java - 你能帮我实现 Clarke 和 Wright 算法吗？

Question

我正在尝试实现 Clarke and Wright algorithm构建初始 VRP 解决方案。 它似乎运行正常，但由于某种原因，我得到的解决方案的质量不是预期的。

这是我计算储蓄元素的代码:

private void computeSavingsElements() {
    for(int i = 0; i<vrp.getDimension(); i++) { 
        for(int j = 0; j <  i; j++) {           
                double savingValue =  vrp.distance(i, 0) + vrp.distance(0, j) - lamda * vrp.distance(i, j);
                SavingsElement savingElement = new SavingsElement (i,j, savingValue);
                savingsElements.add(savingElement);                                                 
        }
    }
    Collections.sort(savingsElements); // sort in ascending order
    Collections.reverse(savingsElements); // but we need descending order
    
}

构造解的方法:

private void constructSolution() {
    List<VRPNode> nodes = this.vrp.getNodesList();
    VRPNode depot = this.vrp.getDepot();
    double vehicleCapacity = this.vrp.getVehicleCapacity();
    
    
    VRPSolution solution = new VRPSolution(vehicleCapacity, depot);
    
    /*
     * In the initial solution, each vehicle serves exactly one customer
     */
    for (VRPNode customer:nodes) {
        if (customer.getId()!=0) { // if not depot
            VRPRoute route = new VRPRoute(vehicleCapacity, depot);
            route.addCustomer(customer);
            solution.addRoute(route);
            route = null; // eliminate obsolete reference to free resources
        }
    }   
    
    //System.out.println("INITIAL SOLUTION: \n"+solution.toString());
    
    int mergesCounter=0;
    for (SavingsElement savingElement : this.savingsElements) {
        if (savingElement.getSavingValue() > 0) { // If serving customers consecutively in a route is profitable
            
            VRPNode i = this.vrp.getNode(savingElement.getNodeId1());
            VRPNode j = this.vrp.getNode(savingElement.getNodeId2());
            
            VRPRoute route1 = solution.routeWhereTheCustomerIsTheLastOne(i);
            VRPRoute route2 = solution.routeWhereTheCustomerIsTheFirstOne(j);
            
            if ((route1!=null) & (route2!=null)) {
                if (route1.getDemand() + route2.getDemand() <= this.vrp.getVehicleCapacity()) { // if merge is feasible
                    /*
                     * Merge the two routes
                     */
                    solution.mergeRoutes(route1, route2);
                    mergesCounter++;
                }
            }
            
            
        }
        
    }   
    //System.out.println("\n\nAfter "+mergesCounter+" Merges"+"\n"+solution.toString());
    this.solutionConstructed = solution;
    
}

对于路线合并:

public void mergeRoutes(VRPRoute a, VRPRoute b) {
    /*
     * Provided that feasibility check has already been performed
     */
    List<VRPNode> customersFromRouteA = new LinkedList<VRPNode>(a.getCustomersInRoute());
    List<VRPNode> customersFromRouteB = new LinkedList<VRPNode>(b.getCustomersInRoute());
    
    /*
     * Remove the old routes
     */
    solutionRoutes.remove(a);
    solutionRoutes.remove(b);
    
    /*
     * Construct a new merged route
     */
    VRPRoute mergedRoute = new VRPRoute(vehicleCapacity,depot);
    
    /*
     * The new route has to serve all the customers 
     * both from route a and b
     */
    for (VRPNode customerFromA:  customersFromRouteA) {
        mergedRoute.addCustomer(customerFromA);
    }
    
    for (VRPNode customerFromB:  customersFromRouteB) {
        mergedRoute.addCustomer(customerFromB);
    }
    
    addRoute(mergedRoute);

    evaluateSolutionCost();
}

似乎正确地计算了节省量并按应有的方式合并了路由，但是构建的解决方案的成本太大了。例如，在给定的实例中，我得到 1220，而它应该是 820。

Answer 1

一个明显的问题是当 j < i 时，您的代码只考虑在路由 i 之后加入路由 j。您还应该考虑以相反的方式加入它们 - 换句话说，在 computeSavingsElements 的内部循环中 j 应该达到客户节点的数量 (vrp.getDimension()).

当然，很难判断您未显示的代码部分是否存在错误，例如数组 routeWhereTheCustomerIsTheLastOne 是否已正确更新？