Error: Unable to perform assignment because the left and right sides have a different number of elements.

Hello all,

i wanted to create a loop, but I always get the following error: Unable to perform assignment because the left and right sides have a different number of elements.

Here is my code:

%% Define all variables:
alpha=0.33;
beta=0.96;
n= 5; %gridsize of kgrid
kgrid= [0 0.1 0.2 0.3 0.4]   ; % Choose a grid of possibloe values of the capital stock.
%To do to find the steady stae of k and choose a grid with k_steadystate in
%between. You might want to use the linspace command. Try different grid
%sizes n.


%% Make an initial guess of the value function

v0= [1 1 1 1 1]; % What is the size of v0 ?

%% Define matrix U

for j=1:n
    for i=1:n
        c= kgrid(i)^(alpha)-kgrid(j);%insert budget constraint
        if c>0
        u(i,j)= log(kgrid(i)^(alpha)-kgrid(j)) ;% for each point in the matrix use a different combination of k and k'   
        else 
            u(i,j)=-inf; %make sure that only c>0 is chosen.
        end
    end
end %what is happening in the loops? What happens if consumption is negative?



%% Now iterate one step to obtain v1

v1=max(u+(beta).*v0'.*ones(n)); 
%Insert the right characters instead of ?try to understand all digits. Check: Is v1 of the same dimension as v0?

%what we have to do step-by-step:
v2=max(u+(beta).*v1'.*ones(n));

epsilon=max(v1-v2);
v3=max(u+(beta).*v2'.*ones(n));
epsilon2=max(v3-v2);
v4=max(u+(beta).*v3'.*ones(n));
epsilon3=max(v4-v3);
v5=max(u+(beta).*v4'.*ones(n));
epsilon4=max(v5-v4);
v6=max(u+(beta).*v5'.*ones(n));
epsilon5=max(v6-v5);

%% Repeat until convergence
%set initial value
V=v1
for iter=1:10     %insert the number of iterations until convergence you may try different numbers.


%calculate next step 
 V(iter+1)=max(u+(beta).*V(iter)'.*ones(n))   
end

I know what to do manually (see "%what we have to do step-by-step:" in the code), but somehow I don't manage to create the loop.

Maybe someone know where I did a mistake.

Thanks in advance!

Answer 1

[madhan ravi]

The error was because V keeps increasing linearly by 5 so cell is needed to contain the values properly (since they have the capaciy to expand) as below:

%% Repeat until convergence
%set initial value
V={v1}
for iter=1:10-1     %insert the number of iterations until convergence you may try different numbers.
%calculate next step 
 V{iter+1}=max(u+(beta).*V{iter}'.*ones(n))  ;
end
V = [V{:}]

[mathxaz]

Thanks a lot, this worked perfectly!

[madhan ravi]

Anytime :)

Answer 2

[shariq khan]

Hey!...pay attention to multiplying mattrix dimension - a[mxn] * b[nxr] = c[mxr] - so you cant multiple v0'(5x1) with ones(5)[5x5] matrix because 1 is not equal to 5

check the dimension multiplication at every level - this is what I encountered in whole program

I didnt solve after c because this is same problem in complete program - as what I perceive

check the attachment

try solving again

[madhan ravi]

@shariq see https://www.mathworks.com/help/matlab/ref/times.html this is called element wise operation (.*) in matlab not matrix multiplication ( * )

[shariq khan]

missed that, thanks :)