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:
  e8 v9 |) g2 O' U
! The Vehicle Routing Problem (VRP);

!************************************;
: ~$ a, C- u; F2 F+ Z, ?! WARNING: Runtimes for this model   ;
+ Z/ g5 i. Y2 L# ?0 a3 v) q! increase dramatically as the number;
. M& j6 h$ N7 M1 W5 k! of cities increase. Formulations   ;
7 a0 j# f5 p1 k3 C7 h! with more than a dozen cities      ;
4 `2 P$ w3 F  B0 S! B0 z1 S! WILL NOT SOLVE in a reasonable     ;
1 V" H& P. _, `! amount of time!                    ;
!************************************;
* f- E' s& p+ v2 c% {% f) J& j
' Y( C3 ?' Y9 s  g SETS:
; t( Z6 N+ Z- `5 ?  ! Q(I) is the amount required at city I,
9 B7 g8 ^( ^& r3 T/ T    U(I) is the accumulated delivers at city I ;
& ]6 ], Q& g% S5 O) B, \! g/ {   CITY/1..8/: Q, U;
+ q3 \6 v+ X( v( M
4 r5 F& F- k9 ~! {7 m" Y  ! DIST(I,J) is the distance from city I to city J
    X(I,J) is 0-1 variable: It is 1 if some vehicle
* \0 y% s6 Z6 y/ e( k! L. x    travels from city I to J, 0 if none;
   CXC( CITY, CITY): DIST, X;
! p% ?. K, [: z7 o5 k6 j( l" F. p ENDSETS
# H$ x2 d: M: W  ^, H* G6 {
/ s0 N) _/ y7 p- g0 J# R% M" O DATA:
! X, e8 `2 u+ u; I$ }2 W  ! city 1 represent the common depo;
0 R; ]' z/ K) m% k' Q; c   Q  =  0    6    3    7    7   18    4    5;

! q6 P2 i9 J! o  ! distance from city I to city J is same from city
+ @# |" H$ h. g& R4 O: |9 t9 s% q    J to city I distance from city I to the depot is
! d/ z8 |. m1 d+ N    0, since the vehicle has to return to the depot;
% y0 j$ k! h5 F
   DIST =  ! To City;
  ! Chi  Den Frsn Hous   KC   LA Oakl Anah   From;
0 o9 Z9 B6 E: s2 d) M9 |/ p      0  996 2162 1067  499 2054 2134 2050!Chicago;
, h* A, y" g2 F; ~: N3 ]      0    0 1167 1019  596 1059 1227 1055!Denver;
      0 1167    0 1747 1723  214  168  250!Fresno;
      0 1019 1747    0  710 1538 1904 1528!Houston;
7 _9 E8 z9 {( D  c) E* z+ Q      0  596 1723  710    0 1589 1827 1579!K. City;
      0 1059  214 1538 1589    0  371   36!L. A.;
      0 1227  168 1904 1827  371    0  407!Oakland;
" n: y8 _1 s: m      0 1055  250 1528 1579   36  407    0;!Anaheim;

  ! VCAP is the capacity of a vehicle ;
" {2 Q$ y2 z$ C   VCAP = 18;
$ J( m- Y5 Q% v2 t3 z- T ENDDATA
( M8 P# T+ a5 R, L
3 v# q7 j5 F* m  ! Minimize total travel distance;
7 i! C$ f, D% P/ S! Z   MIN = @SUM( CXC: DIST * X);
* q) i& d% C; H3 x/ _# }# H
$ I4 S. P- v. P  Y# y  ! For each city, except depot....;
' k# N- q3 m2 x* Y* O6 N   @FOR( CITY( K)| K #GT# 1:
3 v7 N4 p. n) ]$ u9 P, h6 W1 ^4 m
  ! a vehicle does not travel inside itself,...;
     X( K, K) = 0;

  ! a vehicle must enter it,... ;
$ u4 B, K0 B( q% W4 x     @SUM( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR#
. `* y% T+ d$ s5 p6 p9 ?9 w      Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1;

5 M3 v5 W0 H/ s2 Y- L  ! a vehicle must leave it after service ;
7 Z" g* v8 L/ ~1 b7 I     @SUM( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR#
, h" d2 W  Q0 W2 Y1 |: i      Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1;
) O+ |7 k. ?6 T% m, g
5 Y* u) j" K' k% g9 e/ S- t  ! U( K) is at least amount needed at K but can't
0 _. o1 {$ C/ Y1 `( i    exceed capacity;
4 r3 E: N4 |5 [) H2 }     @BND( Q( K), U( K), VCAP);

+ H( ]2 F( I2 w' l! u4 G( 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 *
       ( X( K, I) + X( I, K)) - ( Q( K) + Q( I))
$ h3 b  \1 G4 K0 w        * X( K, I);
8 Z; d4 |4 m; i. A+ U# R6 G     );

  ! If K is 1st stop, then U( K) = Q( K);
     U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K);

5 R0 u# o' |4 ~5 U8 q7 j* x( Y  ! If K is not 1st stop...;
/ E' v& L0 K: z9 U& W" Q     U( K)>= Q( K)+ @SUM( CITY( I)|
: t; D, z1 V, [! F/ C      I #GT# 1: Q( I) * X( I, K));
   );

- P: M* _; Y; |. r: C  ! Make the X's binary;
& r0 y$ ~' j- J$ H   @FOR( CXC: @BIN( X));
6 P5 ]* v8 X/ S" W2 Z9 R! N% r
4 d# a% ~+ s% y; z, `  ! Minimum no. vehicles required, fractional
* ?6 S. E1 m4 |$ g$ F    and rounded;
   VEHCLF = @SUM( CITY( I)| I #GT# 1: Q( I))/ VCAP;
1 r- n% d' g9 |   VEHCLR = VEHCLF + 1.999 -
    @WRAP( VEHCLF - .001, 1);

  ! Must send enough vehicles out of depot;
, Z& U( ?' W6 ]5 V: u( K7 E   @SUM( CITY( J)| J #GT# 1: X( 1, J)) >= VEHCLR;
. }" |$ S4 v% X* M END
8 r  Y: t7 L- v3 p, z 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