IP Address Problems

 Important Points-

 Point-01:

 

For any given IP Address,

• If the range of first octet is [1, 126], then IP Address belongs to class A.
• If the range of first octet is [128, 191], then IP Address belongs to class B.
• If the range of first octet is [192, 223], then IP Address belongs to class C.
• If the range of first octet is [224, 239], then IP Address belongs to class D.
• If the range of first octet is [240, 254], then IP Address belongs to class E.

 

Point-02: 

For any given IP Address,

• IP Address of its network is obtained by setting all its Host ID part bits to 0.

 

Point-03: 

For any given IP Address,

• Direct Broadcast Address is obtained by setting all its Host ID part bits to 1.

 

Point-04: 

• For any given IP Address, limited Broadcast Address is obtained by setting all its bits to 1.
• For any network, its limited broadcast address is always 255.255.255.255

 

Point-05: 

• Class D IP Addresses are not divided into Net ID and Host ID parts.
• Class E IP Addresses are not divided into Net ID and Host ID parts.

 

PRACTICE PROBLEMS BASED ON IP ADDRESS IN NETWORKING-

 

Problem-01: 

For the following IP Addresses-

1. 1.2.3.4
2. 10.15.20.60
3. 130.1.2.3
4. 150.0.150.150
5. 200.1.10.100
6. 220.15.1.10
7. 250.0.1.2
8. 300.1.2.3

 

Identify the Class, Network IP Address, Direct broadcast address and Limited broadcast address of each IP Address. 

Solution-

 

Part-A:

 

Given IP Address is-

1.2.3.4

 

• IP Address belongs to class A
• Network IP Address = 1.0.0.0
• Direct Broadcast Address = 1.255.255.255
• Limited Broadcast Address = 255.255.255.255

 

Part-B:

 

Given IP Address is-

10.15.20.60

 

• IP Address belongs to class A
• Network IP Address = 10.0.0.0
• Direct Broadcast Address = 10.255.255.255
• Limited Broadcast Address = 255.255.255.255

 

Part-C:

 

Given IP Address is-

130.1.2.3

 

• IP Address belongs to class B
• Network IP Address = 130.1.0.0
• Direct Broadcast Address = 130.1.255.255
• Limited Broadcast Address = 255.255.255.255

 

Part-D:

 

Given IP Address is-

150.0.150.150

 

• IP Address belongs to class B
• Network IP Address = 150.0.0.0
• Direct Broadcast Address = 150.0.255.255
• Limited Broadcast Address = 255.255.255.255

 

Part-E:

 

Given IP Address is-

200.1.10.100

 

• IP Address belongs to class C
• Network IP Address = 200.1.10.0
• Direct Broadcast Address = 200.1.10.255
• Limited Broadcast Address = 255.255.255.255

 

Part-F:

 

Given IP Address is-

220.15.1.10

 

• IP Address belongs to class C
• Network IP Address = 220.15.1.0
• Direct Broadcast Address = 220.15.1.255
• Limited Broadcast Address = 255.255.255.255

 

Part-G:

 

Given IP Address is-

250.0.1.2

 

• IP Address belongs to class E
• Network IP Address = Not available
• Direct Broadcast Address = Not available
• Limited Broadcast Address = Not available

 

Part-H:

 

Given IP Address is-

300.1.2.3

 

• This is not a valid IP Address.
• This is because for any given IP Address, the range of its first octet is always [1, 254].
• First and Last IP Addresses are reserved.

 

Problem-02:

 

A device has two or more IP Addresses, the device is called-

1. Workstation
2. Router
3. Gateway
4. All of these

 

Solution-

 

• All the given devices have a network layer.
• So, they will have at least one IP Address.

 

In TCP/IP suite-

• Workstation and gateway have all the 5 layers.
• Router has only 3 layers last layer being network layer.

 

Workstation-

 

• A user may configure more than one IP Addresses in his workstation / computer.
• With more than one IP Address, it remains present in more than one networks.
• So, if one network goes down, it is always reachable from other networks.

 

The following figure shows a host present in more than one networks-

 

 

• It is important to note that IP Addresses are assigned to interfaces.
• When we buy a new laptop, we usually get 2-3 interfaces.
• Thus, a workstation can have more than one IP Addresses.

 

Router-

 

• A router may be connected to various interfaces.
• Each interface has a unique IP Address.
• Thus, a router may also have more than IP Addresses.
• Similar is the case with gateways because gateways are extension of routers.

 

Thus, Option (D) is correct.

 

Problem-03:

 

A host with IP Address 200.100.1.1 wants to send a packet to all the hosts in the same network.

What will be-

1. Source IP Address
2. Destination IP Address

 

Solution-

 

1. Source IP Address = IP Address of the sender = 200.100.1.1
2. Destination IP Address = Limited Broadcast Address = 255.255.255.255

 

Problem-04:

 

A host with IP Address 10.100.100.100 wants to use loop back testing.

What will be-

1. Source IP Address
2. Destination IP Address

 

Solution-

 

• Source IP Address = 10.100.100.100
• Destination IP Address = Loopback Testing Address = 127.0.0.1

 

Problem-05:

 

How many bits are allocated for Network ID and Host ID in 23.192.157.234 address?

 

Solution-

 

Given IP Address belongs to class A.

Thus,

• Number of bits reserved for Network ID = 8
• Number of bits reserved for Host ID = 24

 

Problem-06:

 

Which devices can use logical addressing system?

1. Hub
2. Switch
3. Bridge
4. Router

 

Solution-

 

• Devices which have network layer as the last layer can only use logical addressing system.
• Devices which have data link layer as the last layer can only use physical addressing system.
• IP Addresses are the logical addresses and MAC Addresses are the physical addresses.

 

Option-A:

 

• Hub can neither use physical addressing system nor logical addressing system.
• This is because it has physical layer as the last layer.

 

Option-B:

 

• Switch can use physical addressing system but not logical addressing system.
• This is because it has data link layer as the last layer.

 

Option-C:

 

• Bridge can use physical addressing system but not logical addressing system.
• This is because it has data link layer as the last layer.

 

Option-D:

 

• Router can use physical addressing system as well as logical addressing system.
• This is because it has network layer as the last layer.

 

Thus, option (D) is correct.

 

Problem-07:

 

What is the network ID of the IP Address 230.100.123.70?

 

Solution-

 

• Given IP Address belongs to class D.
• Class D IP Addresses are not divided into the Network ID and Host ID parts.
• Thus, there is no network ID for the given IP Address.

 

Problem-08:

 

Match the following-

 

Column-I:

 

1. 200.10.192.100
2. 7.10.230.1
3. 128.1.1.254
4. 255.255.255.255
5. 100.255.255.255

 

Column-II:

 

1. Class A
2. Limited Broadcast Address
3. Direct Broadcast Address
4. Class C
5. Class B

 

Solution-

 

(I, D), (II, A), (III, E), (IV, B), (V, C)

 

Problem-09:

 

Suppose that instead of using 16 bits for network part of a class B Address, 20 bits have been used. How many class B networks would have been possible?

 

Solution-

 

• Total 20 bits are used for Network ID of class B.
• The first two bits are always set to 10.
• Then, with 18 bits, number of networks possible = 2¹⁸

 

Problem-10:

 

What is the default mask for 192.0.46.10?

 

Solution-

 

• Given IP Address belongs to class C.
• For class C, default mask = 255.255.255.0