On Mon, Feb 21, 2011 at 10:47 AM, John W. Linville <linville@xxxxxxxxxxxxx> wrote: > On Fri, Feb 18, 2011 at 07:44:30PM -0800, Nathaniel Smith wrote: >> On Fri, Feb 18, 2011 at 1:21 PM, John W. Linville >> <linville@xxxxxxxxxxxxx> wrote: >> > + Â Â Â /* grab timestamp info for buffer control estimates */ >> > + Â Â Â tserv = ktime_sub(ktime_get(), skb->tstamp); >> [...] >> > + Â Â Â Â Â Â Â ewma_add(&sta->sdata->qdata[q].tserv_ns_avg, >> > + Â Â Â Â Â Â Â Â Â Â Â Âktime_to_ns(tserv)); >> >> I think you're still measuring how long it takes one packet to get >> from the end of the queue to the beginning, rather than measuring how >> long it takes each packet to go out? > > Yes, I am measuring how long the driver+device takes to release each > skb back to me (using that as a proxy for how long it takes to get > the fragment to the next hop). ÂActually, FWIW I'm only measuring > that time for those skb's that result in a tx status report. > > I tried to see how your measurement would be useful, but I just don't > see how the number of frames ahead of me in the queue is relevant to > the measured link latency? ÂI mean, I realize that having more packets > ahead of me in the queue is likely to increase the latency for this > frame, but I don't understand why I should use that information to > discount the measured latency...? It depends on which latency you want to measure. The way that I reasoned was, suppose that at some given time, the card is able to transmit 1 fragment every T nanoseconds. Then it can transmit n fragments in n*T nanoseconds, so if we want the queue depth to be 2 ms, then we have n * T = 2 * NSEC_PER_MSEC n = 2 * NSEC_PER_MSEC / T Which is the calculation that you're doing: + sta->sdata->qdata[q].max_enqueued = + max_t(int, 2, 2 * NSEC_PER_MSEC / tserv_ns_avg); But for this calculation to make sense, we need T to be the time it takes the card to transmit 1 fragment. In your patch, you're not measuring that. You're measuring the total time between when a packet is enqueued and when it is transmitted; if there were K packets in the queue ahead of it, then this is the time to send *all* of them -- you're measuring (K+1)*T. That's why in my patch, I recorded the current size of the queue when each packet is enqueued, so I could compute T = total_time / (K+1). Under saturation conditions, K+1 will always equal max_enqueued, so I guess in your algorithm, at the steady state we have max_enqueued = K+1 = 2 * NSEC_PER_MSEC / ((K+1) * T) (K+1)^2 = 2 * NSEC_PER_MSEC / T K+1 = sqrt(2 * NSEC_PER_MSEC / T) So I think under saturation, you converge to setting the queue to the square root of the appropriate size? -- Nathaniel -- 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
