| Exam 200-105 | Question id=1916 | Routing Technologies |
Refer to the exhibit.
router#show ip eigrp topology 10.0.0.5 255.255.255.255
IP-EIGRP topology entry for 10.0.0.5/32 State is Passive, Query origin flag is 1, 1 Successor(s), FD is 41152000
Given the output from the show ip eigrp topology command, which router is the feasible successor?
| A. |
10.1.0.3 (Serial0), from 10.1.0.3, Send flag is 0x0
Composite metric is (46866176/46354176), Route is Internal
Vector metric:
Minimum bandwidth is 56 Kbit
Total delay is 45000 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 2
| |
| B. |
10.0.0.2 (Serial0.1), from 10.0.0.2, Send flag is 0x0
Composite metric is (53973248/128256), Route is Internal
Vector metric:
Minimum bandwidth is 48 Kbit
Total delay is 250010 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 1
| |
| C. |
10.1.0.1 (Serial0), from 10.1.0.1, Send flag is 0x0
Composite metric is (46152000/41640000), Route is Internal
Vector metric:
Minimum bandwidth is 64 Kbit
Total delay is 45000 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 2
| |
| D. |
10.1.1.1 (Serial0.1), from 10.1.1.1, Send flag is 0x0
Composite metric is (46763776/46251776), Route is External
Vector metric:
Minimum bandwidth is 56 Kbit
Total delay is 41000 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 2
|
To be the feasible successor, the Advertised Distance (AD) of that route must be less than the Feasible Distance (FD) of the successor. From the output of the show ip eigrp topology 10.0.0.5 255.255.255.255 we learn that the FD of the successor is 41152000.
Now we will mention about the answers, in the Composite metric is (.../...) statement the first parameter is the FD while the second parameter is the AD of that route. So we need to find out which route has the second parameter (AD) less than 41152000 -> only answer B satisfies this requirement with an AD of 128256.