A TCP connection is using an MSS=500 bytes. Which of the following could be a realistic/valid value for the congestion window during its operation?
Can anyone tell me how to compute the result?
So like a realistic/valid value of the congestion window should be a integer multiple times of the maximum segment size(MSS).
Click here to see image
According to this image, the congestion window size increase exponentially, which means * 2 every time. For example, if your MSS is 500 bytes, possible congestion window size should be 500, 1000, 2000, 4000 and something like these.
Hope it's helpful. Cheers!
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I am trying to understand TCP protocol with different simulation scenarios. I have to find out the throughput theoretically by just seeing the congestion window of the TCP.
Simulation scenario,
RTT= 100min (6000sec)
Data rate= 10Mbps
Loss=0;
MSS=1460bytes
SImulation Time= 11445005s
I got this congestion window after simulating. . The TCP is always in the Congestion Avoidance phase.
What I tried is to take the current maximum congestion window value which is in bytes and divide it by the RTT value which is 6000s. is it right? or do you have any suggestions? Please help.
In IPERF we have a option to increase the target bandwidth with the option "-b 100m" but in TCP i dont see a option in both JPERF 2.0.2 and also in cli command. Please let me know how can i increase the bandwidth for my throughput testing since i can only receive the traffic at a rate of 20mbps .
Try setting the TCP window with -w. Multiply your desired throughput by the latency to get a starting point for the window value. If you wanted to get 50mbps on a link with 40ms rtt:
50000000 * .04 = 2000000 bytes
For TCP, you cannot set target bandwidth. As for TCP, its sending rate is regulated by flow and congestion control which is determined by RTT and loss. For example, in slow-start phase, the sender can send double number of packets every RTT. In congestion-avoidance state, the congestion window size will be cut by half (or 1/3 in TCP Cubic) once a loss detected.
However, -w can set the sending/receiving window size. If your window size is too small, the total throughput may be bottle-necked by it. So, usually try a large window size, e.g. 65535. Remember a large window size just makes sure your TCP rate would not be bottle-necked by window size, it does not "guarantee" a large throughput.
I submit text file to server capturing by wireShark. my computer's window size is steady to 17408. but server's window size is increasing 6912, 9856, 12800 ...
I want to know why server's window size is increasing. and first TCP segment data is 502 bytes. and the other TCP segment is 1460 bytes.
why window size is increasing? why first TCP segment data is different the other?
As far as I know, the growth of TCP window size is related to the so-called slow start algorithm, which is described in detail here TCP Slow start
Also in the program Wireshark is a definite option to recompile the TCP-packets in accordance with the useful content L7, so that the size of the package there may be, in principle, any
For example, if a set of TCP-packets, which are encapsulated huge HTTP-request, instead of breaking in fragments, it can be shown in one huge fake TCP-packet
This option is called TCP reassemble
I'm going through some revision and I've been stumped by a TCP question. Maybe someone can give me a quick hint or push in the right direction, just so I can get passed this section.
"Why does the sending entity in TCP need to consider the size of the congestion window when determining the sliding window size? "
"Why does the sending entity in TCP need to consider the size of the congestion window when determining the sliding window size? "
This is because the size of the congestion window represents the possible congestion in the network. This is one of the key features offered by TCP. This window is updated in three stages.
In the first stage, when TCP starts, it starts with congestion windows as 1 MSS (Max Segment Size) and then ramps it up in a slow-start manner. TCP sender starts with this value because it is "estimating" how many packets it can send in the network. This phase is also known as slow-start phase. Btw, even though it is called slow-start, TCP increases the packet by doubling the congestion window and the increase happens upon reception of ACKs.
In the second stage, when the congestion window reaches slow-start (ss) threshold (yep, there is one!), TCP sender grows its cogestion window additively -- this is congestion avoidance phase. Here, the sender becomes more cautious. Once again, the increase happens upon reception of ACKs.
In the third stage, when a packet is dropped (one reason would be that a retransmission timeout happened), then TCP cuts its congestion window back to 1 MSS and restarts to grow it again. This is done because a likely congestion was encountered and so cutting back the congestion window would likely freeup the congestion situation along the path. Unlike other stages, the decrease happens due to lack of reception of ACKs.
TCP can use sliding window method to regulate the packets which need to be sent to the receiver. The receiver can also preserve a sliding window to keep track of which packets have been received and which have acked. When determining the sliding window size at the sender side, we should take the congestion window size into account, as we don't want to overwhelm the network channel. The actual traffic in the network is min{awnd,cwnd}, where awnd is the window size which is advertised by the receiver to the receiver, cwnd stands for the congestion window size, whose maximum value will change according to the network condition.
Consider a single TCP (Reno) connection that uses a 10 Mbps link.
Assume this link does not buffer data and that the receiver's receive buffer is much larger than the congestion window.
Let each TCP segment be of size 1500 bytes and the two-way propagation delay of the connection between sender and receiver be 200 msec.
Also, assume that the TCP connection is always in congestion avoidance phase (ignore slow start).
What is the maximum window size in segments that this TCP connection can achieve?
So we know the throughput of the connection and the delay,
I think we can should be able to manipulate the following formula so that we are able to find the Window Size.
Throughput = Window Size / RTT
Throughput * RTT = Window Size
10 Mbps * 200 msec = Window Size
I am not sure if this is correct. I am having a hard time finding anything else that relates in finding Window Size other than this formula.
The Maximum windows size in terms of segments can be up to 2^30/ MSS, where MSS is the maximum segment size. The 2^30 = (2^16*2^14) comes through this as Michael mentioned you in his answer. If your network bandwidth and delay product exceeds than the TCP receiver window size than the window scaling option is enabled for the TCP connection and most OS support this features. The scaling supports up to 14-bit multiplicative shift for the window size. You can read following for the better explanation:
http://en.wikipedia.org/wiki/TCP_window_scale_option
http://www.ietf.org/rfc/rfc1323.txt
I think what you are asking is how data can I get end to end on the wire. In that case you are close. Throughput*RTT [units: B/S * S] is how much the wire holds. Ignoring PMTU, packet overhead, hardware encoding, etc. then Throughput*RTT/PacketSize would give you the estimate. But hold on, I used RTT. My receive window is really about how much can fit on the wire in one direction so divide that in half.
If your implementation doesn't support window scaling then min that with 2^16. If it does then you min it with 2^30.
A packets will be dropped if the maximum sending rate exceeds link capacity
(max window size*size of 1 segment) / RTT = link capacity
(max window size * 1500*8) / 200*10^-3 = 10 * 10^-6
you can solve this for max window size.
We divide by the RTT because after this time an ACK will be received so the sender can send more segments without the need to increase the window size.