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GATE | GATE CS 2021 | Set 1 | Question 55

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  • Difficulty Level : Medium
  • Last Updated : 24 May, 2021

Consider two hosts P and Q connected through a router R. The maximum transfer unit (MTU) value of the link between P and R is 1500 bytes, and between R and Q is 820 bytes.

A TCP segment of size 1400 bytes was transferred from P to Q through R, with IP identification value as 0×1234. Assume that the IP header size is 20 bytes. Further, the packet is allowed to be fragmented, i.e., Don’t Fragment (DF) flag in the IP header is not set by P.

Which of the following statements is/are correct?
(A) Two fragments are created at R and the IP datagram size carrying the second fragment is 620 bytes.
(B) If the second fragment is lost, R will resend the fragment with the IP identification value 0×1234.
(C) If the second fragment is lost, P is required to resend the whole TCP segment.
(D) TCP destination port can be determined by analysing only the second fragment.

Answer: (A) (C)

Explanation: Two fragments are created at R and the IP datagram size carrying the
First fragment is 20+800 = 820 bytes
Second fragment is 20+600 = 620 bytes.

The loss of a single fragment results in all the fragments having to be resent where a reliable transport layer protocol such as TCP is in use (in fact the sender resends one packet and fragmentation occurs once again). This is the disadvantage of fragmentation.

For the second time in fragmentation, identification value will be changed.

We can determine destination Port number after re-assembling at the destination side.

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