MODEL: ! The Vehicle Routing Problem (VRP); !************************************; ! WARNING: Runtimes for this model ; ! increase dramatically as the number;
! of supplier increase. Formulations ;
! with more than a dozen suppliers ; !
WILL NOT SOLVE in a reasonable ; ! amount of time! ; ! ************************************;
SETS:
! Q(I) is the load capacity of vehicle I,车辆i的承装量，
U(I) is the volume of delivery from supplier I，供应商i的提货量;
Supplier/1..18/: Q, U;
! DIST(I,J) is the distance from supplier I to supplier J
X(I,J) is 0-1 variable: It is 1 if some vehicle travels from supplier I to J, 0 if none;
CXC( supplier, supplier): DIST, X;
ENDSETS DATA:
! supplier 1 represent the common depo;
Q = 0 38.55 2.59 44.35 4.87 1.40 91.15 4.72 0.82 159.73 28.70 83.52 13.10 19.44 64.15 79.87 154.70
! distance from supplier I to supplier J is same from supplier J to supplier I distance from supplier I to the depot is 0,
since the vehicle has to return to the depot;
DIST = ! To supplier; ! a A B C D E F G H I J K L M N O P Q From;
03880038700443002580051100387008000683004460067500698006220026500353001880019000!a;
388000787008000035300281007130041900525007320049100552002260058100131004250039300!A;
387007870002600480008440011003400097100470097200102001050041800695003890038400!B;
443008000026000533008700052003680010020036001000001050009340044200734004120041500!C;
258003530048000533000432004730019800590004840060700666004890043100279001150012000!D;
511002810084400870004320008260052100180008640022700217001910051800182005220052500!E;
870071300110052004730082600032100953006500963001001008940041000678003800038200!F;
380004190034000368001980052100321000662003730069200700006060028200382001350013300!G;
6830052500971001002005900018000953006620001007001000067002600063700363006580066200!H;
446007320047003600484008640065003730010070001014001046009270046200714003960039300!J;
67500491009720010000060700227009630069200100001014000145003030057300389006920069600!K;
69800552001020010500066600217001001007000067001046001450002770068400379006980070200!L;
622002260010500934004890019100894006060026000927003030027700065500220005810058500!M;
265005810041800442004310051800410002820063700462005730068400655000517003970039400!N;
353001310069500734002790018200678003820036300714003890037900220005170003650037300!O;
18800425003890041200115005220038000135006580039600692006980058100397003650001100!P;
19000393003840041500120005250038200133006620039300696007020058500394003730011000!Q;!
VCAP is the capacity of a vehicle ;
VCAP = 8;
ENDDATA
! Minimize total travel distance;
MIN = @SUM( CXC: DIST * X);
! For each supplier, except depot....;
@FOR( supplier( K)| K #GT# 1:
! a vehicle does not traval inside itself,...;
X( K, K) = 0;
! a vehicle must enter it,... ;
@SUM( supplier( 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(supplier( 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( supplier( 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); );
! of supplier increase. Formulations ;
! with more than a dozen suppliers ; !
WILL NOT SOLVE in a reasonable ; ! amount of time! ; ! ************************************;
SETS:
! Q(I) is the load capacity of vehicle I,车辆i的承装量，
U(I) is the volume of delivery from supplier I，供应商i的提货量;
Supplier/1..18/: Q, U;
! DIST(I,J) is the distance from supplier I to supplier J
X(I,J) is 0-1 variable: It is 1 if some vehicle travels from supplier I to J, 0 if none;
CXC( supplier, supplier): DIST, X;
ENDSETS DATA:
! supplier 1 represent the common depo;
Q = 0 38.55 2.59 44.35 4.87 1.40 91.15 4.72 0.82 159.73 28.70 83.52 13.10 19.44 64.15 79.87 154.70
! distance from supplier I to supplier J is same from supplier J to supplier I distance from supplier I to the depot is 0,
since the vehicle has to return to the depot;
DIST = ! To supplier; ! a A B C D E F G H I J K L M N O P Q From;
03880038700443002580051100387008000683004460067500698006220026500353001880019000!a;
388000787008000035300281007130041900525007320049100552002260058100131004250039300!A;
387007870002600480008440011003400097100470097200102001050041800695003890038400!B;
443008000026000533008700052003680010020036001000001050009340044200734004120041500!C;
258003530048000533000432004730019800590004840060700666004890043100279001150012000!D;
511002810084400870004320008260052100180008640022700217001910051800182005220052500!E;
870071300110052004730082600032100953006500963001001008940041000678003800038200!F;
380004190034000368001980052100321000662003730069200700006060028200382001350013300!G;
6830052500971001002005900018000953006620001007001000067002600063700363006580066200!H;
446007320047003600484008640065003730010070001014001046009270046200714003960039300!J;
67500491009720010000060700227009630069200100001014000145003030057300389006920069600!K;
69800552001020010500066600217001001007000067001046001450002770068400379006980070200!L;
622002260010500934004890019100894006060026000927003030027700065500220005810058500!M;
265005810041800442004310051800410002820063700462005730068400655000517003970039400!N;
353001310069500734002790018200678003820036300714003890037900220005170003650037300!O;
18800425003890041200115005220038000135006580039600692006980058100397003650001100!P;
19000393003840041500120005250038200133006620039300696007020058500394003730011000!Q;!
VCAP is the capacity of a vehicle ;
VCAP = 8;
ENDDATA
! Minimize total travel distance;
MIN = @SUM( CXC: DIST * X);
! For each supplier, except depot....;
@FOR( supplier( K)| K #GT# 1:
! a vehicle does not traval inside itself,...;
X( K, K) = 0;
! a vehicle must enter it,... ;
@SUM( supplier( 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(supplier( 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( supplier( 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); );
