How can I set the solution of a model as the upper bound of a variable in the objective of another model?

Answered

Ma. Julie Anne Gala

• February 25, 2024 11:02

Hi!

I have made two models (model_1 and model_2), and this is how their objectives are defined.

VSPz1 = quicksum(VSPx[(1,j,k)] for k in vehicle_types for j in Follow_Trips_VSP[1])

z1 = quicksum(x[(1,j,k)] for k in vehicle_types for j in Follow_Trips[1])

w1 = 5

z2 = quicksum(y[i,r,t,k] for i in N_Depots

                         for r in R

                         for t in Follow_trip2chrg[(r,i)]

                         for k in vehicle_types)

model_1.setObjective(VSPz1, GRB.MINIMIZE)

model_2.setObjective((w1*z1 + z2), GRB.MINIMIZE)

Is there a way I can set the solution of model_1 as the upper bound of z1?

Related to

• Modeling

0

• 

  Marika Karbstein

  Gurobi Staff

  • February 26, 2024 09:15
  • Edited

  If you solved model_1 with \(\texttt{model_1.optimize()}\) make sure it is solved to optimality. Then you can either add the optimal solution value as UB when defining variable z1, e.g.

  if model_1.status == GRB.OPTIMAL:
     z1 = model_2.addVar(lb=model_1.ObjVal) # continuous variable
     model_2.addConstr(z1 = quicksum(x[(1,j,k)] for k in vehicle_types for j in Follow_Trips[1]))

  or add the constraint

  if model_1.status == GRB.OPTIMAL:
     model_2.addConstr(z <= model_1.ObjVal)

  0

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