Yesterday’s sample CCNA exam question attacks EIGRP from a lot of different angles. To answer the question fully, you have to think about the theory behind how EIGRP finds the successor route, any feasible successor routes, what the “feasibility condition” is, a little about how EIGRP calculates metrics, and the requirements for adding multiple equal or unequal metric routes to the routing table. And the question asked all that without showing the output of a single show command, which if included, could have made the question more difficult.
Today, I’ll examine the first two answers, in part just because it takes a while to get through all the little details. I’ll wrap up my comments on the technical parts of the question tomorrow. If you just want to know the answers, I’ve listed them at the bottom of the post, so anyone who didn’t see yesterday’s post (http://www.networkworld.comhttps://www.networkworld.com//community/node/19665) can try out the question first, and answer yesterday’s survey, before seeing the answers.
First, consider answers A and B, as they rely on the same analysis to determine if these answers are correct. The show ip route command lists the local router’s (router A in this case) calculated metric as the 2nd number in brackets. Answers A and B state that this text is for the only route for subnet 1, which means that EIGRP only put one of the three possible routes into the IP routing table. This one route is the EIGRP successor (aka best) route. If this successor route fails, router A can immediately use a feasible successor route if it exists; if no feasible successor route exists, then A must send a Query to its neighbors to verify a loop-free route exists. So, answers A and B boil down to “in these conditions, does a feasible successor route exist, or not?”
Focusing on answer A for a moment, router A’s calculated metric of 12,000 is higher than router B’s calculated metric of 10,000, but lower than the metrics calculated on routers C and D. So, we can infer than router A’s current next-hop router for this route is router B, because the metric for a route through a particular neighbor is a number larger than the metric on that neighbor. As a result, we know that router A’s successor route – the route in router A’s routing table – is the route through router B. Next, the question is whether either of the other two routes (through routers C or D) meet the feasibility condition. Generically, the metric calculated on routers C or D – which is communicated to router A as these routers’ respective “reported distance” – must be better than A’s calculated metric. Whew – more simply stated in this case, are either of C’s and D’s calculated metrics of 13,000 and 15,000, respectively, better than router A’s calculated metric of 12,000? Obviously not, so with the condition in answer A, router A does not have a feasible successor route. As a result, router A must send a query to find a new route when the current route fails.
(Yes, I promise if I do another one of these questions, I’ll number the routers, and letter the answers, so I don’t have to say “router A” and “answer A” in the same thought.)
Answer B follows this same general logic as far as finding a feasible successor. Router A has one successor route, with metric 14,000.Neighboring routers B, C, and D, have calculated metrics of 10,000, 13,000, and 15,000, which they dutifully report to router A. Router A picked one of these three routers as the successor, but no matter which router it is – one of the other two remaining routes will meet the feasibility condition. For example, if router A picked the route through router B as the successor route, then the route through router C, with reported distance 13,000, is a feasible successor route, because 13,000
So far, I’ve chosen answer A as correct, and ruled out answer B as incorrect. As it turns out, A is the only correct answer – more on the other three answers tomorrow.
