Decision Optimization

Here is my coding in CPLEX. My nonlinear objective is actually on the setup cost where the Q[i] is the decision variable. But this decision variable also appears in a holdinginventory cost.

using CP;
execute
{
 cp.param.timeLimit=6400;
}
{string} depot=...;
{string} customer=...;
{string} vehicle=...;
{string} node=...;
int M=12;
int pr=...;
int scale=10;
float dist[node][node]=...;
float fcost[depot]=...;
float D[customer]=...;
float R[customer]=...;
float VC[vehicle]=...;
float VD[depot]=...;
dvar boolean x[node][node][vehicle];
dvar boolean z[depot];
dvar boolean y[depot][customer];
dvar int u[customer][vehicle];

dvar int scaleQ[i in depot] in 0..2400;
dexpr float Q[i in depot]=scaleQ[i]/scale;



dexpr float travel=sum(i in node, j in node, k in vehicle: i!=j)x[i][j][k]*dist[i][j];
dexpr float facility=sum(i in depot)fcost[i]*z[i];
dexpr float setup=17*sum(i in depot, j in customer)(D[j]-R[j])*y[i][j]/Q[i];
dexpr float holdinginventory=(1/pr)*sum(i in depot)(Q[i]*pr - Q[i]*sum(j in customer)(D[j]+R[j])*y[i][j]);
dexpr float cost=travel + facility + setup + holdinginventory;


minimize cost;
subject to {
forall(j in customer) sum(i in node, k in vehicle)x[i][j][k]==1;
forall(k in vehicle) sum(i in depot, j in customer)x[i][j][k]<=1;
forall(k in vehicle) sum(i in node, j in customer)D[j]*x[i][j][k]<=VC[k];
forall(i in node, k in vehicle) (sum(j in node)x[i][j][k])- (sum(j in node)x[j][i][k])==0;
forall(i in depot) sum(j in customer)D[j]*y[i][j]<=VD[i]*z[i];
forall(j in customer) sum(i in depot)y[i][j]==1;
forall(l in customer, j in customer, k in vehicle) u[l][k]-u[j][k]+M*x[l][j][k]<= M-1;
forall(i in depot, j in customer, k in vehicle) sum(h in node)(x[i][h][k]+x[h][j][k])-y[i][j]<=1;
}

Here is the dat file

depot={"D1", "D2"};
customer={"C1", "C2", "C3", "C4", "C5", "C6", "C7", "C8", "C9", "C10", "C11", "C12" };
node={"C1", "C2", "C3", "C4", "C5", "C6", "C7", "C8", "C9", "C10", "C11", "C12", "D1", "D2" };
vehicle={"V1", "V2"};

dist=[[0.0 5.1 10.2 17.1 26.2 4.1 11.7 8.5 14.6 23.4 29.4 29.1 15.0 21.2],
 [5.1 0.0 5.1 12.4 21.4 6.4 8.6 9.2 15.0 20.6 26.2 26.9 13.6 17.0],
 [10.2 5.1 0.0 8.1 16.8 10.8 8.0 12.2 17.1 18.8 23.6 25.6 14.0 13.5],
 [17.1 12.4 8.1 0.0 9.1 16.1 8.1 15.2 17.7 13.0 16.6 20.1 12.8 5.8],
 [26.2 21.4 16.8 9.1 0.0 25.0 16.3 23.5 24.6 14.4 14.1 20.2 19.2 7.2],
 [4.1 6.4 10.8 16.1 25.0 0.0 9.2 4.5 10.4 20.2 26.4 25.5 11.3 19.2],
 [11.7 8.6 8.0 8.1 16.3 9.2 0.0 7.3 10.0 12.0 17.8 18.4 6.1 10.0],
 [8.5 9.2 12.2 15.2 23.5 4.5 7.3 0.0 6.1 16.6 22.8 21.3 7.2 17.0],
 [14.6 15.0 17.1 17.7 24.6 10.4 10.0 6.1 0.0 14.0 20.2 17.0 5.4 17.5],
 [23.4 20.6 18.8 13.0 14.4 20.2 12.0 16.6 14.0 0.0 6.3 7.1 9.5 8.2],
 [29.4 26.2 23.6 16.6 14.1 26.4 17.8 22.8 20.2 6.3 0.0 7.1 15.8 10.8],
 [29.1 26.9 25.6 20.1 20.2 25.5 18.4 21.3 17.0 7.1 7.1 0.0 14.1 15.0],
 [15.0 13.6 14.0 12.8 19.2 11.3 6.1 7.2 5.4 9.5 15.8 14.1 0.0 0.0],
 [21.2 17.0 13.5 5.8 7.2 19.2 10.0 17.0 17.5 8.2 10.8 15.0 0.0 0.0]];
fcost = [100, 100];
D= [20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20];
R= [3, 3, 4, 3, 2, 4, 4, 4, 3, 4, 3, 4];
VC=[140, 140];
VD=[280, 280];
pr=350;

Thank you so much. Finally it works. I got the optimal solution.

Thanks again for your help. Regards.