VRP问题的lingo程序（多旅行商问题）

王冰清

taowenbao

很好的喔~~~~~~~~~~

jiaqing

付钱了。

问安少年

muqian...................

一只想死的鱼

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

yinfeng0814

看看楼主的程序，学习学习

yinfeng0814

5个体力点对新人来说太贵了啊

liyunan220

我下载了 去哪里找啊

liyunan220

MODEL:

! The Vehicle Routing Problem (VRP);

!************************************;
! z2 a7 ]9 r$ y. k1 b+ M! WARNING: Runtimes for this model   ;
! increase dramatically as the number;
& M5 P5 C, U! _# q9 U" l! of cities increase. Formulations   ;
: x6 T+ y$ }* ^# u1 S$ P& [/ S8 `! with more than a dozen cities      ;
, D; r+ J  V# A) i8 ~4 e( V/ _! WILL NOT SOLVE in a reasonable     ;
! amount of time!                    ;
: b5 W2 t, |" G!************************************;

3 g- q! |9 o) I! p! s0 H SETS:
; F+ j2 U8 [% I1 Q# ~  ! Q(I) is the amount required at city I,
0 F3 r$ `' @. z    U(I) is the accumulated delivers at city I ;
   CITY/1..8/: Q, U;

  ! DIST(I,J) is the distance from city I to city J
1 n; S( N2 u! p  y# W    X(I,J) is 0-1 variable: It is 1 if some vehicle
    travels from city I to J, 0 if none;
$ @2 k0 m$ n. v. @1 V7 G: N   CXC( CITY, CITY): DIST, X;
+ p: {$ k) s' Y) `& K5 } ENDSETS

" U, Y! r7 _9 x% ?' z DATA:
9 f" Q, }1 _' S4 C% Z8 g; Y2 X  ! city 1 represent the common depo;
   Q  =  0    6    3    7    7   18    4    5;
+ y+ M& _6 D4 x2 k
  p" A$ o* ~& W) `" y1 V9 |4 n  ! distance from city I to city J is same from city
$ F) Q2 a- F8 c* F+ t/ O    J to city I distance from city I to the depot is
    0, since the vehicle has to return to the depot;
& k7 u' M5 D5 O1 L. w) X
   DIST =  ! To City;
  ! Chi  Den Frsn Hous   KC   LA Oakl Anah   From;
- A! c5 _6 k0 o& u      0  996 2162 1067  499 2054 2134 2050!Chicago;
      0    0 1167 1019  596 1059 1227 1055!Denver;
. X1 O$ g" k. n2 \. m" d      0 1167    0 1747 1723  214  168  250!Fresno;
      0 1019 1747    0  710 1538 1904 1528!Houston;
7 [" K5 I5 O' @4 ]3 Q' X' U      0  596 1723  710    0 1589 1827 1579!K. City;
0 h( v7 n* @; z( ]      0 1059  214 1538 1589    0  371   36!L. A.;
      0 1227  168 1904 1827  371    0  407!Oakland;
      0 1055  250 1528 1579   36  407    0;!Anaheim;
. k6 l7 v4 Z' D, R' c& P8 Y7 M
' j) ~& v# F) a; n' H  ! VCAP is the capacity of a vehicle ;
# v7 l3 ]* Q4 d   VCAP = 18;
ENDDATA

  ! Minimize total travel distance;
: }0 A; @6 Y4 g% ~8 g1 \: c   MIN = @SUM( CXC: DIST * X);

  ! For each city, except depot....;
   @FOR( CITY( K)| K #GT# 1:
: V3 N1 _1 A1 j' W
  ! a vehicle does not travel inside itself,...;
4 m5 J) k0 G! |" _     X( K, K) = 0;
$ {9 Y8 D' N* r1 d
$ o7 _& U" @& M) Z  ! a vehicle must enter it,... ;
     @SUM( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR#
      Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1;

7 |& {' F: F. P  ! a vehicle must leave it after service ;
     @SUM( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR#
! ^7 F0 ~7 |  o      Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1;

  ! U( K) is at least amount needed at K but can't
  |! i! @7 ?" I    exceed capacity;
     @BND( Q( K), U( K), VCAP);

1 P* X, \4 A/ }  \3 N  ! If K follows I, then can bound U( K) - U( I);
- b& t3 U4 a, T8 [     @FOR( CITY( I)| I #NE# K #AND# I #NE# 1:
      U( K) >= U( I) + Q( K) - VCAP + VCAP *
! |! K3 s4 r2 N' P) V/ ?1 ^0 J       ( X( K, I) + X( I, K)) - ( Q( K) + Q( I))
        * X( K, I);
% u1 b* M6 {2 Z9 y     );

  ! If K is 1st stop, then U( K) = Q( K);
  b8 n' g# Z4 ^. J6 k8 ]5 g6 S     U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K);
% M$ q  D" Z0 [9 t6 C3 Y
" H+ k. [' j9 R5 `5 i/ r  ! If K is not 1st stop...;
     U( K)>= Q( K)+ @SUM( CITY( I)|
. d2 Z* u2 l( s( m/ o/ b3 N      I #GT# 1: Q( I) * X( I, K));
, k3 f5 [+ q) s7 B   );
# p6 \" ~. W+ k7 b
1 o& J5 J4 P+ [* p( A* n6 ~$ \: [* T  ! Make the X's binary;
   @FOR( CXC: @BIN( X));

  ! Minimum no. vehicles required, fractional
" R1 E/ d  B0 f0 @. s: q    and rounded;
' i* L, |0 L' }- j   VEHCLF = @SUM( CITY( I)| I #GT# 1: Q( I))/ VCAP;
   VEHCLR = VEHCLF + 1.999 -
0 m$ z3 s% J4 ]' F    @WRAP( VEHCLF - .001, 1);
& V0 u, e) C. G6 ?8 @# H& S
. l: `& }+ j# V( J# P  ! Must send enough vehicles out of depot;
   @SUM( CITY( J)| J #GT# 1: X( 1, J)) >= VEHCLR;
- {3 b  _+ [" x2 c: E END
请问大家里面U(I)的公式如何理解啊 U(I)是城市I 的累积交付量么？谢谢