"Size inputs must be scalar".

Hi guys, can someone please explain me. how I can fix the error "size inputs must be scalar". I read some stuff to sort out this error, even after that I could not solve it. I want to plot loglog dt versus error. for this purpose I want that T(number of time grids) will be different for every iteration. and T could have 10,20,30 values even more. here is my code. its a one dim wave equation. Thanks in advance!

N=10;
X = 100;
dx = pi/X;
x(1,:)=0:dx:pi;
c=1;
 tfinal=5;
 T=10.^(3:5);
 dt=tfinal./T;
 for i=1:length(T)
     t0=0;
     t(:,1)= (0:T)'*dt(i);
     uexac = zeros(T+1,X+1);
     U = zeros(T+1,X+1);
     beta=zeros(N,1);
     alpha=zeros(N,1);
     v0=0;
     u0=sin(3*x);
     U0=u0;
     V0=v0;
     U(1,:)=U0;
     U(2,:)=dt(i)*V0+ U0;
     %exact solution by fourier series
     for n= 1:N
         beta(n)=2/pi*trapz(x,v0.*cos(n*x));
         alpha(n)=2/pi*trapz(x,u0.*sin(n*x));
         an=alpha(n)*cos(n*c*t)+beta(n)/(n*c)*sin(n*c*t);
         uexac=uexac+ an*sin(n*x);
     end
     %approximate solution by finite difference method
     for j = 3:T+1
        U1 = 2*U(j-1,:)-U(j-2,:);

        U1=U1(2:end-1);

        U2 = U(j-1,1:end-2)-2*U(j-1,2:end-1)+U(j-1,3:end);

        U(j,2:end-1) = U1 + U2*c*(dt(i)^2/dx^2);
     end
     % compute error and store
     error(i,1)=(max(max(abs(uexac-U))));
     tt(i,1)=dt(i); %store time step size
 end

Error using zeros
Size inputs must be scalar.

 loglog(tt(:,1),error(:,1),'*');
 xlabel('dt','fontSize',8);
 ylabel('error','fontSize',8);

[Roger Wohlwend]

Your code does not run because several variables are missing: tfinal, X, ... So I cannot reproduce the error.

[sadee]

Roger now I have put the rest variables. You can check

Answer 1

[Henric Rydén]

Here is one problem:

T = [1000       10000      100000] 
uexac = zeros(T+1,X+1);

T is a vector. zeros(m,n) creates an m-by-n matrix so m and n need to be scalars, not vectors.

If you want three vectors, you need three commands. Like this

uexac1 = zeros(T(1)+1,X+1);
uexac2 = zeros(T(2)+1,X+1);
uexac3 = zeros(T(3)+1,X+1);

[sadee]

Thank you Henric for your reply. Actually I want to plot loglog dt versus error as you can see in the end of code. for this purpose I want that T(number of time grids) will be different for every iteration. and it could be 10,20,30 even more, so e.g if T will have 20 values then I can not write U and uexac matrix 20 times. can you tell me please?

[Henric Rydén]

Sure, just overwrite during the loop. I can help you but you need to define all the varibles, and some are still undefined, such as N. Also,

x(1,:)=0:dx:pi;

will not work if x is not defined.

[Roger Wohlwend]

You need an additional loop where you repeat the process for every value of T.

[sadee]

now I have defined rest variables, you can check. from where I have to start my loop. can you please write it here?. I have tried with outer loop with T but it was not working at all

[Roger Wohlwend]

Within the first loop you have to replace T by T(i).

[Henric Rydén]

You still haven't defined N

[sadee]

I have defined N, but it was showing in end of text. Now tried as you said but it's not working

[Roger Wohlwend]

Then show us the updated code and what error occurs.

Answer 2

[Suman Ahmed]

would you please help me to resolve the one error that is size input must be scalar?

A=[-0.0159 0 0.0418 0; 0 -0.011 0 0.0333; 0 0 -0.0419 0; 0 0 0 -0.033];
B=[-0.083325 0; 0 0.0628; 0 0.0479; 0.0312 0];
C=[0.50 0 0 0; 0 0.50 0 0; 0 0 0.50 0; 0 0 0 0];
D=zeros(4,2);
t=100;
t=1:1:t;
Q1=0.00005*ones(1,t);
Q2=0.000004*ones(1,t);
u=[Q1, Q2];
sys=ss(A,B,C,D);
sys_dis=c2d(sys,1);
Ad=sys_dis.a;
Bd=sys_dis.b;
Cd=sys_dis.c;
Dd=sys_dis.d;

Xint=[0; 0 ;0];
X(:,1)=Xint;
Yint=[0; 0 ;0];
y(:,1)=Yint;

for i=1:t-1
    [ y(:,i+1), X(:,i+1) ] = out_cal(Ad,Bd,Cd,Dd,(u(i,:))',X(:,i) );   
end

% defining output for plotting
h1=y(1,:);
h2=y(2,:);
h3=y(3,:);


% Plotting inputs
figure(1);
subplot(2,1,1);
plot(t,Q1,'r-');
title('Time VS Q_1');
ylabel('Q_1');
xlabel('time(s)');

subplot(2,1,2);
plot(t,Q2,'r-');
title('Time VS Q_2');
ylabel('Q_2');
xlabel('time(s)');

% Plotting outputs
figure(2);
subplot(4,1,1);
plot(t,h1,'r-');
title('Time VS level in the tank1');
ylabel('h1_t');
xlabel('time(s)');

subplot(4,1,2);
plot(t,h2,'r-');
title('Time VS level in the tank2');
ylabel('h2_t');
xlabel('time(s)');

subplot(4,1,3);
plot(t,h3,'r-');
title('Time VS level in the tank3');
ylabel('h3_t');
xlabel('time(s)');

%grid

[Steven Lord]

t=100;
t=1:1:t;
Q1=0.00005*ones(1,t);
Q2=0.000004*ones(1,t);

t starts off as the value 100, but then on the next line becomes the vector [1, 2, 3, 4, ... 98, 99, 100]. You then try to use that to build a vector.

Since I know you don't want to try to build a 1-by-1-by-2-by-3-by-4-by-...-by-98-by-99-by-100 array (storing each element of the array in a different atom in this universe, you would still fall well short of the number of atoms needed) I have to assume you want to create 100 separate vectors. If so, use a for loop to process each element of t in turn or create a cell array with 100 elements where element k of that cell array corresponds to element k of the vector t.

If you want to create one vector with 100 elements, delete the line of code where you redefine t from being 100 to being the vector.

[Suman Ahmed]

thanks steven, it works now.

Answer 3

[Elis]

% Introduce your personal data here
Dateofbirth = 960920;  % yymmdd Use your birth date

% Compute the parameters
d1 = mod(Dateofbirth, 123);
c = d1 + 25;
a = 36 * c + 2 * mod(d1, 20);
b = 24 * c + mod(d1, 20);

% Given Data: do not change the order of links and nodes given in these matrices!
Redarc = [1 2           % Red links (first element of row k is the starting node of the k-th arc, second element the final node)
          1 3
          2 4
          2 5
          3 5
          3 6
          4 7
          5 7
          5 8
          6 8
          7 9
          8 9];
Bluearc = [2 1           % Blue links (first element of row k is the starting node of the k-th arc, second element the final node)
           1 3
           4 2
           2 5
           5 3
           3 6
           4 7
           7 5
           5 8
           8 6
           7 9
           9 8];

% Compute node-arc incidence matrix as explained in pages 75-76 of ASKS. Don't remove the last row. Linprog will take care of linear dependency.
numNodes = max(max(Redarc), max(Bluearc));  % Total number of nodes in the networks
numRedArcs = size(Redarc, 1);
numBlueArcs = size(Bluearc, 1);

Ared = zeros(numNodes, numRedArcs);

Error using zeros
Size inputs must be scalar.

Ablue = zeros(numNodes, numBlueArcs);

for i = 1:numRedArcs
    startNode = Redarc(i, 1);
    endNode = Redarc(i, 2);
    Ared(startNode, i) = -1;
    Ared(endNode, i) = 1;
end

for i = 1:numBlueArcs
    startNode = Bluearc(i, 1);
    endNode = Bluearc(i, 2);
    Ablue(startNode, i) = -1;
    Ablue(endNode, i) = 1;
end

%% Part (a): Minimize link load

% Compute the dimensions of Aeq and beq
numArcs = numRedArcs + numBlueArcs;
numConstraints = numArcs + 1;  % Additional constraint for linear dependency

% Define the objective function
c = ones(numArcs, 1);

% Define the equality constraints
Aeq = [Ared; Ablue];
beq = [a; b];

% Add the row for linear dependency
Aeq(numConstraints, :) = 0;
beq(numConstraints) = 0;

% Define the inequality constraints (none in this case)
Ain = [];
bin = [];

% Define the upper and lower bounds (non-negative flow)
ub = Inf(numArcs, 1);
lb = zeros(numArcs, 1);

% Solve the linear programming problem
options = optimoptions('linprog', 'Display', 'off');  % Suppress the output display
Sol = linprog(c, Ain, bin, Aeq, beq, lb, ub, options);

% Extract the optimal flow for red and blue arcs
RedFlow = Sol(1:numRedArcs);
BlueFlow = Sol(numRedArcs + 1:end-1);

% Calculate the link loads (total flow on each link)
LinkLoads = RedFlow + BlueFlow;

% Calculate the maximal load among all links
w = max(LinkLoads);

% Calculate the node loads (total flow on each node)
NodeLoads = Ared(:, 1:numRedArcs) * RedFlow + Ablue(:, 1:numBlueArcs) * BlueFlow;

% Display the results
fprintf('Part (a): Minimize link load\n');
fprintf('Optimal Red Flow: %s\n', mat2str(RedFlow));
fprintf('Optimal Blue Flow: %s\n', mat2str(BlueFlow));
fprintf('Link Loads: %s\n', mat2str(LinkLoads));
fprintf('Maximal Load: %.2f\n', w);
fprintf('Node Loads: %s\n', mat2str(NodeLoads));

%% Part (b): Minimize node load

% Define the objective function
c2 = ones(numNodes, 1);

% Define the equality constraints
Aeq2 = [Ared; Ablue];
beq2 = [a; b];

% Define the inequality constraints (none in this case)
Ain2 = [];
bin2 = [];

% Define the upper and lower bounds (non-negative flow)
ub2 = Inf(numNodes, 1);
lb2 = zeros(numNodes, 1);

% Solve the linear programming problem
Sol2 = linprog(c2, Ain2, bin2, Aeq2, beq2, lb2, ub2, options);

% Extract the optimal flow for red and blue arcs
RedFlow2 = Ared * Sol2;
BlueFlow2 = Ablue * Sol2;

% Calculate the link loads (total flow on each link)
LinkLoads2 = RedFlow2 + BlueFlow2;

% Calculate the maximal load among all links
w2 = max(LinkLoads2);

% Calculate the node loads (total flow on each node)
NodeLoads2 = Sol2;

% Display the results
fprintf('\nPart (b): Minimize node load\n');
fprintf('Optimal Red Flow: %s\n', mat2str(RedFlow2));
fprintf('Optimal Blue Flow: %s\n', mat2str(BlueFlow2));
fprintf('Link Loads: %s\n', mat2str(LinkLoads2));
fprintf('Maximal Load: %.2f\n', w2);
fprintf('Node Loads: %s\n', mat2str(NodeLoads2));

[Walter Roberson]

RedArc and BlueArc are something-by-2.

numNodes = max(max(Redarc), max(Bluearc));  % Total number of nodes in the networks

max of a something-by-2 will proceed along the columns, giving a 1 x 2 result. You then use max() with two parameters where each of the parameters is 1 x 2: the result will be element-by-element maximums between the two vectors, so you will get a 1 x 2 result. So numNodes will be 1 x 2.

When you use max(A,B) then that is not the same as max([A,B]) or max([A;B])