I suspect that they will be going completely switched and have very few
total points on the whole network (depending on what configuration they
go to). A network of OC12s or OC48s in a redundant star will have
significant performance benefits because 1) no routing, or very symmetric
routing. 2) very low latency <8ms coast to coast I'd suspect. 3) priority
queues, quality of service, reserved bandwidth, etc.
#0: well, nobody's physical topology is shaped like a redundant star.
at very fast rates, for (2) above in particular, having the network
topology follow rights-of-way is a good plan
#1: *something* somewhere has to do routing, and has to know the
full details of what the topology looks like and make decisions
based on that information and any constraints. Problem A in
the current Internet is that boxes that have to know everything
are feeling the crunch of complex topology and instability;
I don't see how Internet-II can get around this, and moreover,
I do see how easily additional parallel infrastructure can worsen
Problem A for everyone (and in particular for the Internet-II connectees).
Also, have you noticed Vern Paxson and Van Jacobson saying that
path symmetry really doesn't seem to matter with respect to
TCP throughput, or that path asymmetries even on point-to-point
circuits has existed for some time in the real world? I dunno,
maybe very symmetric routing might help, but aren't you building
some bad failure modes by trying to force this by building a star?
#2: Are you planning on building photon accelerators?
BTW, what kind of delay do you get between Stanford and UCSD if
you have a star topology? You can assume that your physical
topology is built out of the four principal right-of-way carriers'
fabric, whether leased or otherwise.
#3: What type of per-packet and per-flow time budgets do you think
you have at OC12 and OC48? If you can make all this happen
with existing technology (or can make new technology), then there
are at least four companies that will want your resume ASAP, particularly
if you can do this and accomplish your #2 (assuming you accidentally
missed a digit) simultaneously.