Refer to the exhibit. After redistribution is enabled between the routing protocols; PC2, PC3, and PC4 cannot reach PC1. Which action can the engineer take to solve the issue so that all the PCs are reachable?
A.
Set the administrative distance 100 under the RIP process on R2.
B.
Filter the prefix 10.1.1.0/24 when redistributed from OSPF to EIGRP.
C.
Filter the prefix 10.1.1.0/24 when redistributed from RIP to EIGRP.
D.
Redistribute the directly connected interfaces on R2.
100% agree. Labbed it. Reducing AD to 100 will always provide an exit for packets to 10.1.1.0/24. When this route is in EIGRP it's external, so it will be treated with AD=170. Within OSPF AD=110. Reduce RIP down to 100 then on R2 10.1.1.0/24 will always point out towards RIP domain.
After making lab you'll see that problem happens on R2; it redistributes RIP into EIGRP, which then gets redistributed into OSPF at R3. So R2 learns from R3 an OSPF E2 route with an AD of 110. It inserts 10.1.1.0/24 >> R3 into the RIB because the OSPF AD is lower than the RIP learned AD.
Thus lower RIP AD to 100 and it will be preferred over the OSPF route.
What is being redistributed on R2? I may be missing something but when I labbed the below all routers could ping the 10.1.1.1 interface on R1. There were no redistributions on R4.
R2(config)#do sho run | s router
router eigrp 100
network 24.24.24.2 0.0.0.0
redistribute rip metric 1000000 10 255 1 1500
router ospf 100
redistribute rip
network 23.23.23.2 0.0.0.0 area 0
router rip
version 2
redistribute ospf 100 metric 2
redistribute eigrp 100 metric 2
network 10.0.0.0
network 12.0.0.0
neighbor 12.12.12.1
R2(config)#
R3#sho run | s router
router eigrp 100
network 34.34.34.0 0.0.0.255
redistribute ospf 100 metric 1000000 10 255 1 1500
router ospf 100
redistribute eigrp 100
network 10.3.3.3 0.0.0.0 area 0
network 23.23.23.3 0.0.0.0 area
What's result your config?
B correct: It seems there is a loop because of mutual redistributions among RIP, OSPF and EIGRP domains. So we should filter out the prefix 10.1.1.0/24 when redistributed from OSPF to EIGRP (the second redistribution point) to prevent routing loop.
HETKAR and Alnet are correct, and I just want to add simple clarification
D - does not make any sense
C - if you filter prefix 10.1.1.0/24 from RIP to EIGRP, this network becomes unreachable on R3 and R4.
B - there is no redistribution from RIP to OSPF that won't work as OSPF does not learn the prefix from RIP.
A (Correct) When the traffic goes to the R2, the router will have a choice - sent to the R1 (RIP AD is 120) or to R3 (OSPF 110). OSPF is learning the route from R4 via redistribution.
Route with the lover AD will be injected into the routing table.
So if RIP's AD will not be changed to 100, R2 will forward the traffic to R3, so the packet will not reach R1.
I was confused a bit when I saw this scheme first time. R2 OSPF is redistributed into the RIP and RIP redistributed into EIGRP
I hope it helps
When OSPF->EIGRP redistribute, by default OSPF router R3 does not send external-RIP subnet 10.1.1.0/24 (OSPF LSA type2 E2). So no need to worry about OSPF->EIGRP redistribution.
So Option.B gone
Correct Answer is A. Traffic flows for the prefix 10.1.1.10 from R1(RIP)--> R2 (RIP) --> R4 (EIGRP 170) -----> R3 (OSPF 110) --------> R2 (OSPF 110) In R2 OSPF AD(110) is preferred than RIP AD(120). Because of that R2 flushes the RIP prefix(10.1.1.10) out of its routing table. So R4 and R3 will flush that prefix consecutively.
The core issue here is that When the RIP route redistributed from eigrp into OSPF on R3, R2 that is running OSPF will install the route as have better AD, causing the loop. The solution here then is use a distribute list to do not intall the EIGRP route ont the ospf procees and avoid the loop.
Redistribution of subnet 10.1.1.0/24 on from RIP to OSPF on R2 with create an OSPF route to 10.1.1.0/24 with AD of 110. This one is preffered over the RIP route to 10.1.1.0/24 with AD 120.
Redistribution of RIP route 10.1.1.0/24 to EIGRP on R2 creates an external EIGRP route with AD 170. This route will not be chosen anyway.
So if you set the AD of RIP to 100 on R2 that route is chosen to reach 10.1.1.0/24.
It may seem that B solves the problem, which makes it difficult to choose an answer.
But B removes only the EIGRP route that passed from R2 in a clockwise direction.
The main problem is that on R2 there is an external OSPF route(counterclock-wise) that pulls all attempts to get to the PC1.
If in answer B we swapped OSPF and EIGRP, then it would fit. .
if the red arrow in the picture is a symbol to a redistribution, what we should have in R2 are : OSPF is redistributed into RIP, RIP is redistributed into EIGRP and an arrow between OSPF and EIGRP that i can't see where it is pointing at.
A - solves the problem because R2 trusts RI and sends the 10.1.1.0 traffic to the left.
B - doesn't help because the problem is caused by redistributing the route from EIGRP to OSPF, not the other way around.
C - suppresses the 10.1.1.0 route from EIGRP, thereby preventing PC4 from reaching the subnet.
D - just ridiculous nonsense obviously.
Because when R2 is trying to get to 10.1.1.1, it will always go LEFT to the correct destination if it trusts the RIP routes over the OSPF routes. With RIP having AD of 100 it is trusted over OSPF which is 110.
Looks like no answer is correct. Just labbed it. It works when you set an AD of 171 for OSPF routes on R2. But once the RIP distance is set to 100 on R2 instead R1 loses connectivity to PC3 because there is no redistribution from OSPF into RIP and the best route to PC3 in R2s routing table is via OSPF. So all routes from R3 must traverse EIGRP (since this is redistributed into RIP afterwards) which means that these routes must be in R2s routing table available via EIGRP. Nonetheless A makes most sense.
r2#show running-config | s router
router eigrp 1
default-metric 1000000 1 255 1 1500
network 24.24.24.0 0.0.0.255
redistribute rip route-map set-rip-tag
router ospf 1
network 23.23.23.0 0.0.0.255 area 0
distance 171
router rip
version 2
redistribute eigrp 1 metric 1
network 10.0.0.0
network 12.0.0.0
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