[Bloat] Goodput fraction w/ AQM vs bufferbloat

[Jonathan Morton]

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