I have an IPv4 packet, 3523 byte of information Data a MTU (maximum transmission unit) of 516 (496 byte + 20 byte IP header), also an TCP header with 20 byte.

IPv4 packet with 8 Datagrams

How does this result is created from the above information? I always come out at 51 byte Information data for the last package.

3523-(496*7) = 51

Maybe someone can shed some light on this.

  • $\begingroup$ Are you sure your 3523B figure includes the 20B TCP header? I mean, 71-51=20 could be just that. $\endgroup$ – chi Jun 19 '17 at 12:07
  • $\begingroup$ That would mean, that the 20byte tcp header would be part of the 496 byte of data or? $\endgroup$ – joachim Jun 19 '17 at 12:15
  • $\begingroup$ The 20B TCP header is part of the IP packet payload, and gets fragmented with everything else. I.e. in the first IP fragment, the first 20B are the TCP header. So, if the TCP segment payload is 3523B, you have to add 20B to get the IP packet payload, obtaining 3543B, which is then fragmented. So, you have to check whether your "3523B" is the TCP payload or the IP one. If it is the TCP one, you know where the extra 20B are coming from. $\endgroup$ – chi Jun 19 '17 at 12:41
  • $\begingroup$ Task states, a File with 3523 Byte of Data has to be transmitted via Network. So i assume you are correct, thanks for your clarification. $\endgroup$ – joachim Jun 19 '17 at 12:42
  • $\begingroup$ This is not about computer science, but about doing basic arithmetics. $\endgroup$ – Raphael Jun 20 '17 at 4:43

The IP packet has to include the TCP headers in its payload.

Hence we find additional 20 bytes, beyond the 3523 bytes of the sent data (the TCP payload). The total 3543 byes then get fragmented as usual.


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