OPERATING SYSTEMS QUESTION BANK FOR 2nd MID

Banker's Algorithm

#include 
struct file  { 
  int all[10]; 
  int max[10]; 
  int need[10]; 
  int flag; 
  }; 
  void main() 
  {
    struct file f[10];
    int fl;
    int i, j, k, p, b, n, r, g, cnt=0, id, newr; 
    int avail[10],seq[10]; 
    printf("Enter number of processes -- "); 
    scanf("%d",&n);
    printf("Enter number of resources -- "); 
    scanf("%d",&r);  for(i=0;i    {
      printf("Enter details for P%d",i);
      printf("\nEnter allocation\t -- \t");
      for(j=0;j      scanf("%d",&f[i].all[j]);
      printf("Enter Max\t\t -- \t"); 
      for(j=0;j      scanf("%d",&f[i].max[j]);
      f[i].flag=0; 
     
    } 

    printf("\nEnter Available Resources\t -- \t"); 
    for(i=0;i    scanf("%d",&avail[i]);
    printf("\nEnter New Request Details -- "); 
    printf("\nEnter pid \t -- \t"); 
    scanf("%d",&id); 
    printf("Enter Request for Resources \t -- \t"); 
    for(i=0;i    { 
      scanf("%d",&newr); 
      f[id].all[i] += newr;
      avail[i]=avail[i] - newr; 
          } 
    for(i=0;i    { 
      for(j=0;j      { 
       f[i].need[j] = f[i].max[j] - f[i].all[j];
       if (f[i].need[j] < 0)
       f[i].need[j] = 0;
      }
   }

cnt = 0;
fl = 0;
while (cnt != n)
  {
    g = 0;
    for (j = 0; j < n; j++)
      {
 if (f[j].flag == 0)
   {
     b = 0;
     for (p = 0; p < r; p++)
       {
  if (avail[p] >= f[j].need[p])
    b = b + 1;
  else
    b = b - 1;
       }
     if (b == r)
       {
  printf ("\nP%d is visited", j);
  seq[fl++] = j;
  f[j].flag = 1;
  for (k = 0; k < r; k++)
  avail[k] = avail[k] + f[j].all[k];
  cnt = cnt + 1;
  printf ("(");
  for (k = 0; k < r; k++)
    printf ("%3d", avail[k]);
  printf (")");
  g = 1;
       }
   }
      }
    if (g == 0)
      {
 printf ("\n REQUEST NOT GRANTED -- DEADLOCK OCCURRED");
 printf ("\n SYSTEM IS IN UNSAFE STATE");
 goto y;
      }
  }

printf ("\nSYSTEM IS IN SAFE STATE");
printf ("\nThe Safe Sequence is -- (");
for (i = 0; i < fl; i++)
  printf ("P%d ", seq[i]);
printf (")");
y:printf ("\nProcess\t\tAllocation\t\tMax\t\t\tNeed\n");
for (i = 0; i < n; i++)
  {
    printf ("P%d\t", i);
    for (j = 0; j < r; j++)
      printf ("%6d", f[i].all[j]);
    for (j = 0; j < r; j++)
     
printf ("%6d", f[i].max[j]);
    for (j = 0; j < r; j++)
      printf ("%6d", f[i].need[j]);
    printf ("\n");
  }
}

INPUT 

Enter number of processes – 5

Enter number of resources -- 3 

Enter details for P0 

Enter allocation -- 0 1 0 

Enter Max -- 7 5 3

Enter details for P1 

Enter allocation -- 2 0 0 

Enter Max -- 3 2 2 

Enter details for P2

Enter allocation -- 3 0 2 

Enter Max -- 9 0 2 

Enter details for P3 

Enter allocation -- 2 1 1 

Enter Max -- 2 2 2

Enter details for P4 

Enter allocation -- 0 0 2 

Enter Max -- 4 3 3 

Enter Available Resources -- 3 3 2 

Enter New Request Details -- Enter pid -- 1 

Enter Request for Resources -- 1 0 2 

OUTPUT

P1 is visited( 5 3 2) 

P3 is visited( 7 4 3) 

P4 is visited( 7 4 5) 

P0 is visited( 7 5 5) 

P2 is visited( 10 5 7)

 

SYSTEM IS IN SAFE STATE

The Safe Sequence is -- (P1 P3 P4 P0 P2 ) 

Process       Allocation         Max            Need 

P0              0 1 0                  7 5 3           7 4 3 

P1              3 0 2                  3 2 2           0 2 0 

P2              3 0 2                  9 0 2           6 0 0

P3              2 1 1                  2 2 2           0 1 1

P4              0 0 2                  4 3 3           4 31

UNIT-III PPT

Memory Management

Virtual Memory

OS WEEKLY TEST QUESTIONS (12-3-18)

DESCRIPTIVE QUESTIONS

1. A) Define page fault. When does a page fault occur?
    B) Describe the action taken by OS when page fault occurs with diagram.

2. A)Consider the following page reference string

        7,0,2,0,3,0,4,2,3,0,3,2,1,2,0,1,7,0   Assuming three frames, how
     many page faults would occur in each of the following cases:

         i) LRU                ii) FIFO

    B) State and Explain Belady’s anomaly with an example.


Fill in the blanks

1. Virtual memory is
(a) Large secondary memory (b) Large main memory
(c) Illusion of large main memory (d) None of the above

2 . Optimal page – replacement algorithm is difficult to implement, because :
a) it requires a lot of information b) it requires future knowledge of the reference string
c) it is too complex d) it is extremely expensive

3. The two methods how LRU page replacement policy can be implemented in hardware are COUNTER,REFERENCE BIT.

4. LRU stands for__________________________________________.

5. OPTIMAL PAGE REPLACEMENT yields the lowest of all possible page-faults,