Non-equivalent Server Load balancer based on

VPC is not equivalent to VPC. Server Load balancer Principle:SuCcessor FD * VariaNCE> feasible successor FDTopology

After completing the configuration on the three routers, the basic configurations are not listed here, and then the IP address 2.2.2.2 is declared on R2.LookBack 0

Purpose

To go to 2.2.2.2 this network segment to R3, you must first check the route table of R3 by creating the two unequal Load Balancing rules R3> R2 and R3> r1> R2.

On R3, view the detailed neighbor table SH ip address (all ).

We can see that there are two routes in the 2.2.2.0 CIDR block, while S1/0 and E1/0 respectively, FD passing through E1/0 is much larger than FD passing through S1/0, and ad of S0/1 is much larger than FD of E1/0, so no to become a feasible successor, but in load balancing that is not equivalent, feasible successor must be available, here, I am modifying the E1/0 latency of R3 to manually adjust S0/1 to feasible successor. Here is a formula FD me.TrIc = 256*(10 ^ 7/BW + dly/10) Check the default latency of E1/0

Do sh int E1/0: Set the latency to 1000. Here, we set the latency to 8000 and enter the interface R3 (config-If) # delay 8000 to get the detailed neighbor table of R3

Here, S0/1's ad 2297856 is less than the FD 2432000 of E1/0, so S0/1 becomes feasible successor (in actual projects, the latency will not increase .. That's what sb Does. In actual projects, SLB is not implemented through the s port and E port --HereOnly for demonstration) and then calculate the variance value through the principle of non-equivalent load balancingPrinciple: successor FD * variance> feasible successor FD 2432000 *The variance value must be greater than 2809856. Therefore, set this parameter to 2. Enter the OSPF process R3 (config-router) of R3 # variance 2 and check the route table of R3.

The network segment to 2.2.2.0 already has two routes

InPingAfter 2.2.2.2, view the detailed information of 2.2.2.0 route 255.255.0.

We can see that the 23.1.1.2-based data packet passed through 15 data packets 13.1.1.1 passed through 13 Load Balancing because they are not equivalent, so there is a balance algorithm to allocate the data packet path, while FD is smaller, FD is smaller. How is this balancing algorithm calculated .... Don't bother to worry about him... Back pain. Go to rest...