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标题: vrp lingo 源代码 [打印本页]

作者: liyunan220 时间: 2012-12-13 16:48
标题: vrp lingo 源代码
MODEL:

! The Vehicle Routing Problem (VRP);

!************************************;
! WARNING: Runtimes for this model ;
! increase dramatically as the number;
! of cities increase. Formulations ;
! with more than a dozen cities    ;
! WILL NOT SOLVE in a reasonable    ;
! amount of time!                   ;
!************************************;

SETS:
  ! Q(I) is the amount required at city I,城市i的需求量，
U(I) is the accumulated delivers at city I ，城市i的累积交付量;
CITY/1..8/: Q, U;

  ! 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
travels from city I to J, 0 if none;
CXC( CITY, CITY): DIST, X;
ENDSETS

DATA:
  ! city 1 represent the common depo;
Q  =  0 6 3 7 7 18 4 5;

  ! 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;

DIST =  ! To City;
  ! 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;
   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

  ! Minimize total travel distance;
MIN = @SUM( CXC: DIST * X);

  ! For each city, except depot....;
@FOR( CITY( K)| K #GT# 1:

  ! a vehicle does not travel inside itself,...;
   X( K, K) = 0;

  ! 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;

  ! 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;

  ! U( K) is at least amount needed at K but can't
exceed capacity;
   @BND( Q( K), U( K), VCAP);

  ! 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))
      * X( K, I);
   );

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

  ! If K is not 1st stop...;
   U( K)>= Q( K)+ @SUM( CITY( I)|
   I #GT# 1: Q( I) * X( I, K));
);

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

  ! 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;
END

作者: 519574602 时间: 2014-1-4 09:12
学习学习啦

作者: 秋の名山で戦 时间: 2014-1-24 22:22
很好的帖子谢谢楼主分享

作者: 弹你脑瓜崩 时间: 2019-7-4 16:17
谢谢楼主分享，很好的帖子。

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