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: 9 t. H T) a, ]4 y3 E' G) h" l0 r% H* K* x# [; B
! The Vehicle Routing Problem (VRP); 4 j4 ~, I, M* H& Y1 H5 l 7 Y0 J" W, y b9 y!************************************; - @3 v3 I& A# W! WARNING: Runtimes for this model ; $ ?' I* y; i3 P+ N5 }$ r9 J! increase dramatically as the number;5 ]3 w. ]- _& I. e$ s* v& t1 A
! of cities increase. Formulations ; a/ Y' b3 ]0 l L
! with more than a dozen cities ;9 Y# g" V& c- T& N: u+ m
! WILL NOT SOLVE in a reasonable ; , e/ a7 V* r; h3 j& s" w! amount of time! ; , O( S4 `: k: Z7 t!************************************; " A8 t+ q: f" k, k& _5 \& H: }
SETS:& M) D* O% c, @, q7 c0 O2 p
! Q(I) is the amount required at city I, + O0 N$ B" V3 Y8 J; A* p( @! p U(I) is the accumulated delivers at city I ; 5 N; n8 b" K1 D% f CITY/1..8/: Q, U; ( g1 g, J) d: [& [- `6 v4 _/ W, u, T/ r' @8 e
! DIST(I,J) is the distance from city I to city J 2 {9 f8 h9 x$ i X(I,J) is 0-1 variable: It is 1 if some vehicle( P9 V1 R5 g4 s- h3 U" f& O
travels from city I to J, 0 if none; - R% K$ r& {( q CXC( CITY, CITY): DIST, X; 3 s9 A% l- X7 F9 L5 L ` ENDSETS4 ^6 }. D0 C. H Z
4 @$ c& s2 x( T Z DATA: $ T b- n8 P' m$ | ! city 1 represent the common depo; 4 A, v" @# F# i' E+ C. D0 k Q = 0 6 3 7 7 18 4 5; 5 c, A* q5 }: K2 T2 l5 m) t2 [ - t; X( A- K4 w `' Q4 X6 t/ @ ! distance from city I to city J is same from city: @( i, f0 o9 O1 L
J to city I distance from city I to the depot is! N% i n: O% s X; q' _/ s% T/ Q& [* o
0, since the vehicle has to return to the depot; 5 s, u; c* Z* H! m' i/ g( L 8 j/ Y j7 z/ c( x2 w! ]" ] DIST = ! To City;9 D+ _3 \( f, N5 H& c6 a
! Chi Den Frsn Hous KC LA Oakl Anah From; ( z: E7 |4 X! C; x/ p1 e+ t" g 0 996 2162 1067 499 2054 2134 2050!Chicago;1 z: [: L9 A o/ l
0 0 1167 1019 596 1059 1227 1055!Denver;. }; Y% m- z" B8 H p* X) g
0 1167 0 1747 1723 214 168 250!Fresno;+ P5 S4 p! \. `( T K# h, z* G
0 1019 1747 0 710 1538 1904 1528!Houston;3 y8 N2 m! Y# K) U
0 596 1723 710 0 1589 1827 1579!K. City; 0 }/ U: B' c" C 0 1059 214 1538 1589 0 371 36!L. A.;) c1 T0 @2 x; B4 P+ \
0 1227 168 1904 1827 371 0 407!Oakland;( O# c6 z! p* \% L
0 1055 250 1528 1579 36 407 0;!Anaheim;9 I1 D" p: T9 h- A4 y; D
1 P f% T4 b' N3 j9 f+ q" W ! VCAP is the capacity of a vehicle ;' D( ?: F3 v* q4 }! V
VCAP = 18;% t K* K$ ]! m
ENDDATA' I! \8 U+ ]4 ~4 M8 U1 ?3 x
' T9 g. W" A1 ^! E+ ]# |# b
! Minimize total travel distance;+ }- G! k5 k9 x" J, i; T G; U
MIN = @SUM( CXC: DIST * X); & Y; r/ i. a5 \. W4 [; H( P$ Z3 F
! For each city, except depot....; - B2 L. N: g& z' U @FOR( CITY( K)| K #GT# 1:2 G6 c* d: w; F- @1 V! ~. Y
$ }. e: z9 I- ]4 V1 I' F1 I
! a vehicle does not travel inside itself,...;0 F% C% t# c0 y9 X
X( K, K) = 0; ; J+ {* u# }9 t% _: ? P" k( Z& `1 m% j) [6 o
! a vehicle must enter it,... ; 1 M0 B9 r' |6 V @SUM( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR#, w3 A0 L8 m: l2 _
Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1; 5 w; X. Z0 k0 j# }3 E ~: }& N" h F x5 ?$ m
! a vehicle must leave it after service ;' t2 I- o, @+ i0 i' G
@SUM( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR#3 H9 w: h4 m8 a" }5 U) A! s
Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1; 6 H7 ]% c2 v9 W0 r+ G1 j6 U% F
! U( K) is at least amount needed at K but can't 1 k0 V7 m1 X- w# f. h2 y5 W" ~ F exceed capacity; 9 S, R( g8 L' Z7 @# f' y @BND( Q( K), U( K), VCAP);% q4 \5 H# i+ C( O
# D0 L1 S5 t7 b ! If K follows I, then can bound U( K) - U( I);, ~$ I! x t" g1 Z! y
@FOR( CITY( I)| I #NE# K #AND# I #NE# 1: & v4 _1 K+ K& |3 `$ k
U( K) >= U( I) + Q( K) - VCAP + VCAP * 1 L9 B- z- p: {5 v, J: m( v
( X( K, I) + X( I, K)) - ( Q( K) + Q( I))4 I6 {$ ^8 W# S4 s& _9 }3 A+ w
* X( K, I); / ]: N9 J8 f0 M3 k( Y. q0 G ); / v, [4 V1 B3 n f' u; u5 d8 j2 b, }( F0 o6 c: V1 u
! If K is 1st stop, then U( K) = Q( K); + J$ Y4 @* K: `& \" U2 L. y6 F U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K); v7 L# i. D/ ]8 U* D . T" c/ L9 F, d- L0 t& w2 t# W* m ! If K is not 1st stop...;1 N7 `) S x8 a( P* |4 b+ m3 x; |. P
U( K)>= Q( K)+ @SUM( CITY( I)| L1 M. S6 o! `& @4 i _# [# ^ I #GT# 1: Q( I) * X( I, K)); * b G' v, D1 B! u1 Z0 T1 I# P ); v2 f& M# ?: c. b/ | + ^$ b- y6 }# x ! Make the X's binary; 7 t f( `# p* A2 [# V$ z @FOR( CXC: @BIN( X));4 L1 N Q |+ p* O% m; h
