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:
' G$ ~+ _) M; }
! The Vehicle Routing Problem (VRP);
) f0 Z9 i* Y+ Y' I) n& c
: a8 F$ C- o  i2 v6 b/ r% n!************************************;
3 R/ t2 p/ I! I0 u% s! WARNING: Runtimes for this model   ;
1 n* {6 p  k( K8 k# I* ]! increase dramatically as the number;
! of cities increase. Formulations   ;
" N0 ?& h0 B# s' B5 w% g' b* u( v! with more than a dozen cities      ;
! WILL NOT SOLVE in a reasonable     ;
! amount of time!                    ;
4 B# q' y1 r' }: X8 w9 s' `!************************************;
1 q% g! i9 r" L
& W' f! ^  N9 z+ T, N7 @: `" I4 J SETS:
  ! Q(I) is the amount required at city I,
1 W5 |1 h  n$ w& k$ [. l7 @& m    U(I) is the accumulated delivers at city I ;
   CITY/1..8/: Q, U;
* A, C1 d2 Y+ K2 o7 V4 v
0 X( p- C/ F3 F. ~2 \/ x2 V  ! DIST(I,J) is the distance from city I to city J
# V0 V" \- t8 q4 e" D0 ?* `' h- I, F    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
0 ?* I# i4 I6 g3 `. Z
7 b. |. k; D. \% e+ l DATA:
  ! city 1 represent the common depo;
   Q  =  0    6    3    7    7   18    4    5;
7 T  a7 B7 Z; Y: E) t
5 P4 Z2 k6 m. a2 _- B  ! distance from city I to city J is same from city
- j4 U1 m5 m4 q6 X( I    J to city I distance from city I to the depot is
    0, since the vehicle has to return to the depot;

: i( J/ I8 n! ?! y' e, o" k, \4 u   DIST =  ! To City;
( f; K1 N3 D/ }4 G! R. T  ! Chi  Den Frsn Hous   KC   LA Oakl Anah   From;
      0  996 2162 1067  499 2054 2134 2050!Chicago;
4 p  n# @7 T  I$ }      0    0 1167 1019  596 1059 1227 1055!Denver;
      0 1167    0 1747 1723  214  168  250!Fresno;
9 p! i+ c& R( U$ Z6 F( z      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.;
      0 1227  168 1904 1827  371    0  407!Oakland;
      0 1055  250 1528 1579   36  407    0;!Anaheim;

  ! VCAP is the capacity of a vehicle ;
   VCAP = 18;
ENDDATA
1 B9 o1 b' r/ t! S1 Y4 O
6 U4 X( X6 i+ d! P. [) m% |  ! Minimize total travel distance;
8 q* `6 I; h1 G3 C   MIN = @SUM( CXC: DIST * X);

: W" E4 Q) V- k7 x7 I6 X  ! For each city, except depot....;
   @FOR( CITY( K)| K #GT# 1:
2 m, q+ }1 N( V, d: K" g/ v
7 {/ h6 }6 U0 n. w7 q- l7 |  ! a vehicle does not travel inside itself,...;
! }7 D) ]  J5 E7 h     X( K, K) = 0;

  ! a vehicle must enter it,... ;
0 r, \) B% t9 u( o/ [1 D) l     @SUM( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR#
      Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1;

  ! a vehicle must leave it after service ;
3 ^3 b8 L3 ?+ f$ P/ z     @SUM( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR#
7 G1 R) N  ~  G6 w/ P- m* U. k3 R      Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1;
5 R+ Z" Q! ^% H6 @! C) s
  ! U( K) is at least amount needed at K but can't
    exceed capacity;
: C/ g. F$ C: o! f     @BND( Q( K), U( K), VCAP);
' n% d' K$ u; w  |7 U7 [% ?
/ {# N/ l5 o) ?/ @$ M  A  ! If K follows I, then can bound U( K) - U( I);
8 Q- T8 \! V) u  K4 \% T  Z9 `3 x     @FOR( CITY( I)| I #NE# K #AND# I #NE# 1:
      U( K) >= U( I) + Q( K) - VCAP + VCAP *
& }4 s; u: a* o: T) Q       ( X( K, I) + X( I, K)) - ( Q( K) + Q( I))
        * X( K, I);
0 I- P* l- M6 k3 p/ m9 G5 X     );
3 _5 U/ u. ?: Q- s2 }
/ G; n8 Z, }! l  k  ! If K is 1st stop, then U( K) = Q( K);
9 B4 S8 B; P3 {; N' B8 a     U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K);
" |; l" C1 b2 u
  ! If K is not 1st stop...;
     U( K)>= Q( K)+ @SUM( CITY( I)|
      I #GT# 1: Q( I) * X( I, K));
   );

4 i! O$ k8 @5 ~% r  ! Make the X's binary;
   @FOR( CXC: @BIN( X));

1 ?" C5 j  {/ W, _- C! u# u  O  ! Minimum no. vehicles required, fractional
    and rounded;
   VEHCLF = @SUM( CITY( I)| I #GT# 1: Q( I))/ VCAP;
   VEHCLR = VEHCLF + 1.999 -
    @WRAP( VEHCLF - .001, 1);

  ! Must send enough vehicles out of depot;
   @SUM( CITY( J)| J #GT# 1: X( 1, J)) >= VEHCLR;
2 ^1 p- S9 u; u0 j- C1 n& F 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