Hi, I suggest you look at the following paper for a more
general version of this formula (equation 3), which includes the
effect of limited capacity and/or limited receive-window:
http://www.cc.gatech.edu/fac/Constantinos.Dovrolis/Papers/f235-he.pdf
The paper also discusses common mistakes when this formula is used
to predict the throughput of a TCP connection - the basic
idea is that we cannot use the loss rate *before* the start
of a TCP connection to predict what its throughput will be.
A large TCP connection that is not limited by its receive-window
can of course cause an increase in the loss rate of the path
that it traverses (see sections 3.2 - 3.4)
regards
Constantine
On 5/7/2011 8:15 PM, Stephen Hemminger wrote:
On Sat, 7 May 2011 19:39:22 +0300
Jonathan Morton<[email protected]> wrote:
On 7 May, 2011, at 1:10 am, Stephen Hemminger wrote:
Rate<= (MSS/RTT)*(1 / sqrt{p})
where:
Rate: is the TCP transfer rate or throughputd
MSS: is the maximum segment size (fixed for each Internet path, typically 1460
bytes)
RTT: is the round trip time (as measured by TCP)
p: is the packet loss rate.
So if the loss rate is 1.0 (100%), the throughput is MSS/RTT. If the loss rate
is 0, the throughput goes to infinity. That doesn't seem right to me.
If loss rate is 0 there is no upper bound on TCP due to loss.
There are other limits on TCP throughput like window size but not limits
because of loss.
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Constantine
--------------------------------------------------------------
Constantine Dovrolis, Associate Professor
College of Computing, Georgia Institute of Technology
3346 KACB, 404-385-4205, [email protected]
http://www.cc.gatech.edu/~dovrolis/
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