AI, ML, Development + Cisco Learning Blog Learning about Machine Learning, Artificial Intelligence, related devlopment topics and formerly Routing and Switching, Datacenter, Security and other topics, CCIE #23664, Frank Wagner

21. Dezember 2006

Calculation EIGRP delay for load balancing

Filed under: IGP Routing — ocsic @ 12:44
The defaults for K1 to K5 you can see with "sh ip prot"
Three
Rack1R1#sh ip prot
Routing Protocol is "eigrp 100"
Outgoing update filter list for all interfaces is not set
Incoming update filter list for all interfaces is not set
Default networks flagged in outgoing updates
Default networks accepted from incoming updates
EIGRP metric weight K1=1, K2=0, K3=1, K4=0, K5=0 EIGRP maximum hopcount 100 EIGRP maximum metric variance 5 Redistributing: eigrp 100 EIGRP NSF-aware route hold timer is 240s Automatic network summarization is not in effect Maximum path: 4 Routing for Networks: 150.1.1.1/32 164.1.12.1/32 164.1.13.1/32 164.1.18.1/32 Routing Information Sources: Gateway         Distance      Last Update 164.1.12.2            90      01:24:30 164.1.13.3            90      01:24:30 164.1.18.8            90      01:24:30 Distance: internal 90 external 170  As you can see as by default K1 and K3 are one and all other values are 0. So from the formular only (K1*Bandwidth + K3*Delay) will count.  The complete formular for calculating metrics is: (107/Bandwidth + Delay/10)*256
While Bandwith is in Kbps and Delay in ms.

So then you just look at:

Rack1R2#sh ip eigrp top 164.1.26.0/24
IP-EIGRP (AS 100): Topology entry for 164.1.26.0/24
State is Passive, Query origin flag is 1, 1 Successor(s), FD is 281600
Routing Descriptor Blocks:
0.0.0.0 (Ethernet0/0), from Connected, Send flag is 0x0
Composite metric is (281600/0), Route is Internal
Vector metric:
Minimum bandwidth is 10000 Kbit
Total delay is 1000 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 0
164.1.12.1 (Serial0/0.12), from 164.1.12.1, Send flag is 0x0
Composite metric is (3561472/3049472), Route is Internal
Vector metric:
Minimum bandwidth is 1280 Kbit
Total delay is 61000 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 3

Rack1R1#sh ip eigrp topology 164.1.26.0/24
IP-EIGRP (AS 100): Topology entry for 164.1.26.0/24
State is Passive, Query origin flag is 1, 1 Successor(s), FD is 3049472
Routing Descriptor Blocks:
164.1.13.3 (Serial0/1), from 164.1.13.3, Send flag is 0x0
Composite metric is (3049472/2537472), Route is Internal
Vector metric:
Minimum bandwidth is 1280 Kbit
Total delay is 41000  microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 2
164.1.12.2 (Serial0/0), from 164.1.12.2, Send flag is 0x0
Composite metric is (15247360/281600), Route is Internal
Vector metric:
Minimum bandwidth is 256 Kbit
Total delay is 204980 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 1

You can see here Bandwidth and Delay for the local interface and the advertised interface from the EIGRP neighbor.

The question is to make the path from R1 to R3 5 times more choosen than from R1 to R2. That means, the metrics from R1 to R2 has to be 5 times lower than from R1 to R3. Or in other words:
5*MetricR3 = MetricR2

Then you can calculate with the given formular. (Note, for me Delay from R1 to R3 is 41000 and not 40100 like in the example from the Solutionguide.)

5*(107/1280 + 41000/10)*256 = (107/256 + DelayToR2/10)*256

5*(107/1280 + 41000/10) = (107/256 + DelayToR2/10)
5*(7812,5 + 4100) = (39062,5 + DelayToR2/10)
5*(7812 + 4100) = (39062 + DelayToR2/10)
5*11912 = 39062 + DelayToR2/10
59560 = 39062 + DelayToR2/10
20498 = DelayToR2/10
204980 = DelayToR2

Then for my i had to subtract the Delay from R2 (1000ms).

204980 - 1000 = 203980

Dividing it through 10 = 20398

delay for my int s0/0 from R1 to R2 is 20398. So if you see the metric from R1 to R2 and R3:

Rack1R1#sh ip eigrp topology 164.1.26.0/24
IP-EIGRP (AS 100): Topology entry for 164.1.26.0/24
State is Passive, Query origin flag is 1, 1 Successor(s), FD is 3049472
Routing Descriptor Blocks:
164.1.13.3 (Serial0/1), from 164.1.13.3, Send flag is 0x0
Composite metric is (3049472 /2537472), Route is Internal
Vector metric:
Minimum bandwidth is 1280 Kbit
Total delay is 41000 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 2
164.1.12.2 (Serial0/0), from 164.1.12.2, Send flag is 0x0
Composite metric is (15247360/281600), Route is Internal
Vector metric:
Minimum bandwidth is 256 Kbit
Total delay is 204980 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 1

3049472*5 = 15247360

That the feassible successor has a 5 times higher metric than the successor. With variance 5 you let this feassible successor also being installed as a valid path.

Keine Kommentare »

No comments yet.

RSS feed for comments on this post. TrackBack URL

Leave a comment

Powered by WordPress