EIGRP metric calculation

[gongya001]

I

I mutually redistribute ospf and eigrp in R2 and R3. when I use "default-metric 100000 1000 255 1 1500" on both R2 and R3

On R3, I got the following:

R3#show ip eigrp topo | b 192.168.18

P 192.168.18.0/24, 1 successors, FD is 409600

        via 192.168.123.1 (409600/128256), Ethernet0/1

        via 192.168.123.2 (537600/281600), Ethernet0/1

When R3 receives 192.168.18.0 from two source (R1 and R2), it compares AD (R2:537600 and R1:409600) to select a better one.

R3#show ip route 192.168.18.0

Routing entry for 192.168.18.0/24

  Known via "eigrp 10", distance 170, metric 409600, type external

  Redistributing via eigrp 10, ospf 1

  Advertised by ospf 1 subnets

  Last update from 192.168.123.1 on Ethernet0/1, 00:06:32 ago

  Routing Descriptor Blocks:

  * 192.168.123.1, from 192.168.123.1, 00:06:32 ago, via Ethernet0/1

      Route metric is 409600, traffic share count is 1

      Total delay is 6000 microseconds, minimum bandwidth is 10000 Kbit  (this number is decided by metric set on R2)

      Reliability 255/255, minimum MTU 1500 bytes

      Loading 1/255, Hops 1

My question is how I calculate FD from R2 537600  ?

thanks !!

[Jon Marshall]

R2 is receiving an EIGRP external from R1 and an OSPF route (redistributed from R3). 

OSPF AD 110 is better than EIGRP external of 170 so the OSPF route is used and then redistributed into EIGRP and advertised to R1 and R3. 

EIGRP metric calculation is  256 * (bandwidth + delay) where bandwidth is 10,000,000 / the minimum bandwidth of any of the links and delay is the cumulative delay of all links. 

So when you redistribute on R2 you are using a bandwidth of 100,000 and R3 receives this route on it's eth0/1 interface which has a bandwidth of 10,000 and as you use the minimum bandwidth of any link that means - 

bw = 10,000

you are also using a delay of 1000 on R2 and because that is in tens of microseconds you can plug this number directly into the formula because that is what EIGRP uses in it's calculation.  You also need to add the delay on the eth0/1 interface and if you do a "sh int eth0/1" on R3 you will see the delay is 1000 microseconds. As stated above the formula expects tens of microseconds so you need to divide that number by 10 to get to tens of microseconds ie. 

total delay = 1000 + (1000/10) = 1100

and the final calculation is - 256  * (10,000,000 / 10,000 + 1100) = 537600. 

Jon

[gongya001]

thanks so much !!

[gongya001]

thanks so much !!