Output is not as expected

Question

Sahithi Kandhukuri 2024-1-8

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https://ww2.mathworks.cn/matlabcentral/answers/2067531-output-is-not-as-expected

评论： Sam Chak 2024-1-11

I m trying to do ant colony optimisation based upqc but the out is not as expected

I am getting above output i.e, zeroes but the required one is as shown below

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Sam Chak 2024-1-8

在 MATLAB Online 中打开

I haven't tested everything yet, but I experimented with a basic quadratic function and observed that the optimal solution obtained through Ant Colony Optimization does not yield the lowest cost. Until this issue is addressed, it does not make sense to proceed with the current optimization problem.

format long g
%% ACO parameters
n_iter  = 5;                   % number of iteration
NA      = 5;                   % Number of Ants
alpha   = 0.8;
beta    = 0.2;
roh     = 0.7;                  % Evaporation rate
n_param = 1;                    % Number of parameters
LB      = 0.1*ones(1,n_param);  % lower band
UB      = 2.0*ones(1,n_param);  % upper band
n_node  = 1000;                 % number of nodes for each param
%% intilalizing some variables
cost_best_prev      = inf;
ant                 = zeros(NA,n_param);
cost                = zeros(NA,1);
tour_selected_param = zeros(1,n_param);
param_mat           = zeros(n_iter,n_param);
Nodes               = zeros(n_node,n_param);
prob                = zeros(n_node,n_param);
%% Generating nodes
T   =  ones(n_node, n_param).*eps;
dT  = zeros(n_node, n_param);
for i = 1:n_param
    Nodes(:,1) = linspace(LB(i), UB(i), n_node);
end
%% iteration loop
for iter = 1:n_iter
    for tour_i = 1:n_param
        prob(:,tour_i) = (T(:,tour_i).^alpha).*((1./Nodes(:,tour_i)).^beta);
        prob(:,tour_i) = prob(:,tour_i)./sum(prob(:,tour_i));
    end
    for A = 1:NA
        for tour_i = 1:n_param
            node_se1 = rand;
            node_ind = 1;
            prob_sum = 0;
            for j = 1:n_node
                prob_sum = prob_sum + prob(j,tour_i);
                if prob_sum >= node_se1
                    node_ind = j;
                    break
                end
            end
            ant(A,tour_i) = node_ind;
            tour_selected_param(tour_i) = Nodes(node_ind, tour_i);
        %% 
        end
        
        %% Put the Cost function here:
        % cost(A) = cost_func(tour_selected_param);
        cost(A) = costume(tour_selected_param);
        
        disp(['Ant number: ' num2str(A)])
        disp(['Ant Cost: ' num2str(cost(A))])
        disp(['Ant parameters: ' num2str(tour_selected_param)])
        if iter~=1
            disp(['iteration:' num2str(iter)])
            disp('________________')
            disp(['Best cost:' num2str(cost_best)])
            for i=1:n_param
                tour_selected_param(i) = Nodes(ant(cost_best_ind,1), i);
            end
            disp(['Best parameters:' num2str(tour_selected_param)])
        end
    
    end
    [cost_best, cost_best_ind] = min(cost);
    % Elistem
    if (cost_best > cost_best_prev) && (iter ~= 1)
        [cost_worst, cost_worst_ind] = max(cost);
        ant(cost_worst_ind,:)        = best_prev_ant;
        cost_best                    = cost_best_prev;
        cost_best_ind                = cost_worst_ind;
    else
        cost_best_prev = cost_best;
        best_prev_ant  = ant(cost_best_ind,:);
    end
    dT = zeros(n_node,n_param); % Change of Phormone
    for tour_i = 1:n_param
        for A = 1:NA
            dT(ant(A, tour_i), tour_i) = dT(ant(A, tour_i), tour_i) + cost_best/cost(A);
        end
    end
    T = roh.*T + dT;
end
Ant number: 1
Ant Cost: 1.4451
Ant parameters: 1.6672
Ant number: 2
Ant Cost: 1.0667
Ant parameters: 1.2583
Ant number: 3
Ant Cost: 1.5524
Ant parameters: 1.7432
Ant number: 4
Ant Cost: 1.9254
Ant parameters: 1.962
Ant number: 5
Ant Cost: 1.7482
Ant parameters: 1.865
Ant number: 1
Ant Cost: 1.5524
Ant parameters: 1.7432
iteration:2
________________
Best cost:1.0667
Best parameters:1.2583
Ant number: 2
Ant Cost: 1.4451
Ant parameters: 1.6672
iteration:2
________________
Best cost:1.0667
Best parameters:1.6672
Ant number: 3
Ant Cost: 1.0667
Ant parameters: 1.2583
iteration:2
________________
Best cost:1.0667
Best parameters:1.6672
Ant number: 4
Ant Cost: 1.0667
Ant parameters: 1.2583
iteration:2
________________
Best cost:1.0667
Best parameters:1.6672
Ant number: 5
Ant Cost: 1.9254
Ant parameters: 1.962
iteration:2
________________
Best cost:1.0667
Best parameters:1.6672
Ant number: 1
Ant Cost: 1.7482
Ant parameters: 1.865
iteration:3
________________
Best cost:1.0667
Best parameters:1.2583
Ant number: 2
Ant Cost: 1.0667
Ant parameters: 1.2583
iteration:3
________________
Best cost:1.0667
Best parameters:1.2583
Ant number: 3
Ant Cost: 1.7482
Ant parameters: 1.865
iteration:3
________________
Best cost:1.0667
Best parameters:1.865
Ant number: 4
Ant Cost: 1.0667
Ant parameters: 1.2583
iteration:3
________________
Best cost:1.0667
Best parameters:1.865
Ant number: 5
Ant Cost: 1.4451
Ant parameters: 1.6672
iteration:3
________________
Best cost:1.0667
Best parameters:1.865
Ant number: 1
Ant Cost: 1.5524
Ant parameters: 1.7432
iteration:4
________________
Best cost:1.0667
Best parameters:1.2583
Ant number: 2
Ant Cost: 1.4451
Ant parameters: 1.6672
iteration:4
________________
Best cost:1.0667
Best parameters:1.6672
Ant number: 3
Ant Cost: 1.0667
Ant parameters: 1.2583
iteration:4
________________
Best cost:1.0667
Best parameters:1.6672
Ant number: 4
Ant Cost: 1.0667
Ant parameters: 1.2583
iteration:4
________________
Best cost:1.0667
Best parameters:1.6672
Ant number: 5
Ant Cost: 1.4451
Ant parameters: 1.6672
iteration:4
________________
Best cost:1.0667
Best parameters:1.6672
Ant number: 1
Ant Cost: 1.5524
Ant parameters: 1.7432
iteration:5
________________
Best cost:1.0667
Best parameters:1.2583
Ant number: 2
Ant Cost: 1.0667
Ant parameters: 1.2583
iteration:5
________________
Best cost:1.0667
Best parameters:1.2583
Ant number: 3
Ant Cost: 1.9254
Ant parameters: 1.962
iteration:5
________________
Best cost:1.0667
Best parameters:1.962
Ant number: 4
Ant Cost: 1.0667
Ant parameters: 1.2583
iteration:5
________________
Best cost:1.0667
Best parameters:1.962
Ant number: 5
Ant Cost: 1.9254
Ant parameters: 1.962
iteration:5
________________
Best cost:1.0667
Best parameters:1.962
%% ACO Result
k = tour_selected_param
k = 
          1.96196196196196
J = costume(k)
J = 
          1.92537081626171
%% Compare with other Optimizer
[kopt, fval] = fminunc(@costume, 2)
Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.
kopt = 
     1
fval = 
     1
%% Custom Optimization Solution for Targeted User Minimum Equation
function J = costume(k)
    J   = (k - 1)^2 + 1;
end

Sam Chak 2024-1-9

在 MATLAB Online 中打开

HI @Torsten,

Upon reviewing the code, Yarpiz developed an iteration of Ant Colony Optimization (ACO) designed for addressing Combinatorial Optimization Problems, such as the well-known Travelling Salesman and Binary Knapsack problems. Overall, I am of the opinion that the ACO concept can be applied to optimize specific control design parameters within continuous-time dynamical systems.

Hi @Sahithi Kandhukuri

This problem involves numerous design factors that impact the solution, especially when there are no error messages. It's not as straightforward as a simple equation like 1 + 2 = 3. For instance, at the 11:40 timestamp in the video, the optimal solution is presented as follows:

If the Simulink parameters are configured appropriately,

%% Best solution by ACO
k   = [77 230 193 30 330 290];
%% Parameters in Simulink
ke  = k(1,1);
kce = k(1,2);
ku  = k(1,3);
X   = k(1,4);
z_a = abs(k(1,5));
v_a = abs(k(1,6));
%% Fuzzy Logic parameters in Simulink
% negative big
NB  = -X;
% negative medium
NM  = -X/2;
% positive medium
PM  = X/2;
% positive big
PB  = X;
% Sam: Zero
Z   = 0;        % set by User

we would anticipate witnessing identical results (for deterministic systems) as demonstrated in the YouTube video:

However, the outcomes (Load Voltage and Injected Voltage) I achieved in R2023b are evidently different.

Sahithi Kandhukuri 2024-1-11

Will changing the parameters help us in getting those results?

Sam Chak 2024-1-11

In theory, yes. The keys to achieving the same results lie in your contributions to this work. The test suggests that despite using the same optimal values, your Simulink model may differ from the YouTube version.

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