queue waiting times comparison

If those values represent response times in a system, then when I was
responsible for characterizing what the system would do from the
viewpoint of an SLA (service level agreement) with customers using the
system, we usually specified that "90% of the transactions would have
a response time of --- or less".  This took care of most "long tails".
 So it depends on how you are planning to use this data.  We usually
monitored the 90th or 95th percentile to see how a system was
operating day to day.

On Thu, Aug 18, 2011 at 8:52 AM, Petr PIKAL <petr.pikal at precheza.cz> 

Hallo Jim

Thank you and see within text.

jim holtman <jholtman at gmail.com> napsal dne 18.08.2011 14:09:11:

I am not sure why you say that "lapply(ml, mean)" shows (incorrectly)
that the second year has a larger average; it is correct for the 

lapply(ml, my.func)

$y1
    Count      Mean        SD       Min    Median       90%       95%
     Max       Sum
 18.00000  16.83333  12.42980   4.00000  12.50000  37.20000  41.05000
47.00000 303.00000

$y2
    Count      Mean        SD       Min    Median       90%       95%
     Max       Sum
 15.00000  20.06667  25.27694   4.00000  11.00000  45.80000  70.40000
97.00000 301.00000


You have a larger "outlier" in the second year that causes the mean 

be higher.  The median is lower, but I usually look at the 90th
percentile if I am looking at response time from a system and again
the second year has a higher value.

So exactly why do you not "trust" your data?

Well. I trust them, however mean is "correct" central value only when 

are normally distributed or at least symmetrical. As the values are
heavily  distorted I feel that I shall not use mean for comparison of 

sets. Anyway t.test tells me that there is no difference between y2 

y1.

t.test(ml[[1]], ml[[2]])

       Welch Two Sample t-test

data:  ml[[1]] and ml[[2]]
t = -0.452, df = 19.557, p-value = 0.6563
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -18.17781  11.71115
sample estimates:
mean of x mean of y
 16.83333  20.06667

So based on this I probably will never get conclusive result as sd due 

"outliers" will be quite high.

When I do
plot(ecdf(ml[[2]]))
plot(ecdf(ml[[1]]), add=T, col=2)

it seems to me that both sets are almost the same and they differ
substantially only with those "outlier" values.

If I decreased small values of y2 (e.g.)

ml[[2]][ml[[2]]<20] <- ml[[2]][ml[[2]]<20]/2

I get same mean

lapply(ml, mean)
$y1
[1] 16.83333

$y2
[1] 16.1

and t.test tells me that there is no difference between those two 

although I know that most events take half of the time and only few 

longer so for me such set is better (we improved performance for most 

the time however there are still scarce events which take a long 

plot(ecdf(ml[[2]]))
plot(ecdf(ml[[1]]), add=T, col=2)

So still the question stays - what procedure to use for comparison of 

or more sets with such long tailed distribution? - Trimmed mean?, 

...

Thanks.

Regards
Petr

On Thu, Aug 18, 2011 at 7:49 AM, Petr PIKAL <petr.pikal at precheza.cz>

wrote:

Hallo all

I try to find a way how to compare set of waiting times during

different

periods. I tried learn something from queueing theory and used also 

search. There is plenty of ways but I need to find the easiest and

quite

simple.
Here is a list with actual waiting times.

ml <- structure(list(y1 = c(10, 9, 9, 10, 8, 20, 16, 47, 4, 7, 15,
18, 36, 5, 24, 15, 40, 10), y2 = c(97, 10, 26, 11, 11, 10, 5,
13, 19, 5, 5, 59, 4, 16, 10)), .Names = c("y1", "y2"))

par(mfrow=c(1,2))
lapply(ml, hist)

shows that in the first year is more longer waiting times

lapply(ml, mean)

shows (incorrectly) that in the second year there is longer average
waiting time.

lapply(ml, mean)

gives me completely reversed values.

Can you please give me some hints what to use for "correct" and

"simple"

comparison of  waiting times in two or more periods.

Thank you
Petr

-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?