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NEW QUESTION: 1
Refer to the exhibit.

The network is converged.After link-state advertisements are received from Router_A, what information will Router_E contain in its routing table for the subnets 208.149.23.64 and 208.149.23.96?
A. 208.149.23.64[110/13] via 190.173.23.10, 00:00:07, Serial1/0
208.149.23.96[110/13] via 190.173.23.10, 00:00:16, Serial1/0
208.149.23.96[110/13] via 190.173.23.10, 00:00:16, FastEthemet0/0
B. 208.149.23.64[110/13] via 190.173.23.10, 00:00:07, FastEthemet0/0
208.149.23.96[110/13] via 190.173.23.10, 00:00:16, FastEthemet0/0
C. 208.149.23.64[110/3] via 190.172.23.10, 00:00:07, Serial1/0
208.149.23.96[110/3] via 190.173.23.10, 00:00:16, Serial1/0
D. 208.149.23.64[110/1] via 190.172.23.10, 00:00:07, Serial1/0
208.149.23.96[110/3] via 190.173.23.10, 00:00:16, FastEthemet0/0
Answer: B
Explanation:
Explanation
Router_E learns two subnets subnets 208.149.23.64 and 208.149.23.96 via Router_A through FastEthernet interface. The interface cost is calculated with the formula 108 / Bandwidth. For FastEthernet it is 108 / 100 Mbps = 108 / 100,000,000 = 1. Therefore the cost is 12 (learned from Router_A) + 1 = 13 for both subnets -> The cost through T1 link is much higher than through T3 link (T1 cost = 108 / 1.544 Mbps = 64; T3 cost = 108
/ 45 Mbps = 2) so surely OSPF will choose the path through T3 link -> Router_E will choose the path from Router_A through FastEthernet0/0, not Serial1/0.
In fact, we can quickly eliminate answers B, C and D because they contain at least one subnet learned from Serial1/0 -> they are surely incorrect.

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