The following IP datagram is sent to a destination across a network with an MTU of 500. Using the first diagram, shows the header portions of the first two packet fragments that reach the destination (assuming arrival in order). Note all numbers are in base 10. Note that you should put a value in the data size space? Solution packet 1: packet 2: The data is of size 786 as in the total length field the value is 806 which is data+header. we know that the header length is 20bytes the data will be 786. The MTU is 500(both data and header) Therefore the 1st packet can send data upto 480 bytes. which is also equal to fragmentation offset because the next data has to be appended to these 480 bytes for complete information. the left out data is 306, so the total length is 326(306+20).As first packet could not carry the whole message it has to intimate the reciever there is more data to come, to indicate the flag value is set to 1, which is not the case in packet 2 so the flag is set to 0.45TOS500681480.