标题: VRP问题的lingo程序(多旅行商问题) [打印本页] 作者: taowenbao 时间: 2012-9-1 15:21
本主题需向作者支付 5 点体力 才能浏览
作者: 王冰清 时间: 2012-9-1 16:28 作者: taowenbao 时间: 2012-9-6 15:55
很好的喔~~~~~~~~~~作者: jiaqing 时间: 2012-10-27 20:18
付钱了。作者: 问安少年 时间: 2012-10-28 20:14
muqian...................作者: 一只想死的鱼 时间: 2012-10-29 08:52
这个模型不是很清晰吧,你应该把模型写出来,可供参考的。表面上看很普通的程序,而多旅行商和VRP有区别的,你这里的约束也不够。。。太少了作者: yinfeng0814 时间: 2012-12-6 18:33
看看楼主的程序,学习学习作者: yinfeng0814 时间: 2012-12-7 10:22
5个体力点对新人来说太贵了啊作者: liyunan220 时间: 2012-12-13 09:23
我下载了 去哪里找啊作者: liyunan220 时间: 2012-12-18 09:13
MODEL:" J0 _# T" y7 v; o
# c U+ H/ G- I! X$ i
! The Vehicle Routing Problem (VRP); : L3 u% R& l- k6 v8 @: G 5 q* M7 E+ z( c7 h7 r, Z!************************************; 3 m; D+ A( D. s! x5 m- f, r8 v/ A! WARNING: Runtimes for this model ; 0 J1 ?6 v7 x2 m. k) v! increase dramatically as the number;1 g+ i+ e7 L. Z# f L' e
! of cities increase. Formulations ;( O' [& U0 S6 J: t
! with more than a dozen cities ;# r# u$ ?; @! s2 w! {) ^* Y
! WILL NOT SOLVE in a reasonable ;$ ?! V* g7 z$ @1 L$ {4 D
! amount of time! ; 2 W( g9 h4 W! J6 a m!************************************;7 H( E+ s/ |+ b- i: d, b/ U
B, w) @9 G. F SETS: 9 c4 b9 [1 v0 C- C ! Q(I) is the amount required at city I, , O- `. R: w9 T0 A* o+ D U(I) is the accumulated delivers at city I ; 3 Z( d3 Z* @5 m5 o8 C7 y" C CITY/1..8/: Q, U; 9 _5 W0 r7 X+ M% E8 x) K/ W8 v! W( b1 r
! DIST(I,J) is the distance from city I to city J3 f1 C+ M3 `2 G) v
X(I,J) is 0-1 variable: It is 1 if some vehicle . ~6 W7 E9 a1 ^+ g travels from city I to J, 0 if none;( I) s- J9 n* @/ z5 M! x1 ^, G
CXC( CITY, CITY): DIST, X; ( ?& a* Y7 H1 j' g& z' t ENDSETS& x+ D0 ^1 n) X/ e- ]; H0 ~1 _
8 u8 a# V) o$ f9 R! L* X! x DATA: 0 X. _" W* n! D% ^0 ]9 S: P u* H ! city 1 represent the common depo; , y b; A7 U0 k# d& r% M Q = 0 6 3 7 7 18 4 5;; `7 W8 Q- \0 q0 B! G4 W
- T: @4 z q1 z2 b ! distance from city I to city J is same from city' _8 H6 J- t4 a& B! b. r/ d% u# y T3 {
J to city I distance from city I to the depot is/ k+ C* ~; Z( d% f
0, since the vehicle has to return to the depot;( `6 }0 C9 b! R8 G$ z
/ b0 b7 x# L3 A
DIST = ! To City; 2 q0 K% H# K" a3 O1 P- V3 |. Y ! Chi Den Frsn Hous KC LA Oakl Anah From;$ j3 G& C$ w+ i" M4 P6 U3 u4 `
0 996 2162 1067 499 2054 2134 2050!Chicago; 5 n _" ~: ^& K. m3 b8 B" ` 0 0 1167 1019 596 1059 1227 1055!Denver;; d# @- Z! E; z! H
0 1167 0 1747 1723 214 168 250!Fresno;- w; h8 h! }2 V, z5 L/ U/ F" R
0 1019 1747 0 710 1538 1904 1528!Houston; 5 e; Z1 k" {7 S/ A0 {$ u$ o. B 0 596 1723 710 0 1589 1827 1579!K. City; 3 j) g% S6 s* n& T9 Q. @ 0 1059 214 1538 1589 0 371 36!L. A.;% m9 n" r. a5 I& Q" ?0 F- |
0 1227 168 1904 1827 371 0 407!Oakland;$ x0 u1 I8 w; e, u+ k
0 1055 250 1528 1579 36 407 0;!Anaheim; $ W: y* o# D$ V% u) _+ U% x9 Z 0 B" @$ a8 }2 K- W% w5 J# s# } ! VCAP is the capacity of a vehicle ;; x3 A9 ]/ [3 C& ~6 I
VCAP = 18; : ]; s6 Q3 G' \5 {) _% O: P ENDDATA 3 U! h1 h4 q5 l& b) k; n/ n) u3 U! Z$ W' v
! Minimize total travel distance;! c# x3 I& X! p
MIN = @SUM( CXC: DIST * X); # g* }8 h4 u8 v* L; l % w, k" l: n' ?7 z* t; H ! For each city, except depot....;" f7 X: _& ]/ x0 x8 J3 K
@FOR( CITY( K)| K #GT# 1:: W$ P# p0 n% |7 F
) M8 q* `3 ^ r* L9 J ! a vehicle does not travel inside itself,...; 4 y6 J. w/ p$ x e- ~, k( e L; @ X( K, K) = 0; $ r) q4 V; L9 x8 ?8 r$ U( k3 c, c* C1 b7 c. w* x. S
! a vehicle must enter it,... ;/ W+ z; U+ V: T" ]: A& @ `. v' i
@SUM( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR#1 g9 a; J) E. g3 G
Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1;' q$ I2 p/ y4 h9 T, x1 e
- a0 m n& r5 L2 M
! a vehicle must leave it after service ; $ ^! _+ W/ Z Z) e( ? @SUM( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR#+ K; G4 r! g3 m& B# Y7 M
Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1;$ w5 C% R" g( z. F/ h) U
3 F" x# V6 U! u. n& V& Z ! U( K) is at least amount needed at K but can't 1 N$ u3 c' \- Q; ~. _/ J9 @ exceed capacity;- s! {! ?) B+ b0 R, a7 M {0 g
@BND( Q( K), U( K), VCAP);0 `9 f- y! h* A1 W
/ X/ u; i: P7 L& w, j7 W$ Y6 A1 o" ?2 N
! If K follows I, then can bound U( K) - U( I); + X( B# {3 ~( o- g& n, q% L' \1 L @FOR( CITY( I)| I #NE# K #AND# I #NE# 1: 1 V' Q' q S0 W# B U( K) >= U( I) + Q( K) - VCAP + VCAP * / y8 ?0 L) P; X1 v6 J6 v ( X( K, I) + X( I, K)) - ( Q( K) + Q( I)) \/ Z+ h/ x" m * X( K, I);, ?3 Z2 i9 }/ J3 @5 n
);$ B* u9 k2 `' C1 k* r5 U7 Q
3 H- i3 r7 o0 [ ! If K is 1st stop, then U( K) = Q( K); / I8 h2 G, W; n8 ` U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K);: ]: e* o; m% q+ [$ Q7 X
# K& r& w. y4 G7 n9 F. a4 t' \ ! If K is not 1st stop...;( \& Z+ q3 s% ^# ^
U( K)>= Q( K)+ @SUM( CITY( I)| / Z* f1 c3 O! ^0 F, {2 Q I #GT# 1: Q( I) * X( I, K)); 2 g9 H) \) [ F ); 2 f6 j, d3 g. ]/ o( x+ s8 M! x" B, U6 m+ w
! Make the X's binary; 9 ?' p6 l7 A3 R9 `* J$ | v @FOR( CXC: @BIN( X));8 C1 U+ P" [6 h" B/ L1 A3 h( e" L