mongrel unicorn - Mar 2011 - raindrops updates (mainly for Linux users)

I figured I should cross post here.  raindrops (listen queue stats
package for Unicorn/Rainbows!) has gotten some largish updates over the
past few weeks:

raindrops 0.5.0 release announcement:
  http://mid.gmane.org/20110317033547.GA30653 at dcvr.yhbt.net

demo site changes (will probably be in 0.6.0 soon):
  http://mid.gmane.org/20110318094637.GA3526 at dcvr.yhbt.net

Feedback would be greatly appreciated (either here or
raindrops at librelist.com).  There are no backwards-incompatible changes
for existing users as far as I''m aware of.

Thanks for reading!

-- 
Eric Wong

On 03/18/2011 09:09 PM, Eric Wong wrote:
> I figured I should cross post here.  raindrops (listen queue stats
> package for Unicorn/Rainbows!) has gotten some largish updates over the
> past few weeks:
>
> raindrops 0.5.0 release announcement:
>    http://mid.gmane.org/20110317033547.GA30653 at dcvr.yhbt.net
>
> demo site changes (will probably be in 0.6.0 soon):
>    http://mid.gmane.org/20110318094637.GA3526 at dcvr.yhbt.net
>
> Feedback would be greatly appreciated (either here or
> raindrops at librelist.com).  There are no backwards-incompatible changes
> for existing users as far as I''m aware of.
This update is really great! Now we can see ''Mean'' users
awaiting for
their requests to be served - this information is very useful, because 
we now can calculate avarage wait time for each request.

For example we know that an average request time is 0.15 sec, than if 
our mean queue is 32 requests, each request will be served in 
(32+1)*0.15=4.95 seconds. This information is very useful and even 
NewRelics cannot provide it.

I''m going to build graphs for this data. You can see live stats for
700k
request per day project running on 2x Xeon E5410 2.33 Ghz (4 cores each):

http://kinokopilka.tv:9292/queued/%2Fvar%2Fwww%2Fkk%2Fcurrent%2Ftmp%2Fsockets%2Funicorn.sock.html

On 03/23/2011 09:12 PM, Troex Nevelin wrote:
> For example we know that an average response time is 0.15 sec, than if 
> our mean queue is 32 requests, each request will be served in 
> (32+1)*0.15=4.95 seconds.
Probably it should be:
((mean_queue + 1) * avg_response_time) / workers 

so in my example it would be:
((32 + 1) * 0.15) / 8 =~ 0.62 sec

Troex Nevelin <list at mrtech.ru> wrote:
> This update is really great!
Awesome, your feedback is much appreciated.   I''ll forward a link to
your posts to the raindrops list.

-- 
Eric Wong