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New Member

Eigrp metric calculations

Hi.

here is my eigrp topology details. how eigrp get this fd distance, i try to solve this by using famous formula but i didn't get this 156160 value.

R1#sh ip route 1.1.1.1

Routing entry for 1.1.1.1/32

  Known via "eigrp 1", distance 90, metric 156160, type internal

  Redistributing via eigrp 1

  Last update from 172.16.23.2 on FastEthernet1/1, 05:52:13 ago

  Routing Descriptor Blocks:

  * 192.168.1.2, from 192.168.1.2, 05:52:13 ago, via FastEthernet1/0

      Route metric is 156160, traffic share count is 1

      Total delay is 5100 microseconds, minimum bandwidth is 100000 Kbit

      Reliability 255/255, minimum MTU 1500 bytes

      Loading 1/255, Hops 1

    172.16.23.2, from 172.16.23.2, 05:52:13 ago, via FastEthernet1/1

      Route metric is 156160, traffic share count is 1

      Total delay is 5100 microseconds, minimum bandwidth is 100000 Kbit

      Reliability 255/255, minimum MTU 1500 bytes

      Loading 1/255, Hops 1


R1#

P 2.2.2.2/32, 2 successors, FD is 156160

        via 172.16.23.2 (156160/128256), FastEthernet1/1

        via 192.168.1.2 (156160/128256), FastEthernet1/0

R1#sh int fa 1/0 | i BW

  MTU 1500 bytes, BW 100000 Kbit/sec, DLY 100 usec,

R2#sh int fa 1/0 | i BW

  MTU 1500 bytes, BW 100000 Kbit/sec, DLY 100 usec

R2#sh int loopback 0 | i BW

  MTU 1514 bytes, BW 8000000 Kbit/sec, DLY 5000 usec

my calculations

min BW = 100000

Dly = 100+100+5000 = 5200

26931200

1 ACCEPTED SOLUTION

Accepted Solutions
Cisco Employee

Eigrp metric calculations

Hi,

So let's see:

  Routing Descriptor Blocks:

  * 192.168.1.2, from 192.168.1.2, 05:52:13 ago, via FastEthernet1/0

      Route metric is 156160, traffic share count is 1

      Total delay is 5100 microseconds, minimum bandwidth is 100000 Kbit

      Reliability 255/255, minimum MTU 1500 bytes

      Loading 1/255, Hops 1

By default, K1=K3=1, so BW and DLY are taken into account.

Metric = (K1* (10^7 / BW) + K3 * DLY/10) * 256 = (10^7 / 100000 + 510) * 256 = 610 * 256 = 156160

Remember that the metric formula uses tens of microseconds as the delay factor in its computation.

Best regards,

Peter

13 REPLIES
Cisco Employee

Eigrp metric calculations

Hi,

So let's see:

  Routing Descriptor Blocks:

  * 192.168.1.2, from 192.168.1.2, 05:52:13 ago, via FastEthernet1/0

      Route metric is 156160, traffic share count is 1

      Total delay is 5100 microseconds, minimum bandwidth is 100000 Kbit

      Reliability 255/255, minimum MTU 1500 bytes

      Loading 1/255, Hops 1

By default, K1=K3=1, so BW and DLY are taken into account.

Metric = (K1* (10^7 / BW) + K3 * DLY/10) * 256 = (10^7 / 100000 + 510) * 256 = 610 * 256 = 156160

Remember that the metric formula uses tens of microseconds as the delay factor in its computation.

Best regards,

Peter

New Member

Eigrp metric calculations

i saw one article they mention to calculate all link delay,so from r1 to r2 and r2 from loop delay

min BW = 100000

Dly = 100(fa1/0)+100(fa1/0)+5000(loop0) = 5200

256*[10000000 / 100000] + [100+100+5000] /10

256[100]+[5200] /10

135680

Eigrp metric calculations

You are adding an extra delay, you should only be concerned with the exit interfaces.

New Member

Eigrp metric calculations

So i can calculate only from r1  prespective right ?

Metric = 256*[10^7/100000] + [100+100+5000]/10

256*[100]+[5200] /10

256*5300/10

135680


Cisco Employee

Eigrp metric calculations

Hi,

Yes, compute the metric just from R1's perspective.

Metric = 256 * [10^7 / 100000] + 256 * [100 + 5000] / 10 = 256 * 100 + 256 * 510 = 256 * (100+510) = 256*610 = 156160

Best regards,

Peter

New Member

Eigrp metric calculations

Understood. but actually tha interface delay is DLY 100 usec from r1 view,

fastEthernet1/0 is up, line protocol is up

  Hardware is i82543 (Livengood), address is ca00.14f4.001c (bia ca00.14f4.001c)

  Internet address is 192.168.1.1/24

  MTU 1500 bytes, BW 100000 Kbit/sec, DLY 100 usec


Cisco Employee

Eigrp metric calculations

Hi,

but actually tha interface delay is  DLY 100 usec from r1 view, 

In usecs, yes, but the metric formula needs tens of usecs. That is why I wrote 256 * [100 + 5000] / 10 in my computation in the previous response.

Best regards,

Peter

New Member

Eigrp metric calculations

sorry didn't understand this

metric formula needs

tens of usecs

you mean 100^10 ?

Cisco Employee

Eigrp metric calculations

Hi,

metric formula needs

tens of usecs

you mean 100^10 ?

No. The delay on interfaces is expressed in microseconds, e.g. DLY 100 usec means that the delay is 100 microseconds. However, the metric formula needs the delay to be expressed in different units. A single delay unit in the metric formula corresponds to 10 microseconds. Two units would correspond to 20 microseconds. Five units would correspond to 50 microseconds, and 10 units would correspond to 100 microseconds. In general, N delay units correspond to 10*N microseconds, or vice versa, if you have M microseconds, they constitute M/10 units used by the metric formula

Best regards,

Peter

New Member

Eigrp metric calculations

How it change from 100usec to 5100 microsecond ? in formula

Metric = 256*[10^7/BW(min) in Kbps] + [sum of delay in usec/10]

Metric = 256*[10000000/100000(min) in Kbps] + [100usec/10]   <<<< it should be like this

Routing Descriptor Blocks:

  * 192.168.1.2, from 192.168.1.2, 05:52:13 ago, via FastEthernet1/0

      Route metric is 156160, traffic share count is 1

      Total delay is 5100 microseconds, minimum bandwidth is 100000 Kbit

      Reliability 255/255, minimum MTU 1500 bytes

      Loading 1/255, Hops

Cisco Employee

Eigrp metric calculations

Hi,

How it change from 100usec to 5100 microsecond ?

Because you take the sum of delays in microseconds and then divide them by 10. R1->R2 is 100 microseconds, R2->Lo0 is 5000 microseconds. The sum is 5100 microseconds, or 5100/10=510 units for the metric formula.

Best regards,

Peter

New Member

Eigrp metric calculations

Big thanks for your continues reply and  thanks for not scold me

Cisco Employee

Eigrp metric calculations

Hi,

You are counting the delay on the link between R1 and R2 twice - that is the fault in the computation. When R1 needs to reach R2, it goes over a link that has - from its own viewpoint - a delay of 100 usec. R2's delay on its interface would be relevant for the opposite direction of communication.

Best regards,

Peter

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