
配置RIPv1协议,宣告主类网络号下:$ [& o7 e* z# c* c
环境一:
* o4 d5 M& W" _) H8 ]R2#sh ip route
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+ r# e' E# y5 p192.168.10.0/30 is subnetted, 2 subnets
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% ~& N" b5 D; {6 ^192.168.10.4 is directly connected, Serial0/0$ _- P8 Z) @4 j$ F( T" ^
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192.168.10.8 is directly connected, Serial0/1# R t8 O1 |* H) |! }& S8 E' K4 P
# x) L" T2 \, C; r3 w$ nR3#sh ip route. |+ B9 f5 i3 V! {, \% A1 [* z/ f
) u" S7 G4 n& R2 Z- g: ^# ~8 Z192.168.10.0/24 is variably subnetted, 3 subnets, 2 masks! y, H, M6 K% I% k l: x( j$ ?
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192.168.10.32/27 is directly connected, Loopback0 }$ N# A" [, \ q# y2 x
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192.168.10.4/30 [120/1] via 192.168.10.9, 00:00:14, Serial0/0
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) y1 j0 Z% B$ |7 N0 C192.168.10.8/30 is directly connected, Serial0/07 n- s# s' c9 g2 f. T1 W5 q
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环境二:8 n$ u2 |; A' m/ p! x8 g2 n
R1#sh ip route
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192.168.10.0/24 [120/1] via 10.1.1.2, 00:00:19, Serial0/0
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$ ^, I( c0 F% c$ k+ {172.16.0.0/24 is subnetted, 1 subnets
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172.16.10.0 is directly connected, Loopback0
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10.0.0.0/24 is subnetted, 1 subnets
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10.1.1.0 is directly connected, Serial0/0
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& e/ F3 X( V$ G. ]R2#sh ip route
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3 {) c# h& t/ z* x3 e: h [4 q5 {192.168.10.0/26 is subnetted, 1 subnets! x) n% [6 E. `! n
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* t% x9 v& j" Q# c6 [) _: ?5 k192.168.10.64 is directly connected, Loopback0
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172.16.0.0/16 [120/1] via 10.1.1.1, 00:00:00, Serial0/0: P( v1 y! b' ?2 @
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10.0.0.0/24 is subnetted, 1 subnets4 G$ m# G& W' {' I- ?
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/ h' h" r: S) n8 r. t10.1.1.0 is directly connected, Serial0/0/ _+ E( i% K _' B, a/ A# ]
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3 g; m" ]6 v' J0 q* s, z环境一中,R2直连30位掩码,配置RIPv1无法学到R3上192.168.10.33/27位掩码的网络信息。# Z3 @' Y H) j5 N' K' r
环境二中,R1直连24为掩码,同样在RIPv1下能够学到R2上192.168.10.0/24这个主类网络信息。
+ {. M/ V5 d# h, D猜测:在RIPv1中,能学到远端掩码大于直连掩码的网络信息,并且学到的掩码大小不超过直连网络掩码位数。( W& Y7 H# g1 k; F2 O3 n2 M: W' O. Z
1. 求证这个结论是否正确?& f. G4 ]4 V+ @7 H: {: o& z, E
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2. 对于RIP协议与能学到的子网信息的关系,欢迎网友总结与补充。 |
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