Why TIME_WAIT state need to be 2MSL long? - tcp

Question

In a TCP connection the end that performs the active close is required to stay in TIME_WAIT state for 2MSL of time. Why exactly does it need to be 2MSL? Many people said that one MSL is for the final ACK and the other MSL is for retransmitted FIN. But, the RTO of FIN is way shorter than MSL, and the FIN doesn't need to wait for an MSL to be retransmitted. So, their explanation doesn't make sense to me. Can anyone present a specific example of how segments are exchanged during that time?

Answer 1

Figure 1. Packet exchange for TCP connection.
Why does the TIME_WAIT state exist?
The book <<UNIX Network Programming(Volume1,3rd)>> give an answer:
There are two reasons for the TIME_WAIT state:
To implement TCP's full-duplex connection termination reliably
To allow old duplicate segments to expire in the network
I think this is also the answer to this question(Why TIME_WAIT state need to be 2MSL long?)
First look at reason 1. In order to reliably terminate the full-duplex connection, suppose that the last ACK sent by the client in Figure 1 is lost, and the server will retransmit the FIN. In order to receive this timeout and retransmitted FIN, the client needs TIME_WAIT Status; does the TIME_WAIT status have to be 2MSL? In fact, this depends on the server-side FIN timeout retransmission time RTO. If RTO is less than MSL, then TIME_WAIT state MSL is enough. If RTO is greater than 2MSL then TIME_WAIT state 2MSL is not enough, so only when RTO is between MSL and 2MSL , Reason 1 for the existence of the TIME_WAIT state is the reason why the time of TIME_WAIT is 2MSL. In fact, in general, RTO is much smaller than MSL, but considering the worst case, RTO is 2MSL, so the TIME_WAIT state is 2MSL to ensure that the worst case can also receive the FIN that is retransmitted over time.
The time of TIME_WAIT is another important reason for 2MSL. Reason 2, in order to ensure that all packets generated during the duration of this connection disappear from the network, that is, to ensure that when a new TCP connection is established, the old duplicate packets from the connection are already in Disappeared from the network. Some people here may have a question: After the client replies to the last ACK, it feels that all packets can disappear with one MSL. Why do all packets of 2MSL disappear? The reason is:
Suppose that the client sends an ACK just after one MSL time, and the server just starts to retransmit the FIN over time before receiving the ACK, so if the FIN disappears, 2MSL is needed.

Answer 2

Let's recall the process
A FIN B
B ACK A
B FIN A
A ACK B
After above 4 steps, A is now at TIME_WAIT stage while B is at LASK_ACK.
What if step 4(A ACK B) is lost?
B will wait for 1 MSL for step 4, yet failed to get it, so B redo step 3, and step 3 will take 1 MSL to reach A. Hence, A has to wait for 2 MSL for a graceful wave goodbyte.
But there is another question, what if step 4 is lost again? :)

Answer 3

I also have the same question about this. I thought of an assumption.
In extreme senario, suppose the last ACK from client end spends MSL to reach server end. At this point, the end point thinks that this ACK has already perished due to MSL timeout. So server end retransmit the FIN imediately. In order to assure this FIN can reach client end (or if not, we have to assure its perishment), we must have another MSL.

Answer 4

The purpose of TIME-WAIT is to prevent delayed packets from one connection being accepted by a later connection. It can happen as:
A connection from (address a, port p) to (address b, port q) is terminated
A second connection from (address a, port p) to (address b, port q) is established
A duplicate packet from the first connection is delayed in the network and arrives at the second connection when its sequence number is in the second connection’s window.
In such cases the endponts have no idea to identify to identify which connection the packet belongs.

Answer 5

The two reasons for the existence of the TIME-WAIT state and the 2SML timer:
If the last ACK segment is lost, the server TCP, which sets a timer for the last FIN (Finish) bit set, assumes that its FIN is lost and resends it. If the client goes to the CLOSED state and closes the connection before the 2MSL timer expires, it never receives this resent FIN segment, and consequently, the server never receives the final ACK. The server cannot close the connection. The 2MSL timer makes the client wait for a duration that is enough time for an ACK to be lost (one SML) and a FIN to arrive (another SML). If during the TIME-WAIT state, a new FIN arrives, the client sends a new ACK and restarts the 2SML timer.
A duplicate segment from one connection might appear in the next one. Assume a client and a server have closed a connection. After a short period, they open a connection with the same socket addresses (same source and destination IP addresses and the same source and destination port numbers). A duplicated segment from the previous connection may arrive in this new connection and be interpreted as belonging to the new connection if there is not enough time between the two connections. To prevent this problem, TCP requires that an incarnation cannot occur unless a 2MSL amount of time has elapsed.

Answer 6

Tcp must prevent old duplicates from a connection from reappearing at some later time and being misinterpreted as belonging to a new the same connection. To do this, Tcp will not initiate a new connection that is currently in the TIME_WAIT state.
The TIME_WAIT state is twice the MSL, this allows MSL seconds for a packet in one direction to be lost, and another MSL seconds for the reply to be lost.