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中级工程师
   
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发表于 2017-6-16 17:55:45
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Although the “show ip eigrp topology” does not work in the exam but the “show ip eigrp 1 topology” does work so please use this command instead and we will find out the advertised distance on R1.6 y, W& y1 u( e1 D6 Y0 t
3 x$ |) R# p$ Q* [1 m$ M! {There are two parameters in the brackets of 192.168.46.0/24 prefix: (1810944/333056). The first one “1810944” is the Feasible Distance (FD) and the second “333056” is the Advertised Distance (AD) of that route -> A. 333056 is correct.- ], f* N. C0 K) ?- G
% r5 I! y0 H% FJust for your reference, this is the output of the “show ip route” command on R1:
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) G/ a0 g$ s( C D+ {0 j/ K, d! IIn the first line:: f' `& X A' y/ f/ k$ B
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D 192.168.46.0/24 [90/ 1810944] via 192.168.12.2, 00:10:01, Ethernet0/0 # ?! r. L+ a7 S. l
8 `0 N0 |. \ C/ t# a3 SThe first parameter “90” is the EIGRP Administrative Distance. The second parameter “1810944” is the metric of the route 192.168.46.0/24. R1 will use this metric to advertise this route to other routers but the question asks about “the advertised distance for the 192.168.46.0 network on R1” so we cannot use this command to find out the answer.. s6 U( f- h' g* c
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$ {, Q7 A' ^5 X2 L! q _: ]: n% k$ B0 J) b$ `0 d
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2017-6-16 17:55:45
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