I had written the following matlab code to optimize the following LP
max b'x
s.t A'x <= 0;
x <= d;
Also d is
d = {1,2..m}
and A is defined in the code. I am getting the error:
Maximum number of iterations exceeded; increase options.MaxIter.
Upon googling, someone said it is not very good that the error is occuring. and the problem has to reformulated. Any idea how to reformulate it.
The solution is very simple as A > 0, b> 0 and d>0 therefore x = 0
m = 10;
d = [1:1:m];
for j = 1:m,
for i = 1:m,
A(i,j) = 1/(i+j-1);
end
end
for i = 1:m,
b(i)=0;
end
for i = 1:m
lb(i) = -inf;
end
b;
lb = lb';
f = A*d';
[x,fval,exitflag,output] = linprog(-f,A,b,[],[],lb,d); %minimzation problem. Hence -f, A = A'
I'd used the optimist = ('MaxIter', 10000). To stop that error message.
Related
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 don't know where to go with this. I think I have the right stuff down but I don't understand.
https://imgur.com/a/V6gdDdr
It keeps running the loop forever and I don't know why.
n=9;
r =0;
p = 0;
syms x
v=1.7;
while abs(v-r) > 10^(-5)
n=n+1;
r = 0;
a = 0;
b= 1/n;
for i = 1:n
r = r + exp(((i+1)+exp(i))/2)*(b-a)
end
['done']
end
The output should be 85. But I'm getting like a couple thousand. I have tried changing the equation in the for loop but I do not know why it is giving me symbolic errors.
Several problems with the posted code:
The parentheses are misaligned in the r=r + ... statement
Instead of exp(i) and exp(i+1), you should use exp(bi) and exp(b(i+1)) in order to account for the given spacing.
Your value of "v" which is supposed be the exact answer should equal e-1, which is 1.71828. Using the approximate 1.7 is going to be very problematic when you are trying to converge to the exact solution within 5 decimal places.
Your for loop should go from 0 to n-1, this way you don't add up any values past your integral range.
n=9;
r =0;
p = 0;
v=e-1;
while abs(v-r) > 10^(-5)
n=n+1;
r = 0;
a = 0;
b= 1/n;
for i = 0:n-1
r = r + ((exp(b*(i+1))+exp(b*i))/2)*(b-a);
end
end
I am trying to convert my code over to run with parfor, since as it is it takes a long time to run on its own. However I keep getting this error. I have search around on the website and have read people with similar problems, but none of those answers seem to fix my problem. This is my code:
r = 5;
Mu = 12.57e-9;
Nu = 12e6;
I = 1.8;
const = pi*Nu*Mu*r*I;
a = 55;
b = 69;
c = 206;
[m,n,p] = size(Lesion_Visible);
A = zeros(m,n,p);
parpool(2)
syms k
parfor J = 1:m
for I = 1:n
for K = 1:p
if Lesion_Visible(J,I,K) ~= 0
Theta = atand((J-b)/(I-a));
Rho = abs((I-a)/cosd(Theta))*0.05;
Z = abs(c-K)*0.05;
E = vpa(const*int(abs(besselj(0,Rho*k)*exp(-Z*k)*besselj(0,r*k)),0,20),5);
A (J,I,K) = E;
end
end
end
end
I'm trying to calculate the electric field in specific position on an array and matlab give me the error "The variable A in a parfor cannot be classified". I need help. Thanks.
As classification of variables in parfor loop is not permitted, you should try to save the output of each loop in a variable & then save the final output into the desired variable, A in your case!
This should do the job-
parfor J = 1:m
B=zeros(n,p); %create a padding matrix of two dimension
for I = 1:n
C=zeros(p); %create a padding matrix of one dimension
for K = 1:p
if Lesion_Visible(J,I,K) ~= 0
Theta = atand((J-b)./(I-a));
Rho = abs((I-a)./cosd(Theta))*0.05;
Z = abs(c-K).*0.05;
E = vpa(const.*int(abs(besselj(0,Rho.*k).*exp(-Z.*k).*besselj(0,r.*k)),0,20),5);
C(K) = E; %save output of innnermost loop to the padded matrix C
end
end
B(I,:)=C; % save the output to dim1 I of matrix B
end
A(J,:,:)=B; save the output to dim1 J of final matrix A
end
Go through the following for better understanding-
http://www.mathworks.com/help/distcomp/classification-of-variables-in-parfor-loops.html
http://in.mathworks.com/help/distcomp/sliced-variable.html
k = 0.019;
Pstar = 100;
H = 33;
h = 0.1;
X = 36;
N = round(X/h);
t = zeros(1,N+1);
P = zeros(1,N+1);
P(1) = 84;
t(1) = 0;
yHeun = zeros(1,N+1);
yHeun(1)=84;
a = 1; b = 100;
while b-a >0.5
c = (a+b)/2;
for n = 1:N
t(n+1) = t(n) + h;
Inside = nthroot(sin(2*pi*t/12),15);
Harvest = c*0.5*(Inside+1);
P(n+1) = P(n) + h*(k*P(n)*(Pstar-P(n))-Harvest(n));
if P < 0
P = 0;
end
yHeun(n+1) = yHeun(n) + h*0.5*((k*P(n)*(Pstar-P(n))-Harvest(n))+(k*P(n+1)*(Pstar-P(n+1))-Harvest(n+1)));
end
if sign(yHeun(c)) == sign(yHeun(a))
c = a;
else
c = b;
end
end
disp(['The root is between ' num2str(a) ' and ' num2str(b) '.'])
This is the code i'm trying to run and i know it probably sucks but im terrible at coding and every time i try to run the code, it says:
Attempted to access yHeun(50.5); index must be a positive integer or
logical.
Error in Matlab3Q4 (line 30)
if sign(yHeun(c)) == sign(yHeun(a))
I don't have ANY idea how to make yHeun(c or a or whatever) return anything that would be an integer. I dont think i did the while+for loop right either.
Question: "Begin with the higher bound of H being 100 (the high value results in a population of 0 after 36 months), and the lower bound being 1. Put the solver from Problem #3 above in the middle of a while loop and keep bisecting the higher and lower bounds of H until the difference between the higher and lower bound is less than 0.5."
I guess that line 30 (with the error) is this one:
if sign(yHeun(c)) == sign(yHeun(a))
Here, I guess c is equal to 50.5, as a result of c = (a+b)/2 above (BTW you can discover whether I guessed right by debugging - try adding disp(c) before line 30).
To force a number to be an integer, use floor:
c = floor((a+b)/2);
It seems you are trying to use some sort of divide-and-conquer algorithm; it should be enough to stop when b - a is equal to 1.
I'm trying to solve the following question :
maximize x^2-5x+y^2-3y
x+y <= 8
x<=2
x,y>= 0
By using Frank Wolf algorithm ( according to http://web.mit.edu/15.053/www/AMP-Chapter-13.pdf ).
But after running of the following program:
syms x y t;
f = x^2-5*x+y^2-3*y;
fdx = diff(f,1,x); % f'x
fdy = diff(f,1,y); % y'x
x0 = [0 0]; %initial point
A = [1 1;1 0]; %constrains matrix
b = [8;2];
lb = zeros(1,2);
eps = 0.00001;
i = 1;
X = [inf inf];
Z = zeros(2,200); %result for end points (x1,x2)
rr = zeros(1,200);
options = optimset('Display','none');
while( all(abs(X-x0)>[eps,eps]) && i < 200)
%f'x(x0)
c1 = subs(fdx,x,x0(1));
c1 = subs(c1,y,x0(2));
%f'y(x0)
c2 = subs(fdy,x,x0(1));
c2 = subs(c2,y,x0(2));
%optimization point of linear taylor function
ys = linprog((-[c1;c2]),A,b,[],[],lb,[],[],options);
%parametric representation of line
xt = (1-t)*x0(1)+t*ys(1,1);
yt = (1-t)*x0(2)+t*ys(2,1);
%f(x=xt,y=yt)
ft = subs(f,x,xt);
ft = subs(ft,y,yt);
%f't(x=xt,y=yt)
ftd = diff(ft,t,1);
%f't(x=xt,y=yt)=0 -> for max point
[t1] = solve(ftd); % (t==theta)
X = double(x0);%%%%%%%%%%%%%%%%%
% [ xt(t=t1) yt(t=t1)]
xnext(1) = subs(xt,t,t1) ;
xnext(2) = subs(yt,t,t1) ;
x0 = double(xnext);
Z(1,i) = x0(1);
Z(2,i) = x0(2);
i = i + 1;
end
x_point = Z(1,:);
y_point = Z(2,:);
% Draw result
scatter(x_point,y_point);
hold on;
% Print results
fprintf('The answer is:\n');
fprintf('x = %.3f \n',x0(1));
fprintf('y = %.3f \n',x0(2));
res = x0(1)^2 - 5*x0(1) + x0(2)^2 - 3*x0(2);
fprintf('f(x0) = %.3f\n',res);
I get the following result:
x = 3.020
y = 0.571
f(x0) = -7.367
And this no matter how many iterations I running this program (1,50 or 200).
Even if I choose a different starting point (For example, x0=(1,6) ), I get a negative answer to most.
I know that is an approximation, but the result should be positive (for x0 final, in this case).
My question is : what's wrong with my implementation?
Thanks in advance.
i changed a few things, it still doesn't look right but hopefully this is getting you in the right direction. It looks like the intial x0 points make a difference to how the algorithm converges.
Also make sure to check what i is after running the program, to determine if it ran to completion or exceeded the maximum iterations
lb = zeros(1,2);
ub = [2,8]; %if x+y<=8 and x,y>0 than both x,y < 8
eps = 0.00001;
i_max = 100;
i = 1;
X = [inf inf];
Z = zeros(2,i_max); %result for end points (x1,x2)
rr = zeros(1,200);
options = optimset('Display','none');
while( all(abs(X-x0)>[eps,eps]) && i < i_max)
%f'x(x0)
c1 = subs(fdx,x,x0(1));
c1 = subs(c1,y,x0(2));
%f'y(x0)
c2 = subs(fdy,x,x0(1));
c2 = subs(c2,y,x0(2));
%optimization point of linear taylor function
[ys, ~ , exit_flag] = linprog((-[c1;c2]),A,b,[],[],lb,ub,x0,options);
so here is the explanation of the changes
ub, uses our upper bound. After i added a ub, the result immediately changed
x0, start this iteration from the previous point
exit_flag this allows you to check exit_flag after execution (it always seems to be 1 indicating it solved the problem correctly)