Solution

The correct answer is A and C Only

Key Points

• A. Two fragments are created at R and IP datagram size carrying the second fragment is 620 bytes: TRUE
  • Since the TCP segment is of size 1400 bytes and the MTU between P and R is 1500 bytes, P can send it to R without fragmentation. However, when the router R tries to forward them to Q, it will not be able to do so because the MTU between R and Q is 820 bytes.
  • Since the IP header size is 20 bytes, the maximum payload that can be carried in one IP datagram on the R-Q link is 800 bytes.
  • The first fragment will have 800 bytes and the second fragment will have the remaining 600 bytes of the TCP segment. But, we must consider that both fragments will have their own 20 bytes IP header. So, the total size of the second IP datagram will be of 620 bytes (600 + 20 bytes of IP header).
• B. If the second fragment is lost then R resend the fragment with IP Identification value of 0x1234: FALSE
  • In case the second fragment is lost, R does not resend the fragment because the retransmission responsibility belongs to the end systems (here, P), not the routers.
• C. If the second fragment lost then P required to resend the entire TCP segment: TRUE
  • As B, it's the end system's (P) responsibility to detect the loss (via TCP mechanisms) and retransmit the entire TCP segment.
• D. TCP destination port can be determined by analyzing the second fragment only: FALSE
  • Only the first fragment has the entire TCP header because IP fragmentation does not duplicate the TCP headers in the fragments.
  • The TCP header contains information such as source port, destination port, sequence number, and acknowledgment number. The later fragments only contain IP headers and the remaining portion of the data. So, it's not possible to determine TCP port by analyzing the second fragment.