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Facility location problem - Data as arrays or collections Description Solve a facility location problem for which data is given as collections.
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FacilityLocationArray.java // (c) 2023-2024 Fair Isaac Corporation import static com.dashoptimization.objects.Utils.sum; import com.dashoptimization.ColumnType; import com.dashoptimization.XPRSenumerations; import com.dashoptimization.objects.Variable; import com.dashoptimization.objects.XpressProblem; /** * This example demonstrates some modeling devices. We model a very simple * facility location problem: We have customers and facilities. The constraints * are: - each customer must be served from exactly one facility - customers can * only be served from open facilities - customer demand must be satisfied - * facility capacity must not be exceeded We minimize the sum of transport cost * (between customer and facility) and the cost for opening a facility. In this * example data is kept in arrays. */ public class FacilityLocationArray { /** Customer description. */ private static final class Customer { /** Name of customer. */ public final String name; /** Demand for customer. */ public final double demand; public Customer(String name, double demand) { this.demand = demand; this.name = name; } } /** Facility descriptor. */ private static final class Facility { /** Name of facility. */ public final String name; /** Capacity of facility. */ public final double capacity; /** Cost for opening this facility. */ public final double cost; public Facility(String name, double capacity, double cost) { this.name = name; this.capacity = capacity; this.cost = cost; } } /** Customers in this example. */ private static final Customer[] customers = new Customer[] { new Customer("Customer 1", 80), new Customer("Customer 2", 270), new Customer("Customer 3", 250) }; /** Facilities in this example. */ private static final Facility[] facilities = new Facility[] { new Facility("Facility 1", 500, 1000), new Facility("Facility 2", 500, 1000), new Facility("Facility 3", 500, 1000) }; /** Cost for transporting one unit between customer and facility. */ private static final double[][] transportCost = new double[][] { new double[] { 4, 5, 6 }, new double[] { 6, 4, 3 }, new double[] { 9, 7, 4 } }; public static void main(String[] args) { try (XpressProblem prob = new XpressProblem()) { Variable[] y = prob.addVariables(facilities.length).withType(ColumnType.Binary) .withName(f -> facilities[f].name).toArray(); Variable[][] x = prob.addVariables(facilities.length, customers.length).withLB(0.0) .withUB(Double.POSITIVE_INFINITY) .withName((f, c) -> String.format("x[%s,%s]", facilities[f].name, customers[c].name)).toArray(); // for each customer c // sum(f=1..m) x[f,c] = d prob.addConstraints(customers.length, c -> sum(facilities.length, f -> x[f][c]).eq(customers[c].demand)); // for each facility f // sum(c=1..n) x[f,c] <= capacity[j] * y[f] prob.addConstraints(facilities.length, f -> sum(x[f]).leq(y[f].mul(facilities[f].capacity))); // minimize sum(f=1..m) cost[f] * y[f] + // sum(c=1..n) sum(f=1..m) cost[f,c] * x[f,c] prob.setObjective(sum(sum(facilities.length, f -> y[f].mul(facilities[f].cost)), sum(customers.length, c -> sum(facilities.length, f -> x[f][c].mul(transportCost[f][c]))))); prob.writeProb("facilitylocationarray.lp", "l"); prob.optimize(); if (prob.attributes().getSolStatus() != XPRSenumerations.SolStatus.OPTIMAL) throw new RuntimeException("failed to optimize with status " + prob.attributes().getSolStatus()); double[] sol = prob.getSolution(); for (int f = 0; f < facilities.length; ++f) { if (y[f].getValue(sol) > 0.5) { System.out.println("Facility " + facilities[f].name + " is open, serves"); for (int c = 0; c < customers.length; ++c) { if (x[f][c].getValue(sol) > 0.0) System.out.println(" " + customers[c].name + ": " + x[f][c].getValue(sol)); } } } } } } | |||||||||||||
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