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, 00:00:00:07, FastEthernet0/0
208.149.23.96[110/13] via 190.173.23.10, 00:00:00:16, FastEthernet0/0
B.
208.149.23.64[110/1] via 190.173.23.10, 00:00:00:07, Serial1/0
208.149.23.96[110/3] via 190.173.23.10, 00:00:00:16, FastEthernet0/0
C.
208.149.23.64[110/13] via 190.173.23.10, 00:00:00:07, Serial1/0
208.149.23.96[110/13] via 190.173.23.10, 00:00:00:16, Serial1/0
208.149.23.96[110/13] via 190.173.23.10, 00:00:00:16, FastEthernet0/0
D.208.149.23.64[110/13] via 190.173.23.10, 00:00:00:07, Serial1/0
208.149.23.96[110/13] via 190.173.23.10, 00:00:00:16, Serial1/0
The answer is going to be a because your using a link state protocol which means the router is going to take information from the router connected to the shortest path.
I believe that the selection is regarding the cost of the path. Directly trough the Router_C the path will have the T1 cost and trough Router_A will have the FastEthernet cost.
T1 cost is 10 Mb = 19
FaEth cost is 100Mb = 1
T1 path is 1.544 Mbps. The formula is Ref-value/Int-value. 100Mbps is the default reference value. So 100/1.544=64 (always round it to the least number). For T3 path the cost will be 1+2+10=13.
I agree with the answer.