% Power output
power_output = MF_t.*difference.*e;
% POWER OUTPUT OPTIMIZATION USING - GOLDEN SEARCH METHOD
x_lower = 0.2;
x_upper = 1.4;
h = (x_upper - x_lower);
r = (sqrt(5) - 1)*0.5;
x2 = x_lower + (1+r)*h;
x3 = x_upper - (1+r)*h;
tol = 0.05;
count = 1;
while h>-tol
f2 = -power_output(x2)
f3 = -power_output(x3)
if f3<f2
x_lower = x2;
x_upper = x_upper;
h = (x_upper - x_lower);
x2 = x3;
x3 = x_upper - (1-r)*h;
elseif f2<f3
x_lower = x_lower;
x_upper = x3;
h = (x_upper - x_lower);
x3 = x2;
x2 = x_lower + (1-r)*h;
else
x_lower = x2;
x_upper = x3;
h = (x_upper - x_lower);
x2 = x_lower + (1+r)*h;
x3 = x_upper - (1+r)*h;
end
interval = x_upper - x_lower;
count = count + 1
if count > 100
break
end
end
answer = (x_upper + x_lower)/2
value = fanswer
Im trying to run this code, it's telling me that there is an error as my arrays f2 and f3 as "Array indices must be positive integers or logical values". I'm wondering would anyone know if taking the absolute values of my arrays will make my optimisation incorrect
The error message is quite informative in this case. You are trying to access the x2th element of f2. A quick calculation shows that on the first iteration x2 = 2.14. It is not possible to take the 2.14th element of f2 because this is is not an integer, and taking the absolute value will not help.
x2 is not an index, it is a value. On each iteration, the Golden Ratio search requires you to actually evaluate power_output with whatever variable set to x2. So, it looks like you need to do this calculation power_output = MF_t.*difference.*e with x=x2.
Related
I would like some help with my program. I still don’t understand where my problem is, since it’s kind of a big mess. So it consists of the main program:
function x = NewtonM(funcF,JacF)
x= zeros(2,1);
x(1) = 1
x(2) = 5
k = 1;
kmax = 100;
TOL = 10^(-7);
while k < kmax
s = J(x)\(-F(x));
x= x + s
if (norm(s,2)< TOL)
break;
endif
end
and these are the fellow functions:
function y = F(x)
x1 = x(1);
x2 = x(2);
y = zeros(2,1);
y(1) = x1+x2-3;
y(2) = x1^2 + x2^2 -9;
end
function z = Z(x)
x1 = x(1);
x2 = x(2);
z = zeros(3,1);
z(1) = x1+x2-3+10^(-7);
z(2) = (x1+10^(-7))^2 + x2^2 -9;
z(3) = x1^2 + (x2+10^(-7))^2 -9;
end
function J = J(x)
x1 = x(1);
x2 = x(2);
J = zeros(2,2);
J(1,1) = (Z(1)-F(1))/(10^(-7))
J(1,2) = (Z(1)-F(1))/(10^(-7))
J(2,1) = (Z(2)-F(2))/(10^(-7))
J(2,2) = (Z(3)-F(2))/(10^(-7))
end
These are the error messages:
The problem is that you are calling both Z and F with only one input, in the function J.
Then, the first thing you do is try to interpret the input as a 2 valued array (x1,x2) but they don't exist, as you defined x as e.g. 1, by doing Z(1).
I wonder if instead of using Z(1) etc, you meant to do z=Z(x) and then use z(1), inside J.
Assume we have three equations:
eq1 = x1 + (x1 - x2) * t - X == 0;
eq2 = z1 + (z1 - z2) * t - Z == 0;
eq3 = ((X-x1)/a)^2 + ((Z-z1)/b)^2 - 1 == 0;
while six of known variables are:
a = 42 ;
b = 12 ;
x1 = 316190;
z1 = 234070;
x2 = 316190;
z2 = 234070;
So we are looking for three unknown variables that are:
X , Z and t
I wrote two method to solve it. But, since I need to run these code for 5.7 million data, it become really slow.
Method one (using "solve"):
tic
S = solve( eq1 , eq2 , eq3 , X , Z , t ,...
'ReturnConditions', true, 'Real', true);
toc
X = double(S.X(1))
Z = double(S.Z(1))
t = double(S.t(1))
results of method one:
X = 316190;
Z = 234060;
t = -2.9280;
Elapsed time is 0.770429 seconds.
Method two (using "fsolve"):
coeffs = [a,b,x1,x2,z1,z2]; % Known parameters
x0 = [ x2 ; z2 ; 1 ].'; % Initial values for iterations
f_d = #(x0) myfunc(x0,coeffs); % f_d considers x0 as variables
options = optimoptions('fsolve','Display','none');
tic
M = fsolve(f_d,x0,options);
toc
results of method two:
X = 316190; % X = M(1)
Z = 234060; % Z = M(2)
t = -2.9280; % t = M(3)
Elapsed time is 0.014 seconds.
Although, the second method is faster, but it still needs to be improved. Please let me know if you have a better solution for that. Thanks
* extra information:
if you are interested to know what those 3 equations are, the first two are equations of a line in 2D and the third equation is an ellipse equation. I need to find the intersection of the line with the ellipse. Obviously, we have two points as result. But, let's forget about the second answer for simplicity.
My suggestion it's to use the second approce,which it's the recommended by matlab for nonlinear equation system.
Declare a M-function
function Y=mysistem(X)
%X(1) = X
%X(2) = t
%X(3) = Z
a = 42 ;
b = 12 ;
x1 = 316190;
z1 = 234070;
x2 = 316190;
z2 = 234070;
Y(1,1) = x1 + (x1 - x2) * X(2) - X(1);
Y(2,1) = z1 + (z1 - z2) * X(2) - X(3);
Y(3,1) = ((X-x1)/a)^2 + ((Z-z1)/b)^2 - 1;
end
Then for solving use
x0 = [ x2 , z2 , 1 ];
M = fsolve(#mysistem,x0,options);
If you may want to reduce the default precision by changing StepTolerance (default 1e-6).
Also for more increare you may want to use the jacobian matrix for greater efficencies.
For more reference take a look in official documentation:
fsolve Nonlinear Equations with Analytic Jacobian
Basically giving the solver the Jacobian matrix of the system(and special options) you can increase method efficency.
EDIT: The code that I have pasted is too long. Basicaly I dont know how to work with the second code, If I know how calculate alpha from the second code I think my problem will be solved. I have tried a lot of input arguments for the second code but it does not work!
I have written following code to solve a convex optimization problem using Gradient descend method:
function [optimumX,optimumF,counter,gNorm,dx] = grad_descent()
x0 = [3 3]';%'//
terminationThreshold = 1e-6;
maxIterations = 100;
dxMin = 1e-6;
gNorm = inf; x = x0; counter = 0; dx = inf;
% ************************************
f = #(x1,x2) 4.*x1.^2 + 2.*x1.*x2 +8.*x2.^2 + 10.*x1 + x2;
%alpha = 0.01;
% ************************************
figure(1); clf; ezcontour(f,[-5 5 -5 5]); axis equal; hold on
f2 = #(x) f(x(1),x(2));
% gradient descent algorithm:
while and(gNorm >= terminationThreshold, and(counter <= maxIterations, dx >= dxMin))
g = grad(x);
gNorm = norm(g);
alpha = linesearch_strongwolfe(f,-g, x0, 1);
xNew = x - alpha * g;
% check step
if ~isfinite(xNew)
display(['Number of iterations: ' num2str(counter)])
error('x is inf or NaN')
end
% **************************************
plot([x(1) xNew(1)],[x(2) xNew(2)],'ko-')
refresh
% **************************************
counter = counter + 1;
dx = norm(xNew-x);
x = xNew;
end
optimumX = x;
optimumF = f2(optimumX);
counter = counter - 1;
% define the gradient of the objective
function g = grad(x)
g = [(8*x(1) + 2*x(2) +10)
(2*x(1) + 16*x(2) + 1)];
end
end
As you can see, I have commented out the alpha = 0.01; part. I want to calculate alpha via an other code. Here is the code (This code is not mine)
function alphas = linesearch_strongwolfe(f,d,x0,alpham)
alpha0 = 0;
alphap = alpha0;
c1 = 1e-4;
c2 = 0.5;
alphax = alpham*rand(1);
[fx0,gx0] = feval(f,x0,d);
fxp = fx0;
gxp = gx0;
i=1;
while (1 ~= 2)
xx = x0 + alphax*d;
[fxx,gxx] = feval(f,xx,d);
if (fxx > fx0 + c1*alphax*gx0) | ((i > 1) & (fxx >= fxp)),
alphas = zoom(f,x0,d,alphap,alphax);
return;
end
if abs(gxx) <= -c2*gx0,
alphas = alphax;
return;
end
if gxx >= 0,
alphas = zoom(f,x0,d,alphax,alphap);
return;
end
alphap = alphax;
fxp = fxx;
gxp = gxx;
alphax = alphax + (alpham-alphax)*rand(1);
i = i+1;
end
function alphas = zoom(f,x0,d,alphal,alphah)
c1 = 1e-4;
c2 = 0.5;
[fx0,gx0] = feval(f,x0,d);
while (1~=2),
alphax = 1/2*(alphal+alphah);
xx = x0 + alphax*d;
[fxx,gxx] = feval(f,xx,d);
xl = x0 + alphal*d;
fxl = feval(f,xl,d);
if ((fxx > fx0 + c1*alphax*gx0) | (fxx >= fxl)),
alphah = alphax;
else
if abs(gxx) <= -c2*gx0,
alphas = alphax;
return;
end
if gxx*(alphah-alphal) >= 0,
alphah = alphal;
end
alphal = alphax;
end
end
But I get this error:
Error in linesearch_strongwolfe (line 11) [fx0,gx0] = feval(f,x0,d);
As you can see I have written the f function and its gradient manually.
linesearch_strongwolfe(f,d,x0,alpham) takes a function f, Gradient of f, a vector x0 and a constant alpham. is there anything wrong with my declaration of f? This code works just fine if I put back alpha = 0.01;
As I see it:
x0 = [3; 3]; %2-element column vector
g = grad(x0); %2-element column vector
f = #(x1,x2) 4.*x1.^2 + 2.*x1.*x2 +8.*x2.^2 + 10.*x1 + x2;
linesearch_strongwolfe(f,-g, x0, 1); %passing variables
inside the function:
[fx0,gx0] = feval(f,x0,-g); %variable names substituted with input vars
This will in effect call
[fx0,gx0] = f(x0,-g);
but f(x0,-g) is a single 2-element column vector with these inputs. Assingning the output to two variables will not work.
You either have to define f as a proper named function (just like grad) to output 2 variables (one for each component), or edit the code of linesearch_strongwolfe to return a single variable, then slice that into 2 separate variables yourself afterwards.
If you experience a very rare kind of laziness and don't want to define a named function, you can still use an anonymous function at the cost of duplicating code for the two components (at least I couldn't come up with a cleaner solution):
f = #(x1,x2) deal(4.*x1(1)^2 + 2.*x1(1)*x2(1) +8.*x2(1)^2 + 10.*x1(1) + x2(1),...
4.*x1(2)^2 + 2.*x1(2)*x2(2) +8.*x2(2)^2 + 10.*x1(2) + x2(2));
[fx0,gx0] = f(x0,-g); %now works fine
as long as you always have 2 output variables. Note that this is more like a proof of concept, since this is ugly, inefficient, and very susceptible to typos.
I wrote a python function to get the parameters of the following cosine function:
param = Parameters()
param.add( 'amp', value = amp_guess, min = 0.1 * amp_guess, max = amp_guess )
param.add( 'off', value = off_guess, min = -10, max = 10 )
param.add( 'shift', value = shift_guess[0], min = 0, max = 2 * np.pi, )
fit_values = minimize( self.residual, param, args = ( azi_unique, los_unique ) )
def residual( self, param, azi, data ):
"""
Parameters
----------
Returns
-------
"""
amp = param['amp'].value
off = param['off'].value
shift = param['shift'].value
model = off + amp * np.cos( azi - shift )
return model - data
In Matlab how can get the amplitude, offset and shift of the cosine function?
My experience tells me that it's always good to depend as little as possible on toolboxes. For your particular case, the model is simple and doing it manually is pretty straightforward.
Assuming that you have the following model:
y = B + A*cos(w*x + phi)
and that your data is equally-spaced, then:
%// Create some bogus data
A = 8;
B = -4;
w = 0.2;
phi = 1.8;
x = 0 : 0.1 : 8.4*pi;
y = B + A*cos(w*x + phi) + 0.5*randn(size(x));
%// Find kick-ass initial estimates
L = length(y);
N = 2^nextpow2(L);
B0 = (max(y(:))+min(y(:)))/2;
Y = fft(y-B0, N)/L;
f = 5/(x(2)-x(1)) * linspace(0,1,N/2+1);
[A0,I] = max( 2*abs(Y(1:N/2+1)) );
w0 = f(I);
phi0 = 2*imag(Y(I));
%// Refine the fit
sol = fminsearch(#(t) sum( (y(:)-t(1)-t(2)*cos(t(3)*x(:)+t(4))).^2 ), [B0 A0 w0 phi0])
Results:
sol = %// B was -4 A was 8 w was 0.2 phi was 1.8
-4.0097e+000 7.9913e+000 1.9998e-001 1.7961e+000
MATLAB has a function called lsqcurvefit in the optimisation toolbox:
lsqcurvefit(fun,X0,xdata,ydata,lbound,ubound);
where fun is the function to fit, x0 is the initial parameter guess, xdata and ydata are self-explanatory, and lbound and ubound are the lower and upper bounds to the parameters. So, for instance, you might have a function:
% x(1) = amp
% x(2) = shift
% x(3) = offset
% note cosd instead of cos, because your data appears to be in degrees
cosfit = #(x,xdata) x(1) .* cosd(xdata - x(2)) + x(3);
You would then call the lsqcurvefit function as follows:
guess = [7,150,0.5];
lbound = [-10,0,-10]
ubound = [10,360,10]
fit_values = lsqcurvefit(cosfit,guess,azi_unique,los_unique,lbound,ubound);
I'm trying to write a cubic spline interpolation program. I have written the program but, the graph is not coming out correctly. The spline uses natural boundary conditions(second dervative at start/end node are 0). The code is in Matlab and is shown below,
clear all
%Function to Interpolate
k = 10; %Number of Support Nodes-1
xs(1) = -1;
for j = 1:k
xs(j+1) = -1 +2*j/k; %Support Nodes(Equidistant)
end;
fs = 1./(25.*xs.^2+1); %Support Ordinates
x = [-0.99:2/(2*k):0.99]; %Places to Evaluate Function
fx = 1./(25.*x.^2+1); %Function Evaluated at x
%Cubic Spline Code(Coefficients to Calculate 2nd Derivatives)
f(1) = 2*(xs(3)-xs(1));
g(1) = xs(3)-xs(2);
r(1) = (6/(xs(3)-xs(2)))*(fs(3)-fs(2)) + (6/(xs(2)-xs(1)))*(fs(1)-fs(2));
e(1) = 0;
for i = 2:k-2
e(i) = xs(i+1)-xs(i);
f(i) = 2*(xs(i+2)-xs(i));
g(i) = xs(i+2)-xs(i+1);
r(i) = (6/(xs(i+2)-xs(i+1)))*(fs(i+2)-fs(i+1)) + ...
(6/(xs(i+1)-xs(i)))*(fs(i)-fs(i+1));
end
e(k-1) = xs(k)-xs(k-1);
f(k-1) = 2*(xs(k+1)-xs(k-1));
r(k-1) = (6/(xs(k+1)-xs(k)))*(fs(k+1)-fs(k)) + ...
(6/(xs(k)-xs(k-1)))*(fs(k-1)-fs(k));
%Tridiagonal System
i = 1;
A = zeros(k-1,k-1);
while i < size(A)+1;
A(i,i) = f(i);
if i < size(A);
A(i,i+1) = g(i);
A(i+1,i) = e(i);
end
i = i+1;
end
for i = 2:k-1 %Decomposition
e(i) = e(i)/f(i-1);
f(i) = f(i)-e(i)*g(i-1);
end
for i = 2:k-1 %Forward Substitution
r(i) = r(i)-e(i)*r(i-1);
end
xn(k-1)= r(k-1)/f(k-1);
for i = k-2:-1:1 %Back Substitution
xn(i) = (r(i)-g(i)*xn(i+1))/f(i);
end
%Interpolation
if (max(xs) <= max(x))
error('Outside Range');
end
if (min(xs) >= min(x))
error('Outside Range');
end
P = zeros(size(length(x),length(x)));
i = 1;
for Counter = 1:length(x)
for j = 1:k-1
a(j) = x(Counter)- xs(j);
end
i = find(a == min(a(a>=0)));
if i == 1
c1 = 0;
c2 = xn(1)/6/(xs(2)-xs(1));
c3 = fs(1)/(xs(2)-xs(1));
c4 = fs(2)/(xs(2)-xs(1))-xn(1)*(xs(2)-xs(1))/6;
t1 = c1*(xs(2)-x(Counter))^3;
t2 = c2*(x(Counter)-xs(1))^3;
t3 = c3*(xs(2)-x(Counter));
t4 = c4*(x(Counter)-xs(1));
P(Counter) = t1 +t2 +t3 +t4;
else
if i < k-1
c1 = xn(i-1+1)/6/(xs(i+1)-xs(i-1+1));
c2 = xn(i+1)/6/(xs(i+1)-xs(i-1+1));
c3 = fs(i-1+1)/(xs(i+1)-xs(i-1+1))-xn(i-1+1)*(xs(i+1)-xs(i-1+1))/6;
c4 = fs(i+1)/(xs(i+1)-xs(i-1+1))-xn(i+1)*(xs(i+1)-xs(i-1+1))/6;
t1 = c1*(xs(i+1)-x(Counter))^3;
t2 = c2*(x(Counter)-xs(i-1+1))^3;
t3 = c3*(xs(i+1)-x(Counter));
t4 = c4*(x(Counter)-xs(i-1+1));
P(Counter) = t1 +t2 +t3 +t4;
else
c1 = xn(i-1+1)/6/(xs(i+1)-xs(i-1+1));
c2 = 0;
c3 = fs(i-1+1)/(xs(i+1)-xs(i-1+1))-xn(i-1+1)*(xs(i+1)-xs(i-1+1))/6;
c4 = fs(i+1)/(xs(i+1)-xs(i-1+1));
t1 = c1*(xs(i+1)-x(Counter))^3;
t2 = c2*(x(Counter)-xs(i-1+1))^3;
t3 = c3*(xs(i+1)-x(Counter));
t4 = c4*(x(Counter)-xs(i-1+1));
P(Counter) = t1 +t2 +t3 +t4;
end
end
end
P = P';
P(length(x)) = NaN;
plot(x,P,x,fx)
When I run the code, the interpolation function is not symmetric and, it doesn't converge correctly. Can anyone offer any suggestions about problems in my code? Thanks.
I wrote a cubic spline package in Mathematica a long time ago. Here is my translation of that package into Matlab. Note I haven't looked at cubic splines in about 7 years, so I'm basing this off my own documentation. You should check everything I say.
The basic problem is we are given n data points (x(1), y(1)) , ... , (x(n), y(n)) and we wish to calculate a piecewise cubic interpolant. The interpolant is defined as
S(x) = { Sk(x) when x(k) <= x <= x(k+1)
{ 0 otherwise
Here Sk(x) is a cubic polynomial of the form
Sk(x) = sk0 + sk1*(x-x(k)) + sk2*(x-x(k))^2 + sk3*(x-x(k))^3
The properties of the spline are:
The spline pass through the data point Sk(x(k)) = y(k)
The spline is continuous at the end-points and thus continuous everywhere in the interpolation interval Sk(x(k+1)) = Sk+1(x(k+1))
The spline has continuous first derivative Sk'(x(k+1)) = Sk+1'(x(k+1))
The spline has continuous second derivative Sk''(x(k+1)) = Sk+1''(x(k+1))
To construct a cubic spline from a set of data point we need to solve for the coefficients
sk0, sk1, sk2 and sk3 for each of the n-1 cubic polynomials. That is a total of 4*(n-1) = 4*n - 4 unknowns. Property 1 supplies n constraints, and properties 2,3,4 each supply an additional n-2 constraints. Thus we have n + 3*(n-2) = 4*n - 6 constraints and 4*n - 4 unknowns. This leaves two degrees of freedom. We fix these degrees of freedom by setting the second derivative equal to zero at the start and end nodes.
Let m(k) = Sk''(x(k)) , h(k) = x(k+1) - x(k) and d(k) = (y(k+1) - y(k))/h(k). The following
three-term recurrence relation holds
h(k-1)*m(k-1) + 2*(h(k-1) + h(k))*m(k) + h(k)*m(k+1) = 6*(d(k) - d(k-1))
The m(k) are unknowns we wish to solve for. The h(k) and d(k) are defined by the input data.
This three-term recurrence relation defines a tridiagonal linear system. Once the m(k) are determined the coefficients for Sk are given by
sk0 = y(k)
sk1 = d(k) - h(k)*(2*m(k) + m(k-1))/6
sk2 = m(k)/2
sk3 = m(k+1) - m(k)/(6*h(k))
Okay that is all the math you need to know to completely define the algorithm to compute a cubic spline. Here it is in Matlab:
function [s0,s1,s2,s3]=cubic_spline(x,y)
if any(size(x) ~= size(y)) || size(x,2) ~= 1
error('inputs x and y must be column vectors of equal length');
end
n = length(x)
h = x(2:n) - x(1:n-1);
d = (y(2:n) - y(1:n-1))./h;
lower = h(1:end-1);
main = 2*(h(1:end-1) + h(2:end));
upper = h(2:end);
T = spdiags([lower main upper], [-1 0 1], n-2, n-2);
rhs = 6*(d(2:end)-d(1:end-1));
m = T\rhs;
% Use natural boundary conditions where second derivative
% is zero at the endpoints
m = [ 0; m; 0];
s0 = y;
s1 = d - h.*(2*m(1:end-1) + m(2:end))/6;
s2 = m/2;
s3 =(m(2:end)-m(1:end-1))./(6*h);
Here is some code to plot a cubic spline:
function plot_cubic_spline(x,s0,s1,s2,s3)
n = length(x);
inner_points = 20;
for i=1:n-1
xx = linspace(x(i),x(i+1),inner_points);
xi = repmat(x(i),1,inner_points);
yy = s0(i) + s1(i)*(xx-xi) + ...
s2(i)*(xx-xi).^2 + s3(i)*(xx - xi).^3;
plot(xx,yy,'b')
plot(x(i),0,'r');
end
Here is a function that constructs a cubic spline and plots in on the famous Runge function:
function cubic_driver(num_points)
runge = #(x) 1./(1+ 25*x.^2);
x = linspace(-1,1,num_points);
y = runge(x);
[s0,s1,s2,s3] = cubic_spline(x',y');
plot_points = 1000;
xx = linspace(-1,1,plot_points);
yy = runge(xx);
plot(xx,yy,'g');
hold on;
plot_cubic_spline(x,s0,s1,s2,s3);
You can see it in action by running the following at the Matlab prompt
>> cubic_driver(5)
>> clf
>> cubic_driver(10)
>> clf
>> cubic_driver(20)
By the time you have twenty nodes your interpolant is visually indistinguishable from the Runge function.
Some comments on the Matlab code: I don't use any for or while loops. I am able to vectorize all operations. I quickly form the sparse tridiagonal matrix with spdiags. I solve it using the backslash operator. I counting on Tim Davis's UMFPACK to handle the decomposition and forward and backward solves.
Hope that helps. The code is available as a gist on github https://gist.github.com/1269709
There was a bug in spline function, generated (n-2) by (n-2) should be symmetric:
lower = h(2:end);
main = 2*(h(1:end-1) + h(2:end));
upper = h(1:end-1);
http://www.mpi-hd.mpg.de/astrophysik/HEA/internal/Numerical_Recipes/f3-3.pdf