Operating Systems Notes

OS Complete level Notes Pdf -- Click here 

a) Worst-fit b) Best-fit c) First-fit

Write a C program to simulate the following contiguous memory allocation techniques

a) Worst-fit b) Best-fit c) First-fit

DESCRIPTION

One of the simplest methods for memory allocation is to divide memory into several fixed-sized partitions. Each partition may contain exactly one process. In this multiple-partition method, when a partition is free, a process is selected from the input queue and is loaded into the free partition. When the process terminates, the partition becomes available for another process. The operating system keeps a table indicating which parts of memory are available and which are occupied. Finally, when a process arrives and needs memory, a memory section large enough for this process is provided. When it is time to load or swap a process into main memory, and if there is more than one free block of memory of sufficient size, then the operating system must decide which free block to allocate. Best-fit strategy chooses the block that is closest in size to the request. First-fit chooses the first available block that is large enough. Worst-fit chooses the largest available block.

PROGRAM

a)WORST-FIT

#include<stdio.h>

#include<conio.h>

#define max 25

void main()

{

int frag[max],b[max],f[max],i,j,nb,nf,temp;

static int bf[max],ff[max];

clrscr();

printf("\n\tMemory Management Scheme - First Fit");

printf("\nEnter the number of blocks:");

scanf("%d",&nb);

printf("Enter the number of files:");

scanf("%d",&nf);

printf("\nEnter the size of the blocks:-\n");

for(i=1;i<=nb;i++)

{

printf("Block %d:",i);

scanf("%d",&b[i]);

}

printf("Enter the size of the files :-\n");

for(i=1;i<=nf;i++)

{

printf("File %d:",i);

scanf("%d",&f[i]);

}

for(i=1;i<=nf;i++)

{

for(j=1;j<=nb;j++)

{

if(bf[j]!=1)

{

temp=b[j]-f[i];

if(temp>=0)

{

ff[i]=j;

break;

}

}

}

frag[i]=temp;

bf[ff[i]]=1;

}

printf("\nFile_no:\tFile_size :\tBlock_no:\tBlock_size:\tFragement");

for(i=1;i<=nf;i++)

printf("\n%d\t\t%d\t\t%d\t\t%d\t\t%d",i,f[i],ff[i],b[ff[i]],frag[i]);

getch();

}

INPUT

Enter the number of blocks: 3

Enter the number of files: 2

Enter the size of the blocks:-

Block 1: 5

Block 2: 2

Block 3: 7

Enter the size of the files:-

File 1: 1

File 2: 4

OUTPUT

File No          File Size        Block No             Block Size           Fragment

1                              1             1                              5                             4

2                              4             3                              7                             3

b) Best-fit

#include<stdio.h>

#include<conio.h>

#define max 25

void main()

{

int frag[max],b[max],f[max],i,j,nb,nf,temp,lowest=10000;

static int bf[max],ff[max];

clrscr();

printf("\nEnter the number of blocks:");

scanf("%d",&nb);

printf("Enter the number of files:");

scanf("%d",&nf);

printf("\nEnter the size of the blocks:-\n");

for(i=1;i<=nb;i++)

{

printf("Block %d:",i);

scanf("%d",&b[i]);

}

printf("Enter the size of the files :-\n");

for(i=1;i<=nf;i++)

{

printf("File %d:",i);

scanf("%d",&f[i]);

}

for(i=1;i<=nf;i++)

{

for(j=1;j<=nb;j++)

{

if(bf[j]!=1)

{

temp=b[j]-f[i];

if(temp>=0)

if(lowest>temp)

{

ff[i]=j;

lowest=temp;

}

}

}

frag[i]=lowest;

bf[ff[i]]=1;

lowest=10000;

}

printf("\nFile No\tFile Size \tBlock No\tBlock Size\tFragment");

for(i=1;i<=nf && ff[i]!=0;i++)

printf("\n%d\t\t%d\t\t%d\t\t%d\t\t%d",i,f[i],ff[i],b[ff[i]],frag[i]);

getch();

}

INPUT

Enter the number of blocks: 3

Enter the number of files: 2

Enter the size of the blocks:-

Block 1: 5

Block 2: 2

Block 3: 7

Enter the size of the files:-

File 1: 1

File 2: 4

OUTPUT

File No                  File Size               Block No             Block Size           Fragment

1                              1                             2                             2                             1

2                              4                              1                              5                             1

c) First-fit

#include<stdio.h>

#include<conio.h>

#define max 25

void main()

{

int frag[max],b[max],f[max],i,j,nb,nf,temp,highest=0;

static int bf[max],ff[max];

clrscr();

printf("\n\tMemory Management Scheme - Worst Fit");

printf("\nEnter the number of blocks:");

scanf("%d",&nb);

printf("Enter the number of files:");

scanf("%d",&nf);

printf("\nEnter the size of the blocks:-\n");

for(i=1;i<=nb;i++)

{

printf("Block %d:",i);

scanf("%d",&b[i]);

}

printf("Enter the size of the files :-\n");

for(i=1;i<=nf;i++)

{

printf("File %d:",i);

scanf("%d",&f[i]);

}

for(i=1;i<=nf;i++)

{

for(j=1;j<=nb;j++)

{

if(bf[j]!=1) //if bf[j] is not allocated

{

temp=b[j]-f[i];

if(temp>=0)

if(highest<temp)

{

ff[i]=j;

highest=temp;

}

}

}

frag[i]=highest;

bf[ff[i]]=1;

highest=0;

}

printf("\nFile_no:\tFile_size :\tBlock_no:\tBlock_size:\tFragement");

for(i=1;i<=nf;i++)

printf("\n%d\t\t%d\t\t%d\t\t%d\t\t%d",i,f[i],ff[i],b[ff[i]],frag[i]);

getch();

}

INPUT

Enter the number of blocks: 3

Enter the number of files: 2

Enter the size of the blocks:-

Block 1: 5

Block 2: 2

Block 3: 7

Enter the size of the files:-

File 1: 1

File 2: 4

OUTPUT

File No                  File Size               Block No             Block Size           Fragment

1                              1                             3                              7                              6

2                              4                             1                              5                              1

Producer-Consumer problem

Write a C program to simulate producer-consumer problem using semaphores.

Description:

Producer-consumer problem is a common paradigm for cooperating processes. A producer process produces information that is consumed by a consumer process. One solution to the producer-consumer problem uses shared memory.

            To allow producer and consumer processes to run concurrently, there must be available a buffer of items that can be filled by the producer and emptied by the consumer. This buffer will reside in a region of memory that is shared by the producer and consumer processes.

A producer can produce one item while the consumer is consuming another item. The producer and consumer must be synchronized, so that the consumer does not try to consume an item that has not yet been produced.

Program:

#include<stdio.h>

void main()

{

int buffer[10], bufsize, in, out, produce, consume, choice=0;

in = 0;

out = 0;

bufsize = 10;

while(choice !=3)

{

printf("\n 1. Produce \t 2. Consume \t3. Exit");

printf("\n Enter your choice: ");

scanf("%d", &choice);

switch(choice) {

case 1: if((in+1)%bufsize==out)

printf("\n Buffer is Full");

else

{

printf("\nEnter the value: ");

scanf("%d", &produce);

buffer[in] = produce;

in = (in+1)%bufsize;

}

break;

case 2: if(in == out)

printf("\nBuffer is Empty");

else

{

consume = buffer[out];

printf("\nThe consumed value is %d", consume);

out = (out+1)%bufsize;

}

break;

} } }

INPUT AND OUTPUT:

1. Produce 2. Consume 3. Exit

Enter your choice: 2

Buffer is Empty

1. Produce 2. Consume 3. Exit

Enter your choice: 1

Enter the value: 100

1. Produce 2. Consume 3. Exit

Enter your choice: 2

The consumed value is 100

1. Produce 2. Consume 3. Exit

Enter your choice: 3

2^nd one:

1. Produce 2. Consume 3. Exit

Enter your choice: 1

Enter the value: 100

1. Produce 2. Consume 3. Exit

Enter your choice: 1

Enter the value: 300

Enter your choice: 2

The consumed value is 100

1. Produce 2. Consume 3. Exit

Enter your choice: 2

The consumed value is 300

Enter your choice: 3

Deadlock Management Techniques

Deadlock   Management  Techniques

Write a C program to simulate Bankers algorithm for the purpose of  deadlock avoidance.

DESCRIPTION

In a multiprogramming environment, several processes may compete for a finite number of resources.

A process requests resources; if the resources are not available at that time, the process enters a waiting state. Sometimes, a waiting process is never again able to change state, because the resources it has requested are held by other waiting processes. This situation is called a deadlock.

                Deadlock avoidance is one of the techniques for handling deadlocks. This approach requires that the operating system be given in advance additional information concerning which resources a process will request and use during its lifetime. With this additional knowledge, it can decide for each request whether or not the process should wait.

To decide whether the current request can be satisfied or must be delayed, the system must consider the resources currently available, the resources currently allocated to each process, and the future requests and releases of each process. Banker’s algorithm is a deadlock avoidance algorithm that is applicable to a system with multiple instances of each resource type.

Program:

#include<stdio.h>

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];

clrscr();

printf("Enter number of processes -- ");

scanf("%d",&n);

printf("Enter number of resources -- ");

scanf("%d",&r);

for(i=0;i<n;i++)

{

printf("Enter details for P%d",i);

printf("\nEnter allocation\t -- \t");

for(j=0;j<r;j++)

scanf("%d",&f[i].all[j]);

printf("Enter Max\t\t -- \t");

for(j=0;j<r;j++)

scanf("%d",&f[i].max[j]);

f[i].flag=0;

}

printf("\nEnter Available Resources\t -- \t");

for(i=0;i<r;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<r;i++)

{

scanf("%d",&newr);

f[id].all[i] += newr;

avail[i]=avail[i] - newr;

}

for(i=0;i<n;i++)

{

for(j=0;j<r;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");

}

getch();

}

               

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 3 1