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.
O 208.149.23.64 [110/13] via 190.173.23.10, 00:00:07, FastEthernet 0/0
O 208.149.23.96 [110/13] via 190.173.23.10, 00:00:16, FastEthernet 0/0
B.
O 208.149.23.64 [110/1] via 190.172.23.10, 00:00:07, Serial 1/0
O 208.149.23.96 [110/3] via 190.173.23.10, 00:00:16, FastEthernet 0/0
C.
O 208.149.23.64 [110/13] via 190.172.23.10, 00:00:07, Serial 1/0
O 208.149.23.96 [110/13] via 190.172.23.10, 00:00:16, Serial 1/0
O 208.149.23.96 [110/13] via 190.173.23.10, 00:00:16, FastEthernet 0/0
D.
O 208.149.23.64 [110/3] via 190.172.23.10, 00:00:07, Serial 1/0
O 208.149.23.96 [110/3] via 190.172.23.10, 00:00:16, Serial 1/0
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= 13for both subnets – B is not correct.
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 – C & D are not correct.
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.
Should the cost from RA to 208.149.23.64 and .96 network be 13 because its coming out on one of the interface on R_C?
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