On Mon, Jul 21, 2008 at 09:51:24AM -0700, David Miller wrote: > > How so? If the TX hash is well distributed, which it should be, > it is at least going to approximate the distribution provided by > the RX hash. This is a matter of probabilities :) In general, if the TX hash and the RX hash are completely unrelated, then the end result of the hash distribution should be independent of each other (independent in the probablistic sense). That is, for the flows which have been RX hashed into one queue, they should be hashed on average across all queues by the TX hash. Conversely, those that have been hashed into one TX queue would be distributed across all RX queues. Now if you've bound each RX queue to a specific CPU, then this means for a given TX queue, its packets are going to come from all the CPUs (well all those involved in network reception anyway). As each TX queue is designed to be accessed by only one CPU at a time, somebody somewhere has to pay for all this synchronisation :) Cheers, -- Visit Openswan at http://www.openswan.org/ Email: Herbert Xu ~{PmV>HI~} <herbert@xxxxxxxxxxxxxxxxxxx> Home Page: http://gondor.apana.org.au/~herbert/ PGP Key: http://gondor.apana.org.au/~herbert/pubkey.txt -- To unsubscribe from this list: send the line "unsubscribe linux-wireless" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
