An IP datagram of length (including a header of 20 bytes) 1800 bytes needs to cross an Ethernet followed by a WAN to reach its destination. The MTU for Ethernet is 1500 bytes. For the WAN, the MTU is given to be 576 bytes. The more-fragment value, fragment offset value and total length field value stored in the third fragment are _________________ .
(A) 1, 69, and 375 respectively.
(B) 1, 138, and 396 respectively.
(C) 0, 185, and 320 respectively.
(D) 0, 18, and 572 respectively.
Answer: (B)
Explanation: The datagram of 1800 bytes cannot be carried in one unit by Ethernet. Hence 2 fragments are required.
Data header Total Frag 1 1480 20 1500 bytes Frag 2 300 20 320 bytes ------- 1780 bytes of data
At the entry to the WAN, the Router has to further fragment FRAG 1. FRAG 2 goes through the WAN as it is.
Since every sub-fragment must have an IP header of 20 bytes, the WAN can carry a maximum data size of 556 bytes. However 556 is not divisible by 8, as required the Fragment Offset. Hence we decide to send 552 bytes of data in the first sub-fragment of FRAG 1.
Data Header Total Frag 1 A 552 + 20 572 Frag 1 B 552 + 20 572 Frag 1 C 376 + 20 396 ------- 1480 bytes of data
Hence at destination would reach 4 fragments namely FRAG 1A, FRAG 1B, FRAG 1C,
And FRAG 2.
MFB FO TL FRAG 1A 1 0 572 FRAG 1B 1 69 572 FRAG 1C 1 138 396 FRAG 2 0 185 320
SO, option (B) is correct.
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