Hello,
I still have doubt to get Advertised distance and Feasible distance as per topology. As mentioned that Advertised distance: How far the destination is away for your neighbor and Feasible distance: The total distance to the destination.
I followed the given steps:
R4 will advertise the destination network to R3.
R3 will advertise the network to R1 and R2.
R1 will advertise the network to R2.
R2 will advertise the network to R1.
R1 will advertise this network back to R3.
R2 will advertise this network back to R3.
But i am not getting same value as mentioned in result.
As per my understanding the result is mentioned below:
- R4 will advertise the destination Network to R3 which is 5 so AD for R3= 5
- Now R3 will advertise to R1 and R2 which is 9 so as of now AD for R1 and R2 will be 9 (5+4)
- R1 will advertise to R2 what it has learned from R3 so AD for R2= 12 (5+4+3)
- R2 will advertise to R1 what it has learned from R3 so AD fro R1=18 (5+4+9)
- R1 will advertise back to R3 what it has learned from R2 so AD for R3=25 (5+4+9+7)
- R2 will advertise back to R3 what it has learned from R1 so AD for R3=19 (5+4+3+7)
We have two AD values for each router here so now which Advertised distance (AD) would be selected.
And how R4 Feasible distance would be 9 ? as it is directly connected to destination.
R3 must have AD=5 and FD=9 but in case R4 i am confused !
Please help me out to know this concept !