New answers tagged tcp
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In three-way hand shaking, if a port is established how would other packets be connected [duplicate]
The TCP connection state depends on the connection partner. See RFC 793:
Multiplexing:
To allow for many processes within a single Host to use TCP
communication facilities simultaneously, the TCP provides a set of
addresses or ports within each host. Concatenated with the network
and host addresses from the internet communication layer, this forms
a socket....
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In order to see why this is happening we need to consider these two things:
TCP is stateful, in that both end-points of the connection are holding connection state. When the connection is closed, this state has to be deleted at some point. In TCP, connection state can be deleted when (immediately after) switching to CLOSE state.
When TCP receives a segment,...
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RST is sent by the side A doing the active close because it is the side which sends the last ACK. So if it receives FIN from the side B doing the passive close in a wrong state, it sends a RST packet which indicates other side that an error has occured because TCP RST is used in a scenario where TCP connections cannot recover from errors and the connection ...
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See RFC 793:
Reset Generation
As a general rule, reset (RST) must be sent whenever a segment
arrives which apparently is not intended for the current connection.
A reset must not be sent if it is not clear that this is the case.
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It occurred to me now that, ultimately it is the tcp socket which needs to be bound to a VRF. A single process can create multiple sockets each being bound to different VRF. All that the process(BGP in this case) has to do it to receive the packets arriving on a socket.
I hope my understanding is correct. Please share your comments.
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no, SACK will not move the beginning of the sender window (also called SND.UNA)
Let's consider that the receiver has receiver buffer of size 5. If the receiver has received packets 4 and 5 but not 3, his receiver buffer will look like: [space for 3, packet 4, packet 5, space for 6 and 7]. He can't deliver 4 and 5 to application and can't move its window. ...
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[Data link layer] That confuses me a little bit since the physical layer puts the bits on the wire and collects it from there, which is effectively transferring data between nodes on the same network (please help me cut this more clear if you can).
The physical layer puts bits and bytes on the wire - between two nodes. The data link layer frames them and ...
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The "pseudo header" is a way to include vital information from the IP header in the UDP checksum. You could also call it "virtual header".
Basically, the UDP checksum is calculated on the UDP datagram plus selected fields from the IP header. The UDP handler is provided with sufficient information from the underlying network layer to ...
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A UDP checksum value of zero indicates that the checksum option isn't used (checksum value is not to be verified). Therefore, a calculated checksum of zero is replaced by all ones to indicate that case unambiguously.
See RFC 768:
If the computed checksum is zero, it is transmitted as all ones
(the equivalent in one's complement arithmetic). An all ...
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These layers are just abstract concepts. They don't actively do anything by their own. Instead such models are a tool to deal with complexity, get a common understanding of the functionality and to structure the code in a way which can also be understood and managed by others.
This means there is no "transport layer know ...". There is instead a ...
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I would dare say that you are the victim of the simplistic view presented by network 101 (introduction to networking) courses.
Let me start by quoting from ISO 7498 (the OSI reference model).
The purpose of this Reference Model of Open Systems Interconnection is to provide a common basis for the coordination of standards development for the purpose of ...
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There are 2 pairs of FIN/ACKs because in TCP each direction of the stream is closed separately. When A sends a FIN to B, it means that A will not send any more data to B. The connection can remain open and B can send as many data as it wants. Once B sends the second FIN, the data stream is closed in both directions and connection can be terminated.
One ...
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Models are models - concepts for thought. Like philosophies - sometimes one fits better, some other time the other one.
Converting between TCP/IP model and OSI model doesn't make sense. It's like trying to convert a glass that's half full to a glass that's half empty. It's only a matter of perspective.
The OSI and TCP/IP models are actually very similar. OSI ...
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TCP sequence numbers ... exhausted?
... the maximum value of X is 2 ^ (32).
The given answer makes no sense for me. TCP sequence numbers are not exhausted. They simply wrap around. There is no limit to the amount of data which can be send in a single TCP connection. Also, the initial sequence number is random, i.e. not 0 or 1.
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Realtime applications don't care so much about in-order reception as about in-time delivery = low latency. Roughly speaking, there is only a defined, small window in time where data is useful. Late data is simply useless.
If data is lost you might get a small glitch in video (or audio) but the clip continues to run (=the application is designed to cope with ...
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Short answer: Multimedia implements protocols on top of UDP. They implement the required functionality. Actually, they have more functionality, and are more complex than TCP.
Less short answer: see the answer of @Zac67.
Long answer: I hope that the text below will provide you with a better understanding on this issue.
Transmitting Multimedia over IP
First, ...
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TCP is connection-oriented protocol whereas IP isn't connection-oriented protocol.
Correct.
Any packets before sending into transport layer sorted operation must have been done in network layer.
The transport layer can't get its datagrams/segments anywhere by itself. It requires the routing service done in the network layer.
The re-sorting of out-of-order ...
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I still do not understand the question fully and thus point out where misconception comes from. But, does this clarify things?
Note, that TCP is stream based, i.e., on the sender side (left) upper layer passes the stream of bytes which TCP divides into packets. on the receiver TCP reassembles the stream from packets and passes bytes to the upper layer.
I ...
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When TCP Reno sender gets 3 duplicate ACKs, it enters a so called fast retransmit mode. The idea of this mode is that a) since ACKs are delivered it indicates that packets are leaving the network b) if a packet leaves a network, the sender can inject new packet in the network. Duplicate ACK indicates that a packet has left the network. Fast retransmit ...
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Yes, this is theoretically possible. But in practice, a network stack should be designed to minimize the possibility of this happening with a robust random initial sequence number algorithm.
Yes, the host would accept this traffic as it has no way to tell it's from a previous connection. (hackers use the same trick to break connections, but guessing sequence ...
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Identical sequence numbers in different connections (=sockets) are not a problem. Each socket tracks its own sequence numbers. At any point in time, a socket is unambiguously defined by the sourceIP:sourcePort:destinationIP:destinationPort tuple.
To ensure that a previously used source port isn't reused while data may still be 'in flight', the port is ...
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