Refer to the exhibit.
If a connection failure occurs between R1 and R2, which two actions can you take to allow CR-1 to
reach the subnet 192.168.192.0/24 on R2? (Choose two.)
A.
Create a static route on R1 for subnet 192.168.192.0/24 towards R3 and redistribute it into OSPF.
B.
Turn up a BGP session between CR-1 and R1.
C.
Create a static route on R1 for subnet 192.168.192.0/24 towards R3 and redistribute it into BGP.
D.
Turn up an EIGRP session between R1 and R3 with AS 65535.
E.
Create an OSPF virtual link between CR-1 and R2 to bypass R1.
A and B are correct.
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It could be B and C.
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BGP has already know about network 192.168… so static route is not needed.
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shit diagram, I thing the R1-R3 link is iBGP but the BGP oval doesn’t touch it!
A yes
B no, R3 wouldn’t forward iBGP routes
Sloppy question all round
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A yes. It it a route to R2 from CR-1.
Dose R2 have a route to CR-1 ?
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R1, R2 and R3 are all in the same AS so it is iBGP. So R2 will tell both R1 and R3 about it’s route. But when R1’s link to R2 goes down, R1 is no longer learning the route from R2 and since this is iBGP R3 will NOT pass it on to R1 so R1 no longer knows how to get to the subnet hanging off R2. Two things need to happen. R1 needs the static route pointing to R3 to be able to find the subnet off R2 and once it has that THEN turning up eBGP between CR-1 and R1 works. Answers A & B are correct, they don’t work by themselves but they solve the problem when implemented together.
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Well A will work by itself since it is redistributing into OSPF if it specifies redistributing internal ..
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I agreed with Answer A, but itself is not enough, because return traffic from 192.168.192.0/24 could be blocked in R2 or R3, since they dont have routes to CR-1
I disagree about answer B, because it is useless if you implement answr A,
plus R1 and R3 are not BGP peers, so we dont expect any dynamic route exchange
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