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CEF - The Polarization Effect

Hello Friends,

In the last post we discussed about CEF Load Balancing. We also saw that per-destination based load balancing uses hash function to load balance the traffic. Since the hash function is used, it is very important that we also understand the concept of traffic polarization. 

Consider a topology in which there are 4 paths in order to reach A0 to B0.

 

{R1, R2, R4}, {R1. R2, R5}, {R1, R3, R6} and {R1, R3, R7}.

Now, considering R1, per-destination will assign the session from A0 to B0 to one of the two available paths {R1. R2} and {R1, R3}. Lets for this example we pick {R1, R2}. Another way of stating this is that the hash function in R1 picked the first path out of two available paths for the session A0B0. Considering R2 next, if the same hash function is employed, then packets in session A0B0 will again pick the first path {R2,R4} of the two available paths.

Now consider all the session A0B0 to AnBn. At router R1, per-destination load balancing will assign half the sessions to path {R1,R2} and half to path {R1,R3}. However at router R2, all the traffic will be assigned to path {R2,R4} and no traffic will use {R2,R5}. This is because all traffic on path {R1,R2} was assigned to the first path from two available paths. This will happen again at R2. The mirror image effect will happen at R3, with all the traffic using path {R3,R7}. 

This effect is know as traffic polarisation - traffic assigned to paths by a hash function are polarised. When this polarised traffic is passed through the same hash function in a later part of the network, all the traffic will flow one way. This is just like light passing through polarisation filters.

Hope we are now clear on the polarization effect.
Cheers..!!!
Happy reading.

Comments (1) -

  • Lalit

    8/29/2012 4:46:11 PM |

    So does the polarisation issue still exist in latest IOS code?

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