A use case for a longer prefix with the same nexthop:
F
/ \
D E
> >
B C
\ /
A
Suppose A is a customer of B and C.
B has a large address space: 10.1.0.0/16.
B allocates a subset to A: 10.1.1.0/24.
B advertises the longer prefix to its backup provider C.
C propagates it to E and then to F.
B MUST advertise both 10.1.0.0/16 and 10.1.1.0/24 to D.
D MUST propagate both of them to F.
Otherwise, if F only receives 10.1.0.0/16 from D, then
F will have the longer match 10.1.1.0/24 to E,
but E is only the backup route.
Thanks,
Jakob.
A use case for a longer prefix with the same nexthop:
F
/ \
D E
> >
B C
\ /
A
Suppose A is a customer of B and C.
This is possible, but only remotely probable. In the real world, D and E are
likely peered, as are B and C. Further, it's quite possible for F to choose
the path through E anyway, regardless of A's wishes, or even to load share
over to the two paths. If it's really a backup path, and you don't want
traffic on it unless the primary is completely down, then you need to not
advertise it until you actually need it. One of the various principles of
packet based routing is that if you advertise reachability, it means
someone, someplace, might just choose the path you've advertised. You can't
control what other people choose.
Russ
Simpler, with B and C peered:
F
/ \
B---C
\ /
A
If B does not send the /24 to F,
then F will send all the traffic to C,
even if A wanted a load balance.
Maybe I could ask the community:
Why do you advertise longer prefixes with the
same nexthop as the shoter prefix?
Is it this use case, or something else?
Thanks,
Jakob.
If B does not send the /24 to F,
then F will send all the traffic to C,
even if A wanted a load balance.
Maybe I could ask the community:
Why do you advertise longer prefixes with the
same nexthop as the shoter prefix?
Is it this use case, or something else?
it is a common TE use case. but folk watching the water rise
are starting to ask why the whole world should pay for A's TE.
randy
F
/ \
D E
> >
B C
\ /
A
Suppose A is a customer of B and C.
This is possible, but only remotely probable. In the real world, D and
E are likely peered, as are B and C.
"likely?" with what probability? any measurement cite please. nothing
exact; something rough would be fine.
randy
>> F
>> / \
>> D E
>> > >
>> B C
>> \ /
>> A
>>
>> Suppose A is a customer of B and C.
>
> This is possible, but only remotely probable. In the real world, D and
> E are likely peered, as are B and C.
"likely?" with what probability? any measurement cite please. nothing
exact; something rough would be fine.
Well, the average AS Path length is something like 4, and according to the
charts Geoff has presented here and there, the graph is becoming more dense,
as most people interconnect. The odds of finding an end-to-end path (4 hops)
on the global 'net where no-one is peered in the middle seems pretty
unlikely to me. It's not impossible, but it does seem unlikely, just given
the average AS Path length and the density of the graph. For example, I
suppose you could make A/B/C part of the same network which is intentionally
not peered, or B/C two regional providers who are not peered with one
another. You could then make D/E IXPs who have no transit connectivity
between them, and then make F a tier 1 provider... But this really seems
unlikely to me. How would you string together 4 AS' in a row that have no
connectivity to any transit AS, even regional, like this? Two hops I can
see, four I have a hard time seeing.
it is a common TE use case. but folk watching the water rise are starting
to ask why the whole world should pay for A's TE.
Precisely. Tragedy of the commons. To put it in other terms, removing
information reduces optimization -- but if I can get optimization by making
someone else pay for the information, then, well, why not?
Russ
F
/ \
D E
> >
B C
\ /
A
Suppose A is a customer of B and C.
This is possible, but only remotely probable. In the real world, D and
E are likely peered, as are B and C.
"likely?" with what probability? any measurement cite please. nothing
exact; something rough would be fine.
Well, the average AS Path length is something like 4, and according to
the charts Geoff has presented here and there, the graph is becoming
more dense, as most people interconnect. The odds of finding an
end-to-end path (4 hops) on the global 'net where no-one is peered in
the middle seems pretty unlikely to me. It's not impossible, but it
does seem unlikely, just given the average AS Path length and the
density of the graph. For example, I suppose you could make A/B/C part
of the same network which is intentionally not peered, or B/C two
regional providers who are not peered with one another. You could then
make D/E IXPs who have no transit connectivity between them, and then
make F a tier 1 provider... But this really seems unlikely to me. How
would you string together 4 AS' in a row that have no connectivity to
any transit AS, even regional, like this? Two hops I can see, four I
have a hard time seeing.
i was hoping for measurements, not seems unlikely.
as you know, i am sceptical about our internet topology intuitions and
modeling given how good bgp is at hiding information and how poor our
vantage points are. ripe atlas, caida, etc. give us some view, but
views with inconsistencies and contradictions. we could write a paper
on the hazards of as topology. oh, we did.
randy
Am I the only one thinking "RFC4264" here?