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.
- Jonathan
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