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:

* X# O6 y& z$ l) `% j8 t! The Vehicle Routing Problem (VRP);

!************************************;
* b0 w5 G& d) T- G, k! WARNING: Runtimes for this model   ;
! increase dramatically as the number;
+ A: j" _3 N' S! of cities increase. Formulations   ;
! with more than a dozen cities      ;
% X9 E5 G5 I" g% `- u4 s! WILL NOT SOLVE in a reasonable     ;
8 O6 N: Q, w3 k! amount of time!                    ;
& i$ K7 G  g* u+ x) \6 s: z6 E* F: v!************************************;

SETS:
  ! Q(I) is the amount required at city I,
* w$ v) P+ B/ o  u/ S+ z' @- {/ {0 I2 p    U(I) is the accumulated delivers at city I ;
   CITY/1..8/: Q, U;
$ t) C: A0 n" D5 A: D' M
- d4 \2 w0 c6 W4 i8 ~  ! DIST(I,J) is the distance from city I to city J
' A; \8 f' ?# z    X(I,J) is 0-1 variable: It is 1 if some vehicle
    travels from city I to J, 0 if none;
   CXC( CITY, CITY): DIST, X;
' ]& j0 t2 g% B& w( A" {1 Q ENDSETS
" k( Y% d( w- [0 K
DATA:
4 w* r1 ?3 w. Y" b0 @# B% Y. D! y4 Z/ k  ! city 1 represent the common depo;
   Q  =  0    6    3    7    7   18    4    5;

3 ]; v( o( I/ f1 s5 H- g/ o  ! distance from city I to city J is same from city
) T% Q% H6 V: ]1 W    J to city I distance from city I to the depot is
    0, since the vehicle has to return to the depot;
/ s) U; g  N* q2 r
* q& \/ J) W. ~! V! I; O. g9 T   DIST =  ! To City;
8 ~* ^" E! d5 x* I/ a1 s; V3 V, w1 X  ! Chi  Den Frsn Hous   KC   LA Oakl Anah   From;
      0  996 2162 1067  499 2054 2134 2050!Chicago;
      0    0 1167 1019  596 1059 1227 1055!Denver;
% ?: S4 c2 t$ u4 A  R  m      0 1167    0 1747 1723  214  168  250!Fresno;
$ t/ j: |8 @9 U  I1 M7 }$ Q- C      0 1019 1747    0  710 1538 1904 1528!Houston;
9 L1 r/ A0 a% ~. S2 Y( m& x5 X( u      0  596 1723  710    0 1589 1827 1579!K. City;
      0 1059  214 1538 1589    0  371   36!L. A.;
+ W; X) z6 o+ u3 G      0 1227  168 1904 1827  371    0  407!Oakland;
4 ]9 t- L" v/ z9 |7 m3 g$ C      0 1055  250 1528 1579   36  407    0;!Anaheim;

  ! VCAP is the capacity of a vehicle ;
6 a, |! E5 \  `   VCAP = 18;
ENDDATA

8 t4 L/ _; X" I; E  ! Minimize total travel distance;
9 [  _* X+ m! Z* L3 @   MIN = @SUM( CXC: DIST * X);
/ o# B$ M" c: [" v0 p) S
  ! For each city, except depot....;
0 Y7 M" P4 p' n& S+ f% m0 ^7 U   @FOR( CITY( K)| K #GT# 1:
2 \) t% ^- ~1 `/ u, B; a4 \
8 Y! T. `3 r0 D; }# `5 `6 E9 g' f  ! a vehicle does not travel inside itself,...;
  v7 @6 X9 E3 i  P7 P0 Z7 s     X( K, K) = 0;

' Y$ T# g9 y. t: z  k% U  ! a vehicle must enter it,... ;
     @SUM( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR#
5 a$ r0 M# d- L) @      Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1;
- a( {& o" e1 _' E$ ]7 R/ T1 T% ^4 q
  ! a vehicle must leave it after service ;
8 m9 q$ h& O0 F, Q5 D" b! e     @SUM( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR#
      Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1;
1 c; G5 q, P$ r$ o  q; Z7 M7 I1 E
' `' ^* x2 J$ o$ H' w( f$ [  ! U( K) is at least amount needed at K but can't
    exceed capacity;
     @BND( Q( K), U( K), VCAP);
" |, d7 |! t# X/ Y
, ~$ m) ?$ X: B/ b  ! If K follows I, then can bound U( K) - U( I);
     @FOR( CITY( I)| I #NE# K #AND# I #NE# 1:
% [6 _# [) e3 y2 M; ]4 L8 y      U( K) >= U( I) + Q( K) - VCAP + VCAP *
* N* o9 {) M7 Z! z       ( X( K, I) + X( I, K)) - ( Q( K) + Q( I))
$ I# \% ~" m6 R( l+ S        * X( K, I);
     );

  ! If K is 1st stop, then U( K) = Q( K);
     U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K);
# M7 P7 I. L1 N7 j+ p
  ! If K is not 1st stop...;
1 D! l7 Q/ F8 Q# M. U2 M, W# ~6 u     U( K)>= Q( K)+ @SUM( CITY( I)|
      I #GT# 1: Q( I) * X( I, K));
. N( t5 `* P; H2 l   );

  ! Make the X's binary;
   @FOR( CXC: @BIN( X));

  ! Minimum no. vehicles required, fractional
' ]; W1 m7 n6 r% z    and rounded;
+ \, c1 u+ E) v" Z# J   VEHCLF = @SUM( CITY( I)| I #GT# 1: Q( I))/ VCAP;
: E2 ?% K9 y' T! H3 h% j/ J' u   VEHCLR = VEHCLF + 1.999 -
' h  G' D' ]+ G4 g+ Y% n    @WRAP( VEHCLF - .001, 1);

  ! Must send enough vehicles out of depot;
   @SUM( CITY( J)| J #GT# 1: X( 1, J)) >= VEHCLR;
END
# r5 Z' f  \  T, t6 a; J+ t: \" f, _ 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