IPv4 MCQ Quiz in मल्याळम - Objective Question with Answer for IPv4 - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

In class A, the leading 1 bit is fixed as 0 which is used to identify the class.

In class B, 16 bits are reserved for Host Id and 16 bits are reserved for Network Id. Out of these 16 Network Id bits, the leading 2 bits are fixed as 10 which are used to identify the class.

In class C leading three bits are fixed as 110.

In Class D leading four bits are fixed as 1110.

Data = 4400 Byte

Router MTU = 900 byte

Data + header = 900

Data = 900 – 20

∴ Data = 880

Number of fragments = \(\frac{{4400}}{{880}} = 5\)

Since initial fragment offset is 100.

Fragment offset of 2^nd fragment = \(100 + \frac{{880}}{8} = 210\)

Fragment offset of 3^rd fragment = \(210 + \frac{{880}}{8} = 320\)

Fragment offset of 4^th fragment = \(320 + \frac{{880}}{8} = 430\)

Fragment offset of 5^th fragment = \(430 + \frac{{880}}{8} = 540\)

penultimate fragment = 2^nd last fragment = 430

Shortcuts:

penultimate fragment = Initial fragment offset + 110 × (n - 2)

Answer: Option 4

CONCEPT

Minimum header size of IPv4 = 20 byte

Maximum header size of IPv4 = 60 byte

Header length field in the IPv4 header is 4 bits.

Maximum possible value (1111) = 15

The scaling factor of \(\frac{{60}}{{15}} = 4\) is introduced

EXPLANATION: 

IP header: 

But the first 8 bits in the Question are 0100 0010.

The first 4 bits indicate the version which is IPv4 (0100)

Now the next four bit indicates Header length (bits decimal value × 4) which is 2 × 4 = 8 bytes

which is not possible since the minimum header length for IPv4 is 20 bytes.

Hence Receiver will reject the packet.

Therefore option 4 is the correct answer.

Classful address:

In subnet mask:

Network Id of classful address are all ones(1’s) and host Id of classful address are all zeros(0’s).

The domain name system (DNS) maps the name people use to locate a website to the IP address that a computer uses to locate a website.

They maintain a directory of domain names and translate them to Internet Protocol (IP) addresses.

In class A, 8 bits are reserved for Network id

14 bit more is added due to subnetting

Total Network Id = 8 + 14 = 22

Number of bit in host id = x =32 - 22 = 10 

The number of available hosts = y = 2h − 2,

= 210 - 2

= 1024 - 2

= 1022

x + y = 10 + 1022 = 1032

Tips and Tricks:

Since first address and last address cannot be assigned to a Host

IP address:

IP address is an address having information about how to reach a specific host. The 32 bit IP address is divided into five sub classes. These are:

1) Class A

2) Class B

3) Class C

4) Class D

5) Class E

Class D and E are reserved for multicast and experimental purposes.

IPv4 address:

An IPv4 address is a 32 bits address, which is categorized into different IP classes.

IPv4 addresses are divided into two parts: a) Network ID b) Host ID

Class A: This class uses 24 bits for network ID part and 8 bit for host ID.

Class B: This class uses 16 bits for network ID and 16 bits for host ID.

Class C: This class uses 8 bits for network ID and 16 bits for host ID.

Note:

• IPv6 is a128 bits address

Concept:

In subnet mask: Network Id are all ones(1’s) and host Id of IP address are all zeros(0’s).

Calculation:

IP address consist of host id and network id

Host Id + Network id = 32 bits

IP address: 172.26.17.1/25

Network ID =25 bits

Therefore, Host ID= 32- (25) = 7

172.26.17.1 ≡ 172.26.17.00000001

Subnet Mask (binary) = 11111111. 11111111. 11111111.10000000

Subnet Mask = 255.255.255.128.

Key Points

There is a total of 32 bits in the IPv4 address space.

For example, if a network has the address “192.0. 2.0/24”, the number “24” refers to how many bits are contained in the network. From this, the number of bits left for address space can be calculated.

Hence the correct answer is 32.

Additional Information

• IPv6 is written in hexadecimal notation, separated into 8 groups of 16 bits by the colons, thus (8 x 16 = 128) 128 bits in total.
• In a LAN, each node is assigned a physical address, also known as a MAC/Ethernet address. This address is unique to each of the nodes on the LAN and is 6 bytes (48 bits)

Time to Live (TTL)

• Time to Live is 8 bits field present in IPv4 header. Range is (0 to 2⁸ -1) to 0 to 255.
• TTL is a mechanism that limits the lifespan of a packet in a computer network.
• TTL prevents a data packet from circulating indefinitely which is present in IPv4 header
• Every host (operates at Network layer) that passes the packet must reduce the TTL by at least one unit.

Therefore, TTL prevent pocket from wandering around forever.