Calculating the throughput of a given TCP connection











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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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    up vote
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    down vote

    favorite












    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
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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






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      asked 19 hours ago









      DsCpp

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          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 - in 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.






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            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






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              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 - in 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

























                up vote
                3
                down vote














                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 - in 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























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote










                  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 - in 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 - in 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












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                  answered 15 hours ago









                  Zac67

                  23.8k21252




                  23.8k21252






















                      up vote
                      1
                      down vote













                      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

























                        up vote
                        1
                        down vote













                        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























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          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 17 hours ago









                          jonathanjo

                          8,4681629




                          8,4681629






























                               

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