Life teaches us

In life, carrier selection is very important in life.

If we know what we can do then it gives us a direction for our life. Trust yourself to choose what's right for you. For that we set a goal and we start working on it. To achieve that goal we start focusing and our life becomes automatically a healthy and wealthy. Dream a life with thinking of future in mind. Dream creates a desires. Desires create determination, which leads us to our destiny. After our achievements we will get success. Live a life which will set inspiration for others. It will be not so easy there will be many obstacles in your life don't get demotivate by it, fight against it after looking back you will say yourself that this obstacle is very small compared to present. Journey of life will collect many memories and many contacts with a confidence. Mistakes may be happened so don't be afraid learn from it and try not to repeat it. A mistake should be your teacher, not your attacker. A mistake is a lesson, not a loss. It is temporary not a dead end. Life and time are best teachers. Life teaches us the use of time and time teaches us the value of life. Life is like a book, some chapters are happy, some are exciting and some are sad. But if you never turn the page, you will never know what the next chapter holds.

In life there is three needs Roti(food) ,kapda (Clothes) and makaan (house)
but there is one more thing that is education 
Education is never stolen from you
Knowledge is never absolute
Never stop learning because learning never stops.
Example if you once learn cycling or swimming ,you never forget if you do cycling after ten years also.
When you will chase knowledge numbers will always follow.

Sys pro program for CPU scheduling : Priority algo. - Non premptive

PPRIIORITY.C

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/* CPU scheduling : Priority algo. - Non premptive ................... */
// jobQ maintained as ascending order priority Q (1st job-highest priority)
#include<stdlib.h>
#include<stdio.h>
#define newnode (struct node *)malloc(sizeof(struct node))
struct node
{ int jobno,time,proty;  // priority
  struct node *next;
}*JOBQ=NULL;

int job=0;

void erase()
{ struct node *S;
  while(JOBQ!=NULL) { S=JOBQ;  JOBQ=JOBQ->next;  free(S);  }
  job=0;
}

struct node *Append(struct node *F)
{ struct node *s,*t,*s1=NULL;
  t=newnode;        t->next=NULL;
  t->jobno=++job;   t->time=0;

  do  // validation check for job time
  {  printf("\nJOB %d - CPU burst : ",job);
     fflush(stdin); scanf("%d",&t->time);
     if(t->time<=0) printf("\nJob time should be > 0");
  }while(t->time<=0);

  do  // validation check for priority range 1-5
  {  printf("\t Priority : ",job);
     fflush(stdin);  scanf("%d",&t->proty);
     if(t->proty<1 || t->proty >5) printf("\nPriority range is 1(H) to 5(L) ");
  }while(t->proty<1 || t->proty >5);

  if(F==NULL)  F=t;    // JOBQ empty, so this is 1st job
  else    // priority wise insert (acsending oreder priority)
  {  for(s=F; s!=NULL; s1=s, s=s->next)  // find position
     { if(t->proty < s->proty) break;
     }
     t->next=s;  // insert before s
     if(s1==NULL) F=t;  // added before 1st node
     else s1->next=t; // inbetween position
  }
  return F;
}

void CreateJOBQ()
{ struct node *S;   int i,n;
  printf("\nHow many jobs? "); scanf("%d",&n);
  if(n<=0) { printf("\nError : No. of Jobs should be > 0."); return; }

  if(JOBQ!=NULL) erase(JOBQ);

  for(i=1;i<=n;i++)  JOBQ=Append(JOBQ);
}

void DisplayJOBQ()  // sorted on priority
{struct node *T;
  if(JOBQ==NULL) printf("\nJOB Q is Empty!");
  else
  { printf("\n JOB  Time  Priority");
    for(T=JOBQ;T!=NULL;T=T->next) printf("\n J%d : %3d %4d",T->jobno,T->time,T->proty);
  }
}

void delFstNode() // deletes 1st job(highest priority) from JOBQ
{struct node *t;
   t=JOBQ; JOBQ=JOBQ->next;
   free(t);
}

void ExecuteJOBQ()  // Shortest Job First
{ struct node *T,*S=NULL,*t1;  int time=0; char choice;

  if(JOBQ==NULL) { printf("\nJOB Q is Empty!"); return; }

  printf("\nclock CPU   JOBQ ");
  printf("\n%3d  -idle-",time);
  for(T=JOBQ; T!=NULL; T=T->next) printf(" J%d(%d) ",T->jobno,T->time);  // print JOBQ

  while(JOBQ!=NULL)
  {
    printf("\n%3d  J%d(%d) |--",time,JOBQ->jobno,JOBQ->time); // cpu allocated
    time+=JOBQ->time; // increment clock
    delFstNode();  // job finished
    for(T=JOBQ; T!=NULL; T=T->next) printf(" J%d(%d:%d) ",T->jobno,T->proty,T->time);  // print JOBQ

    printf(" --|  Add New Job(y/n)? ");  choice=getche();
    if(choice=='y' || choice=='Y')  JOBQ=Append(JOBQ);
  }
  printf("\n%3d  -idle-",time);
}

char menu()
{ char choice='a';
  clrscr();
  printf("\n< Priority : NonPreemptive > ");
  printf("\nC: Create \nD: Display \nE: Execute \nX: exit");
  printf("\nEnter your choice : ");  choice=getche();
  return toupper(choice);
}

void main()
{
  while(1)
  { switch(menu())
    { case 'X' : erase(); exit(0);  // release memory allocated to JOBQ
      case 'C' : CreateJOBQ();  break;
      case 'D' : DisplayJOBQ(); break;
      case 'E' : ExecuteJOBQ(); break;
      default  : printf("\7 Invalid Choice!");
    }
    printf("\npress any key to continue...");    getch();
  }
}

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ASMB2.C

#include<stdio.h>
#include<string.h>
#include<conio.h>
#include<stdlib.h>

struct symtab
{
    char name[10];
    int used,declared,addr,length,value;
}s[10];

int symtab_cnt;

int check_if_valid_opode(char tok[]);

int search_into_symtab(char *st)
{
    int i;
    for(i=0;i<symtab_cnt;i++)
        if(strcmpi(st,s[i].name)==0)
            return i;
    return -1;
}


char optab[][10]={"stop","add","sub","mult","mover","movem","comp","bc",
             "div","read","print"};
char regtab[][6]={"areg,","breg,","creg,","dreg,"};
char oprtab[][6]={"le,","lt,","eq,","gt,","ge,","any,"};
char adtab[][6]={"start","end","ltorg","origin"};

int optab_cnt=11;
int regtab_cnt=4;
int oprtab_cnt=6;
int adtab_cnt=4;



void display()
{
    int m;
    for(m=0;m<symtab_cnt;m++)
        if(s[m].used == 1 && s[m].declared == 0)
            printf("\nSymb %s used but not defined",s[m].name);
        else if(s[m].used == 0 && s[m].declared == 1)
            printf("\nSymb %s defined but not used",s[m].name);
    getch();
}



main(int argc,char *argv[])
{

    FILE *fp;
    char *w[4],*error[15];
    char str[80],line[80];
    int notok,i,p=0,k,j,cnt=0,h;
    clrscr();
    fp=fopen(argv[1],"r");
    if(fp==NULL)
    {
        printf("\nFile can not be opened...");
        getch();
        exit(1);
    }

    for(i=0;i<4;i++)           //Allocating memory to token variables
        w[i]=(char *)malloc(10);

    cnt=0;
    while(fgets(str,80,fp))
    {
        cnt++;
        printf("%d : %s",cnt,str);
        notok=sscanf(str,"%s%s%s%s",w[0],w[1],w[2],w[3]);
        switch(notok)
        {

            case 2:k=check_if_valid_opcode(w[0]);
                if(k==9||k==10)
                {
                    h=search_into_symtab(w[1]);
                    if(h==-1)
                    {
                        strcpy(s[symtab_cnt].name,w[1]);
                        s[symtab_cnt++].used=1;
                    }
                    continue;
                }
                break;

            case 3:k=check_if_valid_opcode(w[0]);.

                if(k>=1&&k<=8)
                {
                    h=search_into_symtab(w[2]);
                    if(h==-1)
                    {
                        strcpy(s[symtab_cnt].name,w[2]);
                        s[symtab_cnt++].used=1;
                    }
                    continue;
                }

                if(strcmpi(w[1],"ds")==0 || strcmpi(w[1],"dc")==0)
                {
                    h=search_into_symtab(w[0]);
                    if(h==-1)
                    {
                        strcpy(s[symtab_cnt].name,w[0]);
                        s[symtab_cnt++].declared=1;
                    }
                    else
                    {
                        if(s[h].declared==0)
                        {
                            s[h].declared=1;
                        }
                        else
                            printf("Multiple declaration");
                    }
                }
                break;

            case 4:h=search_into_symtab(w[0]);
                if(h==-1)
                {
                    strcpy(s[symtab_cnt].name,w[0]);
                    s[symtab_cnt++].declared=1;
                }
                break;
        }

    }
    display();

    getch();

}


int check_if_valid_opcode(char tok[])
{
    int i;
    for(i=0;i<optab_cnt;i++)
        if((strcmpi(tok,optab[i]))==0)
            return i;
    return -1;
}

C++ Programs Questions and answers Solve it

Solve C++ programs 
Questions and answers:
 

1)Which C++ header file(s) will be essentially required to be included to run /execute the following C++ code:
 void main()
{
     char Msg[ ]="Sunset Gardens";
     for (int I=5;I<strlen(Msg);I++)
     puts(Msg);
 }

Ans : stdio.h, string.h

 

2) Name the header files that shall be need for the following code:
  void main() 
            {
                char text[] =”Something”
               cout<<”Remaining SMS chars: ”<<160-strlen(text)<<endl; 
            }
 
 Ans: iostream.h/iomanip.h , string.h


3) Rewrite the following program after removing the syntactical error(s) if any.

 #include<iostream.h>
Class Item
{ long IId, Qty; public: void Purchase
{ cin>>IId>>Qty;}
 void Sale()
{ cout<<setw(5)<<IId<<”Old:”<< Qty<<endl; cout<< “New :”<<Qty<<endl; }
};
void main()
 { Item I; Purchase(); I.Sale() }

Ans : #include<iostream.h>
class Item // C capital
 { long IId, Qty; public: void Purchase ( ) { cin>>IId>>Qty;}
 // ( ) after function name void Sale( )
{
cout<<setw(5)<<IId<<”Old:”<< Qty<<endl; cout<< “New :”<<Qty<<endl;
}
};
 void main()
{ Item I; I. Purchase(); // object missing I.Sale() ; // ; is missing }


4) Find the output of the following program:
 #include<iostream.h>
#include<ctype.h>
typedef char Str80[80];
 void main()
 {char *Notes;
 Str80 str= “ vR2GooD”;
int L=6;
Notes =Str;
 while(L>=3)
{ Str[L]=(isupper(Str[L])? tolower(Str[L]) : toupper(Str[L]));
cout<<Notes<<endl;
L--;
 Notes++;
}
}

Either the statement is removed or header file included as #include<iomanip.h>

Ans : vR2Good R2GoOd 2GOOd gOOd


5) Observe the following program and find out, which output(s) out id (i) to (iv) will not be expected from program?
 What will be the minimum and maximum value assigned to the variables Chance?
 #include<iostream.h>
 #include<stdlib.h>
void main()
{ randomize();
 int Arr[] = {9,6};, N;
int Chance = random(2)+10;
 for(int c=0;c<2;c++)
 { N= random(2);
 cout<<Arr[N];
 }
}
i) 9#6# ii) 19#17# iii) 19#16# iv) 20#16#

Ans: The output not expected from program are (i),(ii) and (iv) Minimum value of Chance =10
Maximum value of Chance = 11

6) Find the output of the following program:
#include<iostream.h>
class METRO
{ int Mno, TripNo, PassengerCount;
public: METRO(int Tmno=1) { Mno =Tmno; PassengerCount=0;} void Trip(int PC=20) { TripNo++, PassengerCount+=PC};
void StatusShow() { cout<<Mno<< “:”<<TripNo<< “ :”<<PassengerCount<<endl;} };
 void main() { METRO M(5), T; M.Trip(); M.StatusShow(); T.StatusShow(); M.StatusShow(); }
Ans : 5: 1: 20 1: 1: 50 5: 2: 50



7) Find the output for the following program:
 #include<iostream.h>
 #include<ctype.h>
void Encript ( char T[ ])
 { for( int i=0 ; T[i] != ‘ \0’ ; i += 2)
if( T[i] = = ‘A’ || T[i] = = ‘E’ ) T[i] = ‘#’ ;
 else if (islower (T[i] )) T[i] = toupper(T[i]);
 else T[i] = ‘@’;
}
void main()
{ char text [ ] = “SaVE EArTh in 2012”;
encrypt(text);
 cout<<text<<endl;
 }

8) Find the output of the following program:
 #include<iostream.h>
void main( )
{
 int U=10,V=20;
 for(int I=1;I<=2;I++)
{
cout<<”[1]”<<U++<<”&”<<V  5 <<endl;
 cout<<”[2]”<<++V<<”&”<<U + 2 <<endl;
}
 }


9)In the following program, find the correct possible output(s)from the options:
#include<stdlib.h>
 #include<iostream.h>
void main( )
 {
randomize( );
 char City[ ][10]={“DEL”, “CHN”, “KOL”, “BOM”, “BNG”};
int Fly;
for(int I=0; I<3;I++)
{
Fly=random(2) + 1;
cout<<City[Fly]<< “:”;
 }
}
 Outputs: (i) DEL : CHN : KOL: (ii) CHN: KOL : CHN: (iii) KOL : BOM : BNG: (iv) KOL : CHN : KOL:

10)In the following C++ program what is the expected value of Myscore from options (i) to (iv) given below.
 Justify your answer.
#include<stdlib.h>
#include<iostream.h>
 void main( )
{
 randomize( );
 int Score[ ] = {25,20,34,56,72,63},Myscore;
 cout<<Myscore<<endl;
 }

 i) 25 (ii) 34 (iii) 20 (iv) Garbage Value.

Function overloading in C++
A function name having several definitions that are differentiable by the number or types of their arguments is known as function overloading.
Example :

 A same function print() is being used to print different data types:
#include <iostream.h>
class printData { public: void print(int i) { cout << "Printing int: " << i << endl; }
void print(double  f) { cout << "Printing float: " << f << endl; }
void print(char* c) { cout << "Printing character: " << c << endl; }
};
int main(void) { printData pd;
// Call print to print integer pd.print(5);
 // Call print to print float pd.print(500.263);
 // Call print to print character pd.print("Hello C++");
return 0;
}
When the above code is compiled and executed, it produces following result:
Printing int: 5 Printing float: 500.263 Printing character: Hello C++

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College form

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Form of College

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Enter Details FOR JOB

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PHP program of Slip 19 practical of User Information in bill

<?
// This is Slip19_a.php
// It takes User Information And Send it to slip19_b.php
?>
<html>
<body bgcolor=#AAAAAA>
<center> <font color="red" font size=6>U</font>ser Information</center>
<br><br><br>
<form action="slip19_b.php" method="GET">
<p>
<table border=0>
<tr><td>Name:<td> <input type="text" name="u_name">

<tr><td>Address:<td> <input type="text" name="u_address">

<tr><td>Phone No:<td> <input type="text" name="u_no">
<tr><td><td><input type="submit" value="SUBMIT">
</form>
</table>
</form>
</body>
</html>





















<?
// This is Slip19_b.php
// It Takes The Product Information....
//and Passes to Slip19_c.php

?>
<html>
<body bgcolor=#AAAAAAA>
<center><font color="red" font size=6>P</font>roduct Information</center>
<br><br><br>
<form action="slip19_c.php" method="POST">
<p>
<table border=0>
<tr><td>Prod.Name:<td> <input type="text" name="p_name">

<tr><td>Quantity:<td> <input type="text" name="p_qty">

<tr><td>Rate:<td> <input type="text" name="prize">

<tr><td><?
$var1=$_GET['u_name'];
$var2=$_GET['u_address'];
$var3=$_GET['u_no'];
printf("<br><br>Customer Information is:");
printf("<br><br><input type='text' name='user' value='$var1'><br>");
printf("<input type='text' name='add' value='$var2'><br>");
printf("<input type='text' name='phone' value='$var3'><br>");
?>

<tr><td><td><input type="submit" value="Display">

</table>
</form>
</body>
</html>
















<?
// This is Slip19_c.php
//It Genrates the Tabular format bill of the Customer
?>

<html>
<body bgcolor=#AAAAAAA>
<center><font color="red" font size=6>B</font>ill Information</center>
<br><br><br>
<center>
<table border=1 cellspacing=2 cellpadding=2>
<?
$usr_name=$_POST['user'];
$usr_address=$_POST['add'];
$usr_phone=$_POST['phone'];
$prod_name=$_POST['p_name'];
$prod_quantity=$_POST['p_qty'];
$prod_price=$_POST['prize'];
$bill=$prod_quantity*$prod_price;
printf("<br><br><br>");
printf("<th bgcolor=#ABCDEF>Customer Bill<td bgcolor=#ABCDEF>");
printf("<tr><td>Name<td>$usr_name");
printf("<tr><td>Address<td>$usr_address");
printf("<tr><td>Phone No<td>$usr_phone");
printf("<tr><td>Product Name<td>$prod_name");
printf("<tr><td>Product Quantity<td>$prod_quantity");
printf("<tr><td>Product Price<td>$prod_price / Unit");
printf("<tr><td>Total<td>$bill Rs.");
?>
</table>
</form>
</body>
</html>

Conversion of string program java

Conversion of string java program UPPER lower BOLD ITALIC 

import javax.swing .* ;
import java.awt . * ;
import java.awt.event .* ;

class Convt extends JFrame implements ActionListener
{
       JLabel l1;
       JButton b1,b2,b3,b4;
       JTextField t1,t2,t3,t4,t5;
     Convt( )
  {
         setVisible(true);
         setSize(400,400);
         setTitle("Conversion of Text");

         l1=new JLabel("ENTER   THE   STRING :=");
         b1=new JButton("LOWER");
         b2=new JButton("UPPER");
         b3=new JButton("BOLD");
         b4=new JButton("ITALIC");
         b1.addActionListener(this);
         b2.addActionListener(this);
         b3.addActionListener(this);
         b4.addActionListener(this);

         t1=new JTextField(10);
         t2=new JTextField(10);
         t3=new JTextField(10);
         t4=new JTextField(10);
         t5=new JTextField(10);

     JPanel p1=new JPanel( );
      p1.setLayout(new GridLayout(5,2));
      p1.add(l1);
      p1.add(t1);
      p1.add(b1);
      p1.add(t2);
      p1.add(b2);
      p1.add(t3);
      p1.add(b3);
      p1.add(t4);
      p1.add(b4);
      p1.add(t5);

     Container cn=getContentPane( );
     cn.setLayout(new FlowLayout( ));
     cn.add(p1);
   }
    public void actionPerformed(ActionEvent ae)
    {
             String s=t1.getText( );
            if(ae.getSource( )==b1)
        {
             t2.setText(s.toLowerCase( ));
           }
       else
           if(ae.getSource( )==b2)
           {
                 t3.setText(s.toUpperCase( ));
                }
       else
          if(ae.getSource( )==b3)
          {
                 t4.setFont(new Font("Dialog",Font.BOLD,20));
                 t4.setText(s);
               }

        else
             if(ae.getSource( )==b4)
            {
                   t5.setFont(new Font("Dialog",Font.ITALIC,20));
                   t5.setText(s);
                 }
        }
      public static void main(String args[])
      {
             Convt cv=new Convt( );
             cv.validate( );
           }
    }

Write a command line program for line editor of practcal slip

/*
    slip-7
1. Write a command line program for line editor. The file to be edited is taken
 as command line argument, an empty  file is opened for editing if no argument
 is supplied.  It should display '$' prompt to accept the line editing commands.
Implement the following commands.
i)    a                 - to append
ii)    i  n                 - to insert after nth line
iii)    c    n1   n2              - to copy line n1 at  n2 position
iv)    c    n1, n2    n3   - to copy range of line at n3        [30 marks]

*/




#include<stdio.h>
#include<conio.h>
#include<string.h>
#define newnode (struct node *)malloc(sizeof(struct node))

struct node
{
    char line[80];
    struct node *next,*prev;
}*head=NULL;

int  length();
void print();
void append(char *line);
void load(char *);
void add();
void insert(char *,int);
void copy(int,int,int,int);





void load(char *fname)
{
    FILE *fp;
    char line[80];
    fp=fopen(fname,"r");
    if(fp==NULL)
    {
        printf("\nFile cannot boe opened..");
        getch();
        exit(0);
    }
    while(!feof(fp))
    {
        fgets(line,80,fp);
        append(line);
    }

}

void append(char *line)
{
    struct node *f,*s;
    f=newnode;
    strcpy(f->line,line);
    f->next=NULL;
    f->prev=NULL;
    if(head==NULL)
        head=f;
    else
    {
        for(s=head;s->next!=NULL;s=s->next);
        s->next=f;
        f->prev=s;
        s=s->next;
    }
}

void print()
{
    struct node *f;
    int i=1;
    for(f=head;f!=NULL;f=f->next)
        printf("%d %s",i++,f->line);
}

int length()
{
    int cnt=0;
    struct node *s;
    for(s=head;s!=NULL;s=s->next)
        cnt++;
    return cnt;
}

void insert(char *line,int from)
{
    struct node *f,*s;
    int len,i;
    if(from-1>length())
    {
        printf("\nError..");
        return;
    }

    f=newnode;
    f->next=NULL;
    f->prev=NULL;
    strcpy(f->line,line);

    if(from==1)
    {
        f->next=head;
        head->prev=f;
        head=head->prev;
        return;
    }
    else
    {
        s=head;
        for(i=1;i<(from-1);i++)
            s=s->next;
        f->next=s->next;
        s->next->prev=f;
        s->next=f;
        f->prev=s;
    }
}






main(int argc,char *argv[])
{
    int ch;
    int i,notok,from,to,range;
    char cmd[80],tok[4],line[80];
    clrscr();
    for(i=0;i<4;i++)
        tok[i]=(char )malloc(20);

    if(argc==2)
    {
        head=NULL;
        load(argv[1]);
    }
    while(1)
    {
        printf("\n?");
        flushall();
        gets(cmd);
        notok=sscanf(cmd,"%s%s%s%s",tok[0],tok[1],tok[2],tok[3]);
        switch(cmd[0])
        {
            case 'q':exit(1);

            case 'p':print();
                   break;
            case 'i':from=atoi(tok[1]);
                    printf("Input Line:");
                    flushall();
                    fgets(line,80,stdin);
                    insert(line,from);
                    break;

            case 'c':from=atoi(tok[1]);
                   to=atoi(tok[2]);
                   range=atoi(tok[3]);
                   copy(notok,from,to,range);
                   break;



        }
    }
}

void copy(int notok,int from,int to,int range)
{
    int len,i;
    struct node *s,*f,*temp;
    char line[80];
    len=length();
    if(from>len)
    {
        printf("\nError...");
        return;
    }

    if(notok==3)
    {
        s=head;
        for(i=0;i<(from-1);i++)
            s=s->next;
        insert(s->line,to);
    }

    if(notok==4)
    {
        if(to>len || range-1>len)
        {
            printf("\nError,,,");
            return;
        }
        s=head;
        for(i=0;i<(from-1);i++)
            s=s->next;
        for(i=from;i<=to && s!=NULL;i++)
        {
            insert(s->line,range++);
            s=s->next;
        }
    }
}

RoundRobin Program of practical slip

 /* Implementation of RoundRobin method for CPU scheduling using Circular LL */
 


#include<stdlib.h>
#define newnode (struct node *)malloc(sizeof(struct node))
struct node
{ int jobno,time;
  struct node *next;
}*JOBQ=NULL;


void erase()
{ struct node *S,*T;
  if(JOBQ!=NULL)
  { for(T=JOBQ;T->next!=JOBQ; T=T->next) ;  // go to last node
    T->next=NULL;

    while(JOBQ!=NULL)  { S=JOBQ;  JOBQ=JOBQ->next;  free(S);  }
  }
}

void CreateJOBQ()
{ struct node *S;   int i,n;
  printf("\nHow many jobs?"); scanf("%d",&n);


  if(JOBQ!=NULL) erase(JOBQ);
  JOBQ=newnode;
  printf("\nEnter JOB No. & execution time in sec. "); scanf("%d %d",&JOBQ->jobno,&JOBQ->time);

  S=JOBQ;
  for(i=2;i<=n;i++)
  { S->next=newnode;
    S=S->next;
    printf("\nEnter JOB No. & execution time in sec. "); scanf("%d %d",&S->jobno,&S->time);
  }
  S->next=JOBQ;
}

void DisplayJOBQ()
{struct node *T=JOBQ;
  if(JOBQ==NULL) printf("\nJOB Q is Empty!");
  else
  { printf("\nJOB  Rem. Time");
    do{ printf("\n J%d : %d ",T->jobno,T->time);
    T=T->next;
      }while(T!=JOBQ);
  }
}

void ExecuteJOBQ()
{ struct node *T=JOBQ,*S=NULL,*t1;  int flag,timeslot=1,time=0; char choice;
  printf("\nEnter Time Slot (in sec.) "); scanf("%d",&timeslot);
  printf("\nclock CPU   JOBQ ");
  while(JOBQ!=NULL)
  {
    do{ flag=0;
    if(T->time<=timeslot)  // job remaining time < time slot
    { printf("\n%3d  J%d(%d) |--",time,T->jobno,T->time); // cpu allocated
      for(t1=T->next; t1!=T; t1=t1->next) printf(" J%d(%d) ",t1->jobno,t1->time);  // print JOBQ
      time+=T->time; // increment clock
      T->time=0; // job finished, delete from JOBQ
      if(T==JOBQ) // 1st job
      { if(JOBQ->next==JOBQ) // only 1 job in Q
        { JOBQ=NULL; free(T); T=NULL; } // JOBQ became empty
        else  // more jobs in Q
        {  for(t1=JOBQ;t1->next!=JOBQ; t1=t1->next) ;  // go to last node
           JOBQ=JOBQ->next;   t1->next=JOBQ;
           free(T);     T=JOBQ; flag=1;
        }
      }
      else // job other than 1st position
      {   S->next=T->next; free(T);  T=S->next;
      }

    }
    else  // job is executed for given time slot
    {  printf("\n%3d  J%d(%d) |--",time,T->jobno,T->time); // cpu allocated
       for(t1=T->next; t1!=T; t1=t1->next) printf(" J%d(%d) ",t1->jobno,t1->time); // print JOBQ

       T->time-=timeslot;
       time+=timeslot;
       S=T; T=T->next;  // go to next job
    }

    printf(" --|  Add New Job(y/n)? ");  choice=getche();
    if(choice=='y' || choice=='Y')
    { struct node *S1;
      if(JOBQ==NULL) // JOBQ empty
      { JOBQ=newnode; JOBQ->next=JOBQ;    // 1st job in Q
        printf("\nEnter JOB No. & execution time in sec. "); scanf("%d %d",&JOBQ->jobno,&JOBQ->time);
        T=JOBQ; flag=1;
      }
      else
      { for(S1=S;S1->next!=S; S1=S1->next) ;  // go to last node of JOBQ
        S1->next=newnode; S1=S1->next; S1->next=S;    // add new job at the end of JOBQ
        printf("\nEnter JOB No. & execution time in sec. "); scanf("%d %d",&S1->jobno,&S1->time);
      }
    }
      }while(T!=JOBQ || flag);
  }
  printf("\n%3d  -idle-",time);
}

char menu()
{ char choice='a';
  clrscr();
  printf("----- Round Robin ------");
  printf("\nC: Create New JOB Queue");
  printf("\nD: Display JOB Queue");
  printf("\nE: Execute JOB Queue");
  printf("\nX: exit");
  printf("\nEnter your choice : ");
  choice=getche();
  return toupper(choice);
}

void main()
{
  while(1)
  { switch(menu())
    { case 'X' : erase(); exit(0);  // release memory allocated to JOBQ
      case 'C' : CreateJOBQ();  break;
      case 'D' : DisplayJOBQ(); break;
      case 'E' : ExecuteJOBQ(); break;
      default  : printf("\7 Invalid Choice!");
    }
    printf("\npress any key to continue...");
    getch();
  }

}

Earning Money

To earn money is not so easy for everyone it depends on factors
Time
Age
Smartness
Luck
Legacy 

If you want to earn money make proper investments at proper time and at proper place.

Money word says itself:

M - Move

O - Overcome

N - Noble

E -  Energy

Y - Your true value

Earning money should have mind to solve problem which can be generated by creativity to produce a product which is problem solving product.

One will give you money only when he wants his work to be done for that money. One should be experienced in the work which he or she is doing to earn money with doing his marketing.

Money can be earned on :

Seconds basis,

Minutes basis,

Hourly basis,

Daily basis,

Weekly basis,

Monthly basis,

Quarterly basis,

Yearly basis

 It is done by

Manufactured basis,

Selling basis ,

Service basis,

Job basis,

Physical basis,

Mental basis,

Consulting basis,

Mode basis,

Salary basis,

Share market basis,

Interest basis,

Commission  basis

Make study and expert in the field by Blue Ocean Strategy so that you can make money easily on the market. think abut the future what are important and give priority to them accordingly.

Adopt technology and should be easily move with generation otherwise you will stay behind.

Make a statistical study in each of the cases.

Education

Education

Education decides the country growth by the ratio of people educated in a country. Education is the birth right of every child in this world. It gives knowledge with professional courses. Education is not only taught in school but wherever we get to know new concepts, techniques and strategies with various tools. Education should not be put in the category of business. We should spread our education to others. As we educate others we get perfection in it. We get to learn new things again in it. For a healthy and safe lifestyle education plays a important role in it.
It is the innovation of a thing.
Creativity is achieved by education.
Prepare yourself to guide.
Education words itself describes it:

E - Excellence
D - Development
U - Universal
C - Concepts
T - Techniques
I - Information
O - Ocean
N - Notes

If there is educated children then they can solve their own problems by themselves.

DOCUMENTATION - POLICY CONDITION

DOCUMENTATION - POLICY CONDITION

Question 1

Which of the below statement is false with regards to nomination?

I. Policy nomination is not cancelled if the policy is assigned to the insurer in

return for a loan

II. Nomination can be done at the time of policy purchase or subsequently

III. Nomination can be changed by making an endorsement in the policy

IV. A nominee has full rights on the whole of the claim

Question 2

In order for the policy to acquire a guaranteed surrender value, for how long

must the premiums be paid as per law?

I. Premiums must be paid for at least 2 consecutive years

II. Premiums must be paid for at least 3 consecutive years

III. Premiums must be paid for at least 4 consecutive years

IV. Premiums must be paid for at least 5 consecutive years

Question 3

When is a policy deemed to be lapsed?

I. If the premiums are not paid on due date

II. If the premiums are not paid before the due date

III. If the premium has not been paid even during days of grace

IV. If the policy is surrendered

Question 4

Which of the below statement is correct with regards to grace period of an

insurance policy?

I. The standard length of the grace period is one month.

II. The standard length of the grace period is 30 days.

III. The standard length of the grace period is one month or 30 days.

IV. The standard length of the grace period is one month or 31 days.

Question 5

What will happen if the policyholder does not pay the premium by the due date

and dies during the grace period?

I. The insurer will consider the policy void due to non-payment of premium by

the due date and hence reject the claim

II. The insurer will pay the claim and waive off the last unpaid premium

III. The insurer will pay the claim after deducting the unpaid premium

IV. The insurer will pay the claim after deducting the unpaid premium along

with interest which will be taken as 2% above the bank savings interest rate

Question 6

During the revival of a lapsed policy, which of the below aspect is considered

most significant by the insurance company? Choose the most appropriate option.

I. Evidence of insurability at revival

II. Revival of the policy leading to increase in risk for the insurance company

III. Payment of unpaid premiums with interest

IV. Insured submitting the revival application within a specified time frame

Question 7

For an insurance policy nomination is allowed under _________ of the Insurance

Act, 1938.

I. Section 10

II. Section 38

III. Section 39

IV. Section 45

Question 8

Which of the below statement is incorrect with regards to a policy against which

a loan has been taken from the insurance company?

I. The policy will have to be assigned in favour of the insurance company

II. The nomination of such policy will get cancelled due to assignment of the

policy in favour of the insurance company

III. The nominee’s right will affected to the extent of the insurer’s interest in

the policy

IV. The policy loan is usually limited to a percentage of the policy’s surrender

value

Question 9

Which of the below statement is incorrect with regards to assignment of an

insurance policy?

I. In case of Absolute Assignment, in the event of death of the assignee, the

title of the policy would pass to the estate of the deceased assignee.

II. The assignment of a life insurance policy implies the act of transferring the

rights right, title and interest in the policy (as property) from one person to

another.

III. It is necessary that the policyholder must give notice of assignment to the

insurer.

IV. In case of Absolute Assignment, the policy vests absolutely with the assignee

till maturity, except in case of death of the insured during the policy

tenure, wherein the policy reverts back to the beneficiaries of the insured.

Question 10

Which of the below alteration will be permitted by an insurance company?

I. Splitting up of the policy into two or more policies

II. Extension of the premium paying term

III. Change of the policy from with profit policy to without profit policy

IV. Increase in the sum assured

Question 11

Under what circumstances would the policyholder need to appoint an

appointee?

I. Insured is minor

II. Nominee is a minor

III. Policyholder is not of sound mind

IV. Policyholder is not married

Solved DOCUMENTATION – POLICY STAGE

DOCUMENTATION – POLICY STAGE

Question 1

Which of the following documents is an evidence of the contract between

insurer and insured?

I. Proposal form

II. Policy document

III. Prospectus

IV. Claim form

Question 2

If complex language is used to word a certain policy document and it has given

rise to an ambiguity, how will it generally be construed?

I. In favour of insured

II. In favour of insurer

III. The policy will be declared as void and the insurer will be asked to return

the premium with interest to the insured

IV. The policy will be declared as void and the insurer will be asked to return

the premium to the insured without any interest

Question 3

Select the option that best describes a policy document.

I. It is evidence of the insurance contract

II. It is evidence of the interest expressed by the insured in buying an insurance

policy from the company

III. It is evidence of the policy (procedures) followed by an insurance company

when dealing with channel partners like banks, brokers and other entities

IV. It is an acknowledgement slip issued by the insurance company on payment

of the first premium

Question 4

Which of the below statement is correct?

I. The proposal form acceptance is the evidence that the policy contract has

begun

II. The acceptance of premium is evidence that the policy has begun

III. The First Premium Receipt is the evidence that the policy contract has

begun

IV. The premium quote is evidence that the policy contract has begun

Question 5

For the subsequent premiums received by the insurance company after the first

premium, the company will issue __________.

I. Revival premium receipt

II. Restoration premium receipt

III. Reinstatement premium receipt

IV. Renewal premium receipt

Question 6

What will happen if the insured person loses the original life insurance policy

document?

I. The insurance company will issue a duplicate policy without making any

changes to the contract

II. The insurance contract will come to an end

III. The insurance company will issue a duplicate policy with renewed terms and

conditions based on the current health declarations of the life insured

IV. The insurance company will issue a duplicate policy without making any

changes to the contract, but only after a Court order.

Question 7

Which of the below statement is correct?

I. The policy document has to be signed by a competent authority but need

not be compulsorily stamped according to the Indian Stamp Act.

II. The policy document has to be signed by a competent authority and should

be stamped according to the Indian Stamp Act.

III. The policy document need not be signed by a competent authority but

should be stamped according to the Indian Stamp Act.

IV. The policy document neither needs to be signed by a competent authority

nor it needs to be compulsorily stamped according to the Indian Stamp Act.

Question 8

Which of the below forms the first part of a standard insurance policy

document?

I. Policy schedule

II. Standard provisions

III. Specific policy provisions

IV. Claim procedure

Question 9

In a standard insurance policy document, the standard provisions section will

have information on which of the below?

I. Date of commencement, date of maturity and due date of last premium

II. Name of nominee

III. The rights and privileges and other conditions, which are applicable under

the contract

IV. The signature of the authorized signatory and policy stamp

Question 10

“A clause precluding death due to pregnancy for a lady who is expecting at the

time of writing the contract” will be included in which section of a standard

policy document?

I. Policy schedule

II. General provisions

III. Standard provisions

IV. Specific policy provisions

Question 11

What does a first premium receipt (FPR) signify? Choose the most appropriate

option.

I. Free look period has ended

II. It is evidence that the policy contract has begun

III. Policy cannot be cancelled now

IV. Policy has acquired a certain cash value