我正在尝试实施Clarke和Wright算法来构建初始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。 有关算法的说明,请参阅here
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
}
构建解决方案的方法:
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();
}
它似乎正确地计算了节省并且合并了路线。但构建的解决方案的成本太高了。例如,在给定的实例中,我得到1220,而它应该是820.
答案 0 :(得分:0)
一个明显的问题是,当 j <&lt; j 时,您的代码仅考虑在路由 i 之后加入路径 j 我。您还应该考虑以相反的方式加入它们 - 换句话说,在computeSavingsElements j 的内部循环中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?