关于路由考试题库实验题4

panda2046

实验题4 最后有如下一句话：

" ~4 A! v( t& D. j
) N) X7 A# Y) w$ l! eBut please notice that the ip addresses and the subnet masks in your real
3 ?2 ?( A" a) g) xexam might be different so you might use different ones to solve this question)
But in your real exam, if you see the line “10.0.0.0/8 is a summary,….Null0″
then you need to summary using the network 10.0.0.0/8 with the command “ip
; m( }* K0 Q- L# Z/ f# \summary-address eigrp 123 10.0.0.0 255.0.0.0″ . This configuration is less
% m( g" s1 T. M4 \0 foptimize than the first but it summaries into 2 subnets as the question requires。

) P- |" G; U5 ~. e. C! S如果R3出现如下3条10.0.0.0的子路路由怎么办？如何修改呢？谢谢！
! i% A# Q+ L* `  @1 [! }- Y

     172.16.0.0/16 is variably subnetted, 2 subnets, 2 masks
D       172.16.0.0/16 is a summary, 02:10:13, Null0
C       172.16.1.0/24 is directly connected, Loopback1
     10.0.0.0/8 is variably subnetted, 3 subnets, 3 masks
D       10.2.0.0/20 [90/2172416] via 10.2.3.4, 00:00:04, Serial0/0
4 S" e6 P" D  }. F, _6 v/ @5 d. WD       10.0.0.0/8 is a summary, 00:05:49, Null0
3 C+ f' {  N/ s5 F7 Y5 y4 pC       10.2.3.0/24 is directly connected, Serial0/0
0 q6 M! ^  S, C1 C8 z: P8 [r3#
4 j) ]; @, b5 T4 |

Rockyw

you need to summary using the network 10.0.0.0/8 with the command “ip summary-address eigrp 123 10.0.0.0 255.0.0.0″