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, FastEthemet0/0
208.149.23.96[110/13] via 190.173.23.10, 00:00:16, FastEthemet0/0
B.
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
C.
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
D.
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
Explanation:
Explanation/Reference:
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= 13for 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.