Router A has three feasible successors to the 192.168.1.0/24 network, which are listed here:
– Option 1 has a metric of 8123228.
– Option 2 has a metric of 2195467.
– Option 3 has a metric of 8803823.
The variance 4 command was issued on Router A.
How many active entries does Router A have in its routing table for the 192.168.1.0/24 network?
A.
0
B.
1
C.
2
D.
3
For easy understanding of EIGRP Variance have a look at http://www.cisco.com/c/en/us/support/docs/ip/enhanced-interior-gateway-routing-protocol-eigrp/13677-19.html
For this question we have three route with difference metric:
Route 1- 8123228
Route 2- 2195467
Route 3- 8803823
Best is lowest metric. 2 – 2195467.
As the Variance is 4, then ha have to do 4 x 2195467 = 8781904 (Result)
Route 1 will be active route as lower the the result above
Route 3 is higher than result above so will not be active route
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Final answer is C. 2 routes will be active entries.
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BTW, part of that new 326Q 200-310 dumps FYI:
http://www.ciscobraindump.com/?s=200-310
Best Regards!
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