I implemented the following user defined function in MATLAB:
function Q = Calc_Q(Head, freq)
b6 = [3.7572E-07 -1.5707E-05 6.0490E-03 5.0018E-02 2.1180E-01];
b5 = [-9.0927E-06 8.9033E-04 -3.2415E-02 5.4525E-01 -8.1649E+00] / 10e2;
b4 = [7.5172E-06 -5.6565E-04 1.0024E-02 3.5888E-01 3.8894E-02] / 10e5;
b3 = [-4.8767E-06 4.8787E-04 -1.3311E-02 -1.2189E-01 -5.3522E+00] / 10e8;
b2 = [5.9227E-06 -8.1716E-04 3.5392E-02 -4.5413E-01 1.9547E+00] / 10e11;
b1 = [-2.0004E-06 2.9027E-04 -1.3754E-02 2.3490E-01 -1.2363E+00] / 10e14;
a = [polyval(b1,abs(freq)), polyval(b2, abs(freq)), polyval(b3, abs(freq)), polyval(b4, abs(freq)), polyval(b5, abs(freq)), polyval(b6, abs(freq)) - Head];
Q_roots = roots(a);
%Delete roots with imaginary part
i = 1;
while i <= length(Q_roots)
if(imag(Q_roots(i)) ~= 0)
Q_roots(i) = [];
i = i - 1;
end
i = i + 1;
end
%Delete roots with real part greater then 3100
i = 1;
while i <= length(Q_roots)
if(Q_roots(i) >= 3100 || Q_roots(i) < 0)
Q_roots(i) = [];
i = i - 1;
end
i = i +1;
end
if freq < 0
Q = real(Q_roots(1)) * -1;
else
Q = real(Q_roots(1));
end
end
When I Call this function in Matlab it works fine. However if I use this exact code as a MATLAB function in simulink it stop's working. (actually it works, but the ouput is always zero.)
I do have a suspicion of what the problem might be. When running the script in de-bugging mode, I cannot view a result for Q_roots (It just doesn't display anything).
Q_roots = roots(a);
Any ideas ?
The problem is most likely due to your logic that eliminates any roots that do not have exactly zero in the imaginary part. This is a mathematical way of thinking that does not really work well numerically, at least not in general. All the roots are probably being found in both cases (there is no limitation that implies otherwise), but in Simulink and in code generation the problem is treated as a complex one, and some roots might be coming back with tiny imaginary parts. Instead of deleting roots if their imaginary parts are not exactly zero, eliminate the roots with imaginary parts that are numerically insignificant, either very small relative to the real part or very small altogether. Something like
tol = 10*eps(class(Q_roots));
keepers = abs(imag(Q_roots)) < tol*max(abs(real(Q_roots)),1) & ...
real(Q_roots) >= 0 & real(Q_roots) <= 3100;
Q_roots = Q_roots(keepers);
would take care of all the deletions in one fell swoop. I used 10*eps as a tolerance here.
But if you only need the first qualifying root, then you could just do this:
Q = nan('like',a);
tol = 10*eps(class(a));
for k = 1:numel(Q_roots)
r = real(Q_roots(k));
if abs(imag(Q_roots(k))) < tol*max(abs(r),1) && r >= 0 && r <= 3100;
Q = r;
break
end
end
if freq < 0
Q = -Q;
end
Oke I found the problem.
From a different forum:
Hi Cosmin,
I took a look at the implementation of roots for the Embedded MATLAB
Function block (\toolbox\eml\lib\matlab\polyfun\roots.m).
It's stated there:
% Limitations: % Output is always variable size. % Output is
always complex. % Roots may not be in the same order as MATLAB. %
Roots of poorly conditioned polynomials may not match MATLAB. The last
sentence is what makes you the headache (and yes, your polynomial is
badly conditioned!). If you take a look at the plot you will see, that
the curve hardly touches the x-axis.
I have a suggestion though: the value -z/b is a (very) good
approximation of the root you are looking for ...?
Titus
http://www.mathworks.com/matlabcentral/answers/25624-roots-in-simulink
Apparently the root function in simulink does not always found all the roots of a given polynominal.
This is unfortunate and not easily solvable. I did however found a solution.
For all the different polynomials I have to solve, I know the interval of the root I am interested in ( [-3000, 3000]).
I just basically takes steps of 50 from -3000 to 3000, until the function drops below 0. I then know the approximate solution of the root. I use this approximation as seed for the Newton-Raphson method.
Straight implementation of the Newton raphson method with a given seed for all the polynomials I have to solve did not work because sometimes it iterated to a different root (one which i was not interested in.)
Here's the code:
function Q = Calc_Q(Head, freq)
b6 = [3.7572E-07 -1.5707E-05 6.0490E-03 5.0018E-02 2.1180E-01];
b5 = [-9.0927E-06 8.9033E-04 -3.2415E-02 5.4525E-01 -8.1649E+00] / 10e2;
b4 = [7.5172E-06 -5.6565E-04 1.0024E-02 3.5888E-01 3.8894E-02] / 10e5;
b3 = [-4.8767E-06 4.8787E-04 -1.3311E-02 -1.2189E-01 -5.3522E+00] / 10e8;
b2 = [5.9227E-06 -8.1716E-04 3.5392E-02 -4.5413E-01 1.9547E+00] / 10e11;
b1 = [-2.0004E-06 2.9027E-04 -1.3754E-02 2.3490E-01 -1.2363E+00] / 10e14;
%coeff for the polynominal
a = [polyval(b1,abs(freq)), polyval(b2, abs(freq)), polyval(b3, abs(freq)), polyval(b4, abs(freq)), polyval(b5, abs(freq)), polyval(b6, abs(freq)) - Head];
%coeff for the derrivative of polynominal
da = [5*a(1) 4*a(2) 3*a(3) 2*a(4) a(5)];
Q = -3000;
%Search for point where function goes below 0
while (polyval(a, Q) > 0)
Q = Q + 25;
end
error_max = 0.01
iter_counter = 1;
while abs(polyval(a,Q)) >= error_max && iter_counter <= 1000
Q = Q - polyval(a, Q)/polyval(da, Q);
iter_counter = iter_counter + 1;
error = abs(polyval(a,Q));
end
if(freq < 0)
Q = Q * - 1;
end
end
Related
I am solving a problem in my Macroeconomics class. Consider the following equation:
Here, k is fixed and c(k) was defined through the ```interp1''' function in Matlab. Here is my code:
beta = 0.98;
delta = 0.13;
A = 2;
alpha = 1/3;
n_grid = 1000; % Number of points for capital
k_grid = linspace(5, 15, n_grid)';
tol = 1e-5;
max_it = 1000;
c0 = ones(n_grid, 1);
new_k = zeros(n_grid, 1);
dist_c = tol + 1;
it_c = 0;
while dist_c > tol && it_c < max_it
c_handle = #(k_tomorrow) interp1(k_grid, c0, k_tomorrow, 'linear', 'extrap');
for i=1:n_grid
% Solve for k'
euler = #(k_tomorrow) (1/((1-delta)* k_grid(i) + A * k_grid(i)^alpha - k_tomorrow)) - beta*(1-delta + alpha*A*k_tomorrow^(alpha - 1))/c_handle(k_prime);
new_k(i) = fzero(euler, k_grid(i)); % What's a good guess for fzero?
end
% Compute new values for consumption
new_c = A*k_grid.^alpha + (1-delta)*k_grid - new_k;
% Check convergence
dist_c = norm(new_c - c0);
c0 = new_c;
it_c = it_c + 1;
end
When I run this code, for some indexes $i$, it runs fine and fzero can find the solution. But for indexes it just returns NaN and exits without finding the root. This is a somewhat well-behaved problem in Economics and the solution we are looking indeed exists and the algorithm I tried to implement is guaranteed to work. But I don't have much experience with solving this in MATLAB and I guess I have a silly mistake somewhere. Any ideas on how to procede?
This is the typical error message:
Exiting fzero: aborting search for an interval containing a sign change
because complex function value encountered during search.
(Function value at -2.61092 is 0.74278-0.30449i.)
Check function or try again with a different starting value.
Thanks a lot in advance!
The only term that can produce complex numbers is
k'^(alpha - 1) = k'^(-2/3)
You probably want the result according to the real variant of the cube root, which you could get as
sign(k') * abs(k')^(-2/3)
or more generally and avoiding divisions by zero
k' * (1e-16+abs(k'))^(alpha - 2)
I am trying to use the Metropolis Hastings algorithm with a random walk sampler to simulate samples from a function $$ in matlab, but something is wrong with my code. The proposal density is the uniform PDF on the ellipse 2s^2 + 3t^2 ≤ 1/4. Can I use the acceptance rejection method to sample from the proposal density?
N=5000;
alpha = #(x1,x2,y1,y2) (min(1,f(y1,y2)/f(x1,x2)));
X = zeros(2,N);
accept = false;
n = 0;
while n < 5000
accept = false;
while ~accept
s = 1-rand*(2);
t = 1-rand*(2);
val = 2*s^2 + 3*t^2;
% check acceptance
accept = val <= 1/4;
end
% and then draw uniformly distributed points checking that u< alpha?
u = rand();
c = u < alpha(X(1,i-1),X(2,i-1),X(1,i-1)+s,X(2,i-1)+t);
X(1,i) = c*s + X(1,i-1);
X(2,i) = c*t + X(2,i-1);
n = n+1;
end
figure;
plot(X(1,:), X(2,:), 'r+');
You may just want to use the native implementation of matlab mhsample.
Regarding your code, there are a few things missing:
- function alpha,
- loop variable i (it might be just n but it is not suited for indexing since it starts at zero).
And you should always allocate memory in matlab if you want to fill it dynamically, i.e. X in your case.
To expand on the suggestions by #max, the code appears to work if you change the i indices to n and replace
n = 0;
with
n = 2;
X(:,1) = [.1,.1];
It would probably be better to assign X(:,1) to random values within your accept region (using the same code you use later), and/or include a burn-in period.
Depending upon what you are going to do with this, it may also make things cleaner to evaluate the argument to sin in the f function to keep it within 0 to 2 pi (likely by shifting the value by 2 pi if it exceeds those bounds)
for j = 1:4
n = (co(e(j,2),:) - co(e(j,1),:));
n = n./norm(n);
F = e(j,5).*e(j,4)./e(j,3).*-(dot(u,n)).*n
theta = acosd(dot(F,n)/(norm(F)*norm(n)))
if theta == 180
F2 = [F2; -sqrt(F(1)^2 + F(2)^2 + F(3)^2)]
else if theta == 0
F2 = [F2; sqrt(F(1)^2 + F(2)^2 + F(3)^2)]
end
end
end
Essentially, what the issue is that the first value of F (for j = 1) in the if statement is being ignored, even though the value of theta is 180. To check, I've done 'whos theta' and even wrote theta == 180 into the code for which it returned 0. For j = 2:4 though, the code works absolutely fine despite j = 2 yielding a theta value of 180 as well. It's almost as if it's skipping the if statement entirely for j = 1. I believe this is linked to the fact that F is the only matrix vector with three non-zero components, but still don't know how to go about solving this.
theta might a be double and very close to 180 but not exactly 180.00000001 ~= 180. It might be better to compared with a given tolerance.
tol = 1e-3;
if abs(theta-180)<tol
...
end
Matlab also has internal floating-point accuracy eps, it might be an interesting read.
Additionally you 'else-if' statement should elseif but that might be just a typo from copy-paste.
I wanna find a root for the following function with an error less than 0.05%
f= 3*x*tan(x)=1
In the MatLab i've wrote that code to do so:
clc,close all
syms x;
x0 = 3.5
f= 3*x*tan(x)-1;
df = diff(f,x);
while (1)
x1 = 1 / 3*tan(x0)
%DIRV.. z= tan(x0)^2/3 + 1/3
er = (abs((x1 - x0)/x1))*100
if ( er <= 0.05)
break;
end
x0 = x1;
pause(1)
end
But It keeps running an infinite loop with error 200.00 I dunno why.
Don't use while true, as that's usually uncalled for and prone to getting stuck in infinite loops, like here. Simply set a limit on the while instead:
while er > 0.05
%//your code
end
Additionally, to prevent getting stuck in an infinite loop you can use an iteration counter and set a maximum number of iterations:
ItCount = 0;
MaxIt = 1e5; %// maximum 10,000 iterations
while er > 0.05 & ItCount<MaxIt
%//your code
ItCount=ItCount+1;
end
I see four points of discussion that I'll address separately:
Why does the error seemingly saturate at 200.0 and the loop continue infinitely?
The fixed-point iterator, as written in your code, is finding the root of f(x) = x - tan(x)/3; in other words, find a value of x at which the graphs of x and tan(x)/3 cross. The only point where this is true is 0. And, if you look at the value of the iterants, the value of x1 is approaching 0. Good.
The bad news is that you are also dividing by that value converging toward 0. While the value of x1 remains finite, in a floating point arithmetic sense, the division works but may become inaccurate, and er actually goes NaN after enough iterations because x1 underflowed below the smallest denormalized number in the IEEE-754 standard.
Why is er 200 before then? It is approximately 200 because the value of x1 is approximately 1/3 of the value of x0 since tan(x)/3 locally behaves as x/3 a la its Taylor Expansion about 0. And abs(1 - 3)*100 == 200.
Divisions-by-zero and relative orders-of-magnitude are why it is sometimes best to look at the absolute and relative error measures for both the values of the independent variable and function value. If need be, even putting an extremely (relatively) small finite, constant value in the denominator of the relative calculation isn't entirely a bad thing in my mind (I remember seeing it in some numerical recipe books), but that's just a band-aid for robustness's sake that typically hides a more serious error.
This convergence is far different compared to the Newton-Raphson iterations because it has absolutely no knowledge of slope and the fixed-point iteration will converge to wherever the fixed-point is (forgive the minor tautology), assuming it does converge. Unfortunately, if I remember correctly, fixed-point convergence is only guaranteed if the function is continuous in some measure, and tan(x) is not; therefore, convergence is not guaranteed since those pesky poles get in the way.
The function it appears you want to find the root of is f(x) = 3*x*tan(x)-1. A fixed-point iterator of that function would be x = 1/(3*tan(x)) or x = 1/3*cot(x), which is looking for the intersection of 3*tan(x) and 1/x. However, due to point number (2), those iterators still behave badly since they are discontinuous.
A slightly different iterator x = atan(1/(3*x)) should behave a lot better since small values of x will produce a finite value because atan(x) is continuous along the whole real line. The only drawback is that the domain of x is limited to the interval (-pi/2,pi/2), but if it converges, I think the restriction is worth it.
Lastly, for any similar future coding endeavors, I do highly recommend #Adriaan's advice. If would like a sort of compromise between the styles, most of my iterative functions are written with a semantic variable notDone like this:
iter = 0;
iterMax = 1E4;
tol = 0.05;
notDone = 0.05 < er & iter < iterMax;
while notDone
%//your code
iter = iter + 1;
notDone = 0.05 < er & iter < iterMax;
end
You can add flags and all that jazz, but that format is what I frequently use.
I believe that the code below achieves what you are after using Newton's method for the convergence. Please leave a comment if I have missed something.
% find x: 3*x*tan(x) = 1
f = #(x) 3*x*tan(x)-1;
dfdx = #(x) 3*tan(x)+3*x*sec(x)^2;
tolerance = 0.05; % your value?
perturbation = 1e-2;
converged = 1;
x = 3.5;
f_x = f(x);
% Use Newton s method to find the root
count = 0;
err = 10*tolerance; % something bigger than tolerance to start
while (err >= tolerance)
count = count + 1;
if (count > 1e3)
converged = 0;
disp('Did not converge.');
break;
end
x0 = x;
dfdx_x = dfdx(x);
if (dfdx_x ~= 0)
% Avoid division by zero
f_x = f(x);
x = x - f_x/dfdx_x;
else
% Perturb x and go back to top of while loop
x = x + perturbation;
continue;
end
err = (abs((x - x0)/x))*100;
end
if (converged)
disp(['Converged to ' num2str(x,'%10.8e') ' in ' num2str(count) ...
' iterations.']);
end
I have a backwards recursion for a binomial tree. At each node an unknown C enters in such a way that at the starting node we get a formula, A(1,1), that depends upon C. The code is as follows:
A=sym(zeros(1,Steps));
B=zeros(1,Steps);
syms C; % The unknown that enters A at every node
tic
for t=Steps-1:-1:1
% Values needed in A and B
Lambda=1-exp(-(1./S(t,1:t).^b).*h);
Q=((1./D(t))./(1-Lambda)-d)/(u-d);
R=normcdf(a0+a1*Lambda);
% the backward recursion for A and B
A(1:t)=D(t)*C+D(t)*...
(Q.*(1-Lambda).*A(1:t) ...
+ (1-Q).*(1-Lambda).*A(2:t+1));
B(1:t)=Lambda.*(1-R)+D(t)*...
(Q.*(1-Lambda).*B(1:t)...
+ (1-Q.*(1-Lambda).*B(2:t+1)));
end
C = solve(A(1,1)==sym(B(1,1)),C);
This code takes around 4 seconds if Steps = 104. If however we remove C and set matrix A to a regular double matrix, it only takes about 0.02 seconds. Using syms thus increases the calculation time by a factor 200. This seems too much to me. Any suggestions into speeding this up?
I am using Matlab 2013b on a MacBook air 13-inch spring 2013. Furthermore, if you're interested in the code before the above part (not sure whether it is relevant):
a0 = 0.9;
a1 = -3.2557;
b = 1.2594;
S0=18.57;
sigma=0.6579;
h=1/104;
T=1;
Steps=T/h;
f=transpose(normrnd(0.04, 0.001 [1 pl]));
D=exp(-h*f); % discount values
pl=T/h; % pathlength - amount of steps in maturity
u=exp(sigma*sqrt(h));
d=1/u;
u_row = repmat(cumprod([1 u*ones(1,pl-1)]),pl,1);
d_row = cumprod(tril(d*ones(pl),-1)+triu(ones(pl)),1);
path = tril(u_row.*d_row);
S=S0*path;
Unless I'm missing something, there's no need to use symbolic math or use an unknown variable. You can effectively assume that C = 1 in your recursion relation and solve for the actual value at the end. Here's the full code with some other improvements:
rng(1); % Always seed your random number generator
a0 = 0.9;
a1 = -3.2557;
b = 1.2594;
S0 = 18.57;
sigma = 0.6579;
h = 1/104;
T = 1;
Steps = T/h;
pl = T/h;
f = 0.04+0.001*randn(pl,1);
D = exp(-h*f);
u = exp(sigma*sqrt(h));
d = 1/u;
u_row = repmat(cumprod([1 u*ones(1,pl-1)]),pl,1);
d_row = cumprod(tril(d*ones(pl),-1)+triu(ones(pl)),1);
pth = tril(u_row.*d_row);
S = S0*pth;
A = zeros(1,Steps);
B = zeros(1,Steps);
tic
for t = Steps-1:-1:1
Lambda = 1-exp(-h./S(t,1:t).^b);
Q = ((1./D(t))./(1-Lambda)-d)/(u-d);
R = 0.5*erfc((-a0-a1*Lambda)/sqrt(2)); % Faster than normcdf
% Backward recursion for A and B
A = D(t)+D(t)*(Q.*(1-Lambda).*A(1:end-1) + ...
(1-Q).*(1-Lambda).*A(2:end));
B = Lambda.*(1-R)+D(t)*(Q.*(1-Lambda).*B(1:end-1) + ...
(1-Q.*(1-Lambda).*B(2:end)));
end
C = B/A
toc
This take about 0.005 seconds to run on my MacBook Pro. There are certainly other improvements you could make. There are many combinations of variables that are used in multiple places (e.g., 1-Lambda or D(t)*(1-Lambda)), that could be calculated once. Matlab may try to optimize this a bit. And you can try moving Lambda, Q, and R out of the loop – or at least calculate parts of them outside and save the results in arrays.