Help on personalized "slope" function

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ExcelUser

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
 
note: the "SLOPE" function is set in CELL CC187
=(Cost2/Cost1)^(0,693147180559945/LN(Volume2/Volume1))
=RETURN(CC187)
Click to expand...

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
 
ExcelUser said:
My problem is that excel 2003 slope funtion
return me a different value
Click to expand...

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 -----
 
=SLOPE(D6:D7;C6:C7)

note ( Y value first )

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

=0,076
Click to expand...


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
 
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