登录
首页 » matlab » 蚁群算法求解VRPTW问题matlab代码

蚁群算法求解VRPTW问题matlab代码

于 2020-08-23 发布
0 175
下载积分: 1 下载次数: 6

代码说明:

说明:  用于车辆路径问题优化,提高物流运输路线效率,节省成本(It can be used to optimize the vehicle routing problem, improve the efficiency of logistics transportation routes and save costs)

文件列表:

蚁群算法求解VRPTW问题matlab代码\ACO_VRPTW.m, 4549 , 2020-06-03
蚁群算法求解VRPTW问题matlab代码\begin_s.m, 1176 , 2020-04-17
蚁群算法求解VRPTW问题matlab代码\begin_s_v.m, 649 , 2020-04-17
蚁群算法求解VRPTW问题matlab代码\c101.txt, 1899 , 2020-04-19
蚁群算法求解VRPTW问题matlab代码\change.m, 238 , 2020-04-18
蚁群算法求解VRPTW问题matlab代码\costFun.m, 409 , 2020-04-19
蚁群算法求解VRPTW问题matlab代码\deal_vehicles_customer.m, 1107 , 2020-03-24
蚁群算法求解VRPTW问题matlab代码\decode.m, 2086 , 2020-04-17
蚁群算法求解VRPTW问题matlab代码\draw_Best.m, 1766 , 2020-04-19
蚁群算法求解VRPTW问题matlab代码\Judge.m, 1248 , 2020-03-24
蚁群算法求解VRPTW问题matlab代码\JudgeRoute.m, 1505 , 2020-04-17
蚁群算法求解VRPTW问题matlab代码\Judge_Del.m, 452 , 2020-03-24
蚁群算法求解VRPTW问题matlab代码\Judge_TW.m, 919 , 2020-04-17
蚁群算法求解VRPTW问题matlab代码\leave_load.m, 597 , 2020-04-17
蚁群算法求解VRPTW问题matlab代码\linspecer.m, 8449 , 2015-09-16
蚁群算法求解VRPTW问题matlab代码\next_point.m, 4513 , 2020-04-19
蚁群算法求解VRPTW问题matlab代码\next_point_set.m, 2581 , 2020-04-19
蚁群算法求解VRPTW问题matlab代码\part_length.m, 372 , 2020-04-17
蚁群算法求解VRPTW问题matlab代码\roulette.m, 656 , 2020-04-19
蚁群算法求解VRPTW问题matlab代码\travel_distance.m, 734 , 2020-03-24
蚁群算法求解VRPTW问题matlab代码\updateTau.m, 1175 , 2020-04-19
蚁群算法求解VRPTW问题matlab代码\VC_to_Route.m, 210 , 2020-04-19
蚁群算法求解VRPTW问题matlab代码\vehicle_load.m, 975 , 2020-03-24
蚁群算法求解VRPTW问题matlab代码, 0 , 2020-06-03

下载说明:请别用迅雷下载,失败请重下,重下不扣分!

发表评论

0 个回复

  • MVVM-Sample-for-WCF-RIA-Services
    C# WCF Silverlight RIA (rich interface application)
    2014-02-20 02:21:45下载
    积分:1
  • HxDen
    this is a software can edit hex, bin file
    2018-05-10 17:06:57下载
    积分:1
  • 特征提取
    说明:  对接受到的调制信号的特征提取,用于调制识别(automatic modulation recognition)
    2020-01-06 19:35:55下载
    积分:1
  • LCD1602_时钟
    使用51单片机的定时中断原理,在LCD显示时钟(The timing interrupt principle of 51 MCU is used to display the clock in LCD.)
    2018-05-04 14:09:29下载
    积分:1
  • 9283
    因特网隐私管理源码,例程主要针对系统使用网络时留存的网络信息,包括Cookie管理、History管理、Cache管理、Cache对应本地路径。(Internet privacy management source , the routine major network information for the system using the network retained , including the Cookie Manager , History Management , Cache management , Cache corresponding local path.)
    2015-02-04 20:04:52下载
    积分:1
  • RunAsUser
    Run programs as user (Windows)
    2009-06-21 03:41:46下载
    积分:1
  • 周跳探测序(Matlab)
    matlab检测卫星出现周跳的频率和位置(Matlab detection of the frequency and position of the weekly jump)
    2021-05-14 12:30:02下载
    积分:1
  • LabVIEW_Guide
    Labview的详细开发指南,包括:双精度数据不适于相等比较(内置函数);顺序结构是“结构”吗?;到底什么是“节点”;XY GRAPH的输入参数形式 ;整型数据类型和内存映射;状态机的基本概念 ;状态机的基本类型顺序结构 ;状态机的基本类型之标准状态机 ;事件结构中的TIMEOUT进行数据采集合适吗 ;全局变量、移位寄存器和功能型全局变量的性能比较等。(Labview detailed development guidelines, including: double precision data is not suitable for equality comparison (built-in function) sequential structure is " structural" it? In the end what is " node" input parameters of the form XY GRAPH integer data types and memory mapping basic concept of the state machine basic types of sequential structure of the state machine basic types of state machine standard state machine event structure The TIMEOUT data collection suitable for you global variables, shift registers and functional performance comparison of global variables and so on.)
    2014-02-07 14:23:43下载
    积分:1
  • TestDemo
    说明:  这是Biolock的酒店SDK,它可以帮助您从C#Code制作RFID阅读器(This is an hotel SDK for Biolock, this can help you to make RFID Reader from C# Code)
    2020-06-16 10:40:01下载
    积分:1
  • unsustainable under the agreement, users send frames, will first interception Ch...
    非持续CSMA 该协议下,用户在发送帧之前,会先侦听信道的状况,如果没有其他站点发送,它就发送。若信道忙,它就等待一个随机的时间后重复以上动作。 对于某一帧而言,信道是否忙,即是看其绝对时间(FRAMETIME*i+dt)的前一帧时内(FRAMETIME*(i-1)+dt ~ FRAMETIME*i+dt)是否存在发送帧。若无则发送,有则等待一随机时间。 -unsustainable under the agreement, users send frames, will first interception Channel situation, if no other site sent, it sent. If Channel busy, it would wait for a random time to repeat the action. For a frame, whether the busy channel, that is, look at his absolute time (FRAMETIME* i dt) of a time (FRAMETIME* (i-1) dt ~ FRAMETIME dt* i) the existence of this frame. If not then this, then wait for a random time.
    2022-03-16 02:26:57下载
    积分:1
  • 696518资源总数
  • 105949会员总数
  • 22今日下载