Calculating the throughput of a given TCP connection
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 theTCP-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
asked Nov 21 '18 at 11:13
DsCpp
1443
add a comment |
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 theTCP-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
asked Nov 21 '18 at 11:13
DsCpp
1443
add a comment |
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 theTCP-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
asked Nov 21 '18 at 11:13
DsCpp
1443
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 theTCP-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
tcp wireshark
asked Nov 21 '18 at 11:13
DsCpp
1443
asked Nov 21 '18 at 11:13
DsCpp
1443
asked Nov 21 '18 at 11:13
DsCpp
1443
asked Nov 21 '18 at 11:13
DsCpp
1443
asked Nov 21 '18 at 11:13
DsCpp
1443
1443
add a comment |
add a comment |
2 Answers
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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 - 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.
answered Nov 21 '18 at 15:10
Zac67
26.2k21352
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
add a comment |
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
answered Nov 21 '18 at 13:42
jonathanjo
10.9k1834
add a comment |
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2 Answers
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active
oldest
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2 Answers
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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 - 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.
answered Nov 21 '18 at 15:10
Zac67
26.2k21352
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
add a comment |
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.
answered Nov 21 '18 at 15:10
Zac67
26.2k21352
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
add a comment |
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.
answered Nov 21 '18 at 15:10
Zac67
26.2k21352
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.
answered Nov 21 '18 at 15:10
Zac67
26.2k21352
edited Nov 22 '18 at 18:20
answered Nov 21 '18 at 15:10
Zac67
26.2k21352
answered Nov 21 '18 at 15:10
Zac67
26.2k21352
answered Nov 21 '18 at 15:10
Zac67
26.2k21352
26.2k21352
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
add a comment |
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
add a comment |
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
answered Nov 21 '18 at 13:42
jonathanjo
10.9k1834
add a comment |
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
answered Nov 21 '18 at 13:42
jonathanjo
10.9k1834
add a comment |
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
answered Nov 21 '18 at 13:42
jonathanjo
10.9k1834
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
answered Nov 21 '18 at 13:42
jonathanjo
10.9k1834
answered Nov 21 '18 at 13:42
jonathanjo
10.9k1834
answered Nov 21 '18 at 13:42
jonathanjo
10.9k1834
answered Nov 21 '18 at 13:42
jonathanjo
10.9k1834
10.9k1834
add a comment |
add a comment |
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