标签：load enter 版本 不包含 rate 包含 clear oss 分析
2 问题分析
3 基本假设
4 符号说明
5 模型的建立与求解
模型2 区级邮路
clear,clc
%问题一求解
m=[5,5,6];%三个车队的分组数
B = [10 15 6 9 13 6 11 4 13 17 11 2 11 21 13 14;
9 14 5 10 9 10 13 9 15 9 6 7 13 15 10 16];%问题1的16个支局收寄邮件数量
%问题一各个支局之间的距离矩阵
% dist=[0 31 27 49 Inf Inf Inf Inf Inf Inf Inf 52 21 41 Inf Inf 27;
% 31 0 19 Inf 27 32 Inf Inf Inf 47 Inf Inf Inf 50 Inf Inf 44;
% 27 19 0 14 Inf Inf Inf Inf Inf 30 Inf Inf Inf 31 Inf Inf 17;
% 49 Inf 14 0 13 20 Inf Inf 28 15 Inf Inf Inf 15 25 30 11;
% Inf 27 Inf 13 0 9 21 Inf 26 26 Inf Inf Inf 28 29 Inf 27;
% Inf 32 Inf 20 9 0 13 Inf 32 Inf Inf Inf Inf Inf 33 Inf 42;
% Inf Inf Inf Inf 21 13 0 19 Inf Inf Inf Inf Inf Inf Inf Inf Inf;
% Inf Inf Inf Inf Inf Inf 19 0 11 20 Inf Inf Inf Inf 33 21 Inf;
% Inf Inf Inf 28 26 32 Inf 11 0 10 20 Inf Inf 29 14 13 Inf;
% Inf 47 30 15 26 Inf Inf 20 10 0 18 Inf Inf 14 9 20 20;
% Inf Inf Inf Inf Inf Inf Inf Inf 20 18 0 23 Inf Inf 14 Inf 25;
% 52 Inf Inf Inf Inf Inf Inf Inf Inf Inf 23 0 27 22 Inf Inf 21;
% 21 Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf 27 0 Inf Inf Inf 21;
% 41 50 31 15 28 Inf Inf Inf 29 14 Inf 22 Inf 0 11 Inf 18;
% Inf Inf Inf 25 29 33 Inf 33 14 9 14 Inf Inf 11 0 9 27;
% Inf Inf Inf 30 Inf Inf Inf 21 13 20 Inf Inf Inf Inf 9 0 Inf;
% 27 44 17 11 27 42 Inf Inf Inf 20 25 21 21 18 27 Inf 0]
p1=[];p2=[];p3=[];pp1=[];pp2=[];pp3=[];
cost0 = 100;distan0=0;
for i=1:100000
S=randperm(16);
p1=S(1:m(1));
if sum(B(1,p1))>65 | sum(B(1,p1))<46 | sum(B(2,p1))>65 | sum(B(2,p1))<40
continue;
end
p1(m(1)+1)=17;
[cost1, distan1] = vrp_cost(p1);
if distan1/30+5*m(1) > 360
continue;
end
p2=S(m(1)+1:m(1)+m(2));
if sum(B(1,p2))>65 | sum(B(1,p2))<46 | sum(B(2,p2))>65 | sum(B(2,p2))<40
continue;
end
p2(m(2)+1)=17;
[cost2, distan2] = vrp_cost(p2);
if distan2/30+5*m(2) > 360
continue;
end
p3=S(m(1)+m(2)+1:16);
if sum(B(1,p3))>65 | sum(B(1,p3))<46 | sum(B(2,p3))>65 | sum(B(2,p3))<40
continue;
end
p3(m(3)+1)=17;
[cost3, distan3] = vrp_cost(p3);
if distan3/30+5*m(3) > 360
continue;
end
cost=cost1+cost2+cost3;distan=distan1+distan2+distan3;
if cost<cost0
cost10=cost1;cost20=cost2;cost30=cost3;
distan10=distan1;distan20=distan2;distan30=distan3;
cost0=cost;distan0=distan;
pp1=p1;pp2=p2;pp3=p3;
end
end
pp1
cost10
distan10
pp2
cost20
distan20
function [ sum_cost,sum_dist] = vrp_cost(s)
%计算因空车率而减少的收入
%sum_cost为总减少的收入，sum_dist为总路程
B = [10 15 6 9 13 6 11 4 13 17 11 2 11 21 13 14;
9 14 5 10 9 10 13 9 15 9 6 7 13 15 10 16];%问题1的16个支局收寄邮件数量
% 各个支局之间的最短距离
D=[ 0 31 27 38 51 58 71 67 57 47 52 48 21 41 52 61 27;
31 0 19 33 27 32 45 64 53 47 61 57 52 48 56 63 36;
27 19 0 14 27 34 47 49 39 29 42 38 38 29 38 44 17;
38 33 14 0 13 20 33 35 25 15 33 32 32 15 24 30 11;
51 27 27 13 0 9 21 37 26 26 43 45 45 28 29 38 24;
58 32 34 20 9 0 13 32 32 35 47 52 52 35 33 42 31;
71 45 47 33 21 13 0 19 30 39 50 65 65 48 44 40 44;
67 64 49 35 37 32 19 0 11 20 31 54 61 34 25 21 40;
57 53 39 25 26 32 30 11 0 10 20 43 51 24 14 13 30;
47 47 29 15 26 35 39 20 10 0 18 36 41 14 9 18 20;
52 61 42 33 43 47 50 31 20 18 0 23 46 25 14 23 25;
48 57 38 32 45 52 65 54 43 36 23 0 27 22 33 42 21;
21 52 38 32 45 52 65 61 51 41 46 27 0 39 48 57 21;
41 48 29 15 28 35 48 34 24 14 25 22 39 0 11 20 18;
52 56 38 24 29 33 44 25 14 9 14 33 48 11 0 9 27;
61 63 44 30 38 42 40 21 13 18 23 42 57 20 9 0 36;
27 36 17 11 24 31 44 40 30 20 25 21 21 18 27 36 0];
end
C=65;
sum_dist=D(17,s(1));
for i=1:n-1
carry=carry-B(1,s(i))+B(2,s(i));
if carry>C
carry=C;
end
EmpRate=1-carry/C;
sum_dist=sum_dist+D(s(i),s(i+1));
end
clc,clear
%TSP算法求最短路径
A = xlsread(‘shuju1.xls‘,‘A2:C341‘);
distance=zeros(79);
for i=1:340
for j=1:79
for k=1:79
if A(i,1)==j & A(i,2)==k
distance(j,k)=A(i,3);
end
end
end
end
distance=distance+distance‘;
distance(find(distance==0))=inf;
for i=1:79
for j=1:79
if i==j
distance(i,j)=0;
end
end
end
%以下是市局到县局经过的点
x1=[62 8 9 10 74 14 15 11 16];%注：由于X1特殊，此圈中暂时不包含X1点，求出距离后再把X1添上
x2=[79 66 63 18 75 17 27 64 65 67];
x3=[79 28 29 30 31 32 33 68 69 70 76 34 35];
x4=[79 72 73 41 42 43 36 71 77];
x5=[79 61 60 59 53 78 52 58];
%以下是县局内部经过的支局点
xx1=[1:8 12 13 74];
xx1_1=[1 12 13];
xx1_2=[2:7];
xx2=[19:26 75];
xx4=[37:40 77];
xx5=[44:51 54:57 78];
xx5_1=[54:57 78];
xx5_2=[44:51 78];
x=xx1_2;
L=size(x,2);
for i=1:L
for j=1:L
a(i,j) = distance(x(i),x(j));
end
end
版本：2014a
完整代码或代写加1564658423
【路径规划】邮政运输网络中的邮路规划和邮车调度【含Matlab源码 648期】
标签：load enter 版本 不包含 rate 包含 clear oss 分析
原文地址：https://www.cnblogs.com/homeofmatlab/p/14925049.html
