Thanks

1) Today there was activity of Router Replacement
Everyone thinks that this is not a big deal , but as someone said

Experience counts

So Let me tell you what i faced :-
I went to Data Centre 1/2 hour before and mounted the router in rack, then thought that fastethernet cable was lying down crossing 1 rack why not fit the fastethernet cable to the patch panel which is situated in the rack itself……–the one who had done this might be idiot

i thought and then i began testing whether vlan are same , cable is proper…

Oh my god the downtime has started, so i just skipped it and started my activity

First, i shutdown the router then removed the cables and inserted card into new routers(1 Channelized E1/T1 Card, 1 Serial Card, 1 WIC2T card )

But hey which cable are inserted in which card ??

By god’s grace my memory was sharp and i wasted 5 min in remembering the cables and lastly inserted properly.

Now i started my new Router(2651XM) and everything was ok! Then 5 min remaining , i thought why not change the fastethernet cable as i thought earlier ..and i went to do the same.

I inserted cable into proper vlan and then my god what happened i wasn’t able to ping other network behind fastethernet 0/0 ….

Downtime finished and i realized the duplex !!!

I went quickly and typed the following command:

set port duplex 3/3 full
but nothing happened and the port was still in half duplex then i entered the following command and miracles happened..
set port speed 3/3 100

And i remembered in CAT OS set speed first and then duplex
So if you are going for router replacement just ensure that below things are in place:-

1) Configure your router properly beforehand
2) Label the cables on the existing router
3) Check whether the cables will reach to the location where router will be fitted
4) Mount the router in the rack

And ya one more just don’t forget that in CAT OS set speed first and then duplex

AD=advertised distance - Distance advertised by the neighbor to the destination
AD = distance from the neighbor router to the destination

Simple example

R1———R2——–R3———R4—-networkx

Cost from R1 to R2 R3 R4 to network x =9

Cost from R2 to network x is umm 5 so AD is 5

FD = 9

FD and AD are cost calculated individually it doesnt have any relationship and you cant add them….

Here is how it works

metric = [K1 * bandwidth + (K2 * bandwidth) / (256 - load) + K3 * delay] * [K5 / (reliability + K4)]

The default values for K are:

*

K1 = 1
*

K2 = 0
*

K3 = 1
*

K4 = 0
*

K5 = 0

Now what does this K value mean? or do? well it tells the router which metrics you will be using for the calculation… so as you see K1 has a value of 1 and k3 has a value of 1 also so if you look the metric formula you will notice that the metric formula will be reduced to this
metric = bandwidth + delay

If you had a 1 value in K2 then it will use also the LOAD metric for FD calculation…. plus the bandwich and delay that they had cause of the value of 1 of K1 and K3

Now ok we know that so, so good so far.

FD calculation
When the router is calculating the FD it does it by itself. It doesnt see any advertised distance to do it… its individually calculated.

bandwidth = (10000000/bandwidth(i)) * 256
formula was reduced to this cause of the K values

minimum bandwidth = 56k [b]of route 1-4-2-network B[/b]
metric=[(10000000/56) + 2200] x 256=46277376

THE OTHER ROUTE

minimum bandwidth = 128k [b]of route 1-3-2-network B[/b]

[(10000000/128) + 1200] x 256=20307200
So to reach Network A, Router One chooses the route through Router Three
which will be the FD!!
FD= 20307200

ADVERTISE route calculation
We sit on router 3 now and calculate how much it will be in metric to reach network B

minimum bandwidth = 10000k [b]of route 3-2-network B[/b] there is a change of the minimum bandwidth here!
[(10000000/10000) + 200] x 256=307200. DELAY changed here to 200 because 100+100=200 hehe

so the AD here will be 307200

Now the router 1 still calcualates the route to network B through router 4 but this will remain in the topology table here is the calculation

minimum bandwidth = 56k
metric=[(10000000/56) + 2200] x 256=46277376

the AD distance to network B will be the same as you can see it has the same delay and bandwidtch to detiny

so in the topology table we will have

P 172.16.80.0 255.255.255.0, 1 successors, FD is 20307200

via 172.16.80.1 (46277376/307200), Serial1

via 172.16.80.5 (20307200/307200), Serial2

so
(46277376/307200) first number is the distance from the router you are at to the network B, the second number is the advertised distance of the neighbor router to the destination network B

If you had another route for example you had this one also calculated by the router
via 172.16.80.5 (20307200/30307200), Serial2 this one you will NOT see it on the topology table because the Advertised distance is higher than the FD… the advertise distance is 30307200 Remenber for being here it must meet a feasible condition which is “the Advertised distance cant be higher than the FD”
Now with this table EIGRP does the routing table by picking the BEST route to the destination and placing it on the routing table
so in this example you will see in the routing table just this route

D 10.1.1.0 via 172.16.80.5 (90/20307200), Serial2

So as you see you will only see that one because is the best one between all the ones in the topology table…

so you will have this route as a back up in the topology table via 172.16.80.1 (46277376/307200), Serial1
if the via 172.16.80.5 (20307200/307200), Serial2 route fails then it will use the one that had as back up in the topology table which is via 172.16.80.1 (46277376/307200), Serial1

Now what will happen if we got 2 routes with the same metric to the destination in the topology table? ah if this happens it will place BOTH routes in the routing table and load and balance between them….

EIGRP can load and balance 4 routes by default…. this means if we got 4 best routes with the best metric then it will place the 4 in the routing table and load and balance….

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CMDeane