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Rom4ik [11]
3 years ago
7

Contrast the performance of the three techniques for allocating disk blocks (contiguous, linked, and indexed) for both sequentia

l and random file access. [12 marks]
Computers and Technology
2 answers:
sladkih [1.3K]3 years ago
5 0

Answer:

There are three main disk space or file allocation methods.  

• Contiguous Allocation  

• Linked Allocation  

• Indexed Allocation

By the performance level  

Contiguous allocation > linked Allocation > Indexed Allocation

Contiguous is faster than linked allocation and both contiguous and linked are faster than indexed.

As  linked allocation does not support direct access and the indexed allocation , the pointer of it is bigger than linked allocation  

Explanation:

1.Contiguous Allocation :

In this scheme, each file occupies a contiguous set of blocks on the disk.

> It supports both accesses (sequential and direct)

> Its extremely fast as the number of seeks are minimal because of contiguous allocation of file blocks.

but on the other hand it has some disadvantages as

> It can not deal with fragmentation whether its external or internal.

> And also increasing the size of it may get difficult.

2. Linked Allocation:

In this type of allocation, each file is a linked list of disk blocks and the disk blocks can be scattered anywhere on the disk.

> This type is very flexible with the file size.  

> This method also does not suffer from external fragmentation. This makes it better in terms of memory utilization.

> It does not support random or direct access while it slightly support sequential access.

3. Indexed Allocation:

There is a special block known as the Index block contains the pointers to all the blocks occupied by a file in this type of allocation. Each file has its own index block.  

> This type supports direct access therefore provides fast access to the file blocks.  

> It also overcomes with the problem of external fragmentation.

Pie3 years ago
3 0
The allocation methods define how the files are stored in the disk blocks.
There are three main disk space or file allocation methods:
1.Contiguous Allocation-in this scheme,each file occupies a set of blocks on the disk. For example if a file requires x blocks and is given a block y as the starting location,then the blocks assigned to the file be :x,y+1,y+2,......y+x-1.
This means that given the starting block address and the length of the file(in terms of blocks required) we can determine the blocks occupied by the file.
Advantages
-both the sequential and direct accesses are supported
-this is extremely fast since the number of seeks are minimal because of contiguous allocation of file blocks.
2.linked allocation-in this scheme,each file linked list of disk blocks which need not  be contiguous disk blocks can be scattered anywhere on the disk.
Advantages
it is very flexible in terms of file size.file size can be increased easily since the system does not have to look for a contiguous chunk 
of memory.
this method does not suffer from external fragmentation and it makes it relatively better in terms of memory utilization.
3.Indexed Allocation-in this scheme,a special block known as the index block contains the pointers to all the blocks occupied by a file.Each file has its own index block.the entry in the index block contains the disk address of the block
Advantages
it supports direct access to the blocks occupied by the file and therefore provides fast access to the file blocks
it overcomes the problem of external fragmentation.
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Lady_Fox [76]

Answer:

<u>Window.java</u>

  1. public class Window {
  2.    int width;
  3.    int height;
  4.    public Window(int width, int height){
  5.        this.width = width;
  6.        this.height = height;
  7.    }
  8.    public int getWidth(){
  9.        return width;
  10.    }
  11.    public int getHeight(){
  12.        return height;
  13.    }
  14.    public int getClientAreaHeight(){
  15.        return getHeight();
  16.    }
  17. }

<u>Main.java</u>

  1. public class Main {
  2.    public static void main (String [] args) {
  3.        Window win1 = new Window(12, 15);
  4.        System.out.println(win1.getClientAreaHeight());
  5.    }
  6. }

Explanation:

<u>Window.java</u>

There is a Window class with two int type attributes, width and height (Line 1 - 3).

The constructor of this class will take two inputs, width and height and set these input to its attributes (Line 5 - 8). There are two methods getWidth and getHeight which will return the value of attributes width and height, respectively (Line 10 - 16).

The required new method getClientAreaHeight is defined in line 18 -20. This method will call the getHeight method to return the height value of the window (Line 19).

<u>Main.java</u>

We test the Window class by creating one Window instance and call the getClientAreaHeight method and print the return output (Line 1 -6).

We shall see 15 is printed.

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