OSPF redistribution into RIP

[fran19422]

Hello, regarding the following command which redistributes OSPF into RIP:

redistribute ospf 1 metric 10 match internal external 1 external 2

My question is how do you determine the value for metric i.e in this case it is 10, but what factors would you consider when deciding what value to make this metric ?

Thnak you for any help.

[John Blakley]

The metric is used between routes within the same protocol. Suppose you learn the route 192.168.1.0/24 from 2 different sources. The metric is used to be the tie breaker for both sources. If you learn 192.168.1.0/24 from Router A with a metric of 5, and you learn 192.168.1.0/24 from Router B with a metric of 10, the metric of 5 will be put in the routing table pointing to Router B.

With RIP, you have to use a seed metric to tell it what to redistribute the routes as. The maximum number that you can have is 15 in RIP. If it gets to 16, then RIP deems the route as unreachable and removes the route from the routing table.

HTH,

John

[fran19422]

thank you for your help. I understand what you are saying.

However my question is how would you decide what that seed metric is to be ?

i.e. you can choose a seed metric of anything up to 15. What factors determine what number you choose ?

Thank you.

[John Blakley · Accepted solution]

It would depend on what you're wanting to do. Take the following example:

|````````````B

A                 | ---- C

|_______D

Suppose A is running ospf and is adjacent with B and D. A advertises 192.168.1.0/24 to B and D. C is running RIP with B and D. If you wanted to prefer D over B, you could redistribute ospf with a metric of 5 and redistribute ospf on B with a metric of 10. C would put the better metric in its routing table.

HTH,

John

*** Disclaimer: I didn't test the above topology