# Network And Operating System Computer Science Essay

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Subnetting: Parts of a Network are called subnetwork or subnet consisting of host machines. And dividing a network into smaller and manageable pieces called subnetting. Subnet Numbers: All of the IP addresses are 32-bit long and have a subnet field known as subnet numbers. The bits that are designated with the network bits give the required amount of subnet works.

So according to this IP address 192.228.17.56 with the network address of 192.288.17.32 the subnet mask is 255.255.255.224. As it showing down:

192.228.17.32 =11000000.11100000.00010001.00100000

The default of class C mask =11111111.11111111.11111111.00000000

255.255.255.224 =11111111.11111111.11111111.11100000

The subnet bits are 3(because of the first three 1s) and the host bits are 5(number of 0s).

Subnet number = 2x (where x is number of subnet bits in the mask).

So the subnet number for the IP address will be = 23 = 8.

Network addresses: is a unique number to identify a machine in the network. Computers can determine the address of other computers on the network by using the network address in order to send and receive messages among them.

For this IP address 192.228.17.56 we need 32 hosts from which we can use.

So the first subnet is 0. And all of the 8 subnets are:

0 So the first network address is : 192.228.17.0 in binary 11000000.11100100.00010001.00000000

0 + 32 = 32 (add32) So the second network address is 192.228.17.32 in binary 11000000.11100100.00010001.00100000

32 + 32= 64 (still add 32) So the third network address is 192.228.17.64 in binary 11000000.11100100.00010001.01000000

64 + 32 = 96 (still add 32) So the fourth network address is 192.228.17.96 in binary 11000000.11100100.00010001.01100000

96 + 32 = 128 (still add 32) So the fifth network address is 192.228.17.128 in binary 11000000.11100100.00010001.10000000

128 + 32 = 160 (still add 32) So the sixth network address is 192.228.17.160 in binary 11000000.11100100.00010001.10100000

160 + 32 = 192 (still add 32) So the seventh network address is 192.228.17.192 in binary 11000000.11100100.00010001.11000000 192 + 32 = 224 (still add 32) So the last network address is 192.228.17.224 in binary 11000000.11100100.00010001.11100000

Broadcast subnet address: is special IP address that is used to send message to all the machines in a subnet. The Broad cast address cannot be assigned to any host.

And the broadcast address is always the number which come before the next subnet

192.228.17.0

192.228.17.32

192.228.17.64

192.228.17.96

192.228.17.128

192.228.17.160

192.228.17.192

192.228.17.224

## 11000000.11100100.00010001.11111111

NetMasks: it's a mask which has 32bits and used to split an IP address in two parts subnets and specify the networks available hosts, Moreover it's used to transfer message correctly among machines in the same subnet.

As the IP address given 192.228.17.56 is a class C, the default mask is 255.255.255.0, in binary it is = 11111111.11111111.11111111.00000000

So according our IP address the netmask is 255.255.255.224, and it is in binary = 11111111.11111111.11111111.11100000

Host Numbers: it's a unique number of particular host when all hosts sharing the same network prefix. But if any two hosts which are in different networks the same number can be used for both of them.

To find the host numbers for the IP address 192.228.17.56 and the network address 192.228.17.32 firs we have to find the subnet mask. And as we have worked out above the subnet mask is 255.255.255.224 which is in binary:

## 11111111.11111111.11111111.11100000

So we use only the 0s (the bits turned off) for host addressing. So it means we have 5 bits.

Host number = 25 - 2 = 30

It means we have 30 numbers of hosts in each subnet.

Valid host range

192.228.17.0

192.228.17.1 -to- 192.228.17.30

192.228.17.32

192.228.17.33 -to- 192.228.17.62

192.228.17.64

192.228.17.65 -to- 192.228.17.94

192.228.17.96

192.228.17.97 -to- 192.228.17.126

192.228.17.128

192.228.17.129 -to- 192.228.17.158

192.228.17.160

192.228.17.161 -to- 192.228.17.190

192.228.17.192

192.228.17.193 -to- 192.228.17.222

192.228.17.224

192.228.17.225 -to- 192.228.17.254

## Question1 Part(b)

Class C is the third class of the IP addresses which is set under class A and B , also it sets upper than class D and class E which is the last class.

## Class C: 192.0.0.0 to 223.255.255.255

In order to calculate the number of host a simple formula can be used: 2^bits ,Class C IP address has 21 bits for network and 8 bits dedicated for host .Using the formula :

2^ 21 =2,097,152 possible networks.

2^ 8 =256-2=254 possible host where we subtract 2 where 1 IP is for the network Id, and the other 1 is for the broadcast.

Analyzing subnetted IP from question 1(a) 192.228.17.32 we have 8 sub networks

23 = 8 sub networks AND

25 - 2 = 30 possible hosts

But if we use 1 host for Router per subnet than we can see that we are losing 3 IP's per subnet that is 24 IPs for 8 subnets. The main benefit for subnetting a network that it is more manageable and effective.

## Question1 Part(c)

Open Systems Interconnection (OSI)models deals with connecting open systems .It has seven layers or levels. And each layer which sets on the next lower layer to perform more primitive functions. Moreover layer must provide services to the next higher layer. Also change in order of one layer doesn't need to change the rest of them.

Application layer

Presentation layer

Session layer

Transport layer

Network layer

Physical layer

## The architecture of the Open System interconnection Models is shown in above table and it has been briefly explained that which layer deals with exactly what.

1. The Physical Layer

This is the last layer and it's more concerned with transmitting raw bits over a communication/physical channel. And it basically works on mechanical, electrical, functional, procedural, timing interfaces and physical transmission mediums that lies below the physical layer .

Task of this layer is to provide for the reliable transfer of information across physical link. It sends frames which has important synchronization, error control and flow control. This layer is further divided into two layers:

a- Medium Access Control(MAC Layer)

b- Logic Link Control (LLC Layer)

3. The Network Layer

The network layer basically provides the routing facilities for the data packets. It deals with switching or making logical paths and acknowledging the data packets. It provides upper layers with independence from the data transmission and also switching technologies used to connecting the system which is responsible for setting up, maintaining and terminating connections.

4. The Transport Layer

The basic task of this layer is to provide reliable transparent transfer of data between last part of the point, moreover it offers (end to end error recovery and flow controls). Also it determines what kinds of service to provide to session layer.

5. The Session Layer

The session layer manages the session. It creates the session and when the communication between two nodes is done it simply destroys the session. It provides the control structure for the communication between application like; establishes, manages and terminates connections (sessions) between cooperating applications.

6. The presentation layer

This layer presents data in a meaningful format (i.e. compress, encode, and convert data). This also called syntax layer because it is concerned with the syntax of the information presentations to transmitted. It helps the application processes by providing independence from differences in term of data representation syntax.

7. The Application Layer

It contains a variety of user applications (e-mail, ftp, file transfer, ect). Which are commonly needed by the user . One widely used protocol is HTTP which is the basis for World Wide Web. It also provides distributed information services.

Benefits of OSI Reference Model

The idea of hidden objects fits very perfectly with the modern ideas about object oriented programming.

Each layer provides some services without knowing the other layers how these services were implemented.

None of the models is perfect. The OSI Model was never easy to implement. It appeared that OSI Model would take over all the projects and take a grip on the world but it did not happen. These were the reasons considered mainly.

## Question1 Part(d)

Level 1, 2 and 3 devices are Hubs, repeaters, bridges, switches, and routers. These devices are used for different purposes like to connecting hosts to a network making them act like a single segment like hubs do, selecting the traffic path like routers do, and connecting multiple networks like bridges do.

## Router

A router is a specialized network device which determines the next point at network to which data packet to send toward its destination. They Works on OSI layer 3 which is (network layer).

## Hub

Hubs works on OSI layer 1which is (The Physical Layer). Hubs are used to connects multiple Ethernet segments together physically making them act as a single unit or segment. Every node behaves as a single broadcast domain and collision domain when we use a hub device and it allows only one computer to transmit at one time because it's using the same domain. Depending on the network Technology, the hubs connection among the network objects are at basic level 1. By using a hub we get dedicated connections between individual nodes. And it provides shared bandwidth among all the objects.

## Question 2 Part(a)

Head: it's a device which read and writes data from one platter in the disk, and in each side of platter there are usually two heads.

Cylinder: It's a three dimensional concept which has the set of all tracks that are at one arm position makes up a cylinder. There may be thousands of concentric cylinders in a disk drive.

Track: it's concentric rings on a disk .Also it's a part of cylinder. Each track is divided into wedge-shape areas called sector. Moreover it's creating by low level format of a disk.

Sector: it's a small part of the track which holds 512 bytes. Also it can hold the overhead bytes which is used to provide error corrections and timing recovery under the low level format in the lifecycle of the disk drive.

Blocks: are basically sectors or group of sectors; modern disks have arrays of linked list logical blocks where the logical block is the smallest unit of transfer. Always the size of a logical block is smaller than the physical block size.

## Question 2 Part(b)

Dos Boot Record: it's a part which contains the growing boot code written in assembly language and error messages for the disk volume. Each partition has its own boot sector too and each operating system has its own boot sector format. What's more the DOS is the stander of Disc Operating System.

Fat : there are several types of fat file system and each one has different purpose than the rests so for example Fat12 (floppy) .And Fat12,16,32bits which is a cluster address size.

Fat 1: it's a first allocation table ,where cluster stored it's information in. And it's important for blocks which can deal with allocated and unallocated blocks. Moreover it sets next to volume boot sector in the disk.

Fat 2: it's a second copy of fat .And it sets after fat1 directory.

Root Directory: is the main directory where all other directories, sub directories, folders or files are placed. It is the top-level directory in the file-system hierarchy it contains file names and metadata.

Slack space: It's the space left in any of the cluster that is not being used. Or it is Any free space which still among a last byte of the file and a first byte where a next cluster is a shape of internal fragmentation is known as a file slack or slack space.

## Question 2 Part(c)

Differences

Contiguous allocation

## Performance

Needs contiguous memory and memory wasted. Files cannot grow.

It solves the superficial -fragmentation and size-declaration problems of the contiguous allocation. So it's better than contiguous allocation in term of performance .

## Disk space management

External Fragmentation, lots of holes outside files. Wasteful of space (dynamic storage-allocation problem).

Free management system and no wasteful of space.

## File size management

Each file occupies a set of contiguous blocks on the disk.

Each file is a linked list of disk blocks: blocks might be dispersed anywhere on the disk.

## Random access of blocks

Random access.

Simple - only need the starting location which is at (block #) and length (the number of blocks which are required).

No random access.

Allocate as needed, link together. Example, file starts at block 9

## (1).Hard disk size in bytes.

Hard disk size (bytes)

= Number of cylinders Ã- number of heads Ã- sectors/track Ã- 512 =

=512Ã-16Ã-63Ã-512=264241152 bytes

## (2) The minimum physical file size in bytes.

One cluster equal 2 sectors

and each sector has 512 bytes

So in 2 sectors there are

= 2 Ã-512 =1024bytes which is the minimum physical size in bytes.

## (3) The slack space size in byte for a file size 912 bytes?

Slack space=The physical file size in bytes -The file size in bytes(912) =

1024 - 912 bytes = 112 bytes