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NEW QUESTION: 1

A. Option B
B. Option C
C. Option D
D. Option A
Answer: A,B,D

NEW QUESTION: 2
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 190.172.23.10, 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
B. 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
C. 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
D. 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
Answer: D
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 - 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.

NEW QUESTION: 3
DRAG DROP
Drag the Frame Relay acronym on the left to match its definition on the right. (Not all acronyms are used.)

Answer:
Explanation: