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Exam 300-101 topic 2 question 536 discussion

Actual exam question from Cisco's 300-101
Question #: 536
Topic #: 2
[All 300-101 Questions]

Refer to the exhibit.

BigBids Incorporated is a worldwide auction provider. The network uses EIGRP as its routing protocol throughout the corporation. The network administrator does not understand the convergence of EIGRP. Using the output of the show ip eigrp topology all-links command, answer the administrator's question.
Which two networks does the Core1 device have feasible successors for? (Choose two)

  • A. 172.17.0.0/30
  • B. 172.17.1.0/24
  • C. 172.17.2.0/24
  • D. 172.17.3.0/25
  • E. 172.17.3.128/25
Show Suggested Answer Hide Answer
Suggested Answer: F. 10.140.0.0/24 🗳️
Explanation -
To understand the output of the show ip eigrp topology all-links command command, let`s analyze an entry (we choose the second entry because it is better for demonstration than the first one)

The first line tells us there is only 1 successor for the path to 10.140.0.0/24 network but there are 2 lines below. So we can deduce that one line is used for successor and the other is used for another route to that network. Each of these two lines has 2 parameters: the first one (156160 or 157720) is the Feasible
Distance (FD) and the second (128256 or 155160) is the Advertised Distance (AD) of that route.
The next thing we want to know is: if the route via 172.17.10.2 (the last line) would become the feasible successor for the 10.140.0.0/24 network. To figure out, we have to compare the Advertised Distance of that route with the Feasible Distance of the successor`s route, if AD < FD then it will become the feasible successor.
In this case, because AD (155160) < FD (156160) so it will become the feasible successor. Therefore we can conclude the network 10.140.0.0/24 has 1 feasible successor.
Because the question asks about feasible successor so we just need to focus on entries which have more paths than the number of successor. In this case, we find 3 entries that are in blue boxes because they have only 1 successor but has 2 paths, so the last path can be the feasible successor.
By comparing the value of AD (of that route) with the FD (of successor`s route) we figure out there are 2 entries will have the feasible successor: the first and the second entry. The third entry has AD = FD (30720) so we eliminate it.

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diegonamaste
5 years, 2 months ago
A. 172.17.0.0/30 F. 10.140.0.0/24
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