I'm supposed to use eulers method to find the zonal wind field for all latitudes(-90S to 90N) and altitudes (0 to 22km) on earth
%Wind speed = 0 at surface
a = -12;
b_1 = 40;
x = -90:1:90; %latitude
y = 0:1:22; %altitude
z_r = 12;
[X, Y] = meshgrid(x, y);
%dy_dx = (y_2 - y_1)/(x_2 - x_1)
T = a + (b_1*(1-Y/z_r)).(3./2(2./3 + (sin(Xpi/180)).^2).(cos(X*pi/180)).^3);
contourf(X,Y,T,'ShowText','on')
colorbar
title("Temperature field")
xlabel("Latitude (degrees)")
ylabel("Altitude (km)")
%all code works above this line as it should. You can plot this and you will see what is be happening in this simple model.
%gravity in km/s
g = 0.0981
%f = coriolis force
f = (1.458*10^(-4))sin(Xpi/180);
%here I'm trying to use eulers method to find the zonal wind field everywhere on earth
for i = 1:22
for j = 1:180
i(i,j) = 0
end
end
for i = 1:22
for j = 1:180
dtdy(i, j+1) = (T(i, j+1) - T(i, j))./(Y(i, j+1) - Y(i, j))
u(i+1, j) = u(i, j) - ((g./(f*T(i, j))).dtdy(i, j+1)*(X(i+1, j) - X(i, j)))./111.21
end
end
%I get an error saying the matrix dimensions must agree but I'm not very proficient in matlab, so I'm unsure why.
Related
I want to determine the Steepest descent of the Rosenbruck function using Armijo steplength where x = [-1.2, 1]' (the initial column vector).
The problem is, that the code has been running for a long time. I think there will be an infinite loop created here. But I could not understand where the problem was.
Could anyone help me?
n=input('enter the number of variables n ');
% Armijo stepsize rule parameters
x = [-1.2 1]';
s = 10;
m = 0;
sigma = .1;
beta = .5;
obj=func(x);
g=grad(x);
k_max = 10^5;
k=0; % k = # iterations
nf=1; % nf = # function eval.
x_new = zeros([],1) ; % empty vector which can be filled if length is not known ;
[X,Y]=meshgrid(-2:0.5:2);
fx = 100*(X.^2 - Y).^2 + (X-1).^2;
contour(X, Y, fx, 20)
while (norm(g)>10^(-3)) && (k<k_max)
d = -g./abs(g); % steepest descent direction
s = 1;
newobj = func(x + beta.^m*s*d);
m = m+1;
if obj > newobj - (sigma*beta.^m*s*g'*d)
t = beta^m *s;
x = x + t*d;
m_new = m;
newobj = func(x + t*d);
nf = nf+1;
else
m = m+1;
end
obj=newobj;
g=grad(x);
k = k + 1;
x_new = [x_new, x];
end
% Output x and k
x_new, k, nf
fprintf('Optimal Solution x = [%f, %f]\n', x(1), x(2))
plot(x_new)
function y = func(x)
y = 100*(x(1)^2 - x(2))^2 + (x(1)-1)^2;
end
function y = grad(x)
y(1) = 100*(2*(x(1)^2-x(2))*2*x(1)) + 2*(x(1)-1);
end
I am attempting to create a gamma distribution in MATLAB; however, I keep receiving the error:
Index in Position 1 exceeds array bounds (must not exceed 100).
Assuming I am reading this correctly, it is referring to variable M that is simply = 2500 (the number of pseudo-random variables I am using for this project).
I was hoping someone can explain what is wrong with my logic and possibly a solution.
alpha = 0.5;
w = gamma_rdn(M,alpha);
x1 = (0.0001:0.001:1); % For plot
figure(5)
subplot(2,1,1);hist(w);title('Histogram of Gamma RDN');
subplot(2,1,2);plot(x1,pdf('gam',x1,alpha,1));title('Theoretical Gamma Density with \alpha = 0.5');
axis([0 1 0 100]);
% The gamma_rdn function is implemented as follows:
function[w] = gamma_rdn(M,alpha)
% Generate random numbers from the gamma distribution with parameter
% alpha <= 1, beta = 1
pe = exp(1);
w = zeros(M,1);
u = rand(100,1);
b = (alpha + pe)/pe;
i = 0;
j = 0;
while j < M
i = i+1;
y = b*u(i,1);
if y <= 1
z = y^(1/alpha);
i = i+1;
if u(i,1) <= exp(-z)
j = j+1;
w(j,1) = z;
else
i = i+1;
end
else
z = -log((b-y)/alpha);
i = i+1;
if u(i,1) <= z^(alpha - 1)
j = j+1;
w(j,1) = z;
else
i = i+1;
end
end
end
if i > 95
u = rand(100,1);
i = 0;
end
end
Is there a particular reason you chose u = rand(100,1)?
The problem is coming because in while loop, as soon as variable i exceeds 100 (say i=101), y = b*u(i,1) becomes invalid. That is, you are trying to access u(101,1) while the size of u is (100,1).
If there's not particular reason, try a large enough size, like, u = rand(10000,1).
I have the following DP which I am applying on a binarized image (either 0 or 1) in Matlab
[x, y] = size(img);
dp = zeros(x, y);
dp(1,:) = img(1,:);
dp(:,1) = img(:,1);
for i = 2:x
for j = 2:y
if img(i, j) == 0
dp(i, j) = min([dp(i, j - 1), dp(i - 1, j), dp(i - 1, j - 1)]) + 1;
end
end
end
The code for large x and y takes a lot of time maybe because of the if condition and using for loops instead of writing vectorized code.
Can anyone optimize it.?
Or is there any approach which optimizes the above code by exploiting the fact that the matrix img contains either 0 or 1 (fewer 1s than 0s).
Also is it possible to somehow use parallel for loops to speed up.?
As far as I am aware, you cannot really speed up this computation in general. But if you know that there are only very few entries where img(i,j)==0 following approach might save you a little bit of time:
[x, y] = size(img);
dp = zeros(x, y);
dp(1,:) = img(1,:);
dp(:,1) = img(:,1);
[i, j] = find(img(2:end, 2:end) == 0); % Extract only these pixels where we actually need to do something
i = i + 1; %correct for removing the first row and column
j = j + 1;
for k = 1:numel(i);
dp(i(k), j(k)) = min([dp(i(k), j(k) - 1), dp(i(k) - 1, j(k)), dp(i(k) - 1, j(k) - 1)]) + 1;
end
For my studies I had to write a PDE solver for the Poisson equation on a disc shaped domain using the finite difference method.
I already passed the Lab exercise. There is one issue in my code I couldn't fix. Function fun1 with the boundary value problem gun2 is somehow oscillating at the boundary. When I use fun2 everything seems fine...
Both functions use at the boundary gun2. What is the problem?
function z = fun1(x,y)
r = sqrt(x.^2+y.^2);
z = zeros(size(x));
if( r < 0.25)
z = -10^8*exp(1./(r.^2-1/16));
end
end
function z = fun2(x,y)
z = 100*sin(2*pi*x).*sin(2*pi*y);
end
function z = gun2(x,y)
z = x.^2+y.^2;
end
function [u,A] = poisson2(funame,guname,M)
if nargin < 3
M = 50;
end
%Mesh Grid Generation
h = 2/(M + 1);
x = -1:h:1;
y = -1:h:1;
[X,Y] = meshgrid(x,y);
CI = ((X.^2 +Y.^2) < 1);
%Boundary Elements
Sum= zeros(size(CI));
%Sum over the neighbours
for i = -1:1
Sum = Sum + circshift(CI,[i,0]) + circshift(CI,[0,i]) ;
end
%if sum of neighbours larger 3 -> inner note!
CI = (Sum > 3);
%else boundary
CB = (Sum < 3 & Sum ~= 0);
Sum= zeros(size(CI));
%Sum over the boundary neighbour nodes....
for i = -1:1
Sum = Sum + circshift(CB,[i,0]) + circshift(CB,[0,i]);
end
%If the sum is equal 2 -> Diagonal boundary
CB = CB + (Sum == 2 & CB == 0 & CI == 0);
%Converting X Y to polar coordinates
Phi = atan(Y./X);
%Converting Phi R back to cartesian coordinates, only at the boundarys
for j = 1:M+2
for i = 1:M+2
if (CB(i,j)~=0)
if j > (M+2)/2
sig = 1;
else
sig = -1;
end
X(i,j) = sig*1*cos(Phi(i,j));
Y(i,j) = sig*1*sin(Phi(i,j));
end
end
end
%Numberize the internal notes u1,u2,......,un
CI = CI.*reshape(cumsum(CI(:)),size(CI));
%Number of internal notes
Ni = nnz(CI);
f = zeros(Ni,1);
k = 1;
A = spalloc(Ni,Ni,5*Ni);
%Create matix A!
for j=2:M+1
for i =2:M+1
if(CI(i,j) ~= 0)
hN = h;hS = h; hW = h; hE = h;
f(k) = fun(X(i,j),Y(i,j));
if(CB(i+1,j) ~= 0)
hN = abs(1-sqrt(X(i,j)^2+Y(i,j)^2));
f(k) = f(k) + gun(X(i,j),Y(i+1,j))*2/(hN^2+hN*h);
A(k,CI(i-1,j)) = -2/(h^2+h*hN);
else
if(CB(i-1,j) ~= 0) %in negative y is a boundry
hS = abs(1-sqrt(X(i,j)^2+Y(i,j)^2));
f(k) = f(k) + gun(X(i,j),Y(i-1,j))*2/(hS^2+h*hS);
A(k,CI(i+1,j)) = -2/(h^2+h*hS);
else
A(k,CI(i-1,j)) = -1/h^2;
A(k,CI(i+1,j)) = -1/h^2;
end
end
if(CB(i,j+1) ~= 0)
hE = abs(1-sqrt(X(i,j)^2+Y(i,j)^2));
f(k) = f(k) + gun(X(i,j+1),Y(i,j))*2/(hE^2+hE*h);
A(k,CI(i,j-1)) = -2/(h^2+h*hE);
else
if(CB(i,j-1) ~= 0)
hW = abs(1-sqrt(X(i,j)^2+Y(i,j)^2));
f(k) = f(k) + gun(X(i,j-1),Y(i,j))*2/(hW^2+h*hW);
A(k,CI(i,j+1)) = -2/(h^2+h*hW);
else
A(k,CI(i,j-1)) = -1/h^2;
A(k,CI(i,j+1)) = -1/h^2;
end
end
A(k,k) = (2/(hE*hW)+2/(hN*hS));
k = k + 1;
end
end
end
%Solve linear system
u = A\f;
U = zeros(M+2,M+2);
p = 1;
%re-arange u
for j = 1:M+2
for i = 1:M+2
if ( CI(i,j) ~= 0)
U(i,j) = u(p);
p = p+1;
else
if ( CB(i,j) ~= 0)
U(i,j) = gun(X(i,j),Y(i,j));
else
U(i,j) = NaN;
end
end
end
end
surf(X,Y,U);
end
I'm keeping this answer short for now, but may extend when the question contains more info.
My first guess is that what you are seeing is just numerical errors. Looking at the scales of the two graphs, the peaks in the first graph are relatively small compared to the signal in the second graph. Maybe there is a similar issue in the second that is just not visible because the signal is much bigger. You could try to increase the number of nodes and observe what happens with the result.
You should always expect to see numerical errors in such simulations. It's only a matter of trying to get their magnitude as small as possible (or as small as needed).
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.