GATE CSE Computer network fragmentation

#### Show Answer With Best Explanation

Answer: III, IV

Explanation:
Given,
The maximum transfer unit (MTU) value of the link between P and R = 1500 bytes
The maximum transfer unit (MTU) value of the link between R and Q = 820 bytes.

Segment of size = 1400 bytes
IP identification value = 0x1234
IP header size = 20 bytes
Donâ€™t Fragment (DF) flag value = 0

Logic
Since Segment size is 1400 B and MTU value of the link between P and R is 1500 bytes. So, Segment is successfully arrive at router R. MTU value of the link between R and Q is 820 bytes. Now here at router R, Fragmentation is required.

Segment size 1400 Byte is divided into two parts or two fragment
1400 = 800 + 600

1st fragment size = Data + Header = 800 + 20 = 820 Bytes
2nd fragment size = Data + Header = 600 + 20 = 620 Bytes

Option(I): If the second fragment is lost, R will resend the fragment with the IP identification value 0x1234.(False)
If second or any fragment is lost, then whole TCP segment or entire fragment has to be sent again. If any fragment is lost, the sender need to retransmit, But sender does not know about any fragment. Because fragmentation is done at intermediate router. So, sender has to retransmit entire fragment or whole TCP segment and for the next time in fragmentation, IP identification value will be changed.

Option(II): TCP destination port can be determined by analysing only the second fragment.(False)
Each fragment of TCP Segement is send seperately.Each fragment contains TCP destination port. So, TCP destination port can be determined by analysing any fragment.

Option(III): Two fragments are created at R and the IP datagram size carrying the second fragment is 620 bytes.(True)
1st fragment size = Data + Header = 800 + 20 = 820 Bytes
2nd fragment size = Data + Header = 600 + 20 = 620 Bytes

Option(IV): If the second fragment is lost, P is required to resend the whole TCP segment. (True)
If second or any fragment is lost, then whole TCP segment or entire fragment has to be sent again.

So, Option(III) and Option(IV)are correct.

#### Show Answer With Best Explanation

Answer: 144

Explanation:
Given,
IP Packet Length = 4500 Bytes
IP Header = 20 Bytes
MTU = 600 Bytes

Data,
IP Packet = IP Payload + IP Header
IP Payload = IP Packet – IP Header
IP Payload = 4500 – 20 = 4480 Bytes
MTU Payload = 600 – 20 = 580 Bytes
MTU Payload should be multiple of 8 But 580 is not multiple of 8. So, the nearest value of 580 which is multiple of 8 is 576.

Calculation,

The fragmentation offset value stored in the third fragment is 144.

#### Show Answer With Best Explanation

Answer: 13
Explanation: Upload Soon

#### Show Answer With Best Explanation

Answer: III
Explanation: Upload Soon

#### Show Answer With Best Explanation

Answer: I
Explanation: Upload Soon

GATE CSE Computer network fragmentation

#### Show Answer With Best Explanation

Answer: III
Explanation: Upload Soon

error: Content is protected !!
Hi,how Can We Help You ?