facility.m


function facility()

% Copyright 2019, Gurobi Optimization, LLC
%
% Facility location: a company currently ships its product from 5 plants
% to 4 warehouses. It is considering closing some plants to reduce
% costs. What plant(s) should the company close, in order to minimize
% transportation and fixed costs?
%
% Note that this example uses lists instead of dictionaries.  Since
% it does not work with sparse data, lists are a reasonable option.
%
% Based on an example from Frontline Systems:
%   http://www.solver.com/disfacility.htm
% Used with permission.

% define primitive data
nPlants     = 5;
nWarehouses = 4;
% Warehouse demand in thousands of units
Demand      = [15; 18; 14; 20];
% Plant capacity in thousands of units
Capacity    = [20; 22; 17; 19; 18];
% Fixed costs for each plant
FixedCosts  = [12000; 15000; 17000; 13000; 16000];
% Transportation costs per thousand units
TransCosts  = [
    4000; 2000; 3000; 2500; 4500;
    2500; 2600; 3400; 3000; 4000;
    1200; 1800; 2600; 4100; 3000;
    2200; 2600; 3100; 3700; 3200];

% Index helper function
flowidx = @(w, p) nPlants * w + p;

% Build model
model.modelname = 'facility';
model.modelsense = 'min';

% Set data for variables
ncol = nPlants + nPlants * nWarehouses;
model.lb    = zeros(ncol, 1);
model.ub    = [ones(nPlants, 1); inf(nPlants * nWarehouses, 1)];
model.obj   = [FixedCosts; TransCosts];
model.vtype = [repmat('B', nPlants, 1); repmat('C', nPlants * nWarehouses, 1)];

for p = 1:nPlants
    model.varnames{p} = sprintf('Open%d', p);
end

for w = 1:nWarehouses
    for p = 1:nPlants
        v = flowidx(w, p);
        model.varnames{v} = sprintf('Trans%d,%d', w, p);
    end
end

% Set data for constraints and matrix
nrow = nPlants + nWarehouses;
model.A     = sparse(nrow, ncol);
model.rhs   = [zeros(nPlants, 1); Demand];
model.sense = [repmat('<', nPlants, 1); repmat('=', nWarehouses, 1)];

% Production constraints
for p = 1:nPlants
    for w = 1:nWarehouses
        model.A(p, p) = -Capacity(p);
        model.A(p, flowidx(w, p)) = 1.0;
    end
    model.constrnames{p} = sprintf('Capacity%d', p);
end

% Demand constraints
for w = 1:nWarehouses
    for p = 1:nPlants
        model.A(nPlants+w, flowidx(w, p)) = 1.0;
    end
    model.constrnames{nPlants+w} = sprintf('Demand%d', w);
end

% Save model
gurobi_write(model,'facility_m.lp');

% Guess at the starting point: close the plant with the highest fixed
% costs; open all others first open all plants
model.start = [ones(nPlants, 1); inf(nPlants * nWarehouses, 1)];
[~, idx] = max(FixedCosts);
model.start(idx) = 0;

% Set parameters
params.method = 2;

% Optimize
res = gurobi(model, params);

% Print solution
if strcmp(res.status, 'OPTIMAL')
    fprintf('\nTotal Costs: %g\n', res.objval);
    fprintf('solution:\n');
    for p = 1:nPlants
        if res.x(p) > 0.99
            fprintf('Plant %d open:\n', p);
        end
        for w = 1:nWarehouses
            if res.x(flowidx(w, p)) > 0.0001
                fprintf('  Transport %g units to warehouse %d\n', res.x(flowidx(w, p)), w);
            end
        end
    end
else
    fprintf('\n No solution\n');
end

end

Try Gurobi for Free

Choose the evaluation license that fits you best, and start working with our Expert Team for technical guidance and support.

Evaluation License
Get a free, full-featured license of the Gurobi Optimizer to experience the performance, support, benchmarking and tuning services we provide as part of our product offering.
Academic License
Gurobi supports the teaching and use of optimization within academic institutions. We offer free, full-featured copies of Gurobi for use in class, and for research.
Cloud Trial

Request free trial hours, so you can see how quickly and easily a model can be solved on the cloud.

Search