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%
% This is an example call of MIDACO 6.0
% -------------------------------------
%
% MIDACO solves Multi-Objective Mixed-Integer Non-Linear Problems:
%
%
% Minimize F_1(X),... F_O(X) where X(1,...N-NI) is CONTINUOUS
% and X(N-NI+1,...N) is DISCRETE
%
% subject to G_j(X) = 0 (j=1,...ME) equality constraints
% G_j(X) >= 0 (j=ME+1,...M) inequality constraints
%
% and bounds XL <= X <= XU
%
%
% The problem statement of this example is given below. You can use
% this example as template to run your own problem. To do so: Replace
% the objective functions 'F' (and in case the constraints 'G') given
% here with your own problem and follow the below instruction steps.
%
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function example
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%%%%%%%%%%%%%%%%%%%%%%%%% MAIN PROGRAM %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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key = '************************************************************';
problem.func = @problem_function; % Call is [f,g] = problem_function(x)
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%%% Step 1: Problem definition %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% STEP 1.A: Problem dimensions
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problem.o = 1; % Number of objectives
problem.n = 9; % Number of variables (in total)
problem.ni = 3; % Number of integer variables (0 <= nint <= n)
problem.m = 7; % Number of constraints (in total)
problem.me = 3; % Number of equality constraints (0 <= me <= m)
% STEP 1.B: Lower and upper bounds 'xl' & 'xu'
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problem.xl = 0 * ones(1,problem.n);
problem.xu = 10 * ones(1,problem.n);
% binaries
problem.xu(7:9) = 1.0;
% STEP 1.C: Starting point 'x'
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problem.x = problem.xl; % Here for example: 'x' = lower bounds 'xl'
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%%% Step 2: Choose stopping criteria and printing options %%%%%%%%%%%
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% STEP 2.A: Stopping criteria
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option.maxeval = 1000000; % Maximum number of function evaluation (e.g. 1000000)
option.maxtime = 60*60*24; % Maximum time limit in Seconds (e.g. 1 Day = 60*60*24)
% STEP 2.B: Printing options
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option.printeval = 5000; % Print-Frequency for current best solution (e.g. 1000)
option.save2file = 1; % Save SCREEN and SOLUTION to TXT-files [ 0=NO/ 1=YES]
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%%% Step 3: Choose MIDACO parameters (FOR ADVANCED USERS) %%%%%%%%%%%
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option.param( 1) = 0; % ACCURACY
option.param( 2) = 0; % SEED
option.param( 3) = -1.92; % FSTOP
option.param( 4) = 0; % ALGOSTOP
option.param( 5) = 0; % EVALSTOP
option.param( 6) = 0; % FOCUS
option.param( 7) = 0; % ANTS
option.param( 8) = 0; % KERNEL
option.param( 9) = 0; % ORACLE
option.param(10) = 0; % PARETOMAX
option.param(11) = 0; % EPSILON
option.param(12) = 0; % BALANCE
option.param(13) = 0; % CHARACTER
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%%% Step 4: Choose Parallelization Factor %%%%%%%%%%%%%%%%%%%%%%%%%%%%
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option.parallel = 0; % Serial: 0 or 1, Parallel: 2,3,4,5,6,7,8...
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%%% Call MIDACO solver %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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[ solution ] = midaco( problem, option, key);
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%%% End of Example %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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end
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%%%%%%%%%%%%%%%%%%%%% OPTIMIZATION PROBLEM %%%%%%%%%%%%%%%%%%%%%%%%%
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function [ f, g ] = problem_function( x )
% Rename integer variables
y = x(7:9);
x1 = x(1);
x2 = x(2);
x3 = x(3);
x4 = x(4);
x5 = x(5);
x6 = x(6);
b7 = y(1);
b8 = y(2);
b9 = y(3);
% Objective function value F(X)
f = 1.8*x1 + 1.8*x2 + 7*x3 + x4 + 1.2*x5 - 11*x6 ...
+ 3.5*b7 + b8 + 1.5*b9;
% Equality constraints G(X) = 0
g(1) = - log(1 + x1) + x4;
g(2) = - 1.2*log(1 + x2) + x5;
g(3) = - 0.9*x3 - 0.9*x4 - 0.9*x5 + x6;
% Inequality constraints G(X) >= 0
g(4) = -(x6 - b7);
g(5) = -(x4 - 1.111111*b8);
g(6) = - (x5 - 1.111111*b9);
g(7) = - (b8 + b9) + 1;
end
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