# Eigrp Topology Table And Routing Table(how they are done)

NightShade1
Posts:

**431**Member
in CCNA & CCENT

Lets explain some terms first with examples

Feasible distance and Advertise distance

this is simple

FD=fiasable distance=he feasible distance is the best metric to reach the destination or the best metric that was known when the route went active

AD=advertise distance=distance advertised by the neighbor to the destination

AD = distance from the neighbor router to the destiny

Simple example

R1

R2

R3

R4----networkx

Cost from R1 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 has doesnt have any relationship and you cant add them.... i mean doing stuff like FD=AD+R1toR2 metric no.....

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 values means? or does? well it tells the router which metrics you will be using for the calculation... so i 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 forumula 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.

When router is calculating the FD it does it actually dont see any advertised distance to do it... it do it by itself

bandwidth = (10000000/bandwidth(i)) * 256

formula was reduced to this cause of the K values

minimum bandwidth = 56k

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

THE OTHER ROUTE

minimum bandwidth = 128k

[(10000000/12 + 1200] x 256=20307200

So to reach Network A, Router One chooses the route through Router Three

which will be the FD!!

FD= 20307200

We sit on router 3 now and calculate how much it will be the metric to reach network B

minimum bandwidth = 10000k

[(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 calcualate the route to network B through router 4 but this will remain in the topology table here is the calcualtion

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

so

(46277376/307200) first number is the distance from the router you are to the network B the second number is the advertise distance of the neighbor router to the destiny 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 Advertise distance is higher than the FD... the advertise distance is 30307200 Remenber for being here it must meet a feasible condition which is "the Advertise 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

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 witht he same metric to the destination in the topology table? ah if this happen 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....

Any Question or comment are welcome so ask or comment away....

Feasible distance and Advertise distance

this is simple

FD=fiasable distance=he feasible distance is the best metric to reach the destination or the best metric that was known when the route went active

AD=advertise distance=distance advertised by the neighbor to the destination

AD = distance from the neighbor router to the destiny

Simple example

R1

R2

R3

R4----networkx

Cost from R1 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 has doesnt have any relationship and you cant add them.... i mean doing stuff like FD=AD+R1toR2 metric no.....

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 values means? or does? well it tells the router which metrics you will be using for the calculation... so i 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 forumula 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 router is calculating the FD it does it actually dont see any advertised distance to do it... it do it by itself

bandwidth = (10000000/bandwidth(i)) * 256

formula was reduced to this cause of the K values

minimum bandwidth = 56k

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

THE OTHER ROUTE

minimum bandwidth = 128k

**of route 1-3-2-network B**[(10000000/12 + 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 the metric to reach network B

minimum bandwidth = 10000k

**of route 3-2-network 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 calcualate the route to network B through router 4 but this will remain in the topology table here is the calcualtion

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), Serial2via 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 to the network B the second number is the advertise distance of the neighbor router to the destiny 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 Advertise distance is higher than the FD... the advertise distance is 30307200 Remenber for being here it must meet a feasible condition which is "the Advertise 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 witht he same metric to the destination in the topology table? ah if this happen 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....

Any Question or comment are welcome so ask or comment away....

Product Manager - ArubaNetworks

Alternetworks Corp

Alternetworks Corp

0

## Comments

2,112MemberThis is misleading as it uses received metrics, it will use the received delay,if the local interface Bw is greater than that of the received update it will use the Bw value received in the update.

Just to let you know EIGRP doesn't send the calculated metric in the updates, it sends the delay and bandwidth parameters and the calculation is then performed on the received parameters.

431MemberI tried calculating the metric from R1 to R2 then the metric from R2 to net x and i sum it and didnt work Thats why i said which is because its use the lowest BW for the destination....I still think they dont have relationship

because FD is a metric that doesnt depend of the Advertise distance this is my point.You are just looking it with another point of view. The metric depends on his neighbors to give him the delays and all that yeah but it doesnt depend in any way to the calculated Advertise distance...When isaid this i didnt mean that... i was refering that it doesnt use the Advertise distance to calculate the total metric.... i know it uses the recived delay in the update

Could you read and tell me if you got what i really mean and tell me if im still wrong?

Alternetworks Corp