Understanding on Task 2.2 Redistribution
Can someone explain the following solution stated in SG:
If OSPF's external AD is set higher than EIGRP's the RIP routes redistributed into OSPF on R5 will loop back into OSPF by means of mutual redistribution on R1/R2 and preempt RIP routes in R5.
My question is, if the question doesn't require R5 to prefer to reach those prefixes learned from BB2 via RIP.
Would the above solution still needed?
Of course, in order to prevent loop from happening. only thing we need to do is to set external OSPF AD to 171 for R1,R2 and R5.
Can someone care to explain?