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I know it does not affect the answer to the question, but correct me if I am wrong: the exhibit in QID RTE023 should be modified so that the plan proposed in the question would actually work. In the exhibit, the path via RT3 will never be used even after step 2 in the plan (use variance command) because the advertised distance by RT3 is greater than the feasible distance calculated by RT1. Again, this does not affect the answer and also I don't know if the same overlook occurs in a similar question in the actual exam, but thought I would point that out. Maybe I'm overthinking the questions too much? Smile

Reference: http://www.cisco.com/en/US/tech/tk365/te...437d.shtml (see the variance example).
Reference: http://en.wikipedia.org/wiki/EIGRP#Feasi..._Condition (see the feasibility condition section)

Hours together configured this and found the result. now you can identify what the answer is
P 60.0.0.0/8, 1 successors, FD is 263680, serno 318
via 40.0.0.2 (263680/261120), Ethernet0/1, serno 313
via 10.0.0.2 (271360/261120), Ethernet0/0
via 70.0.0.2 (266240/261120), Ethernet0/2
P 70.0.0.0/8, 1 successors, FD is 261120, serno 3
via Connected, Ethernet0/2
via 10.0.0.2 (276480/266240), Ethernet0/0
P 80.0.0.0/8, 1 successors, FD is 266240, serno 229
via 70.0.0.2 (266240/261120), Ethernet0/2
via 40.0.0.2 (268800/266240), Ethernet0/1
via 10.0.0.2 (276480/266240), Ethernet0/0
P 90.0.0.0/8, 1 successors, FD is 271360, serno 320
via 70.0.0.2 (271360/266240), Ethernet0/2, serno 305
via 10.0.0.2 (279040/268800), Ethernet0/0
via 40.0.0.2 (273920/271360), Ethernet0/1
P 100.0.0.0/8, 1 successors, FD is 263680, serno 322
via 40.0.0.2 (263680/261120), Ethernet0/1
via 70.0.0.2 (266240/261120), Ethernet0/2
via 10.0.0.2 (268800/258560), Ethernet0/0
P 101.0.0.0/8, 2 successors, FD is 263680, serno 323
via 70.0.0.2 (263680/258560), Ethernet0/2
via 40.0.0.2 (263680/261120), Ethernet0/1
via 10.0.0.2 (271360/261120), Ethernet0/0

Best of Luck

SK

gvargasdc

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