ccna 200 301 syllabus meaning in marathi

ccna 200 301 syllabus meaning in marathi.active state. Router E advertises its FD to D.Figure 7-2-9 Router E Receives a Reply Figure 7-2-10 Router D Receives All Replies Router D receives a reply from Router E. Router D receives all replies. Router D updates the topology table and calculates that the FD to reach network 10.1.1.0/24 is 5 and there are two equal paths. As shown in Figure 7-2-11. All routers in the network are in a passive state and the network converges again.10.1.1.0/24Figure 7-2-11 Network Convergence Again8 7.3 Advanced Configuration of EIGRP*This section describes the advanced configuration of EIGRP, including manual summarization of EIGRP, non-equal load balancing of EIGRP, authentication of EIGRP, external routing of EIGRP, and bandwidth allocation and adjustment of EIGRP.7.3.1 EIGRP Non-Equivalent Load Balancing IGRP and EIGRP can support non-equivalent load balancing, while protocols such as RIP and OSPF do not have this feature,, 224Hush Chapter 7 EIGRP*** 07 can only support equal-value load balancing. Continue using the topology shown in Figure 7-2-1 here, and continue using the configuration in Notepad as shown in Figure 7-2-2, only only do not misassign the IP address of interface R3 S1/0 again. After the configuration is complete, view the routing table on R1, which displays as follows. Rl#show iproute ' D2.0.0.0/8 [90/2297856] via 12.1.1.2, 00:02:28, Seriall/1 D 23.0.0.0/8 [90/2172416] via 13.1.1.3, 00:02:28, FastEthemetO/O 12.0.0.0/8 is variably subnetted, 2 subnets, 2 masks ...... C12.1.1.0/24 is directly connected, Seriall/1 D12.0.0.0/8 is a summary, 00:02:51, NullO 13.0.0.0/8 is variably subnetted, 2 subnets, 2 masks C13.1.1.0/24 is directly connected, FastEthernetO/O D13.0.0.0/8 is a summary, 00:02:52, NullO R1 has only ... paths to 23.0.0.0/8. If you were running the RIP protocol, two equal paths would emerge here, because RIP simply determines the merit of a route based on hop count. EIGRP, on the other hand, uses a composite metric, which by default is related to bandwidth and delay. In R1's routing table why is it going from R3 to 23.0.0.0/8,but not from R2 to 23.0.0.0/8?Some readers may say that the 100Mb/s link between R1 and R3 is faster than the 1.544Mb/s link between R1 and R2.In fact, it is not because of the link bandwidth.When E1GRP calculates the metric value, it uses the minimum bandwidth, the link between R2 and R3 is also 1.544Mb/s, whether it is from R2 or from R3,the minimum bandwidth of the link is 1.544Mb/so Rltshow ipeigrp topology 23.0.0.0 IP-EIGRP (AS100): Topology entryfor 23.0.0.0/8 State is Passive, Query origin flag is 1,1 Successor(s)rFDis 2172416 Routing Descriptor Blocks: 13.1.1.3 (FastEthernetO/0), from 13.1.1.3, Send flag is 0x0 Composite metric is (2172416/2169856)rVectormetric; RouteisInternal Minimumbandwidthis 1544Kbit Totaldelayis Reliabilityis 20100microseconds Load is 1/255 255/255 MinimumMTU is Hopcountis1 1500 12.1.1.2 (Seriall/1), from 12.1.1.2, Send Composite metric is flagis0x0 (2681856/2169856), Vector metric: Route is Internal Minimum bandwidth is1544Kbit Totaldelayis /ReliabilityisLoadis1/255Mini umMTUis 40000microseconds Hopcountis'1 255/255 1500 Looking at the topology of the 23.0.0.0 network on router R1 can reveal the cause, as shown below. From the output above, you can see that R1 has two paths to 23.0.0.0, which are reachable from either R2 or R3. The metric going from R2 is 2681856,where R2 reports over a distance of 2169856,and the minimum link bandwidth from R1 to the target network is 1.544Mb/s,with a total delay of 40000|is. The metric going from R3 is 2172416,where R3 reports over a distance of 2169856,and the minimum link bandwidth from R1 to the target network is L544Mb/s,and the total delay is 20100|is.The path with the minimum metric value enters the routing table and R3 becomes the successor router for R1 to 23.0.0.0/8. From the above display, you can see that R1 chooses R3 as the successor not because of bandwidth, but because of the delay, which is 1003 by default for the Ethernet interface and 200003 for the serial interface, which can be accessed on router R1 using the show int fa 0/0w command and the " showintsl/1" command on router R1.Divide the larger metric 2681856 by the smaller metric 2172416,2681856/2172416 Xi.2345, an integer of not less than 1.2345, which is 2. Use the variance command to configure the inequality factor, the configuration command is as follows: 7 RKconfigJ #couter; eigrp100'