Good Morning,
This problem has been plaguing me since before Christmas. I have a graph that I apply trendlines to then format the trendline to get a more detailed equation for, 6 Sig Figs rather than excels default mode. However when I come to extract the equation of the trendlines, sometimes I get the full 6 sig fig version of the equation and other times I get the default version. I can see no logic behind why this jumps from case to case, although I'm sure there is a reason for this. The code that I use is below (I'm also sure that there may be far more efficient ways to code this - currently just trying to get it to work)
Set t = My_Chart.Chart.SeriesCollection(X).Trendlines(1)
t.DataLabel.NumberFormat = "0.00000E+00"
Application.Wait (Now + TimeValue("0:00:01"))
Eqn_Len = Len(t.DataLabel.Text)
LHS_Len = Round(Eqn_Len / 2)
RHS_Len = Eqn_Len - LHS_Len
LHS = Left(t.DataLabel.Text, LHS_Len)
RHS = Right(t.DataLabel.Text, RHS_Len)
Equation_Formula = LHS & RHS
Cells(95, 1) = Equation_Formula
Equation_Formula_Main = Replace(Equation_Formula, "y=", "")
Equation_Formula_Main = Replace(Equation_Formula_Main, "x3", "x^3")
Equation_Formula_Main = Replace(Equation_Formula_Main, "x2", "x^2")
Cells(95, 1).Value = Equation_Formula_Main
Equation_Formula_Main = Right(Equation_Formula_Main, Len(Equation_Formula_Main) - 2)
The Trendline I'm using is a 3rd Order Polynomial with about 15 -30 points, dependant on the data. Can anyone see any reason why this would be the case?
To aid, I've added an example.
I have defined all lengths as double, the variables that contain strings as strings and t as a trendline
My current equation = "y = -8.80186E-09X^3 + 1.28815E-04X^2 - 5.17488E-01X + 7.06521E+02. This gives the variable Eqn_Len a value of 42, which is clearly wrong, however the value of Len(t.DataLabel.Text) is equal to 63. Following through the code the variable LHS becomes "y = -9E-09X^3 + 0.0001" and the variable RHS becomes "x^2 - 0.5175X + 706.52. You can see that this equation is excels default version of the trendline rather than the formated version that it should be.
Thank you for any help you may be able to provide plus any light you can shed on this rather small but annoying issue.
Related
I am working on a MonteCarlo simulation model and part of it is to calculate the following formula:
X = Sqr(1-p)Y + Sqr(p)Z,
Where:
Y and Z are randomly obtained values based (idiosyncratic and systematic factors, respectviely) on a standard normal (inv.) distribution, calculated as:
Application.WorksheetFunction.NormInv (Rnd(), mean, sd)
p represents a correlation factor.
My aim is to square root a recalled formula, however when I try the following (inserting the first Sqr), it does not work and gives an error:
Matrix (n, sims) = (R * Sqr(Application.WorksheetFunction.NormInv(Rnd(), mean, sd))) + (Sqr(1 - R) * RandomS(s, x))
where:
R: Correlation factor
RandomS(s,x): generated matrix with Z values.
I don't want to go into too much details about the background and other variables, as the only problem I am getting is with Square Rooting the equation.
Error message I recieve reads:
Run-time error '5':
Invalid procedure call or argument
When I click debug it takes me to the formula, therefore there must be something wrong with the syntax.
Can you help with directly squaring the formula?
Thank you!
Andrew
Square root is simply Sqr.
It works fine in Excel VBA, so for example:
MsgBox Sqr(144)
...returns 12.
Just don't confuse it with the syntax for a worksheet function with is SQRT.
If you're still having an issue with your formula, tit must be with something other than the Square Root function, and I'd suggest you check the values of your variable, and make sure they are properly declared (preferably with Option Explicit at the top of the module).
Also make sure that you're passing Sqr a positive number.
Documentation: Sqr Function
I'm not a math major, but with your formula:
X = Sqr(1-p)Y + Sqr(p)Z,
...you specified how Y and Z are calculated, so calculate them separately to keep it simple:
Dim X as Double, Y as Double, Z as Double
Y = Application.WorksheetFunction.NormInv (Rnd(), mean, sd)
Z = Application.WorksheetFunction.NormInv (Rnd(), mean, sd)
Assuming the comma is not supposed to be in the formula, and having no idea what p is, your final code to calculate X is:
X = Sqr(1-p) * Y + Sqr(p) * Z
I am having a bit of trouble with Sympy's solveset. I am trying to use Sympy to find a solution to an basic circuit analysis question involving three unknown resistors and two equations. I realize that I will have to guess at the value of one of the resistors and then calculate the value of the other two resistors. I am using superposition to solve the circuit.
This is the circuit and superposition I am trying to solve
import sympy
V_outa, V_outb, R_1, R_2, R_3, V_1, V_2 = symbols('V_outa V_outb R_1 R_2 R_3 V_1 V_2')
### Here are a bunch of variable definitions.
R_eq12 = R_2*R_3/(R_2+R_3)
R_eq123 = R_eq12 + R_1
V_outa = V_1 *R_eq12/R_eq123
R_eq13 = R_1*R_3/(R_1+R_3)
R_eq123b = R_2 + R_eq13
V_outb = V_2 * R_eq13/R_eq123b
### Here is my governing equation.
V_out = 0.5* V_outb + (1/6) *V_outa
### Now I can start setting up the equations to solve. This sets the
### coefficient of the V1 term equal to the 1/2 in the governing equation
### and the coefficient of the V2 term equal to 1/6. I have also guessed
### that R_3 is equal to 10 ohms.
eq1 = Eq(1.0/2.0, V_out.coeff(V_2).subs(R_3, 10))
eq2 = Eq(1.0/6.0, V_out.coeff(V_2).subs(R_3, 10))
#### Now when I try to solve eq1 and eq2 with solveset, I get an error.
solveset([eq1, eq2], (R_2, R_3))
And here is the error I get:
ValueError: [Eq(0.500000000000000, 10*R_1/((R_1 + 10)*(10*R_1/(R_1 +
10) + R_2))), Eq(0.166666666666667, 10*R_1/((R_1 + 10)*(10*R_1/(R_1 +
10) + R_2)))] is not a valid SymPy expression
The other thing I don't understand is the set type I get when I try to solve it this way. Could someone also explain what set type this is, and how to make use of it?
expr3 = -1.0/2.0 + V_out.coeff(V_2).subs(R_3, 10)
solveset(expr3, R_2)
A screen shot of the weird set type
Any help would be much appreciated. I know it is a solvable set because Wolfram Alpha had no problems with it.
Thanks!
David
100e^0.25*y = 97.5
Solving for y
Using Excel Solver
I tried using empty column entry for y in 'By changing cells' and Set objective function as LHS of above equation (empty column entry in equation included) equal to value of 97.5 in solver.
It gives no solution
How do I do this?
It's a bit ambiguous what you're asking...
Literal math interpretation: 100*(e^0.25)*y = 97.5
Then y = 97.5 / ( 100 * exp(.25)) = .759
My guess of what you want: 100*e^(0.25*y) = 97.5
Then y = ln(97.5/100) / .25 = -.101
Another possibility: (100 * e)^(0.25 * y) = 97.5
Then y = (ln(97.5) / ln(100*e)) / .25 = 3.268
Whatever it is, this doesn't need solver!
You don't really need the solver. Just re-arrange your formula to solve for Y. Since y = b^x is the same as log(b)Y = x (log of Y, with base b)
Your formula above is the same as:
Y = (log(100e)97.5))/.25
(Read aloud, that's log of 97.5, with base 100e, divided by .25
So, Y = 3.268305672
(Bonus points for someone who can tell me how to format this so the Log looks correct)
The question is "How do I solve this exponential equation on Excel Solver?" which is a fair enough question, as it points to trying to understand how to set up solver.
My interpretation of the equation provided is given in this screenshot ...
The solver dialog box is then setup as follows ...
Of note:
This is a non-linear equation and needs GRG Nonlinear. If you choose LP Simplex, it will not pass the linearity test.
Ensure "Make Unconstrained Variables Non-Negative" is not checked.
It provided this result for me ...
A more precise answer can be obtained by decreasing the "Convergence" value on the GRG Non-Linear Options dialog.
A problem this simple can also be solved using Goal Seek.
I am trying to calculate a few hundred rows of data, solving a system of linear equations using matrices. I am building my matrices based on the row data. I take the inverse of the resultant 3x3 and then multiply twice, once for x's and once for y's. I get 6 variables from this: a,b,c,d,e and f. How can I copy down the example so that is solves for all rows? I am providing the data and then the formulas I am using now. Right now, if I copy down it skips 3 rows or if I copy down with 3 examples it skips 9 rows. I mean I guess I could go an try to insert extra rows into all 300 so I end up with 900 rows, but there has to be an easier way!
I can't figure out how to get my data to populate in here correctly so here is a link: http://codepad.org/qZwua3h9
Note: I split up the matrix rows so you could see them easier, they are not split up on my sheet.
Edit: If anyone can figure out how to paste the example data here I would welcome it so that this post may be of use to someone in the future. I am not sure how long codepad keeps their pastes.
I gave up and came to the conclusion that there is no reasonable amount of effort that will yield the desired results. Not only was the example case only ONE transformation, but the intended case was for 3 transformations - so three times the work. I came up with a Matlab solution in about 15 minutes. I understand that not everyone has access to Matlab though. So, if someone comes up with any reasonable working excel solution, I would welcome the knowledge and mark that answer as the accepted one. Regardless, here is the Matlab script:
M = csvread('pointData.csv');
T1result = zeros(215,6);
T2result = zeros(215,6);
T3result = zeros(215,6);
for i=1:215,
m = [M(i,1) M(i,2) 1; M(i,3) M(i,4) 1; M(i,5) M(i,6) 1];
x = [M(i,7);M(i,9);M(i,11)];
y = [M(i,8);M(i,10);M(i,12)];
xresult = m\x;
yresult = m\y;
T1result(i,:) = [transpose(xresult),transpose(yresult)];
m = [M(i,7) M(i,8) 1; M(i,9) M(i,10) 1; M(i,11) M(i,12) 1];
x = [M(i,13);M(i,15);M(i,17)];
y = [M(i,14);M(i,16);M(i,18)];
xresult = m\x;
yresult = m\y;
T2result(i,:) = [transpose(xresult),transpose(yresult)];
m = [M(i,13) M(i,14) 1; M(i,15) M(i,16) 1; M(i,17) M(i,18) 1];
x = [M(i,19);M(i,21);M(i,23)];
y = [M(i,20);M(i,22);M(i,24)];
xresult = m\x;
yresult = m\y;
T3result(i,:) = [transpose(xresult),transpose(yresult)];
end
LeafId = csvread('extraColumnsForID.csv');
Tresult = [LeafId, T1result, T2result, T3result];
csvwrite('transforms.csv',Tresult);
I'm trying to implement this extenstion of the Karplus-Strong plucked string algorithm, but I don't understand the notation there used. Maybe it will take years of study, but maybe it won't - maybe you can tell me.
I think the equations below are in the frequency domain or something. Just starting with the first equation, Hp(z), the pick direction lowpass filter. For one direction you use p = 0, for the other, perhaps 0.9. This boils down to to 1 in the first case, or 0.1 / (1 - 0.9 z-1) in the second.
alt text http://www.dsprelated.com/josimages/pasp/img902.png
Now, I feel like this might mean, in coding terms, something towards:
H_p(float* input, int time) {
if (downpick) {
return input[time];
} else {
return some_function_of(input[t], input[t-1]);
}
}
Can someone give me a hint? Or is this futile and I really need all the DSP background to implement this? I was a mathematician once...but this ain't my domain.
So the z-1 just means a one-unit delay.
Let's take Hp = (1-p)/(1-pz-1).
If we follow the convention of "x" for input and "y" for output, the transfer function H = y/x (=output/input)
so we get y/x = (1-p)/(1-pz-1)
or (1-p)x = (1-pz-1)y
(1-p)x[n] = y[n] - py[n-1]
or: y[n] = py[n-1] + (1-p)x[n]
In C code this can be implemented
y += (1-p)*(x-y);
without any additional state beyond using the output "y" as a state variable itself. Or you can go for the more literal approach:
y_delayed_1 = y;
y = p*y_delayed_1 + (1-p)*x;
As far as the other equations go, they're all typical equations except for that second equation which looks like maybe it's a way of selecting either HΒ = 1-z-1 OR 1-z-2. (what's N?)
The filters are kind of vague and they'll be tougher for you to deal with unless you can find some prepackaged filters. In general they're of the form
H = H0*(1+az-1+bz-2+cz-3...)/(1+rz-1+sz-2+tz-3...)
and all you do is write down H = y/x, cross multiply to get
H0 * (1+az-1+bz-2+cz-3...) * x = (1+rz-1+sz-2+tz-3...) * y
and then isolate "y" by itself, making the output "y" a linear function of various delays of itself and of the input.
But designing filters (picking the a,b,c,etc.) is tougher than implementing them, for the most part.