82 lines
No EOL
2.3 KiB
Matlab
82 lines
No EOL
2.3 KiB
Matlab
function [x_B,c_B,index_B] = simplexSolve(type, B, D, c_B, c_D, h, x_B, ...
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x_D, index_B, index_D, itMax)
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% Solving a maximization problem with the simplex method
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% Initialize the number of iterations
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nIter = 0;
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% Compute B^{-1}*D and B^{-1}*h
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BiD = B\D;
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Bih = B\h;
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% Compute the reduced cost coefficients
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r_D = c_D - (c_B * BiD);
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% the optimality condition is satisfied if all
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% reduced cost coefficients are positive or negative (depending on the
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% problem)
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tol = max(size(r_D));
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% Check the optimality condition, in order to skip the loop if the
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% solution is already optimal
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optCheck = typeCond(type, sum(r_D <= 0), sum(r_D >= 0));
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while optCheck ~= tol
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% Find the index of the entering variable
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idxIN = find(r_D == typeCond(type, max(r_D), min(r_D)), 1, 'first');
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in = D(:,idxIN);
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c_in = c_D(1,idxIN);
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index_in = index_D(1,idxIN);
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% Evaluate the coefficients ratio for the column corresponding to
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% the entering variable
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ratio = Bih / BiD;
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% Find the smallest positive ratio
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idxOUT = find(ratio == min(ratio(ratio > 0)), 1, 'first');
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out = B(:,idxOUT);
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c_out = c_B(1,idxOUT);
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index_out = index_B(1,idxOUT);
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% TODO: Update the matrices by exchanging the columns
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B(:,idxOUT) = in;
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D(:,idxIN) = out;
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c_B(1,idxOUT) = c_in;
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c_D(1,idxIN) = c_out;
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index_B(1,idxOUT) = index_in;
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index_D(1,idxIN) = index_out;
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% Compute B^{-1}*D and B^{-1}*h
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BiD = B\D;
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Bih = B\h;
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% Compute the reduced cost coefficients
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r_D = c_D - (c_B * BiD);
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% Check the optimality condition, in order to exit the loop if the
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% solution is already optimal
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optCheck = typeCond(type, sum(r_D <= 0), sum(r_D >= 0));
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% Detect inefficient looping if nIter > total number of basic solutions
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nIter = nIter + 1;
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if nIter > itMax
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error('Incorrect loop, more iterations than the number of basic solutions');
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end
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% Compute the new x_B
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x_B = Bih - BiD * x_D;
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end
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end
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function x = typeCond(type, ifMaximum, ifMinimum)
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if strcmp(type, 'max')
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x = ifMaximum;
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elseif strcmp(type, 'min')
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x = ifMinimum;
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else
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error('Incorrect type specified. Choose either a maximisation (max) or minimisation (min) problem.')
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end
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end |