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GATE | Sudo GATE 2020 Mock I (27 December 2019) | Question 37

  • Last Updated : 26 Dec, 2019

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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