out.println("flag value "+flag[k]);
}
}
else continue loop;
}
}
while(check(p,n)); 
for(int i=0;i<n;++i)
p[i].wt=p[i].tt-burst[i];
double s1=0,s2=0;
for(int i=0;i<n;++i)
{
System.out.println("The turn around time of process "+(i+1)+" is "+p[i].tt);
System.out.println("The wait time of process "+(i+1)+" is "+p[i].wt);
}
for(int i=0;i<n;++i)
s1+=p[i].tt;
for(int i=0;i<n;++i)
s2+=p[i].wt;
double att=0,awt=0;
att=s1/n;
awt=s2/n;
System.out.println("The average turn around time is "+att+"\nThe average wait time is "+awt);
}
}
/* Output
Enter the number of processes
4
Enter the burst time of process 1
12
Enter the burst time of process 2
3
Enter the burst time of process 3
7
Enter the burst time of process 4
15
The quantum size is 5 ms
The  burst time of process 1 is 12
The  burst time of process 2 is 3
The  burst time of process 3 is 7
The  burst time of process 4 is 15
The turn around time of process 1 is 32
The wait time of process 1 is 20
The turn around time of process 2 is 8
The wait time of process 2 is 5
The turn around time of process 3 is 25
The wait time of process 3 is 18
The turn around time of process 4 is 37
The wait time of process 4 is 22
The average turn around time is 25.5
The average wait time is 16.25      */

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