Mathematical formulation of a operations research problem

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David Franco
David Franco le 10 Sep 2019
Modifié(e) : David Franco le 2 Déc 2021
How can I implement this objective function and its constraints in Matlab (p-median problem)?
Thank you very much!
  1 commentaire
darova
darova le 10 Sep 2019
I can give you a start
y = zeros(1,10);
y = facility == candidate;

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Hasan Cosgun
Hasan Cosgun le 2 Déc 2021
Modifié(e) : Hasan Cosgun le 2 Déc 2021
nNodes = 52; % no of nodes (customers)
nP = 6; % No of Warehouses/Facilities
% dist: nNodes x nNodes distance matrix
% decision variables
x = optimvar('x',1,nNodes,'Type','integer','LowerBound',0,'UpperBound',1);
y = optimvar('y',nNodes,nNodes,'Type','integer','LowerBound',0,'UpperBound',1);
prob = optimproblem('ObjectiveSense','minimize');
prob.Objective = sum(sum(dist .* y));
onesum = sum(y,2) == 1;
vertxy = optimconstr(nNodes,nNodes);
for i=1:nNodes
for j=1:nNodes
vertxy(i,j) = y(i,j) <= x(j);
end
end
depsum = sum(x) == nP;
prob.Constraints.onesum = onesum;
prob.Constraints.vertxy = vertxy;
prob.Constraints.depsum = depsum;
nSolution = solve(prob);
Just a full start...
  1 commentaire
David Franco
David Franco le 2 Déc 2021
Modifié(e) : David Franco le 2 Déc 2021
Thank you very much @Hasan Cosgun!!! I updated your code with an example:
% P-Median Problem (PMP)
rng default
nNodes = 32; % No of Customers
nP = 6; % No of Facilities
xx = randi(20,nNodes,2);
dist = pdist2(xx,xx,'euclidean');
% decision variables
y = optimvar('y',1,nNodes,'Type','integer','LowerBound',0,'UpperBound',1);
x = optimvar('x',nNodes,nNodes,'Type','continuous','LowerBound',0);
prob = optimproblem('ObjectiveSense','minimize');
prob.Objective = sum(sum(dist .* x));
onesum = sum(x,2) == 1; % Restriction 1
depsum = sum(y) <= nP; % Restriction 2
vertxy = optimconstr(nNodes,nNodes);
for i = 1:nNodes
for j = 1:nNodes
vertxy(i,j) = x(i,j) <= y(j); % Restriction 3
end
end
prob.Constraints.onesum = onesum;
prob.Constraints.depsum = depsum;
prob.Constraints.vertxy = vertxy;
[sol,fval,exitflag,output] = solve(prob);
Solving problem using intlinprog. LP: Optimal objective value is 82.069171. Optimal solution found. Intlinprog stopped at the root node because the objective value is within a gap tolerance of the optimal value, options.AbsoluteGapTolerance = 0 (the default value). The intcon variables are integer within tolerance, options.IntegerTolerance = 1e-05 (the default value).
z = 1:nNodes;
idx = sum(sol.x .* z(ones(nNodes,1),:),2);
h1 = gscatter(xx(:,1),xx(:,2),idx);
hold on
h2 = plot(xx(sol.y == 1,1),xx(sol.y == 1,2),'ko');
legend('Location','eastoutside')
h2.Annotation.LegendInformation.IconDisplayStyle = 'off';

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Plus de réponses (1)

Matt J
Matt J le 11 Mar 2020
Should be very easy with the Problem-Based Optimization Workflow.

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