Talk:Weighted round robin

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Unnecessary transformation[edit]

there is unnecessary transformation going on. mathematically this

for each flow f
   f.normalized_weight = f.weight / f.mean_packet_size

min = findSmallestNormalizedWeight

for each flow f
   f.packets_to_be_served = f.normalized_weight / min

is equal to

 min=find smallest waight
 for each flow f
 (f.weight/f.mean_packet_size)/
 (f.weight(0)/f.mean_packet_size)
  = f.weight/f.weight(0)

is there any reason to to do n times more divisions ? if there are some rounding problems it should be mentionned on main page. thanks. — Preceding unsigned comment added by Xchmelmilos (talkcontribs) 10:30, 4 May 2008‎ (UTC)[reply]

The mean packet size can differ from one flow to another.

f.packets_to_be_served = f.normalized_weight / min

is equal to

f.packets_to_be_served = (f.weight / f.mean_packet_size) / (min_f.weight / min_f.mean_packet_size)

In fact the weakness section of this article mentions that having to know the mean packet size is a problem with this algorithm - having the mean packet size cancel out doesn't make sense. --195.37.186.61 (talk) 23:13, 22 August 2012 (UTC)[reply]

Related links[edit]

The following links might be helpful for improving the article but are inappropriate to be included directly:

--EnOreg (talk) 14:09, 23 December 2011 (UTC)[reply]