[rrd-users] Clarification on Computing Totals

permalink · #1 (**permalink**) 04-15-2008

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Hi,

I have been using RRDtool with great success for a few years, and while=

those of us on the pointy end of the swords can get what we need from t=
he
graphs, upper management too often calls me needing totals within minut=
es
for a conference call they are on, or meeting they are in. To combat th=
is,
the no brainer was to write a script, and though I don't profess to be =
a
real Perl programmer, I can usually get what I need.

In this case though, it's not the code that is confusing me, but someth=
ing
I read in Alex's "Computing Amount of Data Transferred" tutorial. BTW, =
I
thought the tutorial was great! I was in the process of writing logic t=
hat
would based upon the dates selected, choose the appropriate RRA. I was
elated to read that fetch takes care of that for me....that all the dat=
a
will come from the same RRA, hence having the same step size.

My point of confusion is that I would think that to compute the total o=
f,
let's say packets, since that's what I'm dealing with, I would need to
maintain a running total of the (current value * step size). This would=

leave me to believe that if the current value was 10 and the step size =
was
7200 I would have a total of 72000 packets in a two hour period. The ma=
th
would be easy as ( 10 packets/second * 7200 seconds ) =3D ((10 * 7200
packets) * seconds/seconds) =3D 72000 packets.

However, reading the tutorial, I infer that this is not correct, and I =
am
having a mental block figuring out why. I am going to paste the piece o=
f
the tutorial that is confusing me, hopefully that will be acceptable.

"Each row in an RRA represents an amount of time and a rate. The rate i=
s
measured in bytes per second, the time in seconds. Some easy math shows=

that unless the amount of time is zero, you can get rid of it:
(byte/second)*second =3D (byte*second)/second =3D byte*(second/second) =
=3D
byte*1=3Dbyte.

(If "second" is zero, then "second/second" is undefined. This is why it=

doesn't work when second equals zero.)

When dealing with other rates, it works similar. If you have multiplied=
the
input by 3600 (for instance to get messages per hour) then the result o=
f
the computation in the paragraph above is also 3600 times what it shoul=
d
be. Remember that when you continue..."

It is the last paragraph that I am most concerned with, for I have
multiplied my packets per second by the step size to achieve the total =
in
each row - maintaining a running total of the rows, and as I understand=
the
above, this is incorrect. As I understand the tutorial, I should just b=
e
maintaining a running total of the values, and not multiply them in eac=
h
row by the step size. Is that correct?

Thank you,

Jeff Petter

=

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<html><body>
Hi, 
 
I have been using RRDtool with great success for a few=
years, and while those of us on the pointy end of the swords can get w=
hat we need from the graphs, upper management too often calls me needin=
g totals within minutes for a conference call they are on, or meeting t=
hey are in. To combat this, the no brainer was to write a script, and t=
hough I don't profess to be a real Perl programmer, I can usually get w=
hat I need. 
 
In this case though, it's not the code that is confusi=
ng me, but something I read in Alex's "Computing Amount of Data Tr=
ansferred" tutorial. BTW, I thought the tutorial was great! I was =
in the process of writing logic that would based upon the dates selecte=
d, choose the appropriate RRA. I was elated to read that fetch takes ca=
re of that for me....that all the data will come from the same RRA, hen=
ce having the same step size. 
 
My point of confusion is that I would think that to co=
mpute the total of, let's say packets, since that's what I'm dealing wi=
th, I would need to maintain a running total of the (current value * st=
ep size). This would leave me to believe that if the current value was =
10 and the step size was 7200 I would have a total of 72000 packets in =
a two hour period. The math would be easy as ( 10 packets/second * 7200=
seconds ) =3D ((10 * 7200 packets) * seconds/seconds) =3D 72000 packet=
s. 
 
However, reading the tutorial, I infer that this is no=
t correct, and I am having a mental block figuring out why. I am going =
to paste the piece of the tutorial that is confusing me, hopefully that=
will be acceptable. 
 
"Each row in an RRA represents an amount o=
f time and a rate. The rate is measured in bytes per second, the time i=
n seconds. Some easy math shows that unless the amount of time is zero,=
you can get rid of it: (byte/second)*second =3D (byte*second)/second =3D=
byte*(second/second) =3D byte*1=3Dbyte.
(If "second" is zero, then "second/second" is un=
defined. This is why it doesn't work when second equals zero.)
When dealing with other rates, it works similar. If you have multipl=
ied the input by 3600 (for instance to get messages per hour) then the =
result of the computation in the paragraph above is also 3600 times wha=
t it should be. Remember that when you continue...&quo=
t;
It is the last paragraph that I am most concerned w=
ith, for I have multiplied my packets per second by the step size to ac=
hieve the total in each row - maintaining a running total of the rows, =
and as I understand the above, this is incorrect. As I understand the t=
utorial, I should just be maintaining a running total of the values, an=
d not multiply them in each row by the step size. Is that correct?</fon=
t>
Thank you,
Jeff Petter

</body></html>=

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