04-24-2012 01:40 AM - edited 03-07-2019 06:18 AM
I am trying to verify the Unequal cost load sharing configured between two routers, as per the below show
R1#sho ip rou 192.168.17.1
Routing entry for 192.168.17.0/24
Known via "eigrp 15", distance 90, metric 2297856, type internal
Redistributing via eigrp 15
Last update from 172.20.15.10 on Serial0/2, 00:26:51 ago
Routing Descriptor Blocks:
172.20.15.10, from 172.20.15.10, 00:26:51 ago, via Serial0/2
Route metric is 10639872, traffic share count is 13
Total delay is 25000 microseconds, minimum bandwidth is 256 Kbit
Reliability 255/255, minimum MTU 1500 bytes
Loading 1/255, Hops 1
* 172.20.15.6, from 172.20.15.6, 00:26:51 ago, via Serial0/1
Route metric is 2297856, traffic share count is 60
Total delay is 25000 microseconds, minimum bandwidth is 1544 Kbit
Reliability 255/255, minimum MTU 1500 bytes
Loading 1/255, Hops 1
Traffic load share is 60:13 , means after each 60 packet throu one route , 13 packets will be sent throu the other route ,, Can any one confirm, why the rate comming like this , as i am receiving in the other couple router rate of 4:1 as below
R2#sho ip rou 192.168.16.1
Routing entry for 192.168.16.0/24
Known via "eigrp 15", distance 90, metric 2809856, type internal
Redistributing via eigrp 15
Last update from 172.20.15.9 on Serial0/1, 02:43:33 ago
Routing Descriptor Blocks:
172.20.15.9, from 172.20.15.9, 02:43:33 ago, via Serial0/1
Route metric is 11151872, traffic share count is 1
Total delay is 45000 microseconds, minimum bandwidth is 256 Kbit
Reliability 255/255, minimum MTU 1500 bytes
Loading 1/255, Hops 2
* 172.20.15.5, from 172.20.15.5, 02:43:33 ago, via Serial0/0
Route metric is 2809856, traffic share count is 4
Total delay is 45000 microseconds, minimum bandwidth is 1544 Kbit
Reliability 255/255, minimum MTU 1500 bytes
Loading 1/255, Hops 2
Solved! Go to Solution.
04-24-2012 01:56 AM
Well my understanding is, it all depends on the metric.
since the route learnt via 172.20.15.6 has a better metric 2297856 than learnt via 172.20.15.10 (10639872).
The traffic share in this case is: 10639872/2297856 which roughly comes to 4.6 . so my guess is instead of mentioning it in decimals they would have taken the LCM of it and shown as a whole number 60 and 13
when you divide 60/13 you end up with the same result as 10639872/2297856
in the other case the equation (as cited above) comes close to a whole Number (4) so the IOS would have just represented that as '4'
This is just my guess... But will let the experts to comment / validate.
-Vijay
04-24-2012 03:57 AM
Vijay is correct. The metric is used in deciding how many packets to send. The lower the metric means the more packets will be sent in favor of using the higher metric path.
Here's a link to verify:
http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a008009437d.shtml#traffic
John
04-24-2012 01:56 AM
Well my understanding is, it all depends on the metric.
since the route learnt via 172.20.15.6 has a better metric 2297856 than learnt via 172.20.15.10 (10639872).
The traffic share in this case is: 10639872/2297856 which roughly comes to 4.6 . so my guess is instead of mentioning it in decimals they would have taken the LCM of it and shown as a whole number 60 and 13
when you divide 60/13 you end up with the same result as 10639872/2297856
in the other case the equation (as cited above) comes close to a whole Number (4) so the IOS would have just represented that as '4'
This is just my guess... But will let the experts to comment / validate.
-Vijay
04-24-2012 03:57 AM
Vijay is correct. The metric is used in deciding how many packets to send. The lower the metric means the more packets will be sent in favor of using the higher metric path.
Here's a link to verify:
http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a008009437d.shtml#traffic
John
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