Correction accepted. Between the US east and west coasts, the time of flight of packets on fiber or cable is about 23 msec. (Boston-LA, driving route, over fiber, at 207 Mm/sec). So, if all intermediate links are equal in rate, at say, 10 Gb/sec, that means that there should be no more than 10,000,000,000 * 0.023 bits actually in transit on the actual fiber, plus a packet in each router between's outbound queue. Let's say a packet is around 1500 bytes, or 12,000 bits, since that is the MTU we stupidly enforce even today, and there are 10 hops (typical today between Boston and LA.) So we would expect the optimal window size sum, for all flows on any hop of that path to be 10 * 12000 bits + 230,000,000 bits: 230,120,000 bits in flight between BOS and LAX. Divide that by 12,000 bits/packet, and you get about 19,177 packets along that path. At most points along the path, you would expect about 10 different flows or more to be in flight, so there would be, optimally, about 1,918 1500 byte packets. Each flow would get 1 Gb/sec as its share. If the connection is limited to < 1 Gb/sec at either endpoint, then there's no reason for any intermediate node to buffer that much of the flow. This gives a reasonable understanding of where AQM should be ending up in terms of the cwnd needed to sustain max throughput and optimum end-to-end latency (which would be about 23 msec + 10 hops * 12000 / 1 Gb/sec. = 23.012 msec. for a packet to get from one end to the other). On Saturday, May 18, 2019 6:57pm, "Jonathan Morton" <chromati...@gmail.com> said:
> > On 19 May, 2019, at 1:36 am, David P. Reed <dpr...@deepplum.com> > wrote: > > > > Pardon, but cwnd should NEVER be larger than the number of forwarding hops > between source and destination. > > Kleinrock and students recently proved that the optimum cwnd for both > throughput and minimized latency is achieved when there is one packet or less > in > each outbound queue from source to destination (including cross traffic - > meaning > other flows sharing the same outbound queue. > > This argument holds only if time-of-flight *between* nodes is negligible. > Trivially, a geosynchronous satellite hop adds only two nodes but > approximately > half a second to the one-way path delay, with potentially thousands of packets > existing only as radio waves in the distance between, not in a queue. > > - Jonathan Morton > >
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