Calculating the throughput of a given TCP connection












2















Given a TCP session, is there a way to determine the throughput of the sender?
I have a wireshark sniff, and I see that to calculate the throughput of the sender I can use the
TCP-Window-Size-in-bits / Latency-in-seconds = Bits-per-second-throughput formula,
But the window is constantly changing (due to the tcp protocol).
Furthermore, why does the tcp window size is taken into account? isn't that true that sometimes the sender sends more than one segment before receiving the ack?










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    2















    Given a TCP session, is there a way to determine the throughput of the sender?
    I have a wireshark sniff, and I see that to calculate the throughput of the sender I can use the
    TCP-Window-Size-in-bits / Latency-in-seconds = Bits-per-second-throughput formula,
    But the window is constantly changing (due to the tcp protocol).
    Furthermore, why does the tcp window size is taken into account? isn't that true that sometimes the sender sends more than one segment before receiving the ack?










    share|improve this question

























      2












      2








      2








      Given a TCP session, is there a way to determine the throughput of the sender?
      I have a wireshark sniff, and I see that to calculate the throughput of the sender I can use the
      TCP-Window-Size-in-bits / Latency-in-seconds = Bits-per-second-throughput formula,
      But the window is constantly changing (due to the tcp protocol).
      Furthermore, why does the tcp window size is taken into account? isn't that true that sometimes the sender sends more than one segment before receiving the ack?










      share|improve this question














      Given a TCP session, is there a way to determine the throughput of the sender?
      I have a wireshark sniff, and I see that to calculate the throughput of the sender I can use the
      TCP-Window-Size-in-bits / Latency-in-seconds = Bits-per-second-throughput formula,
      But the window is constantly changing (due to the tcp protocol).
      Furthermore, why does the tcp window size is taken into account? isn't that true that sometimes the sender sends more than one segment before receiving the ack?







      tcp wireshark






      share|improve this question













      share|improve this question











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      share|improve this question










      asked Nov 21 '18 at 11:13









      DsCppDsCpp

      1443




      1443






















          2 Answers
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          4















          isn't that true that sometimes the sender sends more than one segment before receiving the ack?




          This is exactly what "window" means. Imagine a protocol requiring acknowledgment but only sending a single data packet/segment each time (window = 1 segment) - there can only be a single segment/ACK pair in each round-trip period, regardless of the actual bandwidth.



          The send window provides a method to send multiple segments consecutively before expecting ACKs. Each segment that's ACKed is removed from the send window and the window advanced to a new segment - hence "sliding window".



          Sending multiple segments "in parallel" with an adaptive, sliding window, you can make use of the total bandwidth. Generally, you cannot transport more data than one full window within each RTT period.



          TCP uses a sliding window that is controlled by the (constantly monitored) RTT value as congestion sensor - from a simplified perspective, when RTT goes up, the window size goes down and vice versa. Naturally, each value may change at any time due to the network load and other variables.






          share|improve this answer


























          • The windows is not only sliding, but also adapting to the dynamic net, i.e. getting bigger/smaller regards the RTT and packet drop statistics, no? (congestion avoidance algorithms)? do we have a better solution to calculate the throughput than just average the window size over time?

            – DsCpp
            Nov 22 '18 at 8:46











          • The windows size adapts, see my last paragraph. "Throughput" is a instantaneous value, just like RTT and window size. If you average one, you average them all.

            – Zac67
            Nov 22 '18 at 18:20



















          2














          All the measurements are changing: the throughput and the latency can be instantaneous measurements too.




          • If you want an average, measure it over a long period

          • Yes, the sender very often will send segments before getting the acknowledgment






          share|improve this answer























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            2 Answers
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            2 Answers
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            active

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            4















            isn't that true that sometimes the sender sends more than one segment before receiving the ack?




            This is exactly what "window" means. Imagine a protocol requiring acknowledgment but only sending a single data packet/segment each time (window = 1 segment) - there can only be a single segment/ACK pair in each round-trip period, regardless of the actual bandwidth.



            The send window provides a method to send multiple segments consecutively before expecting ACKs. Each segment that's ACKed is removed from the send window and the window advanced to a new segment - hence "sliding window".



            Sending multiple segments "in parallel" with an adaptive, sliding window, you can make use of the total bandwidth. Generally, you cannot transport more data than one full window within each RTT period.



            TCP uses a sliding window that is controlled by the (constantly monitored) RTT value as congestion sensor - from a simplified perspective, when RTT goes up, the window size goes down and vice versa. Naturally, each value may change at any time due to the network load and other variables.






            share|improve this answer


























            • The windows is not only sliding, but also adapting to the dynamic net, i.e. getting bigger/smaller regards the RTT and packet drop statistics, no? (congestion avoidance algorithms)? do we have a better solution to calculate the throughput than just average the window size over time?

              – DsCpp
              Nov 22 '18 at 8:46











            • The windows size adapts, see my last paragraph. "Throughput" is a instantaneous value, just like RTT and window size. If you average one, you average them all.

              – Zac67
              Nov 22 '18 at 18:20
















            4















            isn't that true that sometimes the sender sends more than one segment before receiving the ack?




            This is exactly what "window" means. Imagine a protocol requiring acknowledgment but only sending a single data packet/segment each time (window = 1 segment) - there can only be a single segment/ACK pair in each round-trip period, regardless of the actual bandwidth.



            The send window provides a method to send multiple segments consecutively before expecting ACKs. Each segment that's ACKed is removed from the send window and the window advanced to a new segment - hence "sliding window".



            Sending multiple segments "in parallel" with an adaptive, sliding window, you can make use of the total bandwidth. Generally, you cannot transport more data than one full window within each RTT period.



            TCP uses a sliding window that is controlled by the (constantly monitored) RTT value as congestion sensor - from a simplified perspective, when RTT goes up, the window size goes down and vice versa. Naturally, each value may change at any time due to the network load and other variables.






            share|improve this answer


























            • The windows is not only sliding, but also adapting to the dynamic net, i.e. getting bigger/smaller regards the RTT and packet drop statistics, no? (congestion avoidance algorithms)? do we have a better solution to calculate the throughput than just average the window size over time?

              – DsCpp
              Nov 22 '18 at 8:46











            • The windows size adapts, see my last paragraph. "Throughput" is a instantaneous value, just like RTT and window size. If you average one, you average them all.

              – Zac67
              Nov 22 '18 at 18:20














            4












            4








            4








            isn't that true that sometimes the sender sends more than one segment before receiving the ack?




            This is exactly what "window" means. Imagine a protocol requiring acknowledgment but only sending a single data packet/segment each time (window = 1 segment) - there can only be a single segment/ACK pair in each round-trip period, regardless of the actual bandwidth.



            The send window provides a method to send multiple segments consecutively before expecting ACKs. Each segment that's ACKed is removed from the send window and the window advanced to a new segment - hence "sliding window".



            Sending multiple segments "in parallel" with an adaptive, sliding window, you can make use of the total bandwidth. Generally, you cannot transport more data than one full window within each RTT period.



            TCP uses a sliding window that is controlled by the (constantly monitored) RTT value as congestion sensor - from a simplified perspective, when RTT goes up, the window size goes down and vice versa. Naturally, each value may change at any time due to the network load and other variables.






            share|improve this answer
















            isn't that true that sometimes the sender sends more than one segment before receiving the ack?




            This is exactly what "window" means. Imagine a protocol requiring acknowledgment but only sending a single data packet/segment each time (window = 1 segment) - there can only be a single segment/ACK pair in each round-trip period, regardless of the actual bandwidth.



            The send window provides a method to send multiple segments consecutively before expecting ACKs. Each segment that's ACKed is removed from the send window and the window advanced to a new segment - hence "sliding window".



            Sending multiple segments "in parallel" with an adaptive, sliding window, you can make use of the total bandwidth. Generally, you cannot transport more data than one full window within each RTT period.



            TCP uses a sliding window that is controlled by the (constantly monitored) RTT value as congestion sensor - from a simplified perspective, when RTT goes up, the window size goes down and vice versa. Naturally, each value may change at any time due to the network load and other variables.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 22 '18 at 18:20

























            answered Nov 21 '18 at 15:10









            Zac67Zac67

            30.8k21960




            30.8k21960













            • The windows is not only sliding, but also adapting to the dynamic net, i.e. getting bigger/smaller regards the RTT and packet drop statistics, no? (congestion avoidance algorithms)? do we have a better solution to calculate the throughput than just average the window size over time?

              – DsCpp
              Nov 22 '18 at 8:46











            • The windows size adapts, see my last paragraph. "Throughput" is a instantaneous value, just like RTT and window size. If you average one, you average them all.

              – Zac67
              Nov 22 '18 at 18:20



















            • The windows is not only sliding, but also adapting to the dynamic net, i.e. getting bigger/smaller regards the RTT and packet drop statistics, no? (congestion avoidance algorithms)? do we have a better solution to calculate the throughput than just average the window size over time?

              – DsCpp
              Nov 22 '18 at 8:46











            • The windows size adapts, see my last paragraph. "Throughput" is a instantaneous value, just like RTT and window size. If you average one, you average them all.

              – Zac67
              Nov 22 '18 at 18:20

















            The windows is not only sliding, but also adapting to the dynamic net, i.e. getting bigger/smaller regards the RTT and packet drop statistics, no? (congestion avoidance algorithms)? do we have a better solution to calculate the throughput than just average the window size over time?

            – DsCpp
            Nov 22 '18 at 8:46





            The windows is not only sliding, but also adapting to the dynamic net, i.e. getting bigger/smaller regards the RTT and packet drop statistics, no? (congestion avoidance algorithms)? do we have a better solution to calculate the throughput than just average the window size over time?

            – DsCpp
            Nov 22 '18 at 8:46













            The windows size adapts, see my last paragraph. "Throughput" is a instantaneous value, just like RTT and window size. If you average one, you average them all.

            – Zac67
            Nov 22 '18 at 18:20





            The windows size adapts, see my last paragraph. "Throughput" is a instantaneous value, just like RTT and window size. If you average one, you average them all.

            – Zac67
            Nov 22 '18 at 18:20











            2














            All the measurements are changing: the throughput and the latency can be instantaneous measurements too.




            • If you want an average, measure it over a long period

            • Yes, the sender very often will send segments before getting the acknowledgment






            share|improve this answer




























              2














              All the measurements are changing: the throughput and the latency can be instantaneous measurements too.




              • If you want an average, measure it over a long period

              • Yes, the sender very often will send segments before getting the acknowledgment






              share|improve this answer


























                2












                2








                2







                All the measurements are changing: the throughput and the latency can be instantaneous measurements too.




                • If you want an average, measure it over a long period

                • Yes, the sender very often will send segments before getting the acknowledgment






                share|improve this answer













                All the measurements are changing: the throughput and the latency can be instantaneous measurements too.




                • If you want an average, measure it over a long period

                • Yes, the sender very often will send segments before getting the acknowledgment







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 21 '18 at 13:42









                jonathanjojonathanjo

                11.8k1936




                11.8k1936






























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