[BGP] Understanding Internet Routing Architectures book passage

[Americo Massotti]

Hello guys.

I'll reading the Sam Halabi book: Internet Routing Architectures.

I have a problem to understand a paragraph about the chapter on bgp conflict with internal default route.

Based on this topology

Halabi speaks about Rip generated default:

"In Figure, RTC and RTD can learn a 0/0 default or statically configure a 0/0 default toward their respective providers. Under normal conditions, RTD automatically (or via controlled redistribution) injects the 0/0 into RIP. RTC detects the presence of a default originating from RTD, causing RTC to stop generating a default. All traffic is directed toward RTD."

AD of RIP is 120. Why RTC stop to generate its default? The author presume that RTC don't have a static default or don't learn it with any possible routing process which have minor AD?

"In case of a failure in the NY link, RTD stops generating the default into RIP. RTC detects the loss of 0/0 via RIP and injects its own default. Note that RTC receives the 0/0 default via EBGP (from the locally attached external network), RIP, and possibly IBGP if RTD ispassing the 0/0 in the IBGP session. Because of the higher local preference via RTD, RTC prefers the 0/0 via IBGP. Because RTC's IBGP administrative distance is 200, higher than the RIP distance of 120 (see Table 6-1), the 0/0 default via RIP is preferred because it has the lowest administrative distance."

Why? EBGP AD is 20 and IBGP is 200. Why it should prefere IBGP against EBGP?

Thanks a lot for your support.

Best regards.

[Peter Paluch · Accepted solution]

Hi Americo,

I must admit that the quotes from the book you have posted also leave me wondering. Is there perhaps a complete example configuration of RTC and RTD in that book you could share here?

Either way, I do have a hunch about this - but take it only as a personal opinion.

AD of RIP is 120. Why RTC stop to generate its default? The author presume that RTC don't have a static default or don't learn it with any possible routing process which have minor AD?

I think that the answer is not in administrative distances but rather in additional conditions imposed on injecting the default route from RTC. According to the diagram, RTC provides a backup connection, so it should be used only if the primary connection is not available. I would assume that Sam wanted to say this: "If RTC can see a default route from RTD, it won't originate its own because it should be used only as a backup, and by seeing the default route from RTD, RTC knows the primary connection is still there." How that would be configured, however, is another thing, and that's why I asked if there perhaps is a configuration example in the book to double check this.

Why? EBGP AD is 20 and IBGP is 200. Why it should prefere IBGP against EBGP?

This is a different story, and a common confusion. What Sam is saying is:

• BGP alone on RTC sees two default routes, one from RTD over iBGP (with a higher local preference), and one over eBGP. Based on BGP best path algorithm, BGP on RTC decides - thanks to the higher local preference - that the iBGP route is better, and offers this one to the routing table.
• The routing table has two variants to choose from: the iBGP route offered from BGP (AD=200) and the route from RIP (AD=120). It picks RIP, obviously.

Remember: AD is not a parameter in the BGP bestpath algorithm. The AD of a BGP route is the result of the best path selection (whether the eBGP or iBGP route wins), and not an input to it.

Please feel welcome to ask further!

Best regards,
Peter

[Americo Massotti]

Hi Peter.

Thanks a lot for your support.

No, on the book there isn't a config section about this case.

About my second question. Thanks!

It's true, bgp best path selection algorithm "see" local preference before the path type.

About the first question: Do you tell me in which method you could implemented this type of scenario other that AD value?

Thanks once again.

[Dennis Mink]

here is my take:

"In Figure, RTC and RTD can learn a 0/0 default or statically configure a 0/0 default toward their respective providers. Under normal conditions, RTD automatically (or via controlled redistribution) injects the 0/0 into RIP. RTC detects the presence of a default originating from RTD, causing RTC to stop generating a default. All traffic is directed toward RTD."

because, RTD will receive a defaULT ROUTE through its provider and will send it to RTC using IBGP. at this point rtc will have a default route out of eBGP from its provider and an ibgp default route from RTD, so because of iBGP more preferred AD it will not stick the ebGP learned default route in it routing table.

[paul driver]

Hello

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@Dennis Mink wrote:

because, RTD will receive a defaULT ROUTE through its provider and will send it to RTC using IBGP. at this point rtc will have a default route out of eBGP from its provider and an ibgp default route from RTD, so because of iBGP more preferred AD it will not stick the ebGP learned default route in it routing table.

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@Dennis Mink  this isn't correct RTC will choose its own ebgp default and not the ibgp default from RTD - This is providing no path attribute manipulation such as local preference is being performed 

[Dennis Mink]

@paul driver   thanks for that Paul, yes you are right, eBGP has AD20 and ibgp 200

[paul driver]

Hello

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@Americo Massotti wrote:

Hello guys.

I'll reading the Sam Halabi book: Internet Routing Architectures.

I have a problem to understand a paragraph about the chapter on bgp conflict with internal default route.

Based on this topology

Halabi speaks about Rip generated default:

"In Figure, RTC and RTD can learn a 0/0 default or statically configure a 0/0 default toward their respective providers. Under normal conditions, RTD automatically (or via controlled redistribution) injects the 0/0 into RIP. RTC detects the presence of a default originating from RTD, causing RTC to stop generating a default. All traffic is directed toward RTD."

AD of RIP is 120. Why RTC stop to generate its default? The author presume that RTC don't have a static default or don't learn it with any possible routing process which have minor AD?

"In case of a failure in the NY link, RTD stops generating the default into RIP. RTC detects the loss of 0/0 via RIP and injects its own default. Note that RTC receives the 0/0 default via EBGP (from the locally attached external network), RIP, and possibly IBGP if RTD ispassing the 0/0 in the IBGP session. Because of the higher local preference via RTD, RTC prefers the 0/0 via IBGP. Because RTC's IBGP administrative distance is 200, higher than the RIP distance of 120 (see Table 6-1), the 0/0 default via RIP is preferred because it has the lowest administrative distance."

Why? EBGP AD is 20 and IBGP is 200. Why it should prefere IBGP against EBGP?

Not so sure what assumptions being made however it looks like RTD injects a default into rip based on its own received ebgp default now assuming RTC isn't receiving its own ebgp default, it is then open to receive a default route originated from RTD via either rip or ibgp And due to admin distance preference the rip default is chosen because it has lower AD  value (rip120 v ibgp 200)

However if RTC is ALSO receiving a ebgp default then RTC will choose its own ebgp default

As for RTA & RTB, If both RTC & RTD are advertising a rip default then RTA & RTB will choose their respective upstream rtrs advertised rip default.