MCQ

Computer Network MCQ – IP header & IP fragmentation

Networking MCQs questions with answers to prepare for exams, tests, and certifications. These questions are taken from a real written exam and some parts are taken from an interview. So you will find questions on CCNA, TCP/IP, Protocols, IP addressing issues, OSI model, and more. This MCQ will easily prepare anyone to pass their exam.
 

1. What of the following information is required for intermediate routers between the source and destination in the IP header?

A Version

B Protocol

C Identifier

D IP address source

E All the answers are true

E
  • Version: this field is set to “4” in decimal or “0100” in binary. The value indicates the IP version (4 or 6, there is no version 5).
  • Protocol: This indicates what type of protocol is encapsulated in the IP datagram. Some common values include:

    Protocol
    Number(Decimal)
    ICMP 1
    IGMP 2
    TCP 6
    UDP 17

  • Identifier: Sometimes a middle device in the network path cannot handle the datagram at the size at which it was transmitted, and must break it into fragments. If an intermediate system needs to decompose the datagram, it uses this field to help identify the fragments.
  • Source address (32 bits): This is the source IP address of the IP datagram.

 

 

2. Fragmentation of a datagram is necessary only in a _______.

A Datagram-based network

B Virtual circuit network

C Both A and B are true.

D None of the above
 

 
C
Each network has a maximum transmission unit (MTU). If the size of the data packet is larger than the MTU, it will be divided into fragments to transmit it over the network. Therefore, fragmentation may be necessary in both networks.
 

 

 

3. What fields are required in the IP header to allow the destination to perform fragment reassembly?

A Identifier, MF (More Fragments), Offset, Header Length (IHL) and Total Length

B MF (More Fragments), Offset and destination IP address

C MF (More Fragments), datagram length, and destination IP address

D MF (More Fragments), options and offset

A

 

 

4. The IP checksum must be recalculated on each router due to the modification of the following fields ____.

A TTL, Options, Identifier, Offset

B TTL, Options, Datagram length, Offset

C TTL, Options, Data, Offset

D TTL, Header length, Offset, ToS (Type of service)

B
The header checksum allows the IP protocol to detect datagrams with corrupted headers and to remove them. As the TTL changes at each hop, the checksum must be recalculated at each hop. In some cases, this algorithm is replaced by a cyclic redundancy check algorithm.

 

 

5. If the value in the “offset” field of the IP header is 100, the fragment size is ___?

A 100

B 400

C 800

D 200
 

 
C
  • The Fragment Offset field indicates the position of the fragment in the current datagram. All fragments of the datagram, except the last one, must be multiples of 8 bytes long.
  • If the Fragment Offset field value = 100, then the fragment offset = 8 x 100 = 800.
  • Then the size of this fragment is 800 bytes

 

 

6. When the source does not trust the routers to route correctly, what options can be used?

A Loose source routing

B Trace route

C Strict source routing

D Internet Time Stamp

C
Strict source routing: The sender of the data can specify the exact route.

 

7. The checksum calculation in the IP header includes _______?

A IP header only

B IP header and data

C The IP header and the pseudo header

D None of the above

A
The checksum calculation in the IP header includes only the IP header. Errors in the data are handled by the encapsulated protocol.

 

 
8. Suppose a router receives an IP packet containing 600 bytes of data and forwards it to a network with an MTU of 200 bytes. Assume that the IP header is 20 bytes long. What are the fragment offset values for fragmented packets?

A 22, 44, 66

B 0, 22, 44, 66

C 0, 22, 44

D 22, 44, 66, 88
 

 
B
Given :

  • MTU size of the destination network = 200 bytes
  • Length of IP header = 20

So :

  • The maximum data length that can be sent in a fragment = 200 – 20 = 180 bytes.
  • The data size sent in a fragment must be a multiple of 8.
  • Thus, the maximum data that can be sent in a fragment = 176 bytes.

So, 4 fragments are created,

  • The first fragment contains 176 bytes of data.
  • The second fragment contains 176 bytes of data.
  • The 3rd fragment contains 176 bytes of data.
  • The 4th fragment contains 72 bytes of data

Finally,

  • The offset value for the 1st fragment = 0
  • The offset value for the second fragment = 176/8 = 22
  • The offset value for the 3rd fragment = (176 + 176) / 8 = 44
  • The offset value for the 4th fragment = (176 + 176 + 176) / 8 = 66

 

 
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