[183181] in North American Network Operators' Group
Re: Drops in Core
daemon@ATHENA.MIT.EDU (Justin Wilson - MTIN)
Mon Aug 17 10:51:39 2015
X-Original-To: nanog@nanog.org
From: Justin Wilson - MTIN <lists@mtin.net>
In-Reply-To: <571A31DB-4295-4872-AF6F-BA4280CD39A7@ianai.net>
Date: Mon, 17 Aug 2015 10:51:34 -0400
To: "Patrick W. Gilmore" <patrick@ianai.net>
Cc: NANOG list <nanog@nanog.org>
Errors-To: nanog-bounces@nanog.org
I could see it going through several private peering, but not through =
multiple exchanges. =20
Justin Wilson
j2sw@mtin.net
---
http://www.mtin.net Owner/CEO
xISP Solutions- Consulting =E2=80=93 Data Centers - Bandwidth
http://www.midwest-ix.com COO/Chairman
Internet Exchange - Peering - Distributed Fabric
> On Aug 16, 2015, at 8:00 AM, Patrick W. Gilmore <patrick@ianai.net> =
wrote:
>=20
> On Aug 15, 2015, at 1:41 PM, Job Snijders <job@instituut.net> wrote:
>> On Sat, Aug 15, 2015 at 11:01:56PM +0530, Glen Kent wrote:
>=20
>>> Is there a paper or a presentation that discusses the drops in the =
core?
>>>=20
>>> If i were to break the total path into three legs -- the first, =
middle
>>> and the last, then are you saying that the probability of packet =
loss
>>> is perhaps 1/3 in each leg (because the packet passes through
>>> different IXes).
>>=20
>> It is unlikely packets pass through an IXP more then once.
>=20
> =E2=80=9CUnlikely=E2=80=9D? That=E2=80=99s putting it mildly.
>=20
> Unless someone is selling transit over an IX, I do not see how it can =
happen. And I would characterize transit over IXes far more =
pessimistically than =E2=80=9Cunlikely=E2=80=9D.
>=20
>=20
> [Combining responses]
> On Aug 15, 2015, at 1:21 PM, Owen DeLong <owen@delong.com> wrote:
>>=20
>> I would say that the probability of a packet drop at any particular =
peering
>> point is less than the probability at one of the two edges.
>>=20
>> However, given that most packets are likely to traverse multiple =
peering
>> points between the two edges, the probability of a packet drop along
>> the way at one of the several peering points overall is roughly equal
>> to the probability of a drop at one of the two edges.
>=20
> I=E2=80=99m a little confused why most packets are =E2=80=9Clikely to =
traverse multiple peering points=E2=80=9D?
>=20
> Most packets these days are sourced from one of three companies. =
(Which Owen should know well. :) At least one of those companies =
published stats saying the vast majority of packets are =E2=80=9Czero or =
one=E2=80=9D AS hop from the destination. I cannot imagine Google or =
Netflix being 50% behind Akamai on that stat. Which clearly implies most =
packets traverse =E2=80=9Czero or one=E2=80=9D AS hop - i.e. one or zero =
peering points.
>=20
> Finally, I would love to see data backing up the statement that =
packets are more likely to drop at one edge (assuming the destination?) =
than at a peering point.
>=20
> --=20
> TTFN,
> patrick
>=20
Justin Wilson
j2sw@mtin.net
---
http://www.mtin.net Owner/CEO
xISP Solutions- Consulting =E2=80=93 Data Centers - Bandwidth
http://www.midwest-ix.com COO/Chairman
Internet Exchange - Peering - Distributed Fabric
