IP Fragmentation

[Harsh Thanki]

Some people requested the solution of sum 10 from test. PFB

The datagram of 1800 bytes (1780+20) 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:- 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

ID = x

D = 0

M = 0(last frag) 1(more frag)