North American Network Operators Group

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RE: multi-homing fixed

  • From: David Hares
  • Date: Tue Aug 28 12:21:27 2001

Actually, bandwidth to 2 POPs can often be had for close to the same price
as one.  ATM bandwidth pricing is close to linear.  One physical port, 2
VPI/VCI paths, one to each of two POPs, each with half the bandwidth.  If
you were willing to risk one port/one path before you can now do two (or
more).  You're right though if you want sperate last mile paths, but that's
an ILEC pricing problem.  Since we were talking about multi-homing, my
assumption was two ports/two paths were a given.  Destination of the paths
is what I was talking about.

> -----Original Message-----
> From: [email protected] [mailto:[email protected]]On Behalf Of
> Sean M. Doran
> Sent: Tuesday, August 28, 2001 11:56 AM
> To: [email protected]; [email protected]
> Subject: Re: multi-homing fixed
>
>
>
>
> | I've asked this of a number of people now, but how many providers
> | have multiple POP's in a city that are _completely redundant_?
> | That is, they can operate _fully_ with one POP out of service?
>
> The suitcase nuclear bomb that takes out my facility in one location
> is very likely to take out my other facility in the same metro area.
> And probably all the customers who would remotely care.
>
> | Even if they have two pops, many of those cities won't have redundant
> | long haul capacity.
>
> The bigger problem is actually that many of those cities will have zero
> customers who are willing to pay extra to be connected to a router
> in POP-A and a router in POP-B, and I know of no way to make such
> a connection cost the same or less than a single connection to only
> one of those routers.   So, while it's _possible_ to make the Big
> Red Switches'
> states very nearly totally invisible to any customer, it's not often done.
>
> But, hey, deep-pockets people have found out that several suppliers
> are eminently capable of engineering (nearly) all the resilience and
> redundundancy that the deep-pockets can manage to pay for...
>
> 	Sean.