Consider two hosts P and Q are connected through a router R. The maximum…

Solution summary: determine fragmentation, sizes, and which statements are true.

• Compute the original IP datagram size: TCP segment 1400 bytes + IP header 20 bytes = 1420 bytes.

• Maximum data per fragment on the R→Q link: MTU 820 − IP header 20 = 800 bytes of payload per fragment.

• Fragmentation result: first fragment carries 800 bytes payload (800 + 20 = 820 bytes total), second (last) fragment carries remaining 620 bytes payload (620 + 20 = 640 bytes total). The first fragment has fragment offset 0; the second fragment offset = 800/8 = 100.

• IP identification: both fragments use the same identification value 0x1234 so the receiver can reassemble them.

• If the second fragment is lost: the receiver cannot reassemble the full IP datagram and discards the fragments. The router that performed fragmentation does not retransmit fragments; retransmission is handled by the transport layer (TCP) at the sender, so the TCP sender will retransmit the entire TCP segment.

• TCP header fields (including source and destination ports) are located at the start of the TCP segment and therefore appear only in the first fragment. Later fragments do not contain the TCP header and so cannot be used alone to determine the TCP destination port.

Final evaluation: the true statements are that two fragments are created (with payloads 800 and 620 bytes) and that the sender must retransmit the entire TCP segment if a fragment is lost. The statements that the router will resend a lost fragment and that the TCP destination port can be determined from the second fragment are false.

Correct choice: the option that selects the two true statements (two fragments are created; sender must resend the entire TCP segment).