为什么我的解决方案对 sum(n in order) 的值与 sum(n in order:(n-1) in order) 的值相同
Why is does my solution give the same value for sum(n in order) as for sum(n in order:(n-1) in order)
我正在进行合并 model，您可以在其中查看哪些订单已合并，哪些不是基于其假设的数量。但是，我的代码不适用于所有订单组合...这是我的 .mod 文件：
//GENERAL PARAMETERS
int nr_trucks = ...;
//Sets
{int} origin_terminal = ...;
{int} gateway_terminal = ...;
{int} destination_terminal = ...;
{int} order = ...;
{int} cons_truck = ...;
int capacity_cons_truck[cons_truck] = ...;
//Ranges
range truck = 1..nr_trucks;
//ORDER PARAMETERS
int volume[order] = ...;
int non_cons_truck[order] = ...;
//NETWORK PARAMETERS
int fixed_cost_OTG[origin_terminal, gateway_terminal] = ...;
int fixed_cost_GTD[gateway_terminal, destination_terminal] = ...;
//DECISION VARIABLE
dvar int+ x_OTG[origin_terminal, gateway_terminal, truck, order] in 0..1;
dvar int+ x_GTD1[gateway_terminal, destination_terminal, truck, order] in 0..1;
dvar int+ x_GTD2[gateway_terminal, destination_terminal, cons_truck, order] in 0..1;
//OUTPUT VARIABLES
dvar int+ y[gateway_terminal, truck, cons_truck, order] in 0..1;
dvar float exp_fixed_cost_1[origin_terminal, gateway_terminal, truck, order];
dvar float exp_fixed_cost_2[gateway_terminal, destination_terminal, truck, order];
dvar float+ exp_fixed_cost_3[gateway_terminal, destination_terminal, cons_truck];
dvar int+ A[cons_truck];
dvar int+ B[cons_truck];
//PARAMETERS TO WRITE OUTPUT TO EXCEL
tuple tuple_x
{
int terminal;
int terminal2;
int truck;
int order;
int value;
};
tuple tuple_y
{
int terminal;
int truck;
int cons_truck;
int order;
int value;
};
tuple tuple_cost_1_2
{
int terminal;
int terminal2;
int truck;
int order;
float value;
};
tuple tuple_cost_3
{
int terminal;
int terminal2;
int truck;
float value;
};
//MODEL
minimize sum(n in order) exp_fixed_cost_1[1,1,non_cons_truck[n],n]
+ sum(n in order) exp_fixed_cost_2[1,1,non_cons_truck[n],n]
+ sum(h in cons_truck) exp_fixed_cost_3[1,1,h];
subject to
{
//FLOW CONSTRAINTS
forall(n in order) x_OTG[1,1,non_cons_truck[n],n] == 1; //(2)
forall(n in order) x_OTG[1,1,non_cons_truck[n],n] == x_GTD1[1,1,non_cons_truck[n],n] + sum(h in cons_truck) x_GTD2[1,1,h,n]; //(3)
//forall(n in order, h in cons_truck) x_OTG[1,1,non_cons_truck[n],n] + x_GTD2[1,1,h,n] <= 1 + y[1,non_cons_truck[n],h,n]; //(4a) minder strikt?
forall(h in cons_truck, n in order) y[1,non_cons_truck[n],h,n] == x_GTD2[1,1,h,n]; //(4a)
forall(h in cons_truck) sum(n in order) y[1,non_cons_truck[n],h,n]*volume[n] <= capacity_cons_truck[h]; //(4b)
forall(h in cons_truck) sum(n in order) y[1,non_cons_truck[n],h,n] != 1; //(4c) only more than 2 order can be put on a consolidated truck
forall(n in order) sum(h in cons_truck) y[1,non_cons_truck[n],h,n] <= 1; //(5a) order cannot be split over multiple trucks
forall(n in order) sum(h in cons_truck) x_GTD2[1,1,h,n] <= 1; //(5b)
forall(h in cons_truck) sum(n in order) y[1,non_cons_truck[n],h,n] == A[h]; //number of consolidated orders in truck h
forall(h in cons_truck) sum(n in order: (n-1) in order) y[1,non_cons_truck[n],h,n] == B[h]; //all but one consolidated orders in h
//EXPECTED FIXED COSTS
forall(n in order) exp_fixed_cost_1[1,1,non_cons_truck[n],n] >= fixed_cost_OTG[1,1]*x_OTG[1,1,non_cons_truck[n],n]; //(6a)
forall(n in order) exp_fixed_cost_2[1,1,non_cons_truck[n],n] >= fixed_cost_GTD[1,1]*x_GTD1[1,1,non_cons_truck[n],n]; //(6b)
forall(h in cons_truck) exp_fixed_cost_3[1,1,h] >= sum(n in order) x_GTD2[1,1,h,n]*fixed_cost_GTD[1,1] - sum(n in order: (n-1) in order) y[1,non_cons_truck[n],h,n]*fixed_cost_GTD[1,1] ; //(6c)
forall(n in order) exp_fixed_cost_1[1,1,non_cons_truck[n],n] >= 0;
forall(n in order) exp_fixed_cost_2[1,1,non_cons_truck[n],n] >= 0;
forall(n in order, h in cons_truck) exp_fixed_cost_3[1,1,h] >= 0;
}
//OUTPUT TO EXCEL
{tuple_x} output_x_OTG = {<i,j,g,n,x_OTG[i,j,g,n]> | i in origin_terminal, j in gateway_terminal, g in truck, n in order};
{tuple_x} output_xGTD1 = {<i,j,g,n,x_GTD1[i,j,g,n]> | i in gateway_terminal, j in destination_terminal, g in truck, n in order};
{tuple_x} output_xGTD2 = {<i,j,h,n,x_GTD2[i,j,h,n]> | i in gateway_terminal, j in destination_terminal, h in cons_truck, n in order};
{tuple_y} output_y = {<i,g,h,n,y[i,g,h,n]> | i in gateway_terminal, g in truck, h in cons_truck, n in order};
{tuple_cost_1_2} output_exp_fixed_cost_1 = {<i,j,g,n,exp_fixed_cost_1[i,j,g,n]> | i in origin_terminal, j in gateway_terminal, g in truck, n in order};
{tuple_cost_1_2} output_exp_fixed_cost_2 = {<i,j,g,n,exp_fixed_cost_2[i,j,g,n]> | i in gateway_terminal, j in destination_terminal, g in truck, n in order};
{tuple_cost_3} output_exp_fixed_cost_3 = {<i,j,h,exp_fixed_cost_3[i,j,h]> | i in gateway_terminal, j in destination_terminal, h in cons_truck};
这是我的 .dat 文件：
SheetConnection excel("One_to_one_no_leadtime.xlsx");
//GENERAL PARAMETERS
//nr_orders = 3;
nr_trucks = 3;
origin_terminal = {1};
gateway_terminal = {1};
destination_terminal = {1};
order = {1,2,3};
cons_truck = {1,2,3};
//ORDER PARAMETERS
volume = [40,70,20];
non_cons_truck = [1,2,3];
capacity_cons_truck = [100,100,100];
//NETWORK PARAMETERS
fixed_cost_OTG = [[75]];
fixed_cost_GTD = [[75]];
//OUTPUT WEGSCHRIJVEN
output_x_OTG to SheetWrite(excel,"'output'!A3:E1100");
output_xGTD1 to SheetWrite(excel,"'output'!G3:K1100");
output_xGTD2 to SheetWrite(excel,"'output'!M3:Q1200");
output_y to SheetWrite(excel,"'output'!S3:W1000");
output_exp_fixed_cost_1 to SheetWrite(excel,"'cost'!A3:E200");
output_exp_fixed_cost_2 to SheetWrite(excel,"'cost'!G3:K200");
output_exp_fixed_cost_3 to SheetWrite(excel,"'cost'!M3:P200");
如您所见，model 表示当 .dat 文件中的音量 = [40,70,20] 时，A[h] 和 B[h] 相同。为什么 sum(n in order) 等于 sum(n in order: (n-1) in order)？当 volume = [40,40,80] 时不会出现此问题，然后 model 完美运行。我真的不知道发生了什么......我知道这是一个精心设计的 model，但如果有人可以提供帮助，请告诉我！亲切的问候。
你的最优解300中，A和B相同，但公式不同
如果添加新约束
A[1]!=B[1];
然后你得到一个新的解决方案 objective 375
我正在进行合并 model，您可以在其中查看哪些订单已合并，哪些不是基于其假设的数量。但是，我的代码不适用于所有订单组合...这是我的 .mod 文件：
//GENERAL PARAMETERS
int nr_trucks = ...;
//Sets
{int} origin_terminal = ...;
{int} gateway_terminal = ...;
{int} destination_terminal = ...;
{int} order = ...;
{int} cons_truck = ...;
int capacity_cons_truck[cons_truck] = ...;
//Ranges
range truck = 1..nr_trucks;
//ORDER PARAMETERS
int volume[order] = ...;
int non_cons_truck[order] = ...;
//NETWORK PARAMETERS
int fixed_cost_OTG[origin_terminal, gateway_terminal] = ...;
int fixed_cost_GTD[gateway_terminal, destination_terminal] = ...;
//DECISION VARIABLE
dvar int+ x_OTG[origin_terminal, gateway_terminal, truck, order] in 0..1;
dvar int+ x_GTD1[gateway_terminal, destination_terminal, truck, order] in 0..1;
dvar int+ x_GTD2[gateway_terminal, destination_terminal, cons_truck, order] in 0..1;
//OUTPUT VARIABLES
dvar int+ y[gateway_terminal, truck, cons_truck, order] in 0..1;
dvar float exp_fixed_cost_1[origin_terminal, gateway_terminal, truck, order];
dvar float exp_fixed_cost_2[gateway_terminal, destination_terminal, truck, order];
dvar float+ exp_fixed_cost_3[gateway_terminal, destination_terminal, cons_truck];
dvar int+ A[cons_truck];
dvar int+ B[cons_truck];
//PARAMETERS TO WRITE OUTPUT TO EXCEL
tuple tuple_x
{
int terminal;
int terminal2;
int truck;
int order;
int value;
};
tuple tuple_y
{
int terminal;
int truck;
int cons_truck;
int order;
int value;
};
tuple tuple_cost_1_2
{
int terminal;
int terminal2;
int truck;
int order;
float value;
};
tuple tuple_cost_3
{
int terminal;
int terminal2;
int truck;
float value;
};
//MODEL
minimize sum(n in order) exp_fixed_cost_1[1,1,non_cons_truck[n],n]
+ sum(n in order) exp_fixed_cost_2[1,1,non_cons_truck[n],n]
+ sum(h in cons_truck) exp_fixed_cost_3[1,1,h];
subject to
{
//FLOW CONSTRAINTS
forall(n in order) x_OTG[1,1,non_cons_truck[n],n] == 1; //(2)
forall(n in order) x_OTG[1,1,non_cons_truck[n],n] == x_GTD1[1,1,non_cons_truck[n],n] + sum(h in cons_truck) x_GTD2[1,1,h,n]; //(3)
//forall(n in order, h in cons_truck) x_OTG[1,1,non_cons_truck[n],n] + x_GTD2[1,1,h,n] <= 1 + y[1,non_cons_truck[n],h,n]; //(4a) minder strikt?
forall(h in cons_truck, n in order) y[1,non_cons_truck[n],h,n] == x_GTD2[1,1,h,n]; //(4a)
forall(h in cons_truck) sum(n in order) y[1,non_cons_truck[n],h,n]*volume[n] <= capacity_cons_truck[h]; //(4b)
forall(h in cons_truck) sum(n in order) y[1,non_cons_truck[n],h,n] != 1; //(4c) only more than 2 order can be put on a consolidated truck
forall(n in order) sum(h in cons_truck) y[1,non_cons_truck[n],h,n] <= 1; //(5a) order cannot be split over multiple trucks
forall(n in order) sum(h in cons_truck) x_GTD2[1,1,h,n] <= 1; //(5b)
forall(h in cons_truck) sum(n in order) y[1,non_cons_truck[n],h,n] == A[h]; //number of consolidated orders in truck h
forall(h in cons_truck) sum(n in order: (n-1) in order) y[1,non_cons_truck[n],h,n] == B[h]; //all but one consolidated orders in h
//EXPECTED FIXED COSTS
forall(n in order) exp_fixed_cost_1[1,1,non_cons_truck[n],n] >= fixed_cost_OTG[1,1]*x_OTG[1,1,non_cons_truck[n],n]; //(6a)
forall(n in order) exp_fixed_cost_2[1,1,non_cons_truck[n],n] >= fixed_cost_GTD[1,1]*x_GTD1[1,1,non_cons_truck[n],n]; //(6b)
forall(h in cons_truck) exp_fixed_cost_3[1,1,h] >= sum(n in order) x_GTD2[1,1,h,n]*fixed_cost_GTD[1,1] - sum(n in order: (n-1) in order) y[1,non_cons_truck[n],h,n]*fixed_cost_GTD[1,1] ; //(6c)
forall(n in order) exp_fixed_cost_1[1,1,non_cons_truck[n],n] >= 0;
forall(n in order) exp_fixed_cost_2[1,1,non_cons_truck[n],n] >= 0;
forall(n in order, h in cons_truck) exp_fixed_cost_3[1,1,h] >= 0;
}
//OUTPUT TO EXCEL
{tuple_x} output_x_OTG = {<i,j,g,n,x_OTG[i,j,g,n]> | i in origin_terminal, j in gateway_terminal, g in truck, n in order};
{tuple_x} output_xGTD1 = {<i,j,g,n,x_GTD1[i,j,g,n]> | i in gateway_terminal, j in destination_terminal, g in truck, n in order};
{tuple_x} output_xGTD2 = {<i,j,h,n,x_GTD2[i,j,h,n]> | i in gateway_terminal, j in destination_terminal, h in cons_truck, n in order};
{tuple_y} output_y = {<i,g,h,n,y[i,g,h,n]> | i in gateway_terminal, g in truck, h in cons_truck, n in order};
{tuple_cost_1_2} output_exp_fixed_cost_1 = {<i,j,g,n,exp_fixed_cost_1[i,j,g,n]> | i in origin_terminal, j in gateway_terminal, g in truck, n in order};
{tuple_cost_1_2} output_exp_fixed_cost_2 = {<i,j,g,n,exp_fixed_cost_2[i,j,g,n]> | i in gateway_terminal, j in destination_terminal, g in truck, n in order};
{tuple_cost_3} output_exp_fixed_cost_3 = {<i,j,h,exp_fixed_cost_3[i,j,h]> | i in gateway_terminal, j in destination_terminal, h in cons_truck};
这是我的 .dat 文件：
SheetConnection excel("One_to_one_no_leadtime.xlsx");
//GENERAL PARAMETERS
//nr_orders = 3;
nr_trucks = 3;
origin_terminal = {1};
gateway_terminal = {1};
destination_terminal = {1};
order = {1,2,3};
cons_truck = {1,2,3};
//ORDER PARAMETERS
volume = [40,70,20];
non_cons_truck = [1,2,3];
capacity_cons_truck = [100,100,100];
//NETWORK PARAMETERS
fixed_cost_OTG = [[75]];
fixed_cost_GTD = [[75]];
//OUTPUT WEGSCHRIJVEN
output_x_OTG to SheetWrite(excel,"'output'!A3:E1100");
output_xGTD1 to SheetWrite(excel,"'output'!G3:K1100");
output_xGTD2 to SheetWrite(excel,"'output'!M3:Q1200");
output_y to SheetWrite(excel,"'output'!S3:W1000");
output_exp_fixed_cost_1 to SheetWrite(excel,"'cost'!A3:E200");
output_exp_fixed_cost_2 to SheetWrite(excel,"'cost'!G3:K200");
output_exp_fixed_cost_3 to SheetWrite(excel,"'cost'!M3:P200");
如您所见，model 表示当 .dat 文件中的音量 = [40,70,20] 时，A[h] 和 B[h] 相同。为什么 sum(n in order) 等于 sum(n in order: (n-1) in order)？当 volume = [40,40,80] 时不会出现此问题，然后 model 完美运行。我真的不知道发生了什么......我知道这是一个精心设计的 model，但如果有人可以提供帮助，请告诉我！亲切的问候。
你的最优解300中，A和B相同，但公式不同
如果添加新约束
A[1]!=B[1];
然后你得到一个新的解决方案 objective 375