Does the MTU of a TCP packet include the TCP & IPv4 headers? - tcp

Or: What is the maximum payload size for a TCP/IPv4 packet.
Much appreciated!
Cheers.

The MTU of an IP packet is the entire packet, including all IP and higher level headers (such as TCP headers) and payload. Lower level headers (such as Ethernet frames) are not included since they're not IP's concern. However the actual MTU value is influenced by the lower levels, as there's usually a limit, and IP has to stick to it.

MTU = IP Header + Tcp Segment
MSS = Data(doesn't include TCP header)

Related

TCP MSS over WiFi

The MTU for 802.11 is 2296 bytes. Does this mean that if TCP is used over 802.11, the MSS can only be 2296 - 40 = 2256? Can't one use a higher MSS which would then get fragmented over 802.11?
In short, is there a strict limit on the MSS for TCP?
The MTU for 802.11 is 2296 bytes.
Are you sure about that number? This answer says it's 2304.
Does this mean that if TCP is used over 802.11, the MSS can only be
2296 - 40 = 2256?
Assuming that the MTU is 2296, that's correct. You lose at least 20 bytes for an IPv4 header, and 20 more bytes (at least) on a TCP header.
Can't one use a higher MSS which would then get fragmented over
802.11?
Why would you want that? TCP implementations actively try to avoid IP fragmentation using MTU discovery. If a TCP segment is fragmented, and one of the fragments is lost or corrupted, the entire segment would have to be resent since TCP has no concept of IP fragmentation and as far as it is concerned, the whole segment was lost. In general, this is much more wasteful than sending smaller segments that wouldn't be fragmented. This is especially true in wireless networks where frames tend to become corrupt quite often.

Why UDP header has 'length field'?

I couldn't understand why UDP header has 'length' field, and why it is needed?
If the reason is to know where the 'application message(L5 data)' begins in the segment, it can just be gotten from 'UDP data - UDP header length(it is already known value)'.
Because UDP can be transmitted over another protocol than IP.
And also because UDP transmits datagram messages with a length (udp length) which can be sent over multiple IP fragmented packets.
Source: https://notes.shichao.io/tcpv1/ch10/
The UDP header length field is the length of the UDP header plus the UDP data. It is indeed redundant since this length can be calculated from the IP header total length field where the UDP datagram length is the IP total length minus the IP header length.
UDP uses message stream NON FIFO as communication model between the sender and receiver. If the size is not mentioned then it will not be possible to decipher the message at rxr.
Say m1m2m3 is sent then each message to be notified so that u can trace back the message.
Regards

Will TCP header be present in IP fragmented packets

What happen when ICMP is disabled in an router and when packet size greater than MTU how the router fragments that packet?Will TCP header be present in IP fragmented packets?
I don't test this scenario but ... I think that if the IP fragmentation is enabled, your IP packet data part will be fragmented and transferred independently because the fragmentation was used at L3 layer. Without ICMP, the sending computer don't know that the packet size is bigger than the allowed network MTU and it cannot send new fragmented data at L4 layer (TCP header will be presented in each packet). But we are fragmenting on L3 layer and we have to use fragment offset field. I think that the TCP header will be presented only in one packet, followed by fragmented TCP data part.
Try to make an experiment to test this behavior.

IP fragmentation and TCP ACK

I have a question on how TCP_ACK works when the original packet are fragmented.
For example, original packet size is 1,500*N bytes and MTU is 1,500. Then, the packet will be frgmented into (approximately) N packets.
In this case, how does the receiver sends TCP_ACK to the sender?
I checked with wireshark, it seems that the receiver sends TCP_ACK for every two fragmented packet. Is it right?
Could you give me some refereces for this or explanation?
Thanks.
IP layer on the receiver stack reassembles all the IP fragments into a single TCP segment before handing the packet over to TCP. Under normal conditions, TCP should send only one ACK for the entire TCP segment. The ACK # would be the next expected SEQ # as usual.

Is MSS value fixed in SYN packet?

I wonder how MSS is set in SYN packet? Is it a fixed value in one operating system or the value could be changed in the same operating system? I know that the value is different in different operating systems. Also is the MSS value in SYN related to hardware configuration?
Thanks.
RFC 879 describes how MSS is used and specified.
In short, MSS is specified during TCP handshake via SYN packet. However, this value can later be changed by OS itself or by setting a protocol option.
You can set option TCP_MAXSEG via setsockopt.
Whilst the value of MSS in SYN and SYNACK packets are set by the initiator and responder side, respectively, a widely used practice known as MSS clamping can result in the MSS being altered by a network element on the path - this is often used to reduce the MSS of all connections going over some sort of tunnelled link. For example PPPoE is commonly used on residential broadband and requires an MTU of 1492 and corresponding IPv4 MSS of 1452 so whilst the SYN may leave your machine with an MSS of 1460 (assuming you're using Ethernet with an MTU of 1500) but once it passes the MSS clamping ISP router the MSS in SYN packet will subsequently be changed to 1452, and likewise for the incoming/responder's SYNACK packet so the connection proceeds with reduced MSS of 1452. This practice seems to be used instead of Path MTU Discovery which relies upon the use of ICMP Fragmentation Needed responses from the network as these can be lost on poorly configured networks and by certain load balancing techniques.

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