EIGRP Metric Calculation

In this post I will explain about EIGRP metric calculation, and explain what is the meaning of "EIGRP calculates the minimum bandwidth along the path"

in the following topology:

 we are running EIGRP all over the network above, R2 wants to reach loopback of R6 (6.6.6.6/32)

there are two paths, however they are not equal in metric, since EIGRP selects the best metric "lowest metric", then this will mean EIGRP will have a specific way of calculating this metric which is using  two things (delay, Bandwidth)
these two will be put into EIGRP metric calculation formula, and the lowest metric value between the two paths will be installed in the routing table, the second best route will be used as backup and will be installed in EIGRP topology table.

the formula is as this:
256  x  [ (10^7 / minimum bandwidth through the destination)  X   ( Sum of delays / 10 )]


ok, but what is the meaning of minimum bandwidth? well, when EIGRP calculate the metric it will choose the lowest metric to be calculated on the best path selected, for example, the best path in our topology is (R2-R4-R5-R6)

R2 will check the lowest bandwidth on the two paths and then calculate the composite metric using the formula, even if its other interface via R3 had a better Bandwidth configured, the reason is the link from R4 to R5 has a better metric (5500Kbps), this same value which is the minimum bandwidth on the path will be used for the calculation, check highlighted below:

R2#show ip eigrp topology 6.6.6.6/32
EIGRP-IPv4 Topology Entry for AS(100)/ID(2.2.2.2) for 6.6.6.6/32
  State is Passive, Query origin flag is 1, 1 Successor(s), FD is 670208
  Descriptor Blocks:
  24.0.0.4 (Ethernet0/2), from 24.0.0.4, Send flag is 0x0
      Composite metric is (670208/644608), route is Internal
      Vector metric:
        Minimum bandwidth is 5500 Kbit
        Total delay is 8000 microseconds
        Reliability is 255/255
        Load is 1/255
        Minimum MTU is 1500
        Hop count is 3
        Originating router is 6.6.6.6
  23.0.0.3 (Ethernet0/1), from 23.0.0.3, Send flag is 0x0
      Composite metric is (716800/691200), route is Internal
      Vector metric:
        Minimum bandwidth is 5000 Kbit
        Total delay is 8000 microseconds
        Reliability is 255/255
        Load is 1/255
        Minimum MTU is 1500
        Hop count is 3
        Originating router is 6.6.6.6



===============================
below displays interface bandwidth, even though E0/1 that goes to R3 has a better bandwidth, R2 will use the better reported distance value
interface Ethernet0/1
Description TO-R3
 bandwidth 7000
 ip address 23.0.0.2 255.255.255.0
end

R2#show run inter e0/2
Building configuration...

Current configuration : 80 bytes
!
interface Ethernet0/2
Description TO-R4
 bandwidth 6000
 ip address 24.0.0.2 255.255.255.0
end

R2#

So again, R2 will take the minimum bandwidth from each of the paths it has to the destination, then put them separately into the formula mentioned above then the results will be compared and the lowest will be chosen. 


NOTE: it's recommended to use DELAY in the path manipulation in EIGRP instead of Bandwidth for the reasons mentioned above. 

I hope this was helpful.