WRED queue calculations

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Sep 28th, 2007
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Does anyone know how often WRED calculates the average queue length? I know the general formula for AVG_Q(t) for time=t, with exponential weight=n is

AVG_Q(t) = (AVG_Q(t-1) * (1-(1/2^n)) + (CURRENT_Q * 1/2^n)

So, for a standard n=9, the formula becomes:

AVG_Q(t)= (AVG_Q(t-1) * (511/512)) + (CURRENT_Q/512)

I assume that CURRENT_Q is an instantaneous reading at the time of the calculation. Does anyone know how long it takes until AVG_Q(t) becomes AVG_Q(t-1)?? Is it once a second? Every time a new packet arrives?

With the default value of n=9, it takes several thousand iterations for the AVG_Q to match the mean queue depth.... I set up a quick spreadsheet to calculate the AVG_Q with various values of n out to 4096 iterations. If I set CURRENT_Q to a constant value of 40, it takes 2242 iterations for AVG_Q to catch up. If I set CURRENT_Q to a random number between 0 and 40, it takes AVG_Q about 1500 iterations to catch up.

I'm trying to figure out if I want to stay with the default n=9 or tune it lower, but that would depend on how frequently the computation runs. If it's once per incoming packet, that's probably fine -- I often see several hundred packets/second on my WAN interfaces, so 1800 iterations would only take about 10 seconds. Anyone know for certain? Thanks!!



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umedryk Thu, 10/04/2007 - 14:12
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For an explanation of how the Cisco WRED implementation determines parameters to use in the WRED queue size calculations and how to determine optimum values to use for the weight factor,

average = (old_average * (1-2 -n)) + (current_queue_size * 2 -n)

where n is the exponential weight factor, a user-configurable value.

For high values of n, the previous average becomes more important. A large factor smooths out the peaks and lows in queue length. The average queue size is unlikely to change very quickly, avoiding drastic swings in size. The WRED process will be slow to start dropping packets, but it may continue dropping packets for a time after the actual queue size has fallen below the minimum threshold. The slow-moving average will accommodate temporary bursts in traffic

IanTarasevitsch1995 Thu, 10/04/2007 - 22:51
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Yep. Got that part from reading the multiple pages on the Cisco web site. What the text you quoted does not answer is *how often* the calculation occurs. Right now, I'm banking that the average is recomputed every time a new packet arrives. If the process runs slower than that, I probably need to reduce my exponential-weight option. Thanks for answering, though!




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