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:
& c- c, Q7 \) f$ T) u" }7 C4 M
! The Vehicle Routing Problem (VRP);
6 q+ k+ M" N# j! f9 F
!************************************;
! WARNING: Runtimes for this model   ;
! increase dramatically as the number;
, g0 x  f) J0 h. S. v! of cities increase. Formulations   ;
! with more than a dozen cities      ;
: }/ I. j& Q; y8 ?6 b! WILL NOT SOLVE in a reasonable     ;
/ d# s7 u% ~1 Q% A) h0 k! amount of time!                    ;
- K% e3 h8 P- z1 Y% v!************************************;
* L' D, y, ?3 v5 E9 p3 F- f$ l6 D
+ e: K  m9 S+ p( b& F" ^- y SETS:
  ! Q(I) is the amount required at city I,
    U(I) is the accumulated delivers at city I ;
  l4 |7 w" c- _, Z% c; |   CITY/1..8/: Q, U;

  ! DIST(I,J) is the distance from city I to city J
4 s2 t% R5 S' {    X(I,J) is 0-1 variable: It is 1 if some vehicle
6 f, q0 a( D3 {5 Z7 u& R    travels from city I to J, 0 if none;
* Q' T  r4 A9 V) V; q   CXC( CITY, CITY): DIST, X;
* W& g8 @% n" S3 A9 V9 y1 |' S ENDSETS
! I: p4 t6 z$ ?' B2 r
DATA:
+ M  U( v6 n& P  ! city 1 represent the common depo;
   Q  =  0    6    3    7    7   18    4    5;

# i2 U6 }8 V0 x/ I, g# |7 U9 m2 G  ! distance from city I to city J is same from city
    J to city I distance from city I to the depot is
    0, since the vehicle has to return to the depot;

* m1 v1 [( ^/ z6 Q2 ?7 |   DIST =  ! To City;
' i- K. ~/ }& |; ~4 R  ! 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;
      0 1167    0 1747 1723  214  168  250!Fresno;
& F8 _3 k. X' T& @5 K4 Q      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.;
2 ?* a6 c, ]2 {, ^      0 1227  168 1904 1827  371    0  407!Oakland;
      0 1055  250 1528 1579   36  407    0;!Anaheim;

+ ]2 C  u$ p8 N( h  ! VCAP is the capacity of a vehicle ;
   VCAP = 18;
ENDDATA

8 c8 J" V2 t1 G0 \) E% B5 g0 H& ?: p8 A) F  ! Minimize total travel distance;
   MIN = @SUM( CXC: DIST * X);
( w* u4 {2 q( ?  x
2 ]. g* U5 m7 @7 f5 j$ f  ! For each city, except depot....;
& h) @0 ^; \4 u/ S6 E   @FOR( CITY( K)| K #GT# 1:

! S- u6 e+ b3 _5 ~0 P  ! a vehicle does not travel inside itself,...;
3 v! O$ q1 l2 G  {  x: }     X( K, K) = 0;
+ u6 Z6 b4 y9 V; s( Q
2 D- b8 x8 ?7 Z8 X8 q& S  ! a vehicle must enter it,... ;
0 X7 a. T9 q! f; F0 s! |9 k- E     @SUM( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR#
% z$ Y6 c  Y& A      Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1;
& G' d' \1 i; v7 n( \, {2 f# r9 N
9 V$ p( s0 b; c4 E) v3 z  ! 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;

# Q  [- e; g* B/ N6 ?6 k  ! U( K) is at least amount needed at K but can't
( P0 J5 A8 b8 V2 f5 o& h; o2 S    exceed capacity;
     @BND( Q( K), U( K), VCAP);
4 G% M7 Y+ U. ^( |0 C+ p+ Y# ]+ b
/ ^* `6 a' x: J  C9 F3 L  ! If K follows I, then can bound U( K) - U( I);
     @FOR( CITY( I)| I #NE# K #AND# I #NE# 1:
      U( K) >= U( I) + Q( K) - VCAP + VCAP *
/ D/ [4 n9 m5 ^. l+ b! ~  i       ( X( K, I) + X( I, K)) - ( Q( K) + Q( I))
4 F" ]# t8 c7 X7 l% S  A; K$ ~        * X( K, I);
* R* @1 J5 Q' L% y. e     );

  ! If K is 1st stop, then U( K) = Q( K);
# H. O* m6 H) i& t8 D$ W6 D     U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K);
1 P# c: @% |- U: b0 c+ b! \
  ! If K is not 1st stop...;
     U( K)>= Q( K)+ @SUM( CITY( I)|
: y; }4 Z# I$ a& r& @3 b9 B      I #GT# 1: Q( I) * X( I, K));
   );
) O7 m3 a9 M/ S) I
  ! Make the X's binary;
& A# U8 G3 C, \* b   @FOR( CXC: @BIN( X));

  ! Minimum no. vehicles required, fractional
    and rounded;
3 ], i; u. e% O/ ]- {' J   VEHCLF = @SUM( CITY( I)| I #GT# 1: Q( I))/ VCAP;
$ t5 e8 i6 O1 Z6 [   VEHCLR = VEHCLF + 1.999 -
$ R3 M9 H! a' H' r- D    @WRAP( VEHCLF - .001, 1);
6 m0 G& {7 D2 e4 d$ W' k7 n
+ D( }4 V7 q) r/ u6 i7 e$ e  ! Must send enough vehicles out of depot;
   @SUM( CITY( J)| J #GT# 1: X( 1, J)) >= VEHCLR;
, A4 A) z+ b- ]* U7 G END
% B: ]" G; s) }3 O/ _' W 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