ccna 200-125 dumps 2018 free

ccna 200-125 dumps 2018 free

Image R2 becomes the demarcation point between RIP and OSPF domains. So how can I let R1 learn the route from OSPF and let R3 learn from RIP? O - OSPF intra, OI - OSPF inter, OE1 - OSPF ext 1,ccna 200-125 dumps 2018 free, OE2 - OSPF ext 2 ON1 - OSPF NSSA ext 1, ON2 - OSPF NSSA ext 2 Then, in order for R3 to dynamically learn the routing of production and office servers, it is now necessary to redistribute these two static routes into OSPF . Of course, there are tricks when re-publishing. The distribution list is deployed in a link state routing protocol such as OSPF . If the out direction is to be used, it can only be used in such a situation. Image The next outgoing interface of the route is the port of the match . Router(config-router)# router-id xxxx Martian NEXTHOP; ARP The certificate contains the password information when it is verified using plain text. Router ospf 100 Ipv6 address 2001:12::2/64 interface Serial0/1 The configuration of R1 is as follows: Image Notice the Flag field, Set ip next-hop verify-availability 10.1.1.2 10 track 1 Ipv6 nat prefix 2001:2::/96 Image FE80:: 5EFE:202:202 2001:1111:: 5EFE:202:202 Router(config-if)#ipv6 adress 2002::/64 eui-64 !! Also configured with an IPv6 address and populated with EUI-64 Experimental needs Image Image ... partially omitted ... Ip route 0.0.0.0 0.0.0.0 192.168.23.3 Neighbor 2.2.2.2 Image 0600 – 07FF Ipv nd prefix 2012::/64 no-advertise The link layer address corresponding to the Layer 3 address is parsed by the neighbor request ( NS ) and the neighbor advertisement ( NA ) message. ! Phenomenon analysis: These multicast packets generated by R2 , the source address is the address of Fa1/0 port of R2 , and the destination does not say that the IP address of the multicast group is determined . These multicast data arrive at R3 , since R3 is the first hop router, and the Fa0/0 port of R3 is the DR of the local LAN , so When R2 is configured with no validate-update-source , R2 will not verify the RIP update message source, so R2 will accept R1. Exit Known via "ospf 1", distance 110, metric 20 Distance 100 192.168.12.1 0.0.0.0 1 Access-list 1 permit 10.1.1.0 Interface FastEthernet0/0 ipv6 enable 8 8 8 8 RIPng RFC1721 Router bgp 300 address-family ipv6 Isis This comprehensive thinking question helps you sort through several solutions: 2001::1 Ipv6 unicast-routing LSP Seq Num Image Ipv6 enable Interface FastEthernet0/0 !! Core in OSPF BGP table version is 1, main routing table version 1 These modes generally only be on high-end Cisco device applications , such as 7200 series routers . Or 12,000 Series routers . Command : ip route-cache optimum show ip cache optimum In order to get this prefix automatically, just run a protocol between the router and the host. Router using NDP protocol basic configuration None Open Shortest Path First (OSPF) Routing Information Protocol (RIP) Route map reference On Demand stub Routes The last command is to put the host of the IPv4 network 202.101.100.2 into the IPv6 network, so that the IPv6 only user has the access target. Otherwise, what destination address do you use the IPv6 only user to access the IPv4 only network? 2001:1::1 As the example above, it can be further abbreviated as: 2001 : a8 : 207 :: 8207 Route-map test permit/deny 10 IPv6 Address(es): FE80::FFFF:FE10:1 Based on CEF Image Remarks DHCP PD Image The comparison principle of OE2 is first compared to the external metric of OE2 . If they are equal, further comparison of the internal metric is the metric to the ASBR . Image The PC will start sending RS with the following message: After 10.30.0 is passed to R3 , since the RIP updated subnet and the R3 interface subnet belong to the same main class network, R3 receives the update, and uses the interface mask as the mask of the RIP route, so the R3 routing table The route loaded in is: 10.1.20.0/24 ! Exit-address-family 0x7768 Of course it includes information such as object MAC address , contents of the destination interface . In this case we need only one data stream (Flow) in the first packet do process Access-list 2 permit 172.16.20.0 route-map BGP2OSPF permit 10 Load distribution: 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 (refcount 1) FIG Ibid, 4.0 after hang, C sends message poisoning message, theoretically 4.0 routing is C advertised to B , the principle of the horizontal division, B is not the C advertised 4.0 information, but with a level of toxicity reversal split broke this principle, B periodically to C sends 4.0 poison reverse messages to let C know, his neighbors knew a 4.0 hang message and waiting eagerly 4.0 resurrected place full of blood, another benefit of doing so is avoid loops, at least C is not from B to go visit 4.0 a.