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:

7 a& Z/ J" c% I" ?7 q- ]7 w! The Vehicle Routing Problem (VRP);

3 n. s* O1 J! c1 t' l8 f! x5 p!************************************;
! WARNING: Runtimes for this model   ;
. p4 H8 D8 v: z$ A9 e! increase dramatically as the number;
! of cities increase. Formulations   ;
! with more than a dozen cities      ;
! WILL NOT SOLVE in a reasonable     ;
! amount of time!                    ;
!************************************;
8 E- j# b3 B2 o; P/ m( E6 ^  s
/ ~( F9 G' @5 O2 q. k0 E7 c SETS:
$ ]+ j& \  }9 I  w# O2 U# b  ! Q(I) is the amount required at city I,
    U(I) is the accumulated delivers at city I ;
$ Y/ @3 e/ N. Y+ T( D" v% L   CITY/1..8/: Q, U;

: d- G$ B8 c4 A4 g+ h$ u/ s) S4 @  ! DIST(I,J) is the distance from city I to city J
, _5 }& Y6 @% e' }5 M    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;
ENDSETS
  v( w0 y  g/ E  e$ w5 [, ]
3 j! s5 a* m$ U0 k8 h  m0 K DATA:
  ! city 1 represent the common depo;
1 Z! p8 g7 v# M6 I" u. C   Q  =  0    6    3    7    7   18    4    5;

  ! distance from city I to city J is same from city
! t. U9 E! D9 W4 C    J to city I distance from city I to the depot is
    0, since the vehicle has to return to the depot;

   DIST =  ! To City;
  ! Chi  Den Frsn Hous   KC   LA Oakl Anah   From;
& E3 q7 ?6 s4 s, r      0  996 2162 1067  499 2054 2134 2050!Chicago;
      0    0 1167 1019  596 1059 1227 1055!Denver;
3 Y  @( r! T/ q+ H: W6 s+ W      0 1167    0 1747 1723  214  168  250!Fresno;
& g" I' r1 }( n# a, W# o1 j      0 1019 1747    0  710 1538 1904 1528!Houston;
      0  596 1723  710    0 1589 1827 1579!K. City;
      0 1059  214 1538 1589    0  371   36!L. A.;
8 ~9 s. Q: w) D  P      0 1227  168 1904 1827  371    0  407!Oakland;
9 ?! h* D2 S0 D      0 1055  250 1528 1579   36  407    0;!Anaheim;

  ! VCAP is the capacity of a vehicle ;
   VCAP = 18;
ENDDATA
0 r7 r3 n: N8 p3 d+ t: U( i$ n4 {
  ! Minimize total travel distance;
   MIN = @SUM( CXC: DIST * X);
2 ~$ |+ ^, s1 Q5 _8 t
( D4 `5 u+ B, g/ Q  ! For each city, except depot....;
   @FOR( CITY( K)| K #GT# 1:

  ! a vehicle does not travel inside itself,...;
     X( K, K) = 0;
9 A& Y+ {2 p2 E# j! z; V1 _
  ! 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;

: i& ^8 E9 @0 r1 _  ! a vehicle must leave it after service ;
     @SUM( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR#
      Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1;

2 M, }# C; h% S, B3 U+ S. A  ! U( K) is at least amount needed at K but can't
    exceed capacity;
     @BND( Q( K), U( K), VCAP);

- t: }, u" N" Z1 m  ! If K follows I, then can bound U( K) - U( I);
5 ~1 K+ W( D" |6 [     @FOR( CITY( I)| I #NE# K #AND# I #NE# 1:
3 }6 y7 }3 y8 Z) F7 T      U( K) >= U( I) + Q( K) - VCAP + VCAP *
       ( X( K, I) + X( I, K)) - ( Q( K) + Q( I))
        * X( K, I);
* v! `0 H0 ~* E& b: b( \( Z* {2 p: h     );

  ! If K is 1st stop, then U( K) = Q( K);
3 \& L  p; z: t! J+ @+ X( k7 w     U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K);

  ! If K is not 1st stop...;
     U( K)>= Q( K)+ @SUM( CITY( I)|
6 T$ S9 y# I+ S7 n7 i8 H      I #GT# 1: Q( I) * X( I, K));
4 ]9 ]9 a& s- I9 b   );

  ! Make the X's binary;
   @FOR( CXC: @BIN( X));
2 n) L: y* i, q7 U& D
5 Z' R8 R' D, I/ @6 U! Z2 j  ! Minimum no. vehicles required, fractional
0 ~1 x6 N1 g" Z- v* e    and rounded;
: B  _! s2 B7 Y: T( Z$ a   VEHCLF = @SUM( CITY( I)| I #GT# 1: Q( I))/ VCAP;
  q( I. P/ u- C8 H/ p4 o* M, v2 }# h   VEHCLR = VEHCLF + 1.999 -
    @WRAP( VEHCLF - .001, 1);
& C4 O+ O3 b% r
  ! Must send enough vehicles out of depot;
* m0 O, T$ ~! M! e& g# l% P   @SUM( CITY( J)| J #GT# 1: X( 1, J)) >= VEHCLR;
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