One of the questions I get asked several times a week by my clients is as such. While most people never associate latency with the amount of data they can send across a single TCP stream, engineers need to understand this concept.
Luckily, someone has done most of the leg work for us. Many years ago, a college forwarded me a thesis paper written by Jesper Dangaard Brouer from Copenhagen University in Denmark. While this paper mainly focuses on issues concerning ADSL, he does analyze our exact question on page 21. (28 on the pdf)
Throughput = Window Size / RTT
This very simple formula is the key to your single stream transmission rates. Your throughput on a single stream TCP connection is limited to the WindowSize divided by your RTT or latency. So lets try this out…
PING 6.6.7.4 (6.6.7.4) 56(84) bytes of data.
64 bytes from 6.6.7.4: icmp_seq=1 ttl=51 time=69.0 ms
64 bytes from 6.6.7.4: icmp_seq=2 ttl=51 time=69.6 ms
64 bytes from 6.6.7.4: icmp_seq=3 ttl=51 time=67.1 ms
64 bytes from 6.6.7.4: icmp_seq=4 ttl=51 time=69.1 ms
64 bytes from 6.6.7.4: icmp_seq=5 ttl=51 time=69.2 ms
--- 6.6.7.4 ping statistics ---
5 packets transmitted, 5 received, 0% packet loss, time 4005ms
rtt min/avg/max/mdev = 67.113/<b>68.848</b>/69.687/0.954 ms
That shows our connections average RTT is 69ms. And you should by now know that your TCP Window size for Ethernet is 65,535 bytes. When we plug this data into our formula, it shows..
65535 / 69 = 949.78 kBytes/s
Or about 7.4mbit per second. Lets go ahead and see what kind of data we can come up with by running our own test.
Transmitted 8048000 bytes in 8.400634 seconds (902.094 kbytes/s. <b>7216.748</b> kbits/s)
Thats pretty close, 7.2 mbit per second. Now, these results wont always come out as perfect, but it will give you a good idea of what to expect. Just remember, QOS will affect your results, so please disable it while testing. If you have any questions, please feel free to leave a comment. Thanks you!
note: ip address changed to protect public networks
Who writes this crap?
Tony Mattke is a network engineer for a financial institution in Indiana. In the past he has worked for ISPs, data centers, networking manufactures, and the occasional enterprise. For feedback, please leave a comment on the article in question. For everything else including fan mail or death threats, contact him via twitter.
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{ 3 comments… read them below or add one }
Do you mean 7.4 kbit per second?… 949.78 kBytes = 949.78 * 8 -> kbits
No, my statement was correct. The formula outputs the maximum theoretical throughput in bytes.
Your math is in error. Rounding 949 KBytes to 1000 KBytes gives the following:
1000 KB * 8 bits/Byte = 1000 * 8 Kbits = 8000 Kb
1000 K = 1 M so that 8000 Kbits = 8 Mbits
Thus the authors statement 7.4 (or above 8) Mb/s