CPLEX-分支定界算法调用cplex求解VRPTW

前面讲了Cplex直接求解VRPTW的模型，下面我们在分支定界算法中调用Cplex来求解VRPTW
1、分支定界算法
（1）定义：
（2）求解过程：
1）确定一个下界（初始解LB），上界定为无穷大UB
2）把初始问题构建一个节点加入优先队列
3) 判断队列是否为空，如果为空跳转至7，否则取出并弹出队首元素，计算该节点的目标值P
4) 如果P > UB，返回3。否则判断当前节点是否是合法解（对于任意i,j,k,x_ijk均为整数），如果是，跳转5否则跳转6
5) 如果P < UB, 记录UB = P，当前节点为当前最优解BS。返回3
6) 设置两个子节点L, R。L，R的建立方式如上，如果L的目标值L.P <= UB，把L加入队列，如果R的目标值R.P <= UB，把R加入队列。返回3
7) 结束，返回记录的最优节点BS。如果BS为空则无解

2、分支定界求解VRPTW
代码结构如下

Data类的作用是定义变量、变量初始化、读取数据

package com.chb;import java.io.BufferedReader;
import java.io.FileReader;
import java.util.Scanner;//定义参数
class Data{public static final double gap= 1e-6;public static final double big_num = 100000;public static int pointnum=102;          //所有点集合n（包括配送中心和客户点，首尾（0和n）为配送中心）public static double E;             //配送中心时间窗开始时间public static double L;                //配送中心时间窗结束时间public static int carnum;          //车辆数public static double cap;          //车辆载荷public static int[][] point=new int[pointnum][2];;       //所有点的坐标x,ypublic static int[] demand=new int[pointnum];           //需求量public static int[] car=new int[carnum];          //车辆编号public static double[] a=new double[pointnum];               //时间窗开始时间【a[i],b[i]】public static double[] b=new double[pointnum];             //时间窗结束时间【a[i],b[i]】public static double[] s=new double[pointnum];             //客户点的服务时间public static int[][] arcs=new int[pointnum][pointnum];          //arcs[i][j]表示i到j点的弧public static double[][] dist=new double[pointnum][pointnum];      //距离矩阵，满足三角关系,暂用距离表示花费 C[i][j]=dist[i][j]//截断小数3.26434-->3.2public double double_truncate(double v){int iv = (int) v;if(iv+1 - v <= gap)return iv+1;double dv = (v - iv) * 10;int idv = (int) dv;double rv = iv + idv / 10.0;return rv;}  public Data() {super();}//函数功能：从txt文件中读取数据并初始化参数public void read_data(String path,Data data) throws Exception{String line = null;String[] substr = null;Scanner cin = new Scanner(new BufferedReader(new FileReader(path)));  //读取文件//读取1-4行for(int i =0; i < 4;i++){line = cin.nextLine();  }//读取5行line = cin.nextLine();line.trim(); substr = line.split(("\\s+")); carnum = Integer.parseInt(substr[1]); cap = Integer.parseInt(substr[2]);//读取6-9行for(int i =0; i < 4;i++){line = cin.nextLine();}//读取pointnum-1行数据for (int i = 0; i < pointnum - 1; i++) {line = cin.nextLine();line.trim();substr = line.split("\\s+");point[i][0] = Integer.parseInt(substr[2]);point[i][1] = Integer.parseInt(substr[3]);demand[i] = Integer.parseInt(substr[4]);a[i] = Integer.parseInt(substr[5]);b[i] = Integer.parseInt(substr[6]);s[i] = Integer.parseInt(substr[7]);}cin.close();//关闭流//初始化终点参数point[pointnum-1] = point[0];demand[pointnum-1] = 0;a[pointnum-1] = a[0];b[pointnum-1] = b[0];E = a[0];L = b[0];s[pointnum-1] = 0;      double min1 = 1e15;double min2 = 1e15;//距离矩阵初始化for (int i = 0; i < pointnum; i++) {for (int j = 0; j < pointnum; j++) {if (i == j) {dist[i][j] = 0;continue;}dist[i][j] = Math.sqrt(Math.pow(point[i][0]-point[j][0], 2)+Math.pow(point[i][1]-point[j][1], 2));dist[i][j]=double_truncate(dist[i][j]);}}dist[0][pointnum-1] = 0;dist[pointnum-1][0] = 0;//距离矩阵满足三角关系for (int  k = 0; k < pointnum; k++) {for (int i = 0; i < pointnum; i++) {for (int j = 0; j < pointnum; j++) {if (dist[i][j] > dist[i][k] + dist[k][j]) {dist[i][j] = dist[i][k] + dist[k][j];}}}}//初始化为完全图for (int i = 0; i < pointnum; i++) {for (int j = 0; j < pointnum; j++) {if (i != j) {arcs[i][j] = 1;}else {arcs[i][j] = 0;}}}//除去不符合时间窗和容量约束的边for (int i = 0; i < pointnum; i++) {for (int j = 0; j < pointnum; j++) {if (i == j) {continue;}if (a[i]+s[i]+dist[i][j]>b[j] ||demand[i]+demand[j]>cap) {arcs[i][j] = 0;}if (a[0]+s[i]+dist[0][i]+dist[i][pointnum-1]>b[pointnum-1]) {System.out.println("the calculating example is false");}}}for (int i = 1; i < pointnum-1; i++) {if (b[i] - dist[0][i] < min1) {min1 = b[i] - dist[0][i];}if (a[i] + s[i] + dist[i][pointnum-1] < min2) {min2 = a[i] + s[i] + dist[i][pointnum-1];}}if (E > min1 || L < min2) {System.out.println("Duration false!");System.exit(0);//终止程序}//初始化配送中心0，n+1两点的参数arcs[pointnum-1][0] = 0;arcs[0][pointnum-1] = 1;for (int i = 1; i < pointnum-1; i++) {arcs[pointnum-1][i] = 0;}for (int i = 1; i < pointnum-1; i++) {arcs[i][0] = 0;}}
}

Node类的主要作用是记录分支节点

package com.chb;import java.util.ArrayList;public class Node implements Comparable{Data data;int d;double node_cost;             //目标值objectdouble[][][]lp_x;//记录lp解int[][][] node_x_map;//node_xij=1时,node_x_mapijk=1表示必须访问，node_x_mapijk=0表示不能访问int[][] node_x;//0表示弧可以访问，1表示必须访问，-1表示不能访问ArrayList<ArrayList<Integer>> node_routes;       //定义车辆路径链表ArrayList<ArrayList<Double>> node_servetimes; //定义花费时间链表public Node(Data data) {super();this.data = data;node_cost = data.big_num;lp_x = new double [data.pointnum][data.pointnum][data.carnum];node_x_map = new int[data.pointnum][data.pointnum][data.carnum];node_x = new int[data.pointnum][data.pointnum];node_routes = new ArrayList<ArrayList<Integer>>();node_servetimes = new ArrayList<ArrayList<Double>>();}//复制node@SuppressWarnings("unchecked")public Node note_copy() {Node new_node = new Node(data);new_node.d = d;new_node.node_cost = node_cost;for (int i = 0; i < lp_x.length; i++) {for (int j = 0; j < lp_x[i].length; j++) {new_node.lp_x[i][j] = lp_x[i][j].clone();}}for (int i = 0; i < node_x.length; i++) {new_node.node_x[i] = node_x[i].clone();}for (int i = 0; i < node_x_map.length; i++) {for (int j = 0; j < node_x_map[i].length; j++) {new_node.node_x_map[i][j] = node_x_map[i][j].clone();}}for (int i = 0; i < node_routes.size(); i++) {new_node.node_routes.add((ArrayList<Integer>) node_routes.get(i).clone());}for (int i = 0; i < node_servetimes.size(); i++) {new_node.node_servetimes.add((ArrayList<Double>) node_servetimes.get(i).clone());}return new_node;}public int compareTo(Object o){Node node = (Node) o;if(node_cost < node.node_cost)return -1;else if(node_cost == node.node_cost)return 0;elsereturn 1;}
}

BaB_Vrptw类是程序的主函数，其中
init() 这个函数的作用是确定有合法解的最小车辆数量,具体的做法就是建立一个松弛了的cplex模型，并计算使用的车辆数，如果有aa辆未使用车辆就减少aa辆可用车辆，否则减少一辆直到没有可行解。当然，最后我们可使用的车辆是最少的车辆，松弛的模型就是把前面模型把中的x_ijk的整数约束去掉得到的
branch and bound过程：把VRPTW的数学模型松弛的成一个线性规划问题可以求解出VRPTW问题的一个下界，分支的原则就是对于一个选定的x_ijk，且0<x_ijk<1，那么，利用这个x_ijk进行分成两支，左支是不能够走弧ij，右支是必须走弧ij且必须由车辆k经过。即左支对于任意的t，x_ijt = 0。右边则是x_ijk = 1，通过find_arc() 函数找到要分支的弧

package com.chb;
import java.util.ArrayList;
import java.util.PriorityQueue;import ilog.concert.IloException;
import ilog.concert.IloNumExpr;
import ilog.concert.IloNumVar;
import ilog.concert.IloNumVarType;
import ilog.cplex.IloCplex;//类功能：建立模型并求解
public class BaB_Vrptw {Data data;                  //定义类Data的对象Node node1;Node node2;int deep;//深度public PriorityQueue<Node> queue;//分支队列Node best_note;//当前最好分支double cur_best;//最好解int []record_arc;//记录需要分支的节点double x_gap;//很小的数IloCplex model;              //定义cplex内部类的对象     public IloNumVar[][][] x;   //x[i][j][k]表示弧arcs[i][j]被车辆k访问public IloNumVar[][] w;      //车辆访问所有点的时间矩阵double cost;              //目标值objectdouble[][][] x_map;//cplex参数xArrayList<ArrayList<Integer>> routes;       //定义车辆路径链表ArrayList<ArrayList<Double>> servetimes;  //定义花费时间链表public BaB_Vrptw(Data data) {this.data = data;x_gap = data.gap;routes = new ArrayList<>();       //定义车辆路径链表servetimes = new ArrayList<>();    //定义花费时间链表//初始化车辆路径和花费时间链表，链表长度为车辆数kfor (int k = 0; k < data.carnum; k++) {ArrayList<Integer> r = new ArrayList<>();  ArrayList<Double> t = new ArrayList<>();   routes.add(r);                          servetimes.add(t);                          }x_map = new double[data.pointnum][data.pointnum][data.carnum];}public void clear_lp() {data=null;routes.clear();servetimes.clear();x_map=null;}//辅助lp解到node@SuppressWarnings("unchecked")public void copy_lp_to_node(BaB_Vrptw lp, Node node) {node.node_routes.clear();node.node_servetimes.clear();node.node_cost = lp.cost;for (int i = 0; i < lp.x_map.length; i++) {for (int j = 0; j < lp.x_map[i].length; j++) {node.lp_x[i][j] = lp.x_map[i][j].clone();}}for (int i = 0; i < lp.routes.size(); i++) {node.node_routes.add((ArrayList<Integer>) lp.routes.get(i).clone());}for (int i = 0; i < lp.servetimes.size(); i++) {node.node_servetimes.add((ArrayList<Double>) lp.servetimes.get(i).clone());}}//函数功能：根据VRPTW数学模型建立VRPTW的cplex模型//建立模型private void build_model() throws IloException {//modelmodel = new IloCplex();model.setOut(null);
//      model.setParam(IloCplex.DoubleParam.EpOpt, 1e-9);
//      model.setParam(IloCplex.DoubleParam.EpGap, 1e-9);//variablesx = new IloNumVar[data.pointnum][data.pointnum][data.carnum];w = new IloNumVar[data.pointnum][data.carnum];               //车辆访问点的时间//定义cplex变量x和w的数据类型及取值范围for (int i = 0; i < data.pointnum; i++) {for (int k = 0; k < data.carnum; k++) {w[i][k] = model.numVar(0, 1e15, IloNumVarType.Float, "w" + i + "," + k);}for (int j = 0; j < data.pointnum; j++) {if (data.arcs[i][j]==0) {x[i][j] = null;}else{//Xijk,公式(10)-(11)for (int k = 0; k < data.carnum; k++) {x[i][j][k] = model.numVar(0, 1, IloNumVarType.Float, "x" + i + "," + j + "," + k);}}}}//加入目标函数//公式(1)IloNumExpr obj = model.numExpr();for(int i = 0; i < data.pointnum; i++){for(int j = 0; j < data.pointnum; j++){if (data.arcs[i][j]==0) {continue;}for(int k = 0; k < data.carnum; k++){obj = model.sum(obj, model.prod(data.dist[i][j], x[i][j][k]));}}}model.addMinimize(obj);//加入约束//公式(2)for(int i= 1; i < data.pointnum-1;i++){IloNumExpr expr1 = model.numExpr();for (int k = 0; k < data.carnum; k++) {for (int j = 1; j < data.pointnum; j++) {if (data.arcs[i][j]==1) {expr1 = model.sum(expr1, x[i][j][k]);}}}model.addEq(expr1, 1);}//公式(3)for (int k = 0; k < data.carnum; k++) {IloNumExpr expr2 = model.numExpr();for (int j = 1; j < data.pointnum; j++) {if (data.arcs[0][j]==1) {expr2 = model.sum(expr2, x[0][j][k]);}}model.addEq(expr2, 1);}//公式(4)for (int k = 0; k < data.carnum; k++) {for (int j = 1; j < data.pointnum-1; j++) {IloNumExpr expr3 = model.numExpr();IloNumExpr subExpr1 = model.numExpr();IloNumExpr subExpr2 = model.numExpr();for (int i = 0; i < data.pointnum; i++) {if (data.arcs[i][j]==1) {subExpr1 = model.sum(subExpr1,x[i][j][k]);}if (data.arcs[j][i]==1) {subExpr2 = model.sum(subExpr2,x[j][i][k]);}}expr3 = model.sum(subExpr1,model.prod(-1, subExpr2));model.addEq(expr3, 0);}}//公式(5)for (int k = 0; k < data.carnum; k++) {IloNumExpr expr4 = model.numExpr();for (int i = 0; i < data.pointnum-1; i++) {if (data.arcs[i][data.pointnum-1]==1) {expr4 = model.sum(expr4,x[i][data.pointnum-1][k]);}}model.addEq(expr4, 1);}//公式(6)double M = 1e5;for (int k = 0; k < data.carnum; k++) {for (int i = 0; i < data.pointnum; i++) {for (int j = 0; j < data.pointnum; j++) {if (data.arcs[i][j] == 1) {IloNumExpr expr5 = model.numExpr();IloNumExpr expr6 = model.numExpr();expr5 = model.sum(w[i][k], data.s[i]+data.dist[i][j]);expr5 = model.sum(expr5,model.prod(-1, w[j][k]));expr6 = model.prod(M,model.sum(1,model.prod(-1, x[i][j][k])));model.addLe(expr5, expr6);}}}}//公式(7)for (int k = 0; k < data.carnum; k++) {for (int i = 1; i < data.pointnum-1; i++) {IloNumExpr expr7 = model.numExpr();for (int j = 0; j < data.pointnum; j++) {if (data.arcs[i][j] == 1) {expr7 = model.sum(expr7,x[i][j][k]);}}model.addLe(model.prod(data.a[i], expr7), w[i][k]);model.addLe(w[i][k], model.prod(data.b[i], expr7));}}//公式(8)for (int k = 0; k < data.carnum; k++) {model.addLe(data.E, w[0][k]);model.addLe(data.E, w[data.pointnum-1][k]);model.addLe(w[0][k], data.L);model.addLe(w[data.pointnum-1][k], data.L);}//公式(9)for (int k = 0; k < data.carnum; k++) {IloNumExpr expr8 = model.numExpr();for (int i = 1; i < data.pointnum-1; i++) {IloNumExpr expr9 = model.numExpr();for (int j = 0; j < data.pointnum; j++) {if (data.arcs[i][j] == 1) {expr9=model.sum(expr9,x[i][j][k]);}}expr8 = model.sum(expr8,model.prod(data.demand[i],expr9));}model.addLe(expr8, data.cap);}}//函数功能：解模型，并生成车辆路径和得到目标值//获取cplex解public void get_value() throws IloException {routes.clear();servetimes.clear();cost = 0;
//      //初始化车辆路径和花费时间链表，链表长度为车辆数kfor (int k = 0; k < data.carnum; k++) {ArrayList<Integer> r = new ArrayList<>();    ArrayList<Double> t = new ArrayList<>();   routes.add(r);                          servetimes.add(t);                          }for (int i = 0; i < data.pointnum; i++) {for (int j = 0; j < data.pointnum; j++) {for (int k = 0; k < data.carnum; k++) {x_map[i][j][k] = 0.0;}if (data.arcs[i][j]>0.5) {for (int k = 0; k < data.carnum; k++) {x_map[i][j][k]=model.getValue(x[i][j][k]);}}}}//模型可解，生成车辆路径for(int k = 0; k < data.carnum; k++){boolean terminate = true;int i = 0;routes.get(k).add(0);       servetimes.get(k).add(0.0);while(terminate){for (int j = 0; j < data.pointnum; j++) {if (doubleCompare(x_map[i][j][k], 0)==1) {routes.get(k).add(j);servetimes.get(k).add(model.getValue(w[j][k]));i = j;break;}}if (i == data.pointnum-1) {terminate = false;}}}cost = model.getObjValue();}//branch and bound过程public void branch_and_bound(BaB_Vrptw lp) throws IloException {cur_best = 3000;//设置上届deep=0;record_arc = new int[3];node1 = new Node(data);best_note = null;queue = new PriorityQueue<Node>();//初始解（非法解）for (int i = 0; i < lp.routes.size(); i++) {ArrayList<Integer> r = lp.routes.get(i);System.out.println();for (int j = 0; j < r.size(); j++) {System.out.print(r.get(j)+" ");}}lp.copy_lp_to_node(lp, node1);
//      node1.node_cost = lp.cost;
//      node1.lp_x = lp.x_map.clone();
//      node1.node_routes =lp.routes;
//      node1.node_servetimes = lp.servetimes;node2 = node1.note_copy();deep=0;node1.d=deep;queue.add(node1);//branch and bound过程while (!queue.isEmpty()) {Node node = queue.poll();//某支最优解大于当前最好可行解，删除if (doubleCompare(node.node_cost, cur_best)>0) {continue;}else {record_arc = lp.find_arc(node.lp_x);//某支的合法解,0,1组合的解,当前分支最好解if (record_arc[0]==-1) {//比当前最好解好，更新当前解if (doubleCompare(node.node_cost, cur_best)==-1) {lp.cur_best = node.node_cost;System.out.println(node.d+"  cur_best:"+cur_best);lp.best_note = node;}continue;}else {//可以分支node1 = lp.branch_left_arc(lp, node, record_arc);//左支node2 = lp.branch_right_arc(lp, node, record_arc);//右支if (node1!=null && doubleCompare(node1.node_cost, cur_best)<=0) {queue.add(node1);}if (node2!=null && doubleCompare(node2.node_cost, cur_best)<=0) {queue.add(node2);}}}}}//分支设置public void set_bound(Node node) throws IloException {for (int i = 0; i < data.pointnum; i++) {for (int j = 0; j < data.pointnum; j++) {if (data.arcs[i][j]>0.5) {if (node.node_x[i][j]==0) {for (int k = 0; k < data.carnum; k++) {x[i][j][k].setLB(0.0);x[i][j][k].setUB(1.0);}}else if (node.node_x[i][j]==-1) {for (int k = 0; k < data.carnum; k++) {x[i][j][k].setLB(0.0);x[i][j][k].setUB(0.0);}}else {for (int k = 0; k < data.carnum; k++) {if (node.node_x_map[i][j][k]==1) {x[i][j][k].setLB(1.0);x[i][j][k].setUB(1.0);}else {x[i][j][k].setLB(0.0);x[i][j][k].setUB(0.0);}}}}}}}public void set_bound1(Node node) throws IloException {for (int i = 0; i < data.pointnum; i++) {for (int j = 0; j < data.pointnum; j++) {if (data.arcs[i][j]>0.5) {for (int k = 0; k < data.carnum; k++) {if (node.node_x_map[i][j][k]==0) {x[i][j][k].setLB(0.0);x[i][j][k].setUB(1.0);}else if (node.node_x_map[i][j][k]==-1) {x[i][j][k].setLB(0.0);x[i][j][k].setUB(0.0);}else {x[i][j][k].setLB(1.0);x[i][j][k].setUB(1.0);}}}}}}//设置左支public Node branch_left_arc(BaB_Vrptw lp,Node father_node,int[] record) throws IloException {if (record[0] == -1) {return null;}Node new_node = new Node(data);new_node = father_node.note_copy();new_node.node_x[record[0]][record[1]] = -1;for (int k = 0; k < data.carnum; k++) {new_node.node_x_map[record[0]][record[1]][k]=0;}
//      new_node.node_x_map[record[0]][record[1]][record[2]]=-1;//设置左支lp.set_bound(new_node);if (lp.model.solve()) {lp.get_value();deep++;new_node.d=deep;lp.copy_lp_to_node(lp, new_node);System.out.println(new_node.d+" "+lp.cost);}else {new_node.node_cost = data.big_num;}return new_node;
}//设置右支public Node branch_right_arc(BaB_Vrptw lp,Node father_node,int[] record) throws IloException {if (record[0] == -1) {return null;}Node new_node = new Node(data);new_node = father_node.note_copy();new_node.node_x[record[0]][record[1]] = 1;
//      new_node.node_x_map[record[0]][record[1]][record[2]]=1;for (int k = 0; k < data.carnum; k++) {if (k==record[2]) {new_node.node_x_map[record[0]][record[1]][k]=1;}else {new_node.node_x_map[record[0]][record[1]][k]=0;}}//设置右支lp.set_bound(new_node);if (lp.model.solve()) {lp.get_value();deep++;new_node.d=deep;System.out.println(new_node.d+" right: "+lp.cost);lp.copy_lp_to_node(lp, new_node);}else {new_node.node_cost = data.big_num;}return new_node;}//找到需要分支的支点位置public int[] find_arc1(double[][][] x) {int record[] = new int[3];//记录分支顶点double cur_dif = 0;double min_dif = Double.MAX_VALUE;//找出最接近0.5的弧for (int i = 1; i <data.pointnum-1; i++) {for (int j = 1; j < data.pointnum-1; j++) {if (data.arcs[i][j]>0.5) {for (int k = 0; k <data.carnum; k++) {//若该弧值为0或1，则继续if (is_one_zero(x[i][j][k])) {continue;}cur_dif = get_dif(x[i][j][k]);if (doubleCompare(cur_dif, min_dif)==-1) {record[0] = i;record[1] = j;record[2] = k;min_dif = cur_dif;}}}}}//depotif (doubleCompare(min_dif, Double.MAX_VALUE)==0) {for (int i = 1; i < data.pointnum-1; i++) {if (data.arcs[0][i]>0.5) {for (int k = 0; k < data.carnum; k++) {if (is_fractional(x[0][i][k])) {cur_dif = get_dif(x[0][i][k]);if (doubleCompare(cur_dif, min_dif)==-1) {record[0] = 0;record[1] = i;record[2] = k;min_dif = cur_dif;}}}}if (data.arcs[i][data.pointnum-1]>0.5) {for (int k = 0; k < data.carnum; k++) {if (is_fractional(x[i][data.pointnum-1][k])) {cur_dif = get_dif(x[i][data.pointnum-1][k]);if (doubleCompare(cur_dif, min_dif)==-1) {record[0] = i;record[1] = data.pointnum-1;record[2] = k;min_dif = cur_dif;}}}}}}if (doubleCompare(min_dif, data.big_num)==1) {record[0] = -1;record[1] = -1;record[2] = -1;}return record;}
//  找到要分支的弧public int[] find_arc(double[][][] x) {int record[] = new int[3];//记录分支顶点for (int i = 0; i <data.pointnum; i++) {for (int j = 0; j < data.pointnum; j++) {if (data.arcs[i][j]>0.5) {for (int k = 0; k <data.carnum; k++) {//若该弧值为0或1，则继续if (is_one_zero(x[i][j][k])) {continue;}
//                      cur_dif = get_dif(x[i][j][k]);record[0] = i;record[1] = j;record[2] = k;return record;}}}}record[0] = -1;record[1] = -1;record[2] = -1;return record;}//比较两个double数值的大小public int doubleCompare(double a, double b){if(a - b > x_gap)return 1;if(b - a > x_gap)return -1;      return 0;} //判断是否为0到1之间的小数public boolean is_fractional(double v){if( v > (int) v + x_gap && v < (int) v + 1 - x_gap)return true;elsereturn false;}//判断是否为0或者1public boolean is_one_zero(double temp) {if (doubleCompare(temp, 0)==0 || doubleCompare(temp, 1)==0) {return true;}else {return false;}}//获取到0.5的距离public double get_dif(double temp) {double v = (int)temp+0.5;if (v>temp) {return v-temp;} else {return temp-v;}}public BaB_Vrptw init(BaB_Vrptw lp) throws IloException {lp.build_model();if (lp.model.solve()) {lp.get_value();int aa=0;for (int i = 0; i < lp.routes.size(); i++) {if (lp.routes.get(i).size()==2) {aa++;}}System.out.println(aa);if (aa==0) {data.carnum -=1;lp.model.clearModel();lp = new BaB_Vrptw(data);return init(lp);}else {data.carnum -=aa;lp.model.clearModel();lp = new BaB_Vrptw(data);return init(lp);}}else {data.carnum +=1;System.out.println("vehicle number: "+data.carnum);lp.model.clearModel();lp = new BaB_Vrptw(data);lp.build_model();if (lp.model.solve()) {lp.get_value();return lp;}else {System.out.println("error init");return null;}}}public static void main(String[] args) throws Exception {Data data = new Data();//读入不同的文件前要手动修改vetexnum参数，参数值等于所有点个数,包括配送中心String path = "data/c102.txt";//算例地址data.read_data(path,data);System.out.println("input succesfully");System.out.println("cplex procedure###########################");BaB_Vrptw lp = new BaB_Vrptw(data);double cplex_time1 = System.nanoTime();//删除未用的车辆，缩小解空间lp=lp.init(lp);System.out.println(":   "+lp.data.carnum);lp.branch_and_bound(lp);Check check = new Check(lp);check.fesible();double cplex_time2 = System.nanoTime();double cplex_time = (cplex_time2 - cplex_time1) / 1e9;//求解时间，单位sSystem.out.println("cplex_time " + cplex_time + " bestcost " + lp.cur_best);for (int i = 0; i < lp.best_note.node_routes.size(); i++) {ArrayList<Integer> r = lp.best_note.node_routes.get(i);System.out.println();for (int j = 0; j < r.size(); j++) {System.out.print(r.get(j)+" ");}}}
}

Check类的作用是检查解的可行性，包括解是否满足车辆数量约束，是否满足容量约束，时间窗约束等等

package com.chb;import java.util.ArrayList;import ilog.concert.IloException;//类功能：解的可行性判断(可直接跳过此类)
class Check{double epsilon = 0.0001;Data data = new Data();ArrayList<ArrayList<Integer>> routes = new ArrayList<>();ArrayList<ArrayList<Double>> servetimes = new ArrayList<>();public Check(BaB_Vrptw lp) {super();this.data = lp.data;this.routes = lp.routes;this.servetimes = lp.servetimes;}//函数功能：比较两个数的大小public int double_compare(double v1,double v2) {if (v1 < v2 - epsilon) {return -1;}if (v1 > v2 + epsilon) {return 1;}return 0;}//函数功能：解的可行性判断public void fesible() throws IloException {//车辆数量可行性判断if (routes.size() > data.carnum) {System.out.println("error: vecnum!!!");System.exit(0);}//车辆载荷可行性判断for (int k = 0; k < routes.size(); k++) {ArrayList<Integer> route = routes.get(k);double capasity = 0;for (int i = 0; i < route.size(); i++) {capasity += data.demand[route.get(i)];}if (capasity > data.cap) {System.out.println("error: cap!!!");System.exit(0);}}//时间窗、车容量可行性判断for (int k = 0; k < routes.size(); k++) {ArrayList<Integer> route = routes.get(k);ArrayList<Double> servertime = servetimes.get(k);double capasity = 0;for (int i = 0; i < route.size()-1; i++) {int origin = route.get(i);int destination = route.get(i+1);double si = servertime.get(i);double sj = servertime.get(i+1);if (si < data.a[origin] && si >  data.b[origin]) {System.out.println("error: servertime!");System.exit(0);}if (double_compare(si + data.dist[origin][destination],data.b[destination]) > 0) {System.out.println(origin + ": [" + data.a[origin] + ","+data.b[origin]+"]"+ " "+ si);System.out.println(destination + ": [" +data.a[destination] + ","+data.b[destination]+"]"+ " "+ sj);System.out.println(data.dist[origin][destination]);System.out.println(destination + ":" );System.out.println("error: forward servertime!");System.exit(0);}if (double_compare(sj - data.dist[origin][destination],data.a[origin]) < 0) {System.out.println(origin + ": [" + data.a[origin]+ ","+data.b[origin]+"]"+ " "+ si);System.out.println(destination + ": [" + data.a[destination] + ","+data.b[destination]+"]"+ " "+ sj);System.out.println(data.dist[origin][destination]);System.out.println(destination + ":" );System.out.println("error: backward servertime!");System.exit(0);}}if (capasity > data.cap) {System.out.println("error: cap!!!");System.exit(0);}}}
}

运行结果：

注：文本提炼、转载自
https://mp.weixin.qq.com/s?__biz=MzU0NzgyMjgwNg==&mid=2247484626&idx=1&sn=fa7ab4ea6e02084f4e8ed52c6c203eda&chksm=fb49c96bcc3e407db8c7bb9433dfb1fa787518328922bf061a4421c46557089b3429c9aeb391&mpshare=1&scene=1&srcid=0819ACkwrk90AoXHmuTfOewW&sharer_sharetime=1566189897645&sharer_shareid=054592193644de509623829748e83807&key=39e8dcac62579260153e141217c0c82d2c6e85d9ab908f37737e7a53942622438640915d71ffff9fcb2a13fe7ce9bddfcdfbd31c41de4e97aa05544e971a3b9da9c1048280222095be4e5c215e240a6b&ascene=1&uin=MjYzMDA1MzAyMQ%3D%3D&devicetype=Windows+10&version=62060834&lang=zh_CN&pass_ticket=XkzfvjtynhPBmAjeYlUVZf95yJsB8gJ7VNbzHQQ9yaInrn1wKsdFjIxND%2FyV40nF

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