values of parameters dependant on decision variable

Answered

Leon Stefánsson

• October 10, 2022 19:45
• Edited

Hello,

I have a simple problem i wish to solve, where i want to minimize a product of two parameters.

One of the parameters is dependant on the decision variable and the other's indices are dependant open the decision variable. My code below might clarify this better:

###############################################

x = m.addVars(M,J,vtype=GRB.BINARY,name="x")

# create the route matrix R, R[m,N] is 1 if box with ID = m is N'th in the route

R = m.addVars(M,N,vtype=GRB.BINARY,name="R")

m.addConstrs((R[j,n] ==gp.quicksum(x[m,j]*matM[m,n] for m in range(M)) for j in range(J) for n in range(N)),name = "R")

con1 =m.addConstrs((gp.quicksum(x[m,j] for m in range(M)) == 1 for j in range(J)),name="c1")

B = m.addVars(N,vtype=GRB.INTEGER,name="B")

# B lists the order of center points the pickers walks back to.

m.addConstrs((B[n] == gp.quicksum(S[j]*R[j,n] for j in range(J)) for n in range(N)),name="B_con")

m.setObjective(gp.quicksum(D[i,j]*R[j,n] for j in range(J) for n in range(N) for i in B),GRB.MINIMIZE)

 

m.optimize()

##########################################

 

using the last index (for i in B) in the objective function results in an error:

 

------------------------------------------------------------------------------------------------------------------------

IndexError: only integers, slices (`:`), ellipsis (`...`), numpy.newaxis (`None`) and integer or boolean arrays are valid indices

------------------------------------------------------------------------------------------------------------------------

 

After some research I understand this is strictly forbidden, but I'm having trouble working around this issue. Some help would be greatly appreciated.

 

For clarification:The vector S and matrices matM and D are not dependant upon x.

thanks in advance!

 

 

 

0

• 

  Jaromił Najman

  Gurobi Staff

  • October 11, 2022 08:40

  It is not directly possible to use a variable as an index of an expression or another variable. However, it is possible to formulate it in such a way as is discussed in the post use a decision variable as an index.

  In general if possible, I would try to avoid using optimization variables as indices. In your case, what exactly are you trying to achieve by accessing \(\texttt{D[i,j] for i in B}\)? Do you want to actually use the values of the binary variables to access either \(\texttt{D[0,j]}\) or \(\texttt{D[1,j]}\)? You can achieve this by formulating

  \[\begin{align*}
  D_{0} \cdot (1-B) + D_{1} \cdot B
  \end{align*}\]

  with the appropriate indices for \(B\).

  Best regards, 
  Jaromił

  0
• 

  Leon Stefánsson

  • October 11, 2022 09:17

  

  

  Thank you for the answear Jaromil,

  D is not binary, but a distance matrix.

  I found the post yesterday you referred to and after some pondering got around to using a dummy variable

  \begin{align*}
  b \in R^{n x i} \\
  B_n = \sum_{i = 1}^{\I} n*b_{n,i}  \forall n \in N
  \end{align*}
  
  Then I set the objective function as:
  \begin{align*}
  \sum_{i,j,n}^{} D_{i,j}R{j,n}b{n,i}

   

  0

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