Refer to the exhibit. Traffic from R1 to the 10.10.2.0/24 subnet uses 192.168.1.2 as its next hop. A network engineer wants to update the R1 configuration so that traffic with destination 10.10 2.1 passes through router R3, and all other traffic to the 10.10.2.0/24 subnet passes through R2. Which command must be used?
To ensure traffic to the specific host 10.10.2.1 passes through router R3, the command must create a host route with a subnet mask of 255.255.255.255. Administrative Distance (AD) typically doesn't need to be altered since a static route's default AD is 1, which is lower than OSPF's AD of 110. Therefore, the command 'ip route 10.10.2.1 255.255.255.255 192.168.1.4' will achieve the desired result, directing traffic for 10.10.2.1 through R3 without requiring an explicit AD change.
I dont understand why A and D are different. Host routes (/32) win because they have the longest prefix. Here, both A and D include the destination host, but they differ on AD. If you set the AD to 115, its higher than OSPF (110) but it will still prefer the longest prefix route, so i believe that A and D are both correct in this scenario. Correct me if im wrong.
Longest prefix-length takes precedence for *route selection*, while administrative distance takes precedence when deciding *which routes are installed into the routing table*. R1 has a route for the entire 10.10.2.0/24 subnet, learned via OSPF with the default AD of 110. 10.10.2.1/32 falls within the subnet 10.10.2.0/24, and so adding it with an AD of 115 means it won't get installed into the routing table. https://www.cisco.com/c/en/us/support/docs/ip/enhanced-interior-gateway-routing-protocol-eigrp/8651-21.html
It will get installed in the routing table even if it has an AD higher than 110. I simulated the network in packet tracer and both A and D work.
by the way, AD of OSFP should be 110
We don’t need to change the default AD
I feel like the route would work either with or without the AD(administrative distance). I can see the reason why the answer is choosing the option with lower AD than 110(AD of OSPF), to prevent 10.10.2.1 route to R2 like all the other hosts in 10.10.2.0 network. But just like other two comments said, we apply 10.10.2.1/32, the longest mask always win, so AD shouldnt be necessary.
Am i the only person seeing 192.168.1."4100" please, where from the last octect?
No, i do too. A space is missing
In this case, I don't think the AD makes any difference (it can be anyone), because the longest mask always wins, it has the "/32 prefix", sending it to a specific host 10.10.2.1, it will go through R3 with any AD. And any other host on the 10.10.2.0/24 network will automatically go the OSPF route through R2. You can ignore the ADs, they are for distraction.
how can be D the correct answer? the 10.10.2.1 is a user IP address, it should be 10.10.2.0.
by default static route has AD 1, so no need to change AD to 100 if OSPF AD is 115
Answer A The ad should be higher that route 110 ospf learned
definitely not, if the ad was higher, all the traffic to the subnet would go through R2, and the end host would never get traffic
alejandro please don't share your wisdow... its not right to teach people untrue facts.
It would be best for everyone if you just shut up.
Actually there is no answer here because OFSP defaultad is not 115 but 110. So there should be less than 100 AD. But most correct answer is D
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Both A & D would yield the same answer because the Longest prefix match takes precedence over administrative distance. I would pick D since it's a more suitable route. https://packetlife.net/blog/2010/aug/16/route-preference/#:~:text=Prefix%20Length%20%2D%20The%20longest%2Dmatching,lowest%20administrative%20distance%20is%20preferred.
D is correct
i'm on 491 question and didn't see a lot of ACL questions ! everything is just about subnetting