vrp lingo 源代码

liyunan220

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

519574602

学习学习啦

秋の名山で戦

很好的帖子  谢谢楼主分享

秋の名山で戦

很好的帖子  谢谢楼主分享

弹你脑瓜崩

谢谢楼主分享，很好的帖子。