Help on personalized "slope" function

I have just found a personalized slope function and I don't understand 
how it works.

Can you help me ?

note: the "SLOPE" function is set in CELL CC187

Slope(Volume1;Volume2;Cost1;Cost2)
=ARGUMENT("Volume1";1)
=ARGUMENT("Volume2";1)
=ARGUMENT("Cost1";1)
=ARGUMENT("Cost2";1)
=RESULT(1)
=(Cost2/Cost1)^(0,693147180559945/LN(Volume2/Volume1))
=RETURN(CC187)


HOUR 	 		REVENUES   	PERFORMANCE
10			� 88		� 8,8
20			� 123		� 6,2

SLOPE: 	0,5	


Note: PERFORMANCE is da REVENUES/HOUR

Nell'esempio riportato i valori presi dalla funzione sono:

Volume1= 88
Volume2= 123
Cost1= 8,8
Cost2= 6,2


My problem is that excel 2003 slope funtion return me a different value



=SLOPE(D6:D7;C6:C7)

note ( Y value first )

=SLOPE(8,8:6,2;88:123)

=0,076


Why I have a different value ?
How can I check the personalized slope function ?


Thank you in advance
0
ExcelUser
12/19/2009 12:10:06 PM
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 > note: the "SLOPE" function is set in CELL CC187
 > =(Cost2/Cost1)^(0,693147180559945/LN(Volume2/Volume1))
 > =RETURN(CC187)

Hi.  Your macro returns 0.484 for me.  Your format is probably set for 1 
digit, hence the 0.5 display.

Your equation that the macro is using is different than the standard 
equation for Slope.

Consider changing 0,693147180559945  to LN(2)

= = = = = = = = =
HTH
Dana DeLouis



On 12/19/09 7:10 AM, ExcelUser wrote:
> I have just found a personalized slope function and I don't understand
> how it works.
>
> Can you help me ?
>
> note: the "SLOPE" function is set in CELL CC187
>
> Slope(Volume1;Volume2;Cost1;Cost2)
> =ARGUMENT("Volume1";1)
> =ARGUMENT("Volume2";1)
> =ARGUMENT("Cost1";1)
> =ARGUMENT("Cost2";1)
> =RESULT(1)
> =(Cost2/Cost1)^(0,693147180559945/LN(Volume2/Volume1))
> =RETURN(CC187)
>
>
> HOUR REVENUES PERFORMANCE
> 10 � 88 � 8,8
> 20 � 123 � 6,2
>
> SLOPE: 0,5
>
>
> Note: PERFORMANCE is da REVENUES/HOUR
>
> Nell'esempio riportato i valori presi dalla funzione sono:
>
> Volume1= 88
> Volume2= 123
> Cost1= 8,8
> Cost2= 6,2
>
>
> My problem is that excel 2003 slope funtion return me a different value
>
>
>
> =SLOPE(D6:D7;C6:C7)
>
> note ( Y value first )
>
> =SLOPE(8,8:6,2;88:123)
>
> =0,076
>
>
> Why I have a different value ?
> How can I check the personalized slope function ?
>
>
> Thank you in advance
0
Dana
12/19/2009 3:04:02 PM
"ExcelUser" <nospam@nospam.invalid> wrote:
> My problem is that excel 2003 slope funtion
> return me a different value

The SLOPE function computes a __linear__ slope; that is, a __linear__ rate 
of change.  Not all rates of change are linear.

I don't know what your "personalized slope" formula is trying to compute, 
but it is not a linear rate of change.

Moreover, I believe you are misusing at least the SLOPE function, probably 
also the "personalized slope" formula.

If HOUR is A2:A3, REVENUE is B2:B3, and PERFORMANCE is C2:C3, then the 
linear rate of change would be SLOPE(B2:B3;A2:A3).

Similarly, I suspect your use of the "personalize slope" formula should be:

(A3/A2)^(LN(2)/LN(B3/B2))

sustituting LN(2) for the approximation 0,693147180559945.

That is a wild-ass guess, since I have no idea what problem that formula is 
a solution for.

But your original usage, effectively 
(Cost2/Cost1)^(LN(2)/LN(Volume2/Volume1)), does no make sense to me.  Noting 
that Cost2 = Volume2/Hour2, your usage becomes:

(Volume2/Volume1)^(LN(2)/LN(Volume2/Volume1))

While that might be a measure of something (GIGO), I do not believe it is a 
measure of "rate of change" (aka "slope").


----- original message -----

"ExcelUser" <nospam@nospam.invalid> wrote in message 
news:hgifpi$6i1$1@aioe.org...
>I have just found a personalized slope function and I don't understand how 
>it works.
>
> Can you help me ?
>
> note: the "SLOPE" function is set in CELL CC187
>
> Slope(Volume1;Volume2;Cost1;Cost2)
> =ARGUMENT("Volume1";1)
> =ARGUMENT("Volume2";1)
> =ARGUMENT("Cost1";1)
> =ARGUMENT("Cost2";1)
> =RESULT(1)
> =(Cost2/Cost1)^(0,693147180559945/LN(Volume2/Volume1))
> =RETURN(CC187)
>
>
> HOUR REVENUES   PERFORMANCE
> 10 � 88 � 8,8
> 20 � 123 � 6,2
>
> SLOPE: 0,5
>
> Note: PERFORMANCE is da REVENUES/HOUR
>
> Nell'esempio riportato i valori presi dalla funzione sono:
>
> Volume1= 88
> Volume2= 123
> Cost1= 8,8
> Cost2= 6,2
>
>
> My problem is that excel 2003 slope funtion return me a different value
>
>
>
> =SLOPE(D6:D7;C6:C7)
>
> note ( Y value first )
>
> =SLOPE(8,8:6,2;88:123)
>
> =0,076
>
>
> Why I have a different value ?
> How can I check the personalized slope function ?
>
>
> Thank you in advance 

0
Joe
12/19/2009 9:00:05 PM
On 12/19/09 10:04 AM, Dana DeLouis wrote:
> =SLOPE(D6:D7;C6:C7)
>
> note ( Y value first )
>
> =SLOPE(8,8:6,2;88:123)
>
> =0,076


Just something to consider.
Your independent variable appears to be hours, so the 'x in a typical 
slope equation would be the hours.

=SLOPE({8.8,6.2},{10,20})

Again, not the solution.  Just something to consider.
= = = = =
Dana DeLouis
0
Dana
12/19/2009 10:17:24 PM
Reply:

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