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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 188.8.131.52 and 184.108.40.206?
A. 220.127.116.11[110/3] via 18.104.22.168, 00:00:07, Serial1/0 22.214.171.124[110/3] via 126.96.36.199, 00:00:16, Serial1/0
B. 188.8.131.52[110/13] via 184.108.40.206, 00:00:07, FastEthemet0/0 220.127.116.11[110/13] via 18.104.22.168, 00:00:16, FastEthemet0/0
C. 22.214.171.124[110/13] via 126.96.36.199, 00:00:07, Serial1/0 188.8.131.52[110/13] via 184.108.40.206, 00:00:16, Serial1/0 220.127.116.11[110/13] via 18.104.22.168, 00:00:16, FastEthemet0/0
D. 22.214.171.124[110/1] via 126.96.36.199, 00:00:07, Serial1/0 188.8.131.52[110/3] via 184.108.40.206, 00:00:16, FastEthemet0/0
Explanation Router_E learns two subnets subnets 220.127.116.11 and 18.104.22.168 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.