From: Bob Sinclair (bsin@cox.net)
Date: Thu May 26 2005 - 18:07:13 GMT-3
Hi Gladston,
I don't see 14 ms as the standard deviation. I get 9.7. Here is a fairly
good link that describes how to calculate the variance and the standard
deviation of a sample:
http://helios.bto.ed.ac.uk/bto/statistics/tress3.html
Here is my take:
Sum of squared deviations from the Mean (SSDM)=
Sum of squared values (384317)
minus
Average of the sum of values squared ((6063 squared)/98) = 375102
So SSDM = 384317 - 375102 = 9215
Variance = SSDM/(n-1) = 9215/97 = 95
Standard Deviation = Square root of Variance = sqrt (95) = 9.7
Standard Deviation: the square root of the average of the squares of
deviations about the mean of a set of data. In a normal (symmetric and
mound-shaped) distribution, about two-thirds of the scores fall between +1 and
-1 standard deviations from the mean. 95% within 2 standard deviations of the
Mean. 99% within 3 standard deviations.
So, I do not think he has the right value. BUT, maybe it is rounding error,
or maybe it has been way too long since my last statistics course!
HTH,
Bob Sinclair
CCIE #10427, CCSI 30427, CISSP
www.netmasterclass.net
----- Original Message -----
From: gladston@br.ibm.com
To: ccielab@groupstudy.com
Sent: Thursday, May 26, 2005 2:38 PM
Subject: SAA - Jitter
This is the best explanation I found about SAA jitter results.
But I could not understand how it get 14ms. Can you?
Specifically, I did not get this: "If you subtract the square of the average
from this value and take the square root"
=======================
Quoted from Cisco CookBook
Router1#show rtr operational-state 20
Current Operational State
Entry Number: 20
Modification Time: 10:25:36.000 EST Wed Dec 18 2002
Diagnostics Text:
Last Time this Entry was Reset: Never
Number of Octets in use by this Entry: 1742
Number of Operations Attempted: 22
Current Seconds Left in Life: 93400
Operational State of Entry: active
Latest Operation Start Time: 12:10:36.000 EST Wed Dec 18 2002
RTT Values:
NumOfRTT: 98 RTTSum: 6063 RTTSum2: 384317
Applying some simple statistics, you can use the square value to understand
how the actual values are spread around this average. The mean of the squares
of the round trip times is 3,922 milliseconds2 (just dividing the sum of the
squares by the total number of samples). If you subtract the square of the
average from this value and take the square root, you get a statistical
estimate of the variation in milliseconds. The higher this value, the greater
the spread. In this case, you can calculate that this spread is roughly 14
milliseconds. This means that half of the time, the round trip latency is
within the range 61114ms. Note that the 1 symbol is a standard mathematical
notation that, in this case, indicates a range from 47ms (61-14) to 75ms
(61+14).
=========================
=============
quoted
If you subtract the square of the average from this value and take the
square root, you get a statistical estimate of the variation in
milliseconds... In this case, you can calculate that this spread is roughly 14
milliseconds.
==============
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